De Rham derivative satisfies Leibniz rule

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Statement

Let M be a differential manifold. Consider the de Rham derivative on M, defined as a map:

d: C^\infty(M) \to \Gamma(T^*M)

Then d satisfies the Leibniz rule:

d(fg) = g(df) + f(dg)

In other words, d is a derivation from the algebra C^\infty(M) to the module \Gamma(T^*M).

Proof

To verify equality of the two sides, we need to verify that for any vector field X, the two sides evaluate to the same thing for X. The left side, evaluated on X, yields:

X(fg)

The right side, evaluated on X, yields:

g(Xf) + f(Xg)

The equality of these two sides follows from the Leibniz rule for derivations.