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This question is part of Calculation of inner product for Riemannian metrics.. The original question may be too long, so I break it down.

In the following $\{v_i\}$ is the basis.

For an $\langle\ ,\ \rangle$, $\langle a ,b \rangle= \sum g_{ij} v^*_i\otimes v^*_j(a,b)=\sum\langle v_i, v_j\rangle v^*_i\otimes v^*_j(a,b)$.

We may define $\langle a ,b \rangle=a_1b_1+a_2b_2+...+a_nb_n$ where $a_i, b_i$ are the coordinates with $v_1, v_2$ as basis, then $g_{i,j}=\langle v_i, v_j\rangle=\langle (0,\dots,1,\dots,0)_{i\ th\ component\ only\ being\ 1}, (0,\dots1,\dots,0)_{j\ th\ component\ only\ being\ 1}\rangle=\delta_{ij}$.

Or we may define $\langle a ,b \rangle=\lambda_1 a_1b_1+\lambda_2 a_2b_2+...+\lambda_n a_nb_n$, then $g_{i,j}=\langle v_i, v_j\rangle=\langle (0,\dots,1,\dots,0)_{i\ th\ component\ only\ being\ 1}, (0,\dots1,\dots,0)_{j\ th\ component\ only\ being\ 1}\rangle=\lambda_i\lambda_j\delta_{ij}$.

Here I regard the 'braket operation' for $a,b$ and $v_i, v_j$ as the same inner product that we define, is it correct?

From the above calculation we see $g_{ij}$ is closely related to (sort of an extention of) $\delta_{ij}$, is it correct?

Charlie Chang
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The examples you mention are specific cases of what is known as an inner product. In general, given a specific basis, an inner product on the vector space is of the form, $$<x,y>= x^{\top}A y ,$$
where $A$ is a symmetric, positive definite matrix.

So, yes , what we see is a generalisation of the usual dot product, but it can be more general than what you wrote.

janmarqz
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Reader Manifold
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    So my example equals the case where $A$ is diagonal, while in the usual inner product $A = I$ – Charlie Chang Jul 27 '20 at 07:53
  • In a general inner product, $A$ can be any $n×n$ symmetric matrix (where $n$ is the dimension of the vector space) , not just diagonal matrices as given in your examples. In the case of dot product (which I guess is what you mean by the usual inner product), yes, $A =I$. – Reader Manifold Jul 27 '20 at 08:00
  • @Deepak $A$ has to be positive definite as well as symmetric, or at least non-degenerate for pseudo-Riemannian manifolds. – Chrystomath Jul 27 '20 at 08:57
  • @Chrystomath Oh yes, sorry about that. Thanks for pointing it out. – Reader Manifold Jul 27 '20 at 09:31