Transformation law for metric tensor For my topology class I have to calculate the transformation law of the metric tensor $g = g_{ij} \, dx^i \otimes dx^j$ under a coordinate transformation $x \longmapsto y=(x)$.
My approach:
$$ g(y) \enspace = \enspace g_{ij}(y) \, \big( dy^i \otimes dy^j \big) \enspace = \enspace g_{ij}(y) \, \bigg( \frac{\partial y^i}{\partial x^k} dx^k \otimes \frac{\partial y^j}{\partial x^{\ell}} dx^{\ell} \bigg) $$
$$= \enspace g_{ij}(y) \, \frac{\partial y^i}{\partial x^k} \, \frac{\partial y^j}{\partial x^{\ell}} \, \big(  dx^k \otimes dx^{\ell}  \big) \quad .$$
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$\large \textbf{1.)} \enspace$ How do I know how the components $g_{ij}(y)$ transform?
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$\large \textbf{2.)} \enspace$ If I identify the terms $\dfrac{\partial y^i}{\partial x^k}$ with the Jacobian $J^i{}_j$, then I get
$$ g(y) \enspace = \enspace g_{ij}(y) \, J^i{}_k \, J^j{}_{\ell} \, \big(  dx^k \otimes dx^{\ell}  \big) \quad .$$
Is it then true that:
$$ g(y) \enspace = \enspace \big( \boldsymbol{\operatorname{J}}^T \cdot \boldsymbol{\operatorname{g}}(y) \cdot \boldsymbol{\operatorname{J}} \big)_{k\ell} \, \big(  dx^k \otimes dx^{\ell}  \big) \quad ? $$
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$\large \textbf{3.)} \enspace$ If I were to calculate the determinant of $g(y)$, how would I handle the tensor product in a mathematical rigorous way? Assuming that the equation in 2.) is true, would it just be
$$ \det g(y) \enspace = \enspace \det \boldsymbol{\operatorname{J}}^T \cdot \det \boldsymbol{\operatorname{g}}(y) \cdot \det \boldsymbol{\operatorname{J}} \quad ? $$
 A: Keep in mind your goal is to write $g(y) = g(y(x)) = g_{kl}'(x)(dx^k\otimes dx^l)$ and find this $g_{kl}'$. So you're basically done with (1). All that's left to do is read off the coefficients and conclude the transformation law:
$$g_{kl}'(x) = g_{ij}(y(x))\frac{\partial y^i}{\partial x^k}\frac{\partial y^j}{\partial x^l}.$$
(2) Yes, if you put $J^i_{\ k} = \partial y^i/\partial x^k$, it is a basic exercise in indices to check that $g' = J^TgJ$. It might help to consider the following fact about $n\times n$ matrices: $(A^TB)_{ij} = (A^T)_{ik}B_{kj} = A_{ki}B_{kj}$.
With that in hand, we have
$$g_{ij} J^i_{\ k}J^j_{\ l} = J^i_{\ k}g_{ij}J^j_{\ l} = (J^Tg)_{kj}J^j_{\ l} = (J^TgJ)_{kl}.$$
Thus $g_{kl}' = (J^TgJ)_{kl}$, as claimed.
(3) You don't have to worry about tensor product, because when people say $\det(g)$ they're referring to the determinant of the matrix $g_{ij}$. Thus the above calculations show
$$\det(g') = \det(J^TgJ) = \det(J)^2\det(g),$$
where we have used the fact $\det(J^T) = \det(J)$.
Incidentally, this proves the change of variables formula for integration on Riemannian manifolds. The volume form on an $n$-manifold is $\omega = \sqrt{|\det(g)|}\, dx_1\wedge\dotsm\wedge dx_n$. Given a diffeomorphism $x\mapsto y(x)$, the volume form becomes $\omega' = |\det(J)|\sqrt{|\det(g)|}\,dx_1\wedge\dotsm\wedge dx_n$, i.e., we scale by the absolute value of the Jacobian-determinant.
