If we define the inner product as ${\textbf{u}\cdot\textbf{v}=g_{ij}u^{i}v^{j}}$, where ${g_{ij}}$ is the metric tensor, ${S}$ and ${T}$ are transformation matrices, ${S}$-for covariant indices and ${T}$-for contravariant (obviously ${ST=I}$). How do we prove that the inner product is invariant? I tried this way: $${T^{a}_{i}T^{b}_{j}g_{ij}u^{i}v^{j}S^{i}_{a}S^{j}_{b}=\delta^{a}_{a}\delta^{b}_{b}g_{ij}u^{i}v^{j}}$$ which is not good expression moreover in the beginning I multiply by tensor with index that is already dummy which is not right. Is there a way to prove it with this kind of transformations?
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$\begingroup$ It's not clear what you're asking. The equation seems to be nonsense, as you seemed to know yourself, but the words aren't specific enough to know what you want to do. $\endgroup$– BrickCommented Sep 17, 2015 at 21:10
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Start by rewriting the scalar product as a covariant-contravariant contraction, like so: $$ {\bf u}\cdot{\bf v} = g_{ij}u^iv^j = (g_{ij}u^i)v^j = u_jv^j $$ Now transform the components with your $S$ and $T$ matrices, $$ u_jv^j = \left( S_j^a {\bar u}_a \right) \left( T^j_b {\bar v}^b \right) = (S_j^a T^j_b) {\bar u}_a {\bar v}^b = \delta^a_b {\bar u}_a {\bar v}^b = {\bar u}_b {\bar v}^b $$ and reintroduce the metric tensor to obtain $$ g_{ij}u^iv^j = g_{ab}{\bar u}^a {\bar v}^b, \;\;\; \text{or} \;\;\; {\bf u}\cdot{\bf v} = \bar{{\bf u}}\cdot\bar{{\bf v}} $$ I may have redefined the direction of your $S$ and $T$ in the process, but this conveys the gist of it.
