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Mar 25, 2023 at 23:52 comment added Malkoun If you have, say, $n$ vectors on $S^{2n-1} \subset \mathbb{C}^n$, and form the complex $n$ by $n$ matrix, say $A$, with columns these $n$ vectors, then by Hadamard's inequality, $|det(A)| \leq 1$ and equality is then attained when $A$ is unitary. If we instead assume the $n$ vectors lie on $S^{2n-1}_r$, of radius $r > 0$, then by Hadamard's inequality, $|det(A)| \leq r^n$, with equality attained if and only if $A$ is $r$ times a unitary matrix.
Mar 25, 2023 at 17:53 comment added Ben McKay Sorry, I meant to write about the special linear group, $SL_n\mathbb{R}$ or $SL_n\mathbb{C}$. Its elements are matrices of determinant one, but it is not bounded, so matrices with determinant one don't lie on a sphere, in any metric Lipschitz equivalent to the Euclidean one. On the other hand, $SU(2)$ is, as you say, the group of unit quaternions, so these have unit length. Similarly, each unitary group $U(n)$ is contained in a sphere in the usual Euclidean norm on $n\times n$ matrices. But these don't have determinant 1. So the determinant is not relevant.
Mar 25, 2023 at 17:48 comment added Matteo However for SU(2) (excuse me if I use this, but it's the one I'm more familiar with) the constrain gives us a^2 + b^2 + c^2 + d^2 =1 where a, b, c, d are the parameters of the matrix itself. So in SU(2) in particular (despite the name of the question being for SU(N)) the sphere in the parameter space should be limited and confined to radius = 1?
Mar 25, 2023 at 16:00 comment added Matteo If what you say applies for SU(N) (I need to read about it more), is it fixed for SU(2) or not?
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Apr 28, 2023 at 3:06
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S Mar 25, 2023 at 15:09 review First questions
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S Mar 25, 2023 at 15:09 history asked Matteo CC BY-SA 4.0