The question is the following: Let $U:\bf{C}^n\to \bf{C}^n$ be a unitary operator; let $\tilde{U}:\bf{C}^n\to\bf{C}^n$ 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 $\bf{R}^{2n}\to\bf{R}^{2n}$, 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 $det_{\bf{R}}(U)=1$ and $det_{\bf{R}}(\tilde{U})=-1$, 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?

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?