# Proving the Jacobi identity from invariance

1. Apr 17, 2012

### ianhoolihan

"Proving" the Jacobi identity from invariance

Hi all,

In an informal and heuristic manner, I have heard that the "change" in something is the commutator with it, i.e. $\delta A =[J,A]$ for an operator $A$ where the change is due to the Lorentz transformation $U = \exp{\epsilon J} = 1 + \epsilon J + \ldots$ where $J$ is one of the six generators of the Lorentz group (rotation or boost). That is, if we have an operator $\phi\ :\ G\to G$, where $G$ is the vector space spanned by the size generators $J_i,K_i$ of the Lorentz group, (i.e. $G$ is the vector representation of the Lorentz algebra) then
$$\delta (\phi(T)) = \delta\phi (T) + \phi (\delta T)$$
so, using the above definition of "change"
$$[J,\phi(T)] = \delta\phi (T) + \phi ([J,T])$$.

We can then define $\phi$ to be invariant by saying that $\delta\phi = 0$, and hence

$$[J,\phi(T)] = \phi([J,T])$$.

If one does the same for a Lie product $\mu(X,Y) = [X,Y]$ then

$$\delta\mu(Y,Z) =\delta\mu (Y,Z) + \mu(\delta Y, Z) + \mu(Y,\delta Z)$$

We say that $\mu$ is invariant and set $\delta\mu = 0$ and hence

$$[J,\mu(Y,Z)] =\mu([J,Y], Z) + \mu(Y,[J, Z])$$
or

$$[J,[Y,Z]] =[[J,Y], Z] + [Y,[J, Z]]$$

which is the Jacobi identity. This seems great, but I don't understand a few points.

1. I believe the Lie product commutator enters as if we have an operator $A$ on the vectors in the Lorentz group (e.g. Minkowski space), it must change as
$$A\to A' = UAU^{-1} = A + \epsilon [J,A] + \ldots$$
correct? But in the above description with $\phi$ and $\mu$, these are operators on the Lorentz algebra, which I thought would remain unchanged.

2. Is the expression
$$\delta\mu(Y,Z) =\delta\mu (Y,Z) + \mu(\delta Y, Z) + \mu(Y,\delta Z)$$
rigourous? What about terms like $\mu(\delta Y, \delta Z)$? Or are those second order?

Any help would be great,

Ianhoolihan

2. Apr 22, 2012

### samalkhaiat

Re: "Proving" the Jacobi identity from invariance

Can you please clarify the following?
What is $T$? Can you give me a specific example of $\phi (T)$?
Why is it that $\delta$ acts on both $\phi$ and its “argument” $T$? It looks like that you defined $\phi$ to be a Lie algebra representation or a Lie algebra-valued operator! So, if $\exp (\epsilon \phi)$ is not the identity, what does it mean to set $\delta \phi = 0$? The same goes for $\mu$; I can take it to be the linear map $(\mu(X))(Y) = [X,Y]$, defined on the lie algebra such that (representing the lie algebra in itself)
$$\mu([X,Y]) = [\mu(X),\mu(Y)].$$
You can check that such map guarantees the Jacobi identity; expand both sides of
$$\mu([X,Y])(Z) = [\mu(X),\mu(Y)](Z).$$
So, what does it mean to set $\delta \mu = 0$?
In the same way, we can define the action of $\delta_{J}$ on lie algebra element $X$(or, on a lie algebra-valued function $f(x;X)$) by
$$\delta_{J}X = [J,X].$$
This means that $\delta_{J}$ acts as derivation, i.e., it guarantees Jacobi identity. This is because Lie brackets are derivations;
$$\delta_{J}[X,Y] = \delta_{J}(XY) - \delta_{J}(YX),$$
This gives the Jacobi identity:
$$\delta_{J}[X,Y] = [J,XY] - [J,YX] = [X,\delta_{J}Y] + [\delta_{J}X, Y].$$

Sam

3. Apr 22, 2012

### ianhoolihan

Re: "Proving" the Jacobi identity from invariance

Thanks Sam.

$T$ is a vector in the vector space of generators for the algebra. An example would be $T=J_i, K_i$ in the Galilean algebra.

Say, the trivial one: $\phi(T)=T$ or $\phi(T) = c T$ for some constant $c$.

That is my question 2. I can only reason along the lines of the chain/product rule. A variation in an evaluated function must depend on the variation in the function, and the variation in the thing it is acting on. Formalising this is what I'm looking for.

$\exp (\epsilon \phi)$ is not the identity. I believe the point is what I made above --- under a given transformation of the vector space, both the vectors $\phi$ is operating on, and the 1--cochain $\phi$ may depend on the transformation. In differential geometric language, this would be like saying that both the 1--form and vector bases transform, and so may the "coefficients". Setting $\delta \phi = 0$ means the coefficients do not change. I'm not even 50% sure myself, however.

I do not know what you mean by representing a lie algebra in itself, but, as far as I am concerned, the lie product $\mu$ is a bilinear antisymmetric 2--cochain. Yours is a 1--cochain it seems...?

Anyway, it turns out that I wasn't as clued up about Lie algebras etc as I thought. I still am not, but will look again at Kirillov soon. If you've got any comments on this question, however, they'd still be much appreciated.

Ianhoolihan