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$U(1)$ 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 $SU(3)$?

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yes! $SU(3)$! joking, but really the question as it is formulated makes no sense. – domenico fiorenza Jul 2 '11 at 18:31
Copying what someone else said in the post you linked to, $SU(3)$ is diffeomorphic to $SU(3)$. But more seriously, it would depend on what you're looking for. One could ask if $SU(3)$ is diffeomorphic (or even homeomorphic?) to a "more familiar" manifold such as $S^8$ or $T^8$, or perhaps constructed out of a number of these via surgery. Or maybe it's a well-known projective variety? – Jordan Watts Jul 2 '11 at 18:36
domenico beat me to it... – Jordan Watts Jul 2 '11 at 18:37
Maybe something like $\mathbb{R}^n \times \prod_{k=1}^{r}\mathbb{S}^{j_k}$? – Mark Jul 2 '11 at 19:14
Tangential to the question but perhaps useful for some folks: You don't need anything as fancy as $\pi_5$ to see that $SU(3)$ can't be $S^8$. Even-dimensional spheres don't admit even a single nowhere-vanishing tangent vector field, whereas Lie groups are parallelizable. So from this point of view $SU(3)$ is extremely different from $S^8$. – Andreas Blass Aug 1 '13 at 17:34
up vote 36 down vote accepted

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 $\pi_4(SU(3))=\{0\}$).

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so does this mean that SU(3) is no diffeomorphic to any finite product of euclidean space (of finite, possibly zero dimension) and spheres (of finite dimension, possibly with repetitions) ? My knowledge in topology is quite limited, so this question is most likely trivial given the above. – Mark Jul 3 '11 at 19:06
right. assume $SU(3)$ is a product $S^{n_1}\times S^{n_2}\times\cdots S^{n_k}\times \mathbb{R}^m$, with $n_1\leq n_2\leq n_k$. Then compactness of $SU(3)$ gives $m=0$ and connectedness gives $n_1\geq 1$. Then there are just a few possibilities left and looking at homotopy groups excludes all the possibilities. – domenico fiorenza Jul 3 '11 at 19:42
Please recommend a paper or textbook for studying that 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). – Jino Aug 1 '11 at 9:57
Hi Jino, you could try the classic Dale Husemoller's "Fibre bundles", GTM 20. – domenico fiorenza Aug 1 '11 at 21:07
Another method to see that $SU(3)$ is not homeomorphic to $S^3\times S^5$ (an alternative to computing $\pi_4$) is to show that the mod 2 Steenrod operation $Sq^2: H^3 \to H^5$ is non-zero (see Cartan-Serre, Am. J. Math, Vol. 75, No. 3, Jul., 1953; example 12.3). – Tim Perutz Aug 1 '13 at 17:07

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.

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Again, my bad it was a naive question... You answered my question, but I will keep domenico's one since it gives a better description of the fibration. – R.S. Jul 2 '11 at 19:51

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 $S^5$.

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A lot of the properties of $SU(n)$ and $U(n)$ can be summarised in the "commutative diagram" 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}

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It would be nice if I could get my array to work properly... – Jordan Watts Jul 2 '11 at 20:03
they do not have \mathds, I will not spend my LaTeX skills on it... – R.S. Jul 2 '11 at 20:07
I fixed it. I had to use a triple backslash to end a line. – Jordan Watts Jul 2 '11 at 20:14
I also don't like my description of the diagramme. Each row and each column is a "short exact sequence" if you mark a point on each manifold. I put "short exact sequence" in quotes since I'm not sure if the notion of kernel makes sense in the category of pointed topological spaces. – Jordan Watts Jul 2 '11 at 20:23

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 $E$.

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As explained in the survey paper The Geometry of Compact Lie Groups by L.J. Boya, and by several other people here, $SU(3)$ is an $SU(2)\cong S^{3}$ fiber bundle over $S^{5}$, but in this case in particular, you can say a little more: The $SU(2)$-bundles over $S^{5}$ are classified by $\pi_{4}(S^{3})=\mathbb{Z}/2\mathbb{Z}$. We know that $SU(3)$ is not the trivial bundle, hence, it should be the unique square root of the trivial principal bundle.

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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 $S^3$.

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