I know we need prove this property by:
\begin{align*}g_{\rho\sigma} = g_{\mu\nu}\Lambda^\mu_{\ \ \ \rho} \Lambda^\nu_{\ \ \ \sigma} \end{align*}
and
$$\Lambda^\mu_{\ \ \ \nu}=\delta^\mu_{\ \ \ \nu} + \omega^\mu_{{\ \ \ \nu}}$$
then
$$(\delta_{\rho}^{\mu}+\omega_{\rho}^{\mu})g_{\mu\nu}(\delta_{\sigma}^{\nu}+\omega_{\sigma}^{\nu})+o(\omega^{2})=g_{\rho\sigma}$$
then we get this
$$g_{\rho\sigma}+\omega_{\rho}^{\mu}g_{\mu\nu}\delta_{\sigma}^{\nu}+\omega_{\sigma}^{\nu}g_{\mu\nu}\delta_{\rho}^{\mu}+o(\omega^2)=g_{\rho\sigma}$$
just move terms we can get the correct answer right? But I have no idea about the third term:
$$\omega_{\sigma}^{\nu}g_{\mu\nu}\delta_{\rho}^{\mu}$$
I want to know why not in this order:
$$\delta_{\rho}^{\mu}g_{\mu\nu}\omega_{\sigma}^{\nu}$$
Can someone just told me how to get the correct order?
Best Answer
short answer: Since it is a multiplication of normal number then the order doesn't matter.
longer answer: whilst it may not look it $ \delta_{\rho}^{\mu} g_{\mu\nu} \omega_{\sigma}^{\nu}$ is just the sum of several components multiplied together, and since the components of the tensors are just ordinary numbers (not operators, like the derivative) we can swap the order.
Example: (only in x,y)
$$\delta_{\rho}^{\mu}g_{\mu\nu}\omega_{\sigma}^{\nu}=\delta_{\rho}^{0}g_{00}\omega_{\sigma}^{0}+\delta_{\rho}^{0}g_{01}\omega_{\sigma}^{1}+\delta_{\rho}^{1}g_{10}\omega_{\sigma}^{0}+\delta_{\rho}^{1}g_{11}\omega_{\sigma}^{1}=\omega_{\sigma}^{\nu}g_{\mu\nu}\delta_{\rho}^{\mu}$$
From my relatively limited interactions with relativity you don't get a tensor multiplied by another tensor expression like you often get with matrices, instead it is an expression relating components of a tensor to components another tensor
hopefully that helps