Are there intersection theoretic proofs for Ham-Sandwich type theorems?

Solution 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.