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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 $\pi_2(M)$?

Jia Yiyang
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  • Did you define N correctly? It does not look very unitary if, say, A = Id and B = 0. – Johannes Nordström May 28 '14 at 07:19
  • @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 07:22
  • $M$ doesn't look like a subspace of $U(4)$ ($X$ does). – Fernando Muro May 28 '14 at 07:28
  • @FernandoMuro: You might be right. I'm really not sure if it is a subspace, let me edit the post. – Jia Yiyang May 28 '14 at 07:34
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    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 07:40
  • @MarcoGolla: This looks like a promising simplification, but I just started peeking into some general homotopy theory so I'm not able to tell if an answer is immediate using this simplification, would you care to elaborate? – Jia Yiyang May 28 '14 at 07:47
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    $M$ 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 08:30
  • @MarcoGolla: It'd be nice if you can expand it into an answer, it is a shame that I'm quite ignorant about the machinery related. – Jia Yiyang May 28 '14 at 09:12

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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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Marco Golla
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  • 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
  • @FernandoMuro: Thanks. In any case, I think the question was popping up because of the edits (and it's partly my fault -- apologies for that). – Marco Golla May 28 '14 at 13:34
  • @MarcoGolla I mean't in the future. – Fernando Muro May 28 '14 at 14:10
  • Hi Marco, I just tried to follow your argument in detail, and I had difficulty verifying the claim $Y/\sim\cong (U(2)/D)\times U(2)$. For $Y/\sim$ we are identifying $(A,B)\sim(A\Lambda, B\Lambda)$ where $\Lambda$ is any 2 by 2 diagonal unitray matrix, while it seems for $(U(2)/D)\times U(2)$ we are identifying $(A,B)\sim(A\Lambda, B)$, how do I see there is a homeomorphism between them? Is there an explicit construction? – Jia Yiyang May 29 '14 at 07:22
  • Yes, there is an explicit bijection $Y/\sim ; \to (U/D)\times U$, given by $(A,B)\mapsto (A,BA^{-1})$. (Gergő Pintér pointed this out to me.) – Marco Golla May 29 '14 at 15:22
  • Ah, thanks. For $U(2)/D\cong S^2$, I guess an explicit construction would be $A\mapsto A/\sqrt{\det A}\in SU(2)$,then followed by a Hopf map? – Jia Yiyang May 30 '14 at 02:55
  • And sorry if this is naive, I'm also having trouble proving point 1 rigorously, clearly we have the commutative diagram $$\require{AMScd} \begin{CD} Y @>{i}>> X;\ @V{q_2}VV@VV{q_1} V\ Y/\sim @>{f}>>X/\sim \end{CD},$$ where $i$ is inclusion, $q_1,q_2$ are quotient maps, and $f$ is the induced map, it is easy to prove f is bijective and continuous, but I failed to prove it is a homeomorphism. If I can prove $q_1\circ i$ is a quotient map , then it should be done, but my point-set topology failed me. – Jia Yiyang May 30 '14 at 05:34