R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
A LAN A DOLPHSON
S TEVEN S PERBER
Differential modules defined by systems of equations
Rendiconti del Seminario Matematico della Università di Padova, tome 95 (1996), p. 37-57
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of Equations (*).
ALAN
ADOLPHSON (**) -
STEVENSPERBER (***)
1. - Introduction.
Fix
Qi ,
...,Qn ,
convexintegral polytopes (i.e.,
with vertices inZ’)
of dimension n inR n ,
andput Ji
= Let F be a field of charac- teristic 0 and consider indeterminates indexedby
i =1, ... , n and ji
EJi .
Let K = and for i =1,
... , nput
where ji
=(ji ( 1 ),
... ,ji (n )) and xii
=xii ~ 1~ ...
Weregard f
as thegeneric
Laurentpolynomial
with Newtonpolytope equal
to Letwhere
( fi ,
... ,fn) denotes
the ideal ofK[xl,
... , zn ,generat-
ed
by
thef . For g
EK[xl ,
... , zn , we letg
denote itsimage
in L. It is not hard to show
(see
Section7)
that L is a field. Further-more, the
degree
of L over Kequals
the number ofpoints
in the inter- section of the torichypersurfaces fi = 0, i = 1, ... , n. By
Bernstein’s(*) Part of the research for this article was carried out while the authors
were
visiting
theUniversity
of Padua with support from CNR. The authors aregrateful
to theUniversity
of Padua and CNR for theirhospitality.
(**) Indirizzo dell’A.:
Department
ofMathematics,
Oklahoma State Univer-sity,
Stillwater, Oklahoma 74078; e-mail:[email protected].
Partially supported by
NSF grant no. DMS-9305514.(***) Indirizzo dell’A.: School of Mathematics,
University
of Minnesota, Min-neapolis,
Minnesota 55455; e-mail:[email protected].
theorem
[3]
we thus havewhere
M(~1, ... , Qn)
denotes the Minkowski mixed volume ofQl , ... , ~n .
Partial differentiation with
respect
to defines a derivation of the field K that we denoteby 8iji. By
the usualprocedure
ofimplicit
differ-entiation,
these derivations extend in aunique
manner to L . Thus L be-comes a
(left)
module over thealgebra
6D = ofpartial
differ-ential
operators
with rational function coefficients. The purpose of this article is toexplicitly identify
L as a a-module ofhypergeometric type.
We
accomplish
thisby completing
certainaspects
of the work ofKatz
[5], specifically,
we show that L isisomorphic
to a certain Dworkcohomology
space(Theorem 4.1).
Wepostpone stating
this result untilwe have reviewed the definition of Dwork
cohomology. Instead,
westate here a consequence of our
work,
which may be moreimmediately
accessible.
Let el , ... , en be the standard basis of Rn and define
(i
=1,
... , n,jiEJi)
Let B be the lattice of relations of the v
To each b = E B we associate the constant coefficient differential
operator
Write
Eiji
= ...,Eiji (2n)).
We also consider the differential op- erators( k
=1,
...,2 n )
’
Thus for 1~ =
1,
... , n,Put , a
(left)
a-module ofhypergeomet-
ric
type.
Let 31 be the D-submodule of 311generated by
theimage
of allproducts
of the forma1j¡... E Ji
for i =1,
... , n.The rational function field K is
obviously
a (1)-submodule of L. Infact,
it is a direct summand. Let K be analgebraic
closure of K and let G be the set ofimbeddings
of L into K over K. Then the«averaged»
tracemap L - K defined
by
splits
the inclusion K 4 L since the ds commute with theLet C c be the real cone
generated by
=1,
... ,n, ji
EJi I
and let M C
R2n
be the monoid these latticepoints generate:
THEOREM 1.3.
Suppose
Then there is an iso-morphisme of
£o-modules 31 =L/K,
whereL/K
denotes thequotient of
the 6D-module L
by
its (D-submodule K.REMARK. The
hypothesis
of the theorem can be weakened some-what, although
we shall not address thatquestion
here. Some condition isneeded,
however. One can constructexamples
to show that withoutany
hypothesis, 31
andL/K
deed not beisomorphic.
The
isomorphism
of the theorem can be madeexplicit
as follows.Since 31 is
generated
as6D-module by
thec9ljl ...
it suf-fices to
give
the element ofL/K
associated toa1j¡...
Let J E... , be the «toric» Jacobian
of fi , ... , fn , i.e.,
Then J is an invertible element of L. The
isomorphism
isgiven by
Theorem 1.3 will be
proved
in Section 6.The fact that an
algebraic
function satisfies asystem
ofhypergeo-
metric differential
equations (the
case n = 1 ofCorollary
2.30below)
was first
pointed
out to usby
B. Sturmfels. His interest and commentsprovided
theoriginal
stimulus for our consideration of theproblem
treated here.
2. - Results of Katz.
The results of this section all appear in Katz
[5],
modulo the fact thatwe work in the toric case whereas Katz worked in the
projective
case.To our
knowledge
these results have not beenpublished
before(except
for the
hypersurface
case, which is treated in[6]).
Consider the
ring
We make R into a D-module
by letting 3ij,
act aswhere
axr
= 03A iii
-
3A iii
- ° for all r, s,
i, ji .
Inparticular,
Set S =
{ 1,
...,n}
and define for A c SThe same definition makes the
RA
into (D-modules.Katz defined a D-module
homomorphism
O : as follows.Grade the
RA using
thegrading given by
they2’s, namely,
if ev =
(VI,
... ,vn),
defineand let
RA, ~d~
denote theK-subspace of RA spanned by
the monomials ofdegree
d. The map O will be definedinductively
on theR S, (d) .
For each monomialxUyV E
we define and extend to allby K-linearity.
Since every monomial inR S
hasdegree ->
n, we start withd = n. Define
Now suppose O has been defined on
R S~ ~d -1
> and letd > n.
By (2.1)
we haveand since d > n there is some choice of i such that ei E
E R s~ ~d -1> .
. Since we want 0 to be ahomomorphism
ofa)-modules,
set
It remains to
~how
that O iswell-defined, i.e.,
that ifThe first
step
is to reduce to the case d = n +1,
i = 1. If i ~l,
thend~n+2 and
Since
3ij,, commute,
we haveby
the inductionhypothesis
which is the desired result.
Now suppose i = l.
3,
then(2.4)
holds and the sameargument
can be
repeated.
So suppose vi = 2. Foreach r E S, r ;e i, choose jr
EJr .
Then
and
Using again
thecommutativity
of the a’s and the inductionhypothesis,
we are reduced to
showing
thati.e.,
we are reduced toproving (2.3)
when d = n + 1 and i = l. From thedefinition of e when d = n, this means we must prove
Put u’
=~-~-~7 -(1,
... ,... , Xn,
xn 1 ]. Equation (2.5)
may be rewritten asLEMMA 2.7. Let g E
K[xl , xl 19
... , xn , xn1 ]
be a Laurentpolyno-
mial whose
coefficients
areindependent of
andi. e.,
whose co-efficients
are killedby aljl
and8ljí.
ThenPROOF.
Using
the usual rules fordifferentiation,
we are reduced toproving
for i =
1,
..., n. Let A be the Jacobianmatrix, i.e.,
the matrix whose en-try
in row r, column s isair / axs
E ... , xn , The usualprocedure
ofimplicit
differentiationgives
Multiplying by V, gives
Switching
the rolesof j 1, j i ,
we see that both sides of(2.8) satisfy
thesame
system
ofequations (2.10).
Since det A =Jo ~ 0, they
must beequal.
Applying
theproduct
formula for differentiation and Lemma 2.7 to(2.6),
we see that(2.6)
isequivalent
toSolving (2.9) by
Cramer’s Rulegives
where
Crs
denotes the(r, s)-cofactor
ofA, i.e., Cr, equals ( -1 )r + s
timesthe determinant of the matrix obtained from A
by deleting
row r andcolumn s. An
analogous equation
holdswith j1 replaced by ji . Applying aiii
to(2.12), 81j¡
to theanalogous equation
withjl replaced by j{ ,
andusing = 8yj 81j¡’
we conclude thatBut for
l = 1, ... , n, Cl,
satisfies thehypothesis
of Lemma2.7,
hence
Applying
theproduct
formula for differentiation to(2.13)
andusing
this relation
gives
But the column vector
appearing
on both sides of thisequation
is thefirst column of the
adjoint
matrix ofA,
hencetaking
the «innerprod-
uct» with the first row of A
gives
Equation (2.11)
follows sinceJo
is invertible in L.REMARK. If
b E B,
it iseasily
seen that = 0 for allxUyV E
e I~. Since O is a W-module
homomorphism,
it follows that for allxUyV
ee RS,
= 0. Inparticular, ~b (xu / J~ = 0
forWe define differential
operators Dxj D, :
for i =1,
... , nand all A c ,S
by
Note that Put
Since
formally Dxz
=exp ( - g ) o Xi 8/8Xi 0
exp g,Dy2
=exp ( -
o exp g, and =
exp ( - g) o
exp g, all theseoperators
on R commute with one another. In
particular,
theDx2
andDyi
are6D-module
endomorphisms
ofRA .
LEMMA 2.19. The kernel
of
e containsDx, (Rs)
andDy, (RsB1’1) for
i = 1,
..., n.’ ’
PROOF. For this
proof,
we fix a choiceof ji E Ji
for each i =1,
9 ..., n.Let Then
Since
Dyz
commutes with the3’s,
And since O is a 6D-module
homomorphism,
since f
= 0 in L.Now let and write
As before we then have
Thus it suffices to prove
for all We have
But
thus
and
B v ,
We must show this
expression
vanishes.Let G be the n x n matrix whose
entry
in row r, column s isThe usual
procedure
ofimplicit
differentiationgives
where the notation on the
right-hand
sidedesignates
the n x ndiago-
nal matrix whose
diagonal
entries are ... , ThusSince J =
51 ... X-,,70,
we may substitute theresulting expression
for Jinto the
right-hand
side of(2.21). Discarding
the factor( -1 )’~
andputting
tosimplify notation,
weget
Note that
Substituting
into(2.23) gives
theexpression
We rewrite
(2.22)
asor,
equivalently,
But this
equation
saysexactly
that fori, 1
=1,
... , n,Substituting
into(2.24),
we see that thisexpression simplifies
toLet G * be the
adjoint
matrix ofG, i.e.,
theentry
in row r, column s of G * is the(s, r)-cofactor
of G. Since GG * =( det G ) In ,
we have from(2.25)
thatdet G is the
entry
inrow k,
column i of -G * , i.e.,
is thenegative
of the(i, k)-cofactor
of G.Proving (2.26)
vanishes is thusequivalent
toproving
Take i = 1 to fix ideas. For
typographical convenience,
wetemporarily write ak
inplace
of SetWe must show
We rewrite the left-hand side as
Fix a
pair (r, s ),
1 ~ r s ~ n.Taking k
=s, 1
= r in the first doublesum
gives
a contributionwhereas
taking
k =r, 1
= s in the second double sumgives
a contribu-tion
Keeping
in mind thatar as
=as ar ,
it is clear that these two contribu- tions cancel. Hence the entireexpression (2.28)
vanishes. This com-pletes
theproof
of Lemma 2.19.For
A c ,S,
letWe summarize the
results
of this section.THEOREM 2.29. The map e: induces a
surjective
homo-mor~phism of
6D-modulese: w S - L .PROOF. The existence of e follows
immediately
from Lemma 2.19.It is
surjective
becauseyn)
= Yu and L isspanned
as K-vec-tor space
by
theYu,
u E zn .COROLLARY 2.30. For
x u y v
ERS, e(x u y,) satisfies
thefollowing differential
relations.PROOF. The first assertion has
already
been observed in the re-mark
following (2.15).
Note that forApplying
0 to both sides andusing
Lemma 2.19gives
the second andthird assertions.
3. - Reduction to 72.
We shall
eventually
show thathence e is an
isomorphism.
For this it is convenient toreplace
Rby
aclosely
related but moremanageable ring.
Let CcR2n
be the cone gen- eratedby I Eij, i
=1,
... , n,ji
EJi I
andput
Note that the
hypothesis
of Theorem 1.3 isequivalent
to therequire-
ment that R be
generated
assubring
of Rby
the i =1,
... , n,jir=Ji.
For A c S weput RA = R
Note that theD, , 8iji
arestable on all
RA and
that PutThe natural inclusion t:
R 4 R
induces D-modulehomomorphisms iA :
for all A c S.LEMMA 3.1. For all A c
S, iA
issurjective.
PROOF. Given
x u y v
ER A we
need to show thereexists E E RA such
that
Let
L1, ... , l~
be real linear forms in 2n variablesdefining
the coneC,
0 for k =
1, ...,
s.If
v) a
0 for k =1,
..., s, then(u, v)
E C soxu yv
ERA
and there isnothing
to prove.Suppose,
say,L1 (u, v) 0.
Consider- h (Dxl , ... , Dyn),
the differentialoperator
obtainedby replacing
the 2n variables in the linear
form 11 by
the differentialoperators
... ,
9 DYI
... ,DYn.
* A calculationgives
Consider the terms in the double sum. Since
( ji , ei)
E C for all i andji ,
either
11 Vi, ei)
= 0 and the term vanishes orL1 (u
+ji , v
+ei)
>L1 (u, v).
Also, + ji , v
+ei) ~ v)
for 1~ =2,
... , s.Solving (3.3)
forx u y v
we
get
where for all
u’ , v’
we have~’~’(A)e~ l1 (u’ , v’ ) > h (u, v),
andv ’ ) ~ v )
for k =2,
... , s .Iterating
thisprocedure,
we even-tually
arrive at a relation of thistype
withL1 (u’ , v’ ) > 0, lk (u’ , v ’ ) %
~
v)
for k =2,
... , s and allu’ ,
v’ .Repeating
the sameargument successively
for~,
... ,ls,
we arrive at a relation of thetype (3.4)
with...,sandallu’,v’. SincexuyvERA,wehave
by
construction that for allu ’ ,
v ’ . Hence4. - The main theorem.
The results of the
previous
two sectionsgive
ussurjections
THEOREM 4.1. We have ... , hence
by (1.2)
both
is
and O areisomorphism.
We shall prove a
slightly
moregeneral
result.Fix r,
0 % r % n andleir= {1, ...,~}
R - Rbe the map
«set y2
=0», i.e., (Ji
is definedby K-linearity
and the conditionF or A c T, put 9 ;
and defineRA
= cT,
de-fine
RB . Let K. (RA )
be the Koszulcomplex
onRA
definedby
the n +I A I operators
For i
E A,
thesurjective
map induces asurjective
ho-momorphism
ofcomplexes K. (RA ) -~ K.
We denote the kernelof this
homomorphism by
I~. Moregenerally,
suppose K.(Rl)
has been defined for all B c A with Let and define to be the kernel of the
surjection
- K. (RABB{i})
inducedby Oi.
Thus there is a short exact sequence ofcomplexes
When
dimKHz(K.(R!)
is finite for all l and vanishes for all butfinitely many l,
we define an Euler characteristicSince Theorem 4.1 is the
special
case r = n of thefollowing.
THEOREM 4.3. We have
... ;
ir)
denotes the Minkowski mixed volumeof
the npolytopes
ob-tacined
by listing
thepolytope
times.We
begin
with a lemma. Let K’ .(RA),
K’ .(RI)
be defined analo-gously
toK. (RA), K. (RA )
but with theoperators
i =1,
... , n, andD Yi
Ii E A, replaced by
theoperators
ofmultiplication by
convex hull of the
origin
and the and letdenote its volume with
respect
toLebesgue
measure onRn +
IAI.
LEMMA 4.4. For B c A c
T,
B ~A,
we havePROOF. It follows from Kouchnirenko
[9,
Th6or6me6.1]
thatyl f1
++ ... +
Yrh
isnondegenerate,
thus this lemma is aspecial
case of[2,
The-orem
2.17]. (Although
the coefficient field of[2,
Theorem2.17]
isfinite,
the
argument given there
is valid over anyfield.)
We
regard
the1RA
asgraded by
thegrading
defined in Section 2.Since the
operators defining
thecomplex
Kt(RA )
aremultiplication by homogeneous
elements ofdegree 1,
there is an inducedgrading
on thecomplex (The grading
is shifted at eachstep
of thecomplex
sothat the
boundary
maps havedegree zero.)
Thegrading
onRA gives
rise in a natural way to an
increasing filtration,
the k-th term in the fil- trationbeing
the sum of thegraded pieces
ofdegree -
k. This deter-mines in an obvious manner a filtration on the
complex
K.(RA ).
It isclear from the definitions that the
graded complex
is the associ- atedgraded
of the filteredcomplex
I~.(RA ).
Thus there is aconvergent El spectral
sequence[11, Chapter 9]
withAssertion 1 of Lemma 4.4
implies
thatEk,
l = 0 for k + 1 > 0 or k + 10,
hence all the differentials of the
spectral
sequence are zero. It follows thatl for
allk,
l. Thusby
Lemma 4.4 we have the follow-ing.
LEMMA 4.5. For B c A c
T,
B ~A,
we havePROOF OF THEOREM 4.3. The
proof
isby
induction onIT 1.
WhenT = 0,
and =xia/8xi
is the zerooperator
on K. ThusK. (R8)
is thecomplex
where all the maps are zero.
Thus Hi (K. (R0 ))
for all 1 and all as-sertions of the theorem are obvious. Now let T =
{ 1,
...,rl
and consid-er the short exact sequence of
complexes
Using
thelong
exacthomology
sequence andapplying
the inductionhypothesis
to and Lemma 4.5 to K. shows thatK. satisfies assertions
1)
and2)
of Theorem 4.3 and thatLEMMA 4.8. For I c
T,
I #T,
PROOF. We outline the
proof
whenI = ~,
thegeneral
casebeing analogous.
Theprojection
of .L1 T on Rr is thesimplex
The fiber of
L1 T
over(at,
... ,ar)
the Minkowski suma1 Q1
+ ... ++ Thus
The lemma now follows
by applying
the definition of Minkowski mixed volume to expressvol (a1 Q1
+ ... + as apolynomial
in ... , or,and then
evaluating
the aboveintegral.
A
straightforward
calculationusing
this lemma shows that the firstsum on the
right-hand
side of(4.7) equals
Assertion
3)
of Theorem 4.3 is now immediate.5. - Relations between W-spaces.
When A =
,S,
wedrop
thesubscript
S and write inplace
ofWJ.
When B =
0,
wedrop
thesuperscript 0
and writeWA
inplace
of Inparticular, w~
will be denotedsimply
W. This is consistent with ourearlier notation.
For
B, c B2 c A,
the natural inclusion induces a homo-morphism
ofakmodules io f2 -
LEMMA 5.1.
If A,
the above map isinjective.
PROOF. It suffices
by
induction to prove the lemma whenB2
==
B¡
U{i}.
From(4.2)
we have a short exact sequence ofcomplexes
Since
B2 ;z! A,
we have soby
Lemma 4.5 all thesecomplexes
are
acyclic
inpositive
dimension. The associatedlong
exacthomology
sequence thus reduces to the short exact sequence
which establishes the lemma.
From
(4.2)
we have also the short exact sequence ofcomplexes (i e A)
By
Lemma 4.5 the middlecomplex
isacyclic
inpositive dimension,
hence the associatedlong
exacthomology
sequencegives
an exactsequence
and
isomorphisms
for k ~ 1Suppose
A= ~ a 1,
... ,aa) with,
say, a 1 ... a a .Applying (5.5)
in-ductively
we haveisomorphisms
hence the exact sequence
(5.4)
becomesME desc ribe the
image of Ha (K. (R00))
inSince
RS
=K, Ha (K. (RS))
can berepresented
aswhere e, is a formal
symbol.
Write I= I il,
...,i.1,
withil
...ia.
·For
Jc I, IJI [
=k,
write J =... , jk },
withjl
...jk ,
and write...
La _ k .
Definesgn (J, I )
= ± 1by
theequation
To describe the
image
it sufficesby (5.8)
togive
the im-age of
Kei .
Astraightforward
calculationusing
the definition of Koszulcomplexes
and theconnecting homomorphism
shows that theimage
ofKe¡
in is theK-span
of thehomology
class de-fined
by
thecycle
where J =
{ j 1,
... ,jk . Taking k
= a, we see inpartic-
ular that the
image
ofKei
inW~
is theK-span
of thehomology
class de-fined
by
In the
special
case A =S, k
= a = n, this says that theimage
ofHn (K. (R8))
=Kes
inWS
is theK-span
of thehomology
class definedby Jy1...
Yn, where J is as in(1.4).
By
the short exact sequence(5.4)
with A = ,S andrepeated
use ofLemma
5.1,
weget
an exact sequencewhere
[Jyl ... yn]
EWS
denotes thehomology
class definedby
jyl
... yn.6. - Proof of Theorem 1.3.
Assume that
by
Dwork-Loeser[4] (see
also[1,
Theorem4.4]),
the map6D->i-?
definedby K-linearity
and the ruleis a
homomorphism
of W-modules that on passage toquotients
inducesan
isomorphism 311 === W,
where 3ll is as defined in Section 1. Now isa
quotient
of andR S
isgenerated
as 60-moduleby
theproducts
xjl yl ... Xj" Yn,
hence theimage
ofWS
in isgenerated by
these sameproducts.
Under theisomorphism
inducedby (6.1),
thisimage
is thusisomorphic
to 1l1, the 6D-submodule of 3llgenerated by
theimage
of allproducts
of the formanjn .
On the other
hand,
the exact sequence(5.11)
shows that thisimage
is
isomorphic
toWS Ynl-
Under theisomorphism 0 0 ~s : (cf.
Theorem4.1)
thehomology
class[Jy1... yn]
ismapped
to1 E L,
hence[Jyj
...Yn]
=L/K
as 6D-modules. We conclude that 31 =L/K
as 60-modules.7. -
Appendix.
THEOREM 7.1. Let the notation be as in ,Section 1. For r =
1,
... , n,the
quotient ring K[xl ,
9 xn ,9 Xn 111(fl,
... ,fr)
is aregular ring of
dimension n - r.PROOF. For each i =
1,
... , n, fixAi
EJi .
Consider thering
Define an
automorphism q5
of thering H by taking 0
to be theidentity
on F[xl ,
... , 1 Xnby setting
= fi
for i =1,
... , r.Then g5
induces anisomorphism
’
thus
H/( fl,
...,f,)
isclearly
aregular ring.
The theorem now followsby
inverting
the elements ofREFERENCES
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