Another question about unitary and anti-unitary matrices

Question

This question is related to the previous post, "A question about unitary and anti-unitary matrices". Following the suggestion of Lspice, I am posting it as a separate question, as it might be of independent interest.

Let $\cal{U}(n)$ and $\tilde{\cal{U}}(n)$ denote the subspaces of $\mathbf{C}^{n\times n}$ formed by the unitary and the anti-unitary matrices, respectively.

The question is: how separated are $\cal{U}(n)$ and $\tilde{\cal{U}}(n)$, when seen as subspaces of $O(2n)$?

More precisely, what is the value of

$d({\cal U}(n),\tilde{\cal U}(n))=\inf∥U-\tilde{U}∥$,

where the infimum is taken over $U\in \cal{U}(n)$ and $\tilde U\in \tilde{\cal{U}}(n)$, denoting by $∥U∥$ the operator norm of $U$ for the euclidean norm in $R^{2n}$?

It is shown below that $d({\cal U}(n),\tilde{\cal U}(n))\ge \sqrt{2}$. It would be interesting to know if an exact determination is known or can be obtained, and how it depends on $n$.

Answer 1

Here is a simple argument showing that with the above notation, we have $d(\cal{U}(n),\tilde{\cal{U}}(n))\ge \sqrt{2}$.

Let $U$ be a unitary operator, and $\tilde U$ an antiunitary operator in $\mathbf{C}^n$, viewed as elements of $O(2n)$. We note that, by definition of the operator norm $∥ ∥$, we have

$∥U−\tilde{U}∥=\sup_{∥z∥=1} ∥Uz−\tilde{U}z∥=\sup_{∥z∥=1} ∥z−U^{−1}\tilde{U}z∥$.

The operator $U^{−1}\tilde{U}$ is antiunitary, hence of the form $z↦U_0\bar{z}$, where $U_0$ is some unitary operator in $\mathbf{C}^n$.

Hence we must show that for any such $U_0$ we have $\sup_{∥z∥=1} ∥z−U_0\bar{z}∥^2≥2$. Now $U_0$ is diagonalizable, and its eigenvalues have unit modulus. Let $u_0$ be any normalized eigenvector, and let $e^{iϕ_0}$ be its eigenvalue. Setting $z=e^{iα}\bar{u}_0$, we have $∥z−U_0\bar{z}∥^2=∥e^{iα}\bar{u}_0−e^{i(ϕ0−α)}u_0∥^2= ∥\bar{u}_0−e^{i(ϕ0−2α)}u_0∥^2$ $=2−2\Re(e^{i(ϕ_0−2α)}(\bar{u}_0|u_0)).$

It follows that $∥U−\tilde{U}∥^2\ge\sup_{α∈[0,2π]}\big(2−2\Re(e^{iα}(\bar{u}_0|u_0))\big)\ge 2$, Q.E.D.

Possibly one could do better (for example when $n=1$ one finds easily that always $∥U−\tilde{U}∥=2$).