Computing the exterior derivative of a Riemannian metric

Question

Consider a Riemannian metric on a smooth manifold X expressed by local coordinates $(x_1, ..., x_n)$ on a connected open set $U \subseteq X$, $$g = \sum_{i,j=1}^{n} g_{i,j} dx_i \otimes dx_j$$ I am very confused by the $\otimes$ on the right hand side: I'm not clear on how the exterior derivative $d$ would distribute over such products. (Ie. can I use here the rule that $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^k \alpha \wedge d\beta$?) How can I compute the exterior derivative of g?

Answer 1

The exterior derivative operator $d : \Omega^\bullet(X) \to \Omega^\bullet(X)$ is defined on the graded algebra $$\Omega^\bullet(X) = \bigoplus_{k=0}^\infty \Omega^k(X) $$ of linear combinations of differential forms of all degrees. The elements of $\Omega^\bullet(X)$ are assignments of a linear combination of totally covariant, totally alternating tensors to each point of $X$. In this context, the identity that you mention can be established for all tensor fields $\alpha \in \Omega^k(X)$ and $\beta \in \Omega^\ell(X)$.

A Riemannian metric $g$ is by definition a 2-covector field that is totally symmetric, so the object $dg$ is not defined. You can, however, calculate its covariant derivative with respect to a vector field $X$: see this answer of mine to another question.