# How to calculate the first and second homotopy groups of the following space constructed from $U(4)$

In solving a physics problem, I came across a weird topological space constructed from $U(4)$, the group of $4\times4$ unitary matrices. I want to know the first two homotopy groups of it. Here is how it is defined:

Consider the set $X$ of matrices in $U(4)$ $$X=\{F\in U(4):F=N\Lambda\},$$ where $\Lambda$ is any diagonal unitary matrix, and $N$ has the block form $$N=\begin{bmatrix}A&A\\B&-B\end{bmatrix},$$ where $A$ and $B$ are $2\times2$ matrices. This, of course, means $N$ must be also unitary, and $A,B$ must satisfy $AA^\dagger=BB^\dagger=\frac{1}{2} I$. We can form the quotient space $M=X/\sim$, where $F_1\sim F_2$ if $F_1=F_2\Lambda$, $\Lambda$ being any diagonal unitary matrix.

Is there a way to calculate $\pi_1(M)$ and $\pi_2(M)$?

• Did you define N correctly? It does not look very unitary if, say, A = Id and B = 0. – Johannes Nordström May 28 '14 at 7:19
• @JohannesNordström: I meant N has that form and N is unitary at the same time. In other words, $AA^\dagger=BB^\dagger=\frac{1}{2} I$ – Jia Yiyang May 28 '14 at 7:22
• $M$ doesn't look like a subspace of $U(4)$ ($X$ does). – Fernando Muro May 28 '14 at 7:28
• Another description for $M$ is $(U(2)\times U(2))/D$, where $D$ is the subgroup of (unitary) diagonal matrices, acting diagonally on both factors. – Marco Golla May 28 '14 at 7:40
• $M$ should also be the quotient of the set of $N$s by the action of the diagonals (inside that set). An easy computation shows that this is the quotient in my previous comment. In turn, this should be equal to $U(2)/D \times U(2)$, which should be $S^2\times U(2)$. – Marco Golla May 28 '14 at 8:30

Yes, you can compute them, and $\pi_1(M) = \mathbb{Z}$, $\pi_2(M) = \mathbb{Z}$.

The way I see it, there are three steps in the proof. For convenience, let me call $Y = U(2)\times U(2)$ the set of $N$s of the form above.

1. Instead of looking at $X/\sim$, let's look at $Y/\sim$ (where I denote with the same symbol the relation on $X$ and its restriction on $Y$). Topologically, this is exactly the same, so the result is still $M$.

2. Notice that $Y/\sim$ is simply $U(2)\times U(2)$ quotiented by the diagonal action of the diagonal subgroup $D$ of $U(2)$. From this it follows that $M$ is $(U(2)/D)\times U(2)$ (this does not hold at the group level, if you're quotienting by a normal subgroup, but it holds at the topological level).

3. $D = U(1)\times U(1)$, and since $\mathbb{CP}^n = U(n+1)/(U(1)\times U(n))$, so $U(2)/D = S^2$.

From this, since $\pi_1(S^2)$ and $\pi_2(U(2))$ are both trivial (the latter is a general fact for Lie groups, see this question), and $\pi_2(S^2)$ and $\pi_1(U(2))$ are both infinite cyclic.

The first two steps are more or less just general topology, while the third is apparently well-known.

You can also convince yourself that 3. is true since you can see a "fibration" of $U(2)/D$ over $S^1$ such that the generic fiber is $S^1$ and there's one exceptional fiber which is $S^0$ (though this is not a proof, of course).

• Thanks and +1. I'll need to digest this and come back to it after I learn some more homotopy theory – Jia Yiyang May 28 '14 at 10:39
• And just to confirm, this would mean $\pi_1=Z$ and $\pi_2=Z$, right? – Jia Yiyang May 28 '14 at 10:42
• Yes, that follows. Let me add it to the answer. – Marco Golla May 28 '14 at 10:50
• @FernandoMuro: Done. Just curious, what is popping up? – Jia Yiyang May 28 '14 at 12:58
• I meant that the question would continue appearing periodically among the first questions in the forum, and this is not desirable for questions which are already answered. Besides, Marco deserves some credit for this. – Fernando Muro May 28 '14 at 13:31