Two definitions of Euler characteristics

Question

In third chapter of "Differential Topology" book by Guillemin and Pollack ,Euler's characteristic is defined as I(∆,∆) and in the end of chapter, an intuitive proof is given to show how this is same as one calculated by using triangulations. They defined vector field on triangulated manifold which has source on vertex ,sink on face and saddle on edge .Is there any rigorous as well as generalized ( to n dim.) proof of this theorem by using tools from book or other fairly basic tools ?

Answer 1

There is nothing non-rigorous about that proof. It works in arbitrary dimensions once you know that manifolds are triangulable. You need some proof like this because the usual definition of Euler char is in terms of triangulations...

Answer 2

Yes, the "usual proof" involves some (actually, quite a bit of) hand-waving. As an alternative, take a look at a careful write-up of a proof given by Jonathan Libgober, written as a part of his REU:

Euler characteristc, Poincare-Hopf theorem and applications.

With all the details, the proof takes about 16 pages, too long to be reproduced in my answer. An outline is that one first develops Morse theory which connects gradient vector fields of Morse functions and cell decompositions of manifolds. Once the machinery is there, it is fairly easy to see that the index of such a gradient vector field on a manifold $M$ equals $\chi(M)$. Lastly, one argues that any two vector nondegenerate fields have the same index (for this, Guillemin and Pollack suffice).

Lastly, Peter May (at University of Chicago) had done an amazing job with his undergraduate students writing useful REU projects, such as the one quoted above.

Answer 3

One way of dealing with this is to make their argument rigorous, but follow the idea closely. This is a different approach than the one presented by Jonathan Libgober.

1. Construct a smooth function $\rho:[0,2] \to [0,1]$ which has positive derivative on $[0,1)$ and constant equal $1$ on $[1,2]$. For instance, extend the function $1-\exp\left(1-\frac{1}{(1-x)^2}\right) : [0,1]\to [0,1]$ by $1$ on $[1,2]$.
2. For the standard $n$-simplex $\Delta_n = \{ x_0 + \ldots + x_n = 1, x_i \geq 0\}$, consider the function $f_{\Delta_n} = \rho(\|x\|) : \Delta_n \to [0,1]$. It's smooth and its gradient vector field (tangent to $\Delta_n$, i.e., taken within the plane $x_0 + \ldots + x_n = 1$) indeed "looks like" the one drawn by Guillemin and Pollack.
3. Choose a triangulation of your manifold $M$ and define the function $f:M \to [0,1]$ by requiring that the restriction to each simplex $F_i:\Delta_n \to M$ of the triangulation pulls back to $f_{\Delta_n}$.
4. While it's not a smooth function, on each simplex the gradient $\nabla f|_{\mathrm{im} F_i} = \nabla (f_{\Delta_n} \circ F_i^{-1}) $ is well-defined, and agrees on the boundaries of the simplices, so you do get a continuous, piecewise-smooth vector field $\nabla f$.
5. Now there's a little technical but easy work to do: check that Poincare-Hopf (and, actually, the bulk of the book) works for continuous vector fields with isolated critical points (and/or $C^1$ maps between manifolds).
6. Once you have that, you need to show that indices of the critical points are as intended -- that the critical point on a $k$-simplex has index $(-1)^k$. While that seems obvious since each should "look like" the critical point of $-x_1^2 - \ldots - x_k^2 + x_{k+1}^2 + \ldots + x_n^2$, the "look like" part needs to be specified and proven, and I admit I have hard time coming up with an elegant proof. The key insight is that on each individual simplex, the statement is trivially true (in some "relative" sense) by direct computation, since $\nabla f_{\Delta_n}$ is proportional to $\nabla \|x\|^2$. We see that if $p$ is the critical point lying in the interior of a $k$-simplex $\iota:\Delta_k \hookrightarrow \Delta_n$, then on the sphere $S_\varepsilon \cap \iota(\Delta_k)$ the vector field is, up to appropriate scaling, the identity map. On the other hand, on the intersection of $S_\varepsilon$ with the plane perpendicular to $\iota(\Delta_k)$ and with the interior of $\Delta_n$, which is a sphere $S^{k-1} \subset S_\varepsilon \simeq S^{n-1}$, the vector field is - up to positive scaling - the antipodal map. Fragments of these spheres within various $n$-simplices $\Delta_n$ with $p$ on the boundary "patch up" to a $(n-k-1)$-dimensional (piecewise smooth) sphere within $S_\varepsilon$ on which the map is antipodal. Since the complement of $S^{k-1}$ in $S^{n-1}$ contracts to the perpendicular $S^{n-k-1}$ (think of $S^{n-1} \subset \mathbb{R}^n$; we actually have a homeomorphism $S^{n-1} \setminus S^{k-1} \simeq S^{n-k-1} \times \mathbb{R}^k$), algebraic topology tells us that the degree of the map $S^{n-1} \to S^{n-1}$ which is identity on $S^{k-1} \to S^{k-1}$ is exactly the degree of the map on the complement understood as a map between two things homotopy equivalent to $S^{n-k-1}$, which in this case is $(-1)^{n-k} = (-1)^n (-1)^k$, but I don't see a direct differential-topological argument that would prove that.
7. While it seems there's a sign error of $(-1)^n$, there's actually no error here -- Poincare duality does hold for closed manifolds. To prove that there's no error, just use $-f$ in place of $f$. Yes, odd-dimensional closed manifolds turn out to have zero Euler characteristic.

I do agree point (6) is not very clear, but I also do think there should be a shorter proof than what I sketched, and hopefully one that doesn't require using the notion of degree from algebraic topology.

As far as I understand, for (5), it's enough to (a) change the remainder in the Taylor formula $g(t) = g(0) + t g'(0) + t^2 r(t)$ to the Peano form, $g(t) = g(0) + t g'(0) + t R(t)$ where $R(t) \to 0$ as $t \to 0$, and (b) define the degree of a continuous (or: piecewise smooth) map $S^n \to S^n$, which can be done by e.g. smoothing the map and taking the smooth degree. Note that the local Lefschetz number was defined using $df-I$, so for that you only need $f$ to be $C^1$, indeed.