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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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    $\begingroup$ Isnt this just the ideal generated by the commutators? See also en.wikipedia.org/wiki/Symmetric_algebra $\endgroup$
    – Mare
    Nov 14, 2019 at 12:13
  • $\begingroup$ Well, that is obvious, but it is not what I want. There are relations between these generators, which must be take care of. My bimodule $M(V)$ does just that. $\endgroup$ Nov 14, 2019 at 12:19
  • $\begingroup$ So you want to find $M(V)$ as a quotient of bimodules? It sounds like you want to find the kernel of the map $0 \rightarrow \Omega^1(M(V)) \rightarrow P \rightarrow M(V) \rightarrow 0$, where $P$ is a projective cover of $M(V)$ as a bimodule. $\endgroup$
    – Mare
    Nov 14, 2019 at 12:23
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    $\begingroup$ Well, before someone can say whether they'veseen your $M(V)$ before, surely you need to show it? $\endgroup$
    – LSpice
    Nov 14, 2019 at 12:23
  • $\begingroup$ I edited my post with the precise definition of the bimodule $M(V)$ and of the short exact sequence $0\to M(V)\to T(V)\to S(V)\to 0$. I you saw this or something equivalent somewhere, then please let me know. $\endgroup$ Nov 14, 2019 at 18:11

2 Answers 2

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The proof turned out to be longer than expected, so I decided to write it in a separate paper. You may find it on arxiv here: https://arxiv.org/abs/1912.03515

The second section deals with the "semi-symmetric" algebra $S'(V)$. I asked a question about it on May 3. See here: The "semi-symmetric" algebra of a vector space

I need both these results as a prerequisite for a future paper. Again, if anybody saw any of these results, or something similar, then please let me know.

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I suspect that the best way to organise the information that you need will be via the Poincaré-Birkhoff-Witt theorem: the tensor algebra $TV$ is the universal enveloping algebra $ULV$ of the free Lie algebra $LV$ generated by $V$, there is a natural filtration of $TV$ whose associated graded ring is the symmetric algebra $SLV$. We can split $LV$ as $V\oplus \lambda^2 V \oplus L_{\geq 3}V$ and then $SLV=SV\otimes S(\lambda^2 V)\otimes S(L_{\geq 3}V)$.

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  • $\begingroup$ Thank you for your answer. I'll have to look into this Poincaré-Birkhoff-Witt theorem. It's a subject I'm not familiar with. Do you believe it could provide a short proof of the result? My proof is quite technical. I only used elementary properties of the tensor products. At some point I also use the fact that the symmetric group $S_n$ is generated by the transpositions $\sigma_i=(i,i+1)$ with $1\leq i\leq n-1$ and the relations between the generators are $\sigma_i^2=1$, $\sigma_i\sigma_j=\sigma_j\sigma_i$ if $|i-j|\geq 2$ and $(\sigma_i\sigma_{i+1})^3=1$ for $1\leq i\leq n-2$. $\endgroup$ Nov 15, 2019 at 12:08
  • $\begingroup$ Is $\lambda^2 V$ the second exterior power? (Equivalently, the degree $2$ component of the free Lie algebra, spanned by the commutators $x \otimes y - y \otimes x$ for $x, y \in V$,) $\endgroup$ Nov 16, 2019 at 18:53

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