# SU_L x SU_R and SU_V x SU_A equivalence

1. Jul 2, 2014

### whitewolf91

I want to show that the group $SU_L(N)\times SU_R(N)$ is the same as $SU_V(N)\times SU_A(N)$ - i.e. that it is possible to rewrite the transformation:

$$\begin{cases} \psi_L \to \psi'_L=V_L\,\psi_L\\ \psi_R \to \psi'_R=V_R\,\psi_R \end{cases}$$

, where $V_L$ and $V_R$ are $N\times N$ $SU(N)$ matrices and $\psi_L$ and $\psi_R$ are $N$-component vectors, as:

$$\begin{cases} \psi_L \to \psi'_L=V A\,\psi_L\\ \psi_R \to \psi'_R=V A^\dagger\, \psi_R \end{cases}$$

, where again $V$ and $A$ are $N\times N$ $SU(N)$ matrices. The system:

$$\begin{cases} V_L=VA &\\ V_R=VA^\dagger \end{cases}$$

must be solved for $V$ and $A$:

$$A^2=V_R^\dagger V_L$$

So, $A$ is the square root of the $SU(N)$ matrix $V_R^\dagger V_L$.

Problems:

1. The only way I can think of to give meaning to the square root of a matrix is a naive series expansion, but how to prove convergence?

2. Does the series expansion defines an $SU(N)$ matrix?

Random thoughts:

Probably 2. is the less problematic. I believe that it is true that if the square of a matrix equals an $SU(N)$ matrix, that matrix still lives in $SU(N)$. It is not necessary to find the explicit expression of $A$ and $V$ so it would be enough to prove that the square root of an $SU(N)$ matrix is an $SU(N)$ matrix. Another approach is to consider infinitesimal elements of the group, then the previous system can be easily satisfied in the Lie algebra of the group. Is this enough to conclude that there are finite elements in the group that satisfy the same relationships?

Thank you for your time.

2. Jul 2, 2014

### The_Duck

If a matrix A is similar to a diagonal matrix (there is some matrix $P$ and some diagonal matrix $D$ such that $A = P^{-1} D P$) then you can define $A^{1/2} \equiv P^{-1} D^{1/2} P$, where $D^{1/2}$ is defined by just taking the square root of the diagonal entries.

3. Jul 3, 2014

### whitewolf91

An $SU(N)$ matrix is not necessarily hermitian, so it is not guaranteed that it can be diagonalized. If it were hermitian however then $P$ would be unitary and it would be automatic that the square root is in $SU(N)$.

4. Jul 3, 2014

### The_Duck

Unitary matrices are all diagonalizable with a unitary $P$; see e.g. here. So all SU(N) matrices have a square root in SU(N).

5. Jul 3, 2014

6. Jul 3, 2014

7. Jul 3, 2014

8. Jul 3, 2014

### whitewolf91

@The_Duck: thank you, that solves the problem.

@samalkhaiat: I follow you until the very end, when you imply that:

$$e^{i(\alpha+\epsilon \gamma_5)\,\cdot\, \tau/2}=e^{i\alpha\,\cdot\, \tau/2}e^{i\epsilon \gamma_5\,\cdot\, \tau/2}$$

Isn't this false since the exponents are non commuting?

9. Jul 3, 2014

### samalkhaiat

This should be understood as $8 \times 8$ matrix, when you use the BHC identity. Write it as
$$e^{ \frac{i}{2} ( \alpha \cdot \tau ) \otimes I } e^{ \frac{i}{2} ( \epsilon \cdot \tau ) \otimes \gamma_{ 5 } } .$$

10. Jul 4, 2014

### whitewolf91

My reasoning:

$[(\alpha\cdot \tau)\otimes I, (\epsilon\cdot \tau)\otimes\gamma_5]=$

$=[\alpha\cdot\tau,\epsilon\cdot\tau]\otimes\gamma_5=$

$=\alpha_a\epsilon_b[\tau_a,\tau_b]\otimes\gamma_5=$

$=i\alpha_a\epsilon_b\epsilon_{abc}\tau_c\otimes\gamma_5=$

$=i(\alpha \times \epsilon)\cdot \tau \otimes \gamma_5$

, which is zero only if $\alpha$ and $\epsilon$ are parallel.

11. Jul 4, 2014

### vanhees71

One should stress that you shouldn't write $\mathrm{SU}(2)_V \times \mathrm{SU}(2)_A$ (BAD NOTATION!) since this the $\times$ symbol should be reserved for a direct product of groups. That's the case for $\mathrm{SU}(2)_L \times \mathrm{SU}(2)_R$, but $\mathrm{SU}(2)_A$ is not even a sub group but a coset.

12. Jul 5, 2014

### samalkhaiat

I am very sorry, may be I was drunk. For some reason, I thought you were asking about eq(3) with the projection matrices. But you were asking about
$$U_{ L } U_{ R } = \exp ( i \frac{ \alpha \cdot \tau }{ 2 } ) \exp ( i \gamma_{ 5 } \frac{ \epsilon \cdot \tau }{ 2 } )$$
This is not an algebraic equation. As explained in the paragraph bellows it, it means the following

For all $g( \epsilon_{ L } , \epsilon_{ R } ) \in SU_{ L } \times SU_{ R }$, a transformation with $\epsilon_{ L } = \epsilon_{ R }$, corresponds to the vector transformation $g( \alpha )$. And, transformation with $\epsilon_{ L } = - \epsilon_{ R }$ corresponds to the axial transformation $g( \epsilon )$.

As pointed out by vanhees71 , the sign $( \times )$ in $SU \times SU_{ 5 }$ is not a direct product , because the Lie bracket of $SU_{ 5 }$ does not closes on itself. But the group $G = SU \times SU_{ 5 }$ is a direct product of the two independent groups $SU_{ L }$ and $SU_{ R }$, with Lie algebra given as
$$\mathcal{ G } = su_{ L } \oplus su_{ R }$$
Sorry for the confusion

Sam