How can I show that $\epsilon^{ijk} (\partial^\mu\pi^i)\pi^j\partial_\mu\alpha^k(x)=\epsilon^{ijk}(\partial_\mu\pi^i)\pi^j\partial^\mu\alpha^k(x)$.$$\epsilon^{ijk} (\partial^\mu\pi^i)\pi^j\partial_\mu\alpha^k(x)=\epsilon^{ijk}(\partial_\mu\pi^i)\pi^j\partial^\mu\alpha^k(x)~?$$
can I do the following? $\pi$ is the pion triplet and $\alpha$ is an infinitesimal transformation: $\pi'^i=\pi^i-\epsilon^{ijk}\pi^j\alpha^k(x)$.
$$\epsilon^{ijk} (\partial^\mu\pi^i)\pi^j\partial_\mu\alpha^k(x)=\epsilon^{ijk} g^{\mu\nu}(\partial_\nu\pi^i)\pi^jg_{\mu\nu}\partial^\nu\alpha^k(x)=\epsilon^{ijk}(\partial_\nu\pi^i)\pi^j g^{\mu\nu}g_{\mu\nu}\partial^\nu\alpha^k(x)=\epsilon^{ijk}(\partial_\mu\pi^i)\pi^j\partial^\mu\alpha^k(x)$$ since $g^{\mu\nu}g_{\mu\nu}=\delta_\mu^\nu$
