What is the value of homotopy group $\pi_1(SU(2)\times U(1))$

Question

Question background: In particle physics, the weak interaction symmetry group is described by $G=SU(2)\times U(1)$, which is spontaneously broken into $H=U(1)_{em}$. ($U(1)_{em}$ is the electromagnetic interaction group). The generator of $U(1)_{em}$ is $T^3+Y$, where $T^3$ is the diagonal Lie algebra of $SU(2)$, $Y$ is the generator of $U(1)$ before the symmetry breaking.

Now, I want to research the second homotopy group of $$\pi_2(G/H) \tag{1}$$ in the spontaneous symmetry breaking process. In general, we have the relation (for example, see John Preskill's Magnetic Monopoles, page 483, the paragraph below Eq.(53)) $$\pi_2(G/H)=\pi_1(H)/\pi_1(G) \tag{2} $$ And since $\pi_1(U(1))=Z$, I am wondering what's the value for $$\pi_1(G)=\pi_1(SU(2)\times U(1)) $$

More generally,

1. What's the value for $\pi_n(A\times B)$, where $A$ and $B$ are two connected groups? Would this equals to $\pi_n(A)\otimes \pi_n(B)$ or $\pi_n(A)\oplus\pi_n(B)$?
2. I am told that $\pi_2(G/H)=0$ for my above example, so I am interested in how $\pi_1(H)/\pi_1(G)$ leads to zero in my example.

Answer 1

Regarding the first question, as was mentioned in the comments, for path connected spaces $A$ and $B$, we have $\pi_n(A\times B) \cong \pi_n(A)\times\pi_n(B)$, see this question.

As for the second question, again as mentioned in the comments, the quotient $\pi_1(H)/\pi_1(G)$ is meaningless. If $H$ is a closed topological subgroup of closed topological group $G$, then there is a fibration $H \xrightarrow{\iota} G \xrightarrow{p} G/H$ where $\iota$ is the inclusion, and $p$ is the projection. There is an associated long exact sequence in homotopy

$$\dots \to \pi_2(H) \xrightarrow{\iota_*} \pi_2(G) \xrightarrow{p_*} \pi_2(G/H) \to \pi_1(H) \xrightarrow{\iota_*} \pi_1(G) \xrightarrow{p_*} \pi_1(G/H) \to \pi_0(H) \xrightarrow{\iota_*} \pi_0(G) \xrightarrow{p_*} \pi_0(G/H) \to 0.$$

If $H$ and $G$ are Lie groups (as in your case), then $\pi_2(H) = 0$ and $\pi_2(G) = 0$, and if $H$ and $G$ are connected (as in your case), then $\pi_0(H) = 0$ and $\pi_0(G) = 0$. It follows that $\pi_0(G/H) = 0$ (i.e. $G/H$ is connected), and we have an exact sequence

$$0 \to \pi_2(G/H) \to \pi_1(H) \xrightarrow{\iota_*} \pi_1(G) \xrightarrow{p_*} \pi_1(G/H) \to 0.$$

In this situation, we have $\pi_2(G/H) \cong \ker(\iota_* : \pi_1(H)\to \pi_1(G))$. To calculate this kernel, we need to understand the specific $H$ and $G$ you are interested in.

Caveat: I'm not a physicist. There may be different conventions or missing factors, but I believe what I've written below is essentially what is going on.

In your post, the subgroup $H = U(1)_{\text{em}}$ is defined in terms of $T^3$ and $Y$. Here $T^3 = \begin{bmatrix} i & 0\\ 0 & -i\end{bmatrix} \in \mathfrak{su}(2)$, the Lie algebra of $SU(2)$. This spans a one-dimensional Lie subalgebra which is the Lie algebra of the Lie subgroup $\left\{\begin{bmatrix}z & 0\\ 0 & \bar{z}\end{bmatrix}\, \middle|\, z \in U(1)\right\}$ which is a maximal torus of $SU(2)$. On the other hand $Y = i \in \mathfrak{u}(1)$, the Lie algebra of $U(1)$. This generates the whole Lie algebra $\mathfrak{u}(1)$ which is the Lie algebra of $U(1)$.

Now $T^3 + Y \in \mathfrak{su}(2)\oplus\mathfrak{u}(1)$, the Lie algebra of $SU(2)\times U(1)$. As in the previous cases, this spans a one-dimensional Lie subalgebra which is the Lie algebra of a Lie subgroup of $SU(2)\times U(1)$ which is isomorphic to $U(1)$. This is precisely $U(1)_{\text{em}}$ and is nothing more than the diagonal subgroup corresponding to the two copies of $U(1)$ mentioned in the previous paragraph. Explicitly, we have

$$U(1)_{\text{em}} = \left\{\left(\begin{bmatrix} z & 0\\ 0 & \bar{z}\end{bmatrix}, z\right)\, \middle|\, z \in U(1)\right\}.$$

To determine $\ker(\iota_* : \pi_1(H) \to \pi_1(G))$, and hence $\pi_2(G/H)$, we need to understand the homomorphism $\iota_* : \pi_1(H) \to \pi_1(G)$. Since $U(1)_{\text{em}} \cong U(1)$, we have $\pi_1(H) = \pi_1(U(1)_{em}) \cong \mathbb{Z}$. From the answer to the first question, we also have $\pi_1(G) = \pi_1(SU(2)\times U(1)) \cong \pi_1(SU(2))\times\pi_1(U(1)) \cong 0\times\mathbb{Z} \cong \mathbb{Z}$. There are many homomorphisms $\mathbb{Z} \to \mathbb{Z}$, so we have to actually analyse the map $\iota : U(1)_{\text{em}} \to SU(2)\times U(1)$.

Consider the isomorphism $\varphi : U(1) \to U(1)_{\text{em}}$ given by $z \mapsto \left(\begin{bmatrix}z & 0\\ 0 & \bar{z}\end{bmatrix}, z\right)$. If $\operatorname{proj} : SU(2)\times U(1) \to U(1)$ is projection onto the second factor, then we have $\operatorname{proj}\circ\,\iota\circ\varphi = \operatorname{id}_{U(1)}$. Therefore, at the level of fundamental groups, we have $\operatorname{proj}_*\circ\,\iota_*\circ\varphi_* = \operatorname{id}_{\pi_1(U(1))}$. Since $\operatorname{id}_{\pi_1(U(1))}$ is injective and $\varphi_*$ is an isomorphism, it follows that $\iota_* : \pi_1(U(1)_{\text{em}}) \to \pi_1(SU(2)\times U(1))$ is injective, so $\pi_2(G/H) \cong \ker(\iota_* : \pi_1(H) \to \pi_1(G)) = 0$.

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In fact, since $\operatorname{proj}_*$ is also an isomorphism, so too is $\iota_*$, and hence $\pi_1(G/H) = 0$. It also follows from the long exact sequence mentioned above that $\pi_3(G/H) \cong \pi_3(G) \cong \mathbb{Z}$.

We can actually identify the space $G/H$: it is diffeomorphic to $SU(2)$ (and hence $S^3$). The diffeomorphism is given by $[(A, z)] \mapsto A\begin{bmatrix}\bar{z} &0\\ 0 & z\end{bmatrix}$.