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!



The relative homotopy sets $\pi_1(U_n,U_n/O_n)$ and $\pi_1(U_{2n},U_{2n}/USp_{2n})$
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 $diag(-1,1,...,1) \in USp$?


