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)?

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    $\begingroup$ If you don't care about topology, you can simply say that $\mathbb{C}^*$ is injective as an abelian group (because it is divisible). $\endgroup$
    – Roland
    Mar 24, 2020 at 7:39
  • $\begingroup$ @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. $\endgroup$
    – AG learner
    Mar 24, 2020 at 18:29
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    $\begingroup$ 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. $\endgroup$
    – Roland
    Mar 24, 2020 at 19:54
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    $\begingroup$ 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. $\endgroup$
    – Roland
    Mar 24, 2020 at 19:59
  • $\begingroup$ @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 $\mathbb C^n/\Lambda$. $\endgroup$
    – AG learner
    Mar 24, 2020 at 21:21


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