MATHEMATICAE 159 (1999)
The Gaussian measure on algebraic varieties
by
Ilka A g r i c o l a and Thomas F r i e d r i c h (Berlin)
Abstract. We prove that the ring R[M ] of all polynomials defined on a real algebraic variety M ⊂ R
nis dense in the Hilbert space L
2(M, e
−|x|2dµ), where dµ denotes the volume form of M and dν = e
−|x|2dµ the Gaussian measure on M .
1. Introduction. The aim of the present note is to prove that the ring R[M ] of all polynomials defined on a real algebraic variety M ⊂ R
nis dense in the Hilbert space L
2(M, e
−|x|2dµ), where dµ denotes the volume form of M and dν = e
−|x|2dµ is the Gaussian measure on M . For M = R
n, the result is well known since the Hermite polynomials constitute a complete orthonormal basis of L
2(R
n, e
−|x|2dµ).
2. The volume growth of an algebraic variety and some conse- quences. We consider a smooth algebraic variety M ⊂ R
nof dimension d and denote by dµ its volume form. Then M has polynomial volume growth:
there exists a constant C depending only on the degrees of the polynomi- als defining M such that for any euclidian ball B
rwith center 0 ∈ R
nand radius r > 0 the inequality
vol
d(M ∩ B
r) ≤ Cr
dholds (see [Br¨o]). Via the Crofton formulas the above inequality is a con- sequence of Milnor’s results concerning the Betti numbers of an algebraic variety (see [Mi1], [Mi2], in which the stated inequality is already implic- itly contained). This estimate yields first of all that the restrictions to M of polynomials on R
nare square-integrable with respect to the Gaussian measure on M .
1991 Mathematics Subject Classification: 28A75, 58A07, 14P99.
Key words and phrases: Gaussian measure, algebraic variety.
The work was supported by the SFB 288 “Differential geometry and quantum physics”.
[91]
Proposition 1. Let M be a smooth submanifold of the euclidian space R
n. Suppose that M has polynomial volume growth, i.e., there exist constants C and l ∈ N such that for any ball B
r,
vol
d(M ∩ B
r) ≤ Cr
l. Then:
1. The ring R[M ] of all polynomials on M is contained in the Hilbert space L
2(M, e
−|x|2dµ).
2. The functions e
α|x|2for α < 1/2 all belong to L
2(M, e
−|x|2dµ).
P r o o f. Throughout this article, denote the distance of the point x ∈ R
nto the origin by r
2= |x|
2. We shall prove that the integrals
I
m(M ) := \
M
r
me
−r2dµ < ∞, m = 1, 2, . . . , are finite. We have
I
m(M ) = X
∞ j=0\
M ∩(Bj+1−Bj)
r
me
−r2dµ
and consequently we can estimate I
m(M ) as follows:
I
m(M ) ≤ X
∞ j=0(j + 1)
me
−j2[vol(M ∩ B
j+1) − vol(M ∩ B
j)]
≤ X
∞ r=0(r + 1)
me
−r2vol(M ∩ B
r+1).
Using the assumption on the volume growth of M we immediately obtain I
m(M ) ≤ C
X
∞ r=0(r + 1)
m+le
−r2.
Denoting the summands of the latter series by a
r, we readily see that it converges, since
a
r+1a
r= (r + 1)
m+le
−r2−2r−1r
m+le
−r2=
r + 1 r
m+l1
e
2r+1→ 0.
A similar calculation yields the result for the functions e
αr2with α < 1/2.
3. A dense subspace in C
0∞(S
n). The aim of this section is to verify that a certain linear subspace of C
∞0(S
n) is dense therein. Since the family of functions we have in mind cannot be made into an algebra, we have to replace the standard Stone–Weierstraß argument by something different.
The idea for overcoming this problem is to use a combination of the well-
known theorems of Hahn–Banach, Riesz and Bochner.
To begin with, we uniformly approximate the function e
−r2e
ihk,xifor a fixed vector k ∈ R
n.
Lemma 1. Denote by p
m(x) the polynomial p
m(x) =
m−1
X
α=0
i
αhk, xi
α/α!.
Then the sequence e
−r2p
m(x) converges uniformly to e
−r2e
ihk,xion R
n. P r o o f. The inequality
|p
m(x) − e
ihk,xi| ≤ kkk
mkxk
mm! e
kkk·kxkimplies (set y = kkk · kxk)
sup
x∈Rn
|e
−r2p
m(x) − e
−r2e
ihk,xi| ≤ sup
0≤y
y
mm! e
y−y2/kkk2=: C
m.
Therefore, we have to check that for any fixed vector k ∈ R
nthe sequence C
mtends to zero as m → ∞. For simplicity, denote by k the length of the vector k ∈ R
n. A direct calculation yields
C
m= 1 m!
k
24 + k
4
p k
2+ 8m
m× exp
k
24 + k
4 p
k
2+ 8m − 1 k
2k
24 + k
4 p
k
2+ 8m
2. We are only interested in the asymptotics of C
m. We will thus ignore all constant factors not depending on m. In this sense, we obtain
C
m≈ 1 m!
k
24 + k
4 p
k
2+ 8m
mexp
k 8
p
k
2+ 8m − k
2+ 8m 16
. The Stirling formula m! ≈ √
m m
me
−mallows us to rewrite the asymptotics of C
m:
C
m≈ 1
√ m m
mk
24 + k
4
p k
2+ 8m
mexp
k 8
p k
2+ 8m + m 2
. Since
m→∞
lim ( p
k
2+ 8m − √
8m) = 0, we can furthermore replace √
k
2+ 8m by 2 √ 2m:
C
m≈ 1
√ m m
mk
24 + k
4 p
k
2+ 8m
mexp
k 4
√ 2m + m 2
=: e
Cm∗with
C
m∗= m ln
k
24 + k
4
p k
2+ 8m
+ k
2 √ 2
√ m + m
2 − m ln(m) − 1
2 ln(m).
If m is large compared with k, we can estimate ln(k
2/4 + (k/4) √
k
2+ 8m) by
12ln(m) + α for some constant α:
C
m∗/ m
2 ln(m) + αm + k 2 √ 2
√ m + m
2 − m ln(m) − 1 2 ln(m)
≤ − m
2 ln(m) + (α + 1/2)m + k 2 √ 2
√ m
≤ − m
2 ln(m) +
α + 1/2 + k 2 √
2
m
= m
α + 1/2 + k 2 √
2 − 1 2 ln(m)
. Finally, C
m= exp(C
m∗) converges to zero.
We denote the full ring of polynomials on R
nby P.
Proposition 2. The linear space Σ
∞:= P · e
−r2is dense in the space C
∞0(S
n) of all continuous functions on S
n= R
n∪ {∞} vanishing at infinity.
P r o o f. Suppose the closure Σ
∞of the linear space Σ
∞does not coincide with C
∞0(S
n). Then the Hahn–Banach Theorem implies the existence of a continuous linear functional L : C
0(S
n) → R such that
1. L|
Σ∞= 0;
2. L(g
0) 6= 0 for at least one g
0∈ C
∞0(S
n).
According to Riesz’ Theorem (see [Rud, Ch. 6, pp. 129 ff.]), L may be represented by two regular Borel measures µ
+, µ
−on S
n:
L(f ) = \
Sn
f (x) dµ
+(x) − \
Sn
f (x) dµ
−(x).
In particular, µ
+and µ
−are finite. The first property L|
Σ∞= 0 of L implies
\
Sn
e
−r2p(x) dµ
+(x) = \
Sn
e
−r2p(x) dµ
−(x)
for any polynomial p(x). Let us introduce the measures ν
±= e
−r2µ
±on the subset R
n⊂ S
n. Then
\
Rn
p(x) dν
+(x) = \
Rn
p(x) dν
−(x)
holds and remains true for any complex-valued polynomial. We may thus choose p(x) = p
m(x) as in the previous lemma:
p
m(x) =
m−1
X
α=0
i
αhk, xi
α/α!.
But then
\
Sn
p
m(x)e
−r2dµ
+(x) = \
Rn
p
m(x) dν
+(x) = \
Rn
p
m(x) dν
−(x)
= \
Sn
p
m(x)e
−r2dµ
−(x)
together with the uniform convergence of p
m(x)e
−r2to e
ihk,xie
−r2implies
\
Sn
e
ihk,xie
−r2dµ
+(x) = \
Sn
e
ihk,xie
−r2dµ
−(x), i.e.,
\
Rn
e
ihk,xidν
+(x) = \
Rn
e
ihk,xidν
−(x).
Therefore, the Fourier transforms of the measures ν
+and ν
−coincide. Con- sequently, by Bochner’s Theorem (see [Mau, Ch. XIX, pp. 774 ff.]) we con- clude that ν
+= ν
−on R
n. The linear functional L : C
0(S
n) → R must thus be the evaluation of a function at infinity:
L(f ) = cf (∞),
contrary to the existence of a function g
0∈ C
∞0(S
n) satisfying L(g
0) 6= 0.
4. The main result
Theorem 1. Let the closed subset M ⊂ R
nbe a smooth submanifold satisfying the polynomial volume growth condition. Then the ring R[M ] of all polynomials on M is a dense subspace of the Hilbert space L
2(M, e
−r2dµ).
P r o o f. Consider the one-point compactification c M ⊂ S
nof M ⊂ R
n. Then Tietze’s Extension Lemma and Proposition 2 imply that
Σ
∞( c M ) := R[M ] · e
−r2/4is dense in C
∞0( c M ). We introduce the measure dν = e
−r2/2dµ, where dµ is the volume form of M . Since
\
M
dν = \
M
e
−r2/2dµ = \
M
(e
r2/4)
2e
−r2dµ =: V < ∞,
dν defines a regular Borel measure db ν on c M (by setting db ν(∞) = 0). There- fore, the algebra C
∞0( c M ) of all continuous functions on c M vanishing at in- finity is dense in L
2( c M , db ν):
C
∞0( c M ) = L
2( c M , db ν).
For any function f in L
2(M, e
−r2dµ) we have
\
M
|f e
−r2/4|
2e
−r2/2dµ = \
M
|f |
2e
−r2dµ < ∞
and, therefore, f e
−r2/4lies in L
2( c M , db ν). Thus, for a fixed ε > 0, there exists g ∈ C
∞0( c M ) such that
\
M
|f e
−r2/4− g(x)|
2e
−r2/2dµ < ε/2.
According to Proposition 2 we can find a polynomial p(x) ∈ R[M ] approxi- mating g:
sup
x∈ cM
|g(x) − p(x)e
−r2/4|
2< ε/(2V ).
Using the inequality kx + yk
2≤ 2kxk
2+ 2kyk
2we conclude that
\
M
|f (x)e
−r2/4− p(x)e
−r2/4|
2e
−r2/2dµ < ε;
but this is equivalent to
\
M