Continuous rational functions on real and $p$-adic varieties

0.13 MB·English

by  János Kollár

January 20, 2011
CONTINUOUS RATIONAL FUNCTIONS ON
REAL AND p-ADIC VARIETIES
1 JA´NOSKOLLA´R
1
0
2 A rational function on Rn is a quotient of two polynomials

n
p(x ,...,x )
a 1 n
f(x ,...,x ):= .
J 1 n q(x ,...,x )
1 n
9
Strictly speaking, a rationalfunction f is not really a function on Rn in general
1
since it is defined only on the dense open set where q = 0. Nonetheless, even if q
6
] vanishes at some points of Rn, it can happen that there is an everywhere defined
G
continuousfunctionf∗ thatagreeswith f atallpoints where f is defined. Suchan
A f∗ is unique. For this reason, we identify f with f∗ and call f itself a continuous
. rational function on Rn. For instance,
h
t
a p(x1,...,xn)
m x2m+ +x2m
1 ··· n
[
is a continuous rational function on Rn if every monomial in p has degree >2m.

1 The above definition of continuous rational functions makes sense on any real
v
algebraicvariety X, as long as the open set where f is defined is dense in X in the
7
Euclidean topology. This condition always holds on smooth varieties, and, more
3
7 generally, if the singular set is contained in the Euclidean closure of the smooth
3 locus.
. The aim of this note is to consider three basic problems on continuous rational
1
0 functions.
1
1 Question 1. Let X be a real algebraic variety and Z a closed subvariety.
:
v (1) Letf beacontinuousrationalfunctiondefinedonX. Istherestrictionf Z
|
i a rational function on Z?
X
(2) Let g be a continuousrationalfunction defined onZ. Canone extend it to
r
a a continuous rational function on X?
(3) Which systems of linear equations
f (x ,...,x ) y =g (x ,...,x ) i=1,...,m
Pj ij 1 n · j i 1 n
have continuous rational solutions where the g ,f are polynomials (or
i ij
rational functions) on X.
Section 1 contains some examples that I found interesting. Section 2 contains
theproofsofafewresults. TheanswertoQuestion1.1isalwaysyesifX issmooth
(8) but not in general. This leads to the introduction of the notion of hereditarily
rational functions. Thesehavethegoodpropertiesthatonewouldexpectbasedon
the smooth case. In Theorem 9 this concept is used to give a complete answer to
Question 1.2. I do not have a full answer to Question 1.3.
All these results extend to p-adic and other topological fields, see (13).
1
2 JA´NOSKOLLA´R
1. Examples
The following example shows that the answer to Questions 1.1–2 is not always
positive.
Example 2. Consider the surface S and the rational function f given by
S := x3 (1+z2)y3 =0) R3 and f(x,y,z):= x.
(cid:0) − ⊂ y
We claim that
(1) S is a real analytic submanifold of R3,
(2) f is defined away from the z-axis,
(3) f extends to a real analytic function on S, yet
(4) the restriction of f to the z-axis is not rational and
(5) f can not be extended to a continuous rational function on R3.
Proof. Note that
x3 (1+z2)y3 = x (1+z2)1/3y x2+(1+z2)1/3xy+(1+z2)2/3y2 .
− (cid:0) − (cid:1)(cid:0) (cid:1)
The first factor defines S as a real analytic submanifold of R3. The second factor
only vanishes along the z-axis which is contained in S. Therefore x y√31+z2
−
vanishes on S, hence
f = x = 3 1+z2 and so f(0,0,z)= 3 1+z2.
|S y|S p |S p
Assume finally that F is a continuous rational function on R3 whose restriction
to S is f. Then F and f have the same restrictions to the z-axis. We show below
thattherestrictionofanycontinuousrationalfunctionF definedonR3tothez-axis
is a rational function. Thus F z−axis does not equal √31+z2, a contradiction.
|
To see the claim, write F = p(x,y,z)/q(x,y,z) where p,q are polynomials. We
mayassumethattheyarerelativelyprime. SinceF iscontinuouseverywhere,xcan
not divide q. Hence F = p(0,y,z)/q(0,y,z). By canceling common factors,
(x=0)
|
we can write this as F = p (y,z)/q (y,z) where p ,q are relatively prime
(x=0) 1 1 1 1
|
polynomials. As before, y can not divide q1, hence F z−axis = p1(0,z)/q1(0,z) is
|
a rational function. (Note that we seemingly have not used the continuity of F:
for any rational function f(x,y,z) the above procedure defines a rational function
on the z axis. However, if we use x,y in reverse order, we could get a different
−
rational function. This happens, for instance, for f =x2/(x2+y2).)
In the above example, the problems arise since S is not normal. However, the
key properties (2.4–5) can also be realized on a normal hypersurface.
Example 3. Consider
X := (x3 (1+t2)y3)2+z6+y7 =0 R4 and f(x,y,z,t):= x.
(cid:0) − (cid:1)⊂ y
We easily see that the singular locus is the t-axis and so X is normal. Let us blow
up the t-axis. There is one relevant chart, where x = x/y,y = y,z = z/y. We
1 1 1
get the smooth 3-fold
X′ := (x3 (1+t2))2+z6+y =0 R4.
(cid:0) 1− 1 1 (cid:1)⊂
Eachpoint (0,0,0,t)has only 1 preimage in X′, given by √31+t2,0,0,t and the
projection π : X′ X is a homeomorphism. Thus X is(cid:0)a topological m(cid:1) anifold,
→
but it is not a differentiable submanifold of R4.
CONTINUOUS RATIONAL FUNCTIONS 3
Sincef π =x isaregularfunction,weconcludethatf extendstoacontinuous
1
function o◦n X and f(0,0,0,t)= √31+t2. Thus, as before, f can not be extended
to a continuous rational function on R4.
For any m 1 we get the similar examples
≥
X := (x3 (1+t2)y3)2+z6+ +z6 +y7 =0 R3+m and f := x.
m (cid:0) − 1 ··· m (cid:1)⊂ y
For m 1 the X have rational, even terminal singularities.
m
≫
Next we turn to Question 1.3 for a single equation
f (x) y =g(x), ( )
Pi i · i ∗
wheregandthef arepolynomialsinthevariablesx=(x ,...,x ). Suchequations
i 1 n
have a solution where the y are rational functions provided not all of the f are
i i
identically zero. The existence of a solution where the y are continuous functions
i
is studied in [FK11] and [Kol10].
One could then hope to prove that if there is a continuous solution then there
is also a continuous rational solution. In [FK11, Sec.2] we proved that if there is a
continuoussolutionthenthereisalsoacontinuoussemialgebraicsolution. Thenext
example shows, however, that in general there is no continuous rational solution.
Example 4. We claim that the linear equation
x3x y + x3 (1+x2)x3 y =x4 (4.1)
1 2· 1 (cid:0) 1− 3 2(cid:1)· 2 1
has a continuous semialgebraic solution but no continuous rational solution.
A continuous semialgebraic solution is given by
x3
y =(1+x2)1/3 and y = 1 . (4.2)
1 3 2 x2+(1+x2)1/3x x +(1+x2)2/3x2
1 3 1 2 3 2
(Note that x2+(1+x2)1/3x x +(1+x2)2/3x2 1(x2+x2), so y 2x and it
1 3 1 2 3 2 ≥ 2 1 2 | 2|≤ 1
is indeed continuous.) A solution by rational functions is y =x /x and y =0.
1 1 2 2
To see that (4.1) has no continuous rational solution, restrict any solution to
the semialgebraic surface S := (x (1+x2)1/3x = 0). Since x3 (1+x2)x3 is
1 − 3 2 1− 3 2
identicallyzeroonS,weconcludethaty =x4/(x3x ) =(x /x ) . Thelatter
1|S 1 1 2 |S 1 2 |S
is equal to 3 1+x2, thus
p 3
y1|x3−axis = q3 1+x23
which is not a rational function. As we saw in Example 2, this implies that y is
1
not a rational function.
Next we give an example of a continuous rational function f on R3 and of an
irreducible algebraic surface S R3 such that f is zero on a Zariski dense open
S
⊂ |
set yet f is not identically zero.
S
|
Example5. ConsiderthesurfaceS := x2+y2z2 y3 =0 R3. Itssingularlocus
is the z-axisand the Euclideanclosure S(cid:0)∗ S of−the smo(cid:1)ot⊂hlocus is the preimage
⊂
of the inside of the parabola (y z2) R2 under the coordinate projection.
≥ ⊂ yz
Consider the rational function
x2+y2z2 y3
f(x,y,z):=z2 − .
· x2+y2z2+y4
4 JA´NOSKOLLA´R
We see that f vanishes on S∗ and its only possible discontinuities are along the
z-axis. To analyze its behavior there, rewrite it as
y2z2
f =z2 y(1+y) .
− · x2+y2z2+y4
Thefractionisboundedby1,hencef iseverywherecontinuousandf(0,0,z)=z2.
If g(z)is anyrationalfunction without realpoles then g(z)f(x,y,z)vanishes on
S∗ and its restriction to the z-axis is z2g(z).
(The best known example of a surface with a Zariski dense open set which is
not Euclidean dense is the Whitney umbrella W :=(x2 =y2z) R3. We can take
⊂
W := (x = y = 0) and W := W. The Euclidean closure of W W does not
1 2 2 1
\
contain the “handle” (x=y =0,z <0).
In this case, a continuous rational function is determined by its restriction to
W W . TheEuclideanclosureofW W containsthehalfline(x=y =0,z 0),
2 1 2 1
\ \ ≥
and a rational function on a line is determined by its restriction to any interval.)
Thenextexampleshowsthetwonaturalwaysofpullingbackcontinuousrational
functions by a birational morphism can be different.
Example 6 (Two pull-backs). Let π : X′ X be a proper, birational morphism
→
of real varieties. If f is a continuous rational function on X, then one can think of
the composite f π in two different ways.
◦
First, f π is the composite of two continuous maps, hence it is a continuous
function. Se◦cond, one can view f π as a rationalfunction on X′, which may have
a continuous extension to X′. ◦
Tho following example shows that these two notions can be different.
TakeX =R2withf =x3/(x2+y2). Notethatf(0,0)=0. Blowup(x3,x2+y2)
toobtainX′ R2 RP1. The firstprojectionπ :X′ X isanisomorphismaway
from the orig⊂in and×π−1(0,0)=RP1. →
∼
The first interpretation above gives a continuous function f π which vanishes
along π−1(0,0). ◦
The secondinterpretationviews f π as a rationalmap X′ 99KRP1, which is in
fact regular. Its restriction to π−1(0,◦0) is an isomorphism π−1(0,0)=RP1.
∼
This confusion is possible only because X′ :=π−1 R2 (0,0) is not Euclidean
dense in X′. Its Euclidean closure contains0only on(cid:0)e po\int of(cid:1)π−1(0,0) = RP1.
∼
The two versionsof f π agreeon X′, hence also on its Euclideanclosure, but not
◦ 0
everywhere.
2. Hereditarily rational functions
In order to get some results, we have to restrict to those continuous rational
functions for which Question 1.1 has a positive answer. First we show that this
is always the case on smooth varieties and then we prove that for such functions
Question 1.2 also has a positive answer.
Definition 7. LetX bearealalgebraicvarietyandf acontinuousfunctiononX.
We say that f is hereditarily rational if every irreducible, real subvariety Z X
⊂
has a Zariski dense open subvariety Z0 Z such that f Z0 is a regular function
⊂ |
on Z0. (A function f is regular at x X if one can write f = p/q where p,q are
∈
polynomials and q(x)=0. It is called regular if it is regular at every point of X.)
6
CONTINUOUS RATIONAL FUNCTIONS 5
Examples2and3showthatnoteverycontinuousrationalfunctionishereditarily
rational.
If f is rational, there is a Zariski dense open set X0 X such that f X0 is
⊂ |
regular. If f is continuous and hereditarily rational, we can repeat this process
withthe restrictionoff to X X0,andsoon. Thus weconcludethatacontinuous
\
function f is hereditarily rational iff there is a sequence of closed subvarieties =
∅
X−1 X0 Xm = X such that for i = 0,...,m the restriction of f to
⊂ ⊂ ··· ⊂
Xi Xi−1 is regular.
\
If it is convenient, we can also assume that each Xi Xi−1 is smooth of pure
\
dimension i.
The pull-back of a hereditarily rational function by any morphism is again a
(continuous and) hereditarily rational function.
The next result shows that on a smooth variety every continuous rational func-
tion is hereditarily rational.
Proposition 8. Let X be a real algebraic variety and Z an irreducible subvariety
that is not contained in the singular locus of X. Let f be a continuous rational
function on X. Then there is a Zariski dense open subset Z0 Z such that f Z0
⊂ |
is a regular function.
Proof. By replacing X with a suitable open subvariety, we may assume that X
and Z are both smooth.
Assume first that Z has codimension 1. Then the local ring is a principal
X,Z
O
ideal domain [Sha94, Sec.II.3.1]; let t be a defining equation of Z. We can
X,Z
∈O
write f =tmu where m Z and u is a unit. Here m 0 since f does not
X,Z
∈ ∈O ≥
have a pole along Z, hence f is regular along a Zariski dense open subset Z0 Z.
⊂
Thus f Z0 is a regular function.
|
If Z has higher codimension, note that Z is a local complete intersection at
its smooth points [Sha94, Sec.II.3.2]. That is, there is a sequence of subvarieties
Z Z Z Z = X X where each Z is a smooth hypersurface in
0 1 m 0 i
⊃ ⊂ ⊂ ··· ⊂ ⊂
Z fori=0,...,m 1andZ (resp.X )isopenanddenseinZ (resp.X). Wecan
i+1 0 0
−
thus restrict f to Zm−1, then to Zm−2 and so on, until we get that f|Z0 is regular.
(Aswenotedin(4),foranyrationalfunctionf heaboveproceduredefinesaregular
function f , but it depends on the choice of the chain Z Z .) (cid:3)
|Z0 1 ⊂···⊂ m
The following result shows that hereditarily rational functions constitute the
right class for Question 1.2.
Theorem 9. Let Z be a real algebraic variety and f a continuous rational function
on Z. The following are equivalent.
(1) f is hereditarily rational.
(2) For every real algebraic variety X that contains Z as a closed subvariety,
f extends to a (continuous and) hereditarily rational function F on X.
(3) For every real algebraic variety X that contains Z as a closed subvariety,
f extends to a continuous rational function F on X.
(4) Let X be a smooth real algebraic variety that contains Z as a closed sub-
0
variety. Then f extends to a continuous rational function F on X .
0 0
Proof. It is clear that (2) (3) (4) and (4) (1) holds by (8). Thus we
⇒ ⇒ ⇒
need to show that (1) (2).
⇒
6 JA´NOSKOLLA´R
Note first that a regular function on a closed subvariety Z of a real algebraic
variety X always extends to a regular function on the whole variety. (This follows,
for instance, from [BCR98, 4.4.5] or see (14) for a more general argument.)
We reduce the hereditarily rational case to the regular case by a sequence of
blow-ups.
We can embed X into a smooth realalgebraicvariety X′. If we canextend f to
X′ then its restriction to X gives the required extension. Thus we may assume to
start with that X is smooth (or even that X =RN for some N).
Let = Z−1 Z0 Zm = Z be a sequence of closed subvarieties as in
∅ ⊂ ⊂ ··· ⊂
(7). By induction on j we construct composites of blow ups with smooth centers
Π :X X such that
j j
→
(5) Π−j1 is an isomorphism over X \Zj−1,
(6) Ej :=Π−j1(Zj−1) is a simple normal crossing divisor,
(7) E +W X isasimplenormalcrossingsubvariety(11),whereW denotes
j j j j
⊂
the birational transform of Z ,
j
(8) f Π is a regular function on E +W .
j j j
◦
AttheendwehaveΠ :X X withexceptionalsetE =Π−1 (Z )=
Π−1 (Z) such that f mΠ+1 ims+a1r→egular function on E . Tm+hu1s f mΠ+1 mex-
m+1 ◦ m+1 m+1 ◦ m+1
tends to a regular function F on X . By construction F is constant on
m+1 m+1 m+1
thefibersofE Z andtheotherfibersofΠ consistofonepointeachsince
m+1 m+1
→
(X E )=(X Z). Thus by (10), F descends to a continuous rational
m+1\ m+1 ∼ \ m+1
function F on X. Since X is smooth, F is hereditarily rational by (8).
Nowtotheinductiveargument. WewillrepeatedlyuseHironaka-typeresolution
theorems; see [Kol07, Chap.III] for references and relatively short proofs.
We can start induction with X :=X. Assume that we already have Π :X
0 j j
→
X. Then f Π is a continuous function which is regular on (W E ) =
◦ j|Wj j \ j ∼
(Zj Zj−1). Since Wj is smooth, Wj Ej is Euclidean dense in Wj, hence (unlike
\ \
in Example 6) we can identify f Π with the corresponding rational function.
Thus there is a sequence of bl◦ow-ju|Wpsj q : W′ W whose centers are smooth
j j → j
andlie overW E suchthat W′ is smoothand f′ := f Π q is a regular
j∩ j j j (cid:0) ◦ j|Wj(cid:1)◦ j
map to P1. On the other hand, f′ is also the pull-back of the continuous function
j
f , hence it is a regular function. We can perform the same blow-ups on X to
ge|WtjX′ such that W′ X′. j
j j ⊂ j
Ateachstepweblowupoverthe locusE wheref isknowntoberegular. Thus
j
the pull-back of f stays regular over the preimage of E . After a further sequence
j
of blow-ups we may assure the conditions (9.5–7).
Finally, (9.8) follows from (11). (cid:3)
Lemma 10. Let p:Y X be a proper birational morphism of real algebraic vari-
→
eties. Assume that on both of them, the singular set is contained in the Euclidean
closure of the smooth locus. Then composing with p gives a one-to-one correspon-
dencebetweencontinuousrationalfunctionsonX andcontinuousrationalfunctions
on Y that are constant on the fibers of p. (cid:3)
11. Let X be a smooth variety. A closed subvariety Z = Z X has simple
j j
∪ ⊂
normal crossing at a point p X if one can choose local coordinates x ,...,x at
1 m
∈
p such that, in a neighborhoodof p, each irreducible component Z can be defined
j
by equations x = 0: i I(j) for some I(j) 1,...,m . If this holds at every
i
{ ∈ } ⊂{ }
point, we say that Z is a simple normal crossing subvariety.
CONTINUOUS RATIONAL FUNCTIONS 7
Let Z = Z X be a simple normal crossing subvariety and f a function on
j j
∪ ⊂
Z. We claim that the following are equivalent.
(1) f is regular on Z.
(2) f is continuous on Z and f is regular for every j.
|Zj
Itisclearthat(1) (2). Toseetheconverse,itissufficienttoprovethatf isreal
⇒
analytic at each point. Choose local coordinates as above. We can uniquely write
f as a powerseries in the variables x :i 1,...,m I(j) . OverZ Z we
|Zj { i ∈{ }\ } j∩ i
getthesamepowerseries,thus wecanwrite f = a xJ wherea =0ifxJ does
PJ J J
not appear in any of the f and a equals the coefficient of xJ in f whenever
|Zj J |Zj
it is not zero.
12. Itwouldbe desirabletofindanelementaryproofof(9), onethatdoesnotrely
so much on resolution of singularities.
One key question is the following. For any real algebraic variety W one would
like to construct a proper birational morphism W′ W such that the smooth
locus of W′ is Euclidean dense in W′. Resolution of→singularities provides such a
W′ but I do not know any simple construction.
13 (Varieties over other topological fields). Let K be any topological field. The
K-pointsX(K)ofanyK-varietyinheritfromK atopology,calledtheK-topology.
One can then consider rational functions f on X that are continuous on X(K).
Thisdoesnotseemtobe averyinterestingnotioningeneral,unlessK satisfiesthe
following density property.
(DP) If X is smooth, irreducible and = U X is Zariski open then U(K) is
∅ 6 ⊂
dense in X(K) in the K-topology.
Note that if (DP) holds for all smooth curves then it holds for all varieties. It is
easy to see that if K is not discrete and the implicit function theorem holds over
K then K has the above density property. Such examples are the p-adic fields Q ,
p
their finite extensions and, more generally, quotient fields of complete local rings.
The results of this section all hold over such fields of characteristic 0. (The last
assumption is needed only for resolution of singularities).
Theonlystepthatneedsadditionalproofistheassertionthateveryregularfunc-
tion on a subvariety extends to the whole variety. This is proved by a slight mod-
ification of the usual arguments that apply when k is algebraically closed [Sha94,
Sec.I.3.2] or real closed [BCR98, 3.2.3].
Lemma 14 (Extending regular functions). Let k be a field, X an affine k-variety
and Z X a closed subvariety. Let f be a rational function on Z that is regular
⊂
at all points of Z(k). Then there is a rational function F on X that is regular at
all points of X(k) and such that F =f.
Z
|
Proof. If k is algebraically closed, then f is regular on Z hence it extends to a
regular function on X.
If k is not algebraically closed, then, as a auxiliary step, we claim that there
arehomogeneouspolynomialsG(x ,...,x ) inany numberof variableswhose only
1 r
zero on kr is (0,...,0).
Indeed,ifk isreal,thenwecantakeG= x2. ByatheoremofArtin–Schreier,
P i
ifkisnotrealclosed,thenithasfiniteextensionsL/kwhosedegreeisarbitrarylarge
(cf. [Jac80, Sec.11.7]). If c ,...,c is a k-basis of L, then G = Norm ( x c )
1 r L/k i i
P
works.
8 JA´NOSKOLLA´R
Now we construct the extension of f as follows.
For every z Z(k) we can write f = p /q where q (z)= 0. After multiplying
z z z
∈ 6
both p ,q with a suitable polynomial, we can assume that p ,q are regular on Z
z z z z
and then extend them to regularfunctions on X. By assumption, (q =0)
Tz∈Z(k) z
is disjoint from Z(k). Choose finitely many z ,...,z Z(k) such that
1 m
∈
m q =0 = q =0 .
Ti=1(cid:0) zi (cid:1) Tz∈Z(k)(cid:0) z (cid:1)
Let q ,...,q be defining equations of Z X. Set q := q and p := p for
m+1 r ⊂ i zi i zi
i = 1,...,m and p = q for i = m+1,...,r. Write (non-uniquely) G = G x
i i i i
P
and finally set
r G (q ,...,q )p
F := Pi=1 i 1 r i.
G(q ,...,q )
1 r
Since the q have no common zero on X(k), we see that F is regular at all points
i
of X(k).
AlongZ, p =fq fori=1,...,m byconstructionandfori=m+1,...,r since
i i
then both sides are 0. Thus
r G (q ,...,q )fq r G (q ,...,q )q
F Z := Pi=1 i 1 r i Z =f Pi=1 i 1 r i Z =f. (cid:3)
| G(q ,...,q ) | · G(q ,...,q ) |
1 r 1 r
Acknowledgments. I thank Ch. Fefferman, M. Jarden, F. Mangolte and B. To-
taro for useful comments and questions. Partialfinancial support was provided by
the NSF under grant number DMS-0758275.
References
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(2000a:14067)
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(inpreparation), 2011.
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Princeton University, Princeton NJ 08544-1000
[email protected]