show a map of complex projective space is lefschetz

Question

This is a problem from a qualifying exam. Let $A \in GL_{n+1}(\mathbb{C})$. Then $A$ defines a smooth map on $\mathbb{CP}^n$ by $A \cdot [z] = [Az]$ for $[z] \in \mathbb{CP}^n$. We will denote this map by $\tilde{A}$.

Fixed points of $\tilde{A}$ correspond to eigenvectors of $A$ for the following reason. Let $z \in \mathbb{C}^{n+1}$ be an eigenvector of $A$, so that $Az = \lambda z$ for some nonzero $\lambda \in \mathbb{C}$. Then we have that $\tilde{A}[z] = [Az] = [\lambda z] = [z]$. Conversely, if $[z]$ is a fixed point of $\tilde{A}$, then $\tilde{A}[z] = [Az] = [z]$, which means that $Az = \lambda z$ for some nonzero $\lambda \in \mathbb{C}$.

I wish to show that if the eigenvalues of $A$ all have multiplicity 1, then $\tilde{A}$ is Lefschetz map. Recall that $\tilde{A}$ is a Lefschetz map if, for all fixed points $[z]$ of $\tilde{A}$, we have that $d\tilde{A}_{[z]}$ does not have $+1$ as an eigenvalue. This is the infinitesimal analog of the demand that $[z]$ be an isolated fixed point. Intuitively, this makes sense, because $A$ having distinct eigenvalues means that $\mathbb{C}^{n+1}$ has a basis of distinct eigenvectors, and so fixed points of $\tilde{A}$ ought to be separated.

My trouble comes from not knowing the correct way to compute the differential. We have that $T_{[z]}\mathbb{CP}^n \cong \mathbb{C}^n$, but $A$ is map of $\mathbb{C}^{n+1}$, and so computing something like

$$ \lim_{h \to 0} \frac{A(z + hv) - A(z)}{h} $$

doesn't make any sense in $T_{[z]}\mathbb{CP}^n$. If $z_1,\ldots,z_{n+1}$ are the eigenvectors for $A$, then I think that $z_1,\ldots,z_{i-1},z_{i+1},\ldots,z_{n+1}$ should be a basis that can be used for $T_{[z_i]}\mathbb{CP}^n$, but this is more of a hunch than anything I've been able to make rigorous or useful.

Answer 1

In addition to Ted's excellent answer (which I, unfortunately, cannot upvote due to my daily vote limit being transgressed), if you like to think in terms of local coordinates on $\mathbb{CP}^{n}$, then note that $\mathbb{CP}^n$ admits an affine open covering $\mathbb{CP}^n=\bigcup_{i=0}^{n} U_i$ where $U_i=\{[z_0,\dots,z_n]\in \mathbb{CP}^n:z_i\neq 0\}$ ($[z_0,\dots,z_n]$ are the usual homogeneous coordinates on $\mathbb{CP}^n$). Local coordinates on $U_i$ are, e.g., given by $[z_0,\dots,z_n]\to (\frac{z_0}{z_i},\frac{z_1}{z_i},\dots,\widehat{\frac{z_i}{z_i}},\dots,\frac{z_n}{z_i})$; $\widehat{}$ indicates that the coordinate in question is omitted. You can now do explicit calculations in these local coordinates if you wish!

I hope this helps!

Answer 2

Here's a hint: You can think of $T_{[z]}\mathbb CP^n$ as $\mathbb C^{n+1}/\text{Span}(z)$. Thus, to differentiate a map $f$ to projective space $\mathbb CP^n$, thinking of a lifted map $\tilde f$ to $\mathbb C^{n+1}$, you want to compute $d\tilde f_p \pmod{f(p)}$. (You can check this is all well-defined independent of the lift.)