$\chi(X\times Y)=\chi(X)\cdot\chi(Y)$

Question

For a compact oriented manifold $X$ define $\chi (X)=I(\Delta,\Delta)$ where $\Delta$ is the diagonal in $X\times X$ and $I$ is the intersection number. How do I show that $\chi(X\times Y)=\chi(X)\cdot\chi(Y)$?

This boils down to showing that $$I(\{(x,y,x,y)| x\in X, y\in Y\}, \{(x,y,x,y)| x\in X, y\in Y\})=I(\{(x,x)|x\in X\},\{(x,x)|x\in X\})\cdot I(\{(y,y)|y\in Y\},\{(y,y)|y\in Y\})$$

But I'm not sure how to compute the intersection number on the left-hand side to get a product. (To begin with, if I denote by $\Delta$ the diagonal in $X\times Y$, then the submanifolds $\Delta$, $\Delta$ are not transverse, and the inclusion of $\Delta$ into $X\times Y$ isn't transverse to $\Delta$.)

Answer 1

Another proof extending my comment, in case you are not fully glad with Paul's answer using another (equivalent) definition of intersection numbers via the index of vector fields. I hope that you can find the formal similarity in between mine and Paul's.

As the definition of intersection number requires us to perturb the submanifolds ($\Delta$ in this case) to achieve transversality, let me perturb each of them. That is, for $\Delta_X \subseteq X\times X$ we obtain $f_X:\Delta_X\to X\times X$ which is transverse to $\Delta_X$ and similarly $f_Y:\Delta_Y\to Y\times Y$ for $\Delta_Y$. Now, recall that $\Delta=\Delta_{X\times Y}=\Delta_X\times \Delta_Y$ (up to the canonical identification $X\times X\times Y\times Y\cong X\times Y\times X\times Y$, of course!). Again we need to consider a nice perturbation of the diagonal $\Delta$, but now we are in better situation: $f_X\times f_Y:\Delta_X\times \Delta_Y\to X\times Y\times X\times Y$ achieves the required transversality w.r.t. $\Delta$. Hence, from the definition of the intersection number $I(\Delta,\Delta),$

$\begin{align*} I(\Delta,\Delta)&=\sum_{(x,y,x,y)\in \Delta\cap f_X\times f_Y} \left(\mathrm{the\;sign\;of\;the\;intersection\;at\;}(x,y,x,y)\right)\\ &=\sum_{(x,y,x,y)\in \Delta\cap f_X\times f_Y} sgn_{X\times Y\times X\times Y}(x,y,x,y)\;\;\textrm{where }sgn\textrm{ means sign of blahblah from now on}\\ &=\sum_{(x,x)\in \Delta_X\cap f_X}\sum_{(y,y)\in \Delta_Y\cap f_Y} sgn_{X\times Y\times X\times Y}(x,y,x,y)\;\;\;\;\;\;\;\;\;\;\;\;\;\textrm{by the rule of product}\\ &=\sum_{(x,x)\in \Delta_X\cap f_X}\sum_{(y,y)\in \Delta_Y\cap f_Y} sgn_{X\times X}(x,x)sgn_{Y\times Y}(y,y)\;\;\;\;\;\;\;\;\textrm{ by the orientation convention}\\ &=I(\Delta_X,\Delta_X)I(\Delta_Y,\Delta_Y). \end{align*}$

Answer 2

The key idea is that for any manifold $M$ the normal bundle of the diagonal in $M \times M$ is diffeomorphic to the tangent bundle of $M$. Under this identification the diagonal embedding $m \mapsto (m,m)$ corresponds to the zero section of $TM$.

Now, choose vector fields $V$ and $W$ on $X$ and $Y$, respectively, such that $V$ and $W$ each have isolated zeros, which is equivalent to saying they intersect the zero section of the tangent bundle transversely. Consider the vector field $V \times W$ defined by $(V \times W)(x,y) = ((x,y), (V(x),W(y)))$ in $T(X \times Y) \cong \pi_X^*(TX) \oplus \pi_Y^*(TY)$ where $\pi_X \colon X \times Y \to X$ and $\pi_Y \colon X \times Y \to Y$ are the projection maps. Thus the zero set of $V \times W$ satisfies $Z_{V \times W} \cong Z_V \times Z_W$, and in particular the zeros are isolated.

Moreover, for each $(x_i, y_j) \in Z_V \times Z_W$ we have $ind_{x_i, y_j}(V \times W) = ind_{x_i}(V) \cdot ind_{y_j}(W)$. You can prove this in a variety of different ways depending on how you prefer to define the degree of a smooth map, but in any case the result is:

\begin{align*} \chi(X \times Y) &= \sum_{(x_i, y_j) \in Z_{V \times W}} ind_{x_i, y_j}(V \times W) \\ &= \sum_{x_i \in Z_V, y_i \in Z_W} ind_{x_i}(V) \cdot ind_{y_i}(W) \\ &= \left(\sum_{x_i \in Z_V} ind_{x_i}(V)\right) \left(\sum_{y_j \in Z_W} ind_{y_j}(W)\right) \\ &= \chi(X) \chi(Y) \end{align*}