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$\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSp{PSp}\DeclareMathOperator\USp{USp}\DeclareMathOperator\BSp{BSp}\DeclareMathOperator\BUSp{BUSp}\DeclareMathOperator\BPSp{BPSp}$Let $\USp(n,\mathbb{C})$ and $\Sp(n,\mathbb{C})$ be the compact symplectic group and the symplectic group, respectively. This is, $$ \Sp(n,\mathbb{C})=\left\{A\in M(2n,\mathbb{C}): A^{tr}J_{2n}A=J_{2n}\right\} $$ where $M(n,\mathbb{C})$ is the set of $n\times n$ matrices with entries in $\mathbb{C}$, $$ J_{2n}=\begin{pmatrix} 0 & I_{n}\\ -I_{n} & 0\\ \end{pmatrix}, $$ and $\USp(n,\mathbb{C})$ is isomorphic to $\Sp(n,\mathbb{C})\cap \mathrm{U}(2n,\mathbb{C})$. Let $\PSp(n,\mathbb{C})$ be the projective symplectic group, i.e. $\Sp(n,\mathbb{C})/\{\pm I_{2n}\}$.

It can be shown that the inclusion $\USp(n,\mathbb{C}) \to \Sp(n,\mathbb{C})$ is a homotopy equivalence.

[Hatcher, AT, pg 271 and 381] says that the quaternionic Grassmannian $G_{m}(\mathbb{H}^{\infty})$ is a model for $\BUSp(n,\mathbb{C})$, and that the integral cohomology $H^{*}(G_{m}(\mathbb{H}^{\infty}))\cong \mathbb{Z}[\alpha_{1},\alpha_{2},\dots,\alpha_{m}]$ with $|\alpha_{i}|=4i$.

This result makes you think that $\BSp(n,\mathbb{C})$ has a CW structure with cells only in degrees $4i$. I would like to find such structure for $\BSp(n,\mathbb{C})$ and also, know whether this structure induces one of the same kind in $\BPSp(n,\mathbb{C})$.

The closest thing I know about their CW structures is that $G_{m}(\mathbb{H}^{\infty})$ has a CW structure with each $G_{m}(\mathbb{H}^{k})$ a finite subcomplex, [Hatcher, VB and Kthy, pg 31-34].

Thank you for your ideas.

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1 Answer 1

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I'll assume that by $BG$ you mean ``any space of the form $E/G$, where $E$ is a contractible space with free $G$-action''. (The alternative would be to define $BG$ as the geometric realisation of a specific simplicial space, but then $BU(n)$ would not have even cells.) But then you can take $G=Sp(n,\mathbb{C})=\text{Aut}_{\mathbb{H}}(\mathbb{H}^n)$ and $E=\text{Inj}_\mathbb{H}(\mathbb{H}^n,\mathbb{H}^\infty)$ and we get $BSp(n,\mathbb{C})=G^{\mathbb{H}}_n(\mathbb{H}^\infty)$. This has a CW structure based on quaternionic Schubert cells (which have real dimension divisible by $4$) just as in the real and complex cases. We can also describe $G^{\mathbb{H}}_n(\mathbb{H}^\infty)$ as the space of isometric $\mathbb{H}$-linear embeddings $\mathbb{H}^n\to\mathbb{H}^\infty$ mod the action of isometric $\mathbb{H}$-linear automorphisms of $\mathbb{H}^n$, so $G^{\mathbb{H}}_n(\mathbb{H}^\infty)$ qualifies as $BUSp(n,\mathbb{C})$ as well as $BSp(n,\mathbb{C})$.

There is an inclusion $SO(3)\simeq PSU(2)\to PSp(1,\mathbb{C})$, which is a homotopy equivalence and so gives a homotopy equivalence of classifying spaces. It is known that $H^*(BSO(3);\mathbb{Z}/2)\simeq\mathbb{Z}/2[w_2,w_3]$ with $|w_2|=2$ and $|w_3|=3$. This implies that any CW complex homotopy equivalent to $PSp(1,\mathbb{C})$ must have odd-dimensional cells.

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  • $\begingroup$ Thank you so much! I was thinking of $BG$ as the geometric realization actually. Your answer is very helpful. $\endgroup$
    – Faye3
    Commented Oct 12, 2022 at 1:50

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