The fundamental group of $U(n)/O(n)$

Question

Since $O(n)$ is a subgroup of $U(n)$, we can consider the quotient space $L_n=U(n)/O(n)$. The quotient space is homotopic to the ($n$-th) Lagrangian Grassmannian, and it is known that its fundamental group is isomorphic to $\mathbb Z$.

This fact is an easy consequence of homotopy long exact sequence. Namely we have an exact sequence $$\dots \to \pi_1(O(n)) \to \pi_1(U(n)) \to \pi_1(L_n) \to 1.$$ Now $\pi_1(O(n)) = \mathbb Z_2$ ($n\geq 3$) and $\pi_1(U(n))=\mathbb Z$, therefore $\pi_1(L_n)=\mathbb Z$ if $n \geq 3$.

But, since $\pi_1(O(2))=\mathbb Z$, we need to see that the map $\pi_1(O(2)) \to \pi_1(U(2))$ is the zero map.

Now, $$t \mapsto \pmatrix{\cos 2\pi t & -\sin 2\pi t \\ \sin 2\pi t & \cos 2\pi t}$$ gives the generator of $\pi_1(O(2))$.

Can we construct an explicit homotopy from the above loop to the constant loop in $U(2)$?

Answer 1

Yes, you can easily write down such a homotopy: let $\theta(t)$ be the matrix $$\theta(t)=\begin{pmatrix}\cos(2\pi t) & -\sin(2\pi t)\\\sin(2\pi t) & \cos(2\pi t)\end{pmatrix}.$$

For any $t\in \mathbb R$, the matrix $\theta(t)$ is diagonalizable with eigenvalues $e^{\pm 2i\pi t}$ and the change of basis matrix doesn't depend on $t$.

More precisely, let $P$ be the matrix $$P=\frac{1}{\sqrt{2}}\begin{pmatrix}i&1\\1&i\end{pmatrix}$$ then $$\forall t\in\mathbb R, \qquad \theta(t)=P\begin{pmatrix}e^{2i\pi t} & 0 \\ 0 & e^{-2i\pi t}\end{pmatrix}P^{-1}.$$ Furthermore, you can check that $P$ and $\begin{pmatrix}e^{2i\pi t} & 0 \\ 0 & e^{-2i\pi t}\end{pmatrix}$ are both unitary.

This allows us to define the following map : $$\begin{array}{rccc} H: & [0,1]\times [0,1] & \longrightarrow & U(2) \\ & (t,s) & \longmapsto & P\begin{pmatrix}e^{2i\pi st} & 0 \\ 0 & e^{-2i\pi st}\end{pmatrix}P^{-1}\end{array}$$ $H$ takes value in $U(2)$, is continuous and $H(\cdot,0)$ is the constant loop while $H(\cdot,1)$ is your loop.

Edit: Actually, this homotopy is just a homotopy of maps and not of loops.

In fact, this is the best that one can expect because $\pi_1(O(2),I_2)\rightarrow \pi_1(U(2),I_2)$ is a group isomorphism. So you won't get any other information about $\pi_1(L_2)$ in this way and your loop can't be homotopic (as loops) to a constant loop in $U(2)$.

However there is a different, but more general, way to compute $\pi_1(L_n)$: consider the following map $$\begin{array}{cccl} & U(n)& \rightarrow & S^1 \\ & M & \mapsto & {\det}^2(M) \end{array}$$ This map induces a continuous map $f:L_n\rightarrow S^1$ which is a fibration with fibre $SU(n)/SO(n)$. So one can use the long exact sequence in homotopy groups : $$\cdots \rightarrow \pi_1(SU(n)/SO(n)) \rightarrow \pi_1(L_n) \rightarrow \pi_1(S^1) \rightarrow \pi_0(SU(n)/SO(n)) \rightarrow \cdots $$

But,

1. $SO(n)$ is path-connected,
2. $SU(n)$ path-connected and simply-connected,
3. we have the fibration : $SO(n)\rightarrow SU(n) \rightarrow SU(n)/SO(n)$ thus $$\pi_1(SU(n)/SO(n))=1.$$

Hence $$\pi_0(SU(n)/SO(n))=1 \qquad \text{ and } \qquad \pi_1(SU(n)/SO(n))=1.$$ and consequently the map $$\pi_1(L_n)\rightarrow \pi_1(S^1)$$ is a group isomorphism which gives you the result.