Analytic expression of $g^{\sharp}$ The analytical expression of $g^{\flat}\colon TM \to T^*M$ is given by
$$g^{\flat} = g_{ij}x^idx^j$$ with $x^i$ the coefficients of a $X\in TM$ in basis $\frac{\partial}{\partial{x^i}}$. Also, given the invertibility of $G = (g_{ij})$, $g^{\flat}$ is isomorphism and thus $g^{\sharp}\colon T^*M \to TM$ can be defined. According to bibliography, the analytical expression of $g^{\sharp}\colon T^*M \to TM$ is
$$g^{\sharp} = g^{ij}\omega_j\frac{\partial}{\partial{x^i}}$$
with $\omega_i$ the coefficients of a $\omega\in T^*M$ in basis $d\omega^i$. I am trying to obtain the analytic expression of $g^{\sharp}$ starting from the definition of $g^{\flat}$ but I could not come up with an analytical solution. Could you please provide some guide lines?
 A: I think the confusion here is notational. Keep in mind $g^\flat: TM\to T^*M$ is defined by $X\mapsto g(X,\cdot)$. The "analytical expression" is obtained by noting that if $X=x^i\frac{\partial}{\partial x^i}$ and $Y = y^i\frac{\partial}{\partial x^i}$, then
$$g^\flat(X)(Y) = g(X,Y) = g\big(x^i\frac{\partial}{\partial x^i},y^j\frac{\partial}{\partial x^j}\big) = g\big(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\big)x^iy^j = g_{ij}x^idx^j(Y),$$
from which we conclude $g^\flat (X) = g_{ij}x^idx^j$.
Since $g^\sharp$ is the inverse of $g^\flat$, we can obtain a formula for it as follows: write $\omega = \omega_j dx^j$ in local coordinates. We want to find the unique $X$ for which $g(X,\cdot) = \omega$. Since $g(X,Y) = \omega(Y)$ for all $Y$, we must have $g_{ij}x^idx^j(Y) = \omega_j dx^j(Y)$, meaning $g_{ij}x^i=\omega_j$. But this is just a matrix-inversion problem written in Einstein notation; if $g^{ij}$ denotes the inverse of the matrix $g_{ij}$, then $x^i = g^{ij}\omega_j$. Thus
$$g^{\sharp}(\omega) = X = x^i\frac{\partial}{\partial x^i} = g^{ij}\omega_j\frac{\partial}{\partial x^i},$$
which is the formula we wanted.
