Let be a Riemannian manifold with Riemannian connection . A connection is a thing that knows how to differentiate a vector field in the direction of a vector field ; the result is denoted by and is also a vector field. For consistency of notation, it is convenient to write for the derivative of scalar function in the direction , even though this derivative does not need a connection: vector fields are born with the ability to differentiate functions.

The pairs , with a scalar function and a vector field, form a funky nonassociative algebra described in the previous post. And is a derivation on this algebra, because

- by the definition of a connection
- by the metric property of the Riemannian connection.

Recall that the commutator of two derivations is a derivation. Or just check this again: if and are derivations, then

and the difference simplifies to what it should be.

Thus, for any pair of vector fields the commutator is a derivation on . The torsion-free property of the connection tells us how it works on functions:

Subtracting from the commutator, we get a derivation that kills scalar functions,

But a derivation that kills scalar functions is *linear over functions*:

In plain terms, processes any given vector field pointwise, applying some linear operator to the vector at every point of the manifold. No derivatives of are actually taken, either of first or of second order.

Moreover, the derivation property immediately implies that is a skew-symmetric operator: for any vector fields

because kills scalar functions.

The other kind of skew-symmetry was evident from the beginning: by definition.

What is not yet evident is that is also a tensor in and , that is, it does not differentiate the direction fields themselves. To prove this, write where should be thought of as the pointwise second-order derivative in the directions (i.e., the result of plugging two direction vectors into the Hessian matrix). By symmetry, it suffices to show that is a tensor in and . For , this is clear from the definition of connection. Concerning , we have

That’s it, we have a tensor that takes three vector fields and produces another one, denoted . Now I wonder if there is a way to use the language of derivations to give a slick proof of the first Bianchi identity, …

To avoid having two picture-less posts in a row, here is something completely unrelated:

This is the image of the unit circle under the polynomial . Which area is larger: red or green? Answer hidden below.

They are equal.