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The short question is the following: If a positively graded Lie algebra $\mathfrak g$ over a field $F$ is Koszul, is the center of $\mathfrak g$ concentrated in degree $1$?

Let us be more precise. A Lie algebra $\mathfrak g$ over an arbitrary field $F$ is positively graded if it decomposes as a direct sum of (finite dimensional) vector spaces $\mathfrak g_i$, i.e., $\mathfrak g=\bigoplus_{i\geq 1}\mathfrak g_i$, and the Lie brackets induce a graded mapping $\mathfrak g\otimes \mathfrak g\to \mathfrak g$, i.e., $[-,-]:\mathfrak g_i\otimes \mathfrak g_j\to\mathfrak g_{i+j}$. Such a Lie algebra is said to be Koszul if the bigraded cohomology $\text{Ext}^{\bullet,\bullet}_{U(\mathfrak g)}(F,F)$ is concentrated on the diagonal, where $U(\mathfrak g)$ is the universal envelope of $\mathfrak g$, whose grading is induced by that of $\mathfrak g$. In particular, Koszul Lie algebras are generated (as Lie algebras) by elements of degree $1$, and their relations can be chosen to be of degree $2$: This means that there is a graded presentation $0\to (R)\to \mathfrak f\to \mathfrak g\to 0$, where all the maps have of degree $0$, the restriction $\mathfrak f_1\to \mathfrak g_1$ is an isomorphism, and $R\leq \mathfrak f_2$, $(R)$ being the ideal of $\mathfrak f$ generated by $R$. A Lie algebra with such a presentation is said to be quadratic.

(Good reference for quadratic/Koszul algebras is [1]=Quadratic Algebras, Polishchuk, Positselski. Koszul Lie algebras are treated in [2]=Graded Lie algebras of type FP, by Thomas Weigel)

Now, by looking at some examples I convinced myself that the answer to my question is YES, the center of a Koszul Lie algebra is concentrated in degree $1$.

I am now listing here two examples:

  1. Let $\mathfrak g$ be the graded Lie algebra given by the following presentation, $$\langle x_1,y_1,\dots,x_n,y_n\vert [x_i,x_j]=[y_i,y_j]=0, [x_i,y_j]=0,\text{for }i\neq j, [x_i,y_i]=[x_1,y_1]\rangle$$ where all the generators have degree $1$. Then $\mathfrak g$ is quadratic and non-Koszul (see [1], example 1 section 2). Moreover, it has a non-trivial central element of degree $2$.

  2. Consider the class of right-angled Artin Lie algebras (see [2]). Then every Lie algebra in such a class is Koszul and every central element has degree $1$.

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    $\begingroup$ This starts breaking if we allow Lie superalgebras (the free Lie superalgebra on one odd generator has a central element of degree 2), so it would be a little bit surprising that the Koszul property alone is sufficient for this sort of claim - but I do not see any obvious counterexamples. $\endgroup$ Commented Oct 6, 2021 at 15:33

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