We take a fresh look at the vector differential operators grad, div and curl. There are many vector identities relating these. In particular, there are two combinations that always yield zero results:

**Question: Is there a connection between these identities?**

**Adjoint Operators **

A Hilbert space is a function space that is complete and has an inner product: for and in , there is a mapping from to the real numbers: . We consider a linear operator mapping one Hilbert space to another, .

The *adjoint* of is defined as the operator such that

For real-valued functions, we write . In the finite-dimensional case and may be represented by a matrix . Its adjoint is the transformed matrix .

The differential operator for functions on a bounded interval can be represented by a skew-symmetric matrix

with transform .

In the continuous case, we have no matrix representation and we use integration by parts to obtain the adjoint:

Assuming the boundary term vanishes, this means that

so that , just as in the finite-dimensional case.

** Operators on Scalar and Vector Fields **

We let be the space of (smooth) scalar fields over and the space of smooth vector-valued fields on . The vector operators , and define mappings between these function spaces, as shown in the diagram:

We note that

Clearly, some compositions of these operators are well-defined while others are not. The legal combinations of two operators (right operator first) are

The vector Laplacian, mapping to itself, is defined by .

For the function spaces and we define the inner products by integration:

Let us start with the divergence operator, and form :

Assuming that the boundary term (the surface integral) vanishes, this gives

which may be written

This gives us the transpose , and also (these are reminiscent of above). We also note that

so the Laplacian operator is self-adjoint.

Now let us look at the curl operator, and form . We use the vector identity

Assuming that the divergence term integrates to zero, this gives

which may be written

This gives us the transpose , showing that the operator is symmetric.

Finally, using and , we have

so the two vector identities with which we started,

are actually adjoints of each other.

**This answers the question posed at the start of this article:**

** Sources **

Strang, Gilbert, 1986: *Introduction to Applied Mathematics.* Wellesley-Cambridge Press, 758pp. ISBN: 0-961-40880-4.