$SL_2$-action on the free lie algebra on a 2-dimensional vector space

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.

• 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). – YCor Aug 31 '17 at 7:48
• 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 $T(V)$. – Peter Samuelson Aug 31 '17 at 11:11
• @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. – Will Chen Aug 31 '17 at 22:24
• Which artice are you referring to? I am interested in seeing the context where this question came up. Thanks. – Bizfold Sep 27 '18 at 13:09

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.