Residue computation for Hirzebruch-Riemann-Roch

Question

In Vakil's note (https://math.stanford.edu/~vakil/245/245class18.pdf), in order to prove GRR for $\mathbb{P}^n\to pt$ we need to show that the $x^{-1}$ coefficient of $$\frac{e^{nx}}{(1-e^{-x})^{m+1}}$$ equals $\binom{m+n}{m}$. He suggests to apply some residue calculus, that is $$\oint \frac{e^{nx}}{(1-e^{-x})^{m+1}}dx.$$ Applying a change of variable $y=1-e^{-x}$ we get $$\oint \frac{(1-y)^{-(n+1)}}{y^{m+1}}dy$$ if I am not mistaken. So I need to extract now the $m$th coefficient of the series $(\sum y^k)^{n+1}$. Here I am stuck, am I missing something?

Answer 1

As shown in this answer, $$ \binom{-n-1}{k}=(-1)^k\binom{n+k}{k}\tag1 $$ Therefore, $$ (1-y)^{-n-1}=\sum_{k=0}^\infty\binom{n+k}{k}y^k\tag2 $$ Thus, $$ \begin{align} \left[y^{-1}\right]\frac{(1-y)^{-n-1}}{y^{m+1}} &=\left[y^m\right](1-y)^{-n-1}\tag{3a}\\ &=\binom{n+m}{m}\tag{3b} \end{align} $$

Answer 2

As you said, you just need to multiply the series $\sum y^k$ $(n+1)$ times. There are many ways to show what coefficient you get. I present here the combinatorial one.

Note that when you form the product

$(1+y+y^2+ \ldots)(1+y+y^2+ \ldots) \ldots (1+y+y^2 + \ldots)$

You will get a lot of monomials $y^m$. How many times does one get the monomial $y^m$? one has to choose monomials $y^{k_0}$ in the first factor, $y^{k_1}$ in the second factor, $ \ldots , y^{k_n}$ in the last factor, such that $y^{k_0} \ldots y^{ k_n}=y^m$.

Therefore, the $m$-th coefficient is the number of nonnegative solutions to $k_0+ \ldots + k_n=m$. It is well known that this is equal to ${m+n\choose n}$. For example you can find a proof in this Wikipedia page.

Other ways of proving this are by induction in $n$, calculation the $m$-th derivative of $(\sum y^k)^{n+1}$ at $y=0$ via the Leibnitz rule or calculating the $m$-th derivative at $0$ of $(1-y)^{n+1}$... but I think this is the more straightforward one

Answer 3

Here is another way: Since our meromorphic function $f(y)=\frac{(1-y)^{-(n+1)}}{y^{m+1}}$ has pole of order $m+1$ at $y=0$, its residue at $0$ is

$$\text{Res}(f(y),0)=\frac{1}{2\pi i}\oint_{|y|=\varepsilon}\frac{(1-y)^{-(n+1)}}{y^{m+1}}dy=\frac{1}{m!}\frac{d^m}{dy^m}\big ((1-y)^{-(n+1)}\big )|_{y=0}.$$

We know the $m$-th derivative of $(1-y)^{-(n+1)}$ is

$$(n+1)\cdot(n+2)\cdot\cdots\cdot(n+m)\cdot(1-y)^{-(n+m+1)}.$$

So by setting $y=0$, we find our residue equals to

$$\frac{(n+1)(n+2)\cdots(n+m)}{m!}=\frac{(n+m)!}{m!n!}=\binom{m+n}{m}.$$