I'm studying Robert Ghrist papers on integration against Euler Characteristic. I am particularly interested in the relation with Morse Theory. I am trying to understand the proof of Theorem 25.1 (page 52) of this preprint, which I reproduce in an extended version (the parts I understand):

Theorem. Suppose $h\colon M\to \mathbb{R}$ is a function integrable against Euler characteristic and a Morse function on the closed (compact without boundary) manifold $M$. Then:

$$\int_M [h] \: d \chi=\sum_{p\in Cr(h)}(-1)^{n-\mu(p)}h(p)$$

where $Cr(h)$ is the set of critical points of $h$.

Remark. We follow the usual conventions of Milnor's book in Morse theory.

The proof as I understand it. First we use a previous result (Proposition 24.8 of the same paper) which says that:

$$\int_M [h] \: d \chi=\int_{s=0}^{\infty}\chi\{h\ge s\}-\chi\{h<-s\}\: ds.$$

Now, using Theorem 3.2 from Milnor's book I know that $\chi\{h\le s\}$ is piecewise constant and only changes at critical values of $h$. Moreover, the change is due to the addition of a cell of the dimension of the Morse index of that critical value. So, the change in the Euler Characteristic at a critical value $s=h(p)$ is: $$\chi\{h\le s + \epsilon\}-\chi\{h\le s - \epsilon\}=(-1)^{\mu(p)}.$$

At this point I get lost.

So I insert as a picture the proof given in the paper, and a similar proof of the same result in another preprint(page 8): Proof in the first paper

Gap I don't catch in the second paper

Any help would be appreciated! By the way, there is a proof which involves Stratified Morse Theory, but I am interested in this one using elementary methods.


1 Answer 1


I glanced at the paper briefly. Let me try to explain what I understand. Let us suppose that M is compact and without boundary. Let $f: M \to \mathbb{R}$ be a Morse function. Let us further suppose for simplicity that $f$ is positive on $M$, that is it only takes positive values. We could always translate $f$ to make this the case, since $M$ is compact.

In that case $\{h< -s\} = \emptyset$ for all non-negative $s$. So the Euler characteristic of that portion is always zero, so by prop 24.8 we then have

$$\int_m h [d \chi] = \int_{s=0}^\infty \chi(h \geq s) ds$$

Let $p_i$ be the critical points of the Morse function. And assume for simplicity that these take distinct critical values $h(p_i)$. What does $\chi(h \geq s)$ look like? It starts at $s=0$ with the Euler characteristic of the manifold $\chi(M)$. Then it is locally constant with jumps at exactly the critical values $s = h(p_i)$. These jumps are by one unit up or one unit down according to the index via $(-1)^{\mu(p_i) + 1}$. The extra factor of $-1$ is because we are taking away the handle as $s$ increases.

In other words we can write this as

$$ \chi(h \geq s) = \chi(M) + \sum_{i, h(p_i) < s} (-1)^{\mu(p_i + 1)}g_{[h(p_i), \infty)} $$

where $g_{[h(p_i), \infty)}$ is the "jump function" which takes value one on the interval $[h(p_i),\infty)$ and zero otherwise.

It is useful to fix a large $T > \max\{h(p_i)\}$ and consider the definite integral, which we can compute:

$$\int_{s=0}^T \chi(h \geq s) ds = T \cdot \chi(M) - \sum_i (-1)^{\mu(p_i)} \int_{s=0}^T g_{[h(p_i), \infty)} ds $$ $$= T \cdot \chi(M) -\sum_i (-1)^{\mu(p_i)} (T - h(p_i))$$ $$ =T \cdot \chi(M) - T \cdot (\sum_i (-1)^{\mu(p_i)} ) + \sum_i (-1)^{\mu(p_i)} h(p_i) $$ The first two terms cancel and this gives the second claimed formula 25.2. The first one is the same argument with the function $h$ flipped over.

  • $\begingroup$ First of all, thank you very much for your time. Could you elaborate a bit more on flipping over the function $h$, please? It seems to me that if you flip the function, then you are not proving the result for your arbitrary chosen function $h$. Thanks in advance! And, I have thought about this and, It seems we are arriving to the results exchanged, I mean, with the expression for lower integral we arrive to the result for the upper one and the other way around. Am I right? $\endgroup$
    – D1811994
    Jun 4, 2017 at 0:06

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