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, S1 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.


3 Answers 3


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.


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.

  • $\begingroup$ 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. $\endgroup$ Aug 16, 2019 at 1:29

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 $$ \langle \eta(\Delta) \smile \eta(\Delta), \mu \times \mu \rangle = \sum_{i, j} (-1)^{\deg b_i} \langle b_i \smile b_j^\vee , \mu \rangle \cdot \langle b_j \smile b_i^\vee , \mu \rangle $$ $$ \label{eq:8} \tag{8} = \sum_i (-1)^{\deg b_i} = \sum_i (-1)^i \dim H^i(M) $$


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.