Is there a Lie group G which has only two cells?
i.e. $ G = S^n \cup_f~ e^m$ where $f:S^{m-1}\to S^n$ with $m>n$
How many exists that groups? Infinitely many?
If there is no Lie group which has only two cell, how many cells needed to a Lie group?
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1$\begingroup$ $U(1)$, with $m=1$, $n=0$. Can we have a bit of background to this question? $\endgroup$– David Roberts ♦Commented Sep 26, 2011 at 1:27
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$\begingroup$ I want to non-sphere Lie group. Using confibration, I'll compute the homotopy set. $\endgroup$– JinoCommented Sep 26, 2011 at 1:36
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$\begingroup$ See Figures 3 and 4 of my paper arxiv.org/abs/0810.2131 for an answer to your last question for the exceptional Lie groups G_2 and F_4. $\endgroup$– André HenriquesCommented Sep 26, 2011 at 3:56
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1 Answer
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Amongst compact Lie groups, the only examples are $U(1)$ and $SU(2)$, which topologically are $S^1$ and $S^3$. All others are ruled out because their rational cohomology has dimension >3, except $SO(3,\mathbb{R})\cong \mathbb{RP}^3$, which has the same rational cohomology as $S^3$, but is ruled about because its mod 2 cohomology has dimension 4.
On the other hand, there are lots of unipotent Lie groups which are topologically the same as vector spaces.
answered Sep 26, 2011 at 1:36
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$\begingroup$ rational cohomology dimension and number of cells are related? $\endgroup$– JinoCommented Sep 26, 2011 at 1:48
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$\begingroup$ @Jino: if the dimension of your cells are far apart, then computing (co)homology using the cellular complex shows there is a relation. $\endgroup$ Commented Sep 26, 2011 at 2:03
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$\begingroup$ @Webster: Can you recommend a reference which contain your explanation dim > 3? $\endgroup$– JinoCommented Sep 26, 2011 at 2:11
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$\begingroup$ Mariano- The cells being far apart has nothing to do with it; the total dimension of cohomology over any field gives a lower bound on the number of cells you need. Of course, sometimes it's too low because of cancellation; the rational cohomology of $\mathbb{RP}^3$ looks like it might allow you to get away with 2 cells, but the mod 2 cohomology shows you need 4. $\endgroup$– Ben Webster ♦Commented Sep 26, 2011 at 3:29