Topology of SU(3) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T00:55:16Z http://mathoverflow.net/feeds/question/69352 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69352/topology-of-su3 Topology of SU(3) Romero Solha 2011-07-02T18:21:14Z 2011-07-04T17:48:53Z <p>$U(1)$ 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.</p> <p>By one of the tables <a href="http://en.wikipedia.org/wiki/Table_of_Lie_groups" rel="nofollow">here</a> $SU(3)$ is a compact, connected and simply connected 8-dimensional manifold. <a href="http://mathoverflow.net/questions/35125/original-references-for-the-homotopy-groups-pi-5-of-su3-and-pi-4-of-su2" rel="nofollow">This</a> MO post says that its $\pi_5$ is $\mathbb{Z}$ thus it can not be homeomorphic to $S^8$(e.g.: see <a href="http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres" rel="nofollow">this</a> 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?). </p> <p>I guess that this is what the author of <a href="http://mathoverflow.net/questions/57034/manifolds-isomorphic-to-su3-closed" rel="nofollow">this</a> question was trying to know... </p> <p>Anyway, is it known any manifold diffeomorphic to $SU(3)$? </p> http://mathoverflow.net/questions/69352/topology-of-su3/69354#69354 Answer by AndrĂ© Henriques for Topology of SU(3) AndrĂ© Henriques 2011-07-02T19:13:31Z 2011-07-02T19:13:31Z <p>There is not much that can be said about "is it known any manifold diffeomorphic to SU(3)?"...</p> <p>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 $SU(2)$.</p> <p>I hope this helps a bit.</p> http://mathoverflow.net/questions/69352/topology-of-su3/69355#69355 Answer by domenico fiorenza for Topology of SU(3) domenico fiorenza 2011-07-02T19:17:44Z 2011-07-02T22:23:46Z <p>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 <code>$[S^5,BSU(2)]= \pi_5(BSU(2))= \pi_4(SU(2))= \pi_4(S^3)= \mathbb{Z}/2\mathbb{Z}$</code>. 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 $\pi_4(SU(3))={0}$).</p> http://mathoverflow.net/questions/69352/topology-of-su3/69357#69357 Answer by Bruce Westbury for Topology of SU(3) Bruce Westbury 2011-07-02T19:40:57Z 2011-07-02T19:40:57Z <p>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 $S^3$.</p> http://mathoverflow.net/questions/69352/topology-of-su3/69359#69359 Answer by Jordan Watts for Topology of SU(3) Jordan Watts 2011-07-02T19:56:26Z 2011-07-02T20:13:48Z <p>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.</p> <p>\begin{array}{ccccc} SU(n-1) &amp; \to &amp; U(n-1) &amp; \to &amp; S^1 \\ \downarrow &amp; &amp; \downarrow &amp; &amp; \downarrow\\ SU(n) &amp; \to &amp; U(n) &amp; \to &amp; S^1\\ \downarrow &amp; &amp; \downarrow &amp; &amp; \downarrow\\ S^{2n-1} &amp; \to &amp; S^{2n-1} &amp; \to &amp; {*} \end{array}</p> http://mathoverflow.net/questions/69352/topology-of-su3/69384#69384 Answer by Alain Valette for Topology of SU(3) Alain Valette 2011-07-03T08:39:56Z 2011-07-03T08:39:56Z <p>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 $E$. </p> http://mathoverflow.net/questions/69352/topology-of-su3/69485#69485 Answer by Allen Knutson for Topology of SU(3) Allen Knutson 2011-07-04T17:48:53Z 2011-07-04T17:48:53Z <p>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 $S^5$.</p>