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Problem: Prove that (up to isomorphisms) exists a unique Lie algebra $\mathfrak{g}$ with $\dim(\mathfrak{g}) = 3$ such that $\dim([\mathfrak{g},\mathfrak{g }]) = 1$ and $[\mathfrak{g},\mathfrak{g}]\subseteq Z(\mathfrak{g})$.

I don't even know how to start, I guess that is not too difficult but I need an initial guide.

Tryncha
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  • I would start by finding a non-zero vector in $[g,g]$ and extending it to a basis of $g$. – mr_e_man Mar 10 '23 at 04:36
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    Please avoid posting "I don't even know how to start". One can always start with some examples of $3$-dimensional Lie algebras, or have a look at MSE here. If you start doing something, then you have more context for posting where you got stuck. – Dietrich Burde Mar 10 '23 at 13:43
  • @DietrichBurde Thank you, I'll have it in mind. – Tryncha Mar 10 '23 at 20:50

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Some pointers to get you started.

As a vector space $\langle z \rangle = [\mathfrak{g}, \mathfrak{g}]$ for some $z \in \mathfrak{g}$. Note that $z \in Z(\mathfrak{g})$ by assumption.

Extend to a basis $z,x,y$ of $\mathfrak{g}$. What can you say about the commutators between the basis elements?

spin
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