2009-10-16 01:59:19 8 Comments

This is probably quite easy, but how do you show that the Euler characteristic of a manifold *M* (defined for example as the alternating sum of the dimensions of integral cohomology groups) is equal to the self intersection of *M* in the diagonal (of *M* × *M*)?

The few cases which are easy to visualise (ℝ in the plane, S^{1} in the torus) do not seem to help much.

The Wikipedia article about the Euler class mentions very briefly something about the self-intersection and that does seem relevant, but there are too few details.

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## 3 comments

## @cgodfrey 2019-04-11 06:09:03

There is also a nice proof of this fact (which I first saw in Milnor-Stasheff's

Characteristic Classes) which involves decomposing the class $\eta(\Delta) \in H^*(M \times M)$ obtained as the Poincare dual of the diagonal $\Delta \subset M \times M$.I'll assume $M$ is a compact (not necessarily oriented) $n$-manifold and use $\mathbb{Z}/2$-coefficients. If $M$ is oriented one can use coeffecients in any field. In addition I will assume $n$ is even. It simplifies the argument, and by another theorem in that book, the Euler characteristic of an odd dimensional compact manifold is zero -- so we won't miss out on much.

Theorems 11.10 and 11.11 on p. 128 show: for each basis $b_1, \dots, b_r$ for $H^*(M)$ there is a dual basis $b_1^\vee, \dots, b_r^\vee$ characterized as follows: if $\mu \in H_n(M)$ is the fundamental class of $M$, then $$ \begin{equation} \label{eq:1} \tag{1} \langle b_i \smile b_j^\vee , \mu \rangle = \begin{cases} 1, & \text{ if } i = j \\ 0, & \text{ otherwise }\\ \end{cases} \end{equation} $$ Note that $\deg b_i^\vee = n - \deg b_i$ In terms of the $b_i, b_i^\vee$, we have $$ \begin{equation} \label{eq:2}\tag{2} \eta(\Delta) = \sum_{i=1}^r (-1)^{\deg b_i}b_i \times b_i^\vee \in H^n(M \times M) \end{equation} $$ Here's how we use the fact that the diagonal is the diagonal: If $\tau: M \times M \to M \times M$ sends $(x, y) \mapsto (y, x)$, and $\tau^* : H^*(M \times M) \to H^*(M \times M)$ is the induced map on cohomology, one can show $\tau^* \eta(\Delta) = \eta(\Delta)$. Substituting in the above formula, we have $$ \eta(\Delta) = \tau^*\eta(\Delta) = \tau^*(\sum_{i=1}^r (-1)^{\deg b_i}b_i \times b_i^\vee) = \sum_{i=1}^r (-1)^{\deg b_i} (-1)^{\deg b_i \cdot \deg b_i^\vee} b_i^\vee \times b_i $$ $$ \begin{equation} \label{eq:3} \tag{3} = \sum_{i=1}^r (-1)^{\deg b_i (n - \deg b_i + 1)} b_i^\vee \times b_i = \sum_{i=1}^r b_i^\vee \times b_i \end{equation} $$ where I've used that $(-1)^{\deg b_i (n - \deg b_i + 1)} = 1$ since $n$ is even and $\deg b_i \cdot (1 - \deg b_i)$ is always even.

Now when we self-intersect the diagonal, substitute formula \eqref{eq:2} for one copy of $\eta(\Delta)$ and formula \eqref{eq:3} for the other: $$ \begin{equation} \label{eq:4} \tag{4} \eta(\Delta) \smile \eta(\Delta) = (\sum_{i=1}^r (-1)^{\deg b_i}b_i \times b_i^\vee) \smile (\sum_{i=1}^r b_i^\vee \times b_i ) \end{equation} $$ $$ \begin{equation} \label{eq:5} \tag{5} = \sum_{i, j} (-1)^{\deg b_i} b_i \times b_i^\vee \smile b_j^\vee \times b_j \end{equation} $$ $$ \begin{equation} \label{eq:6} \tag{6} = \sum_{i, j} (-1)^{\deg b_i} (-1)^{\deg b_i^\vee \deg b_j^\vee} b_i \smile b_j^\vee \times b_i^\vee \smile b_j \end{equation} $$ $$ \begin{equation} \label{eq:7} \tag{7} = \sum_{i, j} (-1)^{\deg b_i} (-1)^{\deg b_i^\vee \deg b_j^\vee} (-1)^{\deg b_i^\vee \deg b_j} b_i \smile b_j^\vee \times b_j \smile b_i^\vee \end{equation} $$ Long overdue simplification of the factors of $(-1)$: the factor of $(-1)$ on the $i, j$ term of the sum is $$ (-1)^{s} \text{ where } s = \deg b_i + \deg b_i^\vee \cdot (\deg b_j + \deg b_j^\vee) = \deg b_i + \deg b_i^\vee \cdot n $$

Since $n$ is even, $(-1)^s = (-1)^{\deg b_i}$, and we've shown

$$ \eta(\Delta) \smile \eta(\Delta) = \sum_{i, j} (-1)^{\deg b_i} (b_i \smile b_j^\vee) \times (b_j \smile b_i^\vee) $$ Finally, pairing with the fundamental class $\mu \times \mu$ of $M \times M$ and recalling equation \eqref{eq:1}, we see that $$ \langle \eta(\Delta) \smile \eta(\Delta), \mu \times \mu \rangle = \sum_{i, j} (-1)^{\deg b_i} \langle b_i \smile b_j^\vee , \mu \rangle \cdot \langle b_j \smile b_i^\vee , \mu \rangle $$ $$ \label{eq:8} \tag{8} = \sum_i (-1)^{\deg b_i} = \sum_i (-1)^i \dim H^i(M) $$

## @Andy Putman 2009-10-16 02:10:50

The normal bundle to $M$ in $M\times M$ is isomorphic to the tangent bundle of $M$, so a tubular neighborhood $N$ of $M$ in $M\times M$ is isomorphic to the tangent bundle of $M$. A section $s$ of the tangent bundle with isolated zeros thus gives a submanifold $M'$ of $N \subset M\times M$ with the following properties:

1) $M'$ is isotopic to $M$.

2) The intersections of $M'$ with $M$ are in bijection with the zeros of $s$ (and their signs are given by the indices of the zeros).

The desired result then follows from the Hopf index formula.

## @B.B. 2009-10-16 08:01:49

Here's an attempt at a sheaf theoretic argument that I always thought would work, but never actually tried:

The intersection number of two transversal submanifolds $A$ and $B$ of complimentary dimension inside a third manifold $C$ can be computed as $\chi(A \otimes B)$, where I'm using $A$ and $B$ to denote the structure sheaves of the corresponding manifolds, and the tensor product is taking place in $C$-mod. In the case that the intersection is not transversal, this presumably still works provided you take a derived tensor product (take a flat family moving one of the intersectands to a general position, and use invariance of $\chi$ under flat deformation for a perfect complex representing the other intersectand, perhaps?).

Assuming the above, the self-intersection $M.M$ of the diagonal $M$ in $M \times M$ is $\chi(M \otimes^L M)$. As $M$ is smooth, $\text{Tor}^i(M,M) = \Omega^i$. By the additivity of $\chi$, you get:

$M.M = \sum_i \chi(\Omega^i) (-1)^i$

On the other hand, de Rham's theorem (or the Poincare lemma?) identifies the right hand side with $\chi(M, \text{constant sheaf}) = \chi(M)$, so we are done.

## @Phil Tosteson 2019-08-16 01:29:31

This is a pretty idea, but it's not clear to me how to define $\chi(\Omega^i)$ for a smooth manifold. However, when $M$ is a compact complex manifold this argument seems to work nicely.