Lie bracket $[x,x]=0\implies [x,y]=-[y,x]$

Question

Why does $[x,x]=0\implies [x,y]=-[y,x]$ in regard to Lie brackets.

I have tried to play around with bilinearity, but I can't get it to work.

$$[ax+by,z]=a[x,z]+b[y,z],[x,x]=0$$

I have tried subbing in $z=x$ and $z=y$, but I just can't obtain anti-commutativity. Thanks

score 5 · Accepted Answer · answered Aug 08 '15 at 05:20

5

$$[x+y,x+y]=0=[x,x]+[x,y]+[y,x]+[y,y]=[x,y]+[y,x]\Rightarrow [x,y]=-[y,x]$$

answered Aug 08 '15 at 05:20

Moya

5,248

I like to think of this as some sort of Lie algebra analogue of the parallelogram identity for inner products. – Cameron Williams Aug 08 '15 at 05:23
Thanks. This looks like the sort of answer that you see once and know forever, but is there a thought path that would get you to try this? – So many hats Aug 08 '15 at 05:24
Honestly, just experience with algebraic manipulation; it's essentially binomial expansion to be honest. I intuitively approach bilinear forms/operators/etc. as generalizations of multiplication, so you use a similar approach. Note that the two are actually equivalent. – Moya Aug 08 '15 at 05:30

score 0 · Answer 2 · answered Aug 08 '15 at 05:34

0

One can easily show that $[x,x]= x \cdot x - x \cdot x = 0$ and \begin{align} [ax + by, cx + dy] &= ac \, [x,x] + ad \, [x,y] + bc \, [y,x] + bd \, [y,y] \\ &= ad \, [x,y] + bc \, [y,x] \end{align} By letting $c = a$, $d = b$ then $0 = [x,y] + [y,x]$. With this then $$[ax + by, cx + dy] = (ad - bc) \, [x,y].$$

answered Aug 08 '15 at 05:34

Leucippus

26,329

Thanks, although I think this uses the commutator bracket at the top(I think in general the first line is actually an axiom in the general case) – So many hats Aug 08 '15 at 05:37
2

That looks oddly like a determinant ;) – Cameron Williams Aug 08 '15 at 05:39
1

For abstract Lie algebras, $[x,y]$ is defined, but not $x\cdot y$ or $x\cdot x$. – Dietrich Burde Aug 08 '15 at 18:59

Lie bracket $[x,x]=0\implies [x,y]=-[y,x]$

2 Answers2