Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
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

7 Answers 7

up vote 35 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 $\pi_4(SU(3))=\{0\}$).

share|improve this answer
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.

share|improve this answer
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. –  Romero Solha 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 $S^5$.

share|improve this answer

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}

share|improve this answer
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... –  Romero Solha 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 $E$.

share|improve this answer

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 $S^3$.

share|improve this answer

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.