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.

We can define a Lie algebra letting $\mathbb{R}$ be the vector space and also the field. We can then have $[x,y]=xy-yx=0$ for all $x,y$.

Is there a one-dimensional Lie algebra such that $[x,y]$ is not identically zero?

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

No, this won't happen. To see this, let $\mathfrak{g}$ be a one-dimensional Lie algebra over a field $F$. Since $\mathfrak{g}$ is one-dimensional, we can write $\mathfrak{g}=Fz$ for any $z \in \mathfrak{g}-\{0\}$.

Now let $x, y \in \mathfrak{g}$. We will show that $[x,y]=0$. Fix $z \in \mathfrak{g}-\{0\}$ and write $x = az$ and $y = bz$ for some $a, b \in F$. Then $[x,y] = [az,bz] = ab[z,z] = ab*0 = 0 \in \mathfrak{g}$

The second equality follows from the bilinearity of the bracket. The third equality makes use of the alternating axiom of a Lie algebra: $[z,z] = 0 \ \ \forall z \in \mathfrak{g}$.

Thus, any one-dimensional Lie algebra is abelian. That is, the bracket is identically zero. Note, however, that the converse is not true. There are abelian Lie algebras which are not one-dimensional, such as $\mathbb{R}^n$ with the trivial bracket ($n \geq 2$).

share|improve this answer
add comment

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.