7
$\begingroup$

I would be very grateful for a reference to the following results (which are, I think, true, though I never saw it in the literature).

Let $G\subset GL(n,{\Bbb C})$ be $U(n)$, abd $A\in GL(2n,{\Bbb R})$ an endomorphism which satisfies $AGA^{-1}=G$. Then $A\in {\Bbb R}^* \times U(n)$.

Let $G\subset GL(n,{\Bbb C})$ be the group $Sp(n)$ of quaternionic Hermitian matrices, and $A\in GL(2n,{\Bbb R})$ an endomorphism which satisfies $AGA^{-1}=G$. Then $A\in {\Bbb R}^* \times Sp(n)\times Sp(1)$.

Many thanks in advance.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
13
$\begingroup$

First, let me fix a misunderstanding: $\mathrm{Sp}(n)$ does not sit in $\mathrm{GL}(n,\mathbb{C})$, but in $\mathrm{GL}(2n,\mathbb{C})$, so I'll assume that you mean, for the second part that $A$ lies in $\mathrm{GL}(2n,\mathbb{C})$.

These both follow immediately from the facts that all the automorphisms of $\mathrm{SU}(n)$ are either inner or conjugate-inner while the automorphisms of $\mathrm{Sp}(n)$ are are all inner.

It's a bit easier to deal with the $\mathrm{Sp}(n)$ case first since it has no outer automorphisms: If $A\in \mathrm{GL}(2n,\mathbb{C})$ satisfies $A\mathrm{Sp}(n)A^{-1}\subset \mathrm{Sp}(n)$, then consider the homomorphism $\phi:\mathrm{Sp}(n)\to \mathrm{Sp}(n)$ defined by $\phi(g) = AgA^{-1}$. This is a smooth, injective homomorphism, so it must be a smooth automorphism. Since every automorphism of $\mathrm{Sp}(n)$ is of the form $\phi(g) = hgh^{-1}$ for some $h\in\mathrm{Sp}(n)$, it follows that $AgA^{-1} = hgh^{-1}$, so $h^{-1}A$ lies in the commuting ring of $\mathrm{Sp}(n)$ (intersected with the $\mathbb{C}$-linear isomorphisms of $\mathbb{C}^{2n}$), and this is simply the nonzero complex multiples of the identity (since $\mathrm{Sp}(n)$ acts irreducibly on $\mathbb{C}^{2n}$). Thus, $A = \lambda h$ for some $h\in \mathrm{Sp}(n)$ and some nonzero complex scalar $\lambda$.

For the $\mathrm{U}(n)$ case, notice that the problem is equivalent to finding the conjugations that preserve the subgroup $\mathrm{SU}(n)$, so we might as well ask for the $A\in\mathrm{GL}(n,\mathbb{C})$ such that $A\mathrm{SU}(n)A^{-1}= \mathrm{SU}(n)$. Now, there is a slight complication, because for $n>2$, not all of the automorphisms of $\mathrm{SU}(n)$ are inner. For example, the automorphism $\psi(g) = \bar g$ is not inner. Instead, every automorphism is either of the form $\phi(g) = hgh^{-1}$ or $\phi(g) = h\bar gh^{-1}$ for some $h\in\mathrm{SU}(n)$. However, it's easy to see that, for $n>2$, there is no pair $(A,h)$ such that $AgA^{-1} = h\bar gh^{-1}$ for all $g\in\mathrm{SU}(n)$, so we only need to deal with the case $AgA^{-1} = h gh^{-1}$ with $h\in\mathrm{SU}(n)$ and $A\in \mathrm{GL}(n,\mathbb{C})$. Again, we find that $h^{-1}A\in\mathrm{GL}(n,\mathbb{C})$ must commute with all of the elements of $\mathrm{SU}(n)$, and this can happen only if $h^{-1}A$ is a multiple of the identity (again because of the irreducibility of the action of $\mathrm{SU}(n)$ on $\mathbb{C}^n$).

I think that the reason you haven't seen it in the literature is that it is such a direct consequence of well-known facts about Lie groups and representations.

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ "conjugate" means "complex conjugate" (I remained confused for a while about the purported meaning of "conjugate-inner.) $\endgroup$
    – YCor
    Commented Nov 30, 2018 at 18:18
  • $\begingroup$ That's reasonable, thanks. I was thinking of applying Howe duality, but it is too much. I misprinted the question, by the way, it shoud be $A\in GL(2n, {\Bbb R})$ - sorry. $\endgroup$
    – Misha Verbitsky
    Commented Dec 1, 2018 at 1:06
  • $\begingroup$ Of course, it is a question about isometries of a Kahler or a hyperkahler manifold, but you already guessed it I suppose. $\endgroup$
    – Misha Verbitsky
    Commented Dec 1, 2018 at 1:09
  • $\begingroup$ @MishaVerbitsky: Regarding your first comment, which I suppose was meant to refer to the $\mathrm{Sp}(n)$ case: For that, you should have asked for $A\in\mathrm{GL}(4n,\mathbb{R})$, not $A\in\mathrm{GL}(2n,\mathbb{R})$, since $\mathrm{Sp}(n)$ acts irreducibly on $\mathbb{H}^n=\mathbb{R}^{4n}$. In that case, what you find is that $h^{-1}A$ is an invertible element in the commuting ring over $\mathbb{R}$ (instead of $\mathbb{C}$). That commuting ring is indeed $\mathbb{H}$, since $\mathrm{Sp}(n)$ is the group of (right) $\mathbb{H}$-linear orthgonal transformations, thus verifying your claim. $\endgroup$
    – Robert Bryant
    Commented Dec 1, 2018 at 10:39
  • $\begingroup$ @MishaVerbitsky: Certainly, I knew that these facts were basic to the study of the isometry group of an irreducible Kähler or hyperKähler manifold, but I didn't guess that this is what motivated your question. I see that you have now edited your question to allow $A\in\mathrm{GL}(2n,\mathbb{R})$ for the $\mathrm{U}(n)$-case, which makes your proposed answer incorrect. In addition, you now have to allow $A$ to be conjugate-linear (i.e., there are two pieces to the group). For example, complex conjugation preserves the Fubini-Study metric on $\mathbb{CP}^n$. $\endgroup$
    – Robert Bryant
    Commented Dec 1, 2018 at 10:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .