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.<\/p>\n
First order differential operators naturally include vector fields as their homogeneous “top component”.\u00a0 The lowest order component is left multiplication by a smooth function.\u00a0\u00a0 I will attempt to demonstrate that\u00a0 from an algebraic point of view first order differential operators\u00a0 are quite natural and in some sense more fundamental that just the vector fields.<\/p>\n
Geometrically, vector fields are key as they represent infinitesimal diffeomorphisms and are used to construct Lie derivatives as “geometric variations”.\u00a0 This is probably why in introductory geometry textbooks first order differential operators are not described.<\/p>\n
I do not think anything I am about to say is in fact new.\u00a0 I assume the reader has some idea what a differential operator is and that they form a Lie algebra under the commutator bracket.\u00a0 Everything here will be done on supermanifolds.<\/p>\n
I won’t present full proofs, hopefully anyone interested can fill in any gaps.\u00a0 Any serious mistakes then let me know.<\/p>\n
Let \\(M\\) be a supermanifold and let \\(C^{\\infty}(M)\\) denote its algebra of functions.<\/p>\n
Definition<\/strong> A differential operator \\(D\\) is said to be a first order differential operator if and only if<\/p>\n \\(\\left[\u00a0 \\left[ D,f \\right],g \\right]1=0\\),<\/p>\n for all \\(f,g \\in C^{\\infty}(M)\\).<\/p>\n 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).<\/p>\n Let us denote the vector\u00a0 space of\u00a0 first order differential operators as \\(\\mathcal{D}^{1}(M)\\).<\/p>\n Theorem<\/strong> The first order differential operator\u00a0 \\(D \\in\\mathcal{D}^{1}(M) \\) is a vector field if and only if \\(D(1)=0\\).<\/p>\n Proof<\/strong> Writing out the definition of a first order differential operator gives<\/p>\n \\(D(f,g) = D(f)g + (-1)^{\\widetilde{D}\\widetilde{f}}f D(g)- D(1)fg\\),<\/p>\n which reduces to the strict Leibniz rule when \\(D(1)=0\\).\u00a0 QED.<\/p>\n Lemma <\/strong>First order differential operators always decompose as<\/p>\n \\(D = (D-D(1)) + D(1)\\).<\/p>\n The above lemma says that we can write any first order differential operator as the sum of a vector field and a function.<\/p>\n Theorem<\/strong> A first order differential operator \\(D\\) is a zero order operator if and only if \\(D(1) \\neq 0\\) and<\/p>\n \\(\\left[\u00a0 D,f\\right]1 = 0\\),<\/p>\n for all \\(f \\in C^{\\infty}(M)\\).<\/p>\n Proof<\/strong> Writing out the definition of a first order differential operator and using the above Lemma we get<\/p>\n \\(\\left[\u00a0 D,f\\right]1 =\u00a0 (D(f) {-} D(1)f) { -} (-1)^{\\widetilde{D}\\widetilde{f}}f (D {-} D(1)) =0\\).<\/p>\n Thus we decompose the condition into the sum of a function and a vector field.\u00a0 As theses are different they must both vanish separately.\u00a0 In particular \\(D- D(1)\\) must be the zero vector. Then \\(D = D(1)\\) and we have “just” a non-zero function.\u00a0 QED<\/p>\n We assume that the function is not zero, otherwise we can simply consider it to be the zero vector.\u00a0 This avoids the obvious “degeneracy”.<\/p>\n Theorem<\/strong> The space of first order differential operators \\(D \\in\\mathcal{D}^{1}(M) \\) is a bimodule over \\(C^{\\infty}(M)\\).<\/p>\n Proof<\/strong> Let \\(D\\) be a first order differential operator and let \\(k,l \\in C^{\\infty}(M)\\)\u00a0 be functions. Then using all the definitions one arrives at<\/p>\n \\(kDl = k \\left(\u00a0 (-1)^{\\widetilde{l} \\widetilde{D}}(D- D(1)) \u00a0 + D(l) \\right)\\),<\/p>\n which clearly shows that we have a first order differential operator. QED<\/p>\n 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\u00a0 f\\) is not.<\/p>\n Theorem<\/strong> The space of first order differential operators is a Lie algebra with respect to the commutator bracket.<\/p>\n Proof<\/strong> 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<\/p>\n \\(\\left[ D_{1}, D_{2}\u00a0 \\right] = \\left[(D_{1}-D_{1}(1)) , (D_{2} – D_{2}(1))\u00a0 \\right] + (D_{1}-D_{1}(1))(D_{2}(1)){ -} (-1)^{\\widetilde{D_{1}} \\widetilde{D}_{2}} (D_{2}- D_{2}(1)) (D_{1}(1))\\),<\/p>\n which remains a first order differential operator. QED<\/p>\n Note that the above commutator contains the standard Lie bracket between vector fields.\u00a0 So as one expects vector fields are closed with respect to the commutator.<\/p>\n The commutator bracket between first order differential operators is often known as THE Jacobi bracket.<\/p>\n 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.\u00a0 No other order differential operators have these properties.<\/p>\n 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.<\/p>\n","protected":false},"excerpt":{"rendered":" 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”.\u00a0 The lowest order component … Continue reading First Order Differential Operators<\/span>