# Euler Characteristic of a manifold with non-vanishing vector field,

A friend of mine recently asked me if I knew any simple, conceptual argument (even one that is perhaps only heuristic) to show that if a triangulated manifold has a non-vanishing vector field, then Euler's formula (the alternating sum of the number of faces of given dimensions) vanishes. I didn't see how to get started, but it seems like a good MO question.

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How about applying the Lefschetz fixed-point theorem? – Robin Chapman Sep 11 '10 at 17:21
I'm a big fan of the "physics" proof of the Poincare-Hopf formula in chapter 1 of Thurston's book "The Geometry and Topology of 3-manifolds". If you specialize this proof to the case of a nonvanishing vector field, then it becomes particularly easy. – Andy Putman Sep 11 '10 at 17:26

Consider a straight simplex $\Delta^n$ in $\mathbb R^n$ and take a generic constant vector field $v$ (transversal to the faces of $\Delta^n$). Choose all faces of $\Delta^n$ such that the field moves the center of the face inside the simplex. Then the alternating sum of the numbers of these simplices (signed by the parity of the dimension) is zero.

Now, if you have a fine enough triangulation of $M$ and a vector field transversal to all faces, we can apply the above reasoning to the whole manifold.

Edited. There was an explanation here with a mistake (spotted by Sergei) of why each simplex contributes zero, but the statement is correct. The new proof is a follows: $(-1)^{n-1}+(-1)^n=0$.

Proof. Let us say that $v$ is the sunlight. Then it enlightens a part of the simplex $\Delta^{n+1}$. Consider the shade from $\Delta^n$ on some plane below the simplex. The shade is an convex set. It is naturally decomposed into simplices, so the sum of simplices over this shade is $(-1)^{n-1}$ (because the simplices in the boundary of this convex set do not contribute). And we also get $(-1)^n$ for $\Delta^n$.

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Dmitri, that argument seems VERY promising and elegant, but I and my friend (Hermann Karcher) don't understand all the details---in particular just how $(1-1)^n$ gets used. How do the centers of the faces that are moved in get connected with this identity? – Dick Palais Sep 11 '10 at 18:06
The assertion "there will be a unique vertex $A$ of $\Delta^n$ ..." is not correct. You are essentially claiming that a projection of an $n$-simplex to an $(n-1)$-subspace is an $(n-1)$-simplex. But already for $n=3$, the projection of a simplex can be a quadrangle. – Sergei Ivanov Sep 11 '10 at 18:39
The lemma needed is this: if you look at an n-simplex in R^n from any point outside it and remove all the hidden faces, the Euler characteristic summation for what's left (including the interior) is 0. If you want to avoid algebra, you can use induction on dimenson: caluclate the summation over the hidden faces of the simplex using a linear vector field on a projection of the hidden part to R^{n-1}. This method also works for subdivisions into convex cells, not just simplices. – Bill Thurston Sep 11 '10 at 18:57
Many thanks to all. When I asked this question I didn't have much hope for such a definitive answer. I am continually amazed at how many knowledgeable people are involved in MO and as a result how useful it is. – Dick Palais Sep 12 '10 at 21:18
@Dmitri:This is a really beautiful argument! – Vitali Kapovitch Jan 5 '12 at 2:19

This is basically a watered down version of Poincare-Hopf theorem. Assuming that your compact manifold $M$ has a tringulation. At each of the simplex put a vector field $V$ which is zero at the centre of its subsimplices and flows out from the centre of the higher simplex to its boundary simplices. This vector field has finitely many zeroes and the index of $V$ is precisely $\chi(M)$. Now notice that the index of a vector field doesn't depend on the vector field. This is because it's topologically the intersection number of the image of $V$ in $TM$ and $M$ sitting as the zero section and $V$ is homotopic to $M$. In other words, given your non-vanishing vector field $V'$ one can linealy homotope this to $V$ whose index calculates $\chi(M)$.

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Sure, Somnath, but the point was to have, if possible, something that is "elementary" intuitive and that doesn't use much machinery. The collection of comments and answers posted shows that this is possible. – Dick Palais Sep 11 '10 at 23:51