# Why is the Euler class of (the unit tangent bundle over) a manifold equal to its Euler characteristic?

More specifically, I know that the Poincaré-Hopf theorem gives us that $$\chi(M)$$ is equal to the index of a vector field over $$M$$. I'm willing to take this as a blackbox, so the question is reduced to knowing why the latter quantity equals $$e(T^1M)$$.

Unfortunately, when I looked this question up, I noticed that my definition of the Euler class doesn't really match the "usual" definition. In particular, Wikipedia and the question here mention something about the Thom isomorphism, which I don't know anything about. Similarly the question here mentions something about self-intersections, but I'm not sure what this is talking about (self-intersections of what?).

Instead, I was reading a bit about obstruction theory, and I learned to think of the Euler class as a special case of a primary obstruction, where the bundle we're trying to get a section of is $$S^n \rightarrow T^1M \rightarrow M$$. (Note e.g. that a section of this bundle is exactly a non-vanishing vector field on $$M$$.) But the definitions of primary obstruction and the index of a vector field are just too many degrees of freedom apart that I can't make the connection between them. I'm hoping that someone can shed some light on this in particular.

• Actually, the definition of index matches the definition of $e$ using the obstruction theory very nicely. – Moishe Kohan Apr 26 '20 at 21:42
• Hi Moishe, could you elaborate a little bit? I'm afraid that it is a bit difficult from me to see how they match. In particular, the definition of the Euler class is independent of a specific vector field, so I'm not sure how I would talk about the Euler class "in terms of" the vector field whose index we're considering. – fish Apr 26 '20 at 22:57

## 1 Answer

Here's a direct connection between the index of a vector field and the obstruction theory definition of the Euler class. Note, I'm using the definition of the obstruction class from Wikipedia, since you link to it in the question.

Let $$M$$ be an $$(n+1)$$-manifold, because the indexing is slightly easier.

Let $$V$$ be a vector field on $$M$$ with finitely many zeroes. Triangulate $$M$$ such that each zero is in the interior of a different $$(n+1)$$-simplex, oriented so that it agrees with the orientation on $$M$$.

Then $$V|_{M^{[n]}}$$ (meaning the restriction of $$V$$ to the $$n$$-skeleton of $$M$$) is nonvanishing, and thus defines a partial section of $$T^1M$$ on the $$n$$-skeleton. The obstruction cochain will be an $$n$$-cochain with values in $$\pi_n(S^n)\cong \Bbb{Z}$$ (note that the orientation determines the isomorphism of $$\pi_n$$ of the fiber, $$S^n$$ with $$\Bbb{Z}$$). The value of the $$(n+1)$$-cochain on a particular $$(n+1)$$-simplex $$\Delta$$ is given by $$V(\partial \Delta)\in \pi_n(p^{-1}(\Delta))\cong \pi_n(S^n)\cong \Bbb{Z}$$, where $$p:T^1M\to M$$ is the projection.

Now it's worth going over the details of this computation here. If $$\Delta$$ doesn't contain a $$0$$ of $$V$$, then $$V(\Delta)$$ fills in $$V(\partial \Delta)$$, so $$V(\partial\Delta)$$ is $$0$$ in $$\pi_n(p^{-1}(\Delta))$$, and thus the final integer value is $$0$$.

On the other hand, if $$\Delta$$ does contain a $$0$$ of $$V$$, $$x$$, then by definition, the index of $$V$$ at $$x$$ is the degree of a map from $$S^n\to S^n$$ where the first $$S^n$$ is the boundary sphere of a disk $$D^{n+1}$$ in $$M$$ on which $$T^1M$$ is trivializable, and $$V$$'s only zero is $$x$$, and the second $$S^n$$ is the standard sphere in $$\Bbb{R}^n$$. The map is the composite $$\partial D^{n+1} \xrightarrow{V} \partial D^{n+1}\times (\Bbb{R}^{n+1}\setminus\{0\}) \to S^n,$$ where the second map is the projection onto $$\Bbb{R}^{n+1}\setminus\{ 0\}$$ followed by normalization. Note that the first map depends on a choice of trivialization, which should be chosen to be compatible with the orientation.

For convenience, note that we can dispense with requiring $$T^1M$$ to be trivializable, since the degree is a homotopy invariant, so it suffices that $$T^1M$$ is trivializable up to homotopy, which is in fact the case for any choice $$D^{n+1}$$. In particular, it is true for $$D^{n+1}=\Delta$$. Additionally, the degree of the map is usually computed with either homology (or cohomology), but in our case, the degree can also be computed with homotopy groups, since for the sphere we have canonical isomorphisms $$H_n(S^n)\cong \pi_n(S^n)$$ by the Hurewicz theorem.

However, if we use this to translate things over, we see that the value of the obstruction cochain on $$\Delta$$ is the index of the vector field $$V$$ at $$x$$.

Now we're almost done.

The isomorphism of $$H^{n+1}(M)$$ with $$\Bbb{Z}$$ for a closed, connected, oriented manifold $$M$$ is given by evaluating an $$(n+1)$$-cochain at the fundamental class.

If we call the obstruction $$(n+1)$$-cochain $$\delta$$, then the integer we want is $$\sum_{\Delta} \delta(\Delta) = \sum_{x} \operatorname{index}_x V.$$