Let g be a Lie algebra such that [[x,y],y]=0 for all $x,y \in g$. Show that 3[[x,y],z]=0 for all $x,y,z \in g$. [Hint: Observe that the mapping (x,y,z) to [[x,y],z] is skew-symmetric in x,y,z and make use of the jacobi identity.]

So I know that [[x,y],z]=[x,[y,z]]+[y,[z,x]] (using the jacobi identity)

So 3[[x,y],z]=3[x,[y,z]]+3[y,[z,x]], I need to use [[x,y],y]=0, but can't see how to do it. Like I would need to elimate x,y or z from 3[x,[y,z]]+3[y,[z,x]], but it just seems impossible.

Was wondering am I using skew symmetric wrong. I assume that skew symmetric is just this [x,y]=-[y,x].


1 Answer 1


Skew symmetry in general means for a function $f$ of $n$ arguments that whenever $(x_1,\ldots, x_n)$ and $(y_1,\ldots, y_n)$ are permutations of each other, then $f(x_1,\ldots,x_n)=\pm f(y_1,\ldots,y_n)$, where $\pm$ means $+$ or $-$ according whether the permutation is even or odd.

To prove that a function is skew symmetric, it is sufficient to show that $$f(x_1,\ldots,x_{i-1},x_i,x_{i+1},x_{i+2},\ldots x_n) = -f(x_1,\ldots,x_{i-1},x_{i+1},x_i,x_{i+2},\ldots x_n) $$ for all $i$ -- that is, that interchanging any two neighbor arguments will flip the sign of the function result. If the function is known to be linear in each argument, it is also enough to prove that $f(x_1,\ldots,x_n)=0$ whenever two of the $x_i$'s are equal.

Here this concept is being applied to the function $f(x,y,z)=[[x,y],z]$. The hint says to prove, based on $\forall xy: [[x,y],y]=0$, that this $f$ is skew symmetric. (You can do this without the Jacobi identity).

Then, write the Jacobi identity in the form $[[a,b],c]+[[b,c],a]+[[c,a],b]=0$, and use the just proved fact about $f$ to see that the left-hand side is equal to $3[[a,b],c]$.

  • $\begingroup$ I thought all Lie algebras are skew-symmetric. So in $[[a,b],c]+[[b,c],a]+[[c,a],b]=0$ do you just swap and change signs until you get it as $3[[a,b],c]$. I can see that. However, how would on go about proving f is skew-symmetric. $\endgroup$
    – simplicity
    Apr 4, 2012 at 18:31
  • 2
    $\begingroup$ The Lie bracket itself is skew-symmetric by definition, that is $[a,b]=-[b,a]$. But it does not follow from this that the three parameter function $(a,b,c)\mapsto [[a,b],c]$ is also skew-symmetric -- that requires not only $[[a,b],c]=-[[b,a],c]$ (which is true for all Lie algebras) but also, say, $[[a,b],c]=-[[a,c],b]$ (which isn't; consider e.g. $(\mathbf e_1 \times \mathbf e_2)\times \mathbf e_1$ in $\mathbb R^3$). The latter relation can be proved if you assume $[[x,y],y]=0$, however. (Hint: let $x=a$, $y=b+c$ and expand the LHS by linearity!) $\endgroup$ Apr 4, 2012 at 21:14
  • $\begingroup$ Thanks, I understand now. I did the expansion with x=a, y=b+c and can see where you use the fact [[x,y],y]=0. You been a lot of help so thank you. $\endgroup$
    – simplicity
    Apr 4, 2012 at 22:03

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