A Table Theorem for Surfaces with Odd Euler Characteristic

Consider a pair $(h,\theta)$ in $C_{+}^{2}(U_{a}(\Sigma))\times K_{a}(\Sigma)$ and let $\xi$ be its image under $\Psi$. Since $\Psi_{h}$ covers the identity map on $\Sigma$, if $\mathcal{H}_{\theta}$ is a horizontal subspace of $T_{\theta}K_{a}(\Sigma)$, then $d\Psi_{h}(\mathcal{H}_{\theta})$ is also a horizontal subspace of $T_{\xi}Q(\Sigma)$. Therefore, we only need to check that $\Psi$ is a submersion when we restrict the map to a fiber over an arbitrary point $x$. Let $\theta_{i}$ denote the components of $\theta$. Fix all the $\theta_{i}^{\prime}s$ except $\theta_{1}$ and change $\theta_{1}$ along a curve $\delta(t)$ in $U_{a}(\Sigma)$ such that the curve $\gamma(t)=[\delta(t),\theta_{2},\theta_{3},\theta_{4}]$ avoids the diagonal in $sym^{4}(T\Sigma)$. We have

$$\Psi(h,\gamma(t))=[h(\delta(t))\delta(t),h(\theta_{2})\theta_{2},h(\theta_{3})\theta_{3},h(\theta_{4})\theta_{4}].$$

Hence, we get

$$d\Psi(0,\dot{\gamma}(0))=[h(\theta_{1})\cdot\dot{\delta}(0)+\frac{d(h(\delta(t)))}{dt}\Big{|}_{t=0}\cdot\theta_{1},0,0,0]\in\text{Im}(d\Psi),$$

where we used $\delta(0)=\theta_{1}$. Since $\delta(t)$ has constant length, $\dot{\delta}(0)$ is a non-zero vector orthogonal to $\theta_{1}$. Repeating this argument for the other coordinates proves we have vectors of the previous form in the image of $d\Psi$ where all the components are zero except one and the non-zero component is of the form $r_{i}\cdot\omega_{i}+s_{i}\cdot\theta_{i}$ with $r_{i}=h(\theta_{i})$ greater than zero, $\omega_{i}$ a vector orthogonal to $\theta_{i}$, and $s_{i}$ some arbitrary real number. Now fix the four-tuple $\theta$ and pick a function $g$ such that $g(\theta_{1})=1$ and $g$ vanishes on the other three coordinates. Let $g_{t}=h+tg$ and note that for $t$ small enough all the functions $g_{t}$ are positive. We get

$$d\Psi(g,0)=[g(\theta_{1})\cdot\theta_{1},g(\theta_{2})\cdot\theta_{2},g(\theta_{3})\cdot\theta_{3},g(\theta_{4})\cdot\theta_{4}]=[\theta_{1},0,0,0]\in\text{Im}(d\Psi).$$

The same argument shows we have vectors of the form $[0,\dots,\theta_{i},\dots,0]$ in the image of $d\Psi$. We conclude the lemma because the set of vectors

	$\displaystyle[0,\dots,\theta_{i},\dots,0],$
	$\displaystyle[0,\dots,\omega_{i},\dots,0]$

generate all the vertical vectors over $x$ in $Q(\Sigma)$. ∎

This follows from the fact that $\Psi$ is a submersion and $A(\Sigma)$ is a codimension $4$ submanifold. ∎

Fix a point $(h,\theta)$ in $\mathcal{S}$ and consider $dF\colon T_{(h,\theta)}\mathcal{S}\to T_{h}\mathcal{F}$. We need to show $\ker(dF)$ and $\text{coker}(dF)$ are finite dimensional. Let $pr_{1}$ denote the first projection map on $\mathcal{N}$ and note that $\ker(dpr_{1})$ at $(h,\theta)$ is $T_{\theta}K_{a}(\Sigma)$ which has dimension $6$; we conclude that $\dim(\ker(dF))$ is finite. Let $W$ denote $T_{(h,\theta)}\mathcal{N}$ and $V$ denote $T_{(h,\theta)}\mathcal{S}$. We define a map $L\colon W/V\to\text{coker}(dF)$ by

$$[w]\mapsto[dpr_{1}(w)].$$

This map is surjective because $pr_{1}$ is a submersion and we know $\dim(W/V)=\text{codim}(\mathcal{S})=4$. Hence, $\dim(\text{coker}(dF))$ is finite. ∎

Since the square corresponding to $(h_{1},\theta_{1})$ is graceful, if we consider this square in the fiber of $T\Sigma$ over $\pi(\theta_{1})$, the origin cannot lie in the regions $A,B,C,$ and $D$ determined by this square in figure $1$.

Suppose $\gamma(s)=(h_{s},\theta_{s})$ is a path in $\mathcal{S}$ starting from $(h_{1},\theta_{1})$ and ending at $(h_{2},\theta_{2})$. By contradiction, assume the square corresponding to $(h_{2},\theta_{2})$ is not graceful. Hence, the origin will enter one of the regions $A,B,C,$ or $D$. Without loss of generality, let this region be $A$. Then there is a time $s_{0}$ in between the two ends where the origin lies on the line $l$ given in figure $2$.

Now the two vertices $v_{1}$ and $v_{2}$ of the square corresponding to $(h_{s_{0}},\theta_{s_{0}})$ lie on the same side of the origin in $l$ and this contradicts the fact that the curve inscribing this square is star-shaped (this follows from positivity of $h_{s_{0}}$). ∎

Consider a point $(g,\theta)$ in $\mathcal{S}_{0}$. This point corresponds to a graceful square that is inscribed in a star-shaped curve around the origin in some fiber $T_{x}\Sigma$. We can also consider the constant function $1$ and four vertices of a square $\delta_{x}$ on the circle with radius $a$ around the origin in $T_{x}\Sigma$. This gives us a point $(1,\delta_{x})$ in $\mathcal{S}_{0}$ and all the points of this form in $\mathcal{S}_{0}$ can be connected to each other by parallel transport and rotation in their corresponding fibers. Hence, it suffices to prove there is a path between $(g,\theta)$ and $(1,\delta_{x})$ in $\mathcal{S}_{0}$.

Since $(g,\theta)$ corresponds to a graceful square, the origin in $T_{x}\Sigma$ cannot lie in the four region defined in terms of this square given in figure $1$. For any point outside of these four regions, we can find a sufficiently large ellipse going through the four vertices of the square such that the point is inside the ellipse. Consider such an ellipse for the origin; see the figure below.

There is a positive function $h$ defined on the circle with radius $a$ around the origin so that the map

$$\eta\mapsto h(\eta)\cdot\eta$$

takes the circle into the ellipse; extend this function to a positive function on $U_{a}(\Sigma)$ and note that $g$ restricted to the circle of radius $a$ in $T_{x}\Sigma$ will give us the corresponding function for our original curve in $T_{x}\Sigma$. We define a path of positive functions by $h_{t}=th+(1-t)g$. Note that $h_{t}$ is constant for each $\theta_{i}$ in $\theta$ because we have fixed the square and only move the curves. Hence, $(h_{t},\theta)$ gives a path from $(g,\theta)$ to $(h,\theta)$. Now we can translate the ellipse to an ellipse centered around the origin. This gives us a continuous path of functions $g_{t}$ and tuples of points $\theta_{t}$ with $g_{0}=h$ and $\theta_{0}=\theta$ because each ellipse inscribes a unique square and all the ellipses in the translation contain the origin. Thus we get a path from $(g,\theta)$ to $(u,\sigma)$ where $\Psi(u,\sigma)$ is a square inscribed inside an ellipse centered around the origin in $T_{x}\Sigma$. Now we can take $(u,\sigma)$ to $(1,\delta_{x})$ by first a homotopy fixing the four vertices of $(u,\sigma)$ and then scaling and rotation.

∎

Suppose $v$ is a vector in the kernel of $dF$ over $(h,\theta)$. Then we can write $v$ as a pair $(0,\eta)$ where $\eta$ is a vector in $T_{\theta}K_{a}(\Sigma)$ since $F$ is restriction of the first projection map to $\mathcal{S}$. Now since $v$ lies in the tangent space of $\mathcal{S}$ over $(h,\theta)$, we have

$$d\Psi_{(h,\theta)}[(0,\eta)]=d\Psi_{h}[\eta]\in T_{\xi}A.$$

A similar argument shows if a vector $\delta$ is in $T_{\xi}\Psi_{h}(K_{a}(\Sigma))\cap T_{\xi}A$, then $\delta=d\Psi_{h}[\eta]$ for some vector $\eta$ in $T_{\theta}K_{a}(\Sigma)$ and $(0,\eta)$ lies in the kernel of $dF$. ∎

We can identify $T_{h}\mathcal{F}$ with the space of all $C^{2}$ functions on $U_{a}(\Sigma)$; let $g$ be an arbitrary function in $T_{h}\mathcal{F}$. Consider $v=(g,0)$ in $T_{(h,\theta)}\mathcal{N}$ and define $w=d\Psi_{(h,\theta)}[v]$ to be its image in $T_{\xi}Q_{a}(\Sigma)$. We can write $w$ as $w_{A}+w_{K}$ where $w_{A}$ is in $T_{\xi}A(\Sigma)$ and $w_{K}$ is in $T_{\xi}K_{a}(\Sigma)$ because we assumed $\xi$ is a transverse intersection point. There exists a vector $\eta$ in $T_{\theta}K_{a}(\Sigma)$ such that $w_{K}=d\Psi_{h}[\eta]$. Now consider the vector $\tilde{v}=(g,-\eta)$ in $T_{(h,\theta)}\mathcal{N}$; we get

$$d\Psi[\tilde{v}]=d\Psi[v]+d\Psi[(0,-\eta)]=w-w_{K}=w_{A}\in T_{\xi}A(\Sigma).$$

Hence, $\tilde{v}$ lies in $T_{(h,\theta)}\mathcal{S}$ and we have $dF[\tilde{v}]=g$. ∎

Let $x$ be a point on $\Sigma$. Assume $e_{1},e_{2}$ are an orthonormal basis for $T\Sigma$ in a disk $D$ around $x$ and let $v_{1}$ and $v_{2}$ denote the corresponding coordinates for $T\Sigma$ above $D$. Define a local fiber bundle of ellipses around $x$ by

$$E\coloneqq\{(y,v_{1},v_{2})|v_{1}^{2}+2\cdot v_{2}^{2}=1\}.$$

There is a positive function $h$ defined on $U_{a}(S^{2})$ above $D$ such that the map

$$\eta\mapsto h(\eta)\cdot\eta$$

takes $U_{a}(S^{2})$ for each point in $D$ to the ellipse above that point. Extend $h$ to a positive function $\tilde{h}$ in $\mathcal{F}$ and let $\theta$ be the four-tuple of points in $K_{a}(\Sigma)$ over $x$ that corresponds to the unique graceful square inscribed in $E_{x}$. Then $\xi=\Psi(\tilde{h},\theta)$ is a transverse intersection point of $\Psi_{\tilde{h}}(K_{a}(\Sigma))$ and $A(\Sigma)$ because around $x$ all the corresponding curves are ellipses and these meet the space of squares transversely; in fact, $\Psi_{\tilde{h}}(K_{a}(\Sigma))\cap A(\Sigma)$ is locally a two dimensional disk around $\xi$. Therefore, by Lemma 4.4, we know $\ker(dF)$ has dimension two at $(\tilde{h},\theta)$ and by Lemma 4.5, we know $dF$ is surjective at this point. We conclude that $F$ has index two since $\mathcal{S}_{0}$ is connected. ∎

Consider a function $h$ in $\mathcal{F}$. We know $\Psi_{h}(K_{a}(\Sigma))\cap A(\Sigma)$ is non-empty and at least on of these intersection points corresponds to a graceful square since each of the star-shaped curves corresponding to $h$ in fibers of $T\Sigma$ inscribes at least one graceful square. We prove this Theorem in the appendix (Theorem A.5); see also [12, 4] for other versions of this result. Hence, there is a $\theta\in K_{a}(\Sigma)$ such that $(h,\theta)$ lies in $\mathcal{S}_{0}$. ∎

Consider $(h,\theta)$ in $\mathcal{S}_{0}$ and $\xi=\Psi(h,\theta)$ in $A(\Sigma)$. Since $h$ is a regular value of $F$, the kernel of $dF$ at $(h,\theta)$ is two dimensional and this proves $T_{\xi}\Psi_{h}(K_{a}(\Sigma))\cap T_{\xi}A(\Sigma)$ is two dimensional by Lemma 4.4. Both $T_{\xi}\Psi_{h}(K_{a}(\Sigma))$ and $T_{\xi}A(\Sigma)$ are six dimensional subspaces of $T_{\xi}Q_{a}(\Sigma)$ which has dimension ten. We conclude the two subspaces meet transversely because their intersection has dimension two so the subset of graceful squares in $\Psi_{h}(K_{a}(\Sigma))\cap A(\Sigma)$ is a surface. ∎

Consider the regular values of $F$; by Proposition 4.8, we know the intersection is transverse at graceful squares for every element in this subset. The map $F$ is a surjective $C^{\infty}$ Fredholm map of index two between connected second countable Banach manifolds. Thus its regular values are dense in $\mathcal{F}$ by Sard-Smale Theorem; see [11] for more details. ∎

Let $\Sigma_{h}$ denote the subset of graceful squares in $\Psi_{h}(K_{a}(\Sigma))\cap A(\Sigma)$ and note that this is a surface by proposition 4.8. Now consider a sequence $a_{n}=\Psi_{h}(\theta_{n})$ in $\Sigma_{h}$ and since $\theta_{n}$ is in $K_{a}(\Sigma)$, we can assume $\theta_{n}$ converges to a point $\theta$ in $sym^{4}(T\Sigma)$ after passing to a subsequence. The point $\theta$ is not in the fat diagonal of $sym^{4}(T\Sigma)$ because the function $h$ is $C^{2}$ and we can uniformly bound the curvature of all the star-shaped curves corresponding to $h$; thus there is a positive lower bound for the side length of all the squares $a_{n}$ and $\Psi_{h}(\theta)$ is a non-degenerate square. The square $\Psi_{h}(\theta)$ is graceful since it is the limit of a sequence of graceful squares. ∎

Let $\pi$ denote the bundle projection map from $sym^{4}(T\Sigma)$ to $\Sigma$; this is a deformation retract onto $\Sigma$. Consider the restriction of $\pi$ to $\Sigma_{h}$; we will show this map has non-zero mod $2$ degree. Suppose $x\in\Sigma$ is a regular value of this map. Then for every $\theta$ in $\pi^{-1}(x)\cap\Sigma_{h}$, $T_{\theta}\Sigma_{h}$ is a horizontal subspace of $T_{\theta}sym^{4}(T\Sigma)$ by assumption. In particular, $T_{\theta}\Psi_{h}(K_{a}(\Sigma))$ and $T_{\theta}A(\Sigma)$ have no vertical intersection and this proves the manifold of squares in $T_{x}\Sigma$ and fourth symmetric product of the star-shaped curve corresponding to $h$ above $x$ meet transversely at every $\theta$ in $\pi^{-1}(x)\cap\Sigma_{h}$. Therefore, this curve has finitely many graceful squares and the mod $2$ degree of $\pi$ is equal to the number of graceful squares inscribed inside this curve mod $2$. It is proved in the appendix (corollary A.6) that a generic star-shaped curve has an odd number of graceful squares; this was originally proved in [12] for generic smooth curves and it was proved in [10] for generic PL curves. The curve corresponding to $h$ above $x$ is a generic one because of the transversal intersection and we conclude the proposition. ∎

By Proposition 5.2, we know $[\Sigma_{h}]$ is a non-zero homology class. Hence, $c_{*}[\Sigma_{h}]$ is also a non-zero homology class in $T\Sigma$. The mod $2$ intersection number of such homology classes with the zero section in $T\Sigma$ is equal to the second Stiefel–Whitney number of $\Sigma$ and this is equal to $\chi(\Sigma)$ mod $2$; see [7] for more details. We conclude that this intersection number is non-zero because we assumed $\Sigma$ has odd Euler characteristic. ∎

After reordering the four-tuple $\xi$, we can assume

$$\xi_{1}=-\xi_{3},\hskip 5.69054pt\xi_{2}=-\xi_{4},$$

and $\xi_{1}\cdot\xi_{2}=0$ since $\xi_{i}$’s are vertices of a square centered at the origin. Thus we get

$$\theta_{1}\cdot\theta_{2}=\frac{\xi_{1}\cdot\xi_{2}}{h(\theta_{1})h(\theta_{2})}=0.$$

Moreover, we have

$$h(\theta_{1})\theta_{1}=\xi_{1}=-\xi_{3}=-h(\theta_{3})\theta_{3}.$$

Hence, we can write

$$\theta_{3}=\frac{-h(\theta_{1})}{h(\theta_{3})}\theta_{1}.$$

Since $\theta_{1}$ and $\theta_{3}$ have the same length, we deduce that $h(\theta_{1})=h(\theta_{3})$ by positivity of $h$ and $\theta_{1}=-\theta_{3}$. A similar argument shows $h(\theta_{2})=h(\theta_{4})$ and $\theta_{2}=-\theta_{4}$. Therefore, $\theta_{i}$ ’s are vertices of a square on the circle with radius $a$ around the origin in $T_{x}\Sigma$; two vertices of this square are scaled by $h(\theta_{1})$ and the other two by $h(\theta_{2})$. Since the scaled shape is also a square by assumption, we conclude that $h(\theta_{1})=h(\theta_{2})$. ∎

Since $\Sigma$ is compact, it suffices to prove the Theorem for positive functions. Let $f$ be a positive continuous function on $\Sigma$ and consider $\tilde{f}$ on $U_{a}\Sigma$ for $a=\frac{d}{2}$. By Corollary 4.9, we can find a sequence of functions $u_{n}$ in regular values of $F\big{|}_{\mathcal{S}_{0}}$ such that $u_{n}$ converges to $\tilde{f}$ uniformly on $U_{a}\Sigma$. By Lemma 5.4 and Proposition 5.3, we know there is a sequence of graceful squares $\theta_{n}$ inscribed inside the fibers of $U_{a}\Sigma$ such that $u_{n}$ takes the same value on the four vertices of $\theta_{n}$ for every $n$. After passing to a subsequence, we can assume $\theta_{n}$ converges to $\theta$, a square with the same side length as $\theta_{n}$’s and inscribed inside the fiber of $U_{a}\Sigma$ over a point $x$ in $\Sigma$. All the vertices of $\theta$ take the same value under $\tilde{f}$ by uniform convergence and the assumption on $u_{n}$’s. Hence, we conclude $f$ admits a table determined by the four vertices of $\theta$. ∎

Let $f$ be an even function on $S^{2}$ and $g$ a Riemannian metric invariant under the antipodal map. This gives us a function $\bar{f}$ and a Riemannian metric $\bar{g}$ on $\mathbb{R}P^{2}$. Now apply Theorem 1.1 to this Riemannian surface and the function $\bar{f}$. The table for $\bar{f}$ on $\mathbb{R}P^{2}$ lifts to two tables for $f$ on $S^{2}$. ∎