# Is $Ext^1_{Lie-Gr}(\mathbb C, \mathbb C^*)=0$?

I want to know if any Lie group extension

$$1\to \mathbb C^*\to A\xrightarrow{\pi} \mathbb C\to 0 \tag{1}\label{1}$$

is trivial, where the group structure is multiplicative on $$\mathbb C^*$$ and additive on $$\mathbb C$$.

I guess the answer is Yes and it suffices to find a section $$s:\mathbb C\to A$$ as Lie group morphism such that $$\pi\circ s=Id$$. From the point of view of topology, $$A$$ is a principal $$\mathbb C^*$$-bundle over the contractible base $$\mathbb C$$. This implies $$A$$ is trivial as principal $$\mathbb C^*$$-bundle, so there is a section $$s:\mathbb C\to A$$. However, the section need not preserve the group structure. Can I modify the section $$s$$ so as preserve the group structure?

Edited remark: As @Roland's points out, considering the extension $$(1)$$ purely as group extension which lives in $$Ext_{Ab}^1(\mathbb C,\mathbb C^*)$$ is different from an extension as Lie groups (or algebraic groups), because a section in the category of groups does not need to be even continuous.

Second Edition: Here is another idea: If we pull back $$A$$ to its universal covering $$\require{AMScd}$$ $$\begin{CD} \mathbb C @>>> \tilde{A}@>\tilde{\pi}>>\mathbb C\\ @VVV @VVV @VV=V \\ \mathbb C^* @>>> A @>\pi>> \mathbb C \end{CD}$$ If I can show the sequence on the first row is trivial, then a section $$\tilde{s}:\mathbb C\to \tilde{A}$$ will descend to a section $$s:\mathbb C\to A$$, but how do we show $$Ext^1_{Lie-gr}(\mathbb C,\mathbb C)=0$$ (although it sounds trivial)?

• If you don't care about topology, you can simply say that $\mathbb{C}^*$ is injective as an abelian group (because it is divisible). Mar 24, 2020 at 7:39
• @Roland Does that imply $Ext^1(G,\mathbb C^*)=0$ for all group $G$? But take $G$ to be an abelian variety, certainly there are semi-abelian varieties which are nontrivial $\mathbb C^*$-extension of an abelian variety. Mar 24, 2020 at 18:29
• Yes $Ext^1(G,\mathbb{C}^*)=0$ for all abelian group $G$. But this note that this claim is purely a group theoretic one. There is no topology involved, much less algebraic structure. You might as well think that $G$ and $\mathbb{C}^*$ are discrete. Or you might as well think that if $1\to\mathbb{C}^*\to A\to G\to 0$ is a short exact sequence, first there is no given topological/algebraic structure on $A$ (maybe there is one, several or none !), and even if $A$ has one, the claim that the extension splits in the category of groups mean there is a section, but not necessarily continuous/regular. Mar 24, 2020 at 19:54
• This is somehow imprecise in your post, $Ext^1$ in which category ? The category of abelian groups ? commutative topological groups ? commutative algebraic groups ? There are obvious morphisms $Ext^1_{Alg-Gr}(A,B)\to Ext^1_{Top-Gr}(A,B)\to Ext^1_{Ab}(A,B)$ but these morphisms are not isomorphisms in general. Mar 24, 2020 at 19:59
• @Roland Thanks for your comment! I think I really want the extension in the category of Lie groups, and I have edited my question. Actually I was trying to calculate $Ext^1_{Lie-Gr}(J,\mathbb C^*)$, where $J$ is a compact torus $\mathbb C^n/\Lambda$. Mar 24, 2020 at 21:21