Holomorphic Lefschetz formula and basic linear algebra.

Question

There is a well-known way to prove the fact that the field of complex numbers is algebraically closed using Lefschetz fixed point theorem. Let me recall the idea:

1. The existence of a root for any polynomial with complex coefficients is equivalent to the existence of an eigenvector for any endomoprhism of a finite-dimensional vector space over $\mathbb{C}$. Indeed, any polynomial is a characteristic polynomial of certain matrix.

2. Assume, that $V$ is a finite-dimensional complex vector space and $A \colon V \to V$ is an endomorphism without eigenvectors. In particular, $A$ has trivial kernel, so it induces an automorphism of the projectivization: $$\overline{A} \colon \mathbb{P}(V) \to \mathbb{P}(V).$$

3. The group $\mathrm{PGL}(n, \mathbb{C})$ is connected (one doesn't need the main theorem of algebra to prove this), so $\overline{A}$ is homotopic to identity as a diffeomorphism of $\mathbb{P}(V)$. Thus its Lefschetz number equals Euler characteristic of complex projective space, which is non-zero. By Lefschetz fixed point theorem, this implies that $\overline{A}$ has a fixed point.

4. If $l$ is a line fixed by $\overline{A}$, then any vector $v \in l$ is eigen for $A$.

Now, my question is the following: what if we apply the holomorphic Lefschetz theorem to a linear automorphism of a projective space? In this case it is easy to compute the right-hand-side: since the only non-tirivial Dolbeaut cohomology group of projective space is $H^{0,0}(\mathbb{CP}^n)$,and $\overline{A}$ acts on it trivial, the holomorphic Lefschetz number equals $1$.

What is the left-hand-side in this case? It seems to be certain algebraic expression in coefficients of $A$. However, after thinking about it for some time I am not able to write it down explicitly.

Answer 1

I did not know the proof of $\mathbb{C}$ algebraically closed via Lefschetz. But it does not seem quite complete to me. One would need to argue that the fixed points are isolated and nondegenerate, which in general they are not. If you run the same argument over another field, it would only show that there is a cycle of points of degree$$e(\mathbb{P}^n)=n+1$$which is Galois invariant. But maybe this can be fixed with a little more work.

Concerning your question, I just did the following example. Suppose $n=1$, i.e. $A$ a $2\times2$ matrix, and that $A$ has two different eigenvalues $a_1$, $a_2$. Assume we have diagonalized $A$ already. Then the two fixed points are $[1:0]$ and $[0:1]$. The differential at the two points are the numbers $a_1/a_2$ and $a_2/a_1$, respectively. And indeed $${1\over{{1-{{a_1}\over{a_2}}}}} +{1\over{{1-{{a_2}\over{a_1}}}}} = 1.$$ Something similar will hold in higher dimensions. The assumption on $A$ being diagonalizable is necessary to apply Lefschetz.

Talking about the second part of your answer -- yes, I also managed to do this computation by myself and obtained the same result. But it intrigues me what is the generalisations of this identity. I guess there must be something about subdeterminants...

Take a look at the following paper by Atiyah and Bott, A Lefschetz fixed point formula for elliptic complexes: II. Applications, Annals of Maths (2) 88, 1968, 451-491.

https://www.jstor.org/stable/1970721

Look at section 4, particularly examples 2 and 3.