# Two Dimensional Lie Algebra

1. Oct 28, 2012

### valtz

I read in mark wildon book "introduction to lie algebras"
"Let F be any field. Up to isomorphism there is a unique two-dimensional nonabelian
Lie algebra over F. This Lie algebra has a basis {x, y} such that its Lie
bracket is described by [x, y] = x"

and i'm curious,

How can i proof with this bracket [x,y] = x, satisfies axioms of Lie algebra such that
[a,a] = 0 for $a \in L$
and satisfies jacoby identity

cause we only know about bracket of basis vector for L

2. Oct 28, 2012

### lpetrich

Note that the commutator is anti-commuting, that [x,y] = - [y,x]
What happens when y = x?

As to the Jacobi identity, try it with two of the variables equal. See how much constraint the Jacobi identity will make on a 2-generator algebra.

3. Oct 29, 2012

### valtz

can u give me some example for two dimensional lie algebra?

4. Oct 29, 2012

### lpetrich

valtz, the important thing here is working out such things in general, rather than for some specific case. That way, you'll know what's always true without having to go into the details of specific cases.

5. Oct 29, 2012

### Vargo

Basically, you need to fill in the details. Brackets are supposed to be bilinear, anticommutative and satisfy [a,a]=0. So given what you wrote you should be able to derive the following general formula for the bracket of two vectors:
$[ax+by, cx+dy]=ac[x,x]+ad[x,y]+bc[y,x]+bd[y,y] = (ad-bc)x$
Using that general formula, you should be able to prove that [a,a]=0 for all a, and that the Jacoby identity holds.

Here is "real" example of such a 2 dim Lie algebra. Consider smooth functions defined on the real line and let L be the set of linear, first order, differential operators generated by
d/dx and x(d/dx). Notice that their commutator is d/dx.

6. Oct 29, 2012

### lpetrich

Actually, one can derive [a,a] = 0 from antisymmetry.

If [b,a] = - [a,b], then [a,b] + [b,a] = 0
By setting b = a, we get 2[a,a] = 0
yielding [a,a] = 0

Turning to the Jacobi identity, it is
[a,[b,c]] + [b,[c,a]] + [c,[a,b]]= 0

For a = b = c, it's 3[a,[a,a]] = 0
For b = c, it's [a,[b,b]] + [b,[b,a]] + [b,[a,b]] = - [b,[a,b]] + [b,[a,b]] = 0

Thus, the Jacobi identity provides no additional constraints in these cases. However, it will if a,b,c are distinct.

In my earlier posts, I was giving hints in the hope that valtz would then work out the derivations using them.

7. Nov 3, 2012