# What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $G\not\cong G/H \times H$?

What is an example of an abelian Lie group $$G$$ and a closed subgroup $$H$$ such that $$G\not\cong G/H \times H$$?

Would the circle $$S^1$$ in $$\mathbb R^2$$ be an example? what is $$\mathbb R^2/S^1$$?

• Isn't $G=\mathbb{R}$ and $H=\mathbb{Z}$ an example of the non-iso you want? All you need to show is that $\mathbb{R}$ is not iso to $S^1 \times \mathbb{Z}$. That's easy. – Randall Apr 8 '19 at 3:14

EDIT: To be clear I was doing the case when $$H$$ was assumed connected. The disconnected case is handled below by Randall.

Every connected real abelian Lie group $$G$$ is isomorphic to $$\mathbb{R}^m\times (S^1)^n$$ for some $$m$$ and $$n$$. In fact, given $$G$$ you can read off $$m$$ and $$n$$ as $$n=\mathrm{rank}(\pi_1(G))$$ and $$m=\dim G-n$$.

Now, if you have a short exact sequence of abelian Lie groups

$$0\to H\to G\to G/H\to 0$$

Then evidentily $$\dim G=\dim H+\dim G/H$$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence

$$0\to \pi_1(H)\to\pi_1(G)\to\pi_1(G/H)\to 0$$

So, $$\mathrm{rank}(\pi_1(G))=\mathrm{rank}(\pi_1(H))+\mathrm{rank}(\pi_1(G/H))$$. Combining these two gives that $$G\cong H\times G/H$$ as desired

EDIT: Here are more details. To show that $$G\cong H\times (G/H)$$ it suffices to show that

$$\mathrm{rk}(\pi_1(G))=\mathrm{rk}(\pi_1(H\times (G/H))=\mathrm{rk}(\pi_1(G))+\mathrm{rk}(\pi_1(G/H))$$

and

$$\mathrm{dim}(G)-\mathrm{rk}(\pi_1(G))=\dim(G\times (G/H))-\mathrm{rk}(\pi_1(H\times (G/H))$$

The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:

\begin{aligned}\dim(G)-\mathrm{rk}(\pi_1(G)) &= \dim(H)+\dim(G/H)-(\mathrm{rk}(\pi_1(H))+\mathrm{rk}(\pi_1(G/H))\\ &= \dim(G\times G/H))-\mathrm{rank}(\pi_1(G\times (G/H)))\end{aligned}

(Below is for the non-abelian situation) Here's a simple interesting example.

Take $$\mathrm{GL}_2(\mathbb{C})$$ with its center $$Z:=\{\lambda I_2:\lambda\in\mathbb{C}^\times\}$$. Then, $$\mathrm{GL}_2(\mathbb{C})/Z\cong \mathrm{PGL}_2(\mathbb{C})$$. To see that $$\mathrm{GL}_2(\mathbb{C})\not\cong Z\times\mathrm{PGL}_2(\mathbb{C})$$ note that the derived (i.e. commutator) subgroup of the former is $$\mathrm{SL}_2(\mathbb{C})$$ whereas the latter is $$\mathrm{PGL}_2(\mathbb{C})$$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.

• Thank you very much Alex. So the bundle $G\to G/H$ is always trivial! – user328669 Apr 8 '19 at 1:34
• @AmratA No problem. Did you see the updated affirmative answer to the abelian situation? – Alex Youcis Apr 8 '19 at 1:36
• Oh yes, I just did. Thanks again! – user328669 Apr 8 '19 at 1:37
• @AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $G\cong H\times (G/H)$, not that the sequence splits. – Alex Youcis Apr 8 '19 at 1:58
• @AmratA Updated. – Alex Youcis Apr 8 '19 at 2:08

Take $$G = \mathbb{R}$$ and $$H=\mathbb{Z}$$. The quotient $$G/H$$ is the circle $$S^1$$. The question is now to compare $$\mathbb{R}$$ to $$S^1 \times \mathbb{Z}$$. Now, whether you interpret $$\ncong$$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $$\mathbb{R}$$ is connected but $$S^1 \times \mathbb{Z}$$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $$2$$ ($$\mathbb{R}$$ has none, $$S^1 \times \mathbb{Z}$$ has at least one).

• As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this. – Alex Youcis Apr 8 '19 at 3:37