$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Tor{Tor}$Let's suppose $V$ is a $k$-vector space equipped with its standard (left) $\GL (V)$-action. The shuffle algebra is the graded dual of the tensor algebra with the standard Hopf algebra structure, and for ease I'm just going to denote this shuffle algebra by $T^\bullet (V)$.

Notice that the shuffle algebra is equivariant with respect to the induced $\GL (V)$-action, by which I mean $g (v_1 \cdot v_2) = (g v_1) \cdot (g v_2)$ for any $g \in \GL(V)$, $v_i \in T^\bullet (V)$. The decomposables of $T^\bullet (V)$ are going to be all elements in the image of the multiplication map $T^+ (V) \otimes_k T^+ (V) \to T^+ (V)$ ($-^+$ means positive degree); this is evidently a $\GL (V)$-representation by the equivariance mentioned above. The indecomposables are the elements that are not decomposable, unsurprisingly.

My question is this: is it known what the indecomposable elements of the tensor algebra are as a $\GL (V)$-representation? Or, maybe a better way to say this: is there a "nice" (interpretation up to you) description of the representation theoretic structure of the indecomposables?

I can give one possible description that I've come up with but isn't totally satisfying (and likely hard to read): the shuffle algebra may be viewed as the Tor-algebra $\Tor_\bullet^R (k,k)$ of the residue field $k$ over the ring $R = S_\bullet (V) / (V)^2$ (that is, take a polynomial ring and mod out by the irrelevant ideal squared). This is a so-called Golod ring and hence its homotopy Lie algebra is free on (a shift of) $\textrm{Tor}_\bullet^{S_\bullet (V)} (R , k)^*$, which is well-known to be equivariantly isomorphic to a direct sum of Schur modules corresponding to hook shapes $(2,1^i)$. Since the Ext algebra is the universal enveloping algebra of the homotopy Lie algebra in this case, we deduce that the indecomposable elements correspond to a plethysm problem for Schur modules, which I would like to avoid.

Anyway, any help is appreciated.

  • 3
    $\begingroup$ MathJax note: please use Markdown *Markdown*, not $\textit{MathJax trickery}$ $\textit{MathJax trickery}$, to italicise. I have edited accordingly. $\endgroup$
    – LSpice
    Jul 19 at 23:03

1 Answer 1


There is a duality on covariant functors from $k$--vector spaces to itself: $DF(V) = F(V^{\vee})^{\vee}$, where $V^{\vee}$ is the dual of the vector space $V$. $D$ is known to preserve irreducible functors, so in the semisimple case (e.g. if $k$ has characteristic 0), $D$ acts as the identity on appropriately finite objects. (Things are more exciting in characteristic $p$.)

Now towards your question ... In any homogeneous degree, the functors of indecomposables in the shuffle algebra is dual to the functor of primitives in tensor algebra. The functors of primitives in $T_*(V)$ are known: it is $Lie_*(V)$, the free Lie algebra generated by the vector space $V$. (In characteristic $p$, I guess we need the free restricted Lie algebra generated by $V$.)

So one description of the functors of indecomposables in degree $d$ in the shuffle algebra $T^*(V)$ will be $DLie_d(V)$, which in char 0, will be just $Lie_d(V)$.

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    $\begingroup$ And for the decomposition of $Lie_d(V)$ into $GL(V)$ irreps: mathoverflow.net/questions/187545/… $\endgroup$ Jul 20 at 8:17
  • $\begingroup$ (sorry for the delayed response) Thank you, your answer combined with Qiaochu's link completely answer my question, this is perfect. $\endgroup$
    – Rellek
    Jul 25 at 14:45

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