Set of all orthogonal matrices over $\mathbb C$ is compact/not

Question

How to show the fact that the set of all orthogonal matrices over $\mathbb C$ is compact

By an orthogonal matrix over $\mathbb C$ I mean a matrix $A$ satisfying $AA^T=I$ and here $A^T=(a_{ji})$ where $A=(a_{ij})$

It is not the same as unitary matrix where in unitary matrix we take transpose and then conjugate or vice versa

I know that set of all orthogonal matrices over $\mathbb R$ is compact.

I think the closedness of the set will follow from the same arguements as in the above case. But the boundedness part not sure

Answer 1

It is not bounded, unless you are working with $1\times 1$ matrices.

The only complex, $1\times 1$ orthogonal matrix are $(1)$ and $(-1)$.

In $2\times 2$ (and by extension you get $n\times n$) you can consider the family, where $\lambda \geq 1$ is a real parameter, $$ A_\lambda = \begin{pmatrix} \lambda & i \sqrt{\lambda^2 - 1} \\ - i \sqrt{\lambda^2 - 1} & \lambda \end{pmatrix} $$ which you can easily check to be orthogonal. But any matrix norm of $A_\lambda$ would tell you it is of size $\approx \lambda$.