Let $k$ denote a field of characteristic $0$ (assume algebraically closed for convenience). Define $J=k\langle x,y|[x,y]=y^{2}\rangle$. This noncommutative algebra (which can be viewed as a derivation ring over a commutative polynomial ring) is often referred to as the "Jordan plane" in the literature.

It can be see that $J\ncong \mathcal{U}(\mathfrak{g})$ (as algebras) for any Lie algebra $\mathfrak{g}$ (since $J/\langle [J,J]\rangle $ is not semiprime). My question is as follows: does there exist a Lie algebra $\mathfrak{g}$ and ideal $I\subseteq \mathcal{U}(\mathfrak{g})$ such that $J\cong \mathcal{U}(\mathfrak{g})/I$ (as algebras)? I guess that the answer should be "no", but I'm not sure how to go about proving it. I also assume this should be well known, given the depth of literature/ knowledge about the Jordan plane.

EDIT: I'm looking for a finite dimensional Lie algebra $\mathfrak{g}$ with these properties.

  • 1
    $\begingroup$ I'm not sure if protection is the right thing to do here, but the question asker effectively destroyed the question after it was answered. $\endgroup$ Mar 3, 2015 at 19:41
  • $\begingroup$ Jordanblock, welcome to MathOverflow. As you are new here, I thought I should let you know that destruction of one's questions is considered a violation of site rules; the offense is considered even more serious after other users have put some work into responding. You should also know that your question may well be useful to others with this (or a similar) question in the future; it's not a bad question. $\endgroup$
    – Todd Trimble
    Mar 3, 2015 at 20:04

1 Answer 1


Original answer without the finite-dimensionality requirement:

Yes, trivially so. For every algebra $A$, one can consider the underlying Lie-algebra $\mathfrak{a}$ of $A$ and gets a surjection $U(\mathfrak{a}) \twoheadrightarrow A$ because of the universal mapping property of $U(\mathfrak{a})$.

New answer:

An algebra-homomorphism $f:U(\mathfrak{g})\to J$ is surjective iff $f(\mathfrak{g})\subseteq J$ is an algebra-generating-set of $J$.

It is easy to see that $J$ is a $\mathbb{N}$-graded algebra with $\deg(x)=\deg(y)=1$ and $\{y^a x^b \mid a,b\in\mathbb{N}\}$ is a homogenuous $k$-basis of $J$. This proves that if $f$ is surjective, then there must be elements $\hat{x},\hat{y}\in\mathfrak{g}$ with $f(\hat{x})=x+\text{higher degree terms}$ and $f(\hat{y})=y+\text{higher degree terms}$.

Now observe that $x\cdot y^a x^b = y^a x^{b+1} + a y^{a+1} x^b$ holds in $J$ so that $[x,y^a] = a y^{a+1}$. This shows $$[\underbrace{x,[x,\ldots,[x}_{a}, y]]] = a! y^{a+1}$$ $$\implies [\underbrace{f(\hat{x}),[f(\hat{x}),\ldots,[f(\hat{x})}_{a}, f(\hat{y})]]] \equiv a! f(\hat{y})^{a+1} \mod\text{terms of degree}>a+1$$ Since the right hand sides are linearly independent by degree reasons, this shows that the elements $[\underbrace{\hat{x},[\hat{x},\ldots,[\hat{x}}_{a},\hat{y}]]]$ of $\mathfrak{g}$ must also be linearly independent so that $\mathfrak{g}$ cannot be finite-dimensional.

  • $\begingroup$ Edited the question. Should have specified that I wanted the Lie algebra to be finite dimensional. $\endgroup$ Mar 2, 2015 at 18:23
  • $\begingroup$ @Jordanblock: I edited my answer to adress the finite-dimensionality requirement. $\endgroup$ Mar 2, 2015 at 19:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .