Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Trivial here must mean $L = \{0\}$ because if $L \neq \{0\}$ then $U(L)$ is infinite dimensional. –  Jim Feb 22 '13 at 21:36
    
Jes, triavial means $L=\{0\}$ –  ArthurStuart Feb 22 '13 at 21:37
    
@all Thanks! Always learning new things here... –  rschwieb Feb 22 '13 at 22:05
add comment

1 Answer

up vote 1 down vote accepted

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 $x \in L$.

share|improve this answer
    
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
show 12 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.