9

I want to find a compact Riemannian manifold $M$ such that the group $SU(n)/\mathbb{Z}_n$ is the (orientation-preserving) isometry group of $M$. It would be great if you can provide an explicit construction of $M$, $\forall n$. If the question $\forall n$ is too broad, I am mainly interested in the low-dimensional cases of $n=3,4,5,6$. I am aware that for $n=2$, $SU(2)/\mathbb Z_2\cong SO(3)$, which is the (orientation-preserving) isometry group of $S^2$.

The existence of above space $M$ is guaranteed as follows: $SU(n)$ is a compact topological group, and $\mathbb Z_n$ is the center of $SU(n)$, so $SU(n)/\mathbb Z_n$ is a well-defined quotient group. Quotient map preserves compactness. So $SU(n)/\mathbb Z_n$ is a compact group. Finally, this MathOverflow answer states that "every compact group is the full isometry group of a compact Riemannian manifold."

P.S. For you to gauge the level of details of answer: I am a physicist by training and I am not too well-versed in Riemannian geometry (just the basics).

And it would be great if you can provide relevant references.

Leo L.
  • 358
  • 1
  • 9
  • 1
    $SO(3)$ isn't the full isometry group of $S^2$, since it misses all the orientation reversing isometries. Are you likewise interested in cases where $SU(n)/\mathbb{Z}_n$ is not the full isometry group? – Jason DeVito - on hiatus Jun 08 '21 at 02:36
  • @JasonDeVito Yes, I should have added I would like ()/ℤ to be an orientation-preserving isometry group. – Leo L. Jun 08 '21 at 02:58
  • Are you familiar with homogeneous spaces? – AHusain Jun 08 '21 at 03:16
  • @AHusain After reading the Wiki article on it, I understand it's a more general notion where the group $G$ acts on the topological space $X$ transitively. But I don't know properties/theorems about homogeneous spaces. – Leo L. Jun 08 '21 at 03:26
  • 4
    I won't have time to write a full answer until tomorrow, but for $n$ even, the manifold $\mathbb{C}P^{n-1}$ with its Fubini-Study metric is an example of what you want. When $n$ is odd, $SU(n)/\mathbb{Z}_n$ is the identity component of the isometry group, but there is another component of orientation preserving isometries. – Jason DeVito - on hiatus Jun 08 '21 at 03:44

1 Answers1

4

For $G = SU(n)/Z_m$ and $M = \mathbb{C}P^{n-1}$ with its usual Fubini-Study metric, it turns out that $Iso(M)$ contains two connected components with identity component being isomorphic to $G$.

The action is really easy to write down: writing a point $[z_1:...:z_n]\in \mathbb{C}P^{n-1}$ as $\begin{bmatrix} z_1\\ \vdots \\ z_n\end{bmatrix}$, then for any $A\in SU(n)$, we have $A\ast \begin{bmatrix} z_1\\ \vdots \\ z_n\end{bmatrix} = \left[A\begin{pmatrix} z_1\\ \vdots \\ z_n\end{pmatrix}\right].$

The other component of $Iso(M)$ is generated by complex conjugation: $[z_1:...:z_n]\mapsto [\overline{z}_1:...:\overline{z}_n]$. When $n$ is even, this map reverses orientation, while if $n$ is odd, then it preserves orientation.

Thus, when $n$ is even, $\mathbb{C}P^{n-1}$ is an example of what you're looking for: the group of orientation preserving isometries is isomorphic to $G$.

Since you asked for references, I wanted to point out that the answer to this MSE question lists sources where you can find the computation of $Iso(M)$, because $\mathbb{C}P^{n-1}$ with Fubini-Study metric is a symmetric space.

Jason DeVito - on hiatus
  • 51,620