Is $SU(3)$ homeomorphic to $S^n$ for any value of $n$? I was wondering whether any of the $n$-sphere being homeomorphic to $SU(3)$. The motivation for this naive thinking is that $SU(2)$ being homeomorphic to $S^3$.
In general is there any general relation between $SU(m)$ and $S^n$
 A: 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 $1 + t^n$.
