Table Of Content8
9
Whitney’s index formula in higher dimensions
9
1
and Laplace integrals
n
a
J
0 Yu.M.Burman∗†
1
] March 18, 1997
G
D
.
h
t
a
m 1 Introduction
[
1 Let f = (f ,f ) : IR → IR2 be a long curve, i.e. a smooth immersion such
1 2
v
that f (x) = x and f (x) = 0 for x ≤ −1 and x ≥ 1. By the famous theorem
4 1 2
4 of Whitney (see [1]) long curves are classified by a single integral invariant,
0 an index (rotation number). Informally saying, index is a number of full
1
rotations made by the tangent vector D (x) = (f′(x),f′(x)) 6= 0 as x goes
0 f 1 2
8 from −∞ to +∞ (or, equally, from −1 to 1). The explicit formula for the
9
index is as follows:
/
h
t ∞ f′′(x)f′(x)−f′′(x)f′(x)
a I(f) = 1 2 2 1 dx (1)
m (f′(x))2 +(f′(x))2
Z−∞ 1 2
:
v
Xi The integral in the right-hand side of (1) is IR(Df)∗ω where
r z dz −z dRz
a ω(z ,z ) = 2 1 1 2 (2)
1 2 z2 +z2
1 2
is the closed form in IR2 \{0} forming a basis in its de Rham cohomologies
H1(IR2 \{0}) = IR.
Suppose now that f is generic. By Thom’s transversality theorem (see
[2]) f has only a finite number of self-intersection points. All these points are
∗Mathematical College of the Independent University of Moscow
†e-mail: [email protected]
1
simple (i.e. only two branches of the curve meet) and transversal. A point
a ∈ IR2 of simple transversal self-intersection can be equipped with a sign
σ(a) = ±1 by the following rule. Let f(x(1)) = f(x(2)) = a and x(1) < x(2).
By transversality, the tangent vectors D (x(1)) and D (x(2)) are not parallel,
f f
and therefore form a basis in IR2. Choose an orientation of IR2 and take
+1, if the basis (D (x(1)),D (x(2)))
f f
gives a chosen orientation of IR2,
σ(a) = (3)
−1, if the basis (D (x(1)),D (x(2)))
f f
gives an opposite orientation.
Then there holdsWhitney’s index formula (see [1]):
I(f) = σ(a) (4)
a
X
where the sum is taken over all the self-intersection points of the curve f.
Thepaperisdevoted tothegeneralizationof(1)–(4)tothecaseofsmooth
immersions f : IRn → IR2n such that f(x) = (x,0) for |x| large. In Section 2
wedefineanindexandprovea“sumofsigns”formulaforgenericimmersions.
Section 3 contains a proof of an explicit integral formula for the index (for
n even); the proof makes use of asymptotic expansions of certain Laplace
integrals. The formula obtained looks like I(f) = (D )∗ω where ω is
IRn f
an n-form on the Stiefel variety V(n,2n) and D : IRn → V(n,2n) is a
f
R
differential. It is proved in Section 4 that ω is a generator of the de Rham
cohomology group Hn(V(n,2n)).
2 Index and self-intersection points
Recall that the map f = (f ,...,f ) : IRn → IR2n is called an immersion if
1 2n
for any x ∈ IRn the (n×2n)-matrix
∂f
i
D (x) = (5)
f
∂x
(cid:18) j(cid:19)
has the rank n (i.e., belongs to the Stiefel variety V(n,2n)). We will always
suppose that the immersion f is “fixed at infinity”, i.e. f(x) = (x,0) as soon
as x lies outside the unit cube of IRn. Generic immersions IRn → IR2n fixed
2
at infinity have only a finite number of self-intersection points, all of which
are simple and transversal.
The homotopy classes of immersions IRn → IR2n form an abelian group
Imm . Definition of the multiplication resembles that of a homotopy group:
n
take
f (2x +1,x ,...,x ), if x ≤ 0,
(f ∗f )(x) = 1 1 2 n 1
1 2 f (2x −1,x ,...,x ), if x ≥ 0.
2 1 2 n 1
(cid:26)
One of the formulations of the famous Smale classification theorem (see
[3]) is
Theorem 1 The mapping D : Imm → π (V(n,2n)) is a group isomor-
f n n
phism.
So to classify immersions we need to know the group π (V(n,2n)). The
n
answer is the following well-known lemma (see, for example, [4], page 173):
Lemma 1
π (V(n,2n)) = 0 for 1 ≤ i ≤ n−1.
i
ZZ for n = 1 and n even,
π (V(n,2n)) =
n ZZ for odd n ≥ 3.
2
(cid:26)
Thus, smooth immersions f : IRn → IR2n are classified by a single invari-
ant, an index I(f) which is either an integer (for n = 1 and n even) or an
element of ZZ (for odd n ≥ 3). The case n = 1 is covered by the results of
2
Whitney, and in what follows we will always assume that n > 1.
For n even, a point a ∈ IR2n of simple transversal self-intersection of
immersion f can be equipped with a sign σ(a) = ±1 by the following rule:
if f(x(1)) = f(x(2)) = a (and x(1) 6= x(2)), then take
D (x(1))
σ(a) = sgndet f (6)
D (x(2))
f
(cid:18) (cid:19)
(the right-hand side contains a determinant of the matrix 2n × 2n). Geo-
metrically this means that we choose positively oriented bases in the tangent
spacestotheimageoff atpointsx(1) andx(2). Sincef isimmersion, these2n
vectors form a basis in IR2n, and σ(a) is taken ±1 depending on the orienta-
tionofthisbasis(orientationofIR2n isfixedbeforehand). Since nissupposed
3
to be even, the result will not depend on the order in which the preimages
x(1),x(2) ∈ f−1(a) are taken (for n = 1 we supposed that x(1) < x(2), but
there is no way to choose this order for n > 1).
Theorem 2
I(f) = σ(a) (7)
a
X
where the sum is taken over all the self-intersection points of f, and the
summation is performed in ZZ for n even and in ZZ for odd n ≥ 3.
2
Proof Let f be a generic continuous family of smooth immersions IRn →
t
IR2n fixed at infinity. By Thom’s transversality theorem the set of self-
intersection points of f is changes continuously with t, except for a finite
t
number of points t ,...,t where two “catastrophes” may occur: either two
1 k
self-intersection points merge and disappear, or, conversely, emerge. It can
be easily observed that for n even the pair of points to emerge or to perish al-
ways have opposite signs. It follows from this that (for any n) the right-hand
side of (7) depends only on the homotopic class of the immersion f. Thus,
this right-hand side defines a mapping S from Imm to ZZ or ZZ , depending
n 2
on parity of n.
Definition of multiplication in Imm shows that S is a homomorphism.
n
By Theorem 1, to prove equation (7) it is enough to show that S is an
isomorphism, i.e. that its image is the whole ZZ or ZZ . In other words
2
it should be proved that 1 ∈ ImS, i.e. that there exists an immersion
f(n) : IRn → IR2n with only one self-intersection point.
Such immersion is easily constructed by induction on n. For n = 1 take
a smooth plane curve f(1) such that f(1)(−1/2) = f(1)(1/2), and there are
no more self-intersection points. Suppose now that immersion f(n) is already
constructed, andtheself-intersectionisf(n)(−1/2,0,...,0) = f(n)(1/2,0,...,0).
Let ϕ(t) be a smooth function such that ϕ(−1/2) 6= ϕ(1/2) and ϕ(t) = 0 for
|t| ≥ 1. Then it can be easily seen that the formula
f(n+1)(x ,...,x ) = (f(n)(x ,...,x ),...,f(n)(x ,...,x ),x ,
1 n+1 1 1 n n 1 n n+1
(n) (n)
f (x ,...,x ),...,f (x ,...,x ),x ϕ(x ))
n+1 1 n 2n 1 n n+1 1
defines an immersion IRn+1 → IR2n+2 whose sole self-intersection is
f(n+1)(−1/2,0,...,0) = f(n+1)(1/2,0,...,0), so that induction step is made.
4
3 An integral formula for the index
Let us derive an explicit integral formula for the index I(f). For the reasons
explained in the previous Section we restrict ourselves to the case of n even
(an element of ZZ hardly can appear as an integral).
2
Start with a technical result. Let f : IRn → IR2n be an arbitrary smooth
map (even not immersion) fixed at infinity. Consider a mapping A : IR2n →
f
IR2n given by the formula A (x,y) = f(x)−f(y) (x,y ∈ IRn) and recall that
f
degree of a smooth mapping g : M → N (M and N being oriented manifolds
of the same dimension) is defined as
degg = sgn(x).
g(x)=y
X
Here y ∈ N is a generic point, and sgn(x) is taken 1 or −1 depending on
whether the mapping Df : T M → T N preserves or inverts orientation.
x y
The right-hand side of this equation can be called multiplicity of g at y.
Lemma 2 The degree of the mapping A is zero.
f
Proof Prove first that the degree of A is defined. To do this denote E ⊂
f
IR2n thesubspace ofvectors(a ,...,a ,0,...,0). Sincen > 1,anytwopoints
1 n
z ,z ∈ IR2n\E can be connected with the path z that does not intersect E.
0 1 t
Noticenow that since f isfixed at infinity, forevery compact set B ⊂ IR2n\E
its full preimage A−1(B) is compact, and therefore multiplicity of A at z
f f t
does not depend on t. So, multiplicity is the same for all generic points,
and degA is defined. But then degA = 0 because the image of A is not
f f f
the whole IR2n (for example, if z ∈ IR2n is large enough and normal to the
subspace E then certainly A−1(z) = ∅).
f
Consider not A as a change of variable in IR2n and apply it to the 2n-
f
formν = exp(−λ|z|2/2)dz ∧...∧dz . Using Lemma 2, obtainthefollowing
1 2n
result:
D (x)
J(f) = exp −λ|f(x)−f(y)|2/2 det f dxdy
D (y)
ZIRn×IRn (cid:18) f (cid:19)
(cid:0) (cid:1)
= A∗ν = degA ν = 0. (8)
f f
ZIR2n ZIR2n
5
Our idea is now to find the asymptotics of the integral (8) for λ → +∞
using the general results about asymptotics of Laplace integrals. Then we
consider the principal term of the asymptotics and equalize it to zero.
To formulate the result introduce first some notations. Let ϕ = (ϕj) ∈
i
V(n,2n) (i.e. Rkϕ = n, 1 ≤ j ≤ 2n,1 ≤ i ≤ n). For an arbitrary set of
integers J = {j ,...,j } such that 1 ≤ j < ... < j ≤ 2n denote
1 n 1 n
µ(J) = n(n−1)/2+j +...+j ,
1 n
M d=ef det(ϕαs)
J s s = 1,...,n
α ∈/ J
s
Denote also
2n
U d=ef ϕkϕk ,
i j
!
k=1
X
(u (ϕ)) d=ef U−1,
ij
detU = 1/u2(ϕ).
(by the general theorem about Gram determinant, detU equals the sum of
squares of the n-th order minors of the matrix ϕ, and is therefore positive).
Theorem 3 Let n be even, f : IRn → IR2n be a smooth immersion fixed at
infinity, and D : IRn → V(n,2n) be its differential. Then the index of f is
f
given by the formula
I(f) = (D )∗ω (9)
f
ZIRn
where the 2n-form ω at the point ϕ = (ϕj) ∈ V(n,2n) is
i
n
1
ω(ϕ) = − u(ϕ) u (ϕ)...u (ϕ)
2n+1πn/2(n/2)! i1i2 in−1in
i1,.X..,in=1
× (−1)µ(J)M (ϕ)dϕj1 ∧...∧dϕjn . (10)
J i i
σ(1) σ(n)
σX∈Σn J = {jX1,...,jn}
1 ≤ j < ... < j ≤ 2n
1 n
(Σ means the symmetric group).
n
6
To prove Theorem 3 we will need the following asymptotical expansion
found in [5] (Statement 4.1 at page 125):
Lemma 3 Let a,S be smooth real-valued functions at IRk, S achieving its
minimum at the origin (and only at the origin). Let the Hessian G =
∂2S (0) be nondegenerate. Denote L the 2nd order differential operator
∂ui∂uj
L = k (G−1) ∂2 , and R(u) d=ef S(u) − S(0) − 1(Gu,u). Then for
i,j=1 ij∂ui∂uj 2
λ → +∞ there is an aymptotic expansion up to the O(λ−∞):
P
k/2
2π
a(u)exp(−λS(u))du ∼ exp(−λS(0)) |detG|−1/2
λ
ZIRk (cid:18) (cid:19)
∞
1
× Lp a(u)exp(−λR(u)) . (11)
p!(2λ)p u=0
Xp=0 (cid:16) (cid:17)(cid:12)
(cid:12)
(cid:12)
Proof of Theorem 3 The left-hand side J(f) of (8) is exactly the Laplace
integralmentionedinLemma3withk = 2n,u = (x,y),S(x,y) = |f(x)−f(y)|2/2,
D (x)
anda(x,y) = det f . MinimumpointofS isnotuniquebutthegen-
D (y)
f
(cid:18) (cid:19)
eral localization principle for Laplace integrals (see [5]) tells that one should
consider expansions (11) for all such points and sum them up (points other
then local minima make O(λ−∞) contributions).
Local minima of S(x,y) are:
1. Self-intersection points of f: f(x) = f(y) and x 6= y.
2. Diagonal points: x = y.
3. Other (“sporadic”) minima.
At sporadic minima s = S(x,y) > 0. So expansion (11) contains the term
exp(−λs) and therefore is already O(λ−∞). Thus, sporadic minima will be
neglected.
To self-intersection points (x,y) Lemma 3 can be applied directly. The
principal term here is p = 0. Denote g (i = 1,...,2n) the vector
i
(∂fi(x),..., ∂fi(x),
∂x1 ∂xn
7
− ∂fi(y),...,−∂fi(y)) and observe that G = (g ,g ). By the general the-
∂y1 ∂yn ij i j
2
D (x)
orem about Gram determinant we have detG = det f , and the
D (y)
f
(cid:12) (cid:18) (cid:19)(cid:12)
principal term is (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
n n
2π 2π D (x)
|detG|−1/2a(x,y) = sgndet f .
λ λ D (y)
f
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
It follows now from Theorem 2 that total contribution to the principal term
of asymptotics from all the self-intersection points is equal to
n
2π
C = 2 I(f). (12)
self-intersections
λ
(cid:18) (cid:19)
The factor 2 is present because every self-intersection point appears twice as
a minimum of S: the first time as (x,y) and the second time as (y,x).
Consider nowthecontributionfromthediagonal. Theminimum ofS here
is not isolated (and therefore degenerate) but it is easy to see that we should
consider asymptotic expansion of the integral over y (with x fixed) and then
integrate it over x (necessary theorems about uniformity of asymptotic ex-
pansion (11) are contained in [5]). If x is fixed then y = x is a nondegenerate
minimum of S, and G = (h ,h ) where h = (∂f1(x),..., ∂f2n(x)). By the
ij i j i ∂xi ∂xi
general theorem about Gram determinant
−1/2
2n
∂f ∂f
g(x) d=ef |detG|−1/2 = det k(x) k(x)
∂x ∂x
(cid:12) i j !(cid:12)
(cid:12) Xk=1 (cid:12)
(cid:12) (cid:12) −1/2
(cid:12) ∂f (cid:12) 2
= (cid:12) det is(x(cid:12)) . (13)
∂x
1≤i1<X...<in≤2n(cid:12)(cid:12) (cid:18) s (cid:19)(cid:12)(cid:12) !
(cid:12) (cid:12)
The right-hand side contains the sum of squar(cid:12)es of all the n-t(cid:12)h order minors
of the (2n×n)-matrix ∂fi(x) .
∂xj
First we are to find(cid:16)the le(cid:17)ading term of expansion. Let v(t),w(t) be
smooth functions of one variable. Then it is easy to see that
dm
v(t)exp(λw(t)) = exp(λw(t)) λu
dtm
(cid:16) (cid:17) Xu
8
× b v(r)(t)w(k1)(t)...w(ku)(t) (14)
r,k1,...,ku
r,k1,..X.,ku ≥ 0,
r+k +...+k = m
1 u
for some integers b ≥ 0 (which we need not specify). Apply this
r,k1,...,ku
formula with v = a, w = R to expansion (11). Here a(x,y) = O(|x−y|n)
and R(x,y) = O(|x−y|3), and the degree of the operator L is 2 (so, Lp
contains differentiations of the order 2p). Thus, the p-th term of expansion
will contain λu−p−n/2 where
2p = r +k +...+k ,
1 u
r ≥ n,k ≥ 3.
i
Thus 2p ≥ n+3u, and therefore u−p ≤ −n/2. So, the leading term is
πn/2 1
g(x) (Ln/2a(x,y)) dx (15)
(n/2)!λn y=x
ZIRn
(cid:12)
where g(x) is given by (13). (cid:12)
Denote G−1 d=ef (g (x)). Then, by definition,
ij
n ∂2
L = g (y)
ij
∂y ∂y
i j
i,j=1
X
and
n ∂n
Ln/2 = g (y)...g (y) .
i1i2 in−1in ∂y ...∂y
i1,.X..,in=1 i1 in
This yields:
n ∂n ∂fj(x)
(Ln/2a(x,y)) = g (x)...g (x) det ∂xk
(cid:12)y=x i1,.X..,in=1 i1i2 in−1in ∂yi1 ...∂yin ∂∂yfkj(y) !(cid:12)(cid:12)y=x
(cid:12)
(cid:12) n ∂fj(x) (cid:12)
= gi1i2(x)...gin−1in(x) det ∂∂2xfkj (x) . (cid:12)
i1,.X..,in=1 σX∈Σn ∂xk∂xiσ(k) !
Here Σ means the symmetric group. The subscript j in determinants runs
n
from 1 to 2n, the subscript k, from 1 to n. Thus, total contribution to the
9
principal term of the asymptotics of J(f) from the diagonal y = x equals:
C = πn/2 1 g(x) n g (x)...g (x)×
diagonal (n/2)!λn IRn i1,...,in=1 i1i2 in−1in
∂fj(x) (16)
×R σ∈Σn dePt ∂∂2xfkj (x) .
∂xk∂xiσ(k) !
P
But it follows from(8) thatC +C = 0. Thus, combining
diagonal self-intersections
(12) and (16), one has
I(f) = − 1 g(x) n g (x)...g (x)×
2n+1πn/2(n/2)! IRn i1,...,in=1 i1i2 in−1in
R× σ∈ΣnPdet ∂xk∂∂∂∂2xxffkjijσ((xk))(x) !. (17)
P
Notice now that for any 1 ≤ j ,...,j ≤ 2n, 1 ≤ i ,...,i ≤ n
1 n 1 n
det ∂2fjs dx ∧...∧dx = (D )∗dϕj1 ∧...∧dϕjn (18)
∂xs∂xis 1 n f i1 in
(cid:16) (cid:17)
Now to get (9) expand the determinant in (17) in the first n rows.
4 An explicit formula for the generator of
Hn(V (n,2n))
Take up again Lemma 1. By Gurevich theorem (see e.g. [4]) for n even,
Hn (V(n,2n)) = IR is the first nontrivial cohomology group of the Stiefel
DR
variety V(n,2n). Let ω be a closed n-form at V(n,2n) whose integral over
0
the fundamental cycle (generator of the H (V(n,2n),ZZ)) is 1. Then the
n
homotopic class [γ] ∈ π (V(n,2n)) = ZZ of the spheroid γ : Sn → V(n,2n)
n
equals to γ∗ω . Now Theorems 3 and 1 suggest that ω = ω , i.e.,
Sn 0 0
TheoremR4 The cohomology class of the n-form ω given by (10) forms the
basis in Hn (V(n,2n)). The value of this class on the generator of the
DR
H (V(n,2n),ZZ) is 1.
n
The main step in the proof of Theorem 4 is the following
Lemma 4 The form ω is closed.
10
