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I'm wondering if the fact $SU(2)$ group elements have $det = 1$ is connected with the radius of the unitary $S^{3}$ manifold associated.

The context is demonstration of dU being an Haar invariant measure, where U and V are two elements of $SU(2)$. I knew we can write \begin{equation} dU=\frac{1}{\pi^{2}} d^{4}x \delta(x^{2}-1) \end{equation} Where $x^{2}=x_{0}^{2}+x_{1}^{2} +x_{2}^{2}+x_{3}^{2}$

I'm wondering the reason of that constrain, meaning the radius of the sphere being unitary. If that's cause $det=1$ than I can use the properties of determinant to prove it, being: $det(UV) = det(U) det(V) = 1*1 =1$.

Thus I can rewrite $d(UV)$ the exact same way, where the factor $\frac{1}{\pi^{2}}$ is to make unitary the volume of the sphere too.

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  • $\begingroup$ If what you say applies for SU(N) (I need to read about it more), is it fixed for SU(2) or not? $\endgroup$
    – Matteo
    Commented Mar 25, 2023 at 16:00
  • $\begingroup$ 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? $\endgroup$
    – Matteo
    Commented Mar 25, 2023 at 17:48
  • $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Commented Mar 25, 2023 at 17:53
  • $\begingroup$ 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. $\endgroup$
    – Malkoun
    Commented Mar 25, 2023 at 23:52

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