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I have seen so many questions about free Lie algebra and universal enveloping algebra. As in this question (Universal envoloping algebra of a free Lie algebra.), a reader asked for the universal enveloping algebra of a free Lie algebra. I am following the book on Lie algebra written by Humphreys. In this book, he didn't give much about free Lie algebra and its construction. He said if $X$ is a set then to construct its free Lie algebra go through the vector space generated by X as a basis and consider this as a subalgebra of tensor algebra. I can not understand how to do this. Kindly explain this to me. There are so many questions on universal enveloping algebra on this site and I studied them but didn't get my answer.

Eran
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MANI
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First of all, note that an universal enveloping algebra is an associative algebra, wheras Lie algebras are seldom associative. So, it almost never happens that a universal enveloping algebra can also be a Lie algebra.

If $F$ is a set, the free Lie algebra on $F$ (over a field $k$) is basically a Lie algebra over $k$ which contains $F$ and which, besides that, satisfies the minimal possible amount of relations among its elements. So, for instance, if $F=\{X,Y\}$, then the free Lie algebra on $F$ contains $X$ and $Y$. It also contains $[X,Y]$ and $[Y,X]$ and, of course, $[X,Y]=-[Y,X]$, since this relation must hold for any Lie algebra. But it is not true that $[X,Y]=0$, since this doesn't hold in general for Lie algebras.

José Carlos Santos
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  • @JoséCarlosSantos Thanks for your answer, but can you give little more about free Lie algebra, that we can construct it. – MANI Sep 24 '19 at 16:25
  • @MANI Are the answers to this question enough for you? – José Carlos Santos Sep 25 '19 at 09:41
  • @JoséCarlosSantos Thanku sir. I got my answer. – MANI Sep 25 '19 at 11:20
  • @MANI I'm glad I could help. – José Carlos Santos Sep 25 '19 at 11:21
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    Did you mean a Lie algebra is unlikely to be a universal enveloping algebra instead of "it almost never happens that a universal enveloping algebra can also be a Lie algebra"? Because universal enveloping algebras are always Lie algebras under the usual bracket. – nobody Jan 21 '20 at 04:14
  • @justanothermathstudent I was talking about the usual product in a universal enveloping algebra, inherited from the tensor algebra, which makes it an associative algebra. – José Carlos Santos Jan 21 '20 at 14:00
  • @JoséCarlosSantos Okay. Another question: How exactly do we make the statement: "minimal possible amount of relations among its elements" rigorous? – nobody Jan 22 '20 at 16:03
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    Given a set $F$, if $\mathfrak f$ is the free Lie algebra on $F$ and if $\mathfrak g$ is a Lie algebra such that $F\subset\mathfrak g$, then there is a surjective Lie algebra homomorphism from $\mathfrak f$ onto $\mathfrak g$. In other, less precise, words, all relations whiach are valied on $\mathfrak f$ are also valid on $\mathfrak g$. – José Carlos Santos Jan 22 '20 at 18:22