Is there a Poincare-Hopf Index theorem for non compact manifolds?

Question

Does Poincare-Hopf index theorem generalizes in any way to non compact manifolds ? In particular, I am interested in the case of a smooth vector field on a cylinder $\mathbb{T}_1\times\mathbb{R}$? If so, are there some additional assumption that one has to impose on a vector field considered (maybe it should vanish outside some compact set or decay very fast "at infinity"?). Sorry if the question is silly - I know the Hodge index theorem only from very elementary sources (Arnold's book on ODEs and Wikipedia).

Motivation (physical digression): A friend of mine tries to model action of a cardiac tissue in a heart and its soundings. One of his aims is to understand phenomena called "spiral waves" (they are believed to be partially responsible for hearth attacks). I don't know the details but those "spiral waves" can be described by some ODE defined on a domain which is closely related to the real geometry of considered tissue. From the information about indexes of singular points of a corresponding vector field it is possible to deduce some qualitative information about occurrence of this phenomenon.

Answer 1

Every noncompact manifold admits nonzero vector fields, or more generally, vector fields with any specified set of isolated zeros along with the behavior near that zero.

However, if you have information of the behavior of a vector field near infinity, or just in a neighborhood of the boundary of a compact set, there is an index theorem. Perhaps this is the case with your $\mathbb T^! \times \mathbb R$.

In the particular case of a cylinder, there is a simple way to calculate the index. Take any compact subcylinder delimited by two circles. Map the cylinder to the plane minus the origin. Around each of the curves, the vector field has a turning number: as you go around the curve counterclockwise, the vector field turns by some number of rotations (counting counterclockwise as positive. The index of the vector field in the compact subannulus is the difference: the number of turns on the outer boundary minus the number of turns on the inner boundary.

One way to describe a general formula is this: let $N^n$ be manifold, and let $M, \partial M \subset N$ be a compact submanifold. Let $X$ be a vector field that is is nonvanishing in a neighborhood of $\partial M$.

Choose an outward normal vector field $U$ along $\partial M$; now arrange $X$ so that its direction coincides with $U$ only in isolated points, so if we project $X$ to $N$ along $U$, it is a vector field with isolated singularities. Let $i_+(X)$ be the sum of the Poincaré-Hopf indices over all singularities where $X$ is oriented outward. Then the Poincaré-Hopf index $i(X)$ of $X$ in $M$ equals the Euler characteristic of $M$ minus $i_+(X)$.

Here's one proof: triangulate a neighborhood of $M$ so that $\partial M$ is a subcomplex, and so that $X$ is transverse to the triangulation except near the singularities, in the sense that in any simplex, the foliation defined by $X$ is topologically equivalent to the kernel of a linear map in general position of the simplex to $\mathbb R^{n-1}$. Put a $+1$ at the barycenter of each triangle of even dimension, and a $-1$ at the barycenter of each triangle of odd dimension. Think of $X$ as a wind that blows these numbers along, so that after an instant, all numbers (except for exceptions near the zeros of $X$) are inside an $n$-simplex. In any typical simplex, all the signs cancel out. However, along the boundary, some of the numbers are blown away and lost. To regularize the situation, modify $X$ by pushing in the negative normal direction. Now $X$ points inward everywhere except in a neighborhood of points where it coincides with the outward normal. Thus everything cancels out except for local contributions given by $i(X)$ and $i_+(X)$.

Answer 2

Suppose $M$ has empty boundary. Let $U\subset M$ be an open set with compact closure whose topological boundary contains no zero of the continuous vector field $X$ on $M$. Suppose $X$ is smooth and hence generates a local semiflow $f_t$, $t \geq 0$.

For sufficiently small $t>0$ the map $f_t$$\colon U \to M$ is defined and has a "fixed point index" $I(f_t, U)$ (see A. Dold, ``Lectures on Algebraic Topology,'' Die Grundlehren der matematischen Wissenschaften Bd. 52. Springer, New York (1972)). It can be shown that the integer $i(X,U):=I(f_t, U)$ is independent of $t$ and $U$, and is stable under perturbation of $X$.

If $X$ is not smooth, approximate it by a sequence of smooth fields $X_j$ and define $I(X,U):= \lim_{j\to \infty} I(X_j,U)$.

If $X$ has only finitely many zeros in $U$ and none on $U\cap \partial $, then $I(X, U)$ is the sum of their Poincare-Hopf indices.

If $M$ has nonempty boundary, this work if at every boundary point $p$ there is an integral curve $u\colon [0,\epsilon)\to M$ with initial point $u(0)=p$.

Answer 3

As soon as you can construct a vector field with finitely many isolated singularities on a non-compact manifold, you can slide them all the way to infinity and get a vector field with no singular points. If the Poincaré-Hopf index worked, then the Euler characteristic of all non-compact manifolds would vanish.

On the other hand, you can say more useful things. For example, if you can find in your non-compact manifold $M$ a submanifold $M'\subseteq M$ of the same dimension which is compact and with a boundary, such that the inclusion is, say, a homotopy equivalence, then the Poincaré-Hopf theorem works for vector fields with finitely many singular points, all inside the interior of $M'$ and pointing outward on $\partial M'$.

Question: Can one always find such an $M'$ if we start with a non-compact $M$ that has finitely many ends and a vector field with finitely many regular singular points?