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Consider the group $U(2n)$ of unitary matrices. This has two standard and important homomorphic involutions. The most famous one $A \mapsto \overline A$ is complex conjugation, with fixed-set $O(2n)$, the set of real orthogonal matrices. But the one I'm most interested in here is conjugation-by-j: if one considers $\Bbb H^n = (\Bbb C \oplus j \Bbb C)^n = \Bbb R^{4n} $ then the matrix representing right-multiplication-by-j is the $4n \times 4n$ block-diagonal orthogonal matrix, given by summing n copies of $$\begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}.$$

Now conjugation by $j$ induces an outer automorphism of $U(2n)$ of order $2$ whose fixed-set is $\text{Sp}(n)$, the compact group of quaternionic-linear unitary matrices.

It is well-known that unitary groups $U(2n)$ have perfect Morse functions (and I believe $\text{Sp}(n)$ does as well, but I didn't check). I'm interested in the $\Bbb Z/2$-equivariant Morse theory of $U(2n)$, and for computations, it would be nice to have as small a complex as possible.

Question. What is the minimal number of $\Bbb Z/2$-critical orbits for a $j$-invariant Morse function on $U(2n)$? What techniques are available to prove that you've found a Morse function with as few critical orbits as possible?

For my own work I am more immediately interested in the case $n = 1$, where a more concrete formulation can be given: $U(2)$ is equivariantly diffeomorphic to $S^1 \times S^3$ with involution $\iota(\lambda, z, w) = (\bar \lambda, \lambda z, \lambda w)$. One can cook up an invariant Morse function with two critical points on $1 \times S^3$ and four critical orbits on $-1 \times S^3$, but I don't know if this is minimal.

I believe that the sum of the ranks of Bredon equivariant homology (with suitable coefficient systems: I think one needs to assume that $A(G/H)$ is rank 1 for all orbits) gives a lower bound for the number of critical orbits of an invariant Morse function. Based on my calculations, I don't think this is sufficient to show that my '6' is minimal in this case (I only seem to be able to prove a lower bound of $4$), but it's very possible I made some calculation errors.

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  • $\begingroup$ The equivariant Lusternik-Schnirelmann category gives a lower bound on the number of critical orbits. Have you checked out the literature on that topic? $\endgroup$
    – Mark Grant
    Commented Aug 16, 2021 at 21:09
  • $\begingroup$ @MarkGrant I expect this invariant will probably be too weak, since actually it gives a lower bound on the number of critical orbits of any invariant smooth function, and usually one can get away with fewer if you allow for degenerate critical orbits. I really need my functions to be Morse (actually, Morse-Smale, but that's a harder story and probably out of the scope of this question). $\endgroup$
    – mme
    Commented Aug 16, 2021 at 21:14
  • $\begingroup$ Ok. There is also some work of Austin and Braam on equivariant Morse theory. I think the idea was that an equivariant Morse function on M gives a Morse-Bott function in the homotopy orbit space of M $\endgroup$
    – Mark Grant
    Commented Aug 17, 2021 at 7:23
  • $\begingroup$ That is correct but it is mainly useful for compact Lie groups of positive dimension. Their theory recovers Borel cohomology with real coefficients, and for finite groups this is not very interesting: it's $H^*(X)^G$, the fixed set in cohomology of the group action. For U(2) with involution above this gives $H^*(S^3)$ (conjugation by j negates H^1). So I don't think one can get a better bound out of Borel theory than 2 critical orbits! $\endgroup$
    – mme
    Commented Aug 17, 2021 at 12:52
  • $\begingroup$ I see. Some searching also threw up the paper sciencedirect.com/science/article/pii/S0166864101001997, which you've probably found already. $\endgroup$
    – Mark Grant
    Commented Aug 17, 2021 at 13:06

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