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.