Intersection theory proof of the poincare hopf theorem.

Question

Suppose that $M$ is a connected compact oriented smooth manifold, and $X:M\rightarrow TM$ a vector field with isolated zeros. Then if $Z$ is the zero set of $X$ (a $0$-dimensional oriented manifold), we have that its fundamental class $[Z]$ (or rather its image in $H_0(M)$) is poincare dual to the Euler class $e(TM)$, i.e. $e(TM)\cap[M]=[Z]$. Also $$\chi(M)=<e(TM),[M]>=<e(TM)\cup 1,[M]>=<1,e(TM)\cap[M]>=<1,[Z]>,$$ which should be the index sum of $X$. Is this a correct proof of the theorem, and if not, can a proof be made along these lines? References are very welcome

Answer 1

You argument works if the zero set Z is an oriented zero dimensional manifold. Its image in $H_{0}(M)$ is the oriented sum of the points considered as 0-simplices. By oriented sum is meant that each point is multiplied by +1 or -1 depending on whether it is positively or negatively oriented.

But the zero set of the vector field will be an oriented sub manifold only if the zeros of the vector field are non-degenerate, that is: the vector field intersects the zero section transversally. But then the index of each vector field is +1 or -1 depending on orientation so cases where the index is some other integer are not included. For instance the vector field $z$ -> $z^2$ is degenerate and has index 2 at zero in the complex plane.

To account for degenerate zeros, one can perturb the vector field in a small neighborhood of the zero to obtain a new vector field with only non-degenerate zeros in the neighborhood and whose index sum is equal to the index of the degenerate zero. Applying a perturbation as necessary to each zero, your argument then applies to this modified vector field and the theorem follows. A proof can be found in Milnor's Topology from the Differentiable Viewpoint.

• For any oriented n-plane bundle over an n-dimensional manifold, the Euler class of the bundle is Poincare dual to the oriented zero set of a transverse section of the bundle. For the tangent bundle its value on the fundamental cycle is the Euler characteristic. But this follows from other arguments so your proof is assuming this. One way to get at it is to show that the index sum is the same for all vector fields with isolated zeros and then to find an example whose index sum is the Euler characteristic. See Steenrod The Topology of Fiber Bundles for this approach. Another approach can be found in Milnor's Topology from the Differentiable Viewpoint.

• Since you are dealing with smooth manifolds, your proof can be translated into statements about integrals of differential forms. The cup product with 1 in your argument corresponds to multiplication by the constant function f(x) = 1 in de Rham theory. The value of the Euler class on the fundamental cycle its integral over the manifold. The Poincare dual of a smooth embedded oriented sub manifold is the Thom class of its normal bundle and in the case of an oriented zero dimensional sub manifold the Thom class is the sum of finitely many n-dimensional bump forms each with support in a small open neighborhood of one of the zeros. When integrated over the manifold each bump form computes the index of its corresponding zero - which is either +1 or -1 depending on orientation. A reference for this approach is Bott and Tu, Differential Forms in Algebraic Topology.