A Riemannian metrics is a smooth 2-tensor field that is positive definite. And what confuses me about it is calculation of such 2-tensor with respect to dual basis. I will illustrate my confusion with the simplest 2-tensor, the usual inner product and its analog.
I first do the calculation for Euclidean metrics, and then in the second part talk about my confusion about Riemannian metrics.
Given a basis $v_i$, the inner product $\langle \ ,\ \rangle = \sum_{i,j=1}^n g_{i,j}v^*_i\otimes v^*_j$ where $\{v^*_i\}$ is the dual basis, and $g_{i,j}=\langle v_i, v_j\rangle$.
In the $\mathbb{R}^2$ with usual coordianate, set a new basis to be $v_1 = (1,0), v_2 = (\frac{1}{2},\frac{\sqrt{3}}{2})$, then the dual basis should be $v^*_1 = (1,-\frac{1}{\sqrt{3}})^T, v^*_2 = (0,\frac{2}{\sqrt{3}})^T$ (I regard dual vector as transposed vector) so that $v_iv^*_j=\delta_{i,j}$.
Then we try to calculate the inner product $\langle a ,b \rangle$ where $a=b=(1,0)$. With respect to the orthogonal basis $\langle a ,b \rangle_o=1\cdot 1+0\cdot 0 = 1$. ('$_o$' denotes the orthogonal coordinate system.)
Then we try to calculate $\langle a ,b \rangle$ with respect to the same basis but with the formula using the dual basis $\{v^*_i\}$,
--[When I first did the calculation, I made a mistake that I calculated $a, b$'s coordinates with respect to $\{v^*_i\}$ $a=b=(1,\frac{\frac{1}{\sqrt{3}}}{\frac{2}{\sqrt{3}}})_{v*}=(1,\frac{1}{2})_{v*}$, this is wrong since $a,b$ and $v^*_i$ are not in the same space, namely dual basis is not the basis for vector space of $a, b$.]--
$$\langle a ,b \rangle_{o}= \sum\langle v_i, v_j\rangle v^*_i\otimes v^*_j(a,b),$$ which according to definition of tensor product equals \begin{align*} \sum\langle v_i, v_j\rangle v^*_i(a)\otimes v^*_j(b)=\langle(1,0),(1,0)\rangle (\ \langle(1,-\frac{1}{\sqrt{3}}),(1,0)\rangle\ \langle(1,-\frac{1}{\sqrt{3}}), (1,0)\rangle\ )\\ +\langle(1,0),(\frac{1}{2},\frac{\sqrt{3}}{2})\rangle (\ \langle(1,-\frac{1}{\sqrt{3}}),(1,0)\rangle\ \langle(0,\frac{2}{\sqrt{3}}), (1,0)\rangle\ )\\ +\langle(\frac{1}{2},\frac{\sqrt{3}}{2}),(1,0)\rangle (\ \langle(0,\frac{2}{\sqrt{3}}),(1,0)\rangle\ \langle(1,-\frac{1}{\sqrt{3}}),(1,0)\rangle\ )\\ +\langle(\frac{1}{2},\frac{\sqrt{3}}{2}),(\frac{1}{2},\frac{\sqrt{3}}{2})\rangle (\ \langle(0,\frac{2}{\sqrt{3}}),(1,0)\rangle\ \langle(0,\frac{2}{\sqrt{3}}),(1,0)\rangle\ )\\ =1\cdot1\cdot1+\frac{1}{2}\cdot1\cdot0+\frac{1}{2}\cdot0\cdot1+1\cdot0\cdot0\\ =1 \end{align*} (In this calculation , all vectors and dual vectors are written with respect to the usual orthogonal coordinates.)
In this example, $\langle a ,b \rangle$ calculated the usual way in orthogonal coordinate system and using dual basis in orthogonal coordinate systemm are the same, which seems to make sense.
In the following section $a, b$ are not limited to be $(1,0)$.
What confuses me is as follows: With respect to the basis $\{v_i\}$, $a=(a_1,a_2)_v, b=(b_1, b_2)_v$ (I use '$_v$' to distinguish it from the usual orthogonal coordinates), we can define $\langle a ,b \rangle_v$ to be $a_1b_1+a_2b_2$.
Because $v_1, v_2$ are not orthogonal, this is not allowed in Euclidean metrics, where $\sum a_ib_i$ only equals inner product in orthogonal system since the essential way to calculate inner product is based on geometrical fact that inner product equals product of length times $\cos$ of angle, using vector decomposition into linear combination of basis and distribution laws of vector space, i.e. $\langle a,b\rangle = (a_1v_1+a_2v_2)(b_1v_1+b_2b_2)=a_1b_1v_1\cdot v_1+a_2b_2v_2\cdot v_2+\dots$.
However, the definition is proper for the use of Riemannian metrics since it's bilinear and positive definite and it looks like a smooth function.
Then if we, according to the above definition of inner product and using only the $\{v_i\}$ coordinate system, calculate $\langle a, b\rangle_v$ in the same two ways as stated above--namely
replacing $\langle a, b\rangle$, $\langle v_i, v_j\rangle$ with $\langle a, b\rangle_v$ $\langle v_i, v_j\rangle_v$ (this seems natural (though wrong for Euclidean metrics), but is it okay for Riemannian metrics, i.e. to provide consistent calculation results?),
changing coordinates of $a, b, v_i, v^*_i$ accordingly ($v^*_1, v^*_2$ should become $(0,1), (1,0)$ according to $v_iv^*_j=\delta_{i,j}$, even in this non-orthogonal system?),
calculate the tensor $v^*_i(a)$, $v^*_j(b)$ also as 'inner product' (but the new one $\langle \ , \ \rangle_v$) of $v^*_i$ and $a$ (or $b$) (Is this proper?)
--will the two results always match each other? (They should.)
If not, how should one modify the calculation to make the two results consistent?
PS: since there are different coordinate systems and inner products and first order tensors acting on vector (which are also an inner products) used here, I guess in the discussions and calculations above there may be some misuse of coordinate system and inner product, would anyone point that out if any exists?
