Conjugate momentum for scalar field in curved spacetime The Lagrange density $\mathcal{L}$ of a scalar field $\phi$ in curved spacetime is
$$\mathcal{L}=\sqrt{-g}(-\frac{1}{2} g^{\mu\nu}\nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2-\xi R\phi^2)\tag{9.87}$$
where $g$ is metric determinant, $\xi$ is a constant and $R$ is the curvature scalar, and the signature convention is $(-,+,+,+)$.
I read from Sean Carroll's Spacetime and Geometry book, pg 394-395, that the conjugate momentum $\pi$ is given by
$$\pi=\frac{\partial\mathcal{L}}{\partial(\nabla_0\phi)}\tag{9.90}$$
and
$$\pi=\sqrt{-g}\nabla_0\phi.\tag{9.91}$$
How can one show (9.91) without knowing the $g^{\mu\nu}$ components?
 A: OP has a point: Eq. (9.91) should read
$$\pi~=~-\sqrt{-g}g^{0\mu}\nabla_{\mu}\phi~=~-\sqrt{-g}\nabla^0\phi,\tag{9.91'}$$
i.e. the 0-index should be upstairs.
A: To prove equation 9.91 globally we will first prove it locally, the generalization is then straightforward.
We take a point $p$ of the spacetime-manifold $M$, at which a tangent space $T_p$ is defined.
The scalar fields $\phi(x^{\mu})$ are then defined with respect to a coordinate system constructed from the basis vectors of $T_p$.
Now, in general relativity, curved spacetime looks locally like Minkowski space + a gravitational force (Equivalence Principle). So we can construct Riemann Normal Coordinates $x^{\hat{\mu}}(p)$ satisfying:
$$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}, \partial_\hat{\sigma} g_{\hat{\mu} \hat{\nu}}(p) = 0.$$
These coordinates are called locally inertial coordinates (see eq 2.47 in Carroll's book).
Next relabel $g_{\hat{\mu} \hat{\nu}}$ to $g_{\mu\nu}$ to avoid confusion. Then, we show that equation 9.91 holds in these coordinates:
$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$
$$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$
by only looking at the first term, as the rest does not depend on $\nabla_0 \phi$, we further derive:
$$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi;$$
just use the Leibniz rule to prove this statement. Then:
$$\pi = \sqrt{-g} \nabla_0 \phi.$$
Where the metric is put in its canonical form $g_{\mu \nu} = diag(-1, +1, +1, +1)$ as we are using locally inertial coordinates.
Finally, as $\pi = \sqrt{-g} \nabla_0 \phi$ is a tensorial equation (as the covariant derivative of a scalar field is independent of the used coordinate system) so 9.91 is globally true.
It should be noted that $g^{00} = g_{00} = -1$ and by this equation 9.91 is true.
