Lie algebra and enveloping algebra

Question

I have to prove that a Lie algebra over the field $k$ is trivial if and only if the enveloping algebra $U(L)=k$.

I have an idea of proof: If $L=\{0\}$ we have that the tensor algebra $T^m=\{0\}$ for all $m \neq 0$, so we have $U(L)=k$. We have that always exists an injection of $L$ in the enveloping algebra $U(L)$, which has dimension $1$. So $\dim(L) \le 1 $.

How can I finish my proof?

Trivial here must mean $L = {0}$ because if $L \neq {0}$ then $U(L)$ is infinite dimensional. — Jim, Feb 22 '13 at 21:36

Jim · Accepted Answer · 2013-02-23T18:12:11.670

2

To finish note that it is never the case that the identity element $1 \in U(L)$ is contained in $L$ (when we think of $L$ as a linear subspace of $U(L)$) so it cannot be the case that $\dim L = \dim U(L)$ (otherwise $L = U(L)$). Thus $\dim L < 1$. This gives $\dim L = 0$ hence $L = \{0\}$.

The reason $1$ is not in $L \subseteq U(L)$ is because of the way $U(L)$ is constructed. You start with the tensor algebra $T(L)$ where $L$ is in degree $0$. Then you quotient out by the ideal generated by elements of the form $[xy] - x \otimes y + y \otimes x$. Noteice that these elements are sums of monomials of degree $1$ and higher. So the ideal you are quotienting by is contained entirely in $T(L)^+$, the two sided ideal of $T(L)$ generated by homogeneous elements of degree $1$ or greater. So if $x \in L$ the element $1 - x$ does not go to zero in the quotient (it's not contained in $T(L)^+$). So in $U(L)$ the expression $1 = x$ is not true for any $x \in L$.

edited Feb 23 '13 at 18:12

answered Feb 22 '13 at 21:37

Jim

30,682

Coul you explain better... I don't understan. Thanks – ArthurStuart Feb 22 '13 at 21:39
If you are more specific about what confuses you it will be easier to help. Is it that you don't understand why $\dim L < 1$ finishes your argument? Or is it that you don't understand by $1 \notin L$? – Jim Feb 22 '13 at 21:44
I don't understand this two claims. – ArthurStuart Feb 22 '13 at 21:46
I've edited my response to clear both of those up. – Jim Feb 22 '13 at 22:00
And why $dim(L) <1$? – ArthurStuart Feb 22 '13 at 22:05
Because $\dim U(L) = 1$ and $L \subseteq U(L)$ is not all of $U(L)$. – Jim Feb 22 '13 at 22:12
Do you any ideas for this? http://math.stackexchange.com/questions/311363/system-of-roots – ArthurStuart Feb 22 '13 at 22:22
But we have tensorial algebra end not symmetric. In the symmetric algebra $x \otimes y$ is equal to $y \otimes x$. Then $L \subset U(L)$ up to isomorphism. How can I solve this detail? – ArthurStuart Feb 23 '13 at 17:16
Sorry, I just used the wrong word. Everything I said above holds when you replace $S(L)$ with $T(L)$, as I've done in my edit. – Jim Feb 23 '13 at 18:13
And aboul the isomorphism? – ArthurStuart Feb 23 '13 at 18:28
What isomorphism? What about it? – Jim Feb 23 '13 at 19:31
I know that $L \subset U(L)$ up to isomorphism. In $U(L)$ there isn't a concrete copy of $L$. In what sense $1=x$ is false? – ArthurStuart Feb 23 '13 at 19:34
In $U(L)$ there is a canonical copy of $L$. It is the set of cosets $x + I$ where $x \in L$ is an element of $T^1(L) \subseteq T(L)$ and $I$ is the ideal such that $U(L) = T(L)/I$. – Jim Feb 23 '13 at 19:37
Sorry... what is $T^{1}(L)$? – ArthurStuart Feb 23 '13 at 19:41
$T^i(L)$ are tensors $x_1 \otimes \cdots \otimes x_i$. – Jim Feb 23 '13 at 23:11
What is the subject of the sentence: "It is the set of cosets...? – ArthurStuart Feb 24 '13 at 00:49
The canonical copy of $L$ in $U(L)$. – Jim Feb 24 '13 at 04:29

Lie algebra and enveloping algebra

1 Answers1