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The rational global dimension of a graded algebra $A=(A_k)_{k\geq 0}$, with $A_0=\mathbb Q$, denoted here ${\rm gl}\dim A$ is defined to be the greatest integer $k$ (or $\infty$) such that ${\rm Ext}^k_{A}(\mathbb Q,-)\neq 0$.

I'm interested on the case when $A=L$ is a graded Lie algebra to compare $\frac{\ln (\dim L)}{{\rm gl}\dim L}$ and $\ln 2$.

Any comments or references are welcome. Thank you

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    $\begingroup$ Dear MyIsmail, you should add more tags. Even though your problem comes from a topological question it is indeed an algebraic problem. Best $\endgroup$ Apr 12, 2015 at 17:36
  • $\begingroup$ @OliverStraser: Thanks for the suggestion. It is done $\endgroup$
    – MyIsmail
    Apr 12, 2015 at 18:46
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    $\begingroup$ There are two definitions of graded Lie algebra (with grading in nonnegative integers), one being a Lie algebra with a multiplicative grading, and one being a non-associative algebra law and a multiplicative grading, and a "graded-Lie" condition. I'm not sure the OP and the answer agree on the definition, and I'm not sure at all what the OP has in mind, so this should be clarified. (It's quite dire to use such ambiguous terminologies but this is what some people do; the term Lie superalgebra for the second meaning is much better.) $\endgroup$
    – YCor
    Nov 18, 2018 at 16:17

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Not a complete answer, but here are some examples which might be informative:

First consider a finite-dimensional Lie algebra $L$ over an arbitrary field $k$. Assume that $L$ is concentrated in purely even degrees. Set $d = dim(L)$. Let $U(L)$ be the universal enveloping algebra of $L$. Then there is a standard (finite!) $U(L)$-free resolution of the trivial module $k$,

$$P_\bullet: U(L) \otimes \Lambda^d(L) \rightarrow U(L) \otimes \Lambda^{d-1}(L) \rightarrow \cdots \rightarrow U(L) \otimes \Lambda^1(L) \rightarrow U(L) \rightarrow k,$$

where $\Lambda^\bullet(L)$ denotes the exterior algebra on $L$. Since $U(L)$ is a Hopf algebra, it follows for any (graded) $L$-module $M$ that $P_\bullet \otimes M$ is a $U(L)$-projective resolution of $M$. Then $\textrm{Ext}_{U(L)}^i(M,-)=0$ for $i > d$, so $\textrm{gl dim}(L) \leq d = dim_k(L)$. If $L$ is abelian, then one gets

$$\textrm{Ext}_{U(L)}^i(k,k) \cong \textrm{Hom}_k(\Lambda^i(L),k) \cong \Lambda^i(L^*)$$ and hence $\textrm{gl dim}(L) = d$.

On the other hand, let $L$ be a one-dimensional abelian graded Lie algebra spanned by an element $x$ of $\mathbb{Z}$-degree $1$. Assume that $k$ is not of characteristic $2$. Then

$$U(L) = k[x]/(xx-(-1)^{1 \cdot 1}xx-0) = k[x]/(x^2+x^2) = k[x]/(2x^2) = k[x]/(x^2)$$

i.e., $U(L) = \Lambda(x)$ is an exterior algebra generated by $x$. Then a minimal projective resolution of the trivial $L$-module $k$ is given by the periodic resolution $$ k \leftarrow \Lambda(x) \stackrel{x}{\leftarrow} \Lambda(x) \stackrel{x}{\leftarrow} \cdots $$ in which the arrows denote multiplication by $x$. From this it follows that $\textrm{Ext}_{U(L)}^i(k,k) \neq 0$ for all $i \geq 1$, and hence $\textrm{gl dim}(L) = \infty$.

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  • $\begingroup$ Are you sure the OP means Lie superalgebras? $\endgroup$
    – YCor
    Nov 18, 2018 at 16:26

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