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Hi,

I am currently working with Jacobi groups and I am not clear why $R^2 \cdot s_1$ is a central extension of $R^2$ by $s_1$

$R^2 \cdot s_1$ is the following constructed group:

  • The elements of $R^2 \cdot s_1$ are the pairs (x,s) with $x \in R^2 $ and $s \in s^1$

  • The product $(x, s) \cdot (x', s') := (x+x',ss'e^{2\pi i(det(|x | x'|))})$ where $det(|x | x'|) = \lambda \mu ' - \lambda ' \mu$ , $x = (\lambda, \mu)$

With calculation it's clear that $R^2 \cdot s_1$ is a group

  • that's clear to me

$s_1$ is a subgroup of $R^2 \cdot s_1$ because of s->(0, s)

  • that's clear to me

$s_1$ is the center of $R^2 \cdot s_1$

  • that's clear to me

Now the unclear part comes and I don't know how this can be proofed.

$R^2 \cdot s_1$ is a central extension of $R^2$ by $s_1$

It tried properties of central groups, like

$1 -> s_1 -> R^2 \cdot s_1 -> R^2 -> 1$

calculating the direct product

$R^2 \cdot s_1 = s_1 \times R^2$

trying to proof surjectivity

$R^2 \cdot s_1 -> R^2$

and injectivity

$s_1 -> R^2 \cdot s_1$

but without success.

Do you have in clue how to proceed.

Thanks in advance.

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  • $\begingroup$ You need to show $R^2 = R^2\cdot s_1 / s_1$. But it looks to me like $R^2\cdot s_1$ is the semidirect product of $R^2$ and $s_1$. en.wikipedia.org/wiki/Semidirect_product $\endgroup$
    – David Roberts
    Jan 19, 2011 at 6:10
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    $\begingroup$ $(x,s)\mapsto x$ is clearly surjective and $s\mapsto (0,s)$ is injective. Whats the difficulty? $\endgroup$ Jan 19, 2011 at 6:22
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    $\begingroup$ @David: It cannot be a semidirect product, as it is a nontrivial central extension. $\endgroup$
    – BS.
    Jan 20, 2011 at 16:23

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