Let $V$ be a vector space over some arbitrary field. Let $T(V)$ and $S(V)$ be the tensor and symmetric algebras over $V$. We have the projection map $T(V)\to S(V)$, given by $x_1\otimes\cdots\otimes x_n\mapsto x_1\cdots x_n$.

I'm interested in the kernel of this map. More precisely, I want an explicit object $M(V)$, in terms of generators and relations, such that we have an exact sequence $$0\to M(V)\to T(V)\to S(V)\to 0.$$

I did find such an object, a $T(V)$-bilagebra which is an explicit quotient of the bialgebra generated by $\Lambda^2(V)$, i.e. $T(V)\otimes\Lambda^2(V)\otimes T(V)$.

I wonder if somebody saw this somewhere. (With a different notation, not necessarily $M(V)$.) After all, it is a natural question, somebody might have done it before.

I need this result in a paper I'm writing and I'd rather quote it than write the proof myself. The proof is not that obvious and it's not short either. (I guess it's two pages or more.)

The precise result is the following:

$\bf Definition$ We define the $T(V)$-bimodule $M(V)=(T(V)\otimes\Lambda^2(V)\otimes T(V))/W(V)$, where $W(V)$ is the subbimodule of $T(V)\otimes\Lambda^2(V)\otimes T(V)$ generated by $f(x,y,z)$, with $x,y,z\in V$, where $$f(x,y,z)=x\wedge y\otimes z+y\wedge z\otimes x+z\wedge x\otimes y-x\otimes y\wedge z-y\otimes z\wedge x-z\otimes x\wedge y,$$ and the expressions $$[x,y]\otimes\xi\otimes z\wedge t-x\wedge y\otimes\xi\otimes [z,t],$$ with $x,y,z,t\in V$ and $\xi\in T(V)$.

$\bf Theorem$ We have an exact sequence $$0\to M(V)\to T(V)\to S(V)\to 0,$$ where the left map is the morphism of $T(V)$-bimodules given on generators by $x\wedge y\mapsto [x,y]$

(By $[x,y]$ we mean the commutator, $[x,y]=x\otimes y-y\otimes x$.)

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