Can G-Parity Explain the Dominance of Vector Current in Tau Decay?

[GWSguy]

I'm studying the \tau \rightarrow \rho \; \nu_{\tau} decay. I'm asked to calculate the decay width, using a parameterization of the matrix element of the hadronic current. I actually find a matrix element of the form:
\left<\rho |\; \bar{u} \gamma^{\mu} \left( 1-\gamma^{5} \right ) d \; | 0\right>
in which I have both the vector and the axial current (u and d are up and down quarks). The \rho meson is a spin 1 vector meson, so I expect that only the term from the vector current survives. I've infact verified this statement in many articles which report:
\left<\rho |\; \bar{u} \gamma^{\mu} d \; | 0\right> = f_{\rho} m_{\rho} \epsilon ^{\mu}
with \epsilon ^{\mu} the polarization vector of the meson.

The problem is that I'm not able to demonstrate it. How can I formally demonstrate it? Is there any parity argument which allows me to exclude the axial term?

 

[GWSguy]

http://www.physics.gla.ac.uk/~cmcneile/papers/rhodecay.ps

http://arxiv.org/pdf/hep-lat/0301020v4.pdf

These are two examples of the articles I mentioned

 

[GWSguy]

I know understand! The vector current is even under a G-parity transformation. I recall that G-parity is a multiplicative quantum number used for hadrons classification. It is defined as a rotation about the 2-axis, followed by charge conjugation, i.e. C \mathrm{e}^{i\pi I_{2}} where I_{2} is an isospin generator. The \rho is even to. The axial current is odd under such a transformation.
For more details, I suggest Quarks and Leptons by Lev Okun.