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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 $\pi_2(M)$?

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  • $\begingroup$ Did you define N correctly? It does not look very unitary if, say, A = Id and B = 0. $\endgroup$
    – Johannes Nordström
    Commented May 28, 2014 at 7:19
  • $\begingroup$ @JohannesNordström: I meant N has that form and N is unitary at the same time. In other words, $AA^\dagger=BB^\dagger=\frac{1}{2} I$ $\endgroup$
    – Jia Yiyang
    Commented May 28, 2014 at 7:22
  • $\begingroup$ $M$ doesn't look like a subspace of $U(4)$ ($X$ does). $\endgroup$
    – Fernando Muro
    Commented May 28, 2014 at 7:28
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    $\begingroup$ 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. $\endgroup$
    – Marco Golla
    Commented May 28, 2014 at 7:40
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    $\begingroup$ $M$ 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 $S^2\times U(2)$. $\endgroup$
    – Marco Golla
    Commented May 28, 2014 at 8:30

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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).

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  • $\begingroup$ Thanks and +1. I'll need to digest this and come back to it after I learn some more homotopy theory $\endgroup$
    – Jia Yiyang
    Commented May 28, 2014 at 10:39
  • $\begingroup$ And just to confirm, this would mean $\pi_1=Z$ and $\pi_2=Z$, right? $\endgroup$
    – Jia Yiyang
    Commented May 28, 2014 at 10:42
  • $\begingroup$ Yes, that follows. Let me add it to the answer. $\endgroup$
    – Marco Golla
    Commented May 28, 2014 at 10:50
  • $\begingroup$ @FernandoMuro: Done. Just curious, what is popping up? $\endgroup$
    – Jia Yiyang
    Commented May 28, 2014 at 12:58
  • $\begingroup$ 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. $\endgroup$
    – Fernando Muro
    Commented May 28, 2014 at 13:31

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