Third homotopy group of quotient space

Question

The third homotopy group $\pi_3(G)$ of a compact connected simple Lie group $G$ is

$$\pi_3(G)=Z.$$

What if $G$ is not a Lie group but a quotient space of two Lie groups, $G/H$?

I know that if a space is compact, then so are all its quotient spaces. Moreover, if a space is connected, then so are all its quotient spaces. Does this mean that if $G$ is a compact connected simple Lie group with $\pi_3(G)=Z$, then $\pi_3(G/H)=Z$ is also given?

Edit in response to the comment by Arnaud Mortier:

I am especially interested in the case where $G$ is a direct product of two simple Lie groups $G'$, such as $G= G'\times G' = SU(3)\times SU(3)$. Algebraically, the special unitary group $SU(3)$ is a simple Lie group, meaning that its Lie algebra is simple. We know that $\pi_3(SU(3))=Z$. Let us consider the exemplary case where $H=SU(2)\times SU(2)\times U(1)$. Do we get

$$\pi_3(G/H)=\pi_3(G)=\pi_3(SU(3)\times SU(3))=\pi_3(SU(3))\times\pi_3(SU(3))=Z\times Z?$$

Answer 1

No. Looking at the relevant part of the long exact sequence on homotopy groups, we have an exact sequence $$\pi_3(H)\to\pi_3(G)\to \pi_3(G/H)\to \pi_2(H).$$ Since $\pi_2(H)$ is trivial ($\pi_2$ of any Lie group is trivial), this means $\pi_3(G/H)$ will be isomorphic to $\pi_3(G)$ iff the image of $\pi_3(H)$ in $\pi_3(G)$ is trivial. So for instance, taking $G=SU(3)$ and $H=SU(2)$, $\pi_3(H)\cong\mathbb{Z}$ and the map $\pi_3(H)\to \pi_3(G)$ is an isomorphism, so $\pi_3(G/H)$ will be trivial rather than $\mathbb{Z}$ (in fact $G/H\cong S^5$ in that case since $H$ is exactly the stabilizer of a point in the usual action of $G$ on $\mathbb{C}^3$, so the quotient is the orbit of a point which is $S^5$).

Answer 2

Eric's answer is fine, but a bit more can be said. Mainly, it is very rare for $\pi_3(G/H)$ to be isomorphic to $\mathbb{Z}$!

Proposition: Suppose $H\subseteq G$ are compact Lie groups and $H$ is non-abelian. Then $\pi_3(G/H)$ has smaller rank than $\pi_3(G)$.

In particular, if $G$ is simple and $H$ is non-abelian, then $\pi_3(G/H)$ is a finite cyclic group. (On the other hand, if $H$ is abelian, then the long exact sequence in Eric's answer shows that $\pi_3(G/H)$ is canonically isomorphic to $\pi_3(G)$.)

So, why is the proposition true? The key term in the Dynkin index (see https://mathoverflow.net/questions/90124/3rd-homotopy-group-of-a-compact-simple-lie-group).

In more detail, the universal cover of $H$ splits as $\mathbb{R}^n\times H_1\times ... \times H_k$ where each $H_i$ is a compact simply connected simple Lie group. The assumption that $H$ is non-abelian implies that $k\geq 1$. Consider the induced map $H_1\rightarrow H\rightarrow G$ where the first map is the restriction of the covering projection to $H_1$. Since $H_1$ is simple, this gives an induced map $\mathbb{Z}\cong \pi_3(H)\rightarrow \pi_3(G)\cong \mathbb{Z}^s$, where $s$ is the number of (non-abelian) simple factors in the universal cover of $G$. If this map is non-trivial, it follows that the map $\pi_3(H)\rightarrow \pi_3(G)$ is non-trivial, which would then prove the proposition.

But why is this map non-trivial? Well, the map on $\pi_3$ induced by an inclusion of a simple Lie group into another simple group is known as they Dynkin Index. It can be computed in terms of the restriction of the Killing form on $G$ to that of $H$, an in particular, is known to always be a positive integer.