Question

$U(1)$ is diffeomorphic to $S^1$ and $SU(2)$ is to $S^3$, but apparently it is not true that $SU(3)$ is diffeomorphic to $S^8$ (more bellow). Since $SU(3)$ appears in the standard model I would like to understand its topology.

By one of the tables here $SU(3)$ is a compact, connected and simply connected 8-dimensional manifold. This MO post says that its $\pi_5$ is $\mathbb{Z}$ thus it can not be homeomorphic to $S^8$(e.g.: see this wiki article). Even if it was a homotopy sphere Poincaré conjecture would not be helpful (at least in the smooth category: there exists exotic 8-spheres, right?).

I guess that this is what the author of this question was trying to know...

Anyway, is it known any manifold diffeomorphic to $SU(3)$?

Answer 1

Apart from jokes, an answer which may satisfy you is the following: $SU(3)$ is a $S^3$-bundle over $S^5$. To see this just consider the defining representation of $SU(3)$ on $\mathbb{C}^3$; this induces a transitive action of $SU(3)$ on the unit sphere of $\mathbb{C}^3$, which is $S^5$. Since the stabilizer of a point for this action is $SU(2)$ this exhibits $SU(3)$ as an $SU(2)$-bundle over $S^5$, and as you wrote $SU(2)$ is diffeomophic to $S^3$. Now, the next question is: which $SU(2)$-bundle over $S^5$ is $SU(3)$? to answer this, recall that isomorphic classes of principal $SU(2)$-bundles over (a not too wild) topological space $X$ are in bijection with the set $[X,BSU(2)]$ of homotopy classes of maps from $X$ to the classifying space of $SU(2)$. So in the case at hand you are interested in $[S^5,BSU(2)]= \pi_5(BSU(2))= \pi_4(SU(2))= \pi_4(S^3)= \mathbb{Z}/2\mathbb{Z}$. So there are only two $S^3$-bundles over $S^5$, the trivial one and the nontrivial one: $SU(3)$ is the nontrivial one (otherwise one would have $\pi_4(SU(3))=\mathbb{Z}/2\mathbb{Z}$, which is not the case: it is $\pi_4(SU(3))={0}$).

Answer 2

There is not much that can be said about "is it known any manifold diffeomorphic to SU(3)?"...

However, $SU(3)$ is the total space of an $S^3$-fibration (i.e. fibre bundle with fibers $S^3$) over the five-dimensional sphere $S^5$. This comes from the fact that $S^5:=\{(z_1,z_2,z_3)\in \mathbb C^3 : |z_1|^2+|z_2|^2+|z_3|^2=1\}$ has a transitive action by $SU(3)$, and that the stabiliser of any point is isomorphic to $SU(2)$.

I hope this helps a bit.

Answer 3

I often find it more useful to say $SU(3)$ is a $T^2$ bundle over the manifold of flags in ${\mathbb C}^3$ (itself a ${\mathbb CP}^1$-bundle over ${\mathbb CP}^2$). Partly this is because $T^2$'s homotopy groups are easier than those of $S^3$ and $S^5$.

Answer 4

A lot of the properties of $SU(n)$ and $U(n)$ can be summarised in the "commutative diagramme" below, viewed as fibrations. In particular, the diffeomorphisms for $U(1)$ and $SU(2)$ to spheres falls out from it, but fails for higher dimensions. But you can still see various fibrations, as people above mentioned.

\begin{array}{ccccc} SU(n-1) & \to & U(n-1) & \to & S^1 \\ \downarrow & & \downarrow & & \downarrow\\ SU(n) & \to & U(n) & \to & S^1\\ \downarrow & & \downarrow & & \downarrow\\ S^{2n-1} & \to & S^{2n-1} & \to & {*} \end{array}

Answer 5

Take the complete flag variety $B$ of $\mathbb{C}^3$ (consisting of pairs $L\subset P$, where $L$ is a line and $P$ is a plane through the origin): so $B$ is a 3-dimensional complex projective manifold (or 6-dimensional real). To each flag $L\subset P$, associate the set of orthonormal frames, consisting of one unit vector in $L$ and one unit vector in the orthogonal of $L$ in $P$; get in this way the orthonormal frame bundle $E$, a bundle in 2-dimensional real tori over $B$. Then $SU(3)$ is equivariantly diffeomorphic to $E$.

Answer 6

The group $SU(3)$ acts transitively on $S^5$, unit vectors in $\mathbb{C}^3$. The stabiliser of a point is $SU(2)$. This shows $SU(3)$ is the total space of a fibre bundle with base $S^5$ and fibre $S^3$.