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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Conisder 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 simplexes (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 (spoted 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 simeplx. The shade is an convex set. It is naturally decomposed into simplexe, so the sum of simplexes over this shade is $(-1)^{n-1}$ (because the simplexes in the boundary of this convex set do not contrubute). And we also get $(-1)^n$ for $\Delta^n$. |
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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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