6
$\begingroup$

Let $V$ be a $\mathbb{Z}_2$-graded vector space (aka super vector space) and $L(V)$ be the free $\mathbb{Z}_2$-graded Lie algebra (aka super Lie algebra). The free super Lie algebra is also graded by the number of generators (the generators $V$ sit in degree 1, and the Lie bracket itself has degree zero in this sense). So it has a decomposition by degrees, $L(V) = \sum_{n=1}^\infty L^n(V)$. The homogeneous components $L^n(V)$ may be called (super) Lie powers of $V$ (by analogy with symmetric or alternating powers) and have an embedding $L^n(V) \subset V^{\otimes n}$, where the Lie (super)commutators are sent to tensor (super)commutators, $[a,b] = a\otimes b - (-)^{|a||b|} b \otimes a$.

Q: What is the $\mathrm{GL}(V)$-representation theoretic description of $L^n(V)$?

When $V$ is purely even (has no odd component), the question is answered by Klyachko's theorem (as explained in the answers to MO187545). So basically, I'm asking: What is the super Lie algebra analog of Klyachko's theorem? And what is a reference that clearly states it?

$\endgroup$
8
$\begingroup$

Let me work over $\mathbb{C}$ for simplicity. We have

$$L(V) \cong \bigoplus_{n\ge0} V^{\otimes n} \otimes_{S_n} \text{Lie}(n),$$

where $\text{Lie}(n)$ is the $n^{th}$ space of the Lie operad, with specific $S_n$ actions on $\text{Lie}(n)$ and $V^{\otimes n}$ to be precised below; this is, to be clear, true for any $V$, in great generality, and is a special case of a very general fact about free algebras over symmetric operads $O$, namely that they are given by

$$V \mapsto \bigoplus_{n \ge 0} V^{\otimes n} \otimes_{S_n} O(n).$$

This is Proposition 5.2.5 in Loday and Vallette's Algebraic Operads; unfortunately there it is only stated for vector spaces but the argument generalizes with no modifications to $\mathbb{Z}_2$-graded vector spaces, $\mathbb{Z}$-graded vector spaces, or for that matter any cocomplete symmetric monoidal category (which, to be safe, includes the condition that tensor product is cocontinuous in each variable).

In the graded case, the action of $S_n$ on $V^{\otimes n}$ is by permutations with the Koszul sign, namely it is generated by the action

$$v_1 \otimes \cdots v_i \cdots v_j \cdots \otimes v_n \mapsto (-)^{|v_i||v_j|} v_1 \otimes \cdots v_j \cdots v_i \cdots \otimes v_n$$

of permutations $(i j)$ on products of homogeneous elements. Then, $\text{Lie}(n)$ is known to be isomorphic as an $S_n$-representation to the induced representation $\text{Ind}_{C_n}^{S_n} \chi$, where $C_n$ is the cyclic group of order $n$ generated by the cyclic permutation $c = (12\cdots n)$ and $\chi$ is its $1$-dimensional representation $c^k \mapsto e^{ \frac{ 2 \pi i k}{n} }$. This gives

$$\begin{eqnarray*} L^n(V) &\cong& V^{\otimes n} \otimes_{S_n} \text{Ind}_{C^n}^{S_n} \chi \\ &\cong& V^{\otimes n} \otimes_{C_n} \chi \end{eqnarray*}.$$

This is equivalent to Klyachko's theorem in the purely even case, but stating it this way makes it clear that the graded case is taken care of by the action of $S_n$ on $V^{\otimes n}$. Explicitly, if $v_1 \otimes \dots \otimes v_n \in V^{\otimes n}$ and each $v_i$ is homogeneous of degree $|v_i|$ then the cyclic permutation which commutes $v_1$ past all the other vectors produces

$$(-1)^{|v_1|(|v_2| + \dots + |v_n|)} v_2 \otimes \dots \otimes v_n \otimes v_1.$$

For example if the $v_i$ are all odd then the sign is $(-1)^n$ which matches the sign of the $n$-cycle as a permutation in $S_n$.

I guess we can be even more specific as follows. If $V$ is purely even then Schur-Weyl duality gives

$$V^{\otimes n} \cong \bigoplus_{\lambda \vdash n} S_{\lambda}(V) \otimes M^{\lambda}$$

where $S_{\lambda}$ is a Schur functor and $M^{\lambda}$ is a Specht module. This gives that $V^{\otimes n} \otimes_{C_n} \chi$ is a sum of Schur functors where $S_{\lambda}$ appears with multiplicity $\dim M^{\lambda} \otimes_{C_n} \chi$. If $V$ is purely odd then $V^{\otimes n}$ is modified as an $S_n$-representation by the sign representation which has the effect of replacing every $M^{\lambda}$ with $M^{\lambda^T}$ where $\lambda^T$ is the conjugate partition (I forget what the standard notation for this is), or equivalently replacing every $S_{\lambda}$ with $S_{\lambda^T}$. And the general case is a free product of the purely even and purely odd cases.

$\endgroup$
4
  • $\begingroup$ You've stated Klyachko's theorem in the even case, thanks. But the signs are important. What is the thing that needs to be changed to get the graded case? Replace $\mathrm{Lie}(n)$ by $\mathrm{sLie}(n)$ or something like that? Then the representation theoretic information needs to be updated somehow. The $L^2(V)$ case alone is elementary and follows from definitions. What would be really useful is a clear reference! $\endgroup$ – Igor Khavkine Jan 14 at 20:46
  • 1
    $\begingroup$ @Igor: nothing needs to be changed about the Lie operad, the signs are happening entirely in $V$. If $v_1 \otimes \dots \otimes v_n \in V^{\otimes n}$ where each $v_i$ is homogeneous of degree $|v_i|$ then acting by a cyclic permutation amounts to commuting $v_1$ past everything else so we get $(-1)^{|v_1|(|v_2| + \dots + |v_n|)} v_2 \otimes \dots \otimes v_n \otimes v_1$. $\endgroup$ – Qiaochu Yuan Jan 14 at 20:51
  • $\begingroup$ OK, is it correct to distill the added information like this? The isomorphism $L^n(V) \cong V^{\otimes n} \otimes_{S_n} \mathrm{Lie}(n)$ is still true, but the action of $S_n$ on $V^{\otimes n}$ adds the Koszul sign to the usual permutation action? If so, it would help to state that at the top of the answer. And a reference for that isomorphism (including grading) would be really nice! $\endgroup$ – Igor Khavkine Jan 14 at 21:06
  • $\begingroup$ @Igor: yes, I added a bit to the answer clarifying this. It's a completely general fact about free algebras over operads. I don't know a reference off the top of my head but I can try to find one. $\endgroup$ – Qiaochu Yuan Jan 14 at 21:13
0
$\begingroup$

I have found a more narrow reference. It deals quite specifically with my question (confirming Qiauchu Yuan's answer), as well as some generalizations thereof:

Kwon, Jae-Hoon, Free Lie-superalgebras and the representations of $\mathfrak{gl}(m,n)$ and $\mathfrak{q}(n)$ (doi), J. Korean Math. Soc. 42, No. 2, 365-386 (2005). ZBL1168.17301.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.