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.



Euler Characteristic of a manifold with non-vanishing vector field,
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 $\Delta^n$. 


