Compute Euler characteristic

Question

Sorry, this question is a bit vague. Consider a compact manifold (e.g a projective variety) $X$ and the involution $\iota : X \times X \to X \times X, (x,y) \to (y,x)$. We can take the quotient of $(X \times X) / \iota$.

Is $Y = X \times X / \iota$ a projective variety/a manifold if $X$ was ?

Assume the answer is yes, and that I know the cohomology of $X$.

Can I say that the cohomology of $Y$ is generated by the cohomology of $X \times X$ quotiented by the relations $\alpha \times \beta - \beta \times \alpha$ ?

This question comes because I was reading that the Euler characteristic of $Y$ is $6$ where $X = \mathbb CP^2$. My guess above guided me to the right answer but I wanted to know if this was rigourous.

Answer 1

N.H. Yes. In fact, for the manifold question you need $X$ to be an even dimensional oriented manifold. The answer to the projective variety question is yes. Finally your computational method works so long as you are either working with homology with field coefficients, where the field does not have characteristic $2$, or $H_*(X;\mathbb{Z})$ has no torsion.

Lets start with the computation of homology. You need $$x: H_*(X)\otimes H_*(X)\rightarrow H_*(X\times X)$$ to be an isomorphism where $x$ is the exterior product. This will happen as long as there is no torsion in the homology of $X$. The action of $\iota_*:H_*(X\times X)\rightarrow H_*(X\times X)$ corresponds to the map $$flip:H_*(X)\otimes H_*(X)\rightarrow H_*(X)\otimes H_*(X)$$ under the identification via $x$, where $flip(\alpha \otimes \beta)=\beta \otimes \alpha$. The map $flip$ corresponds to an action of $\mathbb{Z}_2$ on $H_*(X)\otimes H_*(X)$, correspondingly $H_*(X)\otimes H_*(X)$ splits as a direct sum of the eigenspaces of the action of $flip$, which are the $+1$ and $-1$ eigenspaces. The $-1$ eigenspace $V_{-1}$ is spanned by everything of the form $\alpha\otimes \beta-\beta\otimes \alpha$, and the $+1$ Eigenspace $V_{+1}$ is spanned by everything of the form $\alpha\otimes \beta +\beta\otimes \alpha$. Clearly $$V_{+1}=H_*(X)\otimes H_*(X)/V_{-1}.$$ This is the space you claim to be isomorphic to the homology of $H_*((X\times X)/\iota)$.

The quotient map $q:X\times X\rightarrow (X\times X)/\iota$ induces a map $$q_*:H_*(X\times X)\rightarrow H_*((X\times X)/\iota).$$ There is a reverse mapping $$q^!:H_*((X\times X)/\iota)\rightarrow H_*(X\times X)$$ coming from Poincare duality ( here I am using $(X\times X)/\iota$ is an orientable manifold. The image of $q^!$ is exactly the image of $V_{+1}$, and $q^!\circ q_*$ is multiplication by $2$. Hence if the coefficients are a field that does not have characteristic $2$, then $$q_*\circ x:V_{+1}\rightarrow H_*((X\times X)/\iota)$$ is an isomorphism.

By playing around you can use the argument above to prove that when $H_*(X;\mathbb{Z})$ has no torsion, the computation you give works.

For any of this to make sense, $(X\times X)/\iota$ needs to be an oriented manifold. The fixed point set of $\iota$ is the diagonal of $X\times X$. Restricting to an equivariant regular neighborhood of the diagonal, you can see that the quotient is still a manifold. Finally it is easy to see when $X$ has even dimensions $\iota$ is an orientation preserving mapping.

Answer 2

Actually, I think that $Y:=(X\times X)/\iota$ is (topological) manifold (possibly with boundary) iff $\dim X \leq 2$. Here is a proof.

Let $\pi:X\times X\rightarrow Y$ denote the projection.

The set of charts $\mathbb{R}^n\cong U\subseteq X$ containing a point $x\in X$ form a neighborhood basis at $x$. It follows that the collection of $U\times U$ forms a neighborhood basis around $(x,x)\in X\times X$. Then, it follows that the collection of $\pi(U\times U)$ forms a neighborhood basis of $\pi(x,x)\in Y$.

Under the diffeomorphism $U\cong\mathbb{R}^n$, $\iota$ swaps the two $\mathbb{R}^n$ factors. That is, $\iota:\mathbb{R}^n\times \mathbb{R}^n\rightarrow \mathbb{R}^n\times \mathbb{R}^n$ is the linear map given by swapping the two factors. As in Charlie Froham's answer, this gives a decomposition $\mathbb{R}^n\times \mathbb{R}^n\cong V_1 \oplus V_{-1}$ into the $\pm 1$-eigenspaces. Both $V_{\pm 1}$ have dimension $n$, being spanned by elements of the form $(y, \pm y)$.

Now $\iota$ acts trivially on $V_1$ and acts as $-1$ on $V_{-1}$. This preserves each sphere $S^{n-1}$ centered at the origin in $V_{-1}$ where it acts as the antipodal map. Thus, $V_{-1}/\iota$ is the cone on $\mathbb{R}P^{n-1}$, $C\mathbb{R}P^{n-1}$.

It follows that for each $U$, $\pi(U\times U)$ is homeomorphic to $V_1 \times C\mathbb{R}P^{n-1}$. This is contractible as $V_1$ is a vector space and the cone is a cone.

Writing $U' = \pi(U\times U)$ and $x' = \pi(x,x)$, consider the local homology $H_\ast(Y, Y\setminus\{x'\})$. Given the chain of subsets $M\setminus U'\subseteq M\setminus \{x'\}\subseteq M$, we use excision to see that $$H_\ast(Y,Y\setminus\{x'\})\cong H_\ast(Y\setminus (Y\setminus U'), (Y\setminus \{x'\})\setminus(M\setminus U')) = H_\ast(U', U'\setminus \{x'\}).$$

Since $U'$ is contractible, we see that $H_\ast(U', U'\setminus\{x'\})\cong H_{\ast-1}(U'\setminus \{x'\})$. But $U'\setminus \{x'\}$ deformation retracts onto $\mathbb{R}P^{n-1}$. This has torsion in its homology for $n > 2$. Thus, $H^\ast(Y,Y\setminus\{x'\})$ has torsion in its homology for $n>2$. But for a topological manifold (possibly with boundary),the local homology is isomorphic to that of a sphere or a disc (if you're at a boundary point).

Thus, for $n > 2$, such spaces can't be topological manifolds, even allowing for boundary.

What about for $\dim X\leq 2$? First, points $(x,y)\in X\times X$ with $x\neq y$ automatically correspond to manifold points in $Y$. This follows, for example, because the action of $\iota$ when restricted to these kinds of points has no fixed points. So we need only think about points $(x,x)$.

When $n = 2$, $V_{-1}$ is homeomorphic to a cone on $\mathbb{R}P^1 \cong S^1$. Of course, a cone on $S^1$ is a disc, so $U'$ is homeomorphic to $\mathbb{R}^2\times D^2 \cong \mathbb{R}^4$. So these points have appropriate neighborhoods.

When $n=1$, $V_{-1}$ is homeomorphic to a cone on $\mathbb{R}P^0 \cong \{pt\}$. The cone is closed line segment. So, $U'\cong \mathbb{R}\times [0,1]$, so we get a manifold with boundary in this case.