I am aware of the group, $SU(2)$, but I came across a space, $BSU(2)$, but I have no idea of what this stands for?
I noticed that $BSU(2) \cong \mathbb{HP}^\infty$, can any one please explain how is this if this is true?
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2No, $BG$ is the classifying space of principal $G$-bundles. – user10354138 Oct 22 '23 at 06:14
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2P.S. $SU(2)$ is simply connected; it has no nontrivial double cover. – Ted Shifrin Oct 22 '23 at 06:18
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I made changes in the question, as it is clear now. – Eden Zane Oct 22 '23 at 06:21
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Recall $SU(2)\cong Sp(1)$, the group of unit quaternions. The (left) quaternionic projective $n$-space $\mathbb{HP}^n$ can be described as the quotient of the unit sphere $S^{4n+3}\subset\mathbb{H}^{n+1}$ by the (free continuous) scalar multiplication of $Sp(1)$ (on the left), hence $Sp(1)\to S^{4n+3}\to\mathbb{HP}^n$. The scalar multiplication respects the inclusion $\mathbb{H}^n\to\mathbb{H}^{n+1}$, so taking limit we have principal bundle $Sp(1)\to\varinjlim_n S^{4n+3}\to\varinjlim_n\mathbb{HP}^n$, i.e., $$ Sp(1)\to S^\infty\to\mathbb{HP}^\infty. $$ Since $S^\infty$ is contractible, we get $BSp(1)\simeq\mathbb{HP}^\infty$ (up to homotopy equivalence).
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Thank you , but I don't fully understand the last line, how does the fact that $S^\infty$ is contractible help confirm $BSp(1) \simeq \mathbb{HP}^\infty$ – Eden Zane Oct 22 '23 at 09:33
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1By definition, we want the total space $ESp(1)$ to be a contractible space with a proper free continuous $Sp(1)$ action (since $Sp(1)$ is compact, any continuous action is automatically proper so I dropped that in the answer), and $S^\infty$ fills that role. General theory says if a universal $G$-bundle $P\to B$ exists, then (a) $B$ can be taken to be a CW-complex; (b) a CW-classifying space $B$ is unique up to canonical homotopy equivalence; and (c) $P$ is unique up to $G$-equivariant homotopy equivalence. – user10354138 Oct 22 '23 at 14:45