How To Divide a Sandwich

Notations

In this article, we only consider continuous maps. So when we write "map", we mean "continuous map". And the coefficient field of all $H_i(X)$ and $C_i(X)$ is $\mathbb{F}_2$ .

Problem

We have a sandwich (of course it is in $\mathbb{R}^3$ ) that contains three parts (the $i$ -th part is a disk $D_i\times [i-1,i]$ , $D_i\subset \mathbb{R}^2$ , $i\in \{ 1,2,3 \}$ ), and we want to cut it into two parts by a plane. Can we make their volumes equal?

We first introduce a theorem:

Borsuk–Ulam Theorem

\begin{gathered} \forall g:S^n\to \mathbb{R}^n \\ \exists x_0\in S^n \ \text{s.t.}\ g(x_0)=g(-x_0) \end{gathered}

Let's see how to prove it.

Lemma 1

The theorem is equivalent to:

\begin{gathered} \not\exists f:S^n \to S^{n-1} \\ \ \text{s.t.}\ f(-x)=-f(x) \end{gathered}

\begin{gathered} \text{If } \exists g \ \text{s.t.}\ \forall x, g(x)\ne g(-x), \\ \text{then let } f(x)=\dfrac{g(x)-g(-x)}{\|g(x)-g(-x)\|} , \text{ which satisfies } f(-x)=-f(x). \\ \text{Conversely, if }\exists f:S^n\to S^{n-1} \text{ s.t. } f(-x)=-f(x), \\ \text{then } g=l\circ f \ \text{s.t.}\ g(x)=-g(-x)\ne g(-x), \\ \text{where } l:S^{n-1}\to \mathbb{R}^n \text{ maps } s\in S^{n-1} \text{ to its coordinate representation in } \{x\in \mathbb{R}^n,\|x\|=1\} \subset \mathbb{R}^n. \end{gathered}

So they are indeed equivalent. We will prove the lemma's form.

Homology

In this section, there will be many definitions to clarify.

We will use simplicial homology to prove the theorem. We will use homology over $\mathbb{Z}/2\mathbb{Z}$ rather than $\mathbb{Z}$ , since the homology of $\mathbb{RP}^n$ is much simpler.

Limited by the length of this article, some basic conclusions about homology in this section will not be proved.

Affine Independence, Simplices, Faces, and Simplicial Complexes

In $\mathbb{R}^m$ , a set of vectors $\{v_1, \dots, v_n\}$ is affinely independent iff for all $\{ c_i \}_{i=1}^n$ such that $\sum_i c_i=0$ , we have $\sum c_i v_i = 0 \iff \forall i, c_i=0$ .

If a set of vectors is affinely independent, then the set $\{ \sum c_i x_i \mid \sum c_i=1, c_i\ge 0 \}$ is an $n$ -simplex represented by $\langle x_1,\dots,x_n\rangle$ . A face of a simplex is $\langle y_i\rangle_{i=1}^k$ where $\{ y_i \}_{i=1}^k \subset \{ x_i \}_{i=1}^n$ .

If a set of simplices composes a set $K$ , and:

$\forall \sigma_1,\sigma_2\in K$ , $\sigma_1\cap \sigma_2$ is a common face of $\sigma_1$ and $\sigma_2$ ,
$\forall \sigma_f$ that is a face of $\sigma\in K$ , we have $\sigma_f\in K$ ,

then $K$ is a simplicial complex (we will just say complex henceforth), and $|K|=\bigcup_{\sigma\in K} \sigma$ is a simplicial polytope.

Simplicial Maps, Stars, and Simplicial Approximations

A simplicial map is a map $V(K)\to V(L)$ , where $V(K)$ is the vertex set of $K$ . A simplicial map can be naturally extended to $|K|$ : it is easy to prove that every point is contained in the interior of only one simplex, so we can define $f(x)$ by the convex combination of that simplex's vertices' values.

$\operatorname{St}(x)$ , the star of $x$ , is $\bigcup_{x\in \sigma} \sigma$ .

A simplicial approximation of $f:|K|\to |L|$ is a simplicial map $g$ s.t. $\forall x,f(\operatorname{St}(x))\subset \operatorname{St}(g(x))$ .

One can prove that, for many (but not all) spaces $X$ , we can find a simplicial complex $K$ such that $|K|\cong X$ . This enables us to study the simplicial complex $K$ instead of $X$ .

Chain Groups, Boundary Operators, and Homology Groups

For a complex $K$ , let $S$ be the set containing all $k$ -simplices. We define $C_k(K,\mathbb{F})=\sum_{s\in S} c_s s, c_s\in \mathbb{F}$ . We will write $C_k(K)$ to represent $C_k(K,\mathbb{F}_2)$ .

Define the operator $d_k:C_k(K)\to C_{k-1}(K)$ by $d_k \langle x_1,\dots, x_n \rangle\mapsto \sum_{i=1}^n (-1)^i \langle x_1,\dots,x_{i-1},x_{i+1},\dots,x_n \rangle$ . We can verify that $d_k\circ d_{k-1}=0$ .

Finally, for the homology groups: we define $H_n(X)=\dfrac{\ker d_n}{\operatorname{im} d_{n+1}}$ .

An Important Lemma

\begin{gathered} X\simeq Y \implies H(X)\cong H(Y) \\ f:X\to Y \text{ induces a homomorphism } \\ f^*:H(X)\to H(Y) \\ f\simeq g \implies f^*=g^* \\ f^* \sum_{i=1}^k c_i\langle v_1\dots v_n\rangle =\sum_{i=1}^k c_i\langle f(v_1)\dots f(v_n)\rangle \end{gathered}

Omitted

This important lemma enables us to only care about the effect of any map on $X\simeq |K|$ on the homology groups, rather than considering a simplicial map on a triangulation.

Degree of a Map

\begin{gathered} \text{For a map } f:S^n\to S^n, f \text{ induces a homomorphism } f^*:H_n(S^n)\to H_n(S^n). \\ \text{Since } H_n(S^n)\cong \mathbb{Z}_2, f^* \text{ is an endomorphism.} \\ \text{So, taking } 1 \text{ as the generator of } H_n(S^n): \\ f^*(1)=m \implies f^*(n\cdot 1)=n\cdot f^*(1) \\ \text{Thus, we define } \deg f=f^*(1). \end{gathered}

To base our proof on some specific spaces, we need the following results:

\begin{gathered} H_n(S^k)=\begin{cases} \mathbb{Z}_2 & k\in \{ 0,n \} \\ 0 & \text{otherwise} \end{cases} \end{gathered}

\begin{gathered} H_n(\mathbb{RP}^k)=\begin{cases} \mathbb{Z}_2 & k\in [ 0,n ] \\ 0 & \text{otherwise} \end{cases} \end{gathered}

Omitted

Lemma 2

\begin{gathered} \forall f:S^n\to S^n \\ f \text{ can be extended to } D^{n+1} \implies \deg f=0 \end{gathered}

The fact that $f$ can be extended indicates that $\exists F:D^{n+1}\to S^n \ \text{s.t.}\ F|_{\partial D^{n+1}}=f$ , making the following diagram commute:

Considering the map it induces on homology, we have:

But $H_n(D^{n+1})=0$ , so $i^*=0$ , which implies $f^*=F^*\circ i^*=F^*\circ 0=0$ . Thus, $\deg f^*=0$ .

Lemma 3

\begin{gathered} \forall f:S^n\to S^n, f(-x)=-f(x) \\ \deg f \equiv 1 \pmod{2} \end{gathered}

Since $f(-x)=-f(x)$ , we can descend $f$ to $\bar{f}:\mathbb{RP}^n\to \mathbb{RP}^n, [x]\mapsto [f(x)]$ . Since $[x]=[y]\land x\ne y \iff x=-y \implies f(x)=-f(y) \iff [f(x)]=[f(y)]$ , we know it is well-defined. This observation (that $f$ can be descended to $\mathbb{RP}^n$ ) reminds us to consider $\mathbb{RP}^n$ . Since $S^n$ is a double cover of $\mathbb{RP}^n$ , we obtain a short exact sequence of chain complexes:

We define $p$ as the simplicial approximation of the covering map ( $\tilde \sigma\mapsto \sigma$ ), and $T:\mathbb{RP}^n\to S^n$ as $T(\sigma) = \sum(p^{-1}(\sigma))$ (the sum of the two lifts in $S^n$ ). Since we work over $\mathbb{Z}_2$ , we have $p\circ T = \sigma+\sigma = 0$ , so the sequence is indeed exact:

Notice that in order to obtain a simplicial approximation of $p$ and $T$ , we have to perform barycentric subdivision on $\mathbb{RP}^n$ and $S^n$ . However, since barycentric subdivision does not change the homotopy type (up to isomorphism), we can write the homology groups simply as $H_i(S^n)$ and $H_i(\mathbb{RP}^n)$ .

Once we obtain the simplicial approximations of $p$ and $T$ , they must commute with the boundary operator $\partial$ . Thus, by the Snake Lemma, we obtain the long exact sequence:

The map $f$ will induce an endomorphism on homology as well. It is easy to verify that $f$ commutes with $T$ and $p$ : $f\circ T(\sigma)=f(\tilde \sigma_1+\tilde \sigma_2)=f(\tilde \sigma_1)+f(\tilde \sigma_2)=\widetilde{f(\sigma_1)}+\widetilde{f(\sigma_2)}=T\circ f(\sigma)$ , and the case for $p$ is similar. This happens naturally because $T$ and $p$ are lift/projection maps, while $f$ preserves the fibers. Since $f^*$ commutes with $T^*$ and $p^*$ , we can show that it must commute with the connecting homomorphism $\delta$ . Thus, we can write the following commutative diagram:

To prove the degree is $1$ , we must prove that $f^*_n$ is an isomorphism. Here, $\bar{f}^*$ plays the role of a bridge.

We will proceed by induction. For $i=0$ , $H_0$ represents the connected components. Since $\mathbb{RP}^n$ is connected, it follows that $\bar{f}^*_0 = \mathrm{id}$ is an isomorphism.

Now consider $i\in [1,n-1]$ . Assume we have proved that for all $k < i$ , $\bar{f}^*_k$ is an isomorphism.

Notice that $H_i(S^n)=0$ for all $i\in [1,n-1]$ . By exactness, we find that $\delta_i$ is an injection, and hence an isomorphism for all $i\in [1,n-1]$ . As for $i=n$ , $\delta_n$ is a surjection, and hence an isomorphism as well. Thus, in this part of the commutative diagram, we see:

The top, bottom, and right edges are all isomorphisms, so the left edge is also an isomorphism.

Finally, look at this part:

Since $H_{n+1}(\mathbb{RP}^n)=0$ , the five maps between the two long exact sequences are all isomorphisms except possibly the middle one. Then, by the Five Lemma, the middle map $f^*_n$ is also an isomorphism.

Thus, $f^*=\mathrm{id}$ on $\mathbb{Z}_2$ , so $\deg f=1$ .

Final Proof

By Lemma 1, we want to prove that there does not exist any map $f:S^n\to S^{n-1}$ such that $f(-x)=-f(x)$ .

We will prove this by contradiction. Assume that such a map $f:S^n\to S^{n-1}$ exists with $f(-x)=-f(x)$ .

We can embed $S^n$ into $\mathbb{R}^{n+1}$ as the unit sphere $\sum_{i=0}^n x_i^2=1$ , and embed $S^{n-1}$ as the equator of $S^n$ given by $x_0=0,\sum_{i=1}^n x_i^2=1$ .

Then we can restrict $f$ to $S^{n-1}$ to obtain a map $f|_{S^{n-1}}:S^{n-1}\to S^{n-1}$ satisfying $f|_{S^{n-1}}(-x)=-f|_{S^{n-1}}(x)$ . By Lemma 3, we know that $\deg f|_{S^{n-1}}=1$ .

However, we can extend $f|_{S^{n-1}}$ to a map $g:D^n\to S^{n-1}$ , where $D^n$ can be embedded as the upper hemisphere of $S^n$ , $\{ x\in S^n \mid x_0 \ge 0 \}$ , by defining $g(x_1, \dots, x_n) = f(\sqrt{1-\sum_{i=1}^n x_i^2},x_1,\dots, x_n)$ . Thus, $\deg f|_{S^{n-1}}=0$ by Lemma 2.

This yields a contradiction. Therefore, such a map $f$ does not exist.

Existence

We can divide the sandwich.

A plane $\sum_{i=1}^k c_ix_i=-c_{k+1}$ in $\mathbb{R}^k$ can be represented by a point $(c_1,\dots,c_{k+1})\in S^k$ satisfying $\sum_{i=1}^{k+1} c_i^2=1$ . Thus, we can define $f_i(c)$ as the volume of the set $(D_i\times [i-1,i])\cap \{ (x_1,\dots, x_k) \mid (x_1,\dots,x_k,1)\cdot (c_1,\dots,c_{k+1})\ge 0 \}$ . Since $\sum_{i=1}^{k+1} c_i^2=1$ , we obtain a map $F:S^k\to \mathbb{R}^k$ , $c\mapsto (f_1(c),\dots, f_k(c))$ .

By the Borsuk–Ulam Theorem, we know that there exists a point $c$ such that $F(c)=F(-c)$ , which means the plane $\sum_{i=1}^k c_ix_i=-c_{k+1}$ divides the $k$ regions into two parts with equal volumes.

Setting $k=3$ yields the sandwich theorem, and setting $k>3$ yields the hyper-sandwich theorem.

Summary

We know we can divide it, but we still do not know how to do it. Sarcastic, right?

How to divide a sandwich