Proof. By the definitions, \(\hat{\phi}(B_D(\mathbf{1}_e)) = \hat{\phi}(\mathbf{1}_{t(e)} - \mathbf{1}_{s(e)}) = \phi(t(e)) - \phi(s(e)) = \partial_{\phi}(\mathbf{1}_e)\). As both sides are \(\mathbb{F}\)-linear and agree on the basis \(\{\mathbf{1}_e\}_{e \in E}\), the equality holds on \(\mathbb{F}^E\). ◻

Proof. By Proposition 5, \(\partial_{\phi} = \hat{\phi}\circ B_D\). Hence \(\xi \in \mathrm{Ker}(\partial_{\phi})\) if and only if \(B_D(\xi) \in \mathrm{Ker}(\hat{\phi})\). The restriction of \(B_D\) to \(\mathrm{Ker}(\partial_{\phi})\) is an \(\mathbb{F}\)-linear surjection onto \(\mathrm{Im}(B_D) \cap \mathrm{Ker}(\hat{\phi})\) with kernel \(\mathrm{Ker}(B_D)\), which yields the short exact sequence \[0 \to \mathrm{Ker}(B_D) \to \mathrm{Ker}(\partial_{\phi}) \xrightarrow{B_D} \mathrm{Im}(B_D) \cap \mathrm{Ker}(\hat{\phi}) \to 0.\] Thus, we have \(\dim_{\mathbb{F}} \mathrm{Ker}(\partial_{\phi}) = \dim_{\mathbb{F}} \mathrm{Ker}(B_D) + \delta_{\phi}\). Substituting Lemma 1 implies 4 . ◻

Proof. By Proposition 5, \(\mathrm{Im}(\partial_{\phi}) = (\hat{\phi}\circ B_D)(\mathbb{F}^E) = \hat{\phi}(\mathrm{Im}(B_D))\). Applying the rank-nullity theorem to the restriction \(\hat{\phi}|_{\mathrm{Im}(B_D)} : \mathrm{Im}(B_D) \to U\), we obtain \[\begin{align} \mathrm{rank}_{\mathbb{F}}(B_D) &= \dim_{\mathbb{F}} \hat{\phi}(\mathrm{Im}(B_D)) + \dim_{\mathbb{F}}\left(\mathrm{Im}(B_D) \cap \mathrm{Ker}(\hat{\phi})\right) \\ &= \mathrm{rank}_{\mathbb{F}}(\partial_{\phi}) + \delta_{\phi}. \end{align}\] The claim now follows from \(\mathrm{rank}_{\mathbb{F}}(B_D) = |X| - c(D)\) in Lemma 1. ◻

Proof. We have \(\mathrm{Im}(\partial_{\phi}) = \mathrm{Span}_{\mathbb{F}}\{\phi(t(e)) - \phi(s(e)) \mid e \in E\}\). For an edge \(e\) in the component \(C\), \[\phi(t(e)) - \phi(s(e)) = \left(\phi(t(e)) - \phi(r_C)\right) - \left(\phi(s(e)) - \phi(r_C)\right) \in U_C.\] Conversely, let \(x \in C\) and take a path \(r_C = x_0, x_1, \ldots, x_{\ell} = x\) from \(r_C\) to \(x\) in the underlying graph of \(D\). For each \(i\), choose an edge \(e_i \in E\) joining \(x_{i-1}\) and \(x_i\), and put \[\varepsilon_i := \begin{cases} +1 & \text{if } s(e_i) = x_{i-1}, \\ -1 & \text{if } s(e_i) = x_i. \end{cases}\] Then \(\varepsilon_i \partial_{\phi}(\mathbf{1}_{e_i}) = \phi(x_i) - \phi(x_{i-1})\) in both cases. Hence \[\partial_{\phi}\left(\sum_{i=1}^{\ell} \varepsilon_i \mathbf{1}_{e_i}\right) = \sum_{i=1}^{\ell} \left(\phi(x_i) - \phi(x_{i-1})\right) = \phi(x) - \phi(r_C),\] so \(\phi(x) - \phi(r_C) \in \mathrm{Im}(\partial_{\phi})\). We conclude \(\mathrm{Im}(\partial_{\phi}) = \sum_C U_C\). ◻

Proof. (1) We have \(\mathrm{Im}(A_{\phi}) = \mathrm{Span}_{\mathbb{F}}\{\phi(x) - \phi(r_C)\} = \sum_C U_C\), so by Proposition 8, \(\mathrm{rank}_{\mathbb{F}}(A_{\phi}) = \dim_{\mathbb{F}}\left(\sum_C U_C\right) = \mathrm{rank}_{\mathbb{F}}(\partial_{\phi})\). The dimension of the domain is \(\sum_C (|C| - 1) = |X| - c(D)\), so Corollary 1 gives \[\dim_{\mathbb{F}} \mathrm{Ker}(A_{\phi}) = \left(|X| - c(D)\right) - \mathrm{rank}_{\mathbb{F}}(A_{\phi}) = \left(|X| - c(D)\right) - \mathrm{rank}_{\mathbb{F}}(\partial_{\phi}) = \delta_{\phi}.\] If \(r_C\) is replaced by another basepoint \(r_C'\), then \[\phi(x) - \phi(r_C') = \left(\phi(x) - \phi(r_C)\right) - \left(\phi(r_C') - \phi(r_C)\right),\] which shows that \(U_C\) is independent of the choice of basepoint. Hence \(\mathrm{rank}_{\mathbb{F}}(A_{\phi})\) is independent of \(\{r_C\}\), and so is \(\dim_{\mathbb{F}} \mathrm{Ker}(A_{\phi}) = \delta_{\phi}\).

(2) The injectivity of \(A_{\phi}\) is equivalent to the \(\mathbb{F}\)-linear independence of \(\{A_{\phi}(\mathbf{1}_x)\}_{C, x \in C \setminus \{r_C\}}\). The claim follows from (1).

(3) This is immediate from (2) and the definition of affine independence. ◻

Proof. For each \(e \in E \setminus T\), by the definition of \(B_D\) and 6 , \[B_D(Z_e^{(\mathrm{top})}) = B_D(\mathbf{1}_e) - B_D(F[s(e), t(e)]) = (\mathbf{1}_{t(e)} - \mathbf{1}_{s(e)}) - (\mathbf{1}_{t(e)} - \mathbf{1}_{s(e)}) = 0,\] so \(Z_e^{(\mathrm{top})} \in \mathrm{Ker}(B_D)\).

By Definition 4, \(F[s(e), t(e)]\) has support in \(T\), so the projection \(\pi : \mathbb{F}^E \to \mathbb{F}^{E \setminus T}\) onto the non-tree coordinates satisfies \(\pi(F[s(e), t(e)]) = 0\). Hence \(\pi(Z_e^{(\mathrm{top})}) = \pi(\mathbf{1}_e) = \mathbf{1}_e\) for \(e \in E \setminus T\). If \(\sum_{e \in E \setminus T} \alpha_e Z_e^{(\mathrm{top})} = 0\) with \(\alpha_e \in \mathbb{F}\), then applying \(\pi\) gives \(\sum_{e \in E \setminus T} \alpha_e \mathbf{1}_e = 0\), and since \(\{\mathbf{1}_e\}_{e \in E \setminus T}\) is the standard basis of \(\mathbb{F}^{E \setminus T}\), all \(\alpha_e = 0\).

The dimension of \(\mathrm{Span}_{\mathbb{F}}\{Z_e^{(\mathrm{top})}\}_{e \in E \setminus T}\) is \(|E \setminus T| = |E| - |X| + c(D)\), which by Lemma 1 equals \(\dim_{\mathbb{F}} \mathrm{Ker}(B_D)\). Therefore, \(\{Z_e^{(\mathrm{top})}\}_{e \in E \setminus T}\) is an \(\mathbb{F}\)-basis of \(\mathrm{Ker}(B_D)\). ◻

Proof. Write \(r = \sum_{x \in X} r_x \mathbf{1}_x\) with \(r_x \in \mathbb{F}\). Note that \(\mathrm{Im}(B_D)\) is spanned by the elements \(B_D(\mathbf{1}_e) = \mathbf{1}_{t(e)} - \mathbf{1}_{s(e)}\). For any \(e\in E\) \(s(e)\) and \(t(e)\) lie in the same weakly connected component. Thus, \(\sum_{x \in C} r_x = 0\) for each weakly connected component \(C\).

For each component \(C\), fix a basepoint \(r_C \in C\). Using the signed path vector \(F[r_C, x] \in \mathbb{F}^T \subset \mathbb{F}^E\) for \(x \in C \setminus \{r_C\}\), define \[\zeta_r := \sum_C \sum_{x \in C \setminus \{r_C\}} r_x\, F[r_C, x] \;\in\; \mathbb{F}^E.\] By 6 , \(B_D(F[r_C, x]) = \mathbf{1}_x - \mathbf{1}_{r_C}\). The condition \(\sum_{x \in C} r_x = 0\) gives \(\sum_{x \in C \setminus \{r_C\}} r_x = -r_{r_C}\), and so \[\begin{align} B_D(\zeta_r) &= \sum_C \sum_{x \in C \setminus \{r_C\}} r_x \left(\mathbf{1}_x - \mathbf{1}_{r_C}\right) \\ &= \sum_C \left(\sum_{x \in C \setminus \{r_C\}} r_x \mathbf{1}_x - \left(\sum_{x \in C \setminus \{r_C\}} r_x\right) \mathbf{1}_{r_C}\right) \\ &= \sum_C \left(\sum_{x \in C \setminus \{r_C\}} r_x \mathbf{1}_x + r_{r_C} \mathbf{1}_{r_C}\right) = \sum_{x \in X} r_x \mathbf{1}_x = r. \end{align}\] Furthermore, by Proposition 5 and \(r \in \mathrm{Ker}(\hat{\phi})\), we have \(\partial_{\phi}(\zeta_r) =\hat{\phi}(r) = 0\). Hence \(\zeta_r \in \mathrm{Ker}(\partial_{\phi})\) and \(B_D(\zeta_r) = r\). ◻

Proof. Every element of \(\mathcal{B}_T^{\mathrm{ext}}\) lies in \(\mathrm{Ker}(\partial_{\phi})\) by Proposition 5, Lemmas 2 and 3. We show \(\mathcal{B}_T^{\mathrm{ext}}\) is linearly independent. Suppose \[\sum_{e \in E \setminus T} \alpha_e Z_e^{(\mathrm{top})} + \sum_{i=1}^{\delta_{\phi}} \gamma_i \zeta_{r_i} = 0, \quad \alpha_e, \gamma_i \in \mathbb{F}.\] Applying \(B_D\) and using \(Z_e^{(\mathrm{top})} \in \mathrm{Ker}(B_D)\) together with \(B_D(\zeta_{r_i}) = r_i\), we obtain \(\sum_{i=1}^{\delta_{\phi}} \gamma_i r_i = 0\). Since \(\{r_1, \ldots, r_{\delta_{\phi}}\}\) is an \(\mathbb{F}\)-basis of \(\mathcal{Z}_{\mathrm{alg}}\), all \(\gamma_i = 0\). Substituting back, \(\sum_{e \in E \setminus T} \alpha_e Z_e^{(\mathrm{top})} = 0\), and by Lemma 2, all \(\alpha_e = 0\).

The topological cycles in \(\mathcal{B}_T^{\mathrm{ext}}\) are indexed by \(E \setminus T\) and the lifts by \(\{1, \ldots, \delta_{\phi}\}\), and the linear independence above shows all these elements are distinct. By the definition of a spanning forest, \(|E \setminus T| = |E| - |X| + c(D)\), so \(|\mathcal{B}_T^{\mathrm{ext}}| = |E| - |X| + c(D) + \delta_{\phi}\). By Theorem 7, this equals \(\dim_{\mathbb{F}} \mathrm{Ker}(\partial_{\phi})\). Hence \(\mathcal{B}_T^{\mathrm{ext}}\) is an \(\mathbb{F}\)-basis of \(\mathrm{Ker}(\partial_{\phi})\). ◻

Proof. Since \(\sigma \mapsto \tau \sigma\) is a bijection of \(S_k\), \[\begin{align} \mathrm{Sym}_k(m_{\tau(1)} \otimes \cdots \otimes m_{\tau(k)}) &= \sum_{\sigma \in S_k} m_{\tau\sigma(1)} \otimes \cdots \otimes m_{\tau\sigma(k)} \\ &= \sum_{\sigma' \in S_k} m_{\sigma'(1)} \otimes \cdots \otimes m_{\sigma'(k)} \\ &= \mathrm{Sym}_k(m_1 \otimes \cdots \otimes m_k). \end{align}\] Thus, the assertion follows. ◻

Proof. Apply Proposition 5 with \(D = \mathcal{H}_{\mathrm{macro}}\), \(\phi(w) = w\), \(U = T(\mathbb{F}^{Q_0})\). ◻

Proof. It follows from Proposition 14 that \[\mathcal{Z}_{\mathrm{top}}(\mathcal{H}) = \mathrm{Ker}(B_{\mathrm{macro}}) \subseteq \mathrm{Ker}(\hat{\phi}\circ B_{\mathrm{macro}}) = \mathrm{Ker}(\partial_{\beta}) = \mathcal{Z}(\mathcal{H}).\] The restriction \(B_{\mathrm{macro}}|_{\mathcal{Z}(\mathcal{H})} : \mathcal{Z}(\mathcal{H}) \to \mathbb{F}^{V_{\mathrm{macro}}}\) has kernel \(\mathcal{Z}_{\mathrm{top}}(\mathcal{H})\), and image \(B_{\mathrm{macro}}(\mathrm{Ker}(\hat{\phi}\circ B_{\mathrm{macro}})) = \mathrm{Im}(B_{\mathrm{macro}}) \cap \mathrm{Ker}(\hat{\phi}) = \mathcal{Z}_{\mathrm{alg}}(\mathcal{H})\). The claim follows from the first isomorphism theorem. ◻

Proof. Since \(A_e\) and \(B_e\) lie in the same weakly connected component of \(\mathcal{H}_{\mathrm{macro}}\), each generator \(B_{\mathrm{macro}}(\mathbf{1}_e) = \mathbf{1}_{B_e} - \mathbf{1}_{A_e}\) satisfies the stated condition, and so does every element of \(\mathrm{Im}(B_{\mathrm{macro}})\). By Lemma 1, \(\mathrm{rank}_{\mathbb{F}}(B_{\mathrm{macro}}) = |V_{\mathrm{macro}}| - c_{\mathrm{macro}}\), which coincides with the dimension of the subspace defined by the condition. The inclusion is therefore an equality. ◻

Proof. Apply Theorem 7 together with Proposition 14, with \(D = \mathcal{H}_{\mathrm{macro}}\), \(X = V_{\mathrm{macro}}\), \(E = Q_1\), \(U = T(\mathbb{F}^{Q_0})\), and \(\phi(w) = w\). ◻

Proof. The bijections \(f\) and \(g\) induce \(\mathbb{F}\)-linear isomorphisms \(\mathbb{F}^{V_{\mathrm{macro}}} \cong \mathbb{F}^{V_{\mathrm{macro}}'}\) and \(\mathbb{F}^{Q_1} \cong \mathbb{F}^{Q_1'}\) intertwining \(B_{\mathrm{macro}}\) with \(B_{\mathrm{macro}}'\) and \(\hat{\phi}\) with \(\hat{\phi}'\). Hence the subspace \(\mathrm{Im}(B_{\mathrm{macro}}) \cap \mathrm{Ker}(\hat{\phi})\) is mapped isomorphically onto \(\mathrm{Im}(B_{\mathrm{macro}}') \cap \mathrm{Ker}(\hat{\phi}')\), and their dimensions agree. ◻

Proof. Apply Corollary 1 with \(D = \mathcal{H}_{\mathrm{macro}}\) and \(\partial_{\phi} = \partial_{\beta}\). By Lemma 1, \(\mathrm{rank}_{\mathbb{F}}(B_{\mathrm{macro}}) = |V_{\mathrm{macro}}| - c_{\mathrm{macro}}\). ◻

Proof. Apply Proposition 9 with \(D = \mathcal{H}_{\mathrm{macro}}\), \(U = T(\mathbb{F}^{Q_0})\), and \(\phi(w) = w\). ◻

Proof. Apply Theorem 11 directly. ◻

Proof. (1) For each \(w \in \mathcal{B}_\mu\), the number of \(\sigma \in S_k\) with \(v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(k)} = w\) equals the order of the Young subgroup \(\prod_v S_{m_v}\), which is \(\prod_v m_v!\). The formula ?? follows.

(2) Since \(\sum_{w \in \mathcal{B}_\mu} w\) is a sum of distinct standard basis elements with coefficient \(1\), it is non-zero in \(T^k(\mathbb{F}^{Q_0})\). Hence \(\mathrm{Sym}_k(v_\mu) = 0\) if and only if \(\prod_v m_v! = 0\) in \(\mathbb{F}\). If \(\mathrm{char}(\mathbb{F}) = 0\), this never holds. If \(\mathrm{char}(\mathbb{F}) = p > 0\), then since \(p\) is prime, \(p \mid \prod_v m_v!\) if and only if \(p \mid m_v!\) for some \(v\), which holds if and only if \(m_v \geq p\).

(3) Let \(\mu \neq \mu'\) be distinct multisets, and put \(k := |\mu|\), \(k' := |\mu'|\). The sets \(\mathcal{B}_\mu\) and \(\mathcal{B}_{\mu'}\) consist of basis monomials whose underlying multisets are \(\mu\) and \(\mu'\) respectively, hence are disjoint. By (1), \(\mathrm{Sym}_k(v_\mu)\) and \(\mathrm{Sym}_{k'}(v_{\mu'})\) have disjoint supports in the standard basis \(\mathcal{B}\), and under the hypothesis (2) ensures both are non-zero. Non-zero elements with pairwise disjoint supports in the standard basis are \(\mathbb{F}\)-linearly independent. Therefore, the assertion follows. ◻

Proof. For each of the standard constructions (1)–(6), we show that the elements of \(V_{\mathrm{macro}}\setminus \{0\}\) form a family of non-zero tensors with pairwise disjoint supports in the standard basis \(\mathcal{B}\). The claim then follows from the fact that non-zero elements with pairwise disjoint supports in the standard basis are \(\mathbb{F}\)-linearly independent. Since \(V_{\mathrm{macro}}\) is a set, multiple edges yielding the same tensor (for example through parallel hyperedges) appear only once in \(V_{\mathrm{macro}}\). In what follows we show that distinct non-zero tensors in \(V_{\mathrm{macro}}\) have pairwise disjoint supports.

For Construction (2), \(V_{\mathrm{macro}}\) is a subset of \(Q_0\), which is the standard basis of \(T^1(\mathbb{F}^{Q_0})\). Distinct vertices are distinct standard basis elements, so each element of \(V_{\mathrm{macro}}\setminus \{0\}\) is non-zero with singleton support, and these supports are pairwise disjoint.

For Construction (1), the source tensor of a non-loop edge \(\{u, v\}\) (\(u \neq v\)) is \(u \otimes v + v \otimes u \in T^2(\mathbb{F}^{Q_0})\), and the source tensor of a loop edge \(\{v\}\) is \(2(v \otimes v) \in T^2(\mathbb{F}^{Q_0})\). In characteristic \(2\), the source tensor of a loop edge equals \(0\) and is therefore excluded from \(V_{\mathrm{macro}}\setminus \{0\}\), although it may still belong to \(V_{\mathrm{macro}}\). The target tensor of every hyperedge is \(1 \in T^0(\mathbb{F}^{Q_0})\). For distinct vertex pairs \(\{u, v\} \neq \{u', v'\}\), the corresponding non-loop source tensors are distinct elements of \(V_{\mathrm{macro}}\) with supports \(\{u \otimes v, v \otimes u\}\) and \(\{u' \otimes v', v' \otimes u'\}\) that are disjoint. For distinct vertices \(v \neq v'\), the loop source tensors (non-zero when \(\mathrm{char}(\mathbb{F}) \neq 2\)) have disjoint singleton supports \(\{v \otimes v\}\) and \(\{v' \otimes v'\}\). Between non-loop and loop source tensors, \(u \neq v\) implies \(u \otimes v, v \otimes u \neq v' \otimes v'\), so the supports are disjoint. Finally, the support \(\{1\}\) of the target tensor lies in \(T^0(\mathbb{F}^{Q_0})\), which is in a different degree from the supports above, hence disjoint.

For Constructions (3) and (5), each \(A_e\) (and each \(B_e\) in Construction (5)) is of the form \(\mathrm{Sym}_k(v_\mu)\) for some multiset \(\mu\). By Proposition 19 (1), the support of \(\mathrm{Sym}_k(v_\mu)\) in the standard basis is \(\mathcal{B}_\mu\), and for distinct multisets \(\mu \neq \mu'\) we have \(\mathcal{B}_\mu \cap \mathcal{B}_{\mu'} = \emptyset\). Since only elements of \(V_{\mathrm{macro}}\setminus \{0\}\) are considered, Proposition 19 (2) restricts attention to non-zero \(\mathrm{Sym}_k(v_\mu)\). Hence distinct symmetric tensors in \(V_{\mathrm{macro}}\setminus \{0\}\) correspond to distinct multisets and have disjoint supports. In Construction (3), the target tensor of every hyperedge is \(1 \in T^0(\mathbb{F}^{Q_0})\), which lies in a different degree from the source supports and is therefore disjoint.

For Constructions (4) and (6), each \(A_e\) (and each \(B_e\) in Construction (6)) is a pure tensor \(u_1 \otimes \cdots \otimes u_k\) associated with an ordered tuple \((u_1, \ldots, u_k)\). This is itself an element of the standard basis \(\mathcal{B}\) and is non-zero by construction. Distinct pure tensors in \(V_{\mathrm{macro}}\) are distinct standard basis elements with disjoint singleton supports. In Construction (4), the target tensor of every hyperedge is \(1 \in T^0(\mathbb{F}^{Q_0})\), which lies in a different degree from the source supports and is therefore disjoint.

This establishes, for each standard construction, that \(V_{\mathrm{macro}}\setminus \{0\}\) is a family of non-zero tensors with pairwise disjoint supports in the standard basis. ◻

Proof. Take an arbitrary \(\eta \in \mathrm{Im}(B_{\mathrm{macro}}) \cap \mathrm{Ker}(\hat{\phi}) \subseteq \mathbb{F}^{V_{\mathrm{macro}}}\), and show \(\eta = 0\).

Expand \(\eta\) in the standard basis: \[\eta = \sum_{w \in V_{\mathrm{macro}}} c_w \mathbf{1}_w \quad (c_w \in \mathbb{F}).\] If \(0 \in V_{\mathrm{macro}}\), note that the corresponding basis vector \(\mathbf{1}_0 \in \mathbb{F}^{V_{\mathrm{macro}}}\) is distinct from the zero element \(0 \in T(\mathbb{F}^{Q_0})\) in the codomain. To unify the notation, write \(c_0\) for the coefficient of \(\mathbf{1}_0\) when \(0 \in V_{\mathrm{macro}}\), and set \(c_0 := 0\) when \(0 \notin V_{\mathrm{macro}}\).

By the definition of the evaluation map \(\hat{\phi}\) and the assumption \(\eta \in \mathrm{Ker}(\hat{\phi})\), \[0 = \hat{\phi}(\eta) = c_0 \cdot 0 + \sum_{w \in V_{\mathrm{macro}}\setminus \{0\}} c_w \cdot w = \sum_{w \in V_{\mathrm{macro}}\setminus \{0\}} c_w \cdot w.\] By Lemma 6, \(V_{\mathrm{macro}}\setminus \{0\}\) is \(\mathbb{F}\)-linearly independent in \(T(\mathbb{F}^{Q_0})\), so \(c_w = 0\) for every \(w \in V_{\mathrm{macro}}\setminus \{0\}\). Hence \(\eta = c_0 \mathbf{1}_0\). If \(0 \notin V_{\mathrm{macro}}\), then \(c_0 = 0\) and we already have \(\eta = 0\).

It remains to show \(c_0 = 0\) when \(0 \in V_{\mathrm{macro}}\). By Lemma 5, any element of \(\mathrm{Im}(B_{\mathrm{macro}})\) has zero coordinate sum on each weakly connected component of \(\mathcal{H}_{\mathrm{macro}}\). Let \(C_0 \subseteq V_{\mathrm{macro}}\) denote the component containing the zero tensor \(0\). The coordinate sum of \(\eta = c_0 \mathbf{1}_0\) on \(C_0\) is \(c_0\). From \(\eta \in \mathrm{Im}(B_{\mathrm{macro}})\) we obtain \(c_0 = 0\), and so \(\eta = 0\).

This proves \(\delta(\mathcal{H}) = 0\). The second assertion is immediate from Theorem 16. ◻

Proof. For each \(i \in \{2, \ldots, k\}\), the equality \(\beta(e_1) = \beta(e_i)\) gives \(A_{e_1} = A_{e_i}\) and \(B_{e_1} = B_{e_i}\). Hence \[\begin{align} &\partial_\beta(\xi_i) = (B_{e_1} - A_{e_1}) - (B_{e_i} - A_{e_i}) = 0, \\ & B_{\mathrm{macro}}(\xi_i) = (\mathbf{1}_{B_{e_1}} - \mathbf{1}_{A_{e_1}}) - (\mathbf{1}_{B_{e_i}} - \mathbf{1}_{A_{e_i}}) = 0, \end{align}\] so \(\xi_i \in \mathrm{Ker}(B_{\mathrm{macro}}) = \mathcal{Z}_{\mathrm{top}}(\mathcal{H}) \subseteq \mathrm{Ker}(\partial_\beta) = \mathcal{Z}(\mathcal{H})\).

For linear independence, suppose \(\sum_{i=2}^k a_i \xi_i = 0\) for some \(a_i \in \mathbb{F}\). Expanding in the basis \(\{\mathbf{1}_e \mid e \in Q_1\}\), the coefficient of \(\mathbf{1}_{e_j}\) on the left side is \(-a_j\) for each \(j \in \{2, \ldots, k\}\). Hence \(a_j = 0\) for \(j = 2, \ldots, k\), and \(\{\xi_2, \ldots, \xi_k\}\) is \(\mathbb{F}\)-linearly independent. ◻

Proof. By \(\beta(e_i) = (w_0, w_i)\), \[\partial_\beta(\xi) = \sum_{i=1}^r \alpha_i (w_i - w_0) = 0,\] so \(\xi \in \mathcal{Z}(\mathcal{H})\).

Next, we show \(B_{\mathrm{macro}}(\xi) \neq 0\). By the definition of \(B_{\mathrm{macro}}\), we obtain \[B_{\mathrm{macro}}(\xi) = \sum_{i=1}^r \alpha_i (\mathbf{1}_{w_i} - \mathbf{1}_{w_0}) = \sum_{i=1}^r \alpha_i \mathbf{1}_{w_i} - \left(\sum_{i=1}^r \alpha_i\right) \mathbf{1}_{w_0}.\] Since \(w_0, w_1, \ldots, w_r\) are distinct elements of \(V_{\mathrm{macro}}\), \(\{\mathbf{1}_{w_0}, \mathbf{1}_{w_1}, \ldots, \mathbf{1}_{w_r}\}\) is linearly independent in \(\mathbb{F}^{V_{\mathrm{macro}}}\). As \((\alpha_1, \ldots, \alpha_r)\) is non-trivial, some \(\alpha_i \neq 0\), and the coefficient of \(\mathbf{1}_{w_i}\) in \(B_{\mathrm{macro}}(\xi)\) is non-zero. Hence \(B_{\mathrm{macro}}(\xi) \neq 0\).

By Proposition 14 and \(\partial_\beta(\xi) = 0\), we have \(B_{\mathrm{macro}}(\xi) \in \mathrm{Ker}(\hat{\phi})\). Combining these, \(\delta(\mathcal{H}) = \dim_{\mathbb{F}}(\mathrm{Im}(B_{\mathrm{macro}}) \cap \mathrm{Ker}(\hat{\phi})) \geq 1\). ◻

Proof. By contraposition, we show \(\delta(\mathcal{H}) = 0\) whenever \(|Q_1| \leq 1\) or \(|V_{\mathrm{macro}}| \leq 2\).

Case \(|Q_1| \leq 1\): If \(|Q_1| = 0\), then \(V_{\mathrm{macro}}= \emptyset\), \(B_{\mathrm{macro}}= 0\), and \(\delta(\mathcal{H}) = 0\) trivially. If \(|Q_1| = 1\) with unique edge \(e\), then \(B_{\mathrm{macro}}(\mathbf{1}_e) = \mathbf{1}_{B_e} - \mathbf{1}_{A_e}\). If \(A_e = B_e\), then \(\mathrm{Im}(B_{\mathrm{macro}}) = 0\) and \(\delta(\mathcal{H}) = 0\). If \(A_e \neq B_e\), then \(\mathrm{Im}(B_{\mathrm{macro}}) = \mathrm{Span}_{\mathbb{F}}\{\mathbf{1}_{B_e} - \mathbf{1}_{A_e}\}\) is one-dimensional, and any non-zero element \(c(\mathbf{1}_{B_e} - \mathbf{1}_{A_e})\) (\(c \in \mathbb{F}\setminus \{0\}\)) is mapped by \(\hat{\phi}\) to \(c(B_e - A_e) \neq 0\), hence does not lie in \(\mathrm{Ker}(\hat{\phi})\). Thus \(\mathrm{Im}(B_{\mathrm{macro}}) \cap \mathrm{Ker}(\hat{\phi}) = 0\) and \(\delta(\mathcal{H}) = 0\).

Case \(|Q_1| \geq 2\) and \(|V_{\mathrm{macro}}| \leq 2\): If \(|V_{\mathrm{macro}}| \leq 1\), then every hyperedge is a loop and \(B_{\mathrm{macro}}= 0\), so \(\delta(\mathcal{H}) = 0\). If \(|V_{\mathrm{macro}}| = 2\), write \(V_{\mathrm{macro}}= \{w_0, w_1\}\) with \(w_0 \neq w_1\). For each \(e \in Q_1\), \(B_{\mathrm{macro}}(\mathbf{1}_e) \in \{0, \pm(\mathbf{1}_{w_1} - \mathbf{1}_{w_0})\}\), so \(\mathrm{Im}(B_{\mathrm{macro}}) \subseteq \mathrm{Span}_{\mathbb{F}}\{\mathbf{1}_{w_1} - \mathbf{1}_{w_0}\}\) and \(\dim_{\mathbb{F}} \mathrm{Im}(B_{\mathrm{macro}}) \leq 1\). On the other hand, \(\dim_{\mathbb{F}} \mathbb{F}^{V_{\mathrm{macro}}} = 2\) and \(\hat{\phi}(\mathbb{F}^{V_{\mathrm{macro}}}) = \mathrm{Span}_{\mathbb{F}}\{w_0, w_1\}\). Since \(w_0 \neq w_1\), at least one of \(w_0, w_1\) is non-zero, so \(\dim_{\mathbb{F}} \hat{\phi}(\mathbb{F}^{V_{\mathrm{macro}}}) \geq 1\) and \(\dim_{\mathbb{F}} \mathrm{Ker}(\hat{\phi}) \leq 1\).

For two subspaces of \(\mathbb{F}^{V_{\mathrm{macro}}}\) each of dimension at most \(1\) to have non-zero intersection, both must be one-dimensional and equal. In that case \(\mathbf{1}_{w_1} - \mathbf{1}_{w_0} \in \mathrm{Ker}(\hat{\phi})\), that is, \(w_1 - w_0 = 0\), contradicting \(w_0 \neq w_1\). Hence \(\mathrm{Im}(B_{\mathrm{macro}}) \cap \mathrm{Ker}(\hat{\phi}) = 0\) and \(\delta(\mathcal{H}) = 0\). ◻

Proof. We have \(\rho \circ \partial_{\beta} = (\rho \circ \hat{\phi}) \circ B_{\mathrm{macro}}\), so applying Theorem 7 to the labeling \(w \mapsto \rho(w)\) (whose linear extension is \(\rho \circ \hat{\phi}\)) gives the claim. ◻

Proof. We have \(\mathcal{Z}(\mathcal{H}) = \mathrm{Ker}(\partial_\beta) \subseteq \mathrm{Ker}(\rho \circ \partial_\beta) = \mathcal{Z}_\rho(\mathcal{H})\). The restriction of \(\partial_\beta\) to \(\mathcal{Z}_\rho(\mathcal{H})\) sends \(\xi \in \mathcal{Z}_\rho(\mathcal{H})\) to an element of \(\mathrm{Im}(\partial_\beta) \cap \mathrm{Ker}(\rho)\), since \(\rho(\partial_\beta(\xi)) = 0\). Its kernel is \(\mathcal{Z}_\rho(\mathcal{H}) \cap \mathrm{Ker}(\partial_\beta) = \mathcal{Z}(\mathcal{H})\). For surjectivity, any \(w = \partial_\beta(\eta) \in \mathrm{Im}(\partial_\beta) \cap \mathrm{Ker}(\rho)\) comes from \(\eta \in \mathcal{Z}_\rho(\mathcal{H})\), since \(\rho(\partial_\beta(\eta)) = \rho(w) = 0\). This establishes the short exact sequence, and the isomorphism of quotients follows from the first isomorphism theorem. The final equality follows from Theorem 25 and Theorem 16. ◻

Proof. We compute \(\rho(\partial_\beta(\mathbf{1}_e))\) for each edge of Construction (1).

For a non-loop edge \(e = \{u, v\}\) (\(u \neq v\)), using \(\mathrm{char}(\mathbb{F}) = 2\), \[\rho(\partial_\beta(\mathbf{1}_e)) = 0 - (u + v) = -(u + v) = u + v.\]

For a loop edge \(e = \{v\}\), \(\beta(e) = (2(v \otimes v),\, 1) = (0, 1)\), so \(\partial_\beta(\mathbf{1}_e) = 1\) and \(\rho(\partial_\beta(\mathbf{1}_e)) = \rho(1) = 0\).

Hence, for any \(\xi = \sum_{e \in Q_1} a_e \mathbf{1}_e \in \mathbb{F}^{Q_1}\), \[\label{eq:rho-circ-Delta} (\rho \circ \partial_\beta)(\xi) = \sum_{\substack{e \in Q_1 \\ e = \{u_e, v_e\} \text{ non-loop}}} a_e (u_e + v_e) \in \mathbb{F}^{Q_0}.\tag{16}\] On the other hand, \(B^{\mathrm{cl}}\) sends a non-loop edge \(\{u, v\}\) to \(u + v\) and a loop edge to the zero vector. Its entries are \(0\) or \(1\), so \(\rho \circ \partial_\beta\) and \(B^{\mathrm{cl}}\) coincide on the basis \(\{\mathbf{1}_e\}_{e \in Q_1}\), hence \(\rho \circ \partial_\beta = B^{\mathrm{cl}}\) as \(\mathbb{F}\)-linear maps. ◻

Proof. \(\mathrm{Ker}(\pi_{\leq k} \circ \partial_{\beta}) \supseteq \mathrm{Ker}(\partial_{\beta})\), and from \(\pi_{\leq k-1} = \pi_{\leq k-1} \circ \pi_{\leq k}\) we get \(\mathrm{Ker}(\pi_{\leq k-1}) \supseteq \mathrm{Ker}(\pi_{\leq k})\). ◻

Proof. Apply Theorem 25 with observation map \(\rho := \pi_{\leq k}\). From \(\rho \circ \partial_{\beta} = (\pi_{\leq k} \circ \hat{\phi}) \circ B_{\mathrm{macro}}\), applying Theorem 7 to the linear extension \(\pi_{\leq k} \circ \hat{\phi}: \mathbb{F}^{V_{\mathrm{macro}}} \to T^{\leq k}(\mathbb{F}^{Q_0})\) gives \[\dim_{\mathbb{F}} \mathrm{Ker}(\rho \circ \partial_{\beta}) = |Q_1| - |V_{\mathrm{macro}}| + c_{\mathrm{macro}}+ \dim_{\mathbb{F}}\left(\mathrm{Im}(B_{\mathrm{macro}}) \cap \mathrm{Ker}(\pi_{\leq k} \circ \hat{\phi})\right).\] The last term equals \(\delta_{\leq k}(\mathcal{H})\) by Definition 15. ◻

Proof. From \(\pi_{\leq k} = \pi_{\leq k} \circ \pi_{\leq k'}\) (\(k \leq k'\)), we have \(\mathrm{Ker}(\pi_{\leq k} \circ \hat{\phi}) \supseteq \mathrm{Ker}(\pi_{\leq k'} \circ \hat{\phi})\). The dimensions of intersections with \(\mathrm{Im}(B_{\mathrm{macro}})\) are therefore monotonically non-increasing. ◻

Proof. For any \(\xi \in \mathbb{F}^{Q_1}\), \(\pi_{\leq k}(\partial_\beta(\xi)) = 0\) if and only if all degree-\(0, 1, \ldots, k\) components vanish, so \(\mathcal{Z}_{\leq k}(\mathcal{H}) = \mathrm{Ker}(\pi_{\leq k} \circ \partial_\beta) = \bigcap_{j=0}^{k} \mathrm{Ker}(\partial_\beta^{(j)})\). Consequently \(\mathcal{Z}_{\leq k}(\mathcal{H}) = \mathcal{Z}_{\leq k-1}(\mathcal{H}) \cap \mathrm{Ker}(\partial_\beta^{(k)})\), and the kernel of the restriction of \(\partial_\beta^{(k)}\) to \(\mathcal{Z}_{\leq k-1}(\mathcal{H})\) is precisely \(\mathcal{Z}_{\leq k}(\mathcal{H})\). The isomorphism follows from the first isomorphism theorem, and the dimension equality from Theorem 29. ◻

Proof. For any \(\xi = \sum_e a_e \mathbf{1}_e \in \mathbb{R}^{Q_1}\) and \(w \in T(\mathbb{R}^{Q_0})\), \[\begin{align} \langle \partial_{\beta}(\xi), w \rangle &= \left\langle \sum_e a_e (B_e - A_e), w \right\rangle = \sum_e a_e \langle B_e - A_e, w \rangle \\ &= \left\langle \xi, \sum_e \langle B_e - A_e, w \rangle \mathbf{1}_e \right\rangle_{\mathbb{R}^{Q_1}}. \end{align}\] Since this holds for every \(\xi\), the definition of the adjoint yields ?? . ◻

Proof. The matrix entry in the standard basis is \[\langle L_{\beta}\mathbf{1}_{e'}, \mathbf{1}_e \rangle = \langle \partial_{\beta} \mathbf{1}_{e'}, \partial_{\beta} \mathbf{1}_e \rangle = \langle B_{e'} - A_{e'}, B_e - A_e \rangle,\] which by the symmetry of the inner product coincides with the right side of ?? . Thus \(L_{\beta}\) is represented by a symmetric matrix. Positive semi-definiteness follows from \[\langle L_{\beta}\xi, \xi \rangle = \langle \partial_{\beta}^{*} \partial_{\beta} \xi, \xi \rangle = \langle \partial_{\beta} \xi, \partial_{\beta} \xi \rangle = \|\partial_{\beta} \xi\|^2 \geq 0\] for any \(\xi \in \mathbb{R}^{Q_1}\). ◻

Proof. The inclusion \(\mathcal{Z}(\mathcal{H}) \subseteq \mathrm{Ker}(L_{\beta})\) is immediate. For the reverse inclusion, let \(\xi \in \mathrm{Ker}(L_{\beta})\). Then \(L_{\beta}(\xi) = 0\) gives \(\langle L_{\beta}(\xi), \xi \rangle = 0\). By the definition of the adjoint, \[\langle L_{\beta}(\xi), \xi \rangle = \langle \partial_\beta^* \partial_\beta(\xi), \xi \rangle = \langle \partial_\beta(\xi), \partial_\beta(\xi) \rangle = \|\partial_\beta(\xi)\|^2,\] so \(\|\partial_\beta(\xi)\|^2 = 0\). By the positive definiteness of the inner product, \(\partial_\beta(\xi) = 0\), that is, \(\xi \in \mathrm{Ker}(\partial_\beta) = \mathcal{Z}(\mathcal{H})\). ◻

Proof. Since \(L_{\beta}\) is symmetric positive semi-definite, it is diagonalizable and \(\mathbb{R}^{Q_1}\) decomposes as an orthogonal direct sum of eigenspaces. The eigenspace at \(0\) is \(\mathrm{Ker}(L_{\beta})\), which coincides with \(\mathcal{Z}(\mathcal{H})\) by Proposition 33, so the multiplicity of \(0\) is \(\dim_{\mathbb{R}} \mathcal{Z}(\mathcal{H})\). The complementary subspace of dimension \(|Q_1| - \dim_{\mathbb{R}} \mathcal{Z}(\mathcal{H})\) is the orthogonal direct sum of the non-zero eigenspaces. Its dimension is \(\mathrm{rank}(L_{\beta})\) and equals the total multiplicity of the non-zero eigenvalues. By Theorem 16, \[\dim_{\mathbb{R}} \mathcal{Z}(\mathcal{H}) = |Q_1| - |V_{\mathrm{macro}}| + c_{\mathrm{macro}}+ \delta(\mathcal{H}),\] and substituting this gives 20 . ◻

Proof. The diagonalizability of \(L_{\beta}\) gives the orthogonal direct sum \(\mathbb{R}^{Q_1} = \mathrm{Ker}(L_{\beta}) \oplus \mathrm{Ker}(L_{\beta})^{\perp}\). By Proposition 33, \(\mathrm{Ker}(L_{\beta}) = \mathcal{Z}(\mathcal{H})\), so \(\xi\) decomposes as \(\xi = P_{\mathcal{Z}} \xi + \xi^{\perp}\) with \(\xi^{\perp} := \xi - P_{\mathcal{Z}} \xi \in \mathcal{Z}(\mathcal{H})^{\perp}\). Since \(L_{\beta}(P_{\mathcal{Z}} \xi) = 0\), \[\|\partial_{\beta} \xi\|^2 = \langle L_{\beta}\xi, \xi \rangle = \langle L_{\beta}\xi^{\perp}, \xi^{\perp} \rangle.\] The eigenvalues of \(L_{\beta}\) restricted to \(\mathcal{Z}(\mathcal{H})^{\perp}\) are positive real numbers in the interval \([\lambda_{\min}^{+}, \lambda_{\max}]\), so \[\lambda_{\min}^{+} \|\xi^{\perp}\|^2 \leq \langle L_{\beta}\xi^{\perp}, \xi^{\perp} \rangle \leq \lambda_{\max} \|\xi^{\perp}\|^2,\] which yields 21 . ◻

Proof. By the orthogonality of distinct degree componants, \(\pi_{\leq k} = \sum_{j=0}^{k} \pi_j\), where each \(\pi_j\) is self-adjoint and satisfies \(\pi_j \pi_{j'} = \delta_{jj'} \pi_j\). Hence \[L_{\leq k} = \partial_{\beta}^{*} \pi_{\leq k} \partial_{\beta} = \partial_{\beta}^{*} \left(\sum_{j=0}^{k} \pi_j\right) \partial_{\beta} = \sum_{j=0}^{k} \partial_{\beta}^{*} \pi_j \partial_{\beta} = \sum_{j=0}^{k} L^{(j)}.\] The equality \(L_{\beta}= L_{\leq K}\) follows from \(\partial_{\beta}(\mathbb{R}^{Q_1}) \subseteq T^{\leq K}(\mathbb{R}^{Q_0})\), giving \(\pi_{\leq K} \partial_{\beta} = \partial_{\beta}\). The orthogonal decomposition ?? is obtained by setting \(k = K\) in 22 . ◻

Proof. (1) Symmetry is immediate from the form \(L_{\leq k} = \partial_{\leq k}^{*} \partial_{\leq k}\), and positive semi-definiteness is clear from 22 .

(2) Since \(L_{\leq k}\) is positive semi-definite, \(\xi \in \mathrm{Ker}(L_{\leq k})\) if and only if \(\langle L_{\leq k} \xi, \xi \rangle = 0\). By 22 this is equivalent to \(\pi_{\leq k} \partial_{\beta}(\xi) = 0\), that is, \(\xi \in \mathrm{Ker}(\pi_{\leq k} \circ \partial_{\beta}) = \mathcal{Z}_{\leq k}(\mathcal{H})\).

(3) By (2), \(\mathrm{rank}(L_{\leq k}) = |Q_1| - \dim \mathcal{Z}_{\leq k}(\mathcal{H})\), and combining with Theorem 29 gives the formula.

(4) For \(k \leq \ell\), applying 22 to the difference \(L_{\leq \ell} - L_{\leq k}\), for any \(\xi \in \mathbb{R}^{Q_1}\), \[\langle (L_{\leq \ell} - L_{\leq k}) \xi, \xi \rangle = \sum_{j=k+1}^{\ell} \|\partial_{\beta}^{(j)}(\xi)\|^2 \geq 0.\] Hence \(L_{\leq \ell} - L_{\leq k}\) is positive semi-definite, that is, \(L_{\leq k} \preceq L_{\leq \ell}\) in the Loewner order. The monotonicity of the eigenvalues follows from the Courant-Fischer min-max principle. ◻

Proof. By Theorem 37 (3), \(\mathrm{rank}(L_{\leq k}) = |V_{\mathrm{macro}}| - c_{\mathrm{macro}}- \delta_{\leq k}(\mathcal{H})\), and under the convention \(k = -1\) this also gives \(\mathrm{rank}(L_{\leq -1}) = 0\). The equation now follows from Proposition 30. ◻

Proof. In Construction (2), \(A_e = s(e)\) and \(B_e = t(e) \in Q_0 \subset T^1(\mathbb{R}^{Q_0})\) for each edge \(e\). The inner product 18 restricted to \(T^1(\mathbb{R}^{Q_0})\) is the standard inner product of \(\mathbb{R}^{Q_0}\), so \(\partial_{\beta}(\mathbf{1}_e) = t(e) - s(e) = B_D(\mathbf{1}_e)\) as \(\mathbb{R}\)-linear maps. The adjoint \(\partial_{\beta}^{*} = B_D^{\top}\) is the transpose with respect to the standard inner product, and \(L_{\beta}= \partial_{\beta}^{*} \partial_{\beta} = B_D^{\top} B_D\). ◻

Proof. In Construction (1), each non-loop edge \(e = \{u, v\}\) (\(u \neq v\)) has \(A_e = u \otimes v + v \otimes u \in T^2(\mathbb{R}^{Q_0})\) and \(B_e = 1 \in T^0(\mathbb{R}^{Q_0})\), so \(\partial_{\beta}(\mathbf{1}_e) = 1 - (u \otimes v + v \otimes u)\). By 18 , the components in \(T^0(\mathbb{R}^{Q_0})\) and \(T^2(\mathbb{R}^{Q_0})\) are orthogonal, and the standard basis \(\{u \otimes v\}_{u, v \in Q_0}\) is orthonormal in \(T^2(\mathbb{R}^{Q_0})\). Hence the diagonal entry is \[(L_{\beta})_{e,e} = \|1\|^2 + \|u \otimes v + v \otimes u\|^2 = 1 + 2 = 3.\] By the absence of parallel edges, distinct edges \(e \neq e'\) correspond to distinct vertex pairs \(\{u, v\} \neq \{u', v'\}\), so the inner product expansion \[\langle u \otimes v + v \otimes u,\, u' \otimes v' + v' \otimes u' \rangle = \delta_{u,u'} \delta_{v,v'} + \delta_{u,v'} \delta_{v,u'} + \delta_{v,u'} \delta_{u,v'} + \delta_{v,v'} \delta_{u,u'}\] vanishes term by term, giving \(\langle A_e, A_{e'} \rangle = 0\). Hence the off-diagonal entry is \[(L_{\beta})_{e,e'} = \langle 1, 1 \rangle - \langle 1, A_{e'} \rangle - \langle A_e, 1 \rangle + \langle A_e, A_{e'} \rangle = 1 - 0 - 0 + 0 = 1,\] and combining diagonal entries \(3\) with off-diagonal entries \(1\) yields \(L_{\beta}= 2 I_m + J_m\). ◻

Proof. \(K_n\) is a loopless simple graph, so Proposition 39 gives \(L_{\beta}= 2 I_m + J_m\). The eigenvalues of \(J_m\) are \(m\) with multiplicity \(1\) (eigenvector \((1, \ldots, 1)^{\top}\)) and \(0\) with multiplicity \(m - 1\) (eigenspace \(\sum_e c_e = 0\)), so the eigenvalues of \(L_{\beta}= 2 I_m + J_m\) are \(m + 2\) and \(2\). Since \(\det(L_{\beta}) = (m + 2) \cdot 2^{m-1} > 0\), \(L_{\beta}\) is invertible, \(\mathrm{Ker}(L_{\beta}) = 0\), and by Proposition 33, \(\mathcal{Z}(K_n) = 0\). ◻

Proof. By the definition of \(\beta_F\), for each \(e \in Q_1\), we have \[\partial_\beta(\mathbf{1}_e) = B_e - A_e = \sum_{v : \mathbb{B}^{\mathrm{oh}}_{v, e} = +1} v - \sum_{v : \mathbb{B}^{\mathrm{oh}}_{v, e} = -1} v = \sum_{v \in Q_0} \mathbb{B}^{\mathrm{oh}}_{v, e} \cdot v.\] By the \(\mathbb{F}\)-linearity of \(\partial_\beta\), for any \(\xi = \sum_e a_e \mathbf{1}_e \in \mathbb{F}^{Q_1}\), \[\partial_\beta(\xi) = \sum_e a_e \sum_v \mathbb{B}^{\mathrm{oh}}_{v, e} v = \sum_v \left(\sum_e \mathbb{B}^{\mathrm{oh}}_{v, e} a_e\right) v = \sum_v (\mathbb{B}^{\mathrm{oh}} \xi)_v \cdot v.\] Since \(Q_0\) is a basis of \(\mathbb{F}^{Q_0}\), the right side vanishes if and only if \(\mathbb{B}^{\mathrm{oh}} \xi = 0\). Hence \(\mathcal{Z}(F(\mathcal{H}_o)) = \mathrm{Ker}(\partial_\beta) = \mathrm{Ker}(\mathbb{B}^{\mathrm{oh}})\). ◻

Proof. By Theorem 43, \(\dim_\mathbb{F}\mathrm{Ker}(\mathbb{B}^{\mathrm{oh}}) = \dim_\mathbb{F}\mathcal{Z}(F(\mathcal{H}_o))\), and applying Theorem 16 to the right side gives the claim. ◻

Proof. A family of one vector is trivially affinely independent.

For two distinct indicator vectors \(x_1, x_2\), an affine dependence relation \(\alpha_1 x_1 + \alpha_2 x_2 = 0\) with \(\alpha_1 + \alpha_2 = 0\) gives \(\alpha_1 (x_1 - x_2) = 0\). Since \(x_1 \neq x_2\), \(\alpha_1 = \alpha_2 = 0\) and the family is affinely independent.

Suppose three distinct indicator vectors \(x_1, x_2, x_3\) are affinely dependent. We derive a contradiction. There exists a non-trivial \((\alpha_i)\) with \(\sum_i \alpha_i x_i = 0\) and \(\sum_i \alpha_i = 0\). If some \(\alpha_i\) is zero, the relation reduces to a non-trivial affine dependence among the remaining two, contradicting the two-vector case. So we may assume \(\alpha_i \neq 0\) for \(i = 1, 2, 3\). From \(\alpha_2 \neq 0\) and \(\alpha_1 + \alpha_3 = -\alpha_2\), \[x_2 = t\, x_1 + (1 - t)\, x_3, \quad t := -\alpha_1/\alpha_2.\] Since \(x_1 \neq x_3\), there is some coordinate \(v\) at which \((x_1)_v\) and \((x_3)_v\) differ, one being \(0\) and the other \(1\). The value \((x_2)_v\) at this coordinate is \(t\) or \(1 - t\), but the indicator property forces \((x_2)_v \in \{0, 1\}\), so \(t \in \{0, 1\}\). However, \(t = 0\) gives \(x_2 = x_3\) and \(t = 1\) gives \(x_2 = x_1\), both contradicting the distinctness of \(x_1, x_2, x_3\). ◻

Proof. The source tensor of each hyperedge is \(\mathbf{1}_S\), and the target tensor is \(\mathbf{1}_{H_i}\). From \(S \neq \emptyset\) and \(S \cap H_i = \emptyset\), we have \(\mathbf{1}_S \neq \mathbf{1}_{H_i}\). Since the \(H_i\) are distinct, the associated macrograph is a star with centre \(\mathbf{1}_S\) and \(r\) leaves \(\mathbf{1}_{H_i}\). Hence \(|V_{\mathrm{macro}}| = r + 1\), \(c_{\mathrm{macro}}= 1\), and Corollary 2 gives \[\begin{align} \delta(F(\mathcal{H}_o)) & = (|V_{\mathrm{macro}}| - c_{\mathrm{macro}}) - \mathrm{rank}_{\mathbb{F}}(\partial_{\beta}) \\ & = r - \dim_{\mathbb{F}} \mathrm{Span}_\mathbb{F}\{\mathbf{1}_{H_i} - \mathbf{1}_S\mid i = 1, \ldots, r\}. \end{align}\] Therefore \(\delta(F(\mathcal{H}_o)) > 0\) if and only if the \(r\) vectors \(\mathbf{1}_{H_i} - \mathbf{1}_S\) are linearly dependent over \(\mathbb{F}\).

Consider a linear dependence relation \(\sum_i \alpha_i (\mathbf{1}_{H_i} - \mathbf{1}_S) = 0\). Since \(H_i \subseteq Q_0 \setminus S\), the support of \(\sum_i \alpha_i \mathbf{1}_{H_i}\) lies in \(Q_0 \setminus S\) and the support of \((\sum_i \alpha_i) \mathbf{1}_S\) lies in \(S\). These are disjoint. Hence the relation is equivalent to \[\sum_i \alpha_i \mathbf{1}_{H_i} = 0 \quad \text{and} \quad \sum_i \alpha_i = 0\] (using \(S \neq \emptyset\), so \(\mathbf{1}_S \neq 0\)), which is precisely an affine dependence relation among the indicator vectors \(\mathbf{1}_{H_1}, \ldots, \mathbf{1}_{H_r}\). This establishes the equivalence of \(\delta(F(\mathcal{H}_o)) > 0\) and the affine dependence of \(\{\mathbf{1}_{H_i}\}\).

By Lemma 7, any three or fewer distinct indicator vectors are affinely independent, so \(r \geq 4\) is neccesary for affine dependence. ◻