Is ${\rm SO}(n, \mathbb{C})$ a bounded subset of $\mathbb{C}^{n\times n}$?

Question

Reason I ask is because although it's bounded for real matrix entries, it's harder to see whether it's bounded for complex entries.

Answer 1

No, it is a non-empty affine variety in $\mathbb C^{n^2}$, and so is unbounded.

Of course, one can see this explicitly in this case. For example, we can change basis so that the quadratic forms is $x_1x_2 + x_3x_4 + \cdots + x_{2m-1}x_{2m}$ (if $n = 2m$ is even) or $x_1x_2 + x_3x_4 + \cdots + x_{2m-1}x_{2m} + x_{2m+1}^2$, if $n = 2m+1$ is odd.

In either case, $SO(n)$ contains a copy of $(\mathbb C^{\times})^m,$ acting via $(\lambda_1, \ldots,\lambda_m) \cdot (x_1,\ldots, x_n) = (\lambda_1 x_1, \lambda_1^{-1} x_2, \ldots, \lambda_m x_{2m -1} ,\lambda_m^{-1} x_{2m}, x_{2m+1})$ (omit the last entry if $n = 2m$ is even); so $(\mathbb C^{\times})^m$ is contained in (in fact equal to) the intersection of $SO(n)$ with the diagonal matrices in $M_n(\mathbb C)$.

Since $\mathbb C^{\times}$ is unbounded, so is $SO(n)$.