During my research I have recently stumbled upon the problem of finding the relative homotopy sets $\pi_1(U_n,U_n/O_n)$ and $\pi_1(U_{2n},U_{2n}/USp_{2n})$ for $n$ large enough to be in the stable regime where Bott periodicity applies.

The subgroup $U_n/O_n\subset U_n$ contains all elements $g\in U_n$ with $g=g^T$ and the Cartan embedding is \begin{align} U_n/O_n&\hookrightarrow U_n\\\ g\cdot O_n&\mapsto gg^T \end{align} Similarly, the subgroup $U_{2n}/USp_{2n}\subset U_{2n}$ contains all elements $g\in U_{2n}$ with $g=Jg^TJ^{-1}$ and the Cartan embedding is \begin{align} U_{2n}/USp_{2n}&\hookrightarrow U_{2n}\\\ g\cdot USp_{2n}&\mapsto gJg^TJ^{-1}, \end{align} with $J:=\begin{pmatrix}0&1_n\newline-1_n&0\end{pmatrix}$.

In order for my other results to be consistent, I expect the result to be \begin{align} X:=\pi_1(U_n,U_n/O_n)&=0\\\ Y:=\pi_1(U_{2n},U_{2n}/USp_{2n})&=\mathbb{Z}_2 \end{align} However, with the following argument I find that both are trivial: Using the exact sequence of homotopy groups/sets, \begin{align} \pi_1(U_n/O_n)&\stackrel{i_1}{\to}\pi_1(U_n)\to X\to\pi_0(U_n/O_n)\\\ \pi_1(U_{2n}/USp_{2n})&\stackrel{i_2}{\to}\pi_1(U_{2n})\to Y\to\pi_0(U_{2n}/USp_{2n}), \end{align} where $i_{1}$ and $i_2$ are induced by the Cartan embeddings above, there is the bijection \begin{align} X=\pi_1(U_n)/\text{img}(i_1)=\mathbb{Z}/\text{img}(i_1)\\\ Y=\pi_1(U_{2n})/\text{img}(i_2)=\mathbb{Z}/\text{img}(i_2) \end{align} since $\pi_0(U_n/O_n)=\pi_0(U_{2n}/USp_{2n})=0$.

The determinant induces a bijection $\pi_1(U_n)\to\pi_1(U_1)=\mathbb{Z}$ with inverse induced by $U_1\hookrightarrow U_n$ (adding ones to the diagonal), so a representative loop of the class $m\in\pi_1(U_n)=\mathbb{Z}$ is $$\phi:t\mapsto\text{diag}(e^{i2\pi m t},1,\dots,1).$$ This has a preimage under $i_1$, namely the loop $$\phi_1:t\mapsto\text{diag}(e^{i\pi m t},1,\dots,1)\cdot O_n.$$ This is indeed a loop since $\phi_1(0)=(1,1,\dots,1)\cdot O_n$ and $\phi_1(1)=(-1,1,\dots,1)\cdot O_n=(1,1,\dots,1)\cdot O_n$.

Similarly, there is a preimage under $i_2$, namely the loop $$\phi_2:t\mapsto\text{diag}(e^{i\pi m t},1,\dots,1)\cdot USp_{2n}.$$ Again, this is a loop since $\phi_2(0)=(1,1,\dots,1)\cdot USp_{2n}$ and $\phi_2(1)=(-1,1,\dots,1)\cdot USp_{2n}=(1,1,\dots,1)\cdot USp_{2n}$. Thus, both $i_1$ and $i_2$ are surjective and $X=Y=0$.

However, for the expected result to hold, $i_1$ should be surjective in the sequence with $X$ and $\text{img}(i_2)=2\mathbb{Z}$ (even integers only) in the sequence with $Y$. This does not seem to be the case - is my argument flawed or is it true that both relative homotopy sets are in fact trivial?

Thank you!


There's a "famous" 24 term sequence of fibrations, involving the spaces of the real and complex K-theory spectra, coming from Bott periodicity. (But apparantly, not famous enough for google to cough up a good reference.) Part of this sequence are homotopy fibration sequences of the form $$ Sp/U \to U/O \xrightarrow{f} U $$ and $$ O/U\to U/Sp\xrightarrow{g} U.$$ ($O$, $U$, and $Sp$ are the infinite versions of the orthogonal, unitary, and unitary-symplectic groups

Assuming that $f$ and $g$ correspond to your Cartan embeddings, you should be able to identify your relative $\pi_1$s with the $\pi_0$s of the homotopy fibers of these fiber sequences. If so, then $$\pi_1(U_n,U_n/O_n) = \pi_1(U,U/O) = \pi_0(Sp/U)=0$$ and $$\pi_1(U_{2n},U_{2n}/USp_{2n}) = \pi_1(U,U/Sp) = \pi_0(O/U) = Z/2,$$ which is the result you were expecting.

If this is all true, then $i_2: \pi_1(U_{2n}/USp_{2n})\to \pi_1(U_{2n})$ isn't really surjective. Maybe $\phi_2$ isn't really a loop: is $diag(-1,1,...,1) \in USp$?

  • $\begingroup$ You are right, it isn't. Somehow I had myself convinced of the opposite by showing that $i_2(\phi_2(1))\in USp_{2n}$, which is obviously not the right thing to check. In fact, even if it were a loop, then it would not even map to the loop $\phi$ in $U_{2n}$ for odd $m$... Thanks for the great answer! $\endgroup$
    – Pierre
    Oct 14 '11 at 5:49
  • $\begingroup$ While your hint that $i_2$ may not really be surjective helped me to resolve my problem, I'm still puzzled about the way you derived the relative $\pi_1$s. Are you saying that $f$ and $g$ are the projections of a Serre fibration? Then I only know the statement that $\pi_n(E,F)=\pi_n(B)$ for a Serre fibration $p:E\to B$ with fiber $F=p^{-1}(b)$ over the basepoint $b$ of $B$, which does not give me your statement. $\endgroup$
    – Pierre
    Sep 4 '12 at 12:51

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