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Is there a Lie Algebra $\mathfrak g$ so that the extension

$$0\xrightarrow{}\mathfrak h\xrightarrow{}\mathfrak g\xrightarrow{}\mathfrak q\xrightarrow{}0$$

does not split, i.e. $\mathfrak g$ is not a semi-direct product of $\mathfrak h$ and $\mathfrak q$? As I understand it, $\mathfrak q$ should not be identifiable with a subalgebra of $\mathfrak g$, right? Since $\mathfrak h$ is already an ideal in $\mathfrak g$ as the kernel.

Mike Pierce
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rabipelais
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1 Answers1

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I guess the Heisenberg algebra will do. This is the algebra $\mathfrak{g}$ with basis $a, b, c$, and products $[a, b] = c$, $[a, c] = [b, c] = 0$. Take $\mathfrak{h} = \langle c \rangle$. Then $\mathfrak{q}$ is a two-dimensional abelian algebra, and all such object in $\mathfrak{g}$ contain $c$.

Andreas Caranti
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  • I think I follow, $\mathfrak g/\mathfrak h$ would be abelian because $\mathfrak h = [\mathfrak g, \mathfrak g]$, and two-dimensional because $\mathfrak h$ is one-dimensional, and it must contain c, otherwise it wouldn't be closed under the lie bracket. But what does that mean? That $\mathfrak q$ cannot be the quotient of the other two and therefore not a subalgebra of $\mathfrak g$? – rabipelais Jan 20 '13 at 23:56
  • Ah, I think I see it, now, how dumb. Because q contains c it is not a direct product to h. Thanks. – rabipelais Jan 21 '13 at 03:08
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    @rabipelais Precisely. Sorry, I could have added a couple of words to make this clear. – Andreas Caranti Jan 21 '13 at 06:50
  • @AndreasCaranti please also see this query http://math.stackexchange.com/questions/1819295/splitting-and-non-splitting-extensions-in-lie-algebras – IgotiT Jun 09 '16 at 04:49