5
$\begingroup$

Let $V$ be a 2-dimensional vector space (over, say, $\mathbb{Q}$). Let $FL$ be the free lie algebra on $V$, then there is a natural action of the group $SL(V)$ on $FL$, such that the action of $-I$ is by multiplication by $(-1)^d$ on the $d$th graded component.

An article I'm reading seems to imply that the representation of $SL(V)$ on each graded piece $FL^d$ of $FL$ decomposes into a direct sum of "$[n]$" for various $n\ge 1$ where $[n]$ is the representation of $SL(V)$ on the symmetric power $Sym^n(V)$.

Is this true? (or could I be misinterpreting something?)

To me, (from my understanding of Serre's "Lie Algebras and Lie Groups") I think of $FL$ as the Lie subalgebra of the tensor algebra $T(V)$ generated in degree 1, and certainly $SL(V)$ acts on the graded pieces of the tensor algebra $T^d(V)$, and hence on $Sym^d(V)$, but $FL^d$ is a subspace of $T^d(V)$, and I don't see why the $SL(V)$ representation on $FL^d$ needs to decompose into representations on the quotient $Sym^d(V)$.

I'm a novice in this area, so I apologize if this question is naive.

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ Every (finite-dim) irreducible rep of $SL_2$ is isomorphic to $[n]$ for some $n$. Hence every finite-dim rep of $SL_2$ is a direct sum of such $[n]$. This is just what they say, and is not related to your specific representation (the $d$-th graded component in the free Lie algebra on 2 generators). $\endgroup$
    – YCor
    Commented Aug 31, 2017 at 7:48
  • 1
    $\begingroup$ The decomposition of $FL^d$ into a sum of $[n]$ for various $n$ is just an isomorphism of abstract $SL_2$ representations, and as YCor says, this type of decomposition holds for any finite dimensional $SL_2$ representation. I don't think the decomposition in this case comes from thinking of $Sym(V)$ as a quotient of $T(V)$. $\endgroup$
    – Peter Samuelson
    Commented Aug 31, 2017 at 11:11
  • $\begingroup$ @YCor Perhaps one of you could post your comment as an answer (and maybe add a reference?) I'd be happy to accept it. Thanks. $\endgroup$
    – Will Chen
    Commented Aug 31, 2017 at 22:24
  • $\begingroup$ Which artice are you referring to? I am interested in seeing the context where this question came up. Thanks. $\endgroup$
    – Bizfold
    Commented Sep 27, 2018 at 13:09

1 Answer 1

Reset to default
3
$\begingroup$

(This was meant to be a comment; at the OP's request I copy it as an answer. I leave it CW)

Every (finite-dim) irreducible rep of $\mathrm{SL}_2$ is isomorphic to $[n]$ for some $n\ge 1$. Hence every finite-dim rep of $\mathrm{SL}_2$ is a direct sum of such $[n]$. This is just what they say, and is not related to your specific representation (the $d$-th graded component in the free Lie algebra on 2 generators).

A reference, among others, for reps of $\mathrm{SL}_2$, is Fulton-Harris' book.

Improve this answer
$\endgroup$

You must log in to answer this question.

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