# De Rham derivative satisfies Leibniz rule

From Diffgeom

## Statement

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

Then satisfies the Leibniz rule:

In other words, is a derivation from the algebra to the module .

## Proof

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

The right side, evaluated on , yields:

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