A question about unitary and anti-unitary matrices

Question

The question is the following: Let $U:\mathbf{C}^n\to \mathbf{C}^n$ be a unitary operator; let $\tilde{U}:\mathbf{C}^n\to\mathbf{C}^n$ be an antiunitary operator. Can one deform $U$ to $\tilde{U}$ within $O(2n)$? That is, seeing $U$ and $\tilde{U}$ as two orthogonal operators $\mathbf{R}^{2n}\to\mathbf{R}^{2n}$, can one find a continuous path $u:[0,1]\to O(2n)$ such that $u(0)=U$ and $u(1)=\tilde{U}$?

According to a remark read in Weinberg's treatise on quantum field theory the answer is no. This is clear if $n$ is odd, for (unless I am mistaken) we have then $\det_{\mathbf{R}}(U)=1$ and $\det_{\mathbf{R}}(\tilde{U})=-1$, so $U$ and $\tilde{U}$ belong to distinct connected components of $O(2n)$. But the argument does not work for $n$ even. Is there a simple alternative argument in that case?

Answer 1

$\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$Well, it seems that the claim is false if $n$ is even: In that case both $U$ and $\tilde{U}$ belong to $\operatorname{SO}(2n)$, which is path connected.

In fact I think I understand better what the remark made by Weinberg in his book really meant: The remark was that a quantum symetry that can be continuously changed into the identity must come from a unitary operator (and not an antiunitary).

Here a 'symmetry' is an automorphism $r$ of projective unitary space $P(\mathbf{C}^n)$, that is a bijective map which preserves the 'projective' inner product $(\ell,\ell')=\abs{(z,z')}/\norm z\norm{z'}$, $z\in \ell, z'\in \ell'$. By a theorem of Wigner any such automorphism comes from a unitary or an antiunitary.

The point is that the continuous path $u:[0,1]\to O(2n)$ that may exist if $n$ is even will always contain operators (for some $t\in \mathopen]0,1\mathclose[$) that are neither unitary nor antiunitary, hence that do not come from any symmetry in the sense of quantum mechanics.

Another question that comes to mind is how separated is the space of unitary operators from the space of antiunitary operators in $O(2n)$. In other words, if $U$ and $\tilde{U}$ are as above, what is the smallest possible value of $\norm{U-\tilde{U}}$ where $\norm U$ denotes the usual operator norm of $U$ for the euclidean norm in $\mathbf{R}^{2n}$?

Answer 2

$\newcommand\norm[1]{\lVert#1\rVert}$Here is a simple argument showing that, with the above notations, if $U$ is a unitary operator, and $\tilde{U}$ an antiunitary operator in $\mathbf{C}^n$, one always has $\norm{U-\tilde{U}}\ge \sqrt{2}$.

We note that, by definition of the operator norm we have $\norm{U-\tilde{U}}=\sup_{\norm z=1} \norm{Uz-\tilde{U}z}=\sup_{\norm z=1}\norm{z-U^{-1}\tilde{U}z}$. The operator $U^{-1}\tilde{U}$ is antiunitary, hence of the form $z\mapsto U_0\bar{z}$, where $U_0$ is some unitary operator in $\mathbf{C}^n$. Hence we must show that for any such $U_0$ we have $\sup_{\norm z=1}\norm{z-U_0\bar{z}}^2\ge 2$. Now $U_0$ is diagonalizable, and its eigenvalues have unit modulus. Let $u_0$ be any normalized eigenvector, and let $e^{i\phi_0}$ be its eigenvalue. Setting $z=e^{i\alpha}\bar{u}_0$, we have $\norm{z-U_0\bar{z}}^2=\norm{e^{i\alpha}\bar{u}_0-e^{i(\phi_0-\alpha)}u_0}^2=\norm{\bar{u}_0-e^{i(\phi_0-2\alpha)}u_0}^2$ $=2-2\Re\{e^{i(\phi_0-2\alpha)}(\bar{u}_0|u_0)\}$.

It follows that $\norm{U-\tilde{U}}^2\ge \sup_{\alpha\in [0,2\pi]}(2-\Re\{e^{i\alpha}(\bar{u_0}|u_0)\})\ge 2$, Q.E.D.

I suspect one could do better (for example when $n=1$ one finds easily that always $\norm{U-\tilde{U}}=2$), but this is not the point here.