Remarks on Riemannian geometry

Let $M$ be a compact $\Coo$ manifold without boundary, of dimension $n$. Compactness is not crucial for some of our arguments (although it may be for others), but is very convenient, since it means that the algebras $C(M)$ and $\Coo(M)$ are unital: the unit is the constant function $1$. For convenience we use the function algebra $A = C(M)$ --a commutative $C^*$-algebra-- at the beginning. We will change to $\sA = \Coo(M)$ later, when the differential structure becomes important.

Any $A$-module (or more precisely, a ``symmetric $A$-bimodule'') which is finitely generated and projective is of the form $\sE = \Ga(M, E)$ for $E \to M$ a (complex) vector bundle. Two important cases are
$$\begin{align*} \gX(M) &= \Ga(M, T_\bC M) = \text{ (continuous) vector fields on... ...sA^1(M) &= \Ga(M, T^*_\bC M) = \text{ (continuous) 1-forms on } M. \end{align*}$$
These are dual to each other: $\sA^1(M) \isom \Hom_A(\gX(M), A)$, where $\Hom_A$ means ``$A$-module maps'' commuting with the action of $A$ (by multiplication).

Since each $g_x$ is positive definite, there are ``musical isomorphisms'' between $\gX(M)$ and $\sA^1(M)$, as $A$-modules

$$\begin{diagram}[small] \gX(M) & \pile{\rTo^{X \to X^\2} \\... ...X^\2(Y) &:= g(X, Y) \\ \al(Y) &=: g(\al^\3, Y). \end{cases}$$

They are mutually inverse, of course. In fact, they can be used to transfer the metric form $\gX(M)$ to $\sA^1(M)$:

$$g(\al, \bt) := g(\al^\3, \bt^\3), \word{for} \al,\bt \in \sA^1(M).$$

One should perhaps write $g^{-1}(\al, \bt)$ --as is done in [GVF]-- since in local coordinates $g_{ij} := g(\del/\del x^i, \del/\del x^j)$ and $g^{rs} := g(dx^r, dx^s)$ have inverse matrices: $[g^{rs}] = [g_{ij}]^{-1}$.

If $f\in C^1(M)$, the gradient of $f$ is $\grad f := (df)^\3$, so that

$$g(\grad f, Y) = df(Y) := Yf.$$