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Let $A$ be an $n \times n$ invertible complex matrix. Let $Gr(k)=Gr(k,\mathbb{C}^n)$ be the complex $k$-Grassmannian, $1\leq k \leq n$. Since $A$ is invertible, it maps a $k$-dimensional subspace to a $k$-dimensional subspace, so it gives a function (which I'll call $A_k$) on $Gr(k)$. The fact that matrices have eigenvalues lets us deduce that $A$ has a $k$-dimensional invariant subspace - i.e. this map $A_k$ has a fixed point.

My question is this: is there a fixed point theorem on $Gr(k)$ (which ideally does not in any way depend on the existence of eigenvalues) that we can invoke to arrive at the same conclusion?

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I think the Lefschetz fixed point theorem still applies. If a self-map M→M of a compact orientable manifold M has no fixed points than the Euler characteristic of M is zero. But if M is a complex Grassmannian then its odd Betti numbers vanish so the Euler characteristic is positive (since b_0=1).

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    $\begingroup$ This is slightly broken as written -- you need $M$ to be homotopic to the identity. Otherwise, consider the $2$-sphere (Euler characteristic $2$) and the antipodal map. But you are fine in this case, since $GL_n(\mathbb{C})$ is connected, $\bigwedge^k M$ is homotopic to $\bigwedge^k \text{Id}$. $\endgroup$
    – David E Speyer
    Commented Feb 3, 2023 at 15:37
  • $\begingroup$ @DavidESpeyer not sure what it means for M to be homotopic to the identity; M is the manifold, not the map. In general the Euler characteristic is a sum over fixed points (assuming that these are isolated), but if there are no fixed points then the sum is automatically zero. $\endgroup$
    – Noam D. Elkies
    Commented Feb 3, 2023 at 16:59
  • $\begingroup$ Sorry, the self map should be homotopic to the identity. $\endgroup$
    – David E Speyer
    Commented Feb 3, 2023 at 19:07
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    $\begingroup$ See also mathoverflow.net/a/24989/1106 $\endgroup$
    – Steven Gubkin
    Commented Feb 3, 2023 at 23:27

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