# Action of orthogonal group on the free Lie algebra

This question is somewhat related and inspired by this post of professor Montgomery.

The free Lie algebra $L(V)$ generated by an $r$-dimensional vector space $V$ is, in the language of https://en.wikipedia.org/wiki/Free_Lie_algebra, the free Lie algebra generated by any choice of basis $e_1,\dots,e_r$ for the vector space $V$. (Work over the field $\mathbb R$ or $\mathbb C$, whichever you prefer.) It is a graded Lie algebra

$$L(V)=V\oplus L_2(V)\oplus L_3(V)\oplus\cdots$$

Let $\mathfrak{so}(r)$ be the Lie algebra of the (special) orthogonal group act on $V$ by matrix multiplication, and extend the action to $L_i(V)$ as a derivation of the algebra, that is $A[X,Y]=[AX,Y]+[X,AY]$. In this way we are inducing representations of $\mathfrak{so}(r)$ into $\mathbb R^{\textrm{dim}(L_i(V))}$, where it is well-known that $$\textrm{dim}(L_i(V))=\frac{1}{i}\sum_{d\mid i}\mu(d)r^{i/d},$$ $\mu$ being the Moebius function. Are there references in the literature on how to decide whether this representation is reducible or not?

My knowledge of representation theory is quite limited, but as far as I understand the matter the answer is easy for $L_2(V)$. Indedd $\textrm{dim}(L_2(V))=\textrm{dim}(\mathfrak{so}(r))$, therefore they are isomorphic as vector spaces, and we can see the action that I described before as the usual adjoint representation of $\mathfrak{so}(r)$. On the other hand, the Lie algebra $\mathfrak{so}(r)$ is simple except for $r=4$, so the representation induced in $\mathbb R^{\frac{r(r-1)}{2}}$ is surely irreducible if $r\neq 4$. On the other hand, the representation is indeed reducible if $r=4$, and there are two invariant subspaces of dimension $3$ in $\mathbb R^6$ (somewhat related to quaternions and anti-quaternions I think).

What about higher layers? Are the references given in professor Montgomery post useful even for this situation? Any help is greatly appreciated.

Thanks for the patience,

Guido

• Under $GL_r$, $L_2(V)$ and $L_3(V)$ are irreducible for all $r\ge 2$. $L_4(V)$ is irreducible for $r=2$ and not for $r\ge 3$. $L_i(V)$ is non-irreducible for all $r\ge 2$ and $i\ge 5$. Reference: R. M. Thrall. On symmetrized Kronecker powers and the structure of the free Lie ring. Amer. J. M￼￼ath., 64:371-388, 1942. (See also arxiv.org/abs/1307.1489, p28.) So under $SO_k$, if we only need to determine whether it's irreducible, the only question concerns $i\le 3$ and $(i,r)=(4,2)$ as otherwise it's not irreducible. – YCor Sep 18 '16 at 17:06
• For $(i,r)=(3,3)$, $L_3(V)$ is 8-dimensional and $SO(3)$ has no even-dimensional irreducible representation. So it's not irreducible in this case. I'm not sure about higher $r$ but it's probably an exercise (the above reference says which irreducible of $GL(V)$ occurs). – YCor Sep 18 '16 at 20:03
• And for $r=2$ and $i\ge 3$, $L_i(V)$ has dimension $\ge 2$ and hence cannot be irreducible over $SO(2)$. This is over the complex numbers. Over the reals, the 2-dimensional $L_3(V)$ is indeed irreducible iff the quadratic form is anisotropic (it's never absolutely irreducible), but $L_i(V)$ for $i\ge 4$ has dimension $\ge 3$ and hence is not irreducible. – YCor Sep 18 '16 at 20:16
• Dear @YCor, thanks for your kind replies. Indeed a part of my question was to decide on the irreducibility of such reps, which now seems to be quinte an exceptional case. Now, do you know if there are similar (combinatorical) techniques to those discussed in the post of prof. Montgomery (e.g. with Tableaux), to describe how to build such reps? BTW if you would like to collect your comments in an answer, I will accept it and give you the bounty. Bests, Guido – guido giuliani Sep 19 '16 at 8:11
• But my answer is not complete, one should figure out what's going on for $L_3(V)$ when $r\ge 4$. – YCor Sep 19 '16 at 8:17