May 29, 2026
We show that the cohomology of the finite element Stokes complex consisting of piecewise polynomials spaces on an Alfeld split mesh from Fu, Guzmán, & Neilan (2020, Math. Comp., 89, 1059–1091) is isomorphic to the cohomologies of the continuous Stokes and de Rham complexes. We also construct novel “minimal” conforming finite element complexes where the \(H^1\)-conforming space is the lowest-order space from Guzmán & Neilan (2018, SIAM J. Numer. Anal., 56, 2826–2844) and the \(L^2\)-conforming space is piecewise constants. These minimal complexes also have cohomologies isomorphic to the continuous Stokes and de Rham complexes. We further construct local, bounded, cochain projections for the minimal complexes. All the results hold for strongly Lipschitz domains with nontrivial topologies and in the presence of mixed boundary conditions.
We consider conforming finite element discretizations of the Stokes complex with mixed boundary conditions. Let \(\Omega \subset \mathbb{R}^3\) be a polyhedral domain whose boundary is partitioned into suitably regular subsets \(\Gamma_0\) and \(\Gamma_1\), and define the following spaces: \[\begin{alignat}{2} \label{eq:hkd-first-def} H^k_{\Gamma_0}(\Omega) &:= \{ \phi \in H^k(\Omega) : D^{\alpha} \phi|_{\Gamma_0} = 0, \;\forall |\alpha| \leq k-1 \}, \qquad & &k \in \mathbb{N}, \\ H^1_{\Gamma_0}(\Omega; \mathop{\mathrm{curl}}) &:= \{ v \in H^1_{\Gamma_0}(\Omega)^3 : \mathop{\mathrm{curl}}v \in H^1_{\Gamma_0}(\Omega)^3 \}, \qquad & & \end{alignat}\tag{1}\] where we may drop the subscript to denote the case \(|\Gamma_0| = 0\) and use the subscript “0” to denote \(|\Gamma_1| = 0\). These spaces fit into the so-called Stokes complex: \[\label{eq:stokes-complex-bcs} \begin{tikzcd} 0 \arrow[r] & H_{\Gamma_0}^2(\Omega) \arrow[r, "\mathop{\mathrm{grad}}"] & H_{\Gamma_0}^1(\Omega; \mathop{\mathrm{curl}}) \arrow[r, "\mathop{\mathrm{curl}}"] & H_{\Gamma_0}^1(\Omega)^3 \arrow[r, "\undefined"] & L^2(\Omega) \arrow[r] & 0. \end{tikzcd}\tag{2}\] In particular, that 2 is a complex means that the composition of any two operators is zero (e.g. \(\mathop{\mathrm{curl}}\mathop{\mathrm{grad}}= 0\)) and that the image of each operator lies in the succeeding space (e.g. \(\mathop{\mathrm{curl}}H_{\Gamma_0}^1(\mathop{\mathrm{curl}};\Omega) \subset H_{\Gamma_0}^1(\Omega)^3\)). Problems involving the spaces in the complex 2 arise in a variety of applications. Many fourth-order problems involve the space \(H^2(\Omega)\), such as the separation of binary alloys [1] or displacement formulations of strain gradient theory [2] to name a few. The space \(H^1(\mathop{\mathrm{curl}}; \Omega)\) appears in displacement formulations of couple stress theory [3]. The final two spaces most famously appear in incompressible flow.
A conforming finite element discretization or subcomplex of 2 is another complex \[\label{eq:stokes-complex-bcs-intro-discrete} \begin{tikzcd} 0 \arrow[r] & V^{0, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{grad}}"] & V^{1, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{curl}}"] & V^{2, h}_{\Gamma_0} \arrow[r, "\undefined"] & V^{3, h}_{\Gamma_0} \arrow[r] & 0, \end{tikzcd}\tag{3}\] where each space \(V^{k, h}_{\Gamma_0}\) is a conforming finite element subspace of the corresponding space in 2 . Choosing finite element discretizations from an underlying complex 3 can offer many benefits. In velocity-pressure formulations of incompressible flow, taking \(V^{2, h}_{\Gamma_0}\) and \(V^{3, h}_{\Gamma_0}\) from 3 lead to mass conserving and “pressure-robust" discretizations [4] and schemes with uniform stability properties for time-dependent or singularly perturbed flows [5], [6]. All four spaces from 3 can also be used to construct an energy and enstrophy stable scheme for the incompressible Navier-Stokes equations [7]. The complex 3 also encodes information that is useful for preconditioning parameter-dependent problems. For example, consider the weighted bilinear forms for \(\alpha, \beta \in \mathbb{R}_+\): \[\label{eq:riesz-maps} \begin{alignat}{2} &(\mathop{\mathrm{grad}}u, \alpha \mathop{\mathrm{grad}}v)_{L^2(\Omega)} + (\mathop{\mathrm{grad}}\mathop{\mathrm{curl}}u, \beta \mathop{\mathrm{grad}}\mathop{\mathrm{curl}}v)_{L^2(\Omega)} \qquad & &\forall u, v \in V_{\Gamma_0}^{1, h}, \\ &(\mathop{\mathrm{grad}}u, \alpha \mathop{\mathrm{grad}}v)_{L^2(\Omega)} + (\undefined u, \beta \undefined v)_{L^2(\Omega)} \qquad & &\forall u, v \in V_{\Gamma_0}^{2, h}, \end{alignat}\tag{4}\] which arise in couple stress theory [3] and augmented Lagrangian preconditioning for incompressible flow [8], [9]. Constructing preconditioners that are robust in the parameters \(\alpha\) and \(\beta\) typically requires knowledge of the kernel of \(\mathop{\mathrm{curl}}: V_{\Gamma_0}^{1, h} \to V_{\Gamma_0}^{2, h}\) and \(\undefined: V_{\Gamma_0}^{2, h} \to V_{\Gamma_0}^{3, h}\) [10], [11], which is precisely encoded in the algebraic structure of the complex 3 .
More specifically, the cohomology of a complex consists of the kernel of an operator modulo the range of the previous operator. For example, the first and second cohomology of the discrete complex 3 are \[\begin{align} \mathfrak{H}^{1, h}_{\Gamma_0} := \frac{ \ker(\mathop{\mathrm{curl}}: V_{\Gamma_0}^{1, h} \to V_{\Gamma_0}^{2, h}) }{ \mathop{\mathrm{im}}(\mathop{\mathrm{grad}}: V_{\Gamma_0}^{0, h} \to V_{\Gamma_0}^{1, h}) } \quad \text{and} \quad \mathfrak{H}^{2, h}_{\Gamma_0} := \frac{ \ker(\undefined: V_{\Gamma_0}^{2, h} \to V_{\Gamma_0}^{3, h}) }{ \mathop{\mathrm{im}}(\mathop{\mathrm{curl}}: V_{\Gamma_0}^{1, h} \to V_{\Gamma_0}^{2, h}) }. \end{align}\] Characterizing the cohomology is thus crucial for constructing robust preconditioners for 4 . The cohomology also plays a key role in finite element exterior calculus (FEEC), including the well-posedness of the Hodge-Laplace problems associated with 3 (see e.g. [12]) which includes incompressible flow as well as other mixed problems involving the spaces and operators in 3 . More generally, the properties of “simpler" complexes like 3 2 , including cohomology, are critical in understanding properties of more complicated sequences via the Bernstein–Gelfand–Gelfand construction [13]–[15]. For a recent review on the importance of cohomology in a variety of applications, we refer to [16]. Nevertheless, the cohomology of any conforming finite element subcomplex of 2 on nontrivial domains and/or with mixed boundary conditions seems to not have been addressed in the literature.
Our first main result shows that if the finite element spaces in 3 are chosen to be the Alfeld-split macroelements in [17], then the cohomology of the discrete complex 3 is isomorphic to the cohomology of the continuous complex 2 . This result extends [17] to the case of nontrivial domains and to the case of mixed boundary conditions. The main technique is applying the recent framework of [18] and modifying the final steps to account for boundary conditions. The second set of results show that, in a certain sense, the Alfeld-split macroelement Stokes complex analog of Whitney forms [19]–[21] and lowest-order complete polynomial complexes can be constructed to form “minimal” subcomplexes of 3 with the same cohomology structure. The final two spaces in these subcomplexes consist of low-order \(H^1(\Omega)^3\)-conforming element from [22] and piecewise constants. For the first two spaces in the subcomplexes, we obtain novel \(H^2(\Omega)\)-conforming and \(H^1(\mathop{\mathrm{curl}}; \Omega)\)-conforming finite elements whose dimension is, in a certain sense, “minimal”.
We assume that the domain \(\Omega \subset \mathbb{R}^3\) and boundary partition \(\partial \Omega = \bar{\Gamma}_0 \cup \bar{\Gamma}_N\) satisfy the assumptions in [23]: \(\Omega\) is a bounded strongly Lipschitz domain and \(\Gamma_0\) and \(\Gamma_1\) are strongly Lipschitz subsets, so that \(\Omega\) and \(\Gamma_0\) form a strong Lipschitz pair in the sense of [24]. We further assume that \(\partial \Omega\) and \(\Gamma_0\) are both the union of a finite number of polygons.
We rewrite the Stokes complex 2 in standard FEEC notation as follows: \[\label{eq:stokes-complex-bcs-feec} \begin{tikzcd} 0 \arrow[r] & V_{\Gamma_0}^0 \arrow[r, "\mathop{\mathrm{d}}^0"] & V_{\Gamma_0}^1 \arrow[r, "\mathop{\mathrm{d}}^1"] & V_{\Gamma_0}^2 \arrow[r, "\mathop{\mathrm{d}}^2"] & V_{\Gamma_0}^3 \arrow[r] & 0, \end{tikzcd}\tag{5}\] where \[\begin{align} {4} V_{\Gamma_0}^0 &:= H^2_{\Gamma_0}(\Omega), \qquad & V_{\Gamma_0}^1 &:= H^1_{\Gamma_0}(\mathop{\mathrm{curl}}; \Omega), \qquad & V_{\Gamma_0}^2 &:= H^1_{\Gamma_0}(\Omega)^3, & \qquad V_{\Gamma_0}^3 &:= L^2(\Omega), \\ \mathop{\mathrm{d}}^0 &:= \mathop{\mathrm{grad}}, \qquad & \mathop{\mathrm{d}}^1 &:= \mathop{\mathrm{curl}}, \qquad & \mathop{\mathrm{d}}^2 &:= \undefined, \qquad & \mathop{\mathrm{d}}^3 &:= 0. \end{align}\] As with the Sobolev spaces, we may drop the subscript “\(D\)” if \(|\Gamma_0| = 0\) or use the subscript “0” if \(|\Gamma_1| = 0\). The Stokes complex 5 may be viewed as a smoother subcomplex of the standard de Rham complex: \[\label{eq:de-rham-complex-bcs-feec} \begin{tikzcd} 0 \arrow[r] & W_{\Gamma_0}^0 \arrow[r, "\mathop{\mathrm{d}}^0"] & W_{\Gamma_0}^1 \arrow[r, "\mathop{\mathrm{d}}^1"] & W_{\Gamma_0}^2 \arrow[r, "\mathop{\mathrm{d}}^2"] & W_{\Gamma_0}^3 \arrow[r] & 0, \end{tikzcd}\tag{6}\] where \[\begin{align} {4} W_{\Gamma_0}^0 &:= H^1_{\Gamma_0}(\Omega), \quad & W_{\Gamma_0}^1 &:= H_{\Gamma_0}(\mathop{\mathrm{curl}}; \Omega), \quad & W_{\Gamma_0}^2 &:= H_{\Gamma_0}(\undefined; \Omega), \quad & W_{\Gamma_0}^3 &:= L^2(\Omega). \end{align}\] Here, \(H_{\Gamma_0}(\mathop{\mathrm{curl}}; \Omega)\), respectively \(H_{\Gamma_0}(\undefined; \Omega)\), is the space of \(L^2(\Omega)^3\) vector fields whose \(\mathop{\mathrm{curl}}\), respectively \(\undefined\), is also square integrable and whose tangential, respectively normal, trace vanishes on \(\Gamma_0\).
The \(k\)th-cohomology, or harmonic form, of the Stokes complex is defined by \[\begin{align} \label{eq:harmonic-forms-bcs} \mathfrak{H}_{\Gamma_0}^k := \frac{ \ker(\mathop{\mathrm{d}}^k : V_{\Gamma_0}^k \to V_{\Gamma_0}^{k+1}) }{ \mathop{\mathrm{im}}(\mathop{\mathrm{d}}^{k-1} : V_{\Gamma_0}^{k-1} \to V_{\Gamma_0}^k) }, \qquad k \in 0:3, \end{align}\tag{7}\] where we adopt the convention that any space or operator with form index \(k < 0\) or \(k > 3\) is the trivial space or operator, and \(m:n := \{ m, m+1,\dots, n-1, n\}\) for nonnegative integers \(m, n\). Theorem 5.124 of [23] shows that the harmonic forms of the Stokes complex and the de Rham complex are finite and isomorphic: \[\begin{align} \label{eq:harmonic-forms-same-dim-diff-smoothness} \dim \mathfrak{H}_{\Gamma_0}^k = \dim \frac{ \ker(\mathop{\mathrm{d}}^k : W_{\Gamma_0}^k \to W_{\Gamma_0}^{k+1}) }{ \mathop{\mathrm{im}}(\mathop{\mathrm{d}}^{k-1} : W_{\Gamma_0}^{k-1} \to W_{\Gamma_0}^k) } \qquad \forall k \in 0:3. \end{align}\tag{8}\]
The dimensions of the harmonic forms correspond to relative Betti numbers. For \(k \in 0:3\), let \(b_k(\Omega, \Gamma_0)\) be the \(k\)-th relative Betti number defined as the dimension of the \(k\)-th singular homology group of \(\Omega\) relative to \(\Gamma_0\) (see e.g. [26]). Applying [27] to 8 then shows that \[\begin{align} \label{eq:harmonic-forms-dim} \dim \mathfrak{H}_{\Gamma_0}^k = b_k(\Omega, \Gamma_0) \qquad \forall k \in 0:3. \end{align}\tag{9}\] Note that \(b_k(\Omega, \emptyset)\) corresponds to the usual Betti number of \(\Omega\); e.g. \(b_0(\Omega)\) is the number of connected components of \(\Omega\). We also have the duality relation \(b_{k}(\Omega, \Gamma_0) = b_{3-k}(\Omega, \Gamma_1)\) for \(k \in 0:3\) thanks to [27]. Combining this with [27], we have \[\begin{align} b_0(\Omega, \Gamma_0) = \begin{cases} 1 & \text{if } |\Gamma_0| = 0, \\ 0 & \text{otherwise}, \end{cases} \quad \text{and} \quad b_3(\Omega, \Gamma_0) = \begin{cases} 1 & \text{if } |\Gamma_1| = 0, \\ 0 & \text{otherwise}. \end{cases} \end{align}\] Unfortunately, \(b_{1}(\Omega, \Gamma_0)\) and \(b_2(\Omega, \Gamma_0)\) do not have such simple expressions.
We first recall a finite element Stokes complex on Alfeld-split meshes from [17]. For a collection of tetrahedra \(\mathcal{S}\) forming a conforming mesh of an open domain \(\mathcal{O}\), we define the following spaces of polynomials for \(p \in \mathbb{N}_0\): \[\begin{align} \mathop{\mathrm{DG}}^p(\mathcal{S}) &:= \{ v \in L^2(\mathcal{O}) : v|_{K} \in \mathcal{P}_p(K) \;\forall K \in \mathcal{S} \} \;\text{and} \; \mathop{\mathrm{CG}}^p(\mathcal{S}) := \mathop{\mathrm{DG}}^p(\mathcal{S}) \cap C(\bar{\mathcal{O}}). \end{align}\] Given a tetrahedron \(K\), let \(K_A\) denote the collection of four tetrahedra in the Alfeld split of \(K\) formed by connecting the four vertices of \(K\) to the barycenter of \(K\), and for \(p \geq 3\), define the following local Alfeld-split macroelement spaces: \[\label{eq:local-full-alfeld} \begin{alignat}{2} V^{0, h}(K_A) &:= \{v \in \mathop{\mathrm{CG}}^{p+2}(K_A) : \mathop{\mathrm{grad}}v \in \mathop{\mathrm{CG}}^{p+1}(K_A)^3\},\\ V^{1, h}(K_A) &:= \{v \in \mathop{\mathrm{CG}}^{p+1}(K_A)^3 : \mathop{\mathrm{curl}}v \in \mathop{\mathrm{CG}}^{p}(K_A)^3\},\\ V^{2, h}(K_A) &:= \mathop{\mathrm{CG}}^p(K_A)^3, \\ V^{3, h}(K_A) &:= \mathop{\mathrm{DG}}^{p-1}(K_A). \end{alignat}\tag{10}\] These local spaces form a complex \[\label{eq:stokes-complex-local-full-alfeld} \begin{tikzcd} \mathbb{R} \arrow[r, "\subset"] & V^{0, h}(K_A) \arrow[r, "\mathop{\mathrm{d}}^0"] & V^{1, h}(K_A) \arrow[r, "\mathop{\mathrm{d}}^1"] & V^{2, h}(K_A) \arrow[r, "\mathop{\mathrm{d}}^2"] & V^{3, h}(K_A) \arrow[r] & 0, \end{tikzcd}\tag{11}\] and the complex is exact for all \(p \geq 3\) [17], which means that the range of each operator is the kernel of the succeeding operator; i.e., all harmonic forms are trivial.
The local spaces 10 may be extended to a multi-element mesh in a typical finite element fashion provided that extra vertex smoothness is imposed on first two spaces. In particular, let \(\mathcal{T}\) be a conforming, shape regular simplicial mesh of \(\Omega\) such that there exists a submesh of triangles \(\mathcal{U}\) that forms a conforming mesh of \(\Gamma_0\). Additionally, we define the global ambient spaces by \[\begin{align} \label{eq:global-ambient-spaces} \mathbb{X}^0 := \mathbb{X}^3 := \mathbb{R} \quad \text{and} \quad \mathbb{X}^1 := \mathbb{X}^2 := \mathbb{R}^3. \end{align}\tag{12}\] \[\label{eq:global-full-alfeld} Then, for k \in 0:1, the global spaces are given by \begin{multline} V^{k, h} := \{ v \in C(\Omega) \otimes \mathbb{X}^k : \mathop{\mathrm{d}}^k v \in C(\Omega) \otimes \mathbb{X}^{k+1}, \; \text{v is C^{2-k} at \Delta_0(\mathcal{T})} \\ \text{ and } v|_{K} \in V^{k, h}(K_A) \;\forall K \in \mathcal{T}\}, \end{multline} where \Delta_{\ell}(\mathcal{S}) denotes the set of all \ell-dimensional subsimplices of a collection of d-dimensional simplices \mathcal{S} with \ell \in 0:d (e.g. \mathcal{T}= \Delta_3(\mathcal{T})). For k \in 2:3, the global spaces are simply \begin{align} V^{k, h} := \{ v \in C^{2-k}(\Omega) \otimes \mathbb{X}^{k} : v|_{K} \in V^{k, h}(K_A) \;\forall K \in \mathcal{T}\}, \end{align} where C^{-1}(\Omega) := L^2(\Omega).\tag{13}\] The corresponding spaces incorporating the boundary conditions are \[\begin{align} \label{eq:global-full-alfeld-bcs} V^{k, h}_{\Gamma_0} := V^{k, h} \cap V_{\Gamma_0}^k, \end{align}\tag{14}\] which form a conforming subcomplex of 5 : \[\label{eq:stokes-complex-full-alfeld-bcs} \begin{tikzcd} 0 \arrow[r] & V^{0, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{d}}^0"] & V^{1, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{d}}^1"] & V^{2, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{d}}^2"] & V^{3, h}_{\Gamma_0} \arrow[r] & 0. \end{tikzcd}\tag{15}\]
Our first main result shows that the discrete harmonic forms \[\begin{align} \label{eq:harmonic-forms-full-alfeld-bcs} \mathfrak{H}^{k, h}_{\Gamma_0} := \frac{\ker(\mathop{\mathrm{d}}^k : V_{\Gamma_0}^{k, h} \to V_{\Gamma_0}^{k+1, h})}{ \mathop{\mathrm{im}}(\mathop{\mathrm{d}}^{k-1} : V_{\Gamma_0}^{k-1, h} \to V_{\Gamma_0}^{k, h})}, \qquad k \in 0:3, \end{align}\tag{16}\] have the same dimension as the continuous harmonic forms in 7 .
Theorem 1. For all \(p \geq 3\), the complexes 5 and 15 have isomorphic cohomologies: \(\dim \mathfrak{H}^{k, h}_{\Gamma_0} = \dim \mathfrak{H}^{k}_{\Gamma_0} = b_k(\Omega, \Gamma_0)\) for all \(k \in 0:3\).
The proof of 1, appearing below in 3.7, applies the general framework of [18] and extends it to the case of mixed boundary conditions for the particular discrete complex 15 . We note that 1 is the extension of [17] to nontrivial domains with mixed boundary conditions.
We now seek a “minimal” conforming finite element subcomplex of the Stokes complex 5 \[\label{eq:stokes-complex-reduced-alfeld-bcs} \begin{tikzcd} 0 \arrow[r] & \tilde{V}^{0, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{d}}^0"] & \tilde{V}^{1, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{d}}^1"] & \tilde{V}^{2, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{d}}^2"] & \tilde{V}^{3, h}_{\Gamma_0} \arrow[r] & 0 \end{tikzcd}\tag{17}\] whose cohomology is isomorphic to that of 5 . We restrict ourselves to subcomplexes of 17 to stay within the setting of Alfeld-split meshes; see [28] for spaces on other types of splits and [28] for minimal two-dimensional elements. One of the key properties of the full polynomial spaces 13 that is crucial in the proof of 1 is that one may choose degrees of freedom for the spaces \(V^{k, h}\) to include the following linear functionals: \[\label{eq:whitney-dofs-currents} \begin{alignat}{2} C(\bar{\Omega}) \otimes \mathbb{X}^0 \ni v &\mapsto \mathcal{I}_{z}^0(v) := v(z) \qquad & &\forall z \in \Delta_0(\mathcal{T}), \\ C(\bar{\Omega}) \otimes \mathbb{X}^1 \ni v &\mapsto \mathcal{I}_{e}^1(v) := \int_{e} v \cdot \hat{t}_e \,\mathrm{d}{s} \qquad & &\forall e \in \Delta_1(\mathcal{T}), \\ C(\bar{\Omega}) \otimes \mathbb{X}^2 \ni v &\mapsto \mathcal{I}_{f}^2(v) := \int_{f} v \cdot \hat{n}_f \,\mathrm{d}{s} \qquad & &\forall f \in \Delta_2(\mathcal{T}), \\ C(\bar{\Omega}) \otimes \mathbb{X}^3 \ni v &\mapsto \mathcal{I}_{K}^3(v) := \int_{K} v \,\mathrm{d}{x} \qquad & &\forall K \in \Delta_3(\mathcal{T}), \end{alignat}\tag{18}\] where \(\hat{n}_f\) and \(\hat{t}_e\) are unit normal and tangent vectors with a fixed global orientation. Note that 18 are simply the canonical set of degrees of freedom for the Whitney forms [19]–[21], the lowest-order conforming discretization of the de Rham complex 6 .
Our starting point is then to choose \[\begin{align} \label{eq:min-l2-space} \tilde{V}^{3, h}(K_A) := \mathop{\mathrm{DG}}^0(K), \quad \tilde{V}^{3, h} := \mathop{\mathrm{DG}}^0(\mathcal{T}), \quad \text{and} \quad \tilde{V}^{3, h}_{\Gamma_0} := \tilde{V}^{3, h} \cap V^3_{\Gamma_0}, \end{align}\tag{19}\] the Whitney forms of index 3. Moving one space to the left in the complex 17 , we seek an \(H^1(\Omega)^3\)-conforming finite element space whose divergence lies in \(\mathop{\mathrm{DG}}^0(\mathcal{T})\) and is large enough to be equipped with the degrees of freedom in 18 . The Guzmán-Neilan element [22] exactly meets these requirements and is defined as follows. For \(f \in \Delta_2(K)\), let \(b_f \in \mathcal{P}_3(K)\) be the face bubble function satisfying \(b_f|_{\partial K \setminus f} = 0\), normalized so that \(\int_f b_f \,\mathrm{d}{s} = |f|\). Henceforth, \(b_f\) is also used to denote the restriction \(b_f|_f\). Moreover, for \(K \in \mathcal{T}\), let \(S_{K} : \mathop{\mathrm{CG}}^{3}(K_A)^3 \to \mathop{\mathrm{CG}}^{3}(K_A)^3\) be any fixed linear operator satisfying the following for all \(v \in \mathop{\mathrm{CG}}^{3}(K_A)^3\): \[\begin{align} \label{eq:generic-gn-extension-operator} S_{K} v|_{\partial K} = v|_{\partial K}, \quad \undefined S_{K} v \in \mathcal{P}_0(K), \;\;\text{and} \;\; | S_K v |_{H^{\ell}(K)} \leq C_{S} |v |_{H^{\ell}(K)}, \;\ell \in 0:1, \end{align}\tag{20}\] where \(C_{S} > 0\) is independent of \(K\). Then, the local and global Guzmán-Neilan spaces are give by \[\begin{align} \tag{21} \mathop{\mathrm{GN}}(K_A) &:= \mathcal{P}_1(K)^3 \oplus \mathop{\mathrm{span}}\{ S_{K}( b_f \hat{n}_f) : f \in \Delta_2(K) \} \qquad \forall K \in \mathcal{T}, \\ \tag{22} \mathop{\mathrm{GN}}(\mathcal{T}) &:= \{v \in C(\Omega)^3 : v|_{K} \in \mathop{\mathrm{GN}}(K_A) \;\forall K \in \mathcal{T}\}. \end{align}\] In particular, \(\mathop{\mathrm{CG}}^1(\mathcal{T})^3 \subset \mathop{\mathrm{GN}}(\mathcal{T}) \subset V^{1, h}\), \(\undefined\mathop{\mathrm{GN}}(\mathcal{T}) \subseteq \mathop{\mathrm{DG}}^0(\mathcal{T})\), and a global set of degrees of freedom are given by [22]: \[\begin{align} \label{eq:gn-dofs} v(z) \qquad \forall z \in \Delta_0(\mathcal{T}) \quad \text{and} \quad \int_f v \cdot n_f \,\mathrm{d}{s} \qquad \forall f \in \Delta_2(\mathcal{T}). \end{align}\tag{23}\]
Remark 2. The definition of \(\mathop{\mathrm{GN}}(K_A)\) in [22] used a particular choice of \(S_K\); however, the only properties used in the analysis are 20 . Thus, we shall use results from [22] for the more generic space here.
The choice \[\begin{align} \label{eq:min-h1-space} \tilde{V}^{2, h}(K_A) := \mathop{\mathrm{GN}}(K_A), \quad \tilde{V}^{2, h} := \mathop{\mathrm{GN}}(\mathcal{T}), \quad \text{and} \quad \tilde{V}^{2, h}_{\Gamma_0} := \tilde{V}^{2, h} \cap V^2_{\Gamma_0} \end{align}\tag{24}\] is then “minimal", as one typically requires vertex degrees of freedom as in 23 to ensure continuity. The construction of the remaining spaces \(\tilde{V}_{\Gamma_0}^{0, h}\) and \(\tilde{V}_{\Gamma_0}^{1, h}\) will be detailed in 4 below. The main result is the following.
Theorem 3. Let \(\tilde{V}^{k, h}(K_A)\) and \(\tilde{V}^{k, h}_{\Gamma_0}\) for \(k \in 2:3\) be given by 19 24 . For each \(K \in \mathcal{T}\) and \(k \in 0:1\), there exists \(\tilde{V}^{k, h}(K_A) \subset V^{k, h}(K_A)\), such that \(\mathcal{P}_{3-k}(K) \otimes \mathbb{X}^k \subset \tilde{V}^{k, h}(K_A)\) and the local complex \[\label{eq:stokes-complex-reduced-alfeld-element} \begin{tikzcd} \mathbb{R} \arrow[r, "\subset"] & \tilde{V}^{0, h}(K_A) \arrow[r, "\mathop{\mathrm{d}}^0"] & \tilde{V}^{1, h}(K_A) \arrow[r, "\mathop{\mathrm{d}}^1"] & \tilde{V}^{2, h}(K_A) \arrow[r, "\mathop{\mathrm{d}}^2"] & \tilde{V}^{3, h}(K_A) \arrow[r] & 0 \end{tikzcd}\tag{25}\] is exact. Moreover, if we define for \(k \in 0:1\) \[\begin{align} \label{eq:min-h2h1curl-space} \tilde{V}^{k, h} := \{ v \in V^{k, h} : v|_{K} \in \tilde{V}^{k, h}(K_A) \;\forall K \in \mathcal{T}\} \quad \text{and} \quad \tilde{V}^{k, h}_{\Gamma_0} := \tilde{V}^{k, h} \cap V^k_{\Gamma_0}, \end{align}\tag{26}\] then the global complexes 5 and 17 have isomorphic cohomologies.
The proof of 3 appears in 5.1 below. We will also show below in 6 [ lem:h1curl-local-unisolvence]{reference-type=“ref” reference=” lem:h1curl-local-unisolvence”} that \(\tilde{V}^{0, h}\) may be equipped with the degrees of freedom \[\begin{align} \label{eq:walkington-dofs} D^{\alpha} v(z) \qquad \forall |\alpha| \leq 2, \;\forall z \in \Delta_0(\mathcal{T}), \end{align}\tag{27}\] while \(\tilde{V}^{1, h}\) may be equipped with \[\tag{28} \begin{alignat}{2} \tag{29} &D^{\alpha} v(z) \qquad & & \forall|\alpha| \leq 1, \;\forall z \in \Delta_0(\mathcal{T}), \\ \tag{30} &\int_{e} v \cdot \hat{t}_e \,\mathrm{d}{s} \qquad & & \forall e \in \Delta_1(\mathcal{T}). \end{alignat}\] In view of the \(C^{2-k}\)-continuity imposed at the mesh vertices of elements in \(V^{k, h}\), \(k \in 0:1\), we see that \(\tilde{V}^{k, h}\) are then “minimal” subspaces of \(V^{k, h}\) whose degrees of freedom can be chosen to include 18 and satisfy \(\mathcal{P}_{3-k}(K) \otimes \mathbb{X}^k \subset \tilde{V}^{k, h}(K_A)\).
In view of this property, the local complex 25 bears resemblance to the complex of complete polynomials used to discretize the de Rham complex 6 . Locally, the complete polynomial complex reads for \(K \in \mathcal{T}\): \[\label{eq:de-rham-complex-complete-poly-element} \begin{tikzcd} \mathbb{R} \arrow[r, "\subset"] & \mathcal{P}_{3}(K) \arrow[r, "\mathop{\mathrm{d}}^0"] & \mathcal{P}_{2}(K)^3 \arrow[r, "\mathop{\mathrm{d}}^1"] & \mathcal{P}_{1}(K)^3 \arrow[r, "\mathop{\mathrm{d}}^2"] & \mathcal{P}_{0}(K) \arrow[r] & 0. \end{tikzcd}\tag{31}\] In fact, 3 shows that 31 is a subcomplex of 25 . The additional complexities of the spaces in 25 only arise due to the additional global continuity imposed by being conforming subspaces of \(V_{\Gamma_0}^{k}\) rather than \(W_{\Gamma_0}^{k}\). Thus, the global complex 17 may be seen as the Alfeld-split macroelement Stokes complex analog of the lowest-order discretization of the de Rham complex with complete polynomials.
Of course, one can further reduce the local complex 31 to the local Whitney complex or lowest-order trimmed polynomial complex: \[\label{eq:de-rham-complex-reduce-poly-element} \begin{tikzcd} \mathbb{R} \arrow[r, "\subset"] & W^{0, h}(K) \arrow[r, "\mathop{\mathrm{d}}^0"] & W^{1, h}(K) \arrow[r, "\mathop{\mathrm{d}}^1"] & W^{2, h}(K) \arrow[r, "\mathop{\mathrm{d}}^2"] & W^{3, h}(K) \arrow[r] & 0, \end{tikzcd}\tag{32}\] where \[\label{eq:whitney-forms-local} \begin{alignat}{2} W^{0, h}(K) &:= \mathcal{P}_1(K), \qquad & W^{1, h}(K) &:= \mathcal{P}_{0}(K)^3 + x \times \mathcal{P}_0(K)^3, \\ W^{2, h}(K) &:= \mathcal{P}_{0}(K)^3 + x \mathcal{P}_0(K), \qquad & W^{3, h}(K) &:= \mathcal{P}_0(K). \end{alignat}\tag{33}\] The following result shows that the two spaces \(\tilde{V}^{k, h}(K_A)\), \(k \in 0:1\), can be reduced further while ensuring that 32 is a subcomplex of the corresponding local complex.
Theorem 4. Let \(\tilde{V}^{k, h}(K_A)\) and \(\tilde{V}_{\Gamma_0}^{k ,h}\) be defined as in 3. Then, the further reduced spaces \[\begin{align} \hat{V}^{0, h}(K_A) &:= \{ v \in \tilde{V}^{0, h}(K_A) : \mathop{\mathrm{hess}}v(z) = 0 \;\forall z \in \Delta_0(K) \}, \\ \hat{V}^{1, h}(K_A) &:= \{ v \in \tilde{V}^{1, h}(K_A) : \mathop{\mathrm{sym}}\mathop{\mathrm{grad}}v(z) = 0 \;\forall z \in \Delta_0(K) \}, \end{align}\] where \(\mathop{\mathrm{hess}}\) denotes the Hessian operator, satisfy \(W^{k, h}(K) \subset \hat{V}^{k, h}(K_A)\), \(k \in 0:1\), and the complex \[\label{eq:stokes-complex-reduced-alfeld-reduced-element} \begin{tikzcd} \mathbb{R} \arrow[r, "\subset"] & \hat{V}^{0, h}(K_A) \arrow[r, "\mathop{\mathrm{d}}^0"] & \hat{V}^{1, h}(K_A) \arrow[r, "\mathop{\mathrm{d}}^1"] & \tilde{V}^{2, h}(K_A) \arrow[r, "\mathop{\mathrm{d}}^2"] & \tilde{V}^{3, h}(K_A) \arrow[r] & 0 \end{tikzcd}\tag{34}\] is exact. Moreover, if we define for \(k \in 0:1\) \[\begin{align} \label{eq:min-h2h1curl-space-reduce} \hat{V}^{k, h} := \{ v \in \tilde{V}^{k, h} : v|_{K} \in \hat{V}^{k, h}(K_A) \;\forall K \in \mathcal{T}\} \quad \text{and} \quad \hat{V}^{k, h}_{\Gamma_0} := \hat{V}^{k, h} \cap V^k_{\Gamma_0}, \end{align}\tag{35}\] then the cohomology of \[\label{eq:stokes-complex-reduced-alfeld-reduced} \begin{tikzcd} 0 \arrow[r] & \hat{V}_{\Gamma_0}^{0, h} \arrow[r, "\mathop{\mathrm{d}}^0"] & \hat{V}_{\Gamma_0}^{1, h} \arrow[r, "\mathop{\mathrm{d}}^1"] & \tilde{V}_{\Gamma_0}^{2, h} \arrow[r, "\mathop{\mathrm{d}}^2"] & \tilde{V}_{\Gamma_0}^{3, h} \arrow[r] & 0 \end{tikzcd}\tag{36}\] is isomorphic to the cohomology of 5 .
On noting that \(\mathop{\mathrm{hess}}W^{0, h}(K) = 0\) and \(\mathop{\mathrm{sym}}\mathop{\mathrm{grad}}W^{1, h}(K) = 0\), the proof of 4 is completely analogous to the proof of 3 and is therefore omitted. One may readily see that if any additional degrees of freedom from 27 or 28 are set to zero, then we would lose the inclusion \(W^{k, h}(K) \subset \hat{V}^{k, h}(K_A)\), and so 36 may be seen as the Alfeld-split macroelement Stokes complex analog of the Whitney complex.
With the cohomology of 17 fully characterized, the last remaining components used extensively in the FEEC literature are bounded commuting cochain projections. One possible avenue is to modify the construction of locally \(L^2\)-bounded cochain projections in [18] to take into account the boundary conditions analogously to the construction of Clément interpolants [29]. Instead, we construct Scott-Zhang [30] type interpolants that also commute.
To describe the result, we define for an open set \(\mathcal{O}\) the following norms: \(\| \cdot \|_{V^0(\mathcal{O})}\) the \(H^2(\mathcal{O})\) norm, \(\| \cdot \|_{V^1(\mathcal{O})}\) the \(H^1(\mathop{\mathrm{curl}}; \mathcal{O})\) norm, \(\| \cdot \|_{V^2(\mathcal{O})}\) the \(H^1(\mathcal{O})^3\) norm, and \(\| \cdot \|_{V^3(\mathcal{O})}\) the \(L^2(\mathcal{O})\) norm. Additionally, given a tetrahedron \(K \in \mathcal{T}\), let \(\omega_K\) denote the 1 element neighborhood of \(K\): \[\begin{align} \omega_K := \mathrm{int} \left( \bigcup \{ \bar{K}' \in \mathcal{T}: \bar{K} \cap \bar{K}' \neq \emptyset \} \right). \end{align}\] The locally bounded cochain projections are summarized in the following result.
Theorem 5. Let \(\tilde{\Pi}^3 := V^3 \to \tilde{V}^{3, h}\) be the \(L^2(\Omega)\)-orthogonal projection. Then, there exist linear projection operators \(\tilde{\Pi}^k : V^k \to \tilde{V}^{k, h}\), \(k \in 0:2\), such that \(\{ \tilde{\Pi}^k \}_{k=0}^{3}\) satisfying the following:
Trace preservation: \(\tilde{\Pi}^k : V^k_{\Gamma_0} \to \tilde{V}_{\Gamma_0}^{k, h}\).
Local boundedness: \[\begin{align} \label{eq:cochain-projections-locally-bounded} \| \tilde{\Pi}^k v\|_{V^k(K)} \leq C \|v\|_{V^k(\omega_K)} \qquad \forall K \in \mathcal{T}, \end{align}\tag{37}\] where \(C\) depends only on \(\Omega\), \(\Gamma_0\), and shape regularity.
Commuting diagram: \[\label{eq:stokes-complex-reduced-commuting-diagram-bcs} \begin{tikzcd}[ampersand replacement=\&] V^{0}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{grad}}"] \arrow[d, "\tilde{\Pi}^0"] \& V^{1}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{curl}}"] \arrow[d, "\tilde{\Pi}^1"] \& V^{2}_{\Gamma_0} \arrow[r, "\undefined"] \arrow[d, "\tilde{\Pi}^2"] \& V^{3}_{\Gamma_0} \arrow[d, "\tilde{\Pi}^3"] \\ \tilde{V}^{0, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{grad}}"] \& \tilde{V}^{1, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{curl}}"] \& \tilde{V}^{2, h}_{\Gamma_0} \arrow[r, "\undefined"] \& \tilde{V}^{3, h}_{\Gamma_0}. \end{tikzcd}\tag{38}\]
The proof of 5 appears below in 6. Owing to the regularity of the spaces appearing in the Stokes complex 5 , the construction of these interpolants involves standard techniques similar to those in [30], [31] in contrast to the more sophisticated techniques employed for discretizations of the de Rham complex 6 ; see e.g. [32]–[34] and references therein.
Remark 6. Since the cohomologies of the first two rows of 38 are isomorphic, the first two sentences of the proof of [12] shows that \(\tilde{\Pi}^k\) is an isomorphism between the cohomologies. Note that we may replace the first row in 38 with the spaces \(\{ V_{\Gamma_0}^{k,h} \}\) (for any \(p \geq 3\)), \(\{ \tilde{V}_{\Gamma_0}^{k, h} \}\), or \(\{ \hat{V}_{\Gamma_0}^{k,h} \}\) (where \(\hat{V}_{\Gamma_0}^{k,h} := \tilde{V}_{\Gamma_0}^{k,h}\) for \(k \in 2:3\)) defined on any other conforming mesh and obtain a commuting diagram. For any of these replacements, the cohomologies of the two complexes in 38 are isomorphic with \(\tilde{\Pi}^k\) again being an isomorphism between the cohomologies.
Remark 7. The operators \(\tilde{\Pi}^k\), \(k \in 0:1\), may be trivially modified so that \(\tilde{\Pi}^k : V^k \to \hat{V}^{k, h}\) and the conclusions of 5 hold with \(\tilde{V}^k\) replaced by \(\hat{V}^k\).
The remainder of the manuscript is organized as follows. In 3, we show how the framework from [18] applies to the complex with full polynomial spaces 15 and modify the framework to account for boundary conditions. Then, in 4, we construct the reduced spaces \(\tilde{V}^{k, h}\), \(k \in 0:1\), show that the degrees of freedom in 27 28 are unisolvent, and demonstrate how boundary conditions may be incorporated into the spaces. The cohomology of the complex 17 is the focus of 5, and the bounded cochain projections are constructed in 6.
We mostly follow the framework in [18], originally developed for determining the cohomology of discrete complexes spaces without boundary conditions, with some modification to handle the case of mixed boundary conditions in the Stokes complex. We first require some additional notation. Give a collection of \(d\)-dimensional simplices \(\mathcal{S}\), let \(\Delta(\mathcal{S}) := \bigcup_{\ell=0}^{d} \Delta_{\ell}(\mathcal{S})\) denote the collection of all subsimplices of \(\mathcal{S}\). Moreover, given a simplex \(\tau\), we say \(\eta \mathop{\mathrm{\unlhd}}\tau\) if \(\eta\) is a subsimplex of \(\tau\) and \(\eta \mathop{\mathrm{\lhd}}\tau\) if \(\eta \mathop{\mathrm{\unlhd}}\tau\) and \(\eta \neq \tau\).
We begin by defining ambient spaces for the various trace operators we will define. For \(K \in \mathcal{T}\), we take \(A^k(K) := V^{k, h}(K_A)\). For \(\tau \in \Delta_{1}(\mathcal{T}) \cup \Delta_{2}(\mathcal{T})\) and \(z \in \Delta_0(\mathcal{T})\), we define \[\begin{align} {2} A^0(\tau) &:= \mathcal{P}_{p+2}(\tau) \oplus \mathcal{P}_{p+1}(\tau)^3 \oplus \bigoplus_{z \in \Delta_0(\tau)} \mathbb{R}^{3 \times 3}_{\mathop{\mathrm{sym}}}, \qquad & A^0(z) &:= \mathbb{R} \oplus \mathbb{R}^3 \oplus \mathbb{R}^{3 \times 3}, \\ A^1(\tau) &:= \mathcal{P}_{p+1}(\tau)^3 \oplus \mathcal{P}_{p}(\tau)^3 \oplus \bigoplus_{z \in \Delta_0(\tau)} \mathbb{R}^{3 \times 3}, \qquad & A^1(z) &:= \mathbb{R}^3 \oplus \mathbb{R}^{3 \times 3}, \\ A^2(\tau) &:= \mathcal{P}_{p}(\tau)^3, \qquad & A^2(z) &:= \mathbb{R}^3, \end{align}\] where \(\mathbb{R}^{3 \times 3}_{\mathop{\mathrm{sym}}}\) denotes the set of \(3\times 3\) symmetric matrices with real entries.
For each \(k \in 0:3\), we show that \(A^k := \{ A^k(\tau) : \tau \in \Delta(\mathcal{T}) \}\) may be equipped with a trace structure in the sense of [18]. That is, for \(\tau \in \Delta(\mathcal{T})\) and \(\eta \mathop{\mathrm{\unlhd}}\tau\), we define linear operators \(\mathop{\mathrm{Tr}}^k_{\eta \mathop{\mathrm{\leftarrow}}\tau} : A^k(\tau) \to A^k(\eta)\) satisfying the following properties:
\(\mathop{\mathrm{Tr}}^k_{\tau \mathop{\mathrm{\leftarrow}}\tau}\) is the identity map for all \(\tau \in \Delta(\mathcal{T})\).
For all \(\sigma \in \mathcal{T}\) and \(\eta \mathop{\mathrm{\unlhd}}\tau \mathop{\mathrm{\unlhd}}\sigma\), there holds \[\begin{align} \label{eq:trace-vanishing-composition} \mathop{\mathrm{Tr}}^k_{\eta \mathop{\mathrm{\leftarrow}}\sigma} u = 0 \implies \mathop{\mathrm{Tr}}^k_{\eta \mathop{\mathrm{\leftarrow}}\tau} \circ \mathop{\mathrm{Tr}}^k_{\tau \mathop{\mathrm{\leftarrow}}\sigma} u = 0 \qquad \forall u \in A^k(\sigma). \end{align}\tag{39}\]
\(\mathop{\mathrm{Tr}}^k := \{ \mathop{\mathrm{Tr}}^k_{\eta \mathop{\mathrm{\leftarrow}}\tau} : \eta \mathop{\mathrm{\unlhd}}\tau, \;\tau \in \Delta(\mathcal{T}) \}\) characterizes \(V^{k, h}\): \[\begin{gather} \label{eq:trace-characterizes-space} V^{k, h} = \{ u \in L^2(\Omega) \otimes \mathbb{X}^k : u|_{K} \in A^k(K) \text{ and } \\ \mathop{\mathrm{Tr}}^k_{\tau \mathop{\mathrm{\leftarrow}}K} u|_{K} = \mathop{\mathrm{Tr}}^k_{\tau \mathop{\mathrm{\leftarrow}}K'} u|_{K'} \;\forall \tau \mathop{\mathrm{\unlhd}}K, K', \;\forall K, K' \in \mathcal{T}\}. \end{gather}\tag{40}\]
Recall that \(\mathbb{X}^k\) are the global ambient spaces defined in 12 . If \((A^k, \mathop{\mathrm{Tr}}^k)\) satisfy (i-ii), then \((A^k, \mathop{\mathrm{Tr}}^k)\) is a trace structure in the sense of [18], while (iii) ensures that \(V^{k, h}\) is the “global space” with respect to the trace structure [18]. In the following subsections, we construct the trace operators \(\mathop{\mathrm{Tr}}^k_{\eta \mathop{\mathrm{\leftarrow}}\tau}\) for \(\eta \neq \tau\), tacitly assuming that (i) always holds.
For \(K \in \mathcal{T}\) and \(\tau \mathop{\mathrm{\lhd}}K\), we define for \(v \in A^0(K)\) \[\begin{align} \mathop{\mathrm{Tr}}^0_{\tau \mathop{\mathrm{\leftarrow}}K} v &= v|_{\tau} \oplus \mathop{\mathrm{grad}}v|_{\tau} \oplus \bigoplus_{z \in \Delta_0(\tau)} \mathop{\mathrm{hess}}v(z), \end{align}\] where \(w|_{\tau} := w(\tau)\) if \(\dim \tau = 0\) and we recall that \(\mathop{\mathrm{hess}}\) denotes the Hessian operator. Given \(\tau \in \Delta_{1}(\mathcal{T}) \cup \Delta_{2}(\mathcal{T})\) and \(\eta \mathop{\mathrm{\lhd}}\tau\), we define for \(\phi \oplus \psi \oplus \bigoplus_{z \in \Delta_0(\tau)} M_z \in A^0(\tau)\) \[\begin{align} \mathop{\mathrm{Tr}}^0_{\eta \mathop{\mathrm{\leftarrow}}\tau} \left( \phi \oplus \psi \oplus \bigoplus_{z \in \Delta_0(\tau)} M_z \right) &= \phi|_{\eta} \oplus \psi|_{\eta} \oplus \bigoplus_{z \in \Delta_0(\eta)} M_z. \end{align}\] Then, one may readily verify that 40 39 hold for \(k = 0\).
For \(K \in \mathcal{T}\) and \(\tau \mathop{\mathrm{\lhd}}K\), we define for \(v \in A^1(K)\) \[\begin{align} \mathop{\mathrm{Tr}}^1_{\tau \mathop{\mathrm{\leftarrow}}K} := \begin{dcases} v|_{\tau} \oplus \mathop{\mathrm{curl}}v|_{\tau} \oplus \bigoplus_{z \in \Delta_0(\tau)} \mathop{\mathrm{grad}}v(z) & \text{if } \dim \tau > 0, \\ v(\tau) \oplus \mathop{\mathrm{grad}}v(\tau) & \text{if } \dim \tau = 0. \end{dcases} \end{align}\] For \(\tau \in \Delta_1(\mathcal{T}) \cup \Delta_2(\mathcal{T})\) and \(\eta \mathop{\mathrm{\lhd}}\tau\), we define for \(\phi \oplus \psi \oplus \bigoplus_{z \in \Delta_0(\tau)} M_z \in A^1(\tau)\) \[\begin{align} \mathop{\mathrm{Tr}}^1_{\eta \mathop{\mathrm{\leftarrow}}\tau} \left( \phi \oplus \psi \oplus \bigoplus_{z \in \Delta_0(\tau)} M_z \right) := \begin{dcases} \phi|_{\eta} \oplus \psi|_{\eta} \oplus \bigoplus_{z \in \Delta_0(\eta)} M_z & \text{if } \dim \eta > 0, \\ \phi(\eta) \oplus M_{\eta} & \text{if } \dim \eta = 0. \end{dcases} \end{align}\] Then, we may readily verify that 39 40 hold for \(k = 1\).
For \(\tau \in \Delta(\mathcal{T})\) and \(\eta \mathop{\mathrm{\lhd}}\tau\), we define for \({v \in A^2(\tau)}\) \[\begin{align} \mathop{\mathrm{Tr}}_{\eta \mathop{\mathrm{\leftarrow}}\tau}^2 v = v|_{\eta}, \end{align}\] while for \(w \in A^3(\tau)\), we set \(\mathop{\mathrm{Tr}}_{\eta \mathop{\mathrm{\leftarrow}}\tau}^3 w = 0\). Then, 39 40 hold for \(k = 2,3\).
Following [18], for \(K \in \mathcal{T}\) and \(\tau \mathop{\mathrm{\unlhd}}K\), we define \(B^k(\tau; K)\) by \[\begin{align} \label{eq:generic-bubble-space} B^k(\tau; K) := \{ v \in \mathop{\mathrm{Tr}}^k_{\tau \mathop{\mathrm{\leftarrow}}K} A^k(K) : \mathop{\mathrm{Tr}}^k_{\eta \mathop{\mathrm{\leftarrow}}\tau} v = 0 \;\forall \eta \mathop{\mathrm{\lhd}}\tau \}. \end{align}\tag{41}\] We now show that each trace structure \((A^k, \mathop{\mathrm{Tr}}^k)\) satisfies the geometric decomposition property [18]: \[\begin{align} \label{eq:geometric-decomposition} \sum_{\tau \mathop{\mathrm{\unlhd}}K} \dim B^k(\tau; K) = \dim A^k(K) \qquad \forall K \in \mathcal{T}. \end{align}\tag{42}\] In the following subsections, let \(K \in \mathcal{T}\), \(f \in \Delta_2(K)\), \(e \in \Delta_1(K)\), and \(z \in \Delta_0(K)\).
Expanding definitions, we see that \[\begin{align} B^0(K; K) &= V_0^{0, h}(K_A), \\ B^0(f; K) &= \left\{ v \oplus w \oplus 0 \in A^0(f) : v \in H^2_0(f), \;w \in H^1_0(f)^3, \right. \\ &\qquad \qquad \left. \text{and } (I - \hat{n}_f \otimes \hat{n}_f) w = \mathop{\mathrm{grad}}_f v \right\}, \\ B^0(e; K) &= \left\{ v \oplus w \oplus 0 \in A^0(e) : v \in H^3_0(e), w \in H^2_0(e)^3, \text{ and } w \cdot \hat{t}_e = \partial_{\hat{t}_e} v \right\}, \\ B^0(z; K) &= \mathbb{R} \oplus \mathbb{R}^3 \oplus \mathbb{R}_{\mathop{\mathrm{sym}}}^{3 \times 3}, \end{align}\] where \(\mathop{\mathrm{grad}}_f\) denotes the surface gradient (viewed as an element of \(\mathbb{R}^3\)). As a consequence, we obtain \[\begin{align} \dim B^0(f; K) &= \dim \mathcal{P}_{p-4}(f) + \dim \mathcal{P}_{p-2}(f), \\ \dim B^0(e; K) &= \dim \mathcal{P}_{p-4}(e) + \dim \mathcal{P}_{p-3}(e)^2, \end{align}\] and so performing a direct calculation and applying [17] shows that 42 holds for \(k=0\).
Expanding definitions gives \[\begin{align} B^1(K; K) &= V_0^{1, h}(K_A), \\ B^1(f; K) &= \left\{ v \oplus w \oplus 0 \in A^1(f) : v,w \in H^1_0(f)^3, \; \text{and } w \cdot \hat{n}_f = \mathop{\mathrm{rot}}_f v \right\}, \\ B^1(e; K) &= \left\{ v \oplus w \oplus 0 \in A^1(e) : v \in H^2_0(e) \text{ and } w \in H^1_0(e) \right\}, \\ B^1(z; K) &= \mathbb{R}^3 \oplus \mathbb{R}^{3 \times 3}, \end{align}\] where \(\mathop{\mathrm{rot}}_f\) is the surface curl defined so that \(\mathop{\mathrm{rot}}_f v|_f := \mathop{\mathrm{curl}}v \cdot \hat{n}_f|_{f}\) for \(v \in C^{\infty}(\mathbb{R}^3)^3\). Note that every \(v \in \mathcal{P}_{p+1}(f)\) satisfies \[\begin{align} v|_{\partial f} = 0 \text{ and } \mathop{\mathrm{rot}}_f v|_{\partial f} = 0 \implies \frac{\partial }{\partial \hat{t}_e \times \hat{n}_f} v \cdot \hat{t}_{e} = 0 \qquad \forall e \in \Delta_1(f), \end{align}\] and so \[\begin{align} \dim B^1(f; K) &= \dim \mathcal{P}_{p-2}(f)^3 - \sum_{e' \in \Delta_1(f)} \dim \mathcal{P}_{p-2}(e') + \dim \mathcal{P}_{p-3}(f)^2 \\ &= \dim \mathcal{P}_{p-2}(f) + \left( p(p-1) - 3(p-1) \right) + \dim \mathcal{P}_{p-3}(f)^2. \end{align}\] Moreover, \(\dim B^1(e; K) = \dim \mathcal{P}_{p-3}(e)^3 + \dim \mathcal{P}_{p-2}(e)^3\), and so performing a direct calculation and applying [17] shows that 42 holds for \(k=1\).
The case \(k=2\) corresponds to the usual bubble spaces: \(B^2(K; K) = V^{2, h}_0(K_A)\), \(B^2(\tau; K) = \mathcal{P}_{p}(\tau)^3 \cap H^1_0(\tau)^3\) for \(\tau \in \Delta_1(K) \cup \Delta_2(K)\), and \(B^2(z; K) = \mathbb{R}^3\). Moreover, \(B^3(K; K) = A^3(K)\), and so 42 holds for \(k=2,3\).
Again let \(K \in \mathcal{T}\), \(f \in \Delta_2(K)\), \(e \in \Delta_1(K)\), and \(z \in \Delta_0(K)\), and consider the following diagram: \[\label{eq:full-spaces-trace-complex} \begin{tikzcd}[ampersand replacement = \&, column sep = 8em] A^0(K) \arrow[r, "\mathop{\mathrm{d}}_K^0"] \arrow[d, "\mathop{\mathrm{Tr}}"] \arrow[dd, bend right=60, "\mathop{\mathrm{Tr}}" description] \arrow[ddd, bend right=70, "\mathop{\mathrm{Tr}}" description] \& A^1(K) \arrow[r, "\mathop{\mathrm{d}}_K^1"] \arrow[d, "\mathop{\mathrm{Tr}}"] \& A^2(K) \arrow[d, "\mathop{\mathrm{Tr}}"] \\ A^0(f) \arrow[r, "\mathop{\mathrm{d}}^{0}_{f}"] \& A^1(f) \arrow{r}{\mathop{\mathrm{d}}^{1}_{f}} \& A^2(f) \\ A^0(e) \arrow{r}{\mathop{\mathrm{d}}^{0}_{e}} \& A^1(e) \arrow{r}{\mathop{\mathrm{d}}^{1}_{e}} \arrow[from=uu, bend right=60, "\mathop{\mathrm{Tr}}" description, pos=0.66] \& A^2(e) \arrow[from=uu, bend right=60, "\mathop{\mathrm{Tr}}" description, pos=0.66] \\ A^0(z) \arrow{r}{\mathop{\mathrm{d}}^{0}_{z}} \& A^1(z) \arrow{r}{\mathop{\mathrm{d}}^{1}_{z}} \arrow[from=uuu, bend right=70, "\mathop{\mathrm{Tr}}" description] \& A^2(z), \arrow[from=uuu, bend right=70, "\mathop{\mathrm{Tr}}" description] \end{tikzcd}\tag{43}\] where the vertical arrows are the corresponding trace operators (with sub and superscripts omitted) and the “differential” operators on the horizontal arrows are defined as follows: Let \(\mathop{\mathrm{d}}^0_K := \mathop{\mathrm{grad}}\) and \(\mathop{\mathrm{d}}^1_K = \mathop{\mathrm{curl}}\). For \(\tau \in \Delta_1(K) \cup \Delta_2(K)\), let \[\begin{align} \mathop{\mathrm{d}}_{\tau}^0 \left( v \oplus w \oplus \bigoplus_{z \in \Delta_0(\tau)} M_z \right) &:= w \oplus 0 \oplus \bigoplus_{z \in \Delta_0(\tau)} M_z, \\ \mathop{\mathrm{d}}_{\tau}^1 \left( v \oplus w \oplus \bigoplus_{z \in \Delta_0(\tau)} M_{\tau} \right) &:= w. \end{align}\] while for \(z \in \Delta_0(K)\), we define \[\begin{align} \mathop{\mathrm{d}}_z^0 \left( c \oplus v \oplus M \right) &= v \oplus M \quad \text{and} \quad \mathop{\mathrm{d}}_z^1 \left( v \oplus M \right) = \sum_{i, j, k = 1}^{3}\epsilon_{ijk} M_{j, i} \hat{e}_k, \end{align}\] where \(\hat{e}_k\) is the standard unit vector in the \(k\)-th direction, and \(\epsilon_{ijk}\) is the permutation symbol. Then, one may verify that 43 commutes and each row is a complex. Thus, \((A^k, \mathop{\mathrm{Tr}}^k, \mathop{\mathrm{d}}^k)\) is a conforming finite element subcomplex with trace structure (FECTS) ([18]) of the de Rham complex.
For \(k \in 0:3\) and \(\tau \in \Delta_k(\mathcal{T})\), we recall the “currents" \(\mathcal{I}_{\tau}^k : C(\bar{\Omega}) \otimes \mathbb{X}^k \to \mathbb{R}\) defined in 18 . In particular, the Stokes formula holds for \(\sigma \in \Delta_{k+1}(\mathcal{T})\) and \(w \in C(\bar{\Omega}) \otimes \mathbb{X}^k\): \[\begin{align} \label{eq:currents-stokes-thm} \mathcal{I}^{k+1}_{\sigma}(\mathop{\mathrm{d}}^k w) = \sum_{\tau \in \Delta_{k}(\sigma)} \mathcal{O}(\tau, \sigma) \mathcal{I}^{k}_{\tau}(w), \end{align}\tag{44}\] where \(\mathcal{O}(\tau, \sigma)\) denotes the orientation of \(\tau\) relative to \(\sigma\). Thus, \((\mathcal{I}, \mathbb{R})\) is a family of generalized currents [18]. Note that [18] assumes that the domain of \(\mathcal{I}_{\tau}^k\) is \(C^{\infty}(\bar{\Omega}) \otimes \mathbb{X}^k\); however, one only needs that \(\mathcal{I}^k_{\tau}\) is well-defined on \(V^{k, h}\).
We also see that for each \(k \in 0:3\) and \(\tau \in \Delta_k(\mathcal{T})\), the functionals \({\tilde{\mathcal{I}}_{\tau}^k : A^k(\tau) \to \mathbb{R}}\) defined by \[\begin{align} {2} \tilde{\mathcal{I}}^0_{\tau} ( c \oplus v \oplus M ) &:= c, \qquad & \tilde{\mathcal{I}}^2_{\tau} (w) &:= \int_{\tau} w \cdot \hat{n}_{\tau} \,\mathrm{d}{s}, \\ \tilde{\mathcal{I}}^1_{\tau} \left(\phi \oplus \psi \oplus \bigoplus_{z \in \Delta_0(\tau)} M_z \right) &:= \int_{\tau} \phi \cdot \hat{t}_{\tau} \,\mathrm{d}{s}, & \qquad \tilde{\mathcal{I}}^3_{\tau} (q) &:= \int_{\tau} q \,\mathrm{d}{x}, \end{align}\] satisfy \[\begin{align} \label{eq:current-trace-compatibility} \tilde{\mathcal{I}}_{\tau}^k( \mathop{\mathrm{Tr}}^k_{\tau \mathop{\mathrm{\leftarrow}}\sigma} w) = \mathcal{I}_{\tau}^k(w) \qquad \forall w \in A^{k}(\sigma), \;\forall \sigma \in \mathcal{T}: \tau \mathop{\mathrm{\unlhd}}\sigma. \end{align}\tag{45}\] Note that above, we are viewing \(A^k(\sigma)\) as defined on all of \(\Omega\), where any \(V^{k}\)-continuous extension is taken.
The bubble spaces 41 do not depend on the parent tetrahedron, so we drop “\(K\)" from the notation. Define the modified bubble spaces \[\begin{align} \tilde{B}^k(\tau) := \begin{cases} B^k(\tau) \cap \ker \tilde{\mathcal{I}}_{\tau}^k & \text{if } k = \dim \tau, \\ B^k(\tau) & \text{otherwise}. \end{cases} \end{align}\] The next result shows that these modified bubble spaces form an exact sequence.
Lemma 1. The following complex is exact for any \(\tau \in \Delta(\mathcal{T})\): \[\label{eq:modified-bubble-complex} \begin{tikzcd}[ampersand replacement = \&] 0 \arrow[r] \& \tilde{B}^0(\tau) \arrow[r, "\mathop{\mathrm{d}}_{\tau}^0"] \& \tilde{B}^1(\tau) \arrow[r, "\mathop{\mathrm{d}}_{\tau}^1"] \& \tilde{B}^2(\tau) \arrow[r, "\mathop{\mathrm{d}}_{\tau}^2"] \& \tilde{B}^3(\tau) \arrow[r] \& 0. \end{tikzcd}\tag{46}\]
Proof. That 46 is a complex follows from the divergence and Stokes theorems. For \(\tau \in \mathcal{T}\), the exactness of 46 follows from [17] (see also [17]). For \(\tau \in \Delta(\mathcal{T}) \setminus \mathcal{T}\), \(\tilde{B}^3(\tau) = \{0\}\), and we also have \[\begin{gather} \dim \tilde{B}^0(\tau) + \dim \tilde{B}^2(\tau) - \dim \tilde{B}^1(\tau) \\ = \dim B^0(\tau) + \dim B^2(\tau) - \dim B^1(\tau) + (-1)^{\dim \tau + 1} = 0. \end{gather}\] The exactness of 46 now follows from a standard counting argument. ◻
Thanks to 45 and 1, the FECTS \((A^k, \mathop{\mathrm{Tr}}^k, \mathop{\mathrm{d}}^k)\) is compatible with respect to the currents \((\mathcal{I}^k, \mathbb{R})\) in the sense of [18].
For \(k \in 0:3\), let \((\cdot,\cdot)_{A^k(\tau)}\) denote the natural \(L^2(\tau)\) inner product on \(A^k(\tau)\). Let \(\mathbb{Q}^k_{\tau} : A^k(\tau) \to \mathop{\mathrm{d}}_{\tau}^{k-1} \tilde{B}^{k-1}(\tau)\) denote the \((\cdot,\cdot)_{A^k(\tau)}\)-orthogonal projection onto \(\mathop{\mathrm{d}}_{\tau}^{k-1} \tilde{B}^{k-1}(\tau)\): \[\begin{align} (\mathbb{Q}^k_{\tau} u, \mathop{\mathrm{d}}_{\tau}^{k-1} v)_{A^k(\tau)} = (u, \mathop{\mathrm{d}}_{\tau}^{k-1} v)_{A^k(\tau)} \qquad \forall v \in \tilde{B}^{k-1}(\tau), \;\forall u \in A^k(\tau). \end{align}\] We define the harmonic inner product \(A^k(\tau)\) [18] as follows: \[\begin{align} \langle u, v \rangle_{A^k(\tau)} := (\mathbb{Q}^k_{\tau} u, \mathbb{Q}^k_{\tau} v)_{A^k(\tau)} + (\mathop{\mathrm{d}}_{\tau}^k u, \mathop{\mathrm{d}}_{\tau}^k v)_{A^{k+1}(\tau)}. \end{align}\] Note that \(\langle \cdot, \cdot \rangle_{A^k(\tau)}\) is an inner product on \(\tilde{B}^k(\tau)\) owing to the exactness of 46 .
For \(v \in V^{k, h}\), property 40 means that for \(\tau \in \Delta(\mathcal{T})\), we may define the trace of \(v\) on \(\tau\) independent of the parent tetrahedron; i.e. \(\mathop{\mathrm{Tr}}^k_{\tau} v := \mathop{\mathrm{Tr}}^k_{\tau \mathop{\mathrm{\leftarrow}}K} v\), where \(K \in \mathcal{T}\) with \(\tau \mathop{\mathrm{\unlhd}}K\) is well-defined independent of the particular choice of \(K\). As shown in [18] and the remaining discussion on [18], a unisolvent set of degrees of freedom on \(V^{k, h}\) are \[\label{eq:global-dofs-bubble} \begin{alignat}{2} &\langle \mathop{\mathrm{Tr}}^k_{\tau} v, \phi \rangle_{_{A^k(\tau)}} \qquad & &\forall \phi \in \tilde{B}^k(\tau), \;\forall \tau \in \Delta(\mathcal{T}), \\ &\mathcal{I}_{\tau}^k(v) \qquad & &\forall \tau \in \Delta_k(\mathcal{T}). \end{alignat}\tag{47}\] Consequently, we define a lift of the modified bubble functions and the skeletal space as follows: \[\begin{align} \mathbb{B}^k(\tau) &:= \{ v \in V^{k, h} : \mathcal{I}^k_{\eta}(v) = 0 \;\forall \eta \in \Delta_k(\mathcal{T}) \text{ and} \\ &\qquad \qquad \langle \mathop{\mathrm{Tr}}_{\eta}^k v, w \rangle_{A^k(\eta)} = 0 \;\forall w \in \tilde{B}^k(\eta), \;\forall \eta \in \Delta(\mathcal{T}) \setminus \{\tau\} \} \qquad \forall \tau \in \Delta(\mathcal{T}),\\ \mathcal{S}^k &:= \{ v \in V^{k, h} : \langle \mathop{\mathrm{Tr}}_{\eta}^k v, w \rangle_{A^k(\eta)} = 0 \;\forall w \in \tilde{B}^k(\eta), \;\forall \eta \in \Delta(\mathcal{T}) \}. \end{align}\] The next lemma summarizes the support properties of these spaces.
Lemma 2. Let \(\tau \in \Delta(\mathcal{T})\).
(i) For \(v \in \mathbb{B}^k(\tau)\), there holds \[\begin{align} \label{eq:modified-bubble-support} \mathop{\mathrm{Tr}}^k_{\eta} v = 0
\qquad \forall \eta \in \left( \bigcup_{\ell=0}^{\dim \tau} \Delta_{\ell}(\mathcal{T}) \cup \bigcup_{\ell=\dim \tau + 1}^{3} \{ \rho \in \Delta_{\ell}(\mathcal{T}) : \tau \mathop{\mathrm{\ntrianglelefteq}}\rho \} \right) \setminus \{\tau\}.
\end{align}\tag{48}\]
(ii) For \(v \in \mathcal{S}^k\), there holds \[\begin{align} \label{eq:skeletal-support} \mathcal{I}^k_{\eta}(v) = 0 \qquad \forall \eta \in \Delta_k(\tau) \implies \mathop{\mathrm{Tr}}^k_{\tau} v = 0. \end{align}\tag{49}\] In particular, \(\mathop{\mathrm{Tr}}^k_{\tau} v = 0\) if \(\tau \in \bigcup_{\ell = 0}^{k-1} \Delta_{\ell}(\mathcal{T})\).
Proof. For (i), let \(\eta\) be as in 48 . Then, we have \(\langle \mathop{\mathrm{Tr}}^{k}_{\rho} v, w \rangle_{A^k(\rho)} = 0\) for all \(w \in \tilde{B}^k(\eta)\) and \(\rho \in \Delta(\eta)\) and \(\mathcal{I}^k_{\rho}(v) = 0\) for all \(\rho \in \Delta_k(\mathcal{T})\). Clearly, \(\mathop{\mathrm{Tr}}^k_{z} v = 0\) for \(z \in \Delta_0(\eta)\) and so \(\mathop{\mathrm{Tr}}^k_{e} v \in \tilde{B}^k(e)\) for all \(e \in \Delta_1(\eta)\). The result now follows by an induction argument since the degrees of freedom \(\langle \cdot, w \rangle_{A^k(\rho)}\) are unisolvent on \(\tilde{B}^k(\rho)\) [18]. (ii) follows from similar arguments. ◻
Thanks to [18], each column of the following diagram is a direct sum decomposition, each row is a complex, and the final row is exact: \[\label{eq:stokes-complex-full-geom-decomp-nobcs} \begin{tikzcd}[ampersand replacement=\&] 0 \arrow[r] \& V^{0, h} \arrow[r, "\mathop{\mathrm{grad}}"] \& V^{1, h} \arrow[r, "\mathop{\mathrm{curl}}"] \& V^{2, h} \arrow[r, "\undefined"] \& V^{3, h} \arrow[r] \& 0 \\[-2em] \& \rotatebox{90}{=} \& \rotatebox{90}{=} \& \rotatebox{90}{=} \& \rotatebox{90}{=} \& \\[-2em] 0 \arrow[r] \& \mathcal{S}^0 \arrow[r, "\mathop{\mathrm{grad}}"] \& \mathcal{S}^1 \arrow[r, "\mathop{\mathrm{curl}}"] \& \mathcal{S}^2 \arrow[r, "\undefined"] \& \mathcal{S}^3 \arrow[r] \& 0 \\[-2em] \& \bigoplus\limits_{\tau \in \Delta(\mathcal{T})} \& \bigoplus\limits_{\tau \in \Delta(\mathcal{T})} \& \bigoplus\limits_{\tau \in \Delta(\mathcal{T})} \& \bigoplus\limits_{\tau \in \Delta(\mathcal{T})} \& \\[-2em] 0 \arrow[r] \& \mathbb{B}^0(\tau) \arrow[r, "\mathop{\mathrm{grad}}"] \& \mathbb{B}^1(\tau) \arrow[r, "\mathop{\mathrm{curl}}"] \& \mathbb{B}^2(\tau) \arrow[r, "\undefined"] \& \mathbb{B}^3(\tau) \arrow[r] \& 0, \end{tikzcd}\tag{50}\] In particular, the cohomology of the first row of 50 is isomorphic to the cohomology of the second row, which in turn is isomorphic to the de Rham cohomology (the cohomology of 6 with \(\Gamma_0 = \emptyset\)) [18].
Recall that \(\mathcal{U}\) is a conforming mesh of \(\Gamma_0\). The supersmoothness of the spaces \(V_{\Gamma_0}^{0, h}\) and \(V_{\Gamma_0}^{1, h}\) at mesh vertices introduces some subtleties that are not naturally reflected in the trace operators. In particular, denote the mesh vertices laying on the “flat” portion of the \(\bar{\Gamma}_{0}\) by \[\begin{align} \Delta_0^{\flat}(\mathcal{U}) := \{ z \in \Delta_0(\mathcal{U}) : \text{ all faces meeting at z are coplanar} \}. \end{align}\] Then, the following result shows that not all Hessian degrees of freedom for functions in \(V_{\Gamma_0}^{0, h}\) or gradient degrees of freedom vector fields in \(V_{\Gamma_0}^{1, h}\) vanish on \(\Delta_0^{\flat}\).
Lemma 3. For \(v \in V_{\Gamma_0}^{0, h}\) and \(w \in V_{\Gamma_0}^{1, h}\), there holds \[\tag{51} \begin{align} \tag{52} \mathop{\mathrm{Tr}}_{z}^0 v &= \begin{dcases} 0 \oplus 0 \oplus \partial_{\hat{n}_{\Gamma}}^2 v(z) \hat{n}_{\Gamma} \otimes \hat{n}_{\Gamma} & \text{if } z \in \Delta_0^{\flat}(\mathcal{U}), \\ 0 & \text{if } z \in \Delta_0(\mathcal{U}) \setminus \Delta_0^{\flat}(\mathcal{U}), \end{dcases} \\ \tag{53} \mathop{\mathrm{Tr}}_{z}^1 w &= \begin{dcases} 0 \oplus \partial_{\hat{n}_{\Gamma}} (w \cdot \hat{n}_{\Gamma})(z) \hat{n}_{\Gamma} \otimes \hat{n}_{\Gamma} & \text{if } z \in \Delta_0^{\flat}(\mathcal{U}), \\ 0 & \text{if } z \in \Delta_0(\mathcal{U}) \setminus \Delta_0^{\flat}(\mathcal{U}), \end{dcases} \end{align}\] where \(\hat{n}_{\Gamma}\) is the outward unit normal vector on \(\partial \Omega\).
Proof. Assume first that \(z \in \Delta_0^{\flat}(\mathcal{U})\) and let \(f \in \mathcal{U}\) be such that \(z \in \Delta_0(f)\). Then, \(v|_f = 0\) and \(\partial_n v|_f = 0\), and so \(v(z) = 0\) and \(\mathop{\mathrm{grad}}v(z) = 0\). Moreover, differentiating in the tangent plane of \(f\) then shows that \(D_f^{\alpha} \mathop{\mathrm{grad}}v(z) = 0\) for all \(|\alpha| \geq 0\), where \(D_f\) denotes the surface differential. In particular, \((I - \hat{n}_{f} \otimes \hat{n}_{f}) \mathop{\mathrm{hess}}v(z) = 0\). Similarly, \(w|_f = 0\) and \(\mathop{\mathrm{curl}}w|_{f} = 0\) and so \(w(z) = 0\), \(\mathop{\mathrm{grad}}_f w(z) = 0\), and \(\mathop{\mathrm{curl}}w(z) = 0\). Consequently, \((I - \hat{n}_{f} \otimes \hat{n}_{f}) \mathop{\mathrm{grad}}w(z) = 0\). The first case of 52 and 53 now follows since \(\hat{n}_f = \hat{n}_{\Gamma}\).
Now suppose that \(z \in \Delta_0(\mathcal{U}) \setminus \Delta_0^{\flat}(\mathcal{U})\). Then, there exist \(f, f' \in \mathcal{U}\) not coplanar such that \(z \in \Delta_0(f)\) and \(z \in \Delta_0(f')\). The same arguments as above show that \(v(z) = \mathop{\mathrm{grad}}v(z) = 0\), \(w(z) = 0\), and \[\begin{align} (I - \hat{\mu} \otimes \hat{\mu}) \mathop{\mathrm{hess}}v(z) = (I - \hat{\mu} \otimes \hat{\mu}) \mathop{\mathrm{grad}}w(z) &= 0, \qquad \forall \mu \in \{ \hat{n}_f, \hat{n}_{f'} \}. \end{align}\] Thus, \(\mathop{\mathrm{hess}}v(z) = \mathop{\mathrm{grad}}w(z) = 0\). ◻
One consequence of 3 is that the “zero trace spaces” defined as the kernel of \(\mathop{\mathrm{Tr}}^k_{f}\) for \(f \in \mathcal{U}\) may not coincide with \(V_{\Gamma_0}^{k, h}\) for \(k=0, 1\); i.e., \[\begin{align} \{ v \in V^{k, h} : \mathop{\mathrm{Tr}}^k_{f} v = 0 \text{ if } f \in \mathcal{U}\} \subseteq V^{k, h}_{\Gamma_0}, \end{align}\] but the inclusion will generally be strict for \(k=0, 1\) if \(\Delta_0^{\flat}(\mathcal{U})\) is nonempty. To rectify this discrepancy, we modify the vertex bubble functions further for \(z \in \Delta_0^{\flat}(\mathcal{U})\): \[\begin{align} \mathbb{B}^{0, nn}(z) &:= \{ v \in \mathbb{B}^{0}(z) : v(z) = 0, \;\mathop{\mathrm{grad}}v(z) = 0, \; (I - \hat{n}_{\Gamma} \otimes \hat{n}_{\Gamma})\mathop{\mathrm{hess}}v(z) = 0 \}, \\ \mathbb{B}^{1, nn}(z) &:= \{ v \in \mathbb{B}^{1}(z) : v(z) = 0, \; (I - \hat{n}_{\Gamma} \otimes \hat{n}_{\Gamma}) \mathop{\mathrm{grad}}v(z) = 0\}. \end{align}\] The corresponding modified bubble spaces, which will be shown to satisfy the boundary conditions, are given by \[\begin{align} \mathbb{B}_{\Gamma_0}^k(\tau) := \begin{cases} \mathbb{B}^k(\tau) & \text{if } \tau \in \Delta(\mathcal{T}) \setminus \Delta(\mathcal{U}), \\ \mathbb{B}^{k, nn}(\tau) & \text{if } \tau \in \Delta_0^{\flat}(\mathcal{U}) \text{ and } k < 2, \\ \{0\} & \text{otherwise}, \end{cases} \qquad \forall \tau \in \Delta(\mathcal{T}), \;\forall k \in 0:3. \end{align}\] The skeletal spaces with boundary conditions are simply \[\begin{align} \mathcal{S}_{\Gamma_0}^k := \{ v \in \mathcal{S}^k : \mathcal{I}^k_{\tau}(v) = 0 \;\forall \tau \in \Delta_k(\mathcal{U}) \} \qquad \forall k \in 0:3. \end{align}\] The next result shows that these spaces do indeed satisfy the boundary conditions.
Lemma 4. For \(k \in 0:3\), \(\mathcal{S}^k_{\Gamma_0} \subset V_{\Gamma_0}^{k, h}\) and \(\mathbb{B}_{\Gamma_0}^k(\tau) \subset V_{\Gamma_0}^{k, h}\) for all \(\tau \in \Delta(\mathcal{T})\).
Proof. Let \(v \in \mathcal{S}^k_{\Gamma_0}\). For any \(f \in \mathcal{U}\), \(\mathcal{I}^k_{\eta}(v) = 0\) for all \(\eta \in \Delta_k(f)\), and so \(\mathop{\mathrm{Tr}}^k_{f} v = 0\) thanks to 2. Thus, \(v \in V_{\Gamma_0}^{k, h}\).
Now let \(\tau \in \Delta(\mathcal{T})\) and \(v \in \mathbb{B}_{\Gamma_0}^k(\tau)\). If \(\tau \notin \Delta(\mathcal{U})\), then \(\mathop{\mathrm{Tr}}^k_{f} v = 0\) for all \(f \in \mathcal{U}\) thanks to 2, and so \(v \in V_{\Gamma_0}^{k, h}\). Now suppose that \(\tau \in \Delta_0^{\flat}(\mathcal{U})\). Applying 2 once again gives \(\mathop{\mathrm{Tr}}_{f}^k v = 0\) for \(f \in \mathcal{U}\) with \(\tau \mathop{\mathrm{\ntrianglelefteq}}f\). Consequently, the final case to verify is \(f \in \mathcal{U}\) with \(\tau \mathop{\mathrm{\unlhd}}f\).
For \(k = 0\), expanding the definition of the harmonic inner products in the condition \(v \in \mathbb{B}_{\Gamma_0}^k(\tau)\) shows that \(v_f := v|_{f} \in \mathcal{P}_{p+2}(f)\) satisfies \[\label{eq:proof:Pp2-face-dofs} \begin{alignat}{2} D_f^{\alpha} v_f (z) &= 0 \qquad & &\forall |\alpha| \leq 2, \;\forall z \in \Delta_0(f), \\ (\partial_{\hat{t}_{\partial f}}v_f, \partial_{\hat{t}_{\partial f}}w)_{L^2(e)} &= 0 \qquad & &\forall w \in \mathcal{P}_{p+2}(e) \cap H^3_0(e), \; \forall e \in \Delta_1(f), \\ (\partial_{\hat{n}_{\partial f}} v_{f}, w )_{L^2(e)} &= 0 \qquad & &\forall w \in \mathcal{P}_{p+1}(e) \cap H^2_0(e), \; \forall e \in \Delta_1(f), \\ (\operatorname{grad}_fv_f, \operatorname{grad}_fw)_{L^2(f)} &=0 \qquad & &\forall w \in \mathcal{P}_{p+2}(f) \cap H^2_0(f), \end{alignat}\tag{54}\] where we recall \(D_f\) denotes the surface differential. Since the above degrees of freedom are unisolvent on \(\mathcal{P}_{p+2}(f)\), \(v_{f} \equiv 0\). Similarly, \(u_f = \partial_{n_f} v|_f \in \mathcal{P}_{p+1}(f)\) satisfies \[\tag{55} \begin{alignat}{2} D_f^{\alpha} u_f (z) &= 0 \qquad & &\forall |\alpha| \leq 1, \;\forall z \in \Delta_0(f), \\ (u_f, w)_{L^2(e)} &= 0 \qquad & &\forall w \in \mathcal{P}_{p+1}(e) \cap H^2_0(e), \; \forall e \in \Delta_1(f), \\ \tag{56} (u_f, w)_{L^2(f)} &=0 \qquad & &\forall w \in \mathcal{P}_{p+1}(f) \cap H^1_0(f), \end{alignat}\] and so \(u_f \equiv 0\). As a result, \(v \in V_{\Gamma_0}^{0, h}\).
For \(k=1\), each component of \(v_f := v|_{f} \in \mathcal{P}_{p+1}(f)^3\) also satisfies \[\tag{57} \begin{alignat}{2} D_f^{\alpha} v_f (z) &= 0 \qquad & &\forall |\alpha| \leq 1, \;\forall z \in \Delta_0(f), \\ (v_f, w)_{L^2(e)} &= 0 \qquad & &\forall w \in \mathcal{P}_{p+1}(e)^3 \cap H^2_0(e)^3, \; \forall e \in \Delta_1(f), \\ \tag{58} (v_f, \operatorname{grad}_f w)_{L^2(f)} &=0 \qquad & &\forall w \in \mathcal{P}_{p+2}(f) \cap H^2_0(f),\\ (\operatorname{rot}_fv_f, \operatorname{rot}_fw)_{L^2(f)} &=0 \qquad & &\forall w \in \mathcal{P}_{p+1}(f)^3 \cap H^1_0(f)^3, \end{alignat}\] and so \(v_f \equiv 0\). Moreover, \(u_f := \mathop{\mathrm{curl}}v|_f \in \mathcal{P}_p(f)^3\) satisfies \[\label{eq:proof:Pp-face-dofs} \begin{alignat}{2} u_f (z) &= 0 \qquad & &\forall z \in \Delta_0(f), \\ (u_f, w)_{L^2(e)} &= 0 \qquad & &\forall w \in \mathcal{P}_{p}(e)^3 \cap H^1_0(e)^3, \;\forall e \in \Delta_1(f), \\ (u_f, w)_{L^2(f)} &=0 \qquad & &\forall w \in \mathcal{P}_{p}(f)^3 \cap H^1_0(f)^3, \end{alignat}\tag{59}\] and so \(u_f \equiv 0\). Consequently, \(v \in V_{\Gamma_0}^{1, h}\). ◻
We also have the analog of the diagram 50 :
Lemma 5. Each column of the diagram below is a direct sum decomposition, each row is a complex, and the final row is exact. \[\label{eq:stokes-complex-full-geom-decomp} \begin{tikzcd}[ampersand replacement=\&] 0 \arrow[r] \& \tilde{V}_{\Gamma_0}^{0, h} \arrow[r, "\mathop{\mathrm{grad}}"] \& \tilde{V}_{\Gamma_0}^{1, h} \arrow[r, "\mathop{\mathrm{curl}}"] \& \tilde{V}_{\Gamma_0}^{2, h} \arrow[r, "\undefined"] \& \tilde{V}_{\Gamma_0}^{3, h} \arrow[r] \& 0 \\[-2em] \& \rotatebox{90}{=} \& \rotatebox{90}{=} \& \rotatebox{90}{=} \& \rotatebox{90}{=} \& \\[-2em] 0 \arrow[r] \& \mathcal{S}_{\Gamma_0}^0 \arrow[r, "\mathop{\mathrm{grad}}"] \& \mathcal{S}_{\Gamma_0}^1 \arrow[r, "\mathop{\mathrm{curl}}"] \& \mathcal{S}_{\Gamma_0}^2 \arrow[r, "\undefined"] \& \mathcal{S}_{\Gamma_0}^3 \arrow[r] \& 0 \\[-2em] \& \bigoplus\limits_{\tau \in \Delta(\mathcal{T})} \& \bigoplus\limits_{\tau \in \Delta(\mathcal{T})} \& \bigoplus\limits_{\tau \in \Delta(\mathcal{T})} \& \bigoplus\limits_{\tau \in \Delta(\mathcal{T})} \& \\[-2em] 0 \arrow[r] \& \mathbb{B}_{\Gamma_0}^0(\tau) \arrow[r, "\mathop{\mathrm{grad}}"] \& \mathbb{B}_{\Gamma_0}^1(\tau) \arrow[r, "\mathop{\mathrm{curl}}"] \& \mathbb{B}_{\Gamma_0}^2(\tau) \arrow[r, "\undefined"] \& \mathbb{B}_{\Gamma_0}^3(\tau) \arrow[r] \& 0. \end{tikzcd}\tag{60}\]
Proof. Step 1: Direct sum decomposition. The column for \(k=3\) is identical to 50 , so consider \(k \in 0:2\). Let \(v \in V_{\Gamma_0}^{k, h}\). Thanks to the direct sum decomposition in 50 , we have \[\begin{align} v = v_S + \sum_{\tau \in \Delta(\mathcal{T})} v_{\tau}, \quad \text{with} \quad v_S \in \mathcal{S}^k, \;v_{\tau} \in \mathbb{B}^k(\tau), \;\tau \in \Delta(\mathcal{T}). \end{align}\] For \(\eta \in \Delta_k(\mathcal{U})\), we have \(\mathcal{I}_{\eta}^{k}(v_S) = \mathcal{I}_{\eta}^{k}(v) = 0\), and so \(v_S \in \mathcal{S}_{\Gamma_0}^k\).
Note that \(\mathbb{B}_{\Gamma_0}^k(\tau) = \mathbb{B}^k(\tau)\) if \(\tau \notin \Delta(\mathcal{U})\), so suppose that \(\tau \in \Delta(\mathcal{U})\).
By 2, we have \[\begin{align} \langle \mathop{\mathrm{Tr}}_{\tau}^k v_{\tau}, w
\rangle_{A^k(\tau)} = \langle \mathop{\mathrm{Tr}}_{\tau}^k v, w \rangle_{A^k(\tau)} \qquad \forall w \in \tilde{B}^k(\tau).
\end{align}\] If \(\dim \tau \geq 1\) or \(k = 2\), then the above quantity vanishes and so \(v_{\tau} \equiv 0\) since the above degrees of freedom
are unisolvent on \(\tilde{B}^k(\tau)\) [18]. For \(\dim \tau = 0\) and \(k \in 0:1\), 3 shows that \(v_{\tau} \in \mathbb{B}^k_{\Gamma_0}(\tau)\). Thus, each column of
60 is a direct sum decomposition.
Step 2: Complex property. The second row of 60 is a complex since the second row of 50 is a complex and 44 holds. Thus, it remains to show that the final row is an exact complex. For \(\tau \in \Delta(\mathcal{T}) \setminus \Delta_0^{\flat}(\mathcal{U})\), we have \(\mathbb{B}_{\Gamma_0}^k(\tau) = \mathbb{B}^k(\tau)\) for all \(k \in 0:3\) or \(\mathbb{B}_{\Gamma_0}^k(\tau) = \{0 \}\) for all \(k
\in 0:3\), both of which are exact complexes thanks to 50 .
Now suppose that \(\tau \in \Delta_0^{\flat}(\mathcal{U})\), for which the bottom row of 60 reads \[\begin{tikzcd}[ampersand replacement=\&] 0 \arrow[r] \& \mathbb{B}^{0, nn}(\tau) \arrow[r, "\mathop{\mathrm{grad}}"] \& \mathbb{B}^{1, nn}(\tau) \arrow[r, "\mathop{\mathrm{curl}}"] \& 0 \arrow[r, "\undefined"] \& 0 \arrow[r] \& 0. \end{tikzcd}\] \(\mathbb{B}^{0, nn}(\tau)\) and \(\mathbb{B}^{1, nn}(\tau)\) clearly both have dimension 1, and so we only need to verify that \(\mathop{\mathrm{grad}}\mathbb{B}^{0, nn}(\tau) = \mathbb{B}^{1, nn}(\tau)\). Since the bottom row of 50 is a complex and \[\begin{align} \mathop{\mathrm{Tr}}_{\tau}^1 \mathop{\mathrm{grad}}v = 0 \oplus \partial_{\hat{n}_{\Gamma}}^2 v(\tau) \hat{n}_{\Gamma} \otimes \hat{n}_{\Gamma} \qquad \forall v \in \mathbb{B}^{0, nn}(\tau), \end{align}\] we have \(\mathop{\mathrm{grad}}\mathbb{B}^{0, nn}(\tau) = \mathbb{B}^{1, nn}(\tau)\). ◻
We now introduce the relative simplicial cochain complex, modifying the presentation in [35]. For \(k \in 0:3\), let \(C_k(\mathcal{V})\) denote the space of simplicial \(k\)-chains (i.e. formal linear combinations of subsimplices of dimension \(k\)) of a mesh \(\mathcal{V}\). Then, \(C_k(\mathcal{U})\) is a subspace of \(C_k(\mathcal{T})\), and we define \(C_k(\mathcal{T}, \mathcal{U}) := C_k(\mathcal{T}) / C_k(\mathcal{U})\). Let \({\partial_k : C_k(\mathcal{T}, \mathcal{U}) \to C_{k-1}(\mathcal{T}, \mathcal{U})}\) denote the standard simplicial boundary operator defined on the quotient space uniquely by the condition \[\begin{align} \partial_k (\tau + C_k(\mathcal{U})) = \sum_{\substack{\eta \in \Delta_{k-1}(\tau) \\ \eta \notin \Delta_{k-1}(\mathcal{U})}} \mathcal{O}(\eta, \tau) (\eta + C_{k-1}(\mathcal{U})) \qquad \forall \tau \in \Delta_k(\mathcal{T}) \setminus \Delta_k(\mathcal{U}). \end{align}\] Let \({\partial}'_k : {C_k(\mathcal{T}, \mathcal{U})}' \to {C_{k+1}(\mathcal{T}, \mathcal{U})}'\) denote the corresponding cochain map defined uniquely by the condition: \[\begin{align} ({\partial}'_k \ell_k)(\omega) = \ell_k( \partial_{k+1} \omega) \qquad \forall \omega \in C_{k+1}(\mathcal{T}, \mathcal{U}), \; \forall \ell_k \in {C_k(\mathcal{T}, \mathcal{U})}'. \end{align}\]
Proof of 1. Let \(k \in 0:3\). A simple consequence of [18] is that the currents \(\{ \mathcal{I}^k_{\eta} : \eta \in \Delta_k(\mathcal{T}) \}\) are a unisolvent set of degrees of freedom on \(\mathcal{S}^k\), and so \(\{ \mathcal{I}^k_{\eta} : \eta \in \Delta_k(\mathcal{T}) \setminus \Delta_k(\mathcal{U}) \}\) are unisolvent on \(\mathcal{S}^k_{\Gamma_0}\). Thus, \(\mathcal{S}^k_{\Gamma_0}\) and \({\mathcal{C}_k(\mathcal{T}, \mathcal{U})}'\) have the same dimension. We define \(\pi^k : \mathcal{S}^k_{\Gamma_0} \to {\mathcal{C}_K(\mathcal{T}, \mathcal{U})}'\) uniquely by the condition \[\begin{align} (\pi^k v)(\tau + C_{k}(\mathcal{U})) = \mathcal{I}^k_{\tau}(v) \qquad \forall \tau \in \Delta_{k}(\mathcal{T}) \setminus \Delta_{k}(\mathcal{U}), \; \forall v \in \mathcal{S}^k_{\Gamma_0}. \end{align}\] By unisolvency, \(\pi^k\) is injective and hence surjective. Thanks to 44 , we also have the following commutativity for all \(v \in \mathcal{S}_{\Gamma_0}^k\) and \(\tau \in \Delta_k(\mathcal{T}) \setminus \Delta_k(\mathcal{U})\): \[\begin{align} {\partial}'_k (\pi^k v)(\tau + C_{k+1}(\mathcal{U})) &= \sum_{\substack{\eta \in \Delta_{k-1}(\tau) \\ \eta \notin \Delta_{k-1}(\mathcal{U})}} \mathcal{O}(\eta, \tau) (\pi^k v)(\eta + C_{k}(\mathcal{U})) \\ &= \sum_{\substack{\eta \in \Delta_{k-1}(\tau) \\ \eta \notin \Delta_{k-1}(\mathcal{U})}} \mathcal{O}(\eta, \tau) \mathcal{I}_{\eta}^k(v) \\ &= \mathcal{I}_{\tau}(\mathop{\mathrm{d}}^k v) \\ &= (\pi^{k+1} \mathop{\mathrm{d}}^k v)(\tau + C_{k+1}(\mathcal{U})), \end{align}\] where we used that \(\mathcal{I}_{\eta}^k(v) = \mathcal{I}_{\rho}^{k+1}(\mathop{\mathrm{d}}^k v) = 0\) for all \(\eta \in \Delta_k(\mathcal{U})\) and \(\rho \in \Delta_{k+1}(\mathcal{U})\). Thus, the diagram \[\begin{tikzcd}[ampersand replacement=\&] 0 \arrow[r] \arrow[d] \& \mathcal{S}_{\Gamma_0}^0 \arrow[r, "\mathop{\mathrm{grad}}"] \arrow[d, "\pi^0"] \& \mathcal{S}_{\Gamma_0}^1 \arrow[r, "\mathop{\mathrm{curl}}"] \arrow[d, "\pi^1"] \& \mathcal{S}_{\Gamma_0}^2 \arrow[r, "\undefined"] \arrow[d, "\pi^2"] \& \mathcal{S}_{\Gamma_0}^3 \arrow[r] \arrow[d, "\pi^3"] \& 0 \arrow[d] \\ 0 \arrow[r] \& {C^0(\mathcal{T}, \mathcal{U})}' \arrow[r, "{\partial}'_0"] \& {C^1(\mathcal{T}, \mathcal{U})}' \arrow[r, "{\partial}'_1"] \& {C^2(\mathcal{T}, \mathcal{U})}' \arrow[r, "{\partial}'_2"] \& {C^3(\mathcal{T}, \mathcal{U})}' \arrow[r] \& 0, \end{tikzcd}\] commutes. Since \(\pi^k\), \(k \in 0:3\) are isomorphisms, the two sequences have isomorphic cohomologies. Additionally, the singular homology group and the relative simplicial homology group have isomorphic cohomologies since \(\Omega\) and \(\Gamma_0\) admit a conforming mesh (see e.g. [36]). ◻
We now turn to the construction of a “minimal” conforming subcomplex of 15 . With \(\tilde{V}^{2, h}(K_A) = \mathop{\mathrm{GN}}(K_A)\) and \(\tilde{V}^{3, h}(K_A) = \mathop{\mathrm{DG}}^0(K)\) as in 2.2, we seek discrete spaces \(\tilde{V}^{k, h}(K_A) \subset V^{k, h}(K_A)\), \(k \in 0:1\), so that 25 is an exact complex. We want the spaces to be sufficiently large so that the inclusions \(\mathcal{P}_{3-k}(K) \otimes \mathbb{X}^k \subset \tilde{V}^{k, h}(K_A)\) hold, but also minimal in the sense that degrees of freedom in 27 for \(k=0\) and 28 for \(k=1\) are unisolvent. We achieve this by defining the spaces \(\tilde{V}^{k, h}(K_A)\) as subspaces of \(V^{k, h}(K_A)\) satisfying particular constraints.
Let \(K \in \mathcal{T}\) and suppose that the bilinear form \({a_K(\cdot,\cdot) : H^1(K)^3 \times H^1(K)^3 \to \mathbb{R}}\) is continuous and satisfies: \[\tag{61} \begin{alignat}{2} \tag{62} a_K(u, \mathop{\mathrm{grad}}\phi) &= 0 \qquad & &\forall u \in \mathcal{P}_2(K)^3, \;\forall \phi \in V^{0, h}_0(K_A), \\ \tag{63} a_K(\mathop{\mathrm{grad}}\psi, \mathop{\mathrm{grad}}\psi) &\geq \gamma_a \|\psi\|_{H^2(K)}^2 \qquad & &\forall \psi \in V^{0, h}_0(K_A), \end{alignat}\] where \(\gamma_a\) is independent of \(K\) and we recall that \(V^{0, h}_0(K_A) = V^{0, h}(K_A) \cap H^2_0(K)\). One bilinear form satisfying 61 is \(a_K(\cdot,\cdot) = (\mathop{\mathrm{grad}}\cdot, \mathop{\mathrm{grad}}\cdot)_{L^2(K)}\). For each face \(f \in \Delta_2(K)\), let \(\ell_f \in \mathcal{P}_3(f)^*\) be any linear functional with the following properties: \[\begin{align} \label{eq:kill-cubic-bubble-dof} \ell_f(p) = 0 \qquad \forall p \in \mathcal{P}_2(f) \quad \text{and} \quad \ell_f(b_f) \neq 0. \end{align}\tag{64}\] For example, one could take \[\begin{align} \ell_f(p) = \int_f (I - \mathbb{P}_{2, f})b_f p \,\mathrm{d}{s} \qquad \forall p \in \mathcal{P}_3(f), \end{align}\] where \(\mathbb{P}_{2, f} : L^1(f) \to \mathcal{P}_2(f)\) is the \(L^2(f)\)-orthogonal projector onto \(\mathcal{P}_2(f)\).
Consider the following subspace of \(V^{0, h}(K_A)\): \[\begin{gather} \label{eq:local-h2-space-def} \tilde{V}^{0, h}(K_A) := \{ v \in V^{0, h}(K_A) : \partial_n v|_{f} \in \mathcal{P}_3(f) \text{ and } \ell_f(\partial_n v|_{f}) = 0 \;\forall f \in \Delta_2(K) \\ \text{and } a_K(\mathop{\mathrm{grad}}v, \mathop{\mathrm{grad}}w) = 0 \;\forall w \in V^{0, h}_0(K_A) \}. \end{gather}\tag{65}\] The following result summarizes the key properties of \(\tilde{V}^{0, h}(K_A)\).
Lemma 6. \(\dim \tilde{V}^{0, h}(K_A) = 40\), the degrees of freedom in 27 with \(\mathcal{T}= K_A\) are unisolvent on \(\tilde{V}^{0, h}(K_A)\), and \(\mathcal{P}_3(K) \subset \tilde{V}^{0, h}(K_A)\).
Proof. We follow similar arguments as in the proof of [37]. Note that the number of independent linear functionals on \(V^{0, h}(K_A)\) in 27 is 40.
Step 1: \(\dim \tilde{V}^{0, h}(K_A) \leq 40\). Assume that \(v \in \tilde{V}^{0, h}(K_A)\) and the degrees of freedom in 27 vanish.
Then, as shown in the proof of [37], \(v|_{\partial K} = 0\) and \[\begin{align}
D_f^{\alpha} (\partial_n v|_{f})(z) = 0 \qquad \forall |\alpha| \leq 1, \; \forall z \in \Delta_0(f), \;\forall f \in \Delta_2(K),
\end{align}\] where we recall that \(D_f\) denotes the surface differential. Thus, \(\partial_n v|_{f} \in \mathop{\mathrm{span}}\{ b_f \}\) and \(\ell_f(\partial_n v|_{f}) = 0\), so \(v \in V^{0, h}_0(K)\). The coercivity of \(a_K(\cdot,\cdot)\) 63
gives \(v \equiv 0\). Consequently, \(\dim \tilde{V}^{0, h}(K_A) \leq 40\).
Step 2: \(\dim \tilde{V}^{0, h}(K_A) \geq 40\). Suppose we are given arbitrary values for the degrees of freedom in 27 \(\{ c_{z}^{\alpha} :
|\alpha| \leq 2, \;z \in \Delta_0(K) \}\). We now show that there exists \(v \in \tilde{V}^{0, h}(K_A)\) with these degrees of freedom by using a minor modification of the degrees of freedom for the space \(V^{0, h}(K_A)\) in [17].
For each \(f \in \Delta_2(K)\), let \(w_{n, f} \in \mathcal{P}_3(f)\) be the unique cubic polynomial satisfying \[\begin{align}
{2} D_f^{\alpha} w_{n, f}(z) &= \sum_{i=1}^{3} c_{z}^{\alpha + e_i} (\hat{n}_f \cdot \hat{e}_i) \qquad & &\forall |\alpha| \leq 1, \;\forall z \in \Delta_0(f), \\ \ell_{f} (w_{n, f}) &= 0. \qquad & &
\end{align}\] Thanks to [17], there exists \(v \in V^{0, h}(K_A)\) satisfying \[\begin{align}
{2} D^{\alpha} v(z) &= c_{z}^{\alpha} \qquad & &\forall |\alpha| \leq 2, \;\forall z \in \Delta_0(K), \\ \int_{e} \mathop{\mathrm{grad}}v \cdot \hat{n}_f \,\mathrm{d}{s} &= \int_{e} w_{n, f} \,\mathrm{d}{s} \qquad & &\forall e \in
\Delta_1(f), \;\forall f \in \Delta_2(K), \\ \int_f (\mathop{\mathrm{grad}}v \cdot \hat{n}_f) \kappa \,\mathrm{d}{s} &= \int_f w_{n, f} \kappa \,\mathrm{d}{s} \qquad & &\forall \kappa \in \mathcal{P}_1(f), \;\forall f \in \Delta_2(K), \\
a_K(\mathop{\mathrm{grad}}v, \mathop{\mathrm{grad}}\phi) &= 0 \qquad & &\forall \phi \in V^{0, h}_0(K_A).
\end{align}\] By construction, the degrees of freedom in 27 of \(v\) match \(\{ c_{z}^{\alpha}\}\) in the sense that \[\begin{align}
{2} D^{\alpha} v(z) &= c_{z}^{\alpha} \qquad & &\forall |\alpha| \leq 2, \;\forall z \in \Delta_0(K).
\end{align}\] We finish the proof by showing that for \(f \in \Delta_2(K)\), \(\partial_{n} v|_{f} = w_{n, f}\) and thus \(v \in \tilde{V}^{0,
h}(K_A)\). By construction, we have that \(v_f := (\partial_n v|_f - w_{n, f}) \in \mathcal{P}_{4}(f)\) satisfies 55 with \(p=3\), and so
\(\partial_n v|_{f} = w_{n, f}\).
Step 3: Inclusion of cubics. Note that for any \(v \in \mathcal{P}_3(K)\), 62 gives \(a_K(\mathop{\mathrm{grad}}v,
\mathop{\mathrm{grad}}w) = 0\) for all \(w \in V^{0, h}_0(K_A)\), while 64 gives \(\ell_f(\partial_n v|_{f}) = 0\) for all \(f \in \Delta_2(K)\), and so \(\mathcal{P}_3(K) \subset \tilde{V}^{0, h}(K_A)\). ◻
Remark 8. An alternative reduction of \(V^{0, h}(K_A)\) was introduced in [37], in which the restriction \(\partial_n v|_f \in \mathcal{P}_3(f)\) for all \(f \in \Delta_2(K)\) was also imposed and \(C^4\) continuity is imposed at the barycenter of \(K\). This supersmooth space locally reproduces \(\mathcal{P}_{4}(K)\) and the degrees of freedom are 27 augmented with one degree of freedom per face for the normal derivative and a single interior degree of freedom.
Note that \(\tilde{V}^{0, h}(K_A)\) chosen as in 65 appears to the left of \(\tilde{V}^{1, h}(K_A)\) in the complex 25 , while \(\mathop{\mathrm{GN}}(K_A)\) appears to the right. To satisfy the complex property, we simply define \(\tilde{V}^{1, h}(K_A)\) to be the subspace of \(V^{1, h}(K_A)\) satisfying the constraints imposed by these two spaces: \[\begin{gather} \label{eq:local-h1curl-space-def} \tilde{V}^{1, h}(K_A) := \{ v \in V^{1, h}(K_A) : v \cdot \hat{n}_f|_{f} \in \mathcal{P}_3(f) \text{ and } \ell_f(v \cdot \hat{n}_f|_{f}) = 0 \;\forall f \in \Delta_2(K), \\ \mathop{\mathrm{curl}}v \in \mathop{\mathrm{GN}}(K_A), \text{ and } a_K(v, \mathop{\mathrm{grad}}\phi) = 0 \;\forall \phi \in V^{0, h}_0(K_A) \}. \end{gather}\tag{66}\]
Lemma 7. \(\dim \tilde{V}^{1, h}(K_A) = 54\), the degrees of freedom in 28 with \(\mathcal{T}= K_A\) are unisolvent, and \(\mathcal{P}_2(K)^3 \subset \tilde{V}^{1, h}(K_A)\).
Proof. Note that there are 54 linearly independent functionals on \(V^{1, h}(K_A)\) in 28 .
Step 1: \(\dim \tilde{V}^{1, h}(K_A) \leq 54\). Let \(v \in \tilde{V}^{1, h}(K_A)\) and suppose that the degrees of freedom in 28 vanish.
Then, \(\mathop{\mathrm{curl}}v\) vanishes on \(\Delta_0(K)\). Since \(\mathop{\mathrm{curl}}v|_{e} \in \mathcal{P}_1(e)^3\) for all edges \(e \in \Delta_1(K)\), \(\mathop{\mathrm{curl}}v\) also vanishes on \(\Delta_1(K)\). Moreover, for \(f \in \Delta_2(K)\), \(\mathop{\mathrm{curl}}v \times \hat{n} \in \mathcal{P}_1(f)^3\), and so \(\mathop{\mathrm{curl}}v \times \hat{n}|_f \equiv 0\). Finally, \[\begin{align} \int_{f}
\mathop{\mathrm{curl}}v \cdot \hat{n}_f \,\mathrm{d}{s} = \sum_{e \in \Delta_1(f)} \mathcal{O}(e, f) \int_{e} v \cdot \hat{t}_{e} \,\mathrm{d}{s} = 0,
\end{align}\] where we recall that \(\mathcal{O}(e, f)\) is the orientation of \(e\) relative to \(f\). Since \(\mathop{\mathrm{curl}}v \cdot \hat{n}_f|_f \in \mathcal{P}_3(f) \cap H^1_0(f)\) and additionally \(\mathop{\mathrm{curl}}v\cdot \hat{n}_f|_{\partial f} = 0\), we have \(\mathop{\mathrm{curl}}v \cdot \hat{n}_f|_{f} \equiv 0\). Thus, \(\mathop{\mathrm{curl}}v|_{\partial K} \equiv 0\) and so \(\mathop{\mathrm{curl}}v \equiv 0\)
since \(\mathop{\mathrm{curl}}v \in \mathop{\mathrm{GN}}(K_A)\).
Thanks to the exactness of 11 , there exists \(\phi \in V^{0, h}(K_A)\) such that \(v = \mathop{\mathrm{grad}}\phi\). By the definition of \(\tilde{V}^{1, h}(K_A)\), \(\phi \in \tilde{V}^{0, h}(K_A)\). Moreover, since \(\int_e v \cdot \hat{t}_e = 0\) for all \(e \in \Delta_1(K)\), the fundamental theorem of calculus shows that \(\phi\) takes the same value at every vertex in \(\Delta_0(K)\). By subtracting a constant from \(\phi\), which does not change \(\mathop{\mathrm{grad}}\phi\), we may assume that \(\phi\) vanishes at the vertices. Thus, 27 vanishes, and so \(\phi \equiv 0\) by 6. Consequently, \(v \equiv 0\) and \(\dim \tilde{V}^{1, h}(K_A) \leq 54\).
Step 2: \(\dim \tilde{V}^{1, h}(K_A) \geq 54\). Suppose we are given arbitrary values for the degrees of freedom in 28 \(\{ c_{z, 1}^{\alpha}, \ldots, c_{z, 3}^{\alpha}, c_e : |\alpha| \leq 1, \;z \in \Delta_0(K), \;e \in \Delta_1(K) \}\). We now show that there exists \(v \in \tilde{V}^{1, h}(K_A)\) with these degrees of freedom. We do this using the degrees of freedom for the space \(V^{1, h}(K_A)\) given in [17] (with slight modification).
Let \(w_c \in \mathop{\mathrm{GN}}(K_A)\) satisfy \[\label{eq:proof:gn-interp-curl} \begin{alignat}{2} w_c(z) &= \sum_{i, j, k=1}^{3}
\epsilon_{ijk} c_{z, j}^{e_i} \hat{e}_k \qquad & &\forall z \in \Delta_0(K), \\ \int_{f} w_c \cdot \hat{n}_f \,\mathrm{d}{s} &= \sum_{e \in \Delta_1(f)} \mathcal{O}(e, f) c_e \qquad & &\forall f \in \Delta_2(K),
\end{alignat}\tag{67}\] where \(\epsilon_{ijk}\) is the permutation symbol. Moreover, for \(f \in \Delta_2(K)\), let \(w_f \in
\mathcal{P}_3(f)\) be the unique cubic polynomial satisfying \[\begin{align}
{2} D_f^{\alpha} w_f(z) &= \sum_{i=1}^{3} c_{z, i}^{\alpha} (\hat{n}_f \cdot \hat{e}_i) \qquad & &\forall |\alpha| \leq 1, \;\forall z \in \Delta_0(f), \\ \ell_f(w_f) \,\mathrm{d}{s} &= 0. \qquad & &
\end{align}\] Thanks to [17], there exists \(v \in V^{1, h}(K_A)\) satisfying \[\begin{align}
{2} D^{\alpha} v(z) \cdot \hat{e}_i &= c_{z, i}^{\alpha} \qquad & &\forall |\alpha| \leq 1, \forall z \in \Delta_0(K), \;i \in 1:3, \\ \int_{e} v \cdot \hat{t}_e \,\mathrm{d}{s} &= c_{e} \qquad & &\forall e \in \Delta_1(K), \\
\int_{e} v \cdot \hat{n}_f \,\mathrm{d}{s} &= \int_{e} w_f \,\mathrm{d}{s} \qquad & & \forall e \in \Delta_1(f), \; \forall f \in \Delta_2(K), \\ \int_{e} \mathop{\mathrm{curl}}v \,\mathrm{d}{s} &= \int_{e} w_{c} \,\mathrm{d}{s} \qquad
& &\forall e \in \Delta_1(K), \\ \int_{f} (v \cdot \hat{n}_f) \kappa \,\mathrm{d}{s} &= \int_{f} w_f \kappa \,\mathrm{d}{s} \qquad & &\forall \kappa \in \mathcal{P}_1(f), \;\forall f \in \Delta_2(K), \\ \int_f \mathop{\mathrm{curl}}v
\times \hat{n}_f \,\mathrm{d}{s} &= \int_f w_{c} \times \hat{n}_f \,\mathrm{d}{s} \qquad & &\forall f \in \Delta_2(K), \\ a_K(v, \mathop{\mathrm{grad}}\phi) &= 0 \qquad & &\forall \phi \in V^{0, h}_0(K_A), \\ \int_{K}
\mathop{\mathrm{curl}}v \cdot u \,\mathrm{d}{x} &= \int_{K} w_c \cdot u \,\mathrm{d}{x} \qquad & &\forall u \in \mathop{\mathrm{curl}}V^{1, h}_0(K_A).
\end{align}\] By construction, we have \[\begin{align}
{2} D^{\alpha} v(z) \cdot \hat{e}_i &= c_{z, i}^{\alpha} \qquad & & \forall|\alpha| \leq 1, \;\forall z \in \Delta_0(K), \;i \in 1:3, \\ \int_{e} v \cdot \hat{t}_e \,\mathrm{d}{s} &= c_e \qquad & & \forall e \in \Delta_1(K),
\end{align}\] so it remains to show that \(v \in \tilde{V}^{1, h}(K_A)\). We first show that \[\begin{align} v \cdot \hat{n}_f|_{f} \in \mathcal{P}_3(f) \quad \text{and} \quad \ell_f(v
\cdot \hat{n}_f|_{f}) = 0 \qquad \forall f \in \Delta_2(K).
\end{align}\] Let \(f \in \Delta_2(K)\). Note that by construction, we have \(u_f := v \cdot \hat{n}_f - w_f \in \mathcal{P}_{3}(f)\) satisfies 55 with \(p=3\), and so \(v \cdot \hat{n}_f = w_f\) on \(f\). Analogous arguments show that \(\mathop{\mathrm{curl}}v = w_c \in \mathop{\mathrm{GN}}(K_A)\). Thus, \(v \in \tilde{V}^{1, h}(K_A)\).
Step 3: Inclusion of quadratics. Note that if \(v \in \mathcal{P}_2(K)^3\), then \(\ell_f(v \cdot \hat{n}_f) = 0\) for all \(f \in
\Delta_2(K)\) by 64 , \(\mathop{\mathrm{curl}}v \in \mathcal{P}_1(K)^3 \subset \mathop{\mathrm{GN}}(K_A)\), and \(a_K(v,
\mathop{\mathrm{grad}}\phi) = 0\) for all \(\phi \in V^{0, h}_0(K_A)\) by 62 ; therefore, \(\mathcal{P}_2(K)^3 \subset \tilde{V}^{1,
h}(K_A)\). ◻
We begin with a simple result showing that reducing the degree of normal components on faces also reduces the degree of the normal components on edges.
Lemma 8. Let \(K \in \mathcal{T}\). For \(v \in \tilde{V}^{0, h}(K_A)\) and \(w \in \tilde{V}^{1, h}(K_A)\), there holds \[\begin{align} \mathop{\mathrm{grad}}v|_e \times \hat{t}_e \in \mathcal{P}_3(e) \quad \text{and} \quad w|_e \times \hat{t}_e \in \mathcal{P}_3(e) \qquad \forall e \in \Delta_1(K). \end{align}\]
Proof. Let \(e \in \Delta_1(f) \cap \Delta_1(f')\) for two distinct faces \(f, f' \in \Delta_2(K)\). Since the normal derivatives \(\partial_{\hat{n}_f} v|_f\) and \(\partial_{\hat{n}_f} v|_{f'}\) are both cubic and \(\{ \hat{n}_f, \hat{n}_{f'} \}\) form a basis for \(\mathbb{R}^3 \times \hat{t}_e\), we have \(\mathop{\mathrm{grad}}v|_e \times \hat{t}_e \in \mathcal{P}_3(e)\). Similar arguments show that \(w|_e \times \hat{t}_e \in \mathcal{P}_3(e)\). ◻
We now proceed space-by-space in 17 and show that the global spaces 26 defined in terms of the local spaces 65 66 properly glue together.
Lemma 9. The space \(\tilde{V}^{0, h}_{\Gamma_0}\) is unisolvent with respect to the degrees of freedom \[\label{eq:min-h2-space-dofs-bcs} \begin{alignat}{2} &D^{\alpha} v(z) \qquad & &\forall |\alpha| \leq 2, \;\forall z \in \Delta_0(\mathcal{T}) \setminus \Delta_0(\mathcal{U}), \\ &\partial_{\hat{n}_{\Gamma}}^2 v(z) \qquad & &\forall z \in \Delta_0^{\flat}(\mathcal{U}), \end{alignat}\tag{68}\] and \(\dim \tilde{V}^{0, h}_{\Gamma_0} = 10(|\Delta_0(\mathcal{T})| - |\Delta_0(\mathcal{U})|) + |\Delta_0^{\flat}(\mathcal{U})|\).
Proof. Step 1: Upper bound. Suppose that \(v \in \tilde{V}^{0, h}_{\Gamma_0}\) and 68 vanish. Thanks to 3, \(D^{\alpha} v(z) = 0\) for all \(|\alpha| \leq 2\) and \(z \in
\Delta_0(\mathcal{T})\). Thus, on each cell \(K \in \mathcal{T}\), the degrees of freedom 27 with \(\mathcal{T}= K_A\) vanish, and so \(v|_K \equiv 0\) by 6. Thus, \(\dim \tilde{V}^{0, h}_{\Gamma_0} \leq
10(|\Delta_0(\mathcal{T})| - |\Delta_0(\mathcal{U})|) + |\Delta_0^{\flat}(\mathcal{U})|\).
Step 2: Lower bound. Suppose we are given values for 68 : \[\begin{align}
\label{eq:proof:min-h2-space-dofs-bcs} \{c_z^{\alpha}, d_{z'} : |\alpha| \leq 2, \;z \in \Delta_0(\mathcal{T}) \setminus \Delta_0(\mathcal{U}), \;z' \in \Delta_0^{\flat}(\mathcal{U})\}.
\end{align}\tag{69}\] For \(K \in \mathcal{T}\), define \(v_K \in \tilde{V}^{0, h}(K_A)\) by \[\begin{align}
{2} D^{\alpha} v_K(z) &= c_z^{\alpha} \qquad & &\forall |\alpha| \leq 2, \;\forall z \in \Delta_0(K) \setminus \Delta_0(\mathcal{U}), \\ \mathop{\mathrm{hess}}v_K(z') &= d_{z'} \hat{n}_{\Gamma} \otimes \hat{n}_{\Gamma} \qquad &
&\forall z' \in \Delta_0(K) \cap \Delta_0^{\flat}(\mathcal{U}),
\end{align}\] with all remaining degrees of freedom in 27 with \(\mathcal{T}= K_A\) set to 0. Let \(v \in L^2(\Omega)\) be defined by \(v|_K := v_K\) for all \(K \in \mathcal{T}\).
We now show that \(v \in \tilde{V}^{0, h}_{\Gamma_0}\). Suppose that \(f \in \Delta_2(\mathcal{T})\) satisfies either (i) there exist distinct \(K, K' \in \mathcal{T}\) with \(f \in \Delta_2(K) \cap \Delta_2(K')\) or (ii) \(f \in \mathcal{U}\) and there exists a unique \(K_f \in \mathcal{T}\) with \(f \in \Delta_2(K)\). Define \(v_f \in \mathcal{P}_5(f)\) and \(u_f \in \mathcal{P}_3(f)\) by (i) \(v_f := v_K|_f - v_{K'}|_f\) and \(u_f := \partial_{\hat{n}_f} v_K|_f - \partial_{\hat{n}_f} v_K'|_f\) or (ii) \(v_f := v_{K_f}|_f\) and \(u_f := \partial_{\hat{n}_f} v_{K_f}|_f\).
In both cases, we have that \(v_f\) belongs to the 2D Bell finite element space thanks to 8 and \(D_f^{\alpha} v_f(z) = 0\) for all \(z \in \Delta_0(f)\). These degrees of freedom are unisolvent [38], and so \(v_f \equiv 0\). Also in both cases, \[\begin{align} D_f^{\beta} u_f(z) = 0 \qquad \forall |\beta| \leq 1, \;\forall z \in \Delta_0(f) \quad \text{and} \quad \ell_f(u_f) = 0. \end{align}\] These degrees of freedom are clearly unisolvent on \(\mathcal{P}_3(f)\), and so \(u_f \equiv 0\). Thus, \(v \in \tilde{V}^{0, h}_{\Gamma_0}\) and the degrees of freedom of \(v\) in 68 match 69 , completing the proof. ◻
Lemma 10. The space \(\tilde{V}^{1, h}_{\Gamma_0}\) is unisolvent with respect to the degrees of freedom \[\label{eq:min-h1curl-space-dofs-bcs} \begin{alignat}{2} &D^{\alpha} v(z) \qquad & &\forall |\alpha| \leq 1, \;\forall z \in \Delta_0(\mathcal{T}) \setminus \Delta_0(\mathcal{U}), \\ &\int_e v \cdot \hat{t}_e \,\mathrm{d}{s} \qquad & &\forall e \in \Delta_1(\mathcal{T}) \setminus \Delta_1(\mathcal{U}), \\ &\partial_{\hat{n}_{\Gamma}} (v \cdot \hat{n}_{\Gamma})(z) \qquad & &\forall z \in \Delta_0^{\flat}(\mathcal{U}), \end{alignat}\tag{70}\] and \(\dim \tilde{V}^{1, h}_{\Gamma_0} = 12(|\Delta_0(\mathcal{T})| - |\Delta_0(\mathcal{U})|) + |\Delta_1(\mathcal{T})| - |\Delta_1(\mathcal{U})| + |\Delta_0^{\flat}(\mathcal{U})|\).
Proof. Step 1: Upper bound. Suppose that \(v \in \tilde{V}^{1, h}_{\Gamma_0}\) and 70 vanish. Thanks to 3, on each cell \(K \in \mathcal{T}\), the degrees of freedom 28 with \(\mathcal{T}=
K_A\) vanish, and so \(v|_K \equiv 0\) by 7. Thus, the dimension count in the statement of the lemma
is an upper bound.
Step 2: Lower bound. Suppose we are given values for 70 : \[\begin{align}
\label{eq:proof:min-h1curl-space-dofs-bcs} \{c_z^{\alpha}, c_e, d_{z'} : |\alpha| \leq 1, \;z \in \Delta_0(\mathcal{T}) \setminus \Delta_0(\mathcal{U}), \;e \in \Delta_1(\mathcal{T}) \setminus \Delta_1(\mathcal{U}), \;z' \in
\Delta_0^{\flat}(\mathcal{U})\}.
\end{align}\tag{71}\] For \(K \in \mathcal{T}\), define \(v_K \in \tilde{V}^{1, h}(K_A)\) by \[\begin{align}
{2} D^{\alpha} v_K(z) &= c_z^{\alpha} \qquad & &\forall |\alpha| \leq 1, \;\forall z \in \Delta_0(K) \setminus \Delta_0(\mathcal{U}), \\ \int_e v_K \cdot \hat{t}_e \,\mathrm{d}{s} &= c_e \qquad & &\forall e \in \Delta_1(K) \setminus
\Delta_1(\mathcal{U}), \\ \mathop{\mathrm{grad}}v_K(z') &= d_{z'} \hat{n}_{\Gamma} \otimes \hat{n}_{\Gamma} \qquad & &\forall z' \in \Delta_0(K) \cap \Delta_0^{\flat}(\mathcal{U}),
\end{align}\] with all remaining degrees of freedom in 28 with \(\mathcal{T}= K_A\) set to 0. Let \(v \in L^2(\Omega)^3\) be defined by \(v|_K := v_K\) for all \(K \in \mathcal{T}\).
We now show that \(v \in \tilde{V}^{1, h}_{\Gamma_0}\). Suppose that \(f \in \Delta_2(\mathcal{T})\) and \(K, K', K_f \in \mathcal{T}\) are as in Step 2 of the proof of 9. Define \(v_f \in \mathcal{P}_4(f)^3\) and \(u_f \in \mathcal{P}_3(f)^3\) by (i) \(v_f := v_K|_f - v_{K'}|_f\) and \(u_f := \mathop{\mathrm{curl}}v_K|_f - \mathop{\mathrm{curl}}v_K'|_f\) or (ii) \(v_f := v_{K_f}|_f\) and \(u_f := \mathop{\mathrm{curl}}v_{K_f}|_f\). We treat both cases simultaneously.
Since \(\mathop{\mathrm{curl}}v \in \mathop{\mathrm{GN}}(K_A)\), \(u_f \in \mathcal{P}_1(f)^3 \oplus \mathop{\mathrm{span}}\{ b_f \hat{n}_f \}\), where we recall that \(b_f\) is the cubic bubble function on \(f\). Clearly, we have \(\mathop{\mathrm{rot}}_f v = u_f \cdot \hat{n}_f\), and so \[\begin{align} u_f(z) = 0 \qquad \forall z \in \Delta_0(f) \quad \text{and} \quad \int_f u_f \cdot \hat{n}_f \,\mathrm{d}{s} = \sum_{e \in \Delta_1(f)} \mathcal{O}(e, f) \int_{e} v \cdot \hat{t}_{e} \,\mathrm{d}{s} = 0, \end{align}\] which are unisolvent on \(\mathcal{P}_1(f)^3 \oplus \mathop{\mathrm{span}}\{ b_f \hat{n}_f \}\), and so \(u_f \equiv 0\).
Turning to \(v_f\), we have \(v_f \cdot \hat{n}_f \in \mathcal{P}_3(f)\) by definition. Since \[\begin{align} D_f^{\alpha} (v_f \cdot \hat{n}_f)(z) = 0 \qquad \forall |\alpha| \leq 1, \;\forall z \in \Delta_0(f) \quad \text{and} \quad \ell_f(v_f \cdot \hat{n}_f) = 0 \end{align}\] and the above degrees of freedom are unisolvent on \(\mathcal{P}_3(f)\), \(v_f \cdot \hat{n}_f \equiv 0\).
We now show that \(v|_{\partial f} \equiv 0\). As \(\mathop{\mathrm{rot}}_f v_f \equiv 0\) and \(v_f \cdot \hat{n}_f \equiv 0\), de Rham’s theorem (see e.g. [39]) shows that there exists \(\phi \in H^1(f)\) such that \(v_f = \mathop{\mathrm{grad}}_f \phi\). Clearly, \(\phi \in \mathcal{P}_5(f)\), and more specifically, \(\phi\) belongs to the Bell finite element space thanks to 8. Arguing as in the proof of 7, we may choose \(\phi\) to vanish at the vertices of \(f\), and so all derivatives up to order 2 of \(\phi\) vanish on \(\Delta_0(f)\). Arguing as in the proof of 9, we have \(\phi \equiv 0\) and so \(v \equiv 0\). Thus, \(v \in \tilde{V}^{1, h}_{\Gamma_0}\) and the degrees of freedom of \(v\) in 70 match 70 . As a result, the dimension count in the statement of the lemma is a lower bound. ◻
Standard arguments also show that \(\tilde{V}_{\Gamma_0}^{2, h}\) may be characterized similarly by omitting degrees of freedom associated to \(\Delta(\mathcal{U})\):
Lemma 11. The space \(\tilde{V}^{2, h}_{\Gamma_0}\) is unisolvent with respect to the degrees of freedom \[\label{eq:min-h1-space-dofs-bcs} \begin{alignat}{2} &v(z) \qquad & &\forall |\alpha| \leq 1, \;\forall z \in \Delta_0(\mathcal{T}) \setminus \Delta_0(\mathcal{U}), \\ &\int_f v \cdot \hat{n}_f \,\mathrm{d}{s} \qquad & &\forall e \in \Delta_2(\mathcal{T}) \setminus \Delta_2(\mathcal{U}), \end{alignat}\tag{72}\] and \(\dim \tilde{V}^{2, h}_{\Gamma_0} = 3(|\Delta_0(\mathcal{T})| - |\Delta_0(\mathcal{U})|) + |\Delta_2(\mathcal{T})| - |\Delta_2(\mathcal{U})|\).
Remark 9. A simple consequence of 9 [ lem:min-h1curl-space-dofs-bcs]{reference-type=“ref” reference=” lem:min-h1curl-space-dofs-bcs”} 11 and their proofs is that \(v \in \tilde{V}^{k, h}\) belongs to \(\tilde{V}^{k, h}_{\Gamma_0}\), \(k \in 0:2\), if and only if \[\begin{align} {2} D^{\alpha} v(z) &= 0 \qquad & &\forall |\alpha| \leq 2-k, \;\forall z \in \Delta_0(\mathcal{U}) \setminus \Delta_0^{\flat}(\mathcal{U}), \\ D^{\alpha} v(z) &= 0 \qquad & &\forall |\alpha| \leq \max\{ 1-k, 0\}, \;\forall z \in \Delta_0^{\flat}(\mathcal{U}), \\ (I - \hat{n}_{\Gamma} \otimes \hat{n}_{\Gamma}) \mathop{\mathrm{grad}}^{2-k} v(z) &= 0 \qquad & & \forall z \in \Delta_0^{\flat}(\mathcal{U}), \\ \mathcal{I}^k_{\tau}(v) &= 0 \qquad & &\forall \tau \in \Delta_k(\mathcal{U}), \end{align}\] where \(\mathop{\mathrm{grad}}^2 := \mathop{\mathrm{hess}}\), \(\mathop{\mathrm{grad}}^0 := I\), and we recall that \(\mathcal{I}^k_{\tau}\) is defined in 18 .
We could proceed as in 3 and apply the framework in [18]. However, the degrees of freedom for the discrete spaces \(\tilde{V}^{k, h}\) are simple enough that we can prove the key results directly. To this end, we define the skeletal spaces for \(k \in 0:2\) by \[\begin{align} \tilde{\mathcal{S}}^{k}_{\Gamma_0} &:= \left\{ v \in \tilde{V}^{k, h}_{\Gamma_0} : D^{\alpha} v(z) = 0 \;\forall |\alpha| \in \max\{1-k, 0\} : 2-k, \;\forall z \in \Delta_0(\mathcal{T}) \right\}, \end{align}\] and set \(\tilde{\mathcal{S}}^3_{\Gamma_0} := \tilde{V}^{3, h}_{\Gamma_0}\). Owing to 9 10 [ lem:min-h1-space-dofs-bcs]{reference-type=“ref” reference=” lem:min-h1-space-dofs-bcs”}, the “currents" \(\mathcal{I}^k_{\tau}(\cdot)\) for \({\tau \in \Delta_k(\mathcal{T}) \setminus \Delta_k(\mathcal{U})}\) defined in 18 are a unisolvent set of degrees of freedom on \(\tilde{\mathcal{S}}^{k}_{\Gamma_0}\).
We also define the lifted (modified) bubble spaces associated with each vertex \(z \in \Delta_0(z)\) for \(k \in 0:2\) by \[\begin{gather} \tilde{\mathbb{B}}^{k}_{\Gamma_0}(z) := \left\{ v \in \tilde{V}^{k, h}_{\Gamma_0} : D^{\alpha} v(z') = 0 \;\forall |\alpha| \leq 2-k, \;\forall z' \in \Delta_0(\mathcal{T}) \setminus \{z\} \right. \\ \left. \vphantom{v \in \tilde{V}^{k, h}_{\Gamma_0}} \text{ and } \mathcal{I}_{\tau}^k(v) = 0 \;\forall \tau \in \Delta_k(\mathcal{T}) \right\}. \end{gather}\] Analogous to 60 , we have an associated diagram: \[\label{eq:stokes-complex-gn-geom-decomp} \begin{tikzcd}[ampersand replacement=\&] 0 \arrow[r] \& \tilde{V}^{0, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{grad}}"] \& \tilde{V}^{1, h}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{curl}}"] \& \tilde{V}^{2, h}_{\Gamma_0} \arrow[r, "\undefined"] \& \tilde{V}^{3, h}_{\Gamma_0} \arrow[r] \& 0 \\[-2em] \& \rotatebox{90}{=} \& \rotatebox{90}{=} \& \rotatebox{90}{=} \& \rotatebox{90}{=} \& \\[-2em] 0 \arrow[r] \& \tilde{\mathcal{S}}^{0}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{grad}}"] \& \tilde{\mathcal{S}}^{1}_{\Gamma_0} \arrow[r, "\mathop{\mathrm{curl}}"] \& \tilde{\mathcal{S}}^{2}_{\Gamma_0} \arrow[r, "\undefined"] \& \tilde{\mathcal{S}}^{3}_{\Gamma_0} \arrow[r] \& 0 \\[-2em] \& \bigoplus\limits_{z \in \Delta_0(\mathcal{T})} \& \bigoplus\limits_{z \in \Delta_0(\mathcal{T})} \& \bigoplus\limits_{z \in \Delta_0(\mathcal{T})} \& \& \\[-2em] 0 \arrow[r] \& \tilde{\mathbb{B}}^{0}_{\Gamma_0}(z) \arrow[r, "\mathop{\mathrm{grad}}"] \& \tilde{\mathbb{B}}^{1}_{\Gamma_0}(z) \arrow[r, "\mathop{\mathrm{curl}}"] \& \tilde{\mathbb{B}}^{2}_{\Gamma_0}(z) \arrow[r, "\undefined"] \& 0 \arrow[r] \& 0. \end{tikzcd}\tag{73}\] Each column is a direct sum decomposition thanks to 9 10 [ lem:min-h1-space-dofs-bcs]{reference-type=“ref” reference=” lem:min-h1-space-dofs-bcs”}. The complex properties are summarized as follows.
Lemma 12. Each row of 73 is a complex, and the final row is exact. Moreover, the cohomologies of the first two rows are isomorphic.
Proof. Note that the degrees of freedom satisfy the following property: If \(v \in \tilde{V}_{\Gamma_0}^{h, k}\), \(k \in 0:2\), then \[\begin{align} \mathcal{I}^{k}_{\tau}(v) = 0 \qquad \forall \tau \in \Delta_{k}(\mathcal{T}) \implies \mathcal{I}^{k+1}_{\eta}(\mathop{\mathrm{d}}^k v) = 0 \qquad \forall \eta \in \Delta_{k+1}(\mathcal{T}), \end{align}\] and similarly for any \(z \in \Delta_0(\mathcal{T})\), there holds \[\begin{align} D^{\alpha} v(z) = 0 \qquad \forall |\alpha| \in \max\{1-k, 0\}:2-k \implies D^{\beta} \mathop{\mathrm{d}}^k v(z) = 0 \qquad \forall |\beta| \leq 1-k. \end{align}\] Thus, each row of 73 is a complex. [lem:min-h2-space-dofs-bcs,lem:min-h1curl-space-dofs-bcs, lem:min-h1-space-dofs-bcs] show that \[\begin{align} \dim \tilde{\mathbb{B}}^{1}_{\Gamma_0} = \dim \tilde{\mathbb{B}}^{0}_{\Gamma_0} + \dim \tilde{\mathbb{B}}^{2}_{\Gamma_0}, \end{align}\] and so exactness of the final row follows from standard arguments. Consequently, the cohomologies of the first two rows of 73 are isomorphic. ◻
For \(k \in 0:3\), let \[\begin{align} W^{k, h} := \{ v \in L^2(\Omega) \otimes \mathbb{X}^{k} : \mathop{\mathrm{d}}^k v \in L^2(\Omega) \otimes \mathbb{X}^{k+1} \text{ and } v|_{K} \in W^{k, h}(K) \;\forall K \in \mathcal{T}\} \end{align}\] denote the Whitney forms on \(\mathcal{T}\) equipped with the canonical degrees of freedom \(\mathcal{I}^k_{\tau}\), \(\tau \in \Delta_k(\mathcal{T})\). Set \(W_{\Gamma_0}^{k, h} := W^{k, h} \cap W^{k}_{\Gamma_0}\) so that \(\{ \mathcal{I}^k_{\tau} : \tau \in \Delta_k(\mathcal{T}) \setminus \Delta_k(\mathcal{U})\}\), are a unisolvent set of degrees of freedom on \(W_{\Gamma_0}^{k, h}\). Then, \(\tilde{\mathcal{S}}_{\Gamma_0}^{k}\) and \(W_{\Gamma_0}^{k, h}\) are isomorphic, and let \(\pi^k : \tilde{\mathcal{S}}_{\Gamma_0}^{k} \to W_{\Gamma_0}^{k, h}\) denote the isomorphism that maps an element of \(\tilde{\mathcal{S}}_{\Gamma_0}^{k}\) to the unique element in \(W_{\Gamma_0}^{k, h}\) with the same degrees of freedom. The generalized Stokes theorem 44 then shows that the following diagram commutes: \[\begin{tikzcd}[ampersand replacement=\&] 0 \arrow[r] \arrow[d] \& \tilde{\mathcal{S}}_{\Gamma_0}^0 \arrow[r, "\mathop{\mathrm{grad}}"] \arrow[d, "\pi^0"] \& \tilde{\mathcal{S}}_{\Gamma_0}^1 \arrow[r, "\mathop{\mathrm{curl}}"] \arrow[d, "\pi^1"] \& \tilde{\mathcal{S}}_{\Gamma_0}^2 \arrow[r, "\undefined"] \arrow[d, "\pi^2"] \& \tilde{\mathcal{S}}_{\Gamma_0}^3 \arrow[r] \arrow[d, "\pi^3"] \& 0 \arrow[d] \\ 0 \arrow[r] \& W_{\Gamma_0}^{0, h} \arrow[r, "\mathop{\mathrm{grad}}"] \& W_{\Gamma_0}^{1, h} \arrow[r, "\mathop{\mathrm{curl}}"] \& W_{\Gamma_0}^{2, h} \arrow[r, "\undefined"] \& W_{\Gamma_0}^{3, h} \arrow[r] \& 0. \end{tikzcd}\] Thus, the two sequences have isomorphic cohomologies, the second of which is isomorphic to 6 (see [35] for the precise application of [35]), which in turn is isomorphic to 5 thanks to 8 .
The exactness of 25 follows on taking \(\mathcal{T}= K_A\) and \(\Gamma_0 = \emptyset\) and noting that \(\ker (\mathop{\mathrm{grad}}: \tilde{V}^{0, h}(K_A) \to \tilde{V}^{1, h}(K_A)) = \mathbb{R}\) and \(b_{k}(K) = 0\) for \(k \in 1:3\) since \(K\) is contractible.
Remark 10. We could have connected the skeletal complex to the relative simplicial cochain complex as we did in 3.7 for the full polynomial complex 15 . Here, we highlight an alternative approach which leverages existing cohomology results for simpler finite element spaces.
For a face \(f\in \Delta_2(\mathcal{T})\) and positive integer \(p \in \mathbb{N}_0\), denote the \(L^2(f)\)-orthogonal projection operator onto \(\mathcal{P}_p(f) \otimes \mathbb{X}^k\) by \(\mathbb{P}^k_{p, f} : L^1(f) \otimes \mathbb{X}^k
\to \mathcal{P}_p(f) \otimes \mathbb{X}^k\). Similarly, for \(K \in \mathcal{T}\), let \(\mathbb{P}^k_{p, K} : L^1(K) \otimes \mathbb{X}^k \to V^{k, h}(K_A)\) denote the \(L^2(K)\)-orthogonal projection operator onto \(V^{k, h}(K_A)\).
Step 1: Construction. Let \(z \in \Delta_0(\mathcal{T})\). We choose \(f_{z} \in \Delta_2(\mathcal{T})\) with \(z
\mathop{\mathrm{\unlhd}}f_{z}\), \(\tau_{z} \in \Delta_2(\mathcal{T}) \cup \mathcal{T}\) with \(z \mathop{\mathrm{\unlhd}}\tau_{z}\), and a basis for \(\mathbb{R}^3\) \(\{ \hat{\nu}_{z, j} \}_{j=1}^{3}\) as follows:
If \(z \in \Delta_0^{\flat}(\mathcal{U})\), let \(f_{z} \in \Delta_2(\mathcal{U})\), \(\tau_z \in \mathcal{T}\), \(\hat{\nu}_{z, j}\), \(j \in 1:2\), span the tangent plane of \(\Gamma_0\) at \(z\), and \(\hat{\nu}_{z, 3}\) be the unit outward normal of \(\partial \Omega\) at \(z\).
If \(z \in \Delta_0(\mathcal{U}) \setminus \Delta_0^{\flat}(\mathcal{U})\), let \(f_{z}, \tau_{z} \in \Delta_2(\mathcal{U})\) be not coplanar, \(\hat{\nu}_{z, j}\), \(j \in 1:2\) span the tangent plane of \(f_z\), and \(\hat{\nu}_{z, 3}\) be in the tangent plane of \(\tau_z\) so that \(\{\hat{\nu}_{z, j}\}_{j=1}^{3}\) is a basis for \(\mathbb{R}^3\).
If \(z \in \Delta_0(\mathcal{T}) \setminus \Delta_0(\mathcal{U})\), let \(f_z, \tau_z \in \Delta_2(\mathcal{T})\) be not coplanar and let \(\{\hat{\nu}_{z, j}\}_{j=1}^{3}\) be as in (b).
We define \(S_z^k \in \mathbb{R}^{3\times 3}_{\mathop{\mathrm{sym}}}\), \(k \in 0:1\), according to \[\begin{align} S_z^k : \hat{\nu}_{z, i} \otimes \hat{\nu}_{z, j} &= (\mathop{\mathrm{sym}}\mathop{\mathrm{grad}}_{f_{z}} \mathbb{P}_{4, f_{z}}^1 \mathop{\mathrm{grad}}^{1-k} v^k)(z) : \hat{\nu}_{z, i} \otimes \hat{\nu}_{z, j}, \\ S_z^k : \hat{\nu}_{z, 3} \otimes \hat{\nu}_{z, 3} &= (\mathop{\mathrm{sym}}\mathop{\mathrm{grad}}_{\tau_z} \mathbb{P}_{4, \tau_{z}}^1 \mathop{\mathrm{grad}}^{1-k} v^k)(z) : \hat{\nu}_{z, 3} \otimes \hat{\nu}_{z, 3}, \end{align}\] for all \(i \in 1:3\) and \(j \in 1:2\), where we recall that \(\mathop{\mathrm{grad}}^0 = I\), \(\mathop{\mathrm{grad}}_{f}\) for \(f \in \Delta_2(\mathcal{T})\) is the surface gradient (here taking values in \(\mathbb{R}^{3 \times 3}\)), and we set \(\mathop{\mathrm{grad}}_K := \mathop{\mathrm{grad}}\) for \(K \in \mathcal{T}\).
We define \(\tilde{\Pi}^k v^k\), \(k \in 0:3\), by assigning the degrees of freedom in 23 28 27 as follows: \[\begin{align} \label{eq:proof:fortin-current-dofs} \mathcal{I}_{\tau}^{k}( \tilde{\Pi}^k v^k) =
\mathcal{I}_{\tau}^{k}(v^k) \qquad \forall \tau \in \Delta_k(\mathcal{T}),
\end{align}\tag{74}\] where we note that the above conditions are well-defined since \(H^2(\Omega)\) is continuously embedded into \(C(\bar{\Omega})\) and \(v \mapsto \int_{e} v \cdot \hat{t}_e \,\mathrm{d}{s}\) is a continuous linear functional on \(H^1(\mathop{\mathrm{curl}}; \Omega)\) [40]. The remaining vertex degrees of freedom for \(z \in \Delta_0(\mathcal{T})\) are given by \[\begin{align}
{2} \mathop{\mathrm{grad}}\tilde{\Pi}^0 v^0(z) &= (\mathbb{P}_{4, f_z}^{1} \mathop{\mathrm{grad}}v^0)(z), \quad & \mathop{\mathrm{hess}}\tilde{\Pi}^0 v^0(z) &= S_z^{0}, \\ \tilde{\Pi}^1 v^1(z) &= (\mathbb{P}_{4, f_z}^{1} v^1)(z), \quad
& \mathop{\mathrm{grad}}\tilde{\Pi}^1 v^1(z) &= S_z^{1} + \frac{1}{2} \mathop{\mathrm{mskw}}(\mathbb{P}_{3, f_z}^{2} \mathop{\mathrm{curl}}v^1)(z), \\ & \quad & \tilde{\Pi}^2 v^2(z) &= (\mathbb{P}_{3, f_z}^{2} v^2)(z),
\end{align}\] where \[\begin{align} \mathop{\mathrm{mskw}}(u) := \begin{pmatrix} 0 & -u_3 & u_2 \\ u_3 & 0 & -u_1 \\ -u_2 & u_1 & 0 \end{pmatrix} \qquad \forall u \in \mathbb{R}^3.
\end{align}\] In particular, we have the formal identity \(\mathop{\mathrm{grad}}v - (\mathop{\mathrm{grad}}v)^T = \mathop{\mathrm{mskw}}(\mathop{\mathrm{curl}}v)\). We also note that above the conditions are
well-defined by the trace theorem.
Step 2: Commutativity. Thanks to 44 74 , we have \[\begin{align} \mathcal{I}_{\tau}^{k+1}( \mathop{\mathrm{d}}^k
\tilde{\Pi}^k v^k) = \sum_{\eta \in \Delta_{k}(\tau)} \mathcal{O}(\eta, \tau) \mathcal{I}_{\tau}^{k}(\tilde{\Pi}^k v^k) &= \sum_{\eta \in \Delta_{k}(\tau)} \mathcal{O}(\eta, \tau) \mathcal{I}_{\tau}^{k}(v^k) \\ &= \mathcal{I}_{\tau}^{k+1}(
\mathop{\mathrm{d}}^k v^k) = \mathcal{I}_{\tau}^{k+1}( \tilde{\Pi}^{k+1} \mathop{\mathrm{d}}^k v^k)
\end{align}\] for all \(\tau \in \Delta_{k+1}(\mathcal{T})\) and \(k \in 0:2\). Moreover, we easily see from the choice of vertex degrees of freedom that for all \(z \in \Delta_0(\mathcal{T})\), there holds \[\begin{align} (\mathop{\mathrm{grad}}^{\ell} \tilde{\Pi}^0 v^0)(z) &= \mathop{\mathrm{grad}}^{\ell-1} (\tilde{\Pi}^1 \mathop{\mathrm{grad}}v^0)(z)
\qquad \ell \in 1:2 \\ (\mathop{\mathrm{curl}}\tilde{\Pi}^1 v^1)(z) &= \tilde{\Pi}^2 \mathop{\mathrm{curl}}v^1 (z),
\end{align}\] and so \(\tilde{\Pi}^{k+1} \mathop{\mathrm{d}}^k v^k = \mathop{\mathrm{d}}^k \tilde{\Pi}^{k} v^k\), \(k \in 0:2\).
Step 3: Trace preservation and projection. Suppose that \(v^k \in V^k_{\Gamma_0}\). Then, we have \(\mathcal{I}^k_{\tau}(\tilde{\Pi}^k v^k) = 0\) for all \(\tau \in \Delta_k(\mathcal{U})\). Moreover, the choice of vertex degrees of freedom ensure that \[\begin{align} (I - \hat{n}_{\Gamma}(z) \otimes \hat{n}_{\Gamma}(z)) S_z^k = 0 \text{ if } z \in
\Delta_0^{\flat}(\mathcal{U}) \quad \text{and} \quad S^k_z = 0 \text{ if } z \in \Delta_0(\mathcal{U}) \setminus \Delta_0^{\flat}(\mathcal{U})
\end{align}\] for \(k \in 0:1\) and additionally \[\begin{align} \mathop{\mathrm{grad}}\tilde{\Pi}^0 v^0(z) = \tilde{\Pi}^1 v^1(z) = \tilde{\Pi}^1 \mathop{\mathrm{curl}}v^1(z) =
\tilde{\Pi}^2 v^2(z) = 0 \qquad \forall z \in \Delta_0(\mathcal{U}).
\end{align}\] Thus, the degrees of freedom in 9 vanish, and so \(\tilde{\Pi}^k v^k \in
V^k_{\Gamma_0}\).
That \(\tilde{\Pi}^k\) is a projection readily follows from the choice of degrees and ranges of the operators \(\mathbb{P}^k_{p, f_z}\) and \(\mathbb{P}^k_{p,
\tau_z}\).
Step 4: Continuity. Continuity of the operators follow from standard scaling and approximation arguments (e.g. analogous arguments to proof of [40] and [30]) and are omitted for brevity.
PB acknowledges that this work has received funding through the UKRI Digital Research Infrastructure Programme through the Science and Technology Facilities Council’s Computational Science Centre for Research Communities (CoSeC)↩︎
YL was partly supported through a Royal Society University Research Fellowship (URF\R1\221398, RF\ERE\221047).↩︎
CP was supported by an appointment to the NRC Research Associateship Program at the U.S. Naval Research Laboratory, administered by the Fellowships Office of the National Academies of Sciences, Engineering, and Medicine. Distribution Statement A. Approved for public release: distribution is unlimited.↩︎
In [23], \(H^k_{\Gamma_0}(\Omega)\) is defined to be the completion of smooth functions whose support is disjoint from \(\Gamma_0\), which is equivalent to 1 thanks to [25] on noting that \(\bar{\Gamma}_0\) is a closed 2-Ahlfors regular set.↩︎