I thought I would share some interesting things about first order differential operators, acting on functions on a supermanifold. One can reduce the theory to operators on manifolds by simply dropping the sign factors and ignoring the parity.

First order differential operators naturally include vector fields as their homogeneous “top component”.  The lowest order component is left multiplication by a smooth function.   I will attempt to demonstrate that  from an algebraic point of view first order differential operators  are quite natural and in some sense more fundamental that just the vector fields.

Geometrically, vector fields are key as they represent infinitesimal diffeomorphisms and are used to construct Lie derivatives as “geometric variations”.  This is probably why in introductory geometry textbooks first order differential operators are not described.

I do not think anything I am about to say is in fact new.  I assume the reader has some idea what a differential operator is and that they form a Lie algebra under the commutator bracket.  Everything here will be done on supermanifolds.

I won’t present full proofs, hopefully anyone interested can fill in any gaps.  Any serious mistakes then let me know.

Let \(M\) be a supermanifold and let \(C^{\infty}(M)\) denote its algebra of functions.

Definition A differential operator \(D\) is said to be a first order differential operator if and only if

\(\left[  \left[ D,f \right],g \right]1=0\),

for all \(f,g \in C^{\infty}(M)\).

We remark that we have a filtration here rather than a grading (nothing to do with the supermanifold grading) as we include zero order operators here (left multiplication by a function).

Let us denote the vector  space of  first order differential operators as \(\mathcal{D}^{1}(M)\).

Theorem The first order differential operator  \(D \in\mathcal{D}^{1}(M) \) is a vector field if and only if \(D(1)=0\).

Proof Writing out the definition of a first order differential operator gives

\(D(f,g) = D(f)g + (-1)^{\widetilde{D}\widetilde{f}}f D(g)- D(1)fg\),

which reduces to the strict Leibniz rule when \(D(1)=0\).  QED.

Lemma First order differential operators always decompose as

\(D = (D-D(1)) + D(1)\).

The above lemma says that we can write any first order differential operator as the sum of a vector field and a function.

Theorem A first order differential operator \(D\) is a zero order operator if and only if \(D(1) \neq 0\) and

\(\left[  D,f\right]1 = 0\),

for all \(f \in C^{\infty}(M)\).

Proof Writing out the definition of a first order differential operator and using the above Lemma we get

\(\left[  D,f\right]1 =  (D(f) {-} D(1)f) { -} (-1)^{\widetilde{D}\widetilde{f}}f (D {-} D(1)) =0\).

Thus we decompose the condition into the sum of a function and a vector field.  As theses are different they must both vanish separately.  In particular \(D- D(1)\) must be the zero vector. Then \(D = D(1)\) and we have “just” a non-zero function.  QED

We assume that the function is not zero, otherwise we can simply consider it to be the zero vector.  This avoids the obvious “degeneracy”.

Theorem The space of first order differential operators \(D \in\mathcal{D}^{1}(M) \) is a bimodule over \(C^{\infty}(M)\).

Proof Let \(D\) be a first order differential operator and let \(k,l \in C^{\infty}(M)\)  be functions. Then using all the definitions one arrives at

\(kDl = k \left(  (-1)^{\widetilde{l} \widetilde{D}}(D- D(1))   + D(l) \right)\),

which clearly shows that we have a first order differential operator. QED

Please note that this is different to the case of vector fields, they only form a left module. That is \(f \circ X\) is a vector field but \(X \circ  f\) is not.

Theorem The space of first order differential operators is a Lie algebra with respect to the commutator bracket.

Proof Let us assume the basic results for the commutator. That is we take for granted that is forms a Lie algebra. The non-trivial thing is that the space of first order differential operators is closed with respect to the commutator. By the definitons we get

\(\left[ D_{1}, D_{2}  \right] = \left[(D_{1}-D_{1}(1)) , (D_{2} – D_{2}(1))  \right] + (D_{1}-D_{1}(1))(D_{2}(1)){ -} (-1)^{\widetilde{D_{1}} \widetilde{D}_{2}} (D_{2}- D_{2}(1)) (D_{1}(1))\),

which remains a first order differential operator. QED

Note that the above commutator contains the standard Lie bracket between vector fields.  So as one expects vector fields are closed with respect to the commutator.

The commutator bracket between first order differential operators is often known as THE Jacobi bracket.

So in conclusion we see that the first order differential operators have a privileged place in geometry. They form a bimodule over the smooth functions and are closed with respect to the commutator.  No other order differential operators have these properties.

They are also very important from other angles including Jacobi algebroids and related structures like Courant algebroids and generalised geometry. But these remain topics for discussion another day.

We all know the fundamental theorems of calculus, if not check Wikipedia.  I now want to  demonstrate what has been called the fundamental misunderstanding of calculus.

Let us consider the two dimensional plane and equip it with coordinates \((x,y)\).  Associated with this choice of coordinates are  the partial derivatives

\(\left( \frac{\partial}{\partial x} , \frac{\partial}{\partial y} \right)\).

You can think about these in terms of the tangent sheaf etc. if so desired, but we will keep things quite simple.

Now let us consider a change of coordinates. We will be quite specific here for illustration purposes

\(x \rightarrow \bar{x} = x +y\),

\(y \rightarrow \bar{y} = y\).

Now think about how these effect the partial derivatives. This is really just a simple change of variables.  Let me now state  the fundamental misunderstanding of  calculus in a way suited to our example:

Misunderstanding: Despite coordinate x changing the partial derivative with respect to x remains unchanged. Despite the coordinate y remaining unchanged the partial derivative with respect to y changes.

This may seem at first counter intuitive, but is correct. Let us prove it.

Note hat we can invert the change of coordinate for x very simply

\(x = \bar{x} {-}\bar{y} \),

using the fact that y does not change. Then one needs to use the chain rule,

\(\frac{\partial}{\partial \bar{x}}  = \frac{\partial x}{\partial \bar{x}}\frac{\partial}{\partial x}+ \frac{\partial y}{\partial \bar{x}}\frac{\partial}{\partial y}   =    \frac{\partial}{\partial x}\),

\(\frac{\partial}{\partial \bar{y}}  = \frac{\partial x}{\partial \bar{y}}\frac{\partial}{\partial x}+ \frac{\partial y}{\partial \bar{y}}\frac{\partial}{\partial y}   =    \frac{\partial}{\partial y} {-} \frac{\partial}{\partial x} \).

There we are. Despite our initial gut feeling that that the partial derivative wrt y should remain unchanged we see that it is in fact the partial derivative wrt x that is unchanged.  This can course some confusion the first time you see it,  and hence the nomenclature the fundamental misunderstanding of calculus.

I apologise for forgetting who first named the misunderstanding.