I read somewhere that $BSU(n)$ has a cellular decomposition that consists of one 4-cell and higher dimensional cells. Can someone tell me why this is the case? In fact I am not sure if this statement is true. For example, $BSU(2)$ is $\mathbb{H}P^\infty$, which I believe has the cellular decomposition $\bigcup_{k=0}^\infty e^{4k}$. In particular, it contains a 0-cell.
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$\begingroup$ This goes under the name of Schubert cells. $\endgroup$– Dave BensonCommented Apr 18, 2023 at 11:49
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$\begingroup$ Are you certain your source deliberately meant to forgo 0-cells, rather than simply assuming but omitting mention of them? $\endgroup$– jdcCommented Apr 18, 2023 at 12:36
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3$\begingroup$ CW complexes without zero cells are empty, so you are certainly correct that there should be one. But also, maybe the focus is on pointed spaces, in which case we pick that zero cell to be the basepoint, and then the first relative cell is in dimension four. $\endgroup$– John Wiltshire-GordonCommented Apr 18, 2023 at 13:40
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$\begingroup$ There's a notion of cell decomposition distinct from that of CW complexes, so there's an outside possibility that this is what was meant, a decomposition of some model of $BSU(n)$ into some union of Euclidean spaces, all of dimension $>4$ (although this would be more of a curiosity than particularly useful). $\endgroup$– jdcCommented Apr 18, 2023 at 14:02
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$\begingroup$ Thanks to all. After a second look it does look like that the author neglected to mention the 0-cell, as what he claims later using this fact hold when there is a 0-cell. However, I still don't see why there are no 2-cells for general $n$. $\endgroup$– user48975Commented Apr 19, 2023 at 7:45
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