# Central charge and Lorentz Algebra

1. Mar 1, 2010

### LAHLH

Hi,

If $$M^{\mu\nu}$$ are the generators of the Lorentz group, i.e. they obey
$$[M^{\mu\nu}, M^{\rho\sigma}]=i\hbar(g^{\mu\rho}M^{\nu\sigma}-g^{\nu\rho}M^{\mu\sigma})+.....(1)$$

and $$L^{\mu\nu}$$ is defined by, $$L^{\mu\nu} := \frac{\hbar}{i} (x^{\mu} \partial^{\nu}-x^{\nu} \partial^{\mu})$$

I have found that L also obeys a similar commutation relation to the generators, namely:
$$[L^{\mu\nu}, L^{\rho\sigma}]\phi=i\hbar(g^{\mu\rho}L^{\nu\sigma}-g^{\nu\rho}L^{\mu\sigma})+......(2)$$

I also know that $$[\phi(x),[M^{\mu\nu}, M^{\rho\sigma}]]=[L^{\mu\nu}, L^{\rho\sigma}]\phi(x) (3)$$

I am now trying to solve a problem that asks to verify equation (1) upto a term on the RHS that commutes with $$\phi(x)$$ and its derivatives, by using equations 2 and 3. I have already proved equation 1 by other means, but no idea how to go about it this way. Apparently the term that might arise on RHS is called the central charge.

Just plugging things in takes me to:

$$\phi(x)[M^{\mu\nu}, M^{\rho\sigma}]-[M^{\mu\nu}, M^{\rho\sigma}]\phi(x)=i\hbar(g^{\mu\rho}L^{\nu\sigma}-g^{\nu\rho}L^{\mu\sigma})+......$$

I'm not sure where one would go from here to verify the generator equation, equation (1).

Thanks for any suggestions

2. Mar 1, 2010

### samalkhaiat

I see a bit of mess in what you trying to do. You need to know the action of the generators on the fields, i.e., the infinitesimal transformation law for the field. Basically, we use Jacobi identities to show the consistency (up to a set of c-numbers) between the Lie algebra and the action of its generators on the fields. Let me explain this for arbitrary Lie group G whose infinitesimal generators $G_{a}$ have the following action on some arbitrary set of fields $\phi_{i}$;

$$\delta \phi_{i} = [iG_{a},\phi_{i}] = (T_{a})^{j}_{i}\phi_{j} \ \ \ \ (1)$$

where $T_{a}$ are a set of operators or/and matrices satisfying the Lie algebra of our group, i.e.,

$$[T_{a},T_{b}] = if_{abc}T_{c} \ \ \ (2)$$

Now, use eq(1) to evaluate the following Jacobi identity

$$[iG_{b},[iG_a,\phi_{i}]] + [\phi_{i},[iG_{b},iG_{a}]] + [iG_{a},[\phi_{i},iG_{b}]] = 0$$

Doing that, you find

$$[[G_{a},G_{b}], \phi_{i}] = \left( T_{a}T_{b}-T_{b}T_{a}\right)^{k}{}_{i}\phi_{k}$$

Then from eq(2), you get

$$[[G_{a},G_{b}],\phi_{i}] = i f_{abc}(T_{c})^{k}{}_{i}\phi_{k}$$

then from eq(1) you find

$$[[G_{a},G_{b}],\phi_{i}] = if_{abc}[iG_{c},\phi_{i}]$$

This equation has the following general solution;

$$[G_{a},G_{b}] = -f_{abc}G_{c} + C_{ab}$$

with $C_{ab}$ are antisymmetric c-numbers, i.e., they commute with all fields.
Now I leave you to do the same thing with Lorentz group.

regards

sam

3. Mar 2, 2010

### LAHLH

Thanks for the reply, I greatly appreciate it.

Not sure I fully understand how to solve my problem yet.

$$\delta \phi_{i} = [iG_{a},\phi_{i}] = (T_{a})^{j}_{i}\phi_{j} \ \ \ \ (1)$$

Does this correspond to my equation:

$$[\phi(x),M^{\mu\nu}]=L^{\mu\nu} \phi(x)$$

With your label 'a' being my $$\mu\nu$$ giving 6 independent labels for my 4-d antisymmetric matrix (so your 'a' goes from 1-6?). I'm not sure why your $$\phi(x)$$ have indices on them, since they are scalar?

Then assuming my correpondance is correct above, your equation:

$$[T_{a},T_{b}] = if_{abc}T_{c} \ \ \ (2)$$

would correspond also to my equation (2):

$$[L^{\mu\nu}, L^{\rho\sigma}]=i\hbar(g^{\mu\rho}L^{\nu\sigma}-g^{\nu\rho}L^{\mu\sigma})+......(2)$$

4. Mar 2, 2010

### LAHLH

$$[\phi(x),[M^{\mu\nu}, M^{\rho\sigma}]]=[L^{\mu\nu}, L^{\rho\sigma}]\phi(x) (3)$$

which I guess somehow links to your:

$$[[G_{a},G_{b}], \phi_{i}] = \left( T_{a}T_{b}-T_{b}T_{a}\right)^{k}{}_{i}\phi_{k}$$

5. Mar 3, 2010

### samalkhaiat

Yes, use the algebra [L,L] = gL + ...., then for each L$\phi$ put the corresponding [$\phi$, M] .

sam