Homotopy type of SU(3)

Question

The quotient $SU(3)/SU(2)$ is a $5$-dimensional sphere, so $SU(3)$ is the total space of a bundle with base $S^5$ and fibre $S^3$. Presumably it is not homeomorphic neither to $S^3\times S^5$ nor $S^8$. What else can be said about it? Is it homeomorphic to something better known than $SU(3)$?

Answer 1

The cohomology of $SU(3)$ can be calulated using the Serre spectral sequence of the fibring $$S^3\rightarrow SU(3)\rightarrow S^5.$$ For dimensional reasons the only possible outcome is that $SU(3)$ has the same homology as $S^3\times S^5$. That is, $$H^*SU(3)\cong \Lambda(x_3,x_5)$$ is an exterior algebra on two class of degrees $3$ and $5$, respectively. This rules out $SU(3)$ from being homotopy equivalent to $S^8$, for instance.

Next, it's possible to distinguish $SU(3)$ from $S^3\times S^5$ by computing the action of the Steenrod algebra on $H^*(SU(3);\mathbb{Z}/2)$. In detail, one can show that $$Sq^2:H^3(SU(3);\mathbb{Z}/2)\rightarrow H^5(SU(3);\mathbb{Z}/2)$$ is nontrivial. Since all Steenrod squares vanish on $H^*(S^3\times S^5;\mathbb{Z}/2)$, we can conclude that $SU(3)$ is not homotopy equivalent to $S^3\times S^5$.

The end result is that $SU(3)$ is neither homotopy equivalent to $S^8$ nor $S^3\times S^5$ (and in particular cannot be homeomorphic to either of these spaces). Really $SU(3)$ is what one might call `something better known'.

Edit to answer the question in the comments: It is true that $SU(3)$ is a principal $SU(2)$-bundle over $S^5\cong SU(3)/SU(2)$. More generally, whenever $G$ is a Lie group and $H\leq G$ a closed subgroup, the projection $G\rightarrow G/H$ has the structure of a principal $H$-bundle. The details for this will be found in any text on principal bundles.

In any case, the bundle in question is classified by a map $\theta:S^5\rightarrow BSU(2)$. Here we have an exceptional isomorphism $SU(2)\cong S^3$, so $BSU(2)\simeq BS^3$ (and the latter can be identified with the infinite quaternionic projective space). Thus $\theta$ is an element in $\pi_5BS^3\cong\pi_4S^3\cong\mathbb{Z}/2$, which is generated by the suspension of the Hopf map.

Since $\pi_5BS^3$ has only two elements, there are only two isomorphism classes of $S^3$ bundles over $S^5$. One, of course, is the trivial bundle, while the other must be nontrivial. According to the above, $SU(3)$ is not even homotopy equivalent to $S^3\times S^5$, so in particular the bundle $$S^3\xrightarrow{i} SU(3)\xrightarrow{p} S^5$$ must be the unique non-trivial $S^3$ bundle over $S^5$. This is equivalent to saying that the classifying map $\theta:S^5\rightarrow BS^3$ is essential; it represents the generator of $\pi_5BS^3$.

We can access the homotopy class of $\theta$ in another way. Writing $S^5\cong \Sigma S^4$ we take the adjoint of $\theta$ and obtain a map $S^4\rightarrow \Omega BS^3$. Composing this with the homotopy equivalence $\Omega BS^3\simeq S^3$ we get a map $\varphi:S^4\rightarrow S^3$. Now consider the long-exact sequence of the bundle $$\dots\rightarrow\pi_5S^5\xrightarrow \Delta \pi_4S^3\xrightarrow{i_*}\pi_4SU_3\xrightarrow{p_*}0=\pi_4S^5$$ where $\Delta$ is a connecting map. We have $\pi_5S^5\cong\mathbb{Z}$, generated by a class $\iota_5$, which is represented by the identity on $S^5$. Then $\Delta$ takes $\iota_5$ to the map adjoint to the classifying map $\theta$. That is, $$\Delta(\iota_5)=\varphi$$ as per the above definition of $\varphi$.

Now we have another way to answer your original question in case we know already that the bundle is non-trivial viz. the classifying map $\theta$ is essential. In this case the adjonint $\varphi$ is also essential, and hence generates $\pi_4S^3$. From the exact sequence above we obtain $$\pi_4SU(3)=0,$$ which is not the same as $\pi_4(S^3\times S^5)\cong\mathbb{Z}/2$.