# Lie algebra of axial charges

Starting from the lagrangian (linear sigma model without symmetry breaking, here $N$ is the nucleon doublet and $\tau_a$ are pauli matrices)

$L=\bar Ni\gamma^\mu \partial_\mu N+ \frac{1}{2} \partial_\mu\sigma\partial^\mu\sigma+\frac{1}{2}\partial_\mu\pi_a\partial^\mu\pi_a+g\bar N(\sigma+i\gamma_5\pi_a \tau_a)N$

we can construct conserved currents using Noether's Theorem applied to $SU(2)_L\otimes SU(2)_R$ symmetry: we get three currents for every $SU(2)$. By adding and subtracting them, we obtain vector and axial currents.
We could have obtained vector charges quickly by observing that they are just isospin charges, so nucleons behave as an $SU(2)$ doublet (fundamental representation), pions as a triplet (adjoint representation) and sigma as a singlet (so basically it does not transform):

$V_a=-i\int d^3x \,\,[iN^\dagger\frac{\tau_a}{2}N+\dot\pi_b(-i\epsilon_{abc})\pi_c]$

But if I wanted to do the same with axial charges, what Lie algebra/representation must I use for pions and sigma?
I mean, axial charges are

$A_a=-i\int d^3x \,\,[iN^\dagger\frac{\tau_a}{2}\gamma_5N+i(\sigma\dot\pi_a-\dot\sigma\pi_a)]$

and I would like to reproduce the second term using a representation of Lie algebra generators of axial symmetry which act on $\sigma$ and $\pi$, but I don't know the algebra (I think it is $SU(2)$), neither the representation to use.
I tried to reproduce that form using the three matrices

$T^1=\begin{bmatrix} 0&-i&0&0\\i&0&0&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}\quad T^2=\begin{bmatrix} 0&0&-i&0\\0&0&0&0\\i&0&0&0\\0&0&0&0 \end{bmatrix}\quad T^3=\begin{bmatrix} 0&0&0&-i\\0&0&0&0\\0&0&0&0\\i&0&0&0 \end{bmatrix}$

which should act on the vector $(\sigma,\pi_1,\pi_2,\pi_3)$, but I calculated their commutator and they don't form an algebra, so I think I'm getting wrong somewhere in my reasoning.

In the linear sigma model, the chiral action on the pion fields can be implemented on the following matrix combination of the fields:

$$U(2) \ni \Sigma = \sigma + i \tau^a \pi_a$$

An element $(U_L = exp(\frac{i}{2}\theta^{(L)}_a \tau^a), U_R = exp(\frac{i}{2}\theta^{(R)}_a \tau^a)) \in SU(2)_L \otimes SU(2)_R$ acts on \Sigma as follows:

$$\Sigma \rightarrow \Sigma' = U_L \Sigma U_R^{\dagger}$$

The kinetic term of the Lagrangian in the matrix representation is given by:

$$L_{kin} = \frac{1}{2} \partial_{\mu}\Sigma \partial^{\mu}\Sigma^{\dagger}$$.

This term is manifestly invariant under all transformations. The interaction term has also a manifestly invariant form:

$$L_{int} = \bar{N}_L \Sigma N_R+ \bar{N}_R \Sigma^{\dagger} N_L$$.

where $N_{L,R} = (1\pm \gamma_5)N$. Thus the whole Lagrangian is invariant under the chiral transformations.

The vector transformation is generated by the subgroup characterized by:

$$\theta^{(L)} = \theta^{(R)} = \theta^{(V)}$$

The axial transformation is generated by the subset characterized by:

$$\theta^{(L)} = -\theta^{(R)} = \theta^{(A)}$$

Substituting in the transformation equations of $\Sigma$ and keeping only the linear terms (this is sufficient for the application of the Noether's theorem), we obtain:

-Vector transformation:

$$\pi_a' = \pi_a +\epsilon_{abc}\theta^{(V)}_b \pi_c$$

$$\sigma' = \sigma$$

-Axial transformation:

$$\pi_a' = \pi_a +\theta^{(A)}_a \sigma$$

$$\sigma' = \sigma + \theta^{(A)}_a \pi_a$$

Now it is not hard to see that these transformations generate the correct contributions of the pionic fields to the currents.

• Thank you, it works but there must be a minus sign in axial transformation law of $\sigma$. Anyway, is there a way to put it in a form like $Q_a=-i\int d^3x\,\,(\pi_i T^a_{ij} \phi_j)$, where here $\pi_i=\dot\phi_i$ is the conjugate momenta? I can't figure out the $\phi_i$ and the $T^a_{ij}$, which should form a representation of $SU(2)$ – gian_25 Mar 9 '14 at 14:20
• @gian_25 The matrix representation is just the 4-dimensional fundamental representation of $SO(4) = SU(2) \times SU(2)$, denoting the indices of the 4-space corresponding to $(\sigma, \pi_1, \pi_2, \pi_3)$ by $0,1,2,3$. Then$(T^{V}_i)_{jk} = i \epsilon_{ijk}, (T^{(A)}_i)_{jk} = i (\delta_{ji}\delta_{k0}+ \delta_{ki}\delta_{i0})$. Please observe that the commutator of two axial generators is a vector generator. – David Bar Moshe Mar 9 '14 at 15:04
• Ok, I think there is still a minus missing and $T^{(A)}_i$ become the same matrices I have written before, but now I understand what their commutators are. Thank you for the answer! – gian_25 Mar 9 '14 at 16:01