2

I’m currently studying Naive Lie Theory by John Stillwell.

As I understand the group $SO(n)$ is the group of rotations in $n$-dimensional space. I understand the set construction of both, but would like to know if there is an analogous geometric interpretation for $SU(n)$.

I (intuitively) assumed that $SU(n)$ would be the rotation group of $\mathbb{R}^{2n}$, but my professor informed me this was incorrect.

I read similar questions such as: Geometric & Intuitive Meaning of $SL(2,R)$, $SU(2)$, etc... & Representation Theory of Special Functions.

However, I am seeking an answer to the more general case.

Raiyan Chowdhury
  • 325
  • $SU(2)$ is geometrically the unit 3-sphere. For higher dimensions we have that the $(2n-1)$-spheres are quotients $SU(n)/SU(n-1)$. – Marius S.L. Sep 28 '20 at 11:12
  • @MariusS.L. I see, so $SU(n)$ on its own has no particular geometric meaning? Also, I'd like to understand the point you made. Where can I find a proof of this? – Raiyan Chowdhury Sep 28 '20 at 14:31
  • 2
    Your intuition isn't wrong, it's just incomplete. An element of $SU(n)$ is naturally a rotation of $\mathbb{R}^{2n}$, however it's not an arbitrary one. They are specifically those rotations which behave well with the additional structure on $\mathbb{C}^n$ given by multiplying by a complex number. – Nate Sep 28 '20 at 14:58
  • @Raiyan Chowdhury I have forgotten where I saw it, but I assume it is not so difficult if you consider the embedding. Another way is to write the sphere as quotient of special orthogonal groups, which are the double covers of $SU(n).$ Should follow with some diagram chasing then. – Marius S.L. Sep 28 '20 at 15:09
  • It’s the group of “complex rotations” (of determinant $1$). It has a geometric meaning, just one that might be unfamiliar. – Qiaochu Yuan Sep 28 '20 at 16:29
  • @Nate Interesting. What would make a rotation not behave “well”? Failure of associativity? Do they all behave well for $SU(1)$ and $SU(2)$? – Raiyan Chowdhury Sep 28 '20 at 18:58
  • 2
    The main thing is that a map being $\mathbb{C}$-linear is a stronger condition than just being $\mathbb{R}$-linear on the underlying real space. A complex linear map $A$ from $\mathbb{C}^n$ to itself (and in particular an element of $SU(n)$) needs to not only be $\mathbb{R}$-linear but it needs to satisfy $A(i v) = i Av$ for all vectors $v$. – Nate Sep 28 '20 at 20:36

0 Answers0