Concrete Question:

Let $f:\mathbb S^n \to \mathbb R^n$ be a $\mathbb Z/2$ smooth equivariant map where the action on the sphere is antipodal, and $\mathbb R^n$ is multiplication of co-ordinates.

Can one show that

$$0 \neq H^n(\bigcap_{i=1}^{n} X_i) \in H^n(S^n)$$

where $$X_i:=f^{-1}\left(\mathbb R \times \dots \underbrace{\{0\}}_{i^\text{th} \text{ component}} \times \dots \times \mathbb R \right) .$$

It's not immediately clear that such a question is well-posed (transverse intersections for example.) One might need to take a subset of the intersection in order to get the Borsuk-Ulam theorem. It's not clear to me how such a proof would go. See the update section for how one can recover the continuous version from the smooth one.

The rest of this question is just expanding on this.


In equivariant topology there is a configuration space/test map paradigm to solve a geometric problem.

Specifically, given a geometric problem $P$, we define the configuration space, $X$, which parametrizes all associated solutions to the problem (such as points, lines, or arcs.) Additionally, we consider a test subspace $Z \subset V$ and a continuous map $f:X \to V$ where $p \in X$ is a solution to a problem if and only if $f(p) \in Z$. With this setup in mind, we further require that $f$ be $G$-equivariant, where $G$ acts on both $X$ and $Y$. From this, the typical method of proof is to show the nonexistence of maps $f:X \to_{G} Y \setminus Z$, ensuring the existence of a geometric solution.

More concretely, in the Ham Sandwich theorem, we let $X=S^n$ parametrize all possible configurations of $n-1$ dimensional hyperplanes (this is done by scaling coefficients in a hyperplane formula and adding two hyperplanes at infinity), we let $Y$ be $\mathbb R^n$, and let $f:X \to Y$ be a function assigning a volume to each "mass" above and below a hyperplane and subtracts them. Here, $G$ is $\mathbf Z_2$, and it acts on the sphere antipodally and $\mathbb R^n$ in the standard way. $f$ is equivariant with respect to these actions.

A solution is $Z=0$, since this would imply a bisection of all the masses. Of course, the Borsuk–Ulam theorem tells us that such an equivariant map cannot exist.


A different way of thinking about the above set up (and I believe it is used in a popular youtube video) is to consider the problem co-ordinate wise:

For example, in the Ham-Sandwich theorem, we have that a solution is an intersection $$\bigcap_{i=1}^{n} X_i$$

where $$X_i:=f^{-1}\left(\mathbb R \times \dots \underbrace{\{0\}}_{i^\text{th} \text{ component}} \times \dots \times \mathbb R \right) $$

My Question

Thinking about the above as an $n-1$-dimensional manifold. Are there theorems guaranteeing that such intersections are not empty when $f$ is equivariant ?(and for my discussion smooth, although I do not know If this is necessary for a proof)

More specifically:

  1. Can one show that $X_i$ intersects all $X_j$ with $j \neq i$ transversely?

  2. (If 1 is true) Regarding $X_i$ as submanifolds, can one show that the homology class of their intersection is nonempty, i.e $0 \neq [X_i \cap X_j] \in H_0(S^n)$?

A Naive Guess

The map $f:S^n \to \mathbb R^n$ that is $\mathbf Z_2$ equivariant can be associated with a $G$-bundle (vector bundles in this case) $ \mathbb RP^n \times_G \mathbb R^n \to \mathbb RP^n$ given by $[x,y] \mapsto [y]$. A section of this bundle is given by $[x] \mapsto [x,f(x)]$ and the nonvanishing of $f$ gives a nonvanishing section in this bundle. One can then show that this is impossible on the level of cohomology by considering Stiefel–Whitney classes (and in particular the top Stiefel–Whitney class.)

My hope is to maybe use something about the cup product structure on $\mathbb RP^n$ (understood as the Poincare dual to intersection) in the above decomposition to get a similar theorem/result.

For example, in the two dimensional version of the ham-sandwich theorem, if we let $[X_i] \in H_{1}(S^n)$, then generically $ [X_1 \cap X_2] \in H_0(X)$ (I'm not sure how to show that they intersect transversally.) Then, using the fact that $[X_1 \cap X_2]^* \in H_2(S^2)$, one can then use $$[X_1]^* \smile [X_2]^*=[X_1 \cap X_2]^*$$ to deduce that $[X_1 \cap X_2]$ is nontrivial, so there is an intersection.

In order to use more information about equivariance, I would consider the images of $X_1,X_2$ in the covering map $S_2 \to \mathbb RP^2$, and actually use cohomology classes $$H^*(\mathbb RP^2,\mathbb Z_2).$$

From here, the problem is essentially just computing $[X_i] \in H_1(\mathbb RP^2)$ and showing that it is indeed nontrivial.

A hopeful, but unfounded idea: The induced map by inclusion $i: X_i \hookrightarrow X$, gives that $$0 \neq w_1(i^*(\mathbb S^n \times_G \mathbb R^n))=i^*(w_1(X))=[X_i],$$ where $w_1$ is the first Stiefel–Whitney class.

From this, we can conclude that $[X_1] \smile [X_2]=x^2 \in \mathbf Z_2[x]/(x^3)$ is nonzero, and hence $H_0(X_1 \cap X_2)$ and deduce the ham sandwich theorem from this.

If there is something drastically different from this, I'd love to hear it as well, this is just kind of where my brain went.


In the paper "geometric proofs of the two-dimensional Borsuk–Ulam theorem", it is shown that

  1. one can recover the continuous version of the theorem from the smooth version (the proof here easily generalizes to higher dimensions.)

  2. Theorem 2.2 in the paper states that a smooth equivariant map with $0$ as a regular value must contain an equator in $f^{-1}(0)$, from which one can deduce the theorem (for two dimensions.)

Therefore, the smooth approach is equivalent, and there is a "geometric" proof showing that the intersection must be nonempty. Part $1$ of my question is unanswered, but it has been shown that each $X_i$ (for two dimensions) contain a subset of manifolds that do intersect transversely.

  • $\begingroup$ According to the pdf you linked to, $V$ is called the test space, not $Z$. $\endgroup$
    – celtschk
    Jun 14, 2018 at 3:38
  • $\begingroup$ @celtschk edited. $\endgroup$ Jun 14, 2018 at 20:14

1 Answer 1


If the question is more generally "can one use cohomology rings to prove the ham sandwich theorem", the answer is yes. Here is one route. Suppose, for the sake of contradiction, there were a map $S^n \to \mathbb{R}^{n} \setminus \{ 0 \}$ as stated. Then the map would pass to the quotients $\mathbb{RP}^n \to \mathbb{RP}^{n-1}$. The condition that antipode in $S^n$ becomes antipode in $\mathbb{R}^n$ means that a path from $x$ to $-x$ in $S^n$ is sent to such a path in $\mathbb{R}^n$, and thus a noncontractible loop in $\mathbb{RP}^n$ is sent to a noncontractible loop in $\mathbb{RP}^{n-1}$. So we would have a map $f : \mathbb{RP}^n \to \mathbb{RP}^{n-1}$ which was nonzero on $\pi_1$.

Now, look at cohomology with $(\mathbb{Z}/2)$-coefficients. We get a map $f^{\ast} : (\mathbb{Z}/2)[\eta]/\eta^n \to (\mathbb{Z}/2)[\eta]/\eta^{n+1}$. The generator $\eta \in H^1(\mathbb{RP}^{\ast}, \mathbb{Z}/2)$ corresponds to the nontrivial map $\pi_1 \to \mathbb{Z}/2$, so $f^{\ast}$ must carry $\eta$ to $\eta$. But we have $\eta^n=0$ on the left hand side and not the right, a contradiction.

An idea which seems closer to your original approach would be to use $(\mathbb{Z}/2)$-equivariant cohomology with $(\mathbb{Z}/2)$-coefficients. The space $\mathbb{R}^n$ is $(\mathbb{Z}/2)$-equivariantly contractible, so $$H^{\ast}_{\mathbb{Z}/2}(\mathbb{R}^n, \mathbb{Z}/2) \cong H^{\ast}_{\mathbb{Z}/2}(\mathrm{point}, \mathbb{Z}/2) \cong H^{\ast}(\mathbb{RP}^{\infty}, \mathbb{Z}/2) \cong (\mathbb{Z}/2)[\eta]$$ where the generator $\eta$ is in degree $1$.

The action on $S^n$ is free, so $$H^{\ast}_{\mathbb{Z}/2}(S^n, \mathbb{Z}/2) \cong H^{\ast}(\mathbb{RP}^n, \mathbb{Z}/2) \cong (\mathbb{Z}/2)[\eta]/\eta^{n+1}.$$ As before, the condition that the map is equivariant means that $f^{\ast} \eta = \eta$. I believe that the class of a hyperplane through $0$ is $\eta$. If I am right, then the computation is that $\eta^n \neq 0$ in $H^n_{\mathbb{Z}/2}(S^n, \mathbb{Z}/2)$, which is true.

  • $\begingroup$ Hi David speyer, I’m aware of the first approach (that is essentially the CS/TM approach specialized to this theorem.) $\endgroup$ Oct 18, 2018 at 15:28
  • $\begingroup$ The second approach Is a nice computation too :). Thank you for answering. $\endgroup$ Oct 18, 2018 at 15:30

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