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 $\mathbb R^2/S^1$?

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    $\begingroup$ 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. $\endgroup$ – Randall Apr 8 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

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


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

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.

  • $\begingroup$ Thank you very much Alex. So the bundle $G\to G/H$ is always trivial! $\endgroup$ – Amrat A Apr 8 at 1:34
  • $\begingroup$ @AmratA No problem. Did you see the updated affirmative answer to the abelian situation? $\endgroup$ – Alex Youcis Apr 8 at 1:36
  • $\begingroup$ Oh yes, I just did. Thanks again! $\endgroup$ – Amrat A Apr 8 at 1:37
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    $\begingroup$ @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. $\endgroup$ – Alex Youcis Apr 8 at 1:58
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    $\begingroup$ @AmratA Updated. $\endgroup$ – Alex Youcis Apr 8 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).

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    $\begingroup$ 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. $\endgroup$ – Alex Youcis Apr 8 at 3:37

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