Is ${\rm SU}(3)$ homeomorphic to $\mathbb S^n$ for any value of $n$?

Question

I was wondering whether any of the $n$-sphere being homeomorphic to ${\rm SU}(3)$. The motivation for this naive thinking is that ${\rm SU}(2)$ being homeomorphic to $\mathbb S^3$.

In general is there any general relation between ${\rm SU}(m)$ and $\mathbb S^n$

The dimension of SU(n) is n^2-1, so there is exactly one sphere it could be homeomorphic to. On the other hand, the only spheres which are Lie groups are S^1 and S^3. — Mariano Suárez-Álvarez, Feb 18 '23 at 06:49
There is a relation, though: the quotient SU(n)/SU(n-1) is the sphere of dimension 2n-1, as explained here. — Mariano Suárez-Álvarez, Feb 18 '23 at 06:54
isn't $S^2$ a lie group? i mean $S^2 \to SO(2)$ is there right? what about $S^5, S^7$ etc. are they too not lie groups? — Eden Zane, Feb 18 '23 at 06:58
No, it isn't. For example, Lie groups have nonzero tangent vector fields, and S^2 famously does not. SO(2) is S^1. — Mariano Suárez-Álvarez, Feb 18 '23 at 06:58
As I said, the only spheres which are Lie groups are those of dimension 1 and 3 (and zero, I guess) Google should find this result and several explanations, even in this site. There are many ways to see this, none of which is particularly easy — it is a rather non trivial theorem. — Mariano Suárez-Álvarez, Feb 18 '23 at 07:00
That this question is closed (and for the stateds reason) is simply absurd. — Mariano Suárez-Álvarez, Feb 19 '23 at 23:40

score 8 · Accepted Answer · answered Feb 18 '23 at 08:46

No, and there's an easier argument than what has been suggested in the comments. $SU(3)$ is $8$-dimensional, so if it were a sphere it could only be $S^8$. However, $S^8$ has Euler characteristic $2$, whereas any compact connected Lie group has Euler characteristic zero by the Poincare-Hopf theorem, because the group multiplication can be used to trivialize the tangent bundle and in particular produce a nonvanishing vector field.

$SU(m), m \ge 4$ is not homotopy equivalent to a sphere either, and there are a few different ways to see this, which are still easier than the argument that has been suggested in the comments. For example, it is possible to compute that every compact, simple, simply connected Lie group $G$ satisfies $H_3(G) \cong \pi_3(G) \cong \mathbb{Z}$, meaning it cannot be a sphere unless it is $3$-dimensional (which is the case iff $G \cong SU(2)$). It is also possible to compute more generally that the Poincare polynomial of $SU(m)$ is $(1 + t^3)(1 + t^5) \dots (1 + t^{2m-1})$, whereas the Poincare polynomial of the $n$-sphere is just $1 + t^n$.

Is ${\rm SU}(3)$ homeomorphic to $\mathbb S^n$ for any value of $n$?

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