4
$\begingroup$

Given a topological space $X$ with involution $\nu$, the $\mathbb Z_2$-index $\text{ind}(X)$ is the minimum integer $n$ such that there exists a map $f:X \to S^n$ which is equivariant with respect to the antipodal map on the sphere $S^n$.

Let $K$ be a (finite) simplicial complex with a fixed-point-free involution such that $\text{ind}(|K|)=d$ and $|K|$ is not homotopy equivalent to $S^d$. Can we always find a (maximal) simplex $\sigma$ such that deleting $\sigma$ and $\nu(\sigma)$ does not decrease the $\mathbb Z_2$-index?

$\endgroup$
2
$\begingroup$

I believe the following should work as a counter-example. Let $K$ be a simplicial torus $T^2$, obtained as the orientation double cover of a triangulated Klein bottle $\overline{K}$. Then $K$ comes equipped with an (orientation-reversing) involution $\nu$. I believe $\operatorname{ind}(|K|)=2$, and that removing any simplex $\sigma$ and its involute $\nu(\sigma)$ decreases the index.

The key observation (for me at least) is this: if $X$ is a (reasonably nice) free $\mathbb{Z}_2$-space and $p:X\to \overline{X}$ is the resulting quotient double cover, then $$ \operatorname{ind}(X) = \operatorname{secat}(p:X\to \overline{X}). $$ Here $\operatorname{secat}(p)$ is the sectional category (or normalized Schwarz genus) of the double cover, which by definition is one less than the smallest number of open sets needed to cover $\overline{X}$, on each of which $p$ admits a section. It follows that $$ \operatorname{cup-length}\ker(p^*:H^*(\overline{X})\to H^*(X))\le \operatorname{ind}(X)\le \dim(\overline{X}). $$

Returning to the example of the torus covering the Klein bottle, the calculation of the index follows from mod 2 cohomology calculations, and the fact that the index decreases on removing any top-dimensional cell follows from homotopy invariance of the sectional category together with the fact that the Klein bottle minus a cell deformation retracts onto its 1-skeleton.

In fact, this should generalize: whenever $X$ is a closed $d$-dimensional free $\mathbb{Z}_2$-manifold with $\operatorname{ind}(X)=d$ we should see this behaviour.

$\endgroup$
  • $\begingroup$ Could you please give a little more detail about the involution $\nu$? Am I right that (forgetting the triangulation) it can be defined as $(x,y,z) \mapsto (-x,-y,-z)$, where the torus is radially symmetric about the $z$-axis? But then the index of $K$ would be $1$ and not $2$. $\endgroup$ – user1272680 Jun 15 '16 at 10:05
  • $\begingroup$ @user1272680: I'm not entirely sure if the orientation-reversing involution you describe has as its quotient the Klein bottle. I prefer to think of the torus as $S^1\times S^1\subset \mathbb{C}^2$ and then my involution is given by $\nu(z,w)=(-z,\bar{w})$. $\endgroup$ – Mark Grant Jun 15 '16 at 10:45
  • $\begingroup$ I think the cohomology calculation I had in mind went as follows: the 1st Stiefel-Whitney class $w_1\in H^1(\mathrm{Klein};\mathbb{Z}/2)$ pulls back to zero in the cohomology of the torus. I think I thought that $w_1^2$ was nonzero, but now I'm not sure. I'll have another think about this when I get time. $\endgroup$ – Mark Grant Jun 15 '16 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.