Let $M$ be a compact orientable $n$ dimensional smooth manifold without boundary, and $e\in H^n_{dR}(M)$ denote the Euler class of the tangent bundle $TM$. We have that \begin{align*} \int_Me=\chi(M) \end{align*} where $\chi(M)$ is the Euler characteristic.

Now, I know of a couple constructions of the Euler class, both of which are mildly involved. One involves the sphere bundle induced by a vector bundle, and the other involves pulling the Thom class back by a section. Both of these makes sense to me, but what I am confused about is whether in the case of the Euler class of the tangent bundle, if it is correct to state that the Euler class is $\chi(M)$ times a generator of $H^n_{dR}(M)$ (by generator, I mean a top form which integrates to $1$).

I feel like this is incorrect since I think that many forms should be able to integrate to $\chi(M)$, and we do so much to construct the Euler class. Then again in the case of a compact orientable smooth manifold we have that integration is an isomorphism from $H^n_{dR}(M)\cong \mathbb{R}$, so I guess any form that happens to be closed and integrates to zero is also actually exact, which is mildly shocking when I think about it.

So, is it correct to say that the Euler class is $\chi(M)$ times a generator of $H^n_{dR}(M)$?

For context, I am reading a paper where the author looks at the Euler class of the vertical tangent bundle of a fibre bundle (with compact orientable boundaryless fibres), and then states that the restriction of the Euler class to the fibre is the Euler characteristic times a generator of the top cohomology of the fibre. Shouldn't this just be the Euler class? Especially since the fibres are embedded submanifolds, so the restriction is just pulling back by the inclusion map, and the functorality of the Euler class should guarantee that this form pulls back to Euler class.

Am I missing something?


1 Answer 1


Indeed, integration gives an isomorphism $\int_M\colon H_{dR}^n(M)\rightarrow\mathbb{R}$ (you need to require $M$ connected, but I assume you just forgot to specify that). This implies that there is one and only one cohomology class $\eta_M\in H_{dR}^n(M)$ such that $\int_M\eta_M=1$. There are many closed top forms that integrate to $1$, but the corollary of this result is that all such top forms differ by an exact form, i.e. define the same cohomology class. I disagree with calling this unique element a "generator", though, cause that conflicts with the algebraic meaning of generator. In any case, since the Euler class $e(M)\in H_{dR}^n(M)$ satisfies $\int_Me(M)=\chi(M)$, it indeed follows that $e(M)=\chi(M)\eta_M$.

The construction of the Euler class is involved and it has to be, because there is an Euler class for every vector bundle. It satisfies nice properties like naturality and has, by virtue of its construction, strong connections to the geometry of the bundle, e.g. the existence of nowhere-vanishing sections. The above gives a very simple description for the Euler class in just one specific vector bundle, namely the tangent bundle. So, instead of thinking of this as trivializing the concept of the Euler class, you should think of it as a remarkable result that connects certain geometric invariants of $M$ that are all individually of interest already with one another.

For your context question, yes. The pullback of the vertical tangent bundle of a fiber bundle to a fiber just becomes the tangent bundle to that fiber, so pulling back the Euler class of the vertical tangent bundle to the fiber yields the Euler class of that fiber, which equals the Euler characteristic of the fiber times the element of the top cohomology of the fiber that integrates to $1$.


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