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 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.

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.