- #1

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- TL;DR Summary
- Are these two (infinite dimensional) Lie algebras isomorphic or not?

Let ##\mathfrak{A}:=\operatorname{span}\left\{D_n:=x^n\dfrac{d}{dx}\, : \,n\in \mathbb{Z}\right\}## and ##\mathfrak{B}:=\operatorname{span}\left\{E_n:=x^n\dfrac{d}{dx}\, : \,n\in \mathbb{N}_0\right\}## with the usual commutation rule.

My question is: How can we prove or disprove the Lie algebra isomorphism ##\mathfrak{A}\cong \mathfrak{B}?##

Multiplication goes: ##[D_n,D_m]=(m-n)D_{n+m-1}## and ##[E_n,E_m]=(m-n)E_{n+m-1}.##

The easy invariants (dimension ##\aleph_0##, center ##\{0\}##, derived algebra ##[\mathfrak{A},\mathfrak{A}]=\mathfrak{A},[\mathfrak{B},\mathfrak{B}]=\mathfrak{B}##, ideals - none) are the same. My suspicion is that they are not isomorphic, since there are infinitely many subalgebras ##\mathfrak{sl}(2)\cong\operatorname{span}\{D_{-n+1},D_1,D_{n+1}\}\leq \mathfrak{A}## and as far as I can see only one ##\mathfrak{sl}(2)\cong \operatorname{span}\{D_0,D_1,D_2\}\leq \mathfrak{B}.## However, this is not obvious (to me) and any manual calculations are a mess of indices. Other common properties (solvability, semisimplicity, Killing-form) aren't of help, either, since we have an infinite-dimensional vector space.

##D_1## is almost a ##1## in both Lie algebras, and presumably their maximal toral subalgebra. So how can we prove, that there aren't any other copies of ##\mathfrak{sl}(2)## in ##\mathfrak{B}## than the obvious one? Or is there an easy invariant I haven't thought of?

My question is: How can we prove or disprove the Lie algebra isomorphism ##\mathfrak{A}\cong \mathfrak{B}?##

Multiplication goes: ##[D_n,D_m]=(m-n)D_{n+m-1}## and ##[E_n,E_m]=(m-n)E_{n+m-1}.##

The easy invariants (dimension ##\aleph_0##, center ##\{0\}##, derived algebra ##[\mathfrak{A},\mathfrak{A}]=\mathfrak{A},[\mathfrak{B},\mathfrak{B}]=\mathfrak{B}##, ideals - none) are the same. My suspicion is that they are not isomorphic, since there are infinitely many subalgebras ##\mathfrak{sl}(2)\cong\operatorname{span}\{D_{-n+1},D_1,D_{n+1}\}\leq \mathfrak{A}## and as far as I can see only one ##\mathfrak{sl}(2)\cong \operatorname{span}\{D_0,D_1,D_2\}\leq \mathfrak{B}.## However, this is not obvious (to me) and any manual calculations are a mess of indices. Other common properties (solvability, semisimplicity, Killing-form) aren't of help, either, since we have an infinite-dimensional vector space.

##D_1## is almost a ##1## in both Lie algebras, and presumably their maximal toral subalgebra. So how can we prove, that there aren't any other copies of ##\mathfrak{sl}(2)## in ##\mathfrak{B}## than the obvious one? Or is there an easy invariant I haven't thought of?