Table Of ContentLectures on algebraic D-modules
Alexander Braverman and Tatyana Chmutova
Contents
Chapter 1. D-modules on a(cid:14)ne varieties 5
1. Lecture 1: Analytic continuation of distributions with respect to a
parameter and -modules (01/31/02) 5
D
2. Lecture 2: Bernstein’s inequality and its applications (02/05/02) 10
3. Lecture 3 (02/07/02) 14
4. Lecture 4 (02/12/02): Functional dimension and homological algebra 21
5. Lecture 5 (02/14/02) 24
6. Lecture 7 (02/21/02): -modules on general a(cid:14)ne varieties 27
D
7. Lecture 8 (02/26/02): Proof of Kashiwara’s theorem and its corollaries 32
8. Lecture 9 (02/28/02): Direct and inverse images preserve holonomicity 35
Chapter 2. D-modules on general algebraic varieties 39
1. Lectures 10 and 11 (03/5/02 and 03/7/02): -modules for arbitrary
D
varieties 39
2. Derived categories. 41
3. Lectures 13 and 16 (03/14/02 and 04/02/02) 46
Chapter 3. The derived category of holonomic -modules 53
D
1. Lecture 17 53
2. Lecture 18: Proof of Theorem 12.2 58
3. Lecture 18 (04/09/02) 60
Chapter 4. -modules with regular singularities 65
D
1. Lectures 14 and 15 (by Pavel Etingof): Regular singularities and the
Riemann-Hilbert correspondence for curves 65
Chapter 5. The Riemann-Hilbert correspondence and perverse sheaves 71
1. Riemann-Hilbert correspondence 71
3
CHAPTER 1
D-modules on a(cid:14)ne varieties
1. Lecture 1: Analytic continuation of distributions with respect to a
parameter and -modules (01/31/02)
D
In this course we shall work over the base (cid:12)eld k of characteristic 0 (in most case
one can assume that k = C). As a motivation for what is going to come let us (cid:12)rst
look at the following elementary problem.
Analytic problem.
Let p C[x ;:::;x ] be a polynomial in n variables (p : Rn C) and let U be a
1 n
connec2ted component of Rn x p(x) = 0 . De(cid:12)ne !
nf j g
p(x) if x U
p (x) = 2
U
0 otherwise.
(
Let us take any (cid:21) C and consider the function p (cid:21).
U
2 j j
It is easy to see that if Re (cid:21) 0 then p (x) (cid:21) makes sense as a distribution on
U
Rn, i.e. p (x) (cid:21)f(x)dx is conv(cid:21)ergent forjany fj(x) C1(Rn) { a smooth function
on Rn witUhjcUompjact support. 2 c
R
Example. Let n = 1, p(x) = x and U = R . Then 1f(x)x(cid:21)dx is de(cid:12)ned for
+ 0
Re (cid:21) 0 (of course the integral is actually well-de(cid:12)ned for Re (cid:21) > 1 but we do not
(cid:21) R (cid:0)
need this). We shall denote the corresponding distribution by x(cid:21).
+
It is easy to see that for Re (cid:21) 0 we have a holomorphic family of distributions
(cid:21)
(cid:21) p (x) (cid:21).
U
7!j j
Question(Gelfand): Can you extend this family meromorphically in (cid:21) to the whole
C?
Let (Rn) be a distribution on Rn. As usual we de(cid:12)ne @E (f) = (@f ).
E 2 D @xi (cid:0)E @xi
Example. Let n = 1, p(x) = x, U = R . We have a distribution x(cid:21) de(cid:12)ned for
+ +
Re (cid:21) 0. We know that d (x(cid:21)+1) = ((cid:21) + 1)x(cid:21). The left hand side is de(cid:12)ned for
(cid:21) dx + +
Re (cid:21) 1. Hence the expression
(cid:21) (cid:0)
1 d
x(cid:21) = (x(cid:21)+1)
+ (cid:21)+1 dx +
gives us an extention of x(cid:21) to Re (cid:21) 1, (cid:21) = 1. Continuing this process by
(cid:21) (cid:0) 6 (cid:0)
induction we get the following
5
Proposition 1.1. x(cid:21) extends to the whole of C meromorphicaly with poles in
+
negative integers. In particular, for every f C1(R)
2 c
1
(cid:11) ((cid:21)) = f(x)x(cid:21)dx
f
Z0
has a meromorphic continuation to the whole of C with poles at 1; 2; 3;:::.
(cid:0) (cid:0) (cid:0)
Example. The proposition works not only for functions with compact support but
also for functions which are rapidly decreasing at + together with all derivatives.
For example we can take f(x) = e(cid:0)x. In this case w1e have 1e(cid:0)xx(cid:21)dx = (cid:0)((cid:21)+1).
0
The proposition implies that (cid:0)((cid:21)) has a meromorphic continuation with poles at
R
0; 1; 2:::.
(cid:0) (cid:0)
Theorem 1.2. [Atiyah, Bernstein-Gelfand] p (cid:21) has a meromorphic continua-
U
tion to the whole C with poles in a (cid:12)nite numbejr of jarithmetic progressions.
The (cid:12)rst proofs of this fact were based on Hironaka’s theorem about resolution
of singularities. We are going to give a completely algebraic proof of Theorem 1.2
which is due to Bernstein. For this let us (cid:12)rst formulate an algebraic statement that
implies Theorem 1.2.
Let = (An) denote the algebra of di(cid:11)erential operators with polynomial coef-
(cid:12)cients DactinDg on = (An) = C[x ;:::;x ]. In other words is the subalgebra of
1 n
O O D
EndCO generated by multiplication by xi and by @@xj.
Theorem 1.3. There exist d [(cid:21)] and b((cid:21)) C[(cid:21)] such that
2 D 2
d(p(cid:21)+1) = b((cid:21))p(cid:21):
Example. Let n = 1 and p(x) = x. Then we can take d = d and b((cid:21)) = (cid:21)+1.
dx
We claim now that Theorem 1.3 implies Theorem 1.2 (note that Theorem 1.3 is a
completely algebraic statement). Indeed, suppose d(p(cid:21)+1) = b((cid:21))p(cid:21). Then d( p (cid:21)+1
U
j j
) = b((cid:21)) p (cid:21). The left hand side is de(cid:12)ned for Re(cid:21) 1, thus the expression
U
j j (cid:21) (cid:0)
1
p (cid:21) = d( p (cid:21)+1)
U U
j j b((cid:21)) j j
givesusameromorphiccontinuationof p (cid:21) toRe (cid:21) 1. So,arguingbyinduction
U
again, we see that p (cid:21) can be merojmorjphicaly ex(cid:21)ten(cid:0)ded to the whole of C with
U
j j
poles at arithmetic progressions (cid:11), (cid:11) 1, (cid:11) 2, ...where (cid:11) is any root of b((cid:21)).
(cid:0) (cid:0)
We now want to reformulate Theorem 1.3 once again. Set ((cid:21)) = C C((cid:21)).
D D (cid:10)
Denote by M the ((cid:21))-module consisting of all formal expressions q(x)p(cid:21)(cid:0)i where
p
D
i Z and q(x) C((cid:21))[x ;:::;x ] subject to the relations qp(cid:21)(cid:0)i+1 = (qp)p(cid:21)(cid:0)i (the
1 n
2 2 g
action of ((cid:21)) is de(cid:12)ned in the natural way).
D
Theorem 1.4. M is (cid:12)nitely generated over ((cid:21)).
p
D
6
Let us show that Theorem 1.3 and Theorem 1.4 are equivalent.
Theorem 1.4 Theorem 1.3: Denote by M the submodule of M generated
i p
)
by p(cid:21)(cid:0)i. Then M M and
i i+1
(cid:26)
M = M : (1.1)
p i
i
[
Assume that M is (cid:12)nitely generated. Then (1.1) implies that there exists j Z such
p
2
that M = M . In other words for i as above the module M is generated by p(cid:21)(cid:0)j.
p j p
Hence there exist d ((cid:21)) such that d(p(cid:21)(cid:0)j) = p(cid:21)(cid:0)j(cid:0)1.
Let (cid:27) be an auto2mDorphism of C((cid:21)) sending (cid:21) to (cid:21)+j 1. Then (cid:27) extends to an
j j
(cid:0)
automorphism of tehe algebra ((cid:21)) (wheich we shall denote by the same symbol) and
D
clearly we have (cid:27) (d)(p(cid:21)+1) = p(cid:21). But (cid:27) (d) can be written as (cid:27) (d) = d , where
j j j b((cid:21))
d [(cid:21)] and b((cid:21)) C[(cid:21)]. Thus we have d(p(cid:21)+1) = b((cid:21))p(cid:21).
2 D 2
e e e
Theorem 1.3 Theorem 1.4: By shifting (cid:21) we see that for every integer i > 0
)
there exists a di(cid:11)erential operator d [(cid:21)] such that
i
2 D
d (p(cid:21)) = b((cid:21) i)p(cid:21)(cid:0)i:
i
(cid:0)
This clearly implies that p(cid:21) generates M .
p
We now want to prove Theorem 1.4. To do this we need to develop some maschin-
ery.
1.5. Filtrations. Let A be an associative algebra over k. Recall that an in-
creasing (cid:12)ltration on A is the collection of k-subspaces F A A (for i 0) such
i
(cid:26) (cid:21)
that
1) A = F A and F A = 0 ;
i i
[ \ f g
2) We have F A F A and F A F A F A. It is also convenient to set
i i+1 i j i+j
(cid:18) (cid:1) (cid:18)
F A = 0.
(cid:0)1
Inthiscaseonemayde(cid:12)netheassociatedgradedalgebragrF AofAinthefollowing
way:
1
grF A = F A=F A:
i i(cid:0)1
i=0
M
We set grF A = F A=F A. Then grFA has a natural structure of a graded algebra
i i i(cid:0)1
(i.e. we have grF A grF A grF A). We shall sometimes drop the super-script F
i (cid:1) j (cid:18) i+j
when it does not lead to a confusion.
Similarly let M be a left module over A. Then an increasing (cid:12)ltration on M
consists of a collection of k-subspaces F M M such that
j
(cid:26)
1) M = F M and F M = 0 ;
j j
[ \ f g
2) We have F M F M and F A F M F M.
j j+1 i j i+j
(cid:18) (cid:1) (cid:18)
7
As before one de(cid:12)nes
1
grF M = F M=F M
j j(cid:0)1
i=0
M
Thus grF M is a graded grA-module.
Definition 1.6. (1) An increasing (cid:12)ltration F M is called a good (cid:12)ltration
j
if grF M is (cid:12)nitely generated as grA-module.
(2) Two (cid:12)ltrations F M and F0M are called equivalent if there exist j and j
j j 0 1
such that
F0 M F M F0 M:
j(cid:0)j0 (cid:18) j (cid:18) j+j1
Proposition 1.7. (1) Let F M be a good (cid:12)ltration on a left A-module M.
j
Then M is (cid:12)nitely generated over A.
(2) If F M is a good (cid:12)ltration on M then there exist j such that for any i 0
j 0
(cid:21)
and any j j F A F M = F M.
0 i j i+j
(cid:21) (cid:1)
(3) Assume that we have two (cid:12)ltrations F and F0 on M such F is good. Assume
also that for any i 0 the F A-module F A is (cid:12)nitely generated. Then there
0 i
(cid:21)
exist j such that F M F0 M for any j.
1 j (cid:26) j+j1
Corollary 1.8. SupposethatF Ais(cid:12)nitelygeneratedoverF Aasaleftmodule.
i 0
Then any two good (cid:12)ltrations on a left A-module M are equivalent.
This clearly follows from the third statement of the theorem.
Proof. (1) By assumption grF M is (cid:12)nitely generated. Let s ;:::s be the
1 k
generators of grF M, s grF M. For any i choose t F M which projects
i 2 ji i 2 ji
to s . It is now easy to see that t generate M.
i i
(2) IfgrFM is(cid:12)nitelygeneratedovergrAthengrFM isgeneratedby j0 grFM
i=0 i
for some j . Then for any j j
0 0
(cid:21) L
gr A grF M = grF M
i (cid:1) j i+j
and hence F A F M +F M = F M.
i j i+j(cid:0)1 i+j
(cid:1)
By induction on i we can assume that F M = F A F M. Then
i+j(cid:0)1 i(cid:0)1 j
(cid:1)
F M = F A F M +F A F M = F A F M.
i+j i j i(cid:0)1 j i j
(cid:1) (cid:1) (cid:1)
(3) First of all we claim that F M is (cid:12)nitely generated over F A for all j. It
j 0
is enough to show that grF M is (cid:12)nitely generated over F A for all j. Let
j 0
m ;:::;m be some generators of grFM. We may assume that they are ho-
1 k
mogeneous, i.e. m grF M for some j . Thus for every j 0 the map
i 2 ji i (cid:21)
k
gr A grF M; (a ;:::;a ) a m +:::a m
j(cid:0)ji ! j 1 k 7! 1 1 k k
i=1
M
8
is surjective. On the other hand since gr A is a quotient of F A (for every
i i
i) it follows that every gr A is (cid:12)nitely generated. Hence grF M is (cid:12)nitely
i j
generated.
Letj beasabove. SinceF0 isa(cid:12)ltration,wehaveF M = F0M F M
0 j0 j \ j0
and since F M is a (cid:12)nitely generated F A-module it follows that F M
j0 0 S j0 (cid:18)
F0 M for some j 0.
j0+j1 1 (cid:21)
Then F M = F A F M F A F0 M F0 M. So we have
i+j0 i (cid:1) j0 (cid:18) i (cid:1) j0+j1 (cid:18) j0+j1+i
proved our proposition for j j . By increasing j cam make it true for any
0 1
(cid:21)
j.
(cid:3)
Here is our main example. Let be the algebra of polynomial di(cid:11)erential opera-
D
tors in n variables. Let’s de(cid:12)ne two (cid:12)ltrations on :
D
Bernstein’s (or arithmetic) (cid:12)ltration: F = k, F = k+span(x ; @ ), F is
0D 1D i @xj iD
the image of F (cid:10)i under the multiplication map.
1
D
Geometric (cid:12)ltration ((cid:12)ltration by order of di(cid:11)erential operator) denoted by
0
D (cid:26)
::::
1
D (cid:26)
= = k[x ;:::;x ]
0 1 n
D O
= span(f ;g @ where g ), is the image of (cid:10)i under the multipli-
D1 2 O @xi 2 O Di D1
cation map.
The following lemma describes the algebra explicitly as a vector space. The
D
proof is left to the reader.
Lemma 1.9. For any d there exists a unique decomposition
2 D
@ @
d = p :::
i1;:::ik@x @x
i1(cid:20)X(cid:1)(cid:1)(cid:1)(cid:20)ik i1 ik
where p k[x ;:::;x ].
i1:::ik 2 1 n
This lemma immediately implies the following
Proposition 1.10. For both (cid:12)ltrations gr = k[x ;:::;x ;(cid:24) ;:::;(cid:24) ]. Here x
(cid:24) 1 n 1 n i
D
are images of x and (cid:24) are images of @ .
i j @xj
Proof. Let us show, that x ’s and (cid:24) ’s commute in gr is for both (cid:12)ltrations.
i j
D
For each i = j we have x @ = @ x in and hence in grF . For i = j we have
6 i@xj @xj i D D
x @ @ x = 1 F and hence equal to 0 in grF for both (cid:12)ltrations.
i@xi (cid:0) @xi i (cid:0) 2 0D D
It follows now easily from the above lemma that gr is a polynomial algebra in
x ’s and (cid:24)’s. D (cid:3)
i
For Bernstein’s (cid:12)ltration the above argument shows a little more { namely that
for all i;j we have [F ;F ] F (note that for the geometric (cid:12)ltration we
i j i+j(cid:0)2
D D (cid:26) D
only have [ ; ] ). We shall need this fact in the next lecture.
i j i+j(cid:0)1
D D (cid:26) D
9
2. Lecture 2: Bernstein’s inequality and its applications (02/05/02)
Let = (An) be the algebra of polynomial di(cid:11)erential operators in n variables.
D D
In the last lecture we have introduced two (cid:12)ltrations on : Bernstein’s (cid:12)ltration
D
F (x and @ are in F ) and geometric (cid:12)ltration (x and @ ). For
iD i @xj 1D Di i 2 D0 @xj 2 D1
both of the (cid:12)ltrations gr = k[x ;:::x ;(cid:24) :::(cid:24) ]. Also dim F < .
(cid:24) 1 n 1 n k i
D D 1
Let A be a (cid:12)ltered algebra such that dim F A < and grA = k[y ;:::;y ]. Let M
k i (cid:24) 1 m
1
be an A-module with a good (cid:12)ltration F. De(cid:12)ne h (M;j) = dim F M.
F k j
Theorem 2.1. There exists apolynomial h (M)(t) (called the Hilbert polynomial
F
of M with respect to (cid:12)ltration F such that h (M;j) = h (M)(j) for any j 0,
F F
(cid:29)
h (M)(j) has a form h (M)(t) = ctd + lower order terms , where d m and
cF Z . F d! f g (cid:20)
+
2
Let us mention that this is actually a theorem from commutative algebra since
h (M;j) = h grM;j where grF is the natural (cid:12)ltration on grM (coming from the
F grF
grading).
Lemma 2.2. c and d in the theorem above do not depend on (cid:12)ltration.
Proof. Let F and F0 be good (cid:12)ltrations. Then there exist j and j such that
0 1
F0 M F M F0 M
j(cid:0)j0 (cid:18) j (cid:18) j+j1
and hence hF0(j j0) hF(j) hF0(j + j1). This can be true only if hF and hF0
(cid:0) (cid:20) (cid:20) (cid:3)
have the same degree and the same leading coe(cid:14)cient.
Definition 2.3. For a (cid:12)nitely generated module M d = d(M) as above is called
the dimension of M (sometimes it is also called Gelfand-Kirillov of functional dimen-
sion of M).
Theorem 2.4. [Bernstein’s inequality] For any (cid:12)nitely generated module M over
= (An) with Bernstein’s (cid:12)ltration we have d(M) n.
D D (cid:21)
BeforeprovingTheorem2.4wewanttoderivesomeveryimportantcorollariesofit
(in particular we are going to explain how this theorem implies the results formulated
in the previous lecture).
Historical remark. This theorem was (cid:12)rst proved in Bernstein’s thesis, then a
simple proof was given by A Joseph. Then O. Gabber proved a very general theorem
which we shall discuss later (this theorem implies that more or less the same is true
for geometric (cid:12)ltration). ADD REFERENCES.
10
