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Background: Consider $SU(2)$ action on the 6-dim flag manifold $M=SU(3)/T^2$ via left multiplication. We view $SU(2)$ as a subgroup of $SU(3)$ corresponding to $2\times 2$-block. The action is just left multiplication by $SU(2)$. The principal orbits are $SU(2)$, the singular orbits are $SU(2)/U(1)=S^2$, and no other orbit types. The orbit space $M/SU(2)$ is topologically a 3-sphere with 3 singular orbits, represented by the 3 permutation matrices of $Id$, up to sign.

Now we want to reverse the procedure. Say we are given a simply connected closed smooth 6-manifold $M^6$ together with a smooth $SU(2)$ action, such that the orbit space structure is the one describe above: a 3-sphere with 3 singular points, the principal orbits being $SU(2)$, the singular orbits being $SU(2)/U(1)=S^2$. We want to recover the topology of $M$ from the orbit structure. For example, is $M$ necessarily homeomorphic to $SU(3)/T^2$, or could it be something else?

(1)Here are some of my thoughts: we can decompose $M$ into 2 parts which are fiber bundles. The first part is the union of all principal orbits, which is a principal $SU(2)$ bundle over $S^3-\{p,q,r\}$; let's call it the "regular bundle". The second part is the union of singular orbits, which is just the union of three $S^2$'s. The regular bundle is necessarily a trivial bundle by observing that there is only one homotopy class of maps from $S^3-\{p,q,r\}$ to $B_{SU(2)}$.

In order to determine the topology of $M$, we also need to look at the gluing map between the regular part and the singular part. Take an equivariant neighborhood of one singular orbit, which by slice theorem can be written as $SU(2)\times_{U(1)}D^4$, then the gluing map can be thought of as an equivariant homeomorphism of the boundary of $SU(2)\times_{U(1)}D^4$, which is $S^2\times S^3$, where $SU(2)$ acts on the $S^3$ factor, say on the right. Thus let $f:S^2\times S^3\to S^2\times S^3$ be a gluing map. Up to homotopy, $f$ is necessarily of the form $f(x,y)=(\phi(x),g(x).y),\ \phi\in O(3),\ g:S^2\to S^3$, since $f$ is $SU(2)$-equivariant. Then the homotopy class of such a map $f$ is determined by whether it is orientation-preseving. Let $+$ denote an orientation preserving gluing map, and $-$ the orientation reversing one.

Since there are 3 singular orbits, we have 3 gluing maps. Up to simultaneous change of orientation of singular orbits, we have 2 possibilities: $(+,+,+),\ (+,+,-)$. If the $SU(2)$-space $M$ is determined by the gluing maps, then up to equivariant homeomorphism there are 2 such spaces, including $SU(3)/T^2$. I don't know whether gluing maps give the complete information and what the other space is.

(2)Other approach: I also used Mayer-Vietoris sequence to compute the cohomology groups of $M^6$. It coincides with that of $SU(3)/T^2$, but I don't know the ring structure. As a reference, the cohomology ring of $SU(3)/T^2$ is $H^*(SU(3)/T^2)=\mathbb{Z} [x,y]/\big(x^2+y^2+xy,\ xy(x+y)\big)$. Since $M$ over the orbit space is a singular fibration, we can't directly use Serre spectral sequence. I'm wondering if there is a way to go about it. If we can get the cohomology ring structure, and the $p_1$ and $w_2$ just from the orbit space struture, we can also determine the topology of $M$ using the classification theorem of simply connected smooth 6-manifolds.

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    $\begingroup$ As this is a long post, could you emphasize the part containing your question, say make it bold? This makes it easier for other people to read. Also, it usually takes a couple of days for someone with enough expertise to come across the question and digest it, 何弃疗? $\endgroup$ – Fan Zheng Feb 5 '18 at 23:28
  • $\begingroup$ It helps to specify what the $SU(2)$ action is in the first paragraph. Presumably $SU(2)$ first gets mapped to $SO(3)$, which then acts on $SU(3)$ by left multiplication, but I am not sure. $\endgroup$ – Fan Zheng Feb 5 '18 at 23:35
  • $\begingroup$ Thanks for the comment, Fan Zheng. The $SU(2)$ action is the subaction of $SU(3)$ on $SU(3)/T^2$, where we view $SU(2)$ as the $2\times 2$-block subgroup of $SU(3)$. I've emphasized the question and other important parts. $\endgroup$ – Yuhang Liu Feb 5 '18 at 23:44
  • $\begingroup$ 关于$SU(3)/T^2$上的复结构,可以把它看成$GL(3,\mathbb C)$商掉上三角矩阵得到。上面的Kahler度量也可从商结构导出(类似$\mathbb{CP}^n$)。你取个领域建立坐标算一下应该就清楚了。 $\endgroup$ – Fan Zheng Mar 5 '18 at 3:25

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