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Is there a way to explicitly compute the homology of the space $$ \varinjlim_{(p,q)} SO(p,q)^+, $$ where each $SO(p,q)$ is the indefinite special orthogonal group, and $SO(p,q)^+$ its identity component, where $\mathbb{N}\times\mathbb{N}$ has the product order, and where we have inclusions $SO(p,q)\hookrightarrow SO(p',q')$ defined by $A\mapsto \mathbb{I}_{p'-p}\oplus A\oplus (-\mathbb{I}_{q'-q})$?

What I tried to argue is:

  1. The maximal connected part of $SO(p,q)^+$ is homotopy equivalent to $SO(p)\times SO(q)$;
  2. There is the chain of isos $$ \begin{align} \textstyle H_n\left( \varinjlim SO(p,q)^+\right) & \cong \textstyle \varinjlim H_n(SO(p,q)^+) \\ &\cong \textstyle \varinjlim H_n\Big(SO(p)\times SO(q)\Big)\\ &\cong \displaystyle \varinjlim \bigoplus_{a+b=n} H_a(SO(p))\otimes_{\mathbb Z}H_b(SO(q))\oplus \text{ some Tor} \end{align} $$ And this seems kinda tame, up to knowing completely the homology of the two factors. Is there a nifty way to conclude something more?
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  • $\begingroup$ I don't understand your "lexicographical ordering". Is there any meaningful map from $SO(p,q)\to SO(r,s)$ assuming only $p<r$? Instead, you should consider $\mathbb N^2$ as a directed system. Apart from that, your chain of isos looks correct to me, and the right hand side stabilises for each individual $n$. $\endgroup$ – Sebastian Goette Feb 9 '16 at 11:13
  • $\begingroup$ Yes, I corrected the typo (I meant the order where $(x,y)\le (x',y')$ iff $x\le x'$ and $y\le y'$. $\endgroup$ – Fosco Loregian Feb 9 '16 at 11:26
  • $\begingroup$ How do i determine the homology of $SO(p)$? $\endgroup$ – Fosco Loregian Feb 11 '16 at 13:53

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