May 23, 2026
We present a method to build free immersions in critical dimension on \(m\)-tori for \(m=2,3,4,5\) by using a factorization trick inspired by tori immersions in critical dimension. As an application, we show that the set of smooth free maps from a closed surface \(M\) to \(\mathbb{R}^5\) is nonempty. In particular, every closed surface embeds freely in \(\mathbb{R}^5\).
It was shown first by John Nash in [1] that free maps of manifolds into Euclidean spaces play an important role in the theory of isometric immersions (e.g. see [2] for several fundamental results involving free maps).
Free maps. Given a smooth \(m\)-dimensional manifold \(M\), set \[\text{\boldsymbol{\boldsymbol{q_m}}}=\frac{m(m+3)}{2}.\] We say that a map \(f\in C^2(M,\mathbb{R}^q)\) is free if, in local coordinates \((x^\alpha)\), \(\alpha=1,\ldots,m\), on \(M\) and \((y^i)\), \(i=1,\ldots,q\), on \(\mathbb{R}^q\), \(q\geq q_m\), the osculating matrix \[\text{\boldsymbol{\boldsymbol{\mathcal{D}f}}}=(\partial_\alpha f^i,\partial_{\alpha\beta}f^i)\] of the first and second derivatives of \(f\) has constant rank equal to \(q_m\). Notice that every free map is an immersion. We denote by \(\operatorname{Free}^k(M,\mathbb{R}^q)\) the set of all free \(C^k\) immersions of \(M\) into \(\mathbb{R}^q\).
Freeness and affine connections. The condition of being free is not intrinsic. Recall indeed that, close to any point where the Jacobian of \(f\) has constant rank \(k\), we can find local coordinates \((x^\alpha)\) on \(M\) and \((y^i)\) on \(\mathbb{R}^q\) where \(f\) writes as the projection \((x^\alpha)\mapsto(x^1,\dots,x^k,0,\dots,0)\), which is clearly not free. This is due to the fact that, given a second pair of charts \((x^{\alpha'})\), \((y^{i'})\), the coordinate transformation of the \(\partial^2_{\alpha'\beta'}f^{i'}\) are affine, rather than linear, in the \(\partial_{\alpha\beta}f^{i}\): \[\begin{align} \partial_{\alpha'}f^{i'}&=\partial_i y^{i'}\partial_{\alpha}f^i\partial_{\alpha'}x^\alpha,\\ \partial_{\alpha'\beta'}f^{i'}&=\partial_i y^{i'}(\partial_{\alpha\beta}f^i\partial_{\alpha'}x^\alpha\partial_{\beta'}x^\beta+\partial_{\alpha}f^i\partial_{\alpha'\beta'}x^\alpha)+\partial_{ij}y^{i'}\partial_{\alpha}f^i\partial_{\alpha'}x^\alpha\partial_\beta f^j\partial_{\beta'}x^\beta. \end{align}\]
In order to have a combination of second derivatives that is tensorial is enough setting an affine connection \(\Gamma\) on \(T\mathbb{R}^q\) and an affine connection \(\hat{\Gamma}\) on \(TM\). Indeed, in this case, the differential \(df\) is a section of the bundle \(T^*M\otimes f^*(TN)\) and one can define the Hessian as the covariant derivative of the differential: \[(\nabla df)^i_{\alpha\beta} = \partial_{\alpha\beta}f^i-\hat{\Gamma}^\gamma_{\beta\alpha}\partial_\gamma f^i+\Gamma^i_{jk}\partial_\alpha f^j \partial_\beta f^k.\] Then, given any other pair of local charts \((x^{\alpha'})\), \((y^{i'})\), we have that \[(\nabla df)^{i'}_{\alpha'\beta'} = \partial_i y^{i'} (\nabla df)^i_{\alpha\beta} \partial_{\alpha'}x^\alpha\partial_{\beta'}x^\beta.\] Hence, freeness is an intrinsic concept within the class of manifolds endowed with an affine connection on their tangent bundle; in particular, it is an intrinsic concept within the class of Riemannian manifolds.
The affine connection on \(M\) does not play a significant role. Other than showing that freeness is intrinsic, the connection on \(M\) can be disregarded. The reason is that freeness is about the span of the first and second derivatives and a change of coordinates in \(M\) would leave that span unchanged. Since every manifold has a Riemannian metric, an affine connection exists on every manifold and so there are no obstructions on this side and we can ultimately forget about this connection.
From a concrete viewpoint, the considerations above show that talking naively about free maps \(M\to\mathbb{R}^q\) makes sense as long as we allow only affine changes on \(\mathbb{R}^q\). We make this assumption throughout the rest of the article.
Existence of free maps. Within their work on elimination of singularities, it was shown by Yakov Eliashberg and Misha Gromov [3] (see also [2]) that the partial differential relation of being a free map \(M\to \mathbb{R}^q\) satisfies the \(h\)-principle if either \(q>q_m\) or if \(q=q_m\) and \(M\) is open (i.e. has no compact connected component). In particular this means that, under those hypotheses, a sufficient condition for free maps \(M\to \mathbb{R}^q\) to arise is that \(M\) be parallelizable.
A canonical free map \(\mathbb{R}^m\to\mathbb{R}^{q_m}\) Since \(\mathbb{R}^m\) is open, it has free maps into \(\mathbb{R}^{q_m}\). In this case, we can easily write explicit ones. Let \((x^\alpha)\), \(\alpha=1,\ldots,m\), be linear coordinates on \(\mathbb{R}^m\). An elementary smooth free map \(F:\mathbb{R}^m\to \mathbb{R}^{q_m}\) is given by any \(q_m\)-ple of all monomials of first and second degree in the \(x^\alpha\), e.g. \[F(x^1,\ldots,x^m)=\bigl(x^1,\ldots,x^m,(x^1)^2/2,\ldots,(x^m)^2/2,x^1x^2,\ldots,x^{m-1}x^m\bigr).\] For a convenient ordering of the derivatives we get that \[\mathcal{D}F=\begin{pmatrix} {\mathbb{1}}_m & *\\ O_{r_m} & {\mathbb{1}}_{r_m} \end{pmatrix},\] where \(r_m=m(m+1)/2\), so that \(\det \mathcal{D}F=1\). Clearly every map \(G:\mathbb{R}^m\to \mathbb{R}^q\), \(q>q_m\), which reduces to \(F\) after some projection of \(\mathbb{R}^q\) into \(\mathbb{R}^{q_m}\) is free as well.
Free maps in critical dimension on compact manifolds. It was pointed out by Eliashberg and Mishachev in [4] and by Gromov in [5] (see also [2] and [6]), that it is unknown whether free maps arise in the “critical dimension” case (i.e. for \(q=q_m\), so that \(\mathcal{D}f\) is a square matrix) when \(M\) is compact, besides the well-known cases of spheres \(S^m\) and projective spaces \(\mathbb{R}\mathrm{P}^m\).
Free maps in critical dimension on \(\mathbb{R}\mathrm{P}^m\) and \(\mathbb{S}^m\). Consider the free embedding \(F\) of \(\mathbb{R}^m\) into \(\mathbb{R}^{q_m}\) shown in Example 1, homogenize its components by introducing a new coordinate \(x^{m+1}\) and add to it a further component \((x^{m+1})^2\). The resulting map \[f(x^1,\ldots,x^{m+1})=\bigl(x^1x^{m+1},\ldots,x^mx^{m+1},(x^{m+1})^2,(x^1)^2,\ldots,x^{m-1}x^m\bigr)\] sends homogeneously \(\mathbb{R}^{m+1}\) into \(\mathbb{R}^{q_m+1}\) and induces canonically the so-called (quadratic) Veronese map \(\nu_2:\mathbb{R}\mathrm{P}^m\longrightarrow \mathbb{R}\mathrm{P}^{q_m}\) given by \[\nu_2([x^1:\cdots:x^{m+1}])= [x^1x^{m+1}:\cdots:x^ix^j:\dots:x^{m-1}x^m]_{1\leq i\leq j\leq m+1}.\] Denote by \(Y^k\) the homogeneous coordinate such that \[Y^k(\nu_2([x^1:\cdots:x^{m+1}]))=(x^k)^2,\;1\leq k\leq m+1.\] Then, since at least one of the coordinates in \([x^1:\cdots:x^{m+1}]\) is non-zero, the image \(\nu_2(\mathbb{R}\mathrm{P}^m)\) does not intersect the set \[Y^1+\dots+Y^{m+1}=0,\] that is the set of points at infinity of the \(q_m\)-dimensional plane represented by the affine chart \[Y^1+\dots+Y^{m+1}\neq0.\] Hence, one can think of \(\nu_2\) as a map to \(\mathbb{R}^{q_m}\).
In the affine chart \(x^{m+1}=1\), this map coincides with the canonical free map from \(\mathbb{R}^m\) to \(\mathbb{R}^{q_m}\) and similarly happens on any other chart \(x^k=1\). Since these charts cover every point of \(\mathbb{R}\mathrm{P}^{m}\), it follows that \(\nu_2\) is free.
Observe finally that \(\nu_2\) extends uniquely to an immersion \(\hat{\nu}_2:S^m\to \mathbb{R}^{q_m}\) invariant under the antipodal map. Then also \(\hat{\nu}_2\in\operatorname{Free}(\mathbb{S}^m,\mathbb{R}^{q_m})\) since locally \(\hat{\nu}_2\) and \(\nu_2\) are the same function.
Free maps on tori. After spheres and projective spaces, the simplest compact manifolds are arguably the tori \(\mathbb{T}^m\). Since tori are parallelizable, the results of Eliashberg and Gromov grant the existence of free immersions \[\mathbb{T}^m\to \mathbb{R}^{q_m+1}\] but says nothing about free immersions in critical dimension.
The 1-torus happens to coincide the 1-sphere and so, because of the example above, we know that it admits free maps in critical dimension. A free map \(\mathbb{S}^1\to \mathbb{R}^q\) is just a loop whose radius of curvature is everywhere finite, so that its velocity and acceleration are always transversal. The Veronese map in this case is equivalent to the canonical embedding of the unit circle on the plane, namely the map \[s(x)=(\cos x,\sin x).\] Indeed \[\mathcal{D}s=\begin{pmatrix} -\sin x & \phantom{-}\cos x\\ -\cos x & -\sin x \end{pmatrix},\] so that \(\det\mathcal{D}s=1\).
An elementary family of free maps can be built using the embedding \(s(x)\) by setting \[f_m(x^1,\dots,x^m) = (s(x^i),s(x^i+x^j))_{1\leq i\leq m,1\leq j<i\leq m}\] This is a free map \(\mathbb{T}^m\to\mathbb{R}^{m(m+1)}\). Indeed, in case \(m=2\), the map is \[f_2(x,y)=(s(x),s(y),s(x+y))\] and a direct calculation, after ordering the rows of \(\mathcal{D}f_2\) as \((\partial_x,\partial_{xx},\partial_y,\partial_{yy},\partial_{xy})\), shows that \[\mathcal{D}f_2=\begin{pmatrix} \mathcal{D}s(x)&O_2&*&*\\ O_2&\mathcal{D}s(y)&*&*\\ 0\;0&0\;0&-\cos(x+y)&-\sin(x+y) \end{pmatrix},\] which clearly has full rank at every point. The case \(m>2\) can be treated in a similar way.
To the knowledge of the author, explicit expressions for free maps \(\mathbb{T}^m\to\mathbb{R}^q\) with \(q<m(m+1)\) are not available in literature. Moreover, as pointed out above, the existence of free maps in critical dimension on \(\mathbb{T}^n\) is not established in literature.
Main results and structure of the article. In Section 2 we show that \(\operatorname{Free}^\infty(\mathbb{T}^m,\mathbb{R}^{q_m})\neq\varnothing\) for \(m=2,3,4,5\) by presenting a method to generate explicit expressions of free maps in critical dimension on \(\mathbb{T}^m\) for such values of \(m\). As a further application of this method, we also show that \(\mathbb{T}^2\) admits “k-free” smooth immersions, namely immersions such that its vectors of partial derivatives from order 1 up to order \(k\) are linearly independent at every point, in critical dimension for \(k=3,4,5,6\).
Then, in Section 3, we show that the existence of free maps in critical dimension on closed surfaces \(M\) and \(N\) implies the existence of similar free maps on their connected sum \(M\# N\) and we use the existence of free maps on \(\mathbb{R}\mathrm{P}^2\) and \(\mathbb{T}^2\) to conclude that free maps arise on any closed surface.
A remark on computations. Except for the \(m=2\) case, computer-aided computations were made to ascertain the freeness of the map via Sturm’s criterion (see Appendix 4). Notice that, since the polynomial involved have rational coefficients and computations have been performed in arbitrary precision symbolic arithmetic, then these results are exact. We provide corresponding SageMath and Mathematica code as ancillary files.
The idea behind the results of this section comes from the following elementary observation about immersions (recall that \(q=m+1\) is the critical dimension for the existence of immersions \(\mathbb{T}^m\to\mathbb{R}^q\)):
for every \(m=2,3,\dots\) there are immersion \(\mathbb{T}^m\to\mathbb{R}^{m+1}\) that are obtained as the image of an immersed loop \(v:\mathbb{S}^1\to\mathbb{R}^{m+1}\) under an action of \(\mathbb{T}^{m-1}\to SO(m+1)\) by successive rotations in suitable coordinate 2-planes.
The fundamental block of these isometric toric actions is the elementary action of \(\mathbb{S}^1\) on the plane given by the matrices \[\text{\boldsymbol{\boldsymbol{A(x)}}}= \begin{pmatrix} \cos x & -\sin x \\ \sin x & \phantom{-}\cos x \end{pmatrix}.\] For instance, the canonical embedding of \(\mathbb{T}^2\) in \(\mathbb{R}^3\) is obtained as the image of the loop \[v(t)=(R+r\cos t,0,r\sin t)\] under the action of \(\mathbb{S}^1\) on \(\mathbb{R}^3\) given by \[A_3(x) = \begin{pmatrix} A(x) & 0\\ 0 & 1 \end{pmatrix}.\] It is natural therefore to pose the following question:
are there integers \(m\) for which there is a loop \(\gamma:\mathbb{S}^1\to\mathbb{R}^{q_m}\) and a torus action \(\mathbb{T}^{m-1}\to SO(q_m)\) such that the image of \(\gamma\) under the action of \(\mathbb{T}^{m-1}\) is a free immersion of \(\mathbb{T}^m\) into \(\mathbb{R}^{q_m}\)?
Notice that the matrix \(A(x)\), which is the building block of the representations of the toric actions we are considering, coincides, modulo an orientation switch, with the osculating matrix of the map \(s(x)=(\sin x,\cos x)\).
The examples exhibited below show that the answer to this question is positive at least for \(m=2,3,4,5\). The method we use, in short, is looking for free maps \(F:\mathbb{T}^m\to\mathbb{R}^{q_m}\) that factor as a \(q_m\times q_m\) matrix \(R\), depending only on the first \(m-1\) variables, acting on a vector \(v\in\mathbb{R}^{q_m}\), depending only on the last variable. A key point of this factorization is that the determinant of the osculating matrix of the corresponding map \(F\) depends only on the last coordinate, making easy to verify whether the determinant has or not zeros.
Our explorations of the cases \(m=6\) and \(m=7\) were unable to find any free map in critical dimension. This is possibly just a sign that the size of the osculating matrix for \(m>5\) is too large for this method to be effective. The simplest next step would be to find some pattern in free maps in critical dimension on low-dimensional tori that could be used to prove by induction the existence of free maps in critical dimension on \(\mathbb{T}^n\) for every \(n\).
We use coordinates \((x,y)\in\mathbb{T}^2\). Define \[F_2:\mathbb{T}^2\longrightarrow \mathbb{R}^5\] by \[F_2(x,y)=R_2(x)v_2(y),\] where \[R_2(x)= \begin{pmatrix} A(x)&O_2&0&\\ O_2&A(2x)&0&\\ 0\;0&0\;0&1 \end{pmatrix}\] and \(v_2(y)\) is an immersion of \(\mathbb{S}^1\) into \(\mathbb{R}^5\) to be determined.
For short, we write \(R_2(x) = \operatorname{diag}(A(x),A(2x),1)\in SO(5).\) Let now \[J= \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}\] and \[X=\operatorname{diag}(J,2J,0).\] Then \[R'_2(x)=R_2(x)X.\] Therefore \[\partial_x F_2=R_2Xv_2,\qquad \partial_y F_2=R_2v_2',\] \[\partial_{xx}F_{2}=R_2X^2v_2,\qquad \partial_{xy}F_{2}=R_2Xv_2', \qquad \partial_{yy}F_{2}=R_2v_2''.\] Set \[D(y) = \det(Xv_2(y),v'_2(y),X^2v_2(y),Xv'_2(y),v''_2(y)).\] Since \(R_2(x)\in SO(5)\), we get \[\det\mathcal{D}F_2 = \det R_2(x) D(y) = D(y).\] Thus, the freeness determinant depends only on \(y\) and this type of map \(F_2\) is free if and only if the five vectors \[v_2',v_2'',Xv_2,Xv_2',X^2v_2\] are linearly independent at every point. A computer-aided search led to the following simple expression: \[v_2(y) = \begin{pmatrix} 2\\ 1+\sin y\\ \cos y\\ \frac{1}{2}+\sin y\\ \cos y \end{pmatrix}.\] The corresponding map on \(\mathbb{T}^2\) is \[F_2(x,y)= \begin{pmatrix} 2\cos x-(1+\sin y)\sin x\\[2pt] 2\sin x+(1+\sin y)\cos x\\[2pt] \cos(2x+y)-\frac{1}{2}\sin(2x)\\[2pt] \sin(2x+y)+\frac{1}{2}\cos(2x)\\ \cos y \end{pmatrix}.\]
A direct computation gives \[D(y) = -4\sin y+\frac{5}{2}\sin(2y)+15\cos y-4\cos(2y)-12.\] Put \[t=\tan\frac{y}{2}.\] Using \[\sin y=\frac{2t}{1+t^2}, \qquad \cos y=\frac{1-t^2}{1+t^2},\] one obtains \[D(y(t)) = -\frac{31t^4+18t^3-2t+1}{(1+t^2)^2}.\] The numerator is strictly positive, since \[31t^4+18t^3-2t+1 = 31\left(t^2+\frac{9}{31}t-\frac{2}{31}\right)^2 + \frac{43}{31}\left(t-\frac{13}{43}\right)^2 + \frac{32}{43}.\] Hence \[31t^4+18t^3-2t+1>0\] for every \(t\in\mathbb{R}\), and therefore \[D(y)<0\] for every \(y\neq \pi\). The remaining point \(y=\pi\), corresponding to \(t=\infty\), is also harmless, since \[D(\pi) = -31.\] Hence, \[\det\mathcal{D}F_2\neq0\] everywhere on \(\mathbb{T}^2\) and therefore \[F_2\in \operatorname{Free}^{\infty}(T^2,\mathbb{R}^5).\]
We use coordinates \((x,y,z)\in T^3\). Set \[R_3(x,y)= \operatorname{diag} \bigl( A(x),\, A(y),\, A(x+y),\, A(x-y),\, 1 \bigr) \in SO(9)\] and define \(F_3:T^3\longrightarrow \mathbb{R}^9\) as \[F_3(x,y,z)=R_3(x,y)v_3(z)\] for some \(v_3(z)\in\mathbb{R}^9\). For \(i=1,2\), define \[X_1= \operatorname{diag}(J,0,J,J,0), \qquad X_2= \operatorname{diag}(0,J,J,-J,0).\] Then \[\frac{\partial R_3}{\partial x}=R_3X_1, \qquad \frac{\partial R_3}{\partial y}=R_3X_2,\] so that \[\partial_xF_3=R_3X_1v_3,\qquad \partial_yF_3=R_3X_2v_3,\qquad \partial_zF_3=R_3v_3',\] \[\partial_{xx}F_3=R_3X_1^2v_3,\qquad \partial_{xy}F_3=R_3X_1X_2v_3,\qquad \partial_{xz}F_3=R_3X_1v_3',\] \[\partial_{yy}F_3=R_3X_2^2v_3,\qquad \partial_{yz}F_3=R_3X_2v_3', \qquad \partial_{zz}F_3=R_3v_3''.\] Since \(R_3(x,y)\in SO(9)\), the determinant of the osculating matrix is \[\det DF_3(x,y,z)=D(z),\] where \[D(z)= \det \bigl( X_1v_3,\, X_2v_3,\, v_3',\, X_1^2v_3,\, X_1X_2v_3,\, X_1v_3',\, X_2^2v_3,\, X_2v_3',\, v_3'' \bigr).\] Thus, the determinant depends only on \(z\).
A computer-aided search of a viable immersion \(v_3\) led to the following expression: \[v_3(z)= \begin{pmatrix} 2\\ 1+\sin z\\ \cos z\\ \frac{1}{2}+\sin z\\ \cos z\\ \sin z\\ \cos z\\ 2-\cos z\\ -2\sin z-5\cos z \end{pmatrix}.\] Notice that we enforced artificially the constraint to keep the first five components equal to the components of the \(v_2\) vector to keep results as homogeneous as possible. Numerical explorations showed that for reasonable choices of the first few components of a map \(v_n\) it is possible to complete the map so that the resulting map \(F_n\) is free.
As in the previous case, after setting \(t=\tan\frac{z}{2}\) we get that \[D(z(t))=\frac{p(t)}{2(1+t^2)^5},\] where \[\begin{align} p(t)= {}& 917t^{10}+752t^9+829t^8+32t^7-1474t^6-440t^5\\ &-202t^4-192t^3+397t^2+232t+45. \end{align}\] It remains to show that \(p\) has no real roots. A Sturm sequence computation for \(p\) gives the following signs at infinity: \[\begin{array}{c|ccccccccccc} & p_0&p_1&p_2&p_3&p_4&p_5&p_6&p_7&p_8&p_9&p_{10}\\ \hline +\infty &+&+&-&-&-&-&+&+&-&+&-\\ -\infty &+&-&-&+&-&+&+&-&-&-&- \end{array}\]
Since the number of sign changes is \[V(-\infty)=5, \qquad V(+\infty)=5\] then, by Sturm’s theorem, \[\#\{t\in\mathbb{R}:p(t)=0\} = V(-\infty)-V(+\infty) = 0.\] Since \[p(0)=45>0\] and \(p\) has no real roots, we have \[p(t)>0 \qquad \forall t\in\mathbb{R}.\] Moreover, the leading coefficient is \[917>0,\] so the point \(z=\pi\), corresponding to \(t=\infty\), is also covered. Therefore \[D(z)>0 \qquad \text{ for all }z\in S^1.\]
Hence, \(\mathcal{D}F_3\) is non-degenerate at every point, namely
\[F_3\in \operatorname{Free}^{\infty}(T^3,\mathbb{R}^9).\]
We use coordinates \((x_1,x_2,x_3,z)\in \mathbb{T}^4.\)
Set \[R_4(x_1,x_2,x_3) =\] \[diag\left( A(x_1), A(x_2), A(x_3), A(x_1+x_2), A(x_1+x_3), A(x_2+x_3), A(x_1+x_2+x_3) \right)\] \[\in SO(14)\] and define \[F_4:\mathbb{T}^4\longrightarrow \mathbb{R}^{14}\] by \[F_4(x_1,x_2,x_3,z)=R_4(x_1,x_2,x_3)v_4(z).\]
For \(i=1,2,3\), define \[X_i=diag(w_{1i}J,w_{2i}J,\ldots,w_{7i}J),\] where \[\begin{align} w_1&=(1,0,0),& w_2&=(0,1,0),& w_3&=(0,0,1),\\ w_4&=(1,1,0),& w_5&=(1,0,1),& w_6&=(0,1,1),& w_7&=(1,1,1). \end{align}\] Equivalently, \[\bigl\langle w_1,x\bigr\rangle=x_1,\quad \bigl\langle w_2,x\bigr\rangle=x_2,\quad \bigl\langle w_3,x\bigr\rangle=x_3,\] \[\bigl\langle w_4,x\bigr\rangle=x_1+x_2,\quad \bigl\langle w_5,x\bigr\rangle=x_1+x_3,\quad \bigl\langle w_6,x\bigr\rangle=x_2+x_3,\quad \bigl\langle w_7,x\bigr\rangle=x_1+x_2+x_3.\] Thus \[R_4(x_1,x_2,x_3) = \operatorname{diag} \bigl( A(\langle w_1,x\rangle), A(\langle w_2,x\rangle), \dots, A(\langle w_7,x\rangle) \bigr),\] where \(x=(x_1,x_2,x_3)\). Then \[\frac{\partial R_4}{\partial x_i}=R_4X_i.\] Therefore the fourteen derivative vectors of \(F\) are obtained by applying the common matrix \(R_4\) to the fourteen vectors \[X_1v_4,\;X_2v_4,\;X_3v_4,\;v_4',\] \[X_1^2v_4,\;X_1X_2v_4,\;X_1X_3v_4,\;X_2^2v_4,\;X_2X_3v_4,\;X_3^2v_4,\] \[X_1v_4',\;X_2v_4',\;X_3v_4',\;v_4''.\] Thus \[\det\mathcal{D}F_4(x_1,x_2,x_3,z) = \det(R_4) D(z),\] where \[\begin{align} D(z)= \det\big(& X_1v_4,\;X_2v_4,\;X_3v_4,\;v_4', X_1^2v_4,\;X_1X_2v_4,\;X_1X_3v_4,\\ &X_2^2v_4,\;X_2X_3v_4,\;X_3^2v_4, X_1v_4',\;X_2v_4',\;X_3v_4',\;v_4'' \big). \end{align}\] Since \(R_4\in SO(14)\), we have \(\det(R_4)=1\). Hence the full freeness determinant is exactly \(D(z)\), and in particular depends only on \(z\).
A computer-aided search of viable immersion \(v_4\) led to the following expression: \[v_4(z)= \begin{pmatrix} 2\\ 1+\sin z\\ \cos z\\ \frac{1}{2}+\sin z\\ \cos z\\ \sin z\\ \cos z\\ 2-\cos z\\ -2\sin z-5\cos z\\ \sin z\\ \cos z\\ \sin z\\ \cos z\\ h(z) \end{pmatrix},\] where \[h(z)=-20+9\sin z+13\cos z-7\sin(2z)+3\cos(2z).\] As we pointed out in case of \(\mathbb{T}^3\), we enforced the first 9 components to coincide with the components of \(v_3\) just to display homogeneous results in the article. Many choices of different nine initial components lead to different free maps.
After setting, as usual, \(t=\tan\frac{z}{2}\), we get that \[D(z(t))=\frac{q(t)}{(1+t^2)^{11}},\] where \[\begin{align} q(t)= {}& -60010t^{22}-561400t^{21}-3039880t^{20}-337260t^{19}\\ &+18967810t^{18}+11879260t^{17}-23665760t^{16}\\ &-17732160t^{15}-42514660t^{14}-536720t^{13}\\ &+22966960t^{12}+23893640t^{11}+32807300t^{10}\\ &+18046840t^9-9136320t^8-11620000t^7\\ &-19941970t^6-7234520t^5+1406360t^4\\ &+511940t^3-71910t^2-7540t-160. \end{align}\] A Sturm computation (more detail are included in Appendix 5) gives \[V(-\infty)=11,\qquad V(+\infty)=11.\] Therefore \(q\) has no real roots. Since \[q(0)=-160<0\] and the leading coefficient is also negative, it follows that \[q(t)<0 \qquad\text{for all }t\in\mathbb{R}.\] The point \(z=\pi\), corresponding to \(t=\infty\), is covered by the leading coefficient. Hence \[D(z)<0 \qquad\text{for all }z\in \mathbb{S}^1.\] Therefore the osculating matrix has full rank everywhere, namely \[F_4\in \operatorname{Free}^{\infty}(T^4,\mathbb{R}^{14}).\]
We use coordinates \((x_1,x_2,x_3,x_4,z)\in \mathbb{T}^5.\) We use the ten weights \[\begin{align} &w_1=(1,0,0,0),\qquad w_2=(0,1,0,0),\qquad w_3=(0,0,1,0),\\ &w_4=(1,1,0,0),\qquad w_5=(1,0,1,0),\qquad w_6=(0,1,1,0),\\ &w_7=(0,0,1,1),\qquad w_8=(0,0,0,1),\\ &w_9=(1,0,0,1),\qquad w_{10}=(0,1,0,1). \end{align}\] and define \[R_5(x_1,x_2,x_3,x_4) = \operatorname{diag} \bigl( A(\langle w_1,x\rangle), A(\langle w_2,x\rangle), \dots, A(\langle w_{10},x\rangle) \bigr) \in SO(20),\] where \(x=(x_1,x_2,x_3,x_4)\). Equivalently, \[\begin{align} R_5(x_1,x_2,x_3,x_4) = \operatorname{diag}\bigl( &A(x_1),A(x_2),A(x_3), A(x_1+x_2),A(x_1+x_3),A(x_2+x_3),\\ &A(x_3+x_4),A(x_4),A(x_1+x_4),A(x_2+x_4) \bigr). \end{align}\]
Now set \[F_5:\mathbb{T}^5\longrightarrow\mathbb{R}^{20}, \qquad F_5(x_1,x_2,x_3,x_4,z) = R_5(x_1,x_2,x_3,x_4)v_5(z).\]
For \(i=1,2,3,4\), put \[X_i= \operatorname{diag}(w_{1i}J,w_{2i}J,\ldots,w_{10,i}J).\] Then \[\frac{\partial R_5}{\partial x_i}=R_5X_i.\] Therefore the twenty derivative vectors of \(F_5\) are obtained by applying the common matrix \(R_5\) to the twenty vectors \[X_1v_5,\quad X_2v_5,\quad X_3v_5,\quad X_4v_5,\quad v_5',\] \[X_iX_jv_5,\qquad 1\le i\le j\le 4,\] \[X_1v_5',\quad X_2v_5',\quad X_3v_5',\quad X_4v_5',\quad v_5''.\] Hence \[\det DF_5(x_1,x_2,x_3,x_4,z)=D(z),\] where \[D(z)= \det\Bigl( X_1v_5,X_2v_5,X_3v_5,X_4v_5,v_5', (X_iX_jv_5)_{1\le i\le j\le4}, X_1v_5',X_2v_5',X_3v_5',X_4v_5',v_5'' \Bigr).\]
Let \[s=\sin z,\qquad c=\cos z.\]
A computer-aided search of viable immersed loops \(v\) led to the following expression: \[v_5(z)= \begin{pmatrix} 2\\ 1+s\\ c\\ \frac{1}{2}+s\\ c\\ s\\ c\\ 2-c\\ -2s-5c\\ s\\ c\\ s\\ c\\ h(z)\\ 10-7s\\ 2+4s-6c\\ -5+3s+5c\\ -9+15s+2c\\ -1+8s\\ -2+6c \end{pmatrix},\] where \(h(z)\) is defined in the previous subsection.
Putting \[t=\tan\frac{z}{2},\] and clearing denominators, one obtains a polynomial \(p(t)\) such that \[D(z(t))=\frac{p(t)}{(1+t^2)^N}\] for a positive integer \(N\). A Sturm computation (more detail are included in Appendix 6) gives that \(p\) has no real roots. Moreover, \[p(0)=D(0)=626467660>0,\] and the value at the point \(z=\pi\), corresponding to \(t=\infty\), is \[D(\pi)=77499567520>0.\] Thus \[D(z)>0 \qquad\text{for all }z\in \mathbb{S}^1.\] Therefore the osculating matrix of \(F_5\) has full rank everywhere, and hence \[F_5\in \operatorname{Free}^{\infty}(T^5,\mathbb{R}^{20}).\]
There is an obstruction to this method for \(m>5\). The ansatz used in this section does not work on \(\mathbb{T}^m\) for \(m>5\). Indeed, let \(k=m-1\). Then the ansatz has the form \[F(x_1,\ldots,x_k,z)=R(x_1,\ldots,x_k)v(z),\] where \(a:\mathbb{S}^1\to\mathbb{R}^{q_m}\) is a loop and \(R\) is a block-diagonal orthogonal matrix made of planar rotations: \[R(x)= \operatorname{diag} \bigl( A(\langle w_1,x\rangle), \ldots, A(\langle w_r,x\rangle) \bigr),\] possibly with one additional fixed coordinate if \(q_m\) is odd. Here \[w_j=(w_{j1},\ldots,w_{jk})\in\mathbb{Z}^k, \qquad r=\left\lfloor \frac{q_m}{2}\right\rfloor .\] For \(i=1,\ldots,k\), set \[X_i= \operatorname{diag} \bigl( w_{1i}J,\ldots,w_{ri}J \bigr), \qquad J= \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}.\] Then \[\frac{\partial R}{\partial x_i}=RX_i.\] Since \(R\) is invertible, the rank of the osculating matrix of \(F\) is the same as the rank of the matrix whose columns are obtained after removing the common left factor \(R\).
In particular, the pure second derivatives in the \(x\)-variables are \[\partial_{x_i x_j}F=R X_iX_j v(z), \qquad 1\leq i\leq j\leq k.\] Thus, after removing the common factor \(R\), the relevant columns are \[X_iX_j v(z), \qquad 1\leq i\leq j\leq k.\] Now look at one rotation block. If \(a_j(z)\in\mathbb{R}^2\) denotes the \(j\)-th two-dimensional component of \(v(z)\), then \[\bigl(X_iX_\ell v(z)\bigr)_j = -w_{ji}w_{j\ell}\,a_j(z).\] Hence, for each block \(j\), the pure second derivatives contribute only one scalar row, namely the quadratic vector \[\bigl(w_{j1}^2,\, w_{j1}w_{j2},\, \ldots,\, w_{jk}^2\bigr) \in \operatorname{Sym}^2(\mathbb{R}^k).\] Therefore the rank of the family \[\{X_iX_j v(z)\}_{1\leq i\leq j\leq k}\] is at most the number \(r\) of rotation blocks.
But there are \[\dim \operatorname{Sym}^2(\mathbb{R}^k) = \frac{k(k+1)}{2} = \frac{m(m-1)}{2}\] pure second derivatives in the \(x\)-variables. For the osculating matrix to have full rank, these columns must in particular be linearly independent. Hence a necessary condition for this ansatz to work is \[r\geq \frac{m(m-1)}{2}.\] Since \[r=\left\lfloor \frac{q_m}{2}\right\rfloor = \left\lfloor \frac{m(m+3)}{4}\right\rfloor,\] we obtain the necessary condition \[\left\lfloor \frac{m(m+3)}{4}\right\rfloor \geq \frac{m(m-1)}{2}.\] For \(m>5\), however, \[\frac{m(m+3)}{4} < \frac{m(m-1)}{2},\] because this inequality is equivalent to \[m+3<2m-2,\] that is, to \(m>5\). Therefore the necessary condition fails for every \(m>5\).
Consequently, the simple block-rotation ansatz \[F(x,z)=R(x)v(z),\] with \(a\) depending on only one variable and \(R\) built only out of planar rotation blocks, cannot produce free maps \[T^m\longrightarrow \mathbb{R}^{q_m}\] in critical dimension for \(m>5\).
A natural extension of the ansatz used so far is to replace \(v:\mathbb{S}^1\to\mathbb{R}^{q_m}\) by some \(u:\mathbb{S}^s\to\mathbb{R}^{q_m}\) with \(s>1\). As an example, below we use this extended method with \(s=2\) to build a free map \(\mathbb{T}^4\to\mathbb{R}^{14}\).
We use coordinates \((x_1,x_2,u,v)\in \mathbb{T}^4\) and consider the five weights \[w_1=(-2,-1),\; w_2=(2,0),\; w_3=(1,-2),\; w_4=(-1,-2),\; w_5=(0,-2).\] Define \[R_{10}(x_1,x_2) = \operatorname{diag} \bigl( R_{-2x_1-x_2}, R_{2x_1}, R_{x_1-2x_2}, R_{-x_1-2x_2}, R_{-2x_2} \bigr) \in SO(10).\] Equivalently, \[R_{10}(x_1,x_2) = \operatorname{diag} \bigl( R_{\langle w_1,x\rangle}, R_{\langle w_2,x\rangle}, R_{\langle w_3,x\rangle}, R_{\langle w_4,x\rangle}, R_{\langle w_5,x\rangle} \bigr), \qquad x=(x_1,x_2).\]
We now construct \(b:\mathbb{T}^2\to\mathbb{R}^{10}\) and \(c:\mathbb{T}^2\to\mathbb{R}^4\). Set \[c(u,v)=(\cos u,\sin u,\cos v,\sin v).\] Write \[b(u,v)=(b_1(u,v),\ldots,b_5(u,v)), \qquad b_r(u,v)\in\mathbb{R}^2,\] and define \[b_r(u,v)=\rho_r(u) \begin{pmatrix} \cos(\mu_r v)\\ \sin(\mu_r v) \end{pmatrix},\] where \[(\mu_1,\mu_2,\mu_3,\mu_4,\mu_5)=(2,2,-2,-3,1).\] We take \[\rho_r(u)=e^{Q_r(u)}.\] Let \[q_r(u)=Q_r'(u).\] The logarithmic derivatives are chosen as \[\begin{align} q_1&=2\cos u+3\sin u,\\ q_2&=-4\cos u-2\sin u-2\cos(2u)+3\sin(2u),\\ q_3&=-3\cos u+2\sin u+4\cos(2u)-\sin(2u),\\ q_4&=\cos u-2\sin u-\cos(2u)+2\sin(2u),\\ q_5&=-2\cos u-2\sin u-2\cos(2u)-\sin(2u). \end{align}\] Equivalently, one may choose the following periodic primitives: \[\begin{align} Q_1&=2\sin u-3\cos u,\\ Q_2&=-4\sin u+2\cos u-\sin(2u)-\frac{3}{2}\cos(2u),\\ Q_3&=-3\sin u-2\cos u+2\sin(2u)+\frac{1}{2}\cos(2u),\\ Q_4&=\sin u+2\cos u-\frac{1}{2}\sin(2u)-\cos(2u),\\ Q_5&=-2\sin u+2\cos u-\sin(2u)+\frac{1}{2}\cos(2u). \end{align}\] Thus, each \(\rho_r\) is smooth, positive and periodic.
Set now \[F(x_1,x_2,u,v) = \operatorname{diag}\bigl(R_{10}(x_1,x_2),I_4\bigr) \begin{pmatrix} b(u,v)\\ c(u,v) \end{pmatrix}.\]
For \(i=1,2\), set \[X_i=\operatorname{diag}(w_{1i}J,w_{2i}J,w_{3i}J,w_{4i}J,w_{5i}J).\] Then \[\frac{\partial R_{10}}{\partial x_i}=R_{10}X_i.\] The four columns supplied by the \(I_4\)-block are \[c_u,\qquad c_v,\qquad c_{uu},\qquad c_{vv}.\] They are linearly independent, since \[\det(c_u,c_v,c_{uu},c_{vv})=-1.\] Therefore, after expanding the full \(14\times14\) osculating determinant along the \(I_4\)-block, it is enough to check the \(10\times10\) determinant \[\Delta_b = \det\bigl( X_1b,\;X_2b,\; X_1^2b,\;X_1X_2b,\;X_2^2b,\; X_1b_u,\;X_1b_v,\; X_2b_u,\;X_2b_v,\; b_{uv} \bigr).\]
A direct block computation gives \[\Delta_b = 4784\,e^{2(Q_1+\cdots+Q_5)} P(q_1,\ldots,q_5),\] where, after setting \[t=\tan\frac{u}{2},\] one has \[P(q_1(u),\ldots,q_5(u)) = \frac{16N(t)}{(1+t^2)^6},\] with \[\begin{align} N(t)= {}& 148t^{12}-142t^{11}-385t^{10}+4821t^9+4271t^8\\ &-42986t^7+39802t^6+17416t^5-20370t^4\\ &-5368t^3+3111t^2+1203t+111. \end{align}\] A Sturm sequence computation gives \[V(-\infty)=6,\qquad V(+\infty)=6.\] Hence, \(N\) has no real roots. Since \[N(0)=111>0\] and the leading coefficient of \(N\) is \[148>0,\] we have \[N(t)>0 \qquad\text{for all }t\in\mathbb{R}.\] The point \(u=\pi\), corresponding to \(t=\infty\), is also covered by the positive leading coefficient. Therefore \[\Delta_b>0 \qquad\text{for all }(u,v)\in \mathbb{T}^2.\] Consequently the full osculating determinant of \(F\) is everywhere nonzero, and so \[F\in \operatorname{Free}^{\infty}(T^4,\mathbb{R}^{14}).\]
All our attempts at finding free maps on \(\mathbb{T}^6\) and \(\mathbb{T}^7\) with this extended method failed so far. If the reason is that the set of free maps for \(m>5\) is “very small”, a systematic numerical exploration might be able to find them.
Given a manifold \(M\) of dimension \(m\), a smooth map \(F:M\to\mathbb{R}^q\) is \(k\)-free if, in local coordinates, the partial derivatives of \(F\) up to order \(k\) are linearly independent at every point. In particular, 2-free maps are simply free maps. The critical dimension for the existence of \(k\)-free maps on \(M\), namely the number of partial derivative of order from 1 up to \(k\), is \[\text{\boldsymbol{\boldsymbol{q_{m,k}}}} = \sum_{j=1}^{k} \binom{m+j-1}{j} = \binom{m+k}{k}-1.\]
In [7], S.I. Rodriguez Costa showed that there exist \(k\)-free maps on \(\mathbb{S}^1\) in critical dimension for every \(k\). Here, we show that our method allows to find \(k\)-free maps on \(\mathbb{T}^2\) in critical dimension.
We discuss first the case \(k=3\). The critical dimension for a 3-free map on \(\mathbb{T}^2\) is \[q_{2,3}=\binom{5}{2}-1 = 9.\]
We look for a map \(\mathbb{T}^2\to\mathbb{R}^9\) that, projected on the first five components, coincides with the free map \(\mathbb{T}^2\to\mathbb{R}^5\) illustrated in Section 2.1. Define \[R_3(x)=\operatorname{diag}\bigl(A(x),A(2x),1,A(3x),A(4x)\bigr)\in SO(9).\] Equivalently, if \[X=\operatorname{diag}(J,2J,0,3J,4J),\] then \[R'_3(x)=R_3(x)X.\]
Set \[F_3:T^2\longrightarrow \mathbb{R}^9, \qquad F(x,y)=R_3(x)v_3(y).\]
Since \(R_3(x)\in SO(9)\), the common factor \(R_3(x)\) can be removed from the osculating determinant. The relevant determinant is therefore \[D(y)= \det\bigl( Xv_3,\;v'_3,\;X^2v_3,\;Xv'_3,\;v''_3,\;X^3v_3,\;X^2v'_3,\;Xv''_3,\;v'''_3 \bigr).\] Indeed, \[\partial_xF_3=R_3Xv_3,\qquad \partial_yF_3=R_3v'_3,\] \[\partial_{xx}F_3=R_3X^2v_3,\qquad \partial_{xy}F_3=R_3Xv'_3,\qquad \partial_{yy}F_3=R_3v''_3,\] and \[\partial_{xxx}F_3=R_3X^3v_3,\qquad \partial_{xxy}F_3=R_3X^2v'_3,\qquad \partial_{xyy}F_3=R_3Xv''_3,\qquad \partial_{yyy}F_3=R_3v'''_3.\] Thus \[\det\bigl(\partial_xF,\partial_yF,\partial_{xx}F,\partial_{xy}F,\partial_{yy}F, \partial_{xxx}F,\partial_{xxy}F,\partial_{xyy}F,\partial_{yyy}F\bigr)=D(y).\]
A computer-aided search led to the expression \[v_3(y)= \begin{pmatrix} 2\\ 1+\sin y\\ \cos y\\ \frac{1}{2}+\sin y\\ \cos y\\ \sin(2y)\\ \cos(2y)\\ \sin(2y)\\ 7-2\sin(2y) \end{pmatrix}.\]
We now verify that \(D(y)\) is nowhere zero. Put \[t=\tan\frac{y}{2}.\] Using \[\sin y=\frac{2t}{1+t^2}, \qquad \cos y=\frac{1-t^2}{1+t^2},\] one obtains \[D(y(t))=\frac{288P(t)}{(1+t^2)^8},\] where \[\begin{align} P(t)= {}& 6372t^{16}+17377t^{15}+165036t^{14}+378109t^{13}\\ &+1389496t^{12}+1516089t^{11}+5179932t^{10} +402061t^9\\ &+7821744t^8-1544773t^7+6119524t^6-1243329t^5\\ &+2688648t^4-602325t^3+445140t^2-7113t+3372. \end{align}\] A Sturm sequence computation gives \[V(-\infty)=8,\qquad V(+\infty)=8.\] Therefore \(P\) has no real roots. Since \[P(0)=3372>0\] and the leading coefficient of \(P\) is \[6372>0,\] we get \[P(t)>0 \qquad\text{for all }t\in\mathbb{R}.\] The point \(y=\pi\), corresponding to \(t=\infty\), is covered by the positive leading coefficient. Hence \[D(y)>0 \qquad\text{for all }y\in S^1.\]
Thus, the nine vectors \[\partial_xF,\;\partial_yF,\;\partial_{xx}F,\;\partial_{xy}F,\;\partial_{yy}F, \partial_{xxx}F,\;\partial_{xxy}F,\;\partial_{xyy}F,\;\partial_{yyy}F\] are linearly independent everywhere, i.e. \(F\) is a 3-free immersion of \(\mathbb{T}^2\) into \(\mathbb{R}^9\).
We now consider the cases \(k=4,5,6\). We choose the following versions for the rotation matrices: \[\begin{align} R_4(x)&=\operatorname{diag}\bigl(A(x),\dots,A(7x)\bigr)\in SO(14)\\ R_5(x)&=\operatorname{diag}\bigl(A(x),\dots,A(10x)\bigr)\in SO(20)\\ R_6(x)&=\operatorname{diag}\bigl(A(x),\dots,A(13x),1\bigr)\in SO(27). \end{align}\] Let \(v_6(y)\) a vector such that \(R_6(x)v_6(y)\) is a \(6\)-free map. With the choices above, the projection to the first 14 components (resp. 20 components) of \(v_6(y)\) is such that \(R_4(x)v_4(y)\) (resp. \(R_5(x)v_5(y)\)) is a 4-free map (resp. 5-free map). Hence, here we present only the results regarding \(k=6\).
A viable loop \(v_6(y)\) is the following: \[v_6(y)= \begin{pmatrix} \cos y\\ \sin y\\ \cos y\\ \sin y\\ \cos y\\ \sin y\\ \cos 2y\\ \sin 2y\\ \cos 2y\\ \sin 2y\\ 71\cos y\\ -42\\ -34\cos y\\ 58+42\cos y\\ \cos y\\ \sin y\\ -\sin 3y\\ \cos 2y-\cos3y\\ 5\sin y-2\cos2y\\ -7+2\sin2y\\ \cos y\\ \sin y\\ \cos y\\ \sin y\\ \alpha(y)\\ \beta(y)\\ \gamma(y) \end{pmatrix},\] where \[\begin{align} \alpha(y)=&\;8035\cos y+341\sin2y-108\cos2y-317\cos3y-17\sin4y\\ &-16\cos4y-15\sin5y+5\cos5y,\\ \beta(y)=&\;2850+4887\cos y+33\sin2y-1244\cos2y+314\sin3y\\ &-11\cos3y-11\sin4y-16\cos4y-4\sin5y-15\cos5y,\\ \gamma(y)=&\;-349\cos3y+85\sin4y+177\cos4y+30\sin5y-5\cos5y. \end{align}\] Let \[X=\operatorname{diag}(J,2J,\dots,13J,0).\] The reader can verify with symbolic computing that \[D_6(y)=\det(X^pv_6^{(q)}(y))_{1\leq p+q\leq6} = \frac{P(t)}{(1+t^2)^{23}},\] where \(P\in\mathbb{Z}[t]\) is a polynomial of degree 46. Via Sturm’s algorithm, it can be shown that this polynomial has no zeros.
Based on the results in [7] and in this article, we pose the following conjecture:
for every \(k=2,3,...\), there exist smooth \(k\)-free maps \(\mathbb{T}^2\to\mathbb{R}^{q_{2,k}}\).
The main result of this section is that, given two compact closed smooth surfaces \(M\) and \(N\), the existence of smooth free maps in critical dimension on \(M\) and \(N\) implies the existence of free maps in critical dimension on their connected sum \(M\# N\). Due to the classification theorem for 2-manifolds, this will, in turn, allow us to show that free maps arise on every compact closed surface.
It is a standard result that every closed compact orientable (resp. non-orientable) surface is the connected sum of some finite number of copies of \(\mathbb{T}^2\) (resp. \(\mathbb{R}P^2\)). Hence, in order to prove the existence of smooth free maps in any closed compact surface, it is enough to show that one can “glue together” nicely enough any two free maps on any given pair of surfaces.
We shall use the following linear variant of the standard quadratic free map: \[\widetilde{P}:\mathbb{R}^2\to\mathbb{R}^5, \qquad \text{\boldsymbol{\boldsymbol{\widetilde{P}(x,y)}}} = \left( x,\;y,\;\frac{x^2-y^2}{2},\;xy,\;\frac{x^2+y^2}{2} \right).\] This differs from \[P(x,y)=\left(x,y,\frac{x^2}{2},xy,\frac{y^2}{2}\right)\] by an invertible linear change in the last three target coordinates. Hence \(\widetilde{P}\) is free. In polar coordinates, \[x=r\cos\theta,\qquad y=r\sin\theta,\] we write \[\text{\boldsymbol{\boldsymbol{\widetilde{W}(r,\theta)}}} = \widetilde{P}(r\cos\theta,r\sin\theta).\] Explicitly, \[\widetilde{W}(r,\theta) = \left( r\cos\theta,\, r\sin\theta,\, \frac{r^2}{2}\cos2\theta,\, \frac{r^2}{2}\sin2\theta,\, \frac{r^2}{2} \right).\]
We also introduce the signed-cylinder model \[\text{\boldsymbol{\boldsymbol{C(u,\theta)}}} = \left( \cos\theta,\, \sin\theta,\, u\cos2\theta,\, u\sin2\theta,\, -\frac{u^2}{2} \right).\] A direct computation gives \[\det\mathcal{D}C = 2.\] Thus \(C\) is free on a full signed cylinder \[(-\epsilon,\epsilon)\times S^1.\]
For \(\lambda>0\), define the anisotropic target dilation \[D_\lambda(X_1,X_2,X_3,X_4,X_5) = (\lambda X_1,\lambda X_2,\lambda^2X_3,\lambda^2X_4,\lambda^2X_5).\] We define the scaled signed-cylinder collar as \[\text{\boldsymbol{\boldsymbol{C_\lambda(u,\theta)}}}=D_\lambda C(u,\theta),\] that is, \[C_\lambda(u,\theta) = \left( \lambda\cos\theta,\, \lambda\sin\theta,\, \lambda^2u\cos2\theta,\, \lambda^2u\sin2\theta,\, -\frac{\lambda^2u^2}{2} \right).\]
The idea of the proof is the following. In Lemma 1 we show that there is a free map on the cylinder that is equal to \(C(u,\theta)\) at one end of the cylinder and to \(W(2u,\theta)\) at the other end. Then, in Lemma 2, we show that, given any free map \(f\) on a surface \(M\), we can modify \(f\) so that, near some disc, after replacing the disc with a cylinder collar, the modified free map coincides with \(C_\lambda(u,\theta)\) for some \(\lambda>0\). At this point, in Theorem 1 we show how to glue two free maps on two surfaces \(M\) and \(N\) so to obtain a free map on \(M\#N\).
Lemma 1 (Interpolation from the polar collar to the signed cylinder). Let \(u\in[0,1]\) and \(\theta\in\mathbb{S}^1\). There exists a smooth free map \[A:[0,1]\times S^1\to\mathbb{R}^5\] such that, near \(u=0\), \[A(u,\theta)=C(u,\theta),\] and, near \(u=1\), \[A(u,\theta)=\widetilde{W}(2u,\theta).\] Consequently, for every \(\lambda>0\), the map \[A_\lambda:=D_\lambda\circ A\] is a free interpolation between \(C_\lambda\) near \(u=0\) and \[\widetilde{W}(2\lambda u,\theta)\] near \(u=1\).
Proof. Consider maps of the form \[F(u,\theta) = \left( a(u)\cos\theta,\, a(u)\sin\theta,\, b(u)\cos2\theta,\, b(u)\sin2\theta,\, c(u) \right).\] Set \[b(u)=u\,a(u).\] Then a direct calculation gives \[\det\mathcal{D}F = -2a(u)^3H(u),\] where \[H(u) = \bigl(a-3ua'\bigr)c'' + \bigl(3ua''-2a'\bigr)c'.\]
Choose \[a(u)= \frac{ 508u^9+2909u^8+2292u^7+986u^6 -53420u^5+76121u^4-28396u^3+1000 }{1000},\] and \[c(u)= \frac{ u^2\bigl( 7214u^7+3187u^6+1311u^5-291u^4 -144573u^3+223056u^2-87404u-500 \bigr) }{1000}.\] Then \[\begin{align} &a(0)=1,\;\; \;\;a'(0)=0,\;\; \;\;a''(0)=\phantom{-}0,\\ &b(0)=0,\;\;\;\; b'(0)=1,\;\; \;\;b''(0)=\phantom{-}0,\\ &c(0)=0,\;\;\;\; c'(0)=0,\;\;\;\; c''(0)=-1, \end{align}\] Notice that \[C(u,\theta) = \left( \alpha(u)\cos\theta,\, \alpha(u)\sin\theta,\, \beta(u)\cos2\theta,\, \beta(u)\sin2\theta,\, \gamma(u) \right)\] for \[\alpha(u)=1,\;\;\beta(u)=u,\;\;\gamma(u)=-u^2/2.\] Since the 2-jets of \(\alpha,\beta,\gamma\) at \(u=0\) coincide with the 2-jets of, respectively, \(a,b,c\) at the same point, it follows that \(j^2F=j^2C\) along \(\{0\}\times\mathbb{S}^1\), where the 2-jets are taken with respect to the variables \((u,\theta)\).
At the other endpoint, \[a(1)=2,\qquad a'(1)=2,\qquad a''(1)=0,\] \[b(1)=2,\qquad b'(1)=4,\qquad b''(1)=4,\] \[c(1)=2,\qquad c'(1)=4,\qquad c''(1)=4.\] Now, notice that \[\widetilde{W}(2u,\theta) = \left( \hat{\alpha}(u)\cos\theta,\, \hat{\alpha}(u)\sin\theta,\, \hat{\beta}(u)\cos2\theta,\, \hat{\beta}(u)\sin2\theta,\, \hat{\gamma}(u) \right).\] for \[\hat{\alpha}(u) = 2u,\;\; \hat{\beta}(u) = 2u^2,\;\; \hat{\gamma}(u) = 2u^2.\] Since the 2-jets of \(\hat{\alpha},\hat{\beta},\hat{\gamma}\) at \(u=1\) coincide with the 2-jets of, respectively, \(a,b,c\) at the same point, it follows that \(j^2F=j^2W\) along \(\{1\}\times\mathbb{S}^1\), where the 2-jets are taken with respect to the variables \((u,\theta)\).
For the chosen \(a(u)\) and \(c(u)\), one can verify that \[H(u)=-\frac{K(u)}{500000},\] where \[\begin{align} K(u)={}& 164912040u^{16}+2918300000u^{15}+3807304560u^{14} +2517895353u^{13}\\ &-89566180787u^{12}+151601083986u^{11} -80205126406u^{10}-10857198663u^9\\ &+119308348533u^8-243323486472u^7 +220917024720u^6\\ &-93123128268u^5+15542937404u^4 +1388938000u^3\\ &-1338336000u^2+262212000u+500000. \end{align}\] By Sturm’s theorem, \[a(u)>0 \text{ and } K(u)>0\] for all \(u\in[0,1]\). Hence, \[\det\mathcal{D}F = \frac{a(u)^3K(u)}{250000}>0\] on \([0,1]\times S^1\).
At this stage, \(F\) agrees with the each endpoint model to order \(2\) at one of the two boundaries of the collar. It remains to make the agreement with the endpoint models exact in neighborhoods of the two boundary circles. Let \[h(u)=(a(u),b(u),c(u)).\] Let \[h_0(u)=\left(1,u,-\frac{u^2}{2}\right),\] and \[h_1(u)=\left(2u,2u^2,2u^2\right).\] By the endpoint conditions already verified, we have that \[h(u)-h_0(u)=O(u^3)\qquad \text{as }u\to 0,\] together with the corresponding estimates for the first and second derivatives, and similarly \[h(u)-h_1(u)=O((1-u)^3)\qquad \text{as }u\to 1,\] again with the corresponding first and second derivative estimates.
Now, choose a smooth function \(\rho:\mathbb{R}\to[0,1]\) such that \[\rho(s)=0 \quad \text{for } s\le 1, \qquad \rho(s)=1 \quad \text{for } s\ge 2.\] For \(0<\delta<1/4\), set \[\rho_0^\delta(u)=\rho\left(\frac{u}{\delta}\right), \qquad \rho_1^\delta(u)=\rho\left(\frac{1-u}{\delta}\right).\] First define \[h^{(0)}_\delta(u) = h_0(u)+\rho_0^\delta(u)\bigl(h(u)-h_0(u)\bigr).\] Thus \(h^{(0)}_\delta=h_0\) for \(0\le u\le \delta\), and \(h^{(0)}_\delta=h\) for \(u\ge 2\delta\). Then define \[\widetilde{h}_\delta(u) = h_1(u)+\rho_1^\delta(u) \bigl(h^{(0)}_\delta(u)-h_1(u)\bigr).\] Since the supports of the two transition regions are disjoint, this gives \[\widetilde{h}_\delta=h_0 \quad\text{for }0\le u\le \delta,\] \[\widetilde{h}_\delta=h \quad\text{for }2\delta\le u\le 1-2\delta,\] and \[\widetilde{h}_\delta=h_1 \quad\text{for }1-\delta\le u\le 1.\]
Moreover, \[\|\widetilde{h}_\delta-h\|_{C^2([0,1])}\longrightarrow 0 \qquad\text{as }\delta\to 0.\] Indeed, on the transition region near \(u=0\), the differences \(h-h_0\), \((h-h_0)'\), and \((h-h_0)''\) are respectively \(O(\delta^3)\), \(O(\delta^2)\), and \(O(\delta)\), while derivatives of \(\rho_0^\delta\) contribute at worst factors of \(\delta^{-1}\) and \(\delta^{-2}\). Hence the resulting \(C^2\)-error is \(O(\delta)\). The same argument applies near \(u=1\).
Replacing \(h=(a,b,c)\) by \(\widetilde{h}_\delta =(\widetilde{a}_\delta,\widetilde{b}_\delta,\widetilde{c}_\delta)\), and setting \[\widetilde{F}_\delta(u,\theta) = \bigl( \widetilde{a}_\delta(u)\cos\theta,\, \widetilde{a}_\delta(u)\sin\theta,\, \widetilde{b}_\delta(u)\cos 2\theta,\, \widetilde{b}_\delta(u)\sin 2\theta,\, \widetilde{c}_\delta(u) \bigr),\] we obtain maps \(\widetilde{F}_\delta\) converging to \(F\) in the \(C^2\)-topology as \(\delta\to0\). Since \(\det\mathcal{D}F\) has a positive minimum on the compact cylinder \([0,1]\times S^1\), freeness is preserved for all sufficiently small \(\delta\). Choose such a \(\delta\) and set \(A=F_\delta\). Then \(A\) is free, agrees exactly with the signed-cylinder model \(C\) near \(u=0\), and agrees exactly with the polar model \(W(2u,\theta)\) near \(u=1\). ◻
Lemma 2 (Normalization to the signed-cylinder collar). Let \(M\) be a smooth surface, let \(p\in M\), and let \[f\in\operatorname{Free}^\infty(M,\mathbb{R}^5).\] Then, for every sufficiently small \(\lambda>0\), after possibly postcomposing \(f\) with an affine automorphism of \(\mathbb{R}^5\), there exists a disk \(D\subset M\) centered at \(p\) and a free map \[\widehat f:M\setminus\operatorname{int}D\to\mathbb{R}^5\] such that:
\(\widehat f\) agrees with \(f\), up to the chosen affine automorphism, outside an arbitrarily small neighborhood of \(D\);
near \(\partial D\), in collar coordinates \((u,\theta)\in[0,\epsilon)\times S^1\), \[\widehat f(u,\theta)=C_\lambda(u,\theta).\]
Proof. Choose local coordinates \((x,y)\) centered at \(p\). Since \(f\) is free, the matrix \[\mathcal{D}f(0)\] is invertible. Hence, after translating the target and applying an invertible linear map of \(\mathbb{R}^5\), we may assume that \[j^2_0f=j^2_0\widetilde{P}.\] Thus \[f(x,y)=\widetilde{P}(x,y)+R(x,y),\] where \[\partial_\alpha R(x,y)=O\bigl(|(x,y)|^{3-|\alpha|}\bigr), \qquad |\alpha|\le 2.\]
Fix \(\lambda>0\) sufficiently small. On a small annulus around the circle \[r=2\lambda\] we interpolate between \(f\) and \(\widetilde{P}\) by \[f_\lambda(x,y) = \widetilde{P}(x,y) + \chi(r/\lambda)\bigl(f(x,y)-\widetilde{P}(x,y)\bigr),\] where \(r=\sqrt{x^2+y^2}\), and \(\chi\) is a smooth cutoff which is \(0\) near \(r=2\lambda\) and \(1\) farther out.
Since \(f-\widetilde{P}=O(r^3)\) and the derivatives of the cutoff of order \(j\) are \(O(\lambda^{-j})\), one has \[\|j^2f_\lambda-j^2\widetilde{P}\|_{C^0}=O(\lambda)\] on the interpolation annulus. For \(\lambda\) sufficiently small, \(f_\lambda\) is therefore free, since \(\widetilde{P}\) is free and freeness is an open condition on \(2\)-jets.
Thus, after this first modification, the map equals the polar standard model \[\widetilde{W}(r,\theta) = \widetilde{P}(r\cos\theta,r\sin\theta)\] near the circle \(r=2\lambda\).
Now attach the annular interpolation \(A_\lambda\) from the previous lemma. Near its outer end, it equals \[\widetilde{W}(2\lambda u,\theta),\] so near \(u=1\) it agrees with the polar standard model near \(r=2\lambda\). Near its inner end, it equals the signed-cylinder model \[C_\lambda(u,\theta).\]
Replacing the corresponding punctured neighborhood by this annular collar gives a free map on \(M\setminus\operatorname{int}D\), equal to the original map outside a small neighborhood of \(p\), and equal to \(C_\lambda\) near \(\partial D\). ◻
Theorem 1 (Connected-sum stability in minimal dimension for surfaces). Let \(M\) and \(N\) be smooth compact connected surfaces. If \[\operatorname{Free}^\infty(M,\mathbb{R}^5)\neq\varnothing \qquad\text{and}\qquad \operatorname{Free}^\infty(N,\mathbb{R}^5)\neq\varnothing,\] then \[\operatorname{Free}^\infty(M\#N,\mathbb{R}^5)\neq\varnothing.\]
Proof. Choose \[f\in\operatorname{Free}^\infty(M,\mathbb{R}^5), \qquad g\in\operatorname{Free}^\infty(N,\mathbb{R}^5).\] Choose points \(p\in M\), \(q\in N\). Applying the signed-cylinder normalization lemma to \(f\) and \(g\), with the same value of \(\lambda>0\), we obtain disks \[D_M\subset M, \qquad D_N\subset N,\] and free maps \[\widehat f:M\setminus\operatorname{int}D_M\to\mathbb{R}^5, \qquad \widehat g:N\setminus\operatorname{int}D_N\to\mathbb{R}^5,\] such that near the two boundary circles \[\widehat f(u,\theta)=C_\lambda(u,\theta),\] and \[\widehat g(u,\theta)=C_\lambda(u,\theta),\] where \(u\ge 0\) is the inward collar coordinate on each punctured surface.
Let \[S=\operatorname{diag}(1,1,-1,-1,1).\] Then \[\det S=1,\] so postcomposition by \(S\) preserves freeness. Moreover \[S C_\lambda(u,\theta)=C_\lambda(-u,\theta).\] Replace \(\widehat g\) by \(S\widehat g\). Then near \(\partial D_N\), \[S\widehat g(u,\theta)=C_\lambda(-u,\theta).\]
Now form the connected sum \(M\#N\) by identifying the two boundary circles using the chosen angular parameter \(\theta\). Choosing the boundary angular parameters appropriately realizes the desired connected-sum identification.
Define a map \[h:M\#N\to\mathbb{R}^5\] by \[h= \begin{cases} \widehat f & \text{on }M\setminus\operatorname{int}D_M,\\ S\widehat g & \text{on }N\setminus\operatorname{int}D_N. \end{cases}\]
Near the seam, introduce a signed collar coordinate \(\tau\) by \[\tau=u \quad\text{on the \(M\)-side}, \qquad \tau=-u \quad\text{on the \(N\)-side}.\] Then on both sides of the seam, \[h(\tau,\theta)=C_\lambda(\tau,\theta).\] Therefore \(h\) is smooth across the seam. It is free near the seam because \(C_\lambda\) is free, and it is free away from the seam because it agrees locally with either \(\widehat f\) or \(S\widehat g\), both of which are free. Hence, \[h\in\operatorname{Free}^\infty(M\#N,\mathbb{R}^5).\] ◻
We are now able to prove our main result.
Theorem 2. Let \(M\) be a compact closed surface. Then \(\mathrm{Free}^\infty(M,\mathbb{R}^5) \neq \emptyset\).
Proof. Assume first that \(M\) is connected. By the classification theorem for closed surfaces, \(M\) is diffeomorphic to exactly one of the following: \[\mathbb{S}^2,\qquad \#_g T^2\quad (g\ge 1),\qquad \#_k \mathbb{R}\mathrm{P}^2\quad (k\ge 1).\]
We have:
\(\operatorname{Free}^\infty(\mathbb{S}^2,\mathbb{R}^5)\ne\varnothing\), by the Veronese map;
\(\operatorname{Free}^\infty(T^2,\mathbb{R}^5)\ne\varnothing\), by the explicit map above;
\(\operatorname{Free}^\infty(\mathbb{R}\mathrm{P}^2,\mathbb{R}^5)\ne\varnothing\), by the Veronese map.
Iterated application of Theorem 1 gives the result for connected \(M\). If \(M\) is disconnected, apply the connected case to each connected component. ◻
Corollary 1. Let \(M\) be a compact closed surface. Then there is a free smooth embedding of \(M\) into \(\mathbb{R}^5\).
Proof. By Theorem 2, the open set \(\operatorname{Free}^\infty(M,\mathbb{R}^5)\) is nonempty. Since freeness is a \(C^2\)-open condition and \(M\) is compact, \(\operatorname{Free}^\infty(M,\mathbb{R}^5)\) is open in the \(C^\infty\)-topology.
By Whitney approximation and transversality, since \(5=2\cdot 2+1,\) embeddings \(M\to\mathbb{R}^5\) are \(C^\infty\)-dense in \(C^\infty(M,\mathbb{R}^5)\). Hence any free map \(M\to\mathbb{R}^5\) admits a sufficiently small smooth perturbation which is still free and is an embedding. ◻
The author warmly thanks M. Gromov and Y. Eliashberg for their valuable comments on earlier versions of the manuscript. The author was partially supported by NSF grant # 2308225.
Let \(p\in\mathbb{Q}[t]\) be a nonzero polynomial. Its Sturm sequence is the finite sequence \[S_0=p,\qquad S_1=p', \qquad S_{i+1}=-\operatorname{rem}(S_{i-1},S_i),\] where \(\operatorname{rem}(S_{i-1},S_i)\) denotes the Euclidean remainder of \(S_{i-1}\) upon division by \(S_i\). The construction stops when the remainder is zero. For \(a\in\mathbb{R}\cup\{\pm\infty\}\), let \(V(a)\) be the number of sign changes in the list \[S_0(a),S_1(a),\dots,S_N(a),\] after omitting zero entries. Sturm’s theorem (e.g. see [8]) states that the number of distinct real roots of \(p\) in an interval \((a,b)\), assuming \(a\) and \(b\) are not roots of \(p\), is \[V(a)-V(b).\] In particular, the total number of real roots of \(p\) is \[V(-\infty)-V(+\infty).\] Thus, if \(V(-\infty)=V(+\infty)\), then \(p\) has no real roots. If in addition \(p(t_0)>0\) for some \(t_0\in\mathbb{R}\), then \(p(t)>0\) for every \(t\in\mathbb{R}\); similarly, if \(p(t_0)<0\), then \(p(t)<0\) everywhere. All computations used below are carried out exactly in \(\mathbb{Q}[t]\).
In Section 2.3, after the substitution \[t=\tan\frac{z}{2},\] the freeness determinant takes the form \[D(z(t))=\frac{q(t)}{(1+t^2)^{11}},\] where \[\begin{align} q(t)= {}& -60010t^{22}-561400t^{21}-3039880t^{20}-337260t^{19} +18967810t^{18} \\ &+11879260t^{17}-23665760t^{16} -17732160t^{15}-42514660t^{14}-536720t^{13} \\ &+22966960t^{12}+23893640t^{11}+32807300t^{10} +18046840t^9-9136320t^8 \\ &-11620000t^7-19941970t^6-7234520t^5 +1406360t^4+511940t^3 \\ &-71910t^2-7540t-160 . \end{align}\] Since \((1+t^2)^{11}>0\), it is enough to prove that \(q(t)\) does not vanish on \(\mathbb{R}\).
We use the Sturm sequence \[S_0=q,\qquad S_1=q', \qquad S_{i+1}=-\operatorname{rem}(S_{i-1},S_i),\] computed in \(\mathbb{Q}[t]\). The sequence has \(23\) terms, with degrees \[22,21,20,\ldots,1,0.\] The signs of the leading terms give the following signs at \(-\infty\) and \(+\infty\):
\[\begin{array}{c|c|c|c} i & \deg S_i & \operatorname{sign}S_i(-\infty) & \operatorname{sign}S_i(+\infty)\\ \hline 0 & 22 & - & -\\ 1 & 21 & + & -\\ 2 & 20 & + & +\\ 3 & 19 & - & +\\ 4 & 18 & + & +\\ 5 & 17 & - & +\\ 6 & 16 & - & -\\ 7 & 15 & - & +\\ 8 & 14 & - & -\\ 9 & 13 & + & -\\ 10 & 12 & - & -\\ 11 & 11 & - & +\\ 12 & 10 & + & +\\ 13 & 9 & + & -\\ 14 & 8 & - & -\\ 15 & 7 & - & +\\ 16 & 6 & + & +\\ 17 & 5 & + & -\\ 18 & 4 & - & -\\ 19 & 3 & + & -\\ 20 & 2 & + & +\\ 21 & 1 & + & -\\ 22 & 0 & + & + \end{array}\]
Therefore the number of sign changes is \[V(-\infty)=11, \qquad V(+\infty)=11.\] By Sturm’s theorem, \[\#\{t\in\mathbb{R}:q(t)=0\} = V(-\infty)-V(+\infty) = 0.\] Thus \(q\) has no real roots.
Finally, \[q(0)=-160<0.\] Since \(q\) has no real roots, its sign is constant on \(\mathbb{R}\), and therefore \[q(t)<0 \qquad \forall t\in\mathbb{R}.\] The point \(z=\pi\), corresponding to \(t=\infty\), is controlled by the leading coefficient: \[-60010<0.\] Hence \[D(z)<0 \qquad \forall z\in \mathbb{S}^1.\] This proves that \(\det\mathcal{D}F_4=D(z)\) is nowhere zero.
In the \(\mathbb{T}^5\) construction, after setting \[t=\tan\frac{z}{2},\] we have \[\sin z=\frac{2t}{1+t^2}, \qquad \cos z=\frac{1-t^2}{1+t^2}.\] The determinant \(D(z)\) can therefore be written as a rational function of \(t\).
The entries of \(\mathcal{D}F_5\) have denominators bounded by powers of \(1+t^2\). Multiplying each row by \((1+t^2)^2\), computing the determinant, and then cancelling the harmless positive factor \((1+t^2)^{25}\), one obtains \[D(z(t))=\frac{p(t)}{(1+t^2)^{15}},\] where \[\begin{align} p(t)={}& 77499567520t^{30} +393071545000t^{29} +624105453260t^{28} -2123936935740t^{27} \\ &-1759943802760t^{26} +2441233882140t^{25} -2897450936620t^{24} -9631185563480t^{23} \\ &+45848535094560t^{22} -49554696590640t^{21} -2415739796580t^{20} +52612311592380t^{19} \\ &+111907002620280t^{18} -437235030859820t^{17} +509646314333860t^{16} -506960386499280t^{15} \\ &+429448025715360t^{14} +65367491720040t^{13} -290352458559260t^{12} +311545262726140t^{11} \\ &-230021273661720t^{10} +25849442057540t^9 +67010870215740t^8 -63725239598680t^7 \\ &+30209374034080t^6 -4416239459680t^5 +651580051380t^4 -436319537660t^3 \\ &+16504531240t^2 +11135564620t +626467660 . \end{align}\] Since \((1+t^2)^{15}>0\), it is enough to prove that \(p(t)>0\) for all \(t\in\mathbb{R}\).
We use the Sturm sequence \[S_0=p,\qquad S_1=p', \qquad S_{i+1}=-\operatorname{rem}(S_{i-1},S_i),\] computed exactly in \(\mathbb{Q}[t]\). The sequence has \(31\) terms, with degrees \[30,29,28,\ldots,1,0.\] The signs at \(-\infty\) and \(+\infty\) are as follows: \[\begin{array}{c|c|c|c} i & \deg S_i & \operatorname{sign}S_i(-\infty) & \operatorname{sign}S_i(+\infty)\\ \hline 0 & 30 & + & +\\ 1 & 29 & - & +\\ 2 & 28 & + & +\\ 3 & 27 & + & -\\ 4 & 26 & - & -\\ 5 & 25 & + & -\\ 6 & 24 & + & +\\ 7 & 23 & + & -\\ 8 & 22 & + & +\\ 9 & 21 & - & +\\ 10 & 20 & + & +\\ 11 & 19 & - & +\\ 12 & 18 & - & -\\ 13 & 17 & - & +\\ 14 & 16 & + & +\\ 15 & 15 & + & -\\ 16 & 14 & - & -\\ 17 & 13 & - & +\\ 18 & 12 & + & +\\ 19 & 11 & + & -\\ 20 & 10 & - & -\\ 21 & 9 & - & +\\ 22 & 8 & + & +\\ 23 & 7 & + & -\\ 24 & 6 & - & -\\ 25 & 5 & - & +\\ 26 & 4 & - & -\\ 27 & 3 & + & -\\ 28 & 2 & - & -\\ 29 & 1 & - & +\\ 30 & 0 & - & - \end{array}\] Thus \[V(-\infty)=15, \qquad V(+\infty)=15.\] By Sturm’s theorem, \[\#\{t\in\mathbb{R}:p(t)=0\} = V(-\infty)-V(+\infty) = 0.\] Therefore \(p\) has no real roots.
Moreover, \[p(0)=626467660>0.\] Since \(p\) has no real roots, it follows that \[p(t)>0 \qquad \forall t\in\mathbb{R}.\] The point \(z=\pi\), corresponding to \(t=\infty\), is controlled by the leading coefficient: \[77499567520>0.\] Hence \[D(z)>0 \qquad \forall z\in \mathbb{S}^1.\] This proves that \(\det\mathcal{D}F_5=D(z)\) is nowhere zero.