Let $V$ be a finite dimensional vector space. Let $T(V)$ be the tensor algebra over $V$.
Do we have $T(V) \cong S(Lie(V))$ as a graded vector space? Here $S(Lie(V))$ is the symmetric algebra of the free Lie algebra over $V$.
Thank you very much.
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asked Jan 18, 2017 at 7:10
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5$\begingroup$ Yes, by PBW, in characteristic zero. $\endgroup$– Qiaochu YuanCommented Jan 20, 2017 at 9:09
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$\begingroup$ Is there something like "the free algebra over a vector space"? If you consider $V$ just as a set, the answer to your question is obviously no: $T(V)$ depends of course on the vector space structure. $\endgroup$– abxCommented Jan 20, 2017 at 10:02
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3$\begingroup$ To complete Qiaochu's comment, the key fact is that $T(V) $ is isomorphic as an Hopf algebra to the enveloping algebra of the free Lie algebra on $V$. $\endgroup$– AdrienCommented Jan 20, 2017 at 10:08
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$\begingroup$ I think the question is a bit ambiguous: here each $K$ of $T$, $S$ and $\mathrm{Lie}$ is considered as a mapping a graded vector space $W$ to a graded algebra $K(W)$ with $K(W)_1=W$ as graded vector space, and the original vector space $V$ is entirely graded in degree 1 (but this should work without this restriction, namely in the "universal" case when $V$ of dimension $d$ is graded in $\mathbf{Z}^d$ with independent weights. (As asked now, another interpretation would be to put $\mathrm{Lie}(V)$ in degree 1, but then the result fails since $T(V)_1$ is finite-dim but not $S(Lie(V))_1$.) $\endgroup$– YCorCommented Apr 26, 2020 at 5:36
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