$SU(2)$ infinitesimal transformation of pion triplet lagrangian raising and lowering indices

Question

How can I show that $$\epsilon^{ijk} (\partial^\mu\pi^i)\pi^j\partial_\mu\alpha^k(x)=\epsilon^{ijk}(\partial_\mu\pi^i)\pi^j\partial^\mu\alpha^k(x)~?$$

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 $g^{\mu\nu}g_{\mu\nu}=\delta_\mu^\nu$

Answer 1

Note that your attempt is not technically correct, because you can't repeat a dummy index, so you should use (for example) $g_{\alpha\nu}$ instead of $g_{\mu\nu}$, since $\mu$ is already in use.

Apart from that,

since $g^{\mu\nu}g_{\mu\nu}=\delta^\nu_\mu$

is not correct either, it would be $g^{\mu\nu}g_{\mu\nu}=\delta^\mu_\mu=\delta^0_0+\delta^1_1+\delta^2_2+\delta^3_3=4.$ What you were refering to is $g^{\mu\nu}g_{\alpha\nu}=\delta^\mu_\alpha.$

Besides, you could skip one step using the $g^{\mu\nu}$ to raise the second index, i.e. $$(\partial^\mu\pi^i)\partial_\mu\alpha^k=g^{\mu\nu}(\partial_{\nu}\pi^i)\partial_\mu\alpha^k=(\partial_{\nu}\pi^i)\partial^\nu\alpha^k,$$ and then rename the indices.