An intro to D-modules Daniel Smolkin BIKES, February 9, 2016 1 The Weyl algebra Let K[x1; ··· ; xn] (henceforth abbreviated K[X]) be a polynomial ring. The nth Weyl algebra is a K-subalgebra of EndK (K[X]) generated by: • x^i, wherex ^i(g) = xig • @ @xi @ Typically we denotex ^i simply by xi and by @i. So a typical element of A2 might look @xi like 2 3 2 σ = @1 @2x1x2 − 4x1x2 where 3 3 2 @ 2 3 2 2 @ 3 5 2 5 2 σ(x1) = 2 x1x2x1 − 4x1x2x1 = 2 x1x2 − 4x1x2 = 6x1 − 4x1x2 @ x1@x2 @ x1@x2 We denote this Weyl algebra An. Note that the Weyl algebra is NOT commutative! It comes with a commutator [·; ·] defined by [a; b] = ab − ba. Here are some commutators to know: • [@i; xi] = 1, because for any f 2 K[X], the product rule says [@i; xi]f = (@ixi − xi@i) f = @i(xif) − xi(@if) = f + xi@if − xi@if = 1 · f @f • More generally, [@i; f] = (this is an element of the Weyl algebra!) @xi • Thus, [@i; xj] = δij • [xi; xj] = 0 for all i; j 1.1 Canonical form α1 αn Writing x1 ··· xn is clunky, so we adopt the following notation: given α = (α1; ··· ; αn) 2 n α α1 αn α N , by x we mean x1 ··· xn . Similarly for @ . Such an α is called a multi-index. Also, by jαj we mean α1 + ··· αn. If char k = 0, The Weyl algebra An has K-vector space basis α β n given by x @ j α; β 2 N . For instance, 2 2 2 2 2 @i xi = @i(xi @i + 2xi) = @ixi @i + 2@ixi = (xi @i + 2xi)@i + 2xi@i + 2 p What goes wrong if char k = p? Well, in that case @i = 0! Indeed, if a ≥ p, then p a @i xi = 0. This canonical form is our friend and most proofs will rely on looking at it and applying some commutators cleverly. 1 1.2 Degree of an operator α β α β Given a monomial x @ 2 An, we can define degx @ = jαj + jβj. Given any element of An, we define its degree to be the largest degree among the degrees of its monomials. It takes a bit of work, but one can show Proposition 1.1. We have the following: • deg(D + D0) ≤ maxfdeg D; deg D0g • deg(DD0) = deg D + deg D0 • deg[D; D0] ≤ deg D + deg D0 − 2 Proof. 1 is clear. For 2 and 3, Induce on degD + degD0. Also, from 1, it's enough to assume that D and D0 are monomials. This shows that An is a domain. Also, we have the following neat fact: Proposition 1.2. An has no nontrivial two-sided ideals (so it's a simple algebra) Proof. Proof by contradiction. Suppose I is a nonzero two-sided ideal. Choose D of minimal degree k in I. Then [D; f] 2 I for all f. Now, deg D > 0, so there is some α and β such α β that x @ has a nonzero coefficient in the expansion of D, say βi > 0. But then [xi;D] is nonzero and has degree ≤ k − 1. Thus, β = 0. But if αi > 0, then [D; @i] has degree ≤ k − 1 and is not zero. 2 Modules over An We haven't even defined a D-module yet! A D-module is a module over An (or the ring of differential operators of any ring). 3 Graded and filtered modules L Recall that a ring R is graded if R = Ri with Ri · Rj ⊆ Ri+j. If R is a graded ring, then i L a graded R-module is a module M such that M = i Mi where RiMj ⊆ Mi+j. Now, we'd like to make An into a graded algebra. There's one problem though: we can't say something like \@1x1 degree 2" because @1x1 = x1@1 + 1. So instead we have to be content with a filtration. Recall that a filtration of a K-algebra R is an ascending chain of K-vector spaces F0 ⊂ F1 ⊂ F2 ::: such that FiFj ⊂ Fi+j. One filtration on An that's of particular interest to us is the Bernstein filtration, denoted α β Bi. We define Bi to be the k-vectorspace generated by x @ j jαj + jβj ≤ i . This filtration is cool because every Bi is a finite-dimensional vector space over k. One more construction: let R be a ring and fFig a filtration of R. For all n, let σn : Fn ! Fn=Fn−1 be the usual quotient map. This map σ is called the symbol map in this context. Then the graded ring associated to F , denoted grF (R), is given by F M gr (R) = Fi=Fi−1 i Note that any homogenous element of degree k in this ring can be written as σk(q) for some q 2 Fk. B In the case of the Weyl algebra, we write Sn to denote gr An. 2 Lemma 3.1. Sn is a polynomial ring in 2n variables. β Proof. It's easy to see that gr An is generated by σ1(xi) and σ1(δi) over k. Let yi = σ(xi) 2 Sn and yn+i = @i 2 Sn. We wish to show that the yi commute. For this it's enough to show that yiyn+i = yn+iyi. But this is clear: σ2(@ixi) = σ2(xi@i + 1) = σ2(x2@i). Thus we can write a surjective map from a polynomial ring k[z1; ··· ; z2n] ! Sn by zj 7! yj, for 1 ≤ j ≤ 2n. It remains to show that this map, ', is injective. To that end, suppose some homogenous polynomial F is sent to zero. If '(F ) = 0, then '(F ) = σk(d) where d = F (x1; ··· ; xn;@1; ··· ;@n). But if σk(d) = 0, then d can also be written as a sum of monomomials of degree less than k. So d must be zero to begin with, since canonical form is a k-basis of An. Similarly define a filtered module and graded module associated to a filtration: if M is an An module, then a filtration of M with respect to B is an ascending chain of k-vector S spaces Γ0 ⊆ Γ1 ⊆ · · · with M = i Γi and BiΓi ⊆ Γi+j. A filtration on M induces filtrations on its submodules and quotients: let Γ be a filtration on M with respect to B and let N ⊆ M be a submodule. Then Γi \ N and Γi + N=N are filtrations on N and M=N, repesctively, that agree with B. Now, let M be a left An module. Then Γ Lemma 3.2. if gr M is a (left) noetherian Sn module, then M is a noetherian An module 0 Proof. Pick any N ⊆ M and let Γ0 be the induced grading on N. Then grΓ N ⊆ grΓM is a finitely generated submodule. We wish to show that N is a finitely generated submodule of M. Γ0 Since gr N is finitely generated, we can enumerate its generators f1; ··· ; fs. Then 0 there is some m such that grΓ N is generated by elements of degree ≤ m. I claim that N is generated by elements of degree ≤ m. Suppose this weren't the case: then choose 0 a homogenous element f 2 N of minimal degree such that f 62 An · Γm. Then f has P some degree k > m. Then σk(f) = σk−ri (ai)fi for some ai 2 Sn. But this implies P ^ ^ f − aifi 2 ker σk = Γk−1, where fi = µ(fi). This contradicts the fact that deg f = k. 0 0 Thus N is generated by elements in Γm. But Γm is a finite-dimensional k-vector space with a finite dimensional basis. So this basis generates N. The converse doesn't hold. If grΓM is noetherian, we call Γ a good filtration of M. Theres a lemma saying all good filtrations are sorta the same. Namely: Lemma 3.3. A filtration Γ of M is good if and only if there exists some k0 such that Γi+k = Biγk for all k > k0. and Lemma 3.4. Let Γ and Ω be two filtrations of M. If Γ is good, then there is some k1 such that Γj ⊆ Ωj+k1 for all j. Proof. This follows from the above: choose some k0 such that Γi+k0 = BiΓk0 . Then there is some k1 such that Ωk1 ⊇ Γk0 , since Γk0 is a finite-dimensional k-vector space. Then for any j, Γj ⊆ Γj+k0 = BjΓk0 ⊆ BjΩk0 ⊆ Ωk0+j 3 Note that any Noetherian An module has a good filtration: if M is generated by u1; ··· uk, Pk set Γj = i=1 Bjui. 4 Dimension and multiplicity Here's a result from commutative algebra: L Theorem 4.1. Let M = i≥0 Mi be a graded module over a polynomial ring K[x1; ··· ; xn]. There exists a polynomial h(t) 2 Q[t] such that s X dimk Mi = h(s) i=0 for s >> 0. Γ Let M be a module over An and let h be the hilbert polynomial of gr M, where Γ is a good filtration of M with respect to An. We define the dimension of M to be the degree d of h, and we define the multiplicity of M to be d!ad, where ad is the leading coefficient. It's not hard to see this is indepedent of good filtration using 4.3. Example: d(K[x1; ··· ; xn] = n and d(An) = 2n.