Triangles, Differential Equations, a Theoremand the Jackpot

Dave Auckly

Mathematics Department, Kansas State University, Manhattan, KS 66506dav@math.ksu.edu

Introduction

The Atiyah-Singer index theorem is a wonderful theorem that has manyvaried applications. The goal of this paper is to introduce this theorem andsome of its applications to a broad audience in time to celebrate its goldenanniversary. The theorem was announced in a paper published in 1963 [3].There is now a vast literature related to this theorem and its generalizations.We are going to describe a result about triangles, discuss linear differentialequations, discuss the difference between circulation and rotation, state atheorem (guess which one), and hit a jackpot.

Triangles

Consider the area of a sector of a sphere of radius R having angle θ as in thefigure below. The area of such a sector is Sθ(R) = 2R2θ.

Figure 1: A sector

Now consider a spherical triangle on a sphere of radius R as in the upperleft of the following figure. The great circles that make up the edges of the

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Figure 2: Spherical Triangle

triangle divide the sphere into six truncated sectors and two copies of thetriangle. Denote the angles of one of the two congruent spherical trianglesby α, β, and γ. Let A denote the area of either triangle. The above figureproves:

2A +2 (Sα(R)− A)+2 (Sβ(R)− A) +2 (Sγ(R)− A)

= 4πR2 .

Substituting the formula for the area of a sector, combining like terms anddividing by 4πR2 gives:

α + β + γ − A · 1

R2= π .

Notice that we can recover the formula for the sum of the angles in a planetriangle by constructing a suitable family of spherical triangles and takingthe limit as R→∞.

The fraction 1R2 is known as the sectional curvature. It is denoted by

K. With this notation the formula for the sum of the angles in a spherical

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triangle reads:α + β + γ = π + A ·K . (1)

Now consider a triangulated smooth surface (not necessarily a sphere) asin figure 3. Summing the angle formula (1) over all triangles leads to:

Figure 3: Triangulated surface

α1 +β1 +γ1

+α2 +β2 +γ2

· · ·+αn +βn +γn

=

π + A1K1

+π + A2K2

·+π + AnKn

Of course A = A1 + A2 + · · · + An is the total area of the surface. Thequantity

K =(A1K1 + A2K2 + · · ·+ AnK3)

(A1 + A2 + · · ·+ An),

is the average curvature.The sum of all of the angles that meet at a given vertex is 2π. If we group

all of the angles by vertex and let V denote the number of vertices and Fthe number of faces we obtain:

2πV = πF +(A1K1 + A2K2 + · · ·+ AnK3)

(A1 + A2 + · · ·+ An)(A1 + A2 + · · ·+ An)

= πF +KA = πF +

∫surface

K dArea .

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This is a proof for constant positive sectional curvature and it certainlyshows that the result is plausible in general. Rearranging a bit gives:

V − 1

2F =

1

2π

∫surface

K dArea .

There is something interesting about this formula. The left hand sidedoes not depend on the curvature or geometry of the surface and the righthand side does not depend on the combinatorics of the triangulation. Thismeans that the expression is independent of both. It is a topological invariantknown as the Euler characteristic.

We can rewrite the left hand side a bit in order to make it manifestlyinvariant under two basic moves that change the combinatorics. Imagineputting a red dot on each side of each edge. Thus there are 2E dots whereE denotes the number of edges. Since each face has three edges, there wouldbe three dots in each face, so there are 3F dots and 3F = 2E. Adding andsubtracting E gives:

V − 1

2F = V − E +

3

2F − 1

2F = (V + F )− E .

The quantity χ(surface) := (V +F )−E is known as the Euler characteristic.Written in this way one can see that it will stay the same when an edge issubdivided into two by adding a new vertex. (The faces will no longer all betriangles.) It will also stay the same if a face is divided into two by addinga new edge connecting two vertices. One can use these moves to obtain analternate proof that the Euler characteristic is an invariant.

To summarize we have given a plausibility argument for the Gauss-Bonnettheorem:

χ(surface) =1

2π

∫surface

K dArea . (2)

The Gauss-Bonnet theorem is good motivation for the Atiyah-Singer indextheorem and follows from it as a corollary. An appropriate reference for adiscussion of curvature and a proof of the Gauss-Bonnet theorem is the bookby Singer and Thorpe, [19].

Differential Equations

The Gauss-Bonnet theorem actually states something about differential equa-tions, so we now turn to a discussion of linear PDEs. The existence and

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uniqueness of solutions of differential equations are two of the first issuesaddressed in the theory. Consider the equation grad(u) = V as a linearpartial differential equation for u. Since it is linear, solutions will be uniqueprovided the only solution to grad(u) = 0 is u = 0. Of course any constantfunction has trivial gradient, so without imposing any boundary conditionssolutions to our equation would not be unique. In fact there could even bemore functions with trivial gradient depending on the topology of the do-main. To understand this, let C∞N (Σ) denote the space of smooth functionssatisfying Neumann boundary conditions on a domain Σ, i.e. functions w sothat n · grad(w) = 0. Here n is the outer unit normal vector to the bound-ary. Of course the boundary conditions are redundant in this case, but theybecome useful later. Set

H0(Σ) := {w ∈ C∞N (Σ) | grad(w) = 0}.

If Σ is the union of two disjoint intervals in R1, say [0, 1] and [2, 3], thenH0(Σ) would be isomorphic to R2 with the isomorphism given by taking theordered pair consisting the values of the function on each interval. In generaldim(H0(Σ)) is just the number of components of Σ.

Now turn to the existence of solutions of grad(u) = V. In this case onehas to understand what are known as compatibility conditions. Recall thatthe rotation (or curl) of a vector field

V = f i + gj + hk

is given by

curl(V) := det

i j k∂∂x

∂∂y

∂∂z

f g h

.

Vector fields with vanishing curl are called irrotational. One computes thatthe curl of a gradient is trivial, thus one can only solve grad(u) = V providedthat the compatibility condition curl(V) = 0 is satisfied. This leaves us tosolve the compatibility equation or more generally curl(V) = W, as well as tounderstand how the original problem may be solved when the compatibilitycondition is satisfied.

Solving curl(V) = W appears to be analogous to our original problemsolving grad(u) = V, and indeed it is. One can see that the correspondingcompatibility conditions in three dimensions are given by div(W) = 0. Wedo not, however, need to follow this thought any further.

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The second question of solving the original gradient problem is moreinteresting to us. Since the gradient is essentially just a derivative, it is notsurprising that a function with given gradient may be found by integration.The formula u(δ(1)) − u(δ(0)) =

∫δ

grad(u) · ds, where ds := i dx + j dy isthe usual line element, and δ : [0, 1] → Σ will determine the value of thefunction u in terms the value at any point the of domain (assuming theregion is path connected). This meshes well with our analysis of the kernelof the gradient operator. To summarize, the solution to grad(u) = V isu(x) = u(x∗) +

∫δV · ds. However, there is a problem – different paths may

lead to different values of the function u. This may be measured by the valueof the integral around a loop, consisting of taking one path to the point anda second path back.

The circulation around a loop γ is given by∫γV·ds. The circulation of the

gradient of any function about any closed curve is zero. In fact, any vectorfield with vanishing circulation is the gradient of a function as the formula inthe previous paragraph shows. A non-constant periodic integral curve of avector field is a good example of a curve with non-vanishing circulation. Wesee that a gradient field can have no such curves. This makes sense since itis impossible for one to come back to the starting point of a walk by alwayswalking up.

Figure 4: The gradient of a function has no circulation.

The concepts of rotation and circulation are related, but they are not thesame. A vector field may be irrotational and still have non-trivial circulation.Even though such vector fields are not the gradient of any function, they maybe viewed as the “gradient” of a multi-valued function, such as the helicoid

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in figure 5. Here the walk up the “spiral staircase edge” projects to a circlein the plane that is an integral curve of the “gradient” of the helicoid. Thecirculation around this curve is non-trivial even though the vector field isirrotational.

Figure 5: The “gradient” of a helicoid has non-zero circulation.

Our discussion is about to take flight. It is reasonable to assume that air-flow below one third of the speed of sound is irrotational. Thus for example,one curve may circulate around the wing of a plane and the other may wraparound the trailing vortex. As these two curves bound a common (bent)cylinder the corresponding circulations must be the same. The lift acting onthe airplane is proportional to this circulation [1]. Thus, one could estimatethe lift of a plane based on a pair of timed photos taken from behind theplane.

To see that the circulation of an irrotational vector field about two curvesthat cobound a surface must be the same, one applies Stokes’ theorem:∫

γ1

V · ds −∫γ2

V · ds =

∫∂Σ′

V · ds =

∫Σ′

curl(V) · k dArea = 0 .

Let Σ be a bounded 2-dimensional domain with smooth boundary. Forc-ing the velocity field to be tangent to the boundary is natural in the contextof fluid flows. Let XT (Σ) denote the space of smooth vector fields on Σ thatsatisfy V · n = 0. Since we are restricting to a 2-dimensional domain, the curlis encoded in the one component k · curl. We can now analyze the equationcurl(V) = 0 in much the same way we analyzed the equation grad(u) = 0.Once again, the topology of the domain will determine the answer. Since

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cropduster.large.jpg (JPEG Image, 521x318 pixels) http://www.av8n.com/irro/gallery/cropduster.large.jpg

1 of 1 8/12/2010 5:51 PM

Figure 6: The circulation of the trailing vortex is proportional to the lift.

the curl of the gradient is trivial, we are really interested in irrotational vec-tor fields that are not gradient fields. The key is to construct the followingquotient space that declares that the gradient of any function is trivial:

H1(Σ) := {V ∈ XT (Σ) |k · curl(V) = 0}/Image(grad : C∞N (Σ)→ XT (Σ)) .

This now shows why we chose Neumann boundary conditions for our originalgradient equation. The gradient of a function in C∞N (Σ) is tangent to theboundary and thus satisfies the boundary conditions that are natural for fluidflow.

It turns out that dim(H1(Σ)) is the number of “holes” in the domainΣ. As an example consider the domain Σ displayed in figure 7. The figureindicates a pair of simple closed curves in the domain. We will denote thesecurves by γ+ and γ−. We will see that H1(Σ) is two-dimensional in thisexample because Σ has two “holes,” one around (1, 0) and one around (−1, 0).In fact the following function is an isomorphism:

Ψ : H1(Σ)→ R2; Ψ(V) := (

∫γ−

V · ds,∫γ+

V · ds) .

To see that Ψ is injective, assume that Ψ(V) = 0. It is plausible (and infact true) that any closed curve in Σ together with some number of copies ofγ− and γ+ bounds a parameterized subsurface of Σ, say Σ′. Stokes’ theoremthen implies that the circulation of any irrotational vector field about any

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domain3.png

Figure 7: The domain Σ

curve in Σ is zero. Thus the vector field V is the gradient of some function,and thus represents zero in H1(Σ).

The vector fields V± := 12π

(x∓1)j−yi(x∓1)2+y2

may be used to show that Ψ is sur-jective. They do not satisfy the proper boundary conditions, but this can befixed by adding appropriate gradients.

The curl of any vector field w = k · curl(V) satisfies Dirichlet boundaryconditions on Σ, i.e. restricts to zero on the boundary for any V ∈ XT (Σ).The space of functions satisfying Dirichlet boundary conditions on Σ is de-noted C∞D (Σ). To finish our analysis, we should address the “two-dimensionalholes:”

H2(Σ) := C∞D (Σ)/Image(k · curl : XT (Σ)→ C∞D (Σ)) .

To analyze this we will use the notion of the adjoint of a linear operator.The adjoint of a linear operator D : E → F between inner product spaces isthe operator D∗ : F → E such that 〈Dx, y〉 = 〈x,D∗y〉. Clearly the adjointof D∗ is D. The space F/Image(D) is known as the cokernel of D. It isisomorphic to the kernel of D∗. This is especially easy to see in the casewhen the cokernel is trivial. A vector y ∈ F is perpendicular to the image

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of D if and only if 0 = 〈Dx, y〉 = 〈x,D∗y〉 for every x ∈ E, and this is onlytrue when y is in the kernel of D∗.

It is clear that H2(Σ) is the cokernel of k · curl. The pairing on XT (Σ)is given by 〈V,W〉 :=

∫Σ

V ·W dArea. The pairing on C∞D (Σ) is given by〈u,w〉 :=

∫Σuw dArea. The following computation shows that the adjoint of

k · curl is grad(·)×k. Since the only function with trivial gradient satisfyingDirichlet boundary conditions is zero, the cokernel of k · curl is trivial.

The curl satisfies the following product rule:

curl(wV) = w curl(V) + grad(w)×V .

For w ∈ C∞D (Σ) we have

0 =

∫∂Σ

wV · ds =

∫Σ

curl(wV) · k dArea

=

∫Σ

w curl(V) · k dArea +

∫Σ

(grad(w)×V) · k dArea

=

∫Σ

w curl(V) · k dArea−∫

Σ

(grad(w)× k) ·V dArea .

A complex of operators

E0D0−−−→ E1

D1−−−→ E2 ,

i.e. a sequence of operators satisfying D1D0 = 0, may be collapsed to asingle operator D : E0 ⊕ E2 → E1 given by D = D0 + D∗1. One may seethat ker(D) ∼= H0 ⊕H2 where H0 = ker(D0) and H2 = coker(D1). Indeed,D0(u) + D∗1(w) = 0 implies D1D

∗1w = 0, so 〈D∗1w,D∗1w〉 = 〈D1D

∗1w,w〉 =

0 giving D∗1w = 0 and therefore D0u = 0. Similarly, one may see thatcoker(D) ∼= H1, where H1 = ker(D1)/image(D0).

Apply all of our discussion so far to the operator obtained by collapsingthe grad-curl complex,

D : C∞N (Σ)⊕ C∞D (Σ)→ XT (Σ) ,

given by D(u,w) := grad(u) + grad(w) × k. For a domain with g holes wehave

dim(ker(D)) = 1dim(coker(D)) = g .

(3)

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Figure 8: The surface Σ

We wish to combine this with the Gauss-Bonnet theorem. Since theGauss-Bonnet theorem applies to closed surfaces we need to construct aclosed surface. We do so by doubling the surface Σ into a surface Σ asin figure 8. We recover Σ by projecting Σ to the xy-plane. This projectionidentifies a point (x, y, z) ∈ Σ with the point τ(x, y, z) := (x, y,−z). Smooth

functions u on Σ such that τ ∗u := u ◦ τ = u correspond exactly to functionson Σ satisfying Neumann boundary conditions, i.e. C∞N (Σ). Similarly func-

tions w on Σ such that w ◦ τ = −w correspond exactly to functions on Σsatisfying Dirichlet boundary conditions, i.e. C∞D (Σ). It is also possible toidentify the set of vector fields on Σ satisfying the given boundary conditionswith τ -invariant vector fields on Σ.

There is a differential operator denoted D taking pairs of functions on Σto vector fields on Σ lifting the operator D acting on the analogous objects onΣ. One can check that Dτ ∗ = τ ∗D for this operator. It follows that τ acts onthe kernel and cokernel of D. Since τ 2 = 1 the only possible eigenvalues are±1. Any vector field can be written as V = 1

2(V + τ ·V )+ 1

2(V − τ ·V ) where

the first term is in the +1-eigenspace and the second is in the −1-eigenspace.Furthermore, there is a duality that carries one eigenspace to the other. Asimilar thing holds for the pairs of functions. The important consequenceto notice now is that the +1-eigenspace of the kernel of D on Σ may beidentified with H0(Σ)⊕H2(Σ) and this is isomorphic to the −1-eigenspace.

Similarly, the +1-eigenspace of the cokernel of D on Σ may be identified withH1(Σ) and this is isomorphic to the −1-eigenspace. It follows from this and

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equation (3) that

dim(ker(D)) = 2

dim(coker(D)) = 2g .(4)

Definition 1. The index of an operator D is

index(D) = dim(ker(D))− dim(coker(D)) .

Thus index(D) = 2 − 2g. By counting the vertices, edges and faces ofany polygonal decomposition of a surface of genus g, one can compute thatsuch a surface has Euler characteristic equal to 2− 2g. Thus the index is theEuler characteristic and the Gauss-Bonnet theorem (equation (2)) gives:

index(D) =1

2π

∫Σ

K dArea .

The structure of this formula generalizes. The index of any ellipticdifferential operator may be expressed as an integral over the basemanifold.

Theorem

We state the Atiyah-Singer Index theorem here and will explain terms usedin the theorem later.

Theorem 1 (Atiyah-Singer). If E and F are complex vector bundles over aclosed oriented manifold X and D : C∞(E)→ C∞(F ) is an elliptic differen-tial operator, then

index(D) =

∫X

(−1)n(n+1)/2ch(E − F ) td(TCX) (e(TX))−1 .

Experts will know that a slightly simpler formula may be written byreplacing the integral over the base manifold by an integral over the co-tangent space of the base manifold. We prefer this form because it fits withthe nice slogan that “the index of an elliptic differential operator may beexpressed as the integral over something on the base.”

We now sketch an argument that the index of an elliptic differential oper-ator can be expressed as the integral of some expression that only depends on

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the geometry of the functions in the domain and codomain of the operator.This argument can be understood using only concepts from linear algebra.

The first step is a simple computation using the definition of the adjointthat shows ker(D) = ker(D∗D) and ker(D∗) = ker(DD∗) provided that theinner products are positive definite. Notice that D∗Dx = λx implies thatDD∗(Dx) = λDX, thus

D : ker(D∗D − λ)→ ker(DD∗ − λ) .

If λ 6= 0 this is an isomorphism. Indeed, the same argument shows that D∗

takes ker(DD∗ − λ) to ker(D∗D− λ) and D∗Dx = λx implies that λ−1D∗ isthe inverse to D on this subspace.

It follows that

index(D) = dim(ker(D))− dim(ker(D∗))= dim(ker(D∗D))− dim(ker(DD∗))=∑{uλ} e

−tλ −∑{vµ} e

−tµ

= Tr(e−tD∗D)− Tr(e−tDD

∗) .

(5)

Here {uλ} and {vµ} are orthonormal bases of eigenfunctions of D∗D and DD∗

respectively. A straightforward exercise shows that distinct eigenspaces ofD∗D are perpendicular. It takes a bit more work to show that the eigenspacesare finite dimensional when D is what is known as an elliptic operator andto show that the sums in the second to last line above converge. Notice thatall of the terms in the sums corresponding to non-zero eigenvalues cancelbecause the λ eigenspaces of D∗D and DD∗ are isomorphic for all non-zeroλ. Notice this expression is independent of t, so we could use any t value ortake the limit as t→ 0+.

The operator takes vector-valued functions to vector-valued functions.Given an inner product on the vector space one defines an inner product onthe space of functions via

〈u,w〉C∞(E) :=

∫X

〈u(x), w(x)〉E d volX .

A vector u(y) can be considered as an operator taking R to E. The adjointof this operator is denoted by u(y)†.

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Setting u0(x) =∑{uµ} aµuµ(x) one has

u(x, t) := e−tD∗Du0(x)

= e−tD∗D∑{uµ}

aµuµ(x)

=∑{uµ}

aµe−µtuµ(x)

=∑{uµ}

∑{uλ}

aµe−λtuλ(x)

∫X

uλ(y)†uµ(y) d voly

=

∫X

∑{uλ}

e−tλuλ(x)uλ(y)†

∑{uµ}

aµuµ(y) d voly

=

∫X

kD∗D

t (x, y)u0(y) d voly ,

since the integral in the fourth line is equal to one when λ = µ and zerootherwise (as {uλ} is a orthonormal basis). The expression

kD∗D

t (x, y) :=∑{uλ}

e−tλuλ(x)uλ(y)† ,

is called the heat kernel of the operator D∗D. It is apparent from the firstline that ∂

∂tu(x, t) = −D∗Du(x, t). Also u(x, 0) = u0(x). In other words we

see that u(x, t) is the solution to this partial differential equation with initialdata u0(x). This equation is called a heat equation because the scaler versionof it models heat flow.

The t → 0+ limit of kD∗D

t (x, y) must be a delta function. In particularthis only depends on the geometry of the space of functions and not on the

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specific form of the differential operator. Now,

Tr(e−tD∗D) =

∑{uλ}

e−tλ

=∑{uλ}

e−tλ∫X

uλ(y)†uλ(y) d voly

=∑{uλ}

e−tλ∫X

Tr(uλ(y)uλ(y)†) d voly

=

∫X

Tr(∑{uλ}

e−tλuλ(y)uλ(y)†) d voly

=

∫X

Tr(kD∗D

t (x, y)) d voly .

Combining this and the analogous expression for DD∗ with the expressionfor the index in equation (5) gives

index(D) =

∫X

limt→0+

(Tr(kD

∗Dt (x, y))− Tr(kDD

∗

t (x, y)))d voly .

To summarize, in this section we stated the amazing theorem of Atiyahand Singer relating the index of a differential operator to an integral of anexpression depending only on the geometry of the space of functions in thedomain and codomain of the operator. The outline that we gave just usedelementary linear algebra. To identify the integrand and turn the outline intoa rigorous proof there are a number of technical convergence questions thatone must answer. The first outline of a proof of the index theorem by Atiyahand Singer used something called cobordism theory to reduce the proof to acheck of several special cases [3]. The Topology Seminar at Harvard in theFall of 1962 buzzed with excitement about this result. Mimeographed notesfrom the seminar authored by Bott, Atiyah, and Singer may still be found insome well stocked libraries. Atiyah and Singer’s paper with a full proof, useda different technique known as K-theory [4]. Their original cobordism proofwas presented by Palais [16]. A heat equation proof of the index theorem wasconjectured by Kotake [14], and proofs and refinements of the heat equationproof were given in a number of papers, [12, 2]. A simplified argument toidentify the integrand based on ideas about supersymmetry in physics leadsto the short proof of the index theorem given by Getzler [11]. There is now

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a vast literature on the index theorem and its generalizations with manyexcellent articles and books. Analytical details following the outline that wehave given here appear for example in the textbook [20].

We still need to describe some of the terms that appear in the IndexTheorem. We defer a discussion of complex vector bundles until our firstapplication. The space C∞(E) is a twisted version of smooth complex vectorvalued functions, that we will better understand via our first application.A differential operator is elliptic essentially when it has a finite dimensionalkernel and cokernel. The exact definition is that the symbol of the operatormust be invertible. In the case of our grad-curl complex, we can explain thesymbol a bit. People often write grad(u) = ∇u and curl(V) = ∇ × V. Ifwe replace ∇ with an arbitrary vector W we get the expressions Wu andW×V these are the symbols of the gradient and curl operators respectively.Thus the symbol of the collapsed grad-curl operator D(u,w) = grad(u) +grad(w)× k is

σW(D)(u,w) = Wu+ wW × k .

One can check that this is an isomorphism from R2 to itself provided W 6= 0.In the introduction we noticed that the Gauss-Bonnet formula related

combinatorial data (numbers of vertices, edges, and faces) to geometric data(curvature and area). This implied that the resulting expression could notdepend on the combinatorics or on the geometry. The same thing is true ofthe index theorem. This means that the integral should not depend on thecoordinates that are used and should have coordinate independent descrip-tions. We described the curvature of a two-dimensional surface. For higherdimensional spaces the curvature may be encoded in something known asthe curvature tensor. It is essentially a matrix with entries representingtwo-dimensional sections of the space. It is known that any coordinate inde-pendent invariant of a matrix (linear transformation) is a symmetric functionof the eigenvalues. The elementary symmetric functions of the eigenvalues ofa matrix are the coefficients of the characteristic polynomial of the matrix.

The symmetric functions of the eigenvalues of the curvature ten-sor are known as characteristic classes.

The Chern character and Todd class appearing in the Index Formula are

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given by

ch(E) = tr(exp(− 1

2πiF )) =

∑exk ,

td(E) = det(− 1

2πiF (1− exp(

1

2πiF ))−1) =

∏ xk(1− e−xk)

.

Here the xk are the eigenvalues of the normalized curvature operator. Theyare called the Chern roots. The Euler class e(E) is a symmetric functionknown as the Pfaffian of the Chern roots. Up to a sign it is the square rootof the determinant of the normalized curvature matrix.

It is time to look at some of the amazing and varied applications of theindex theorem.

Knotty

Things get a bit knotty now. The index theorem has been used to establishmany results in knot theory and geometric topology. A knot is a smoothembedding of a circle into R3 (one glues the ends of the string togetherto keep the knot from coming undone). By scaling we may assume thatthe image of the knot is contained in [−1, 1]3. A knot is trivial if it is theboundary of a smooth disk in R3.

A non-trivial knot may still bound a smooth disk in [−1, 1]4. Knotswith this property are called slice. An example of a slice knot togetherwith a movie of the slice disk is displayed in figure 9. The left side is ananalogue in one dimension less. In this case the “knot” is just the waist ofthe pants contained in [−1, 1]2. The Jordan curve implies that this waist isthe boundary of a disk in [−1, 1]2. It is also the boundary of a pair of pantsin [−1, 1]3. The image on the left is a movie representation of this pair ofpants embedded in [−1, 1]3. To make it into a disk in [−1, 1]3 a ”sock” can besewed on the the bottom of each pant leg. The right side has the analogousmovie picture in the four dimensional case. The top frame of this movie isthe granny knot, and the entire movie displays the pants that granny wears.The slice disk is obtained by sewing a pair of socks onto granny’s pants togive her a pair of footies.

There are algebraic invariants that can demonstrate that certain knots(including the trefoil) are not slice, however, we will use the index theoremto prove that the trefoil knot in figure 10 is not slice. There are easier ways

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Figure 9: Movie of a slice disk

to prove that this knot is not slice, but the ideas that we use are useful formore complicated knots and the index theorem is a very effective tool tostudy such problems. In particular, one of the early applications of the indextheorem to topology was the work by Casson and Gordon establishing thatthere are algebraically slice knots that are not slice, [9]. We are essentiallyusing their argument here. More recently Jae Choon Cha has used theseideas to obtain very subtle invariants of concordance in the formidable paper[10].

Figure 10: The trefoil knot

The proof that the trefoil is not slice is an indirect proof. Assuming that itis the boundary of a smooth disk, we construct a special type of manifold andapply the index theorem to get a contradiction. The construction begins with

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Figure 11: A branched cover

something called a 5-fold branched cover. A branched cover is a map fromone manifold to another that is locally modeled on the map Rk×C→ Rk×Cgiven by (v, z) 7→ (v, zd). Figure 11 displays a d = 5-fold branched cover ofD3 ∼= D2 × [−1, 1] to D3 ∼= D2 × [−1, 1] branched along an embedded D1.This is a good schematic for the 5-fold branched cover, branched over theassumed slice disk. We will call this hypothetical cover R.

The 5-fold branched cover of the three sphere S3 branched along thetrefoil knot is the boundary of a spin manifold with signature −8, denotedVres. We will explain the signature and the notion of a spin manifold soon.The important points up to here are that the hypothetical cover R has no2-cycles, so the manifold X = Vres ∪R will be a closed, spin 4-manifold withsignature −8.

We now describe the signature of a 4-dimensional manifold. In 4-dimensionssomething called the Hodge star operator sends a 2-form dx1 ∧ dx2 to the“perpendicular” form dx3 ∧ dx4. The second cohomology,

H2(X) =Ker(d : C∞(∧2X)→ C∞(∧3X))

Im(d : C∞(∧1X)→ C∞(∧2X)),

can then be decomposed into positive and negative eigenspaces of the Hodgestar operator. These are denoted by H2

+(X) and H2−(X). (There are var-

ious technicalities here.) By definition the signature of W is Sign(X) :=dim(H2

+(X)) − dim(H2−(X)). This looks like the index of a differential op-

erator. This is indeed the case, the signature is the index of the signatureoperator. An application of the index theorem gives

Sign(X) =1

3

∫X

p1(X) .

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In four dimensions one may use the quaternions to describe spin struc-tures. The quaternions are analogous to the complex numbers. They can beconstructed by adjoining three elements, i, j, and k to the real numbers suchthat i2 = j2 = k2 = ijk = −1. A spin structure on a 4-manifold consistsof a pair of quaterionic vector bundles, W+ and W− together with so calledclifford multiplication operations c : T ∗X ×W± → W∓. It is known that asimply-connected 4-manifold admits a spin structure exactly when the selfintersection number of every cycle is even. One defines an elliptic differentialoperator (the Dirac operator) ∂/ : C∞(W+) → C∞(W−) as the compositionof the covariant derivative (think gradient) and clifford multiplication. Anapplication of the index theorem gives

Index(∂/) =1

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∫X

p1(X) .

Thus,

Sign(X) = 8 Index(∂/) = 8 (dimC(ker(∂/)− dimC(coker(∂/)) .

Since the Dirac operator is quaterionic linear, if follows that the signatureof any spin 4-manifold is divisible by 16. This result is known as Rochlin’stheorem. It also establishes that the trefoil knot is not slice, since the exis-tence of a slice disk would imply the existence of a closed, spin 4-manifoldwith signature negative eight.

The book by Shanahan gives a nice explanation of the signature theorem,the spin index theorem, and how these results follow from the Atiyah-Singerindex theorem [18]. The Casson-Gordon result uses an equivariant version ofthe index theorem and more complicated covers, but the general idea is thesame.

We now return to the question of constructing the manifold Vres. Thestarting point is to consider the singular algebraic variety

V := {(x, y, z) ∈ C3 |x2 + y3 + z5 = 0} .

This has an isolated singularity. By definition the link of this singularity isthe boundary of a neighborhood of the singularity, i.e.

Σ = Σ(2, 3, 5) := {(x, y, z) ∈ C3 |x2 + y3 + z5 = 0 and max(|x|, |y|) = 1} .

The variety V is just an open cone on Σ(2, 3, 5), as one can see via themap [0,∞) × Σ(2, 3, 5) → V by (t, (x, y, z)) 7→ (t15x, t10y, t6z). The closed

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cone on Σ(2, 3, 5) is denoted by V . It is just the set of points in V withmax(|x|, |y|) ≤ 1. The map

Σ→ S3 := {(x, y) ∈ C2 |max(|x|, |y|) = 1} ,

given by (x, y, z) 7→ (x, y) is a 5-fold branched cover. The branch set isT (2, 3) := {(x, y) ∈ S3 |x2 + y3 = 0}. The torus becomes a standard toruswhen S3−{(0, 1)} is identified with R3. The branch set can be parameterizedas x = ieit/2, y = eit/3, and this wraps twice around the torus in one directionwhile wrapping three times around the torus in the other direction. Thus,the branch set is a trefoil knot.

The variety V is singular, but it is possible to construct a smooth varietythat agrees with V away from zero. This smooth variety is called the reso-lution, and is denoted by Vres. It turns out that the signature of Vres is −8.This fact is at the heart of something known as the McKay correspondence.A proof that the signature is −8 may be found in [13].

Jackpot

You just won at a slot machine! The casino always empties the slot when itreaches $1, 000.00. The machine only has quarters and nickels. One possiblejackpot is 2 quarters and 3 nickels (what a jip!), one possibility is 4, 000quarters and there are many more possibilities. All possibilities totaling $1.00or less are displayed in figure 12. How many different jackpots are there?

Figure 12: Jackpots no more than $1.00

The index theorem can answer this question. To see how, let q denote the

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number of quarters and n denote the number of nickels. The possibilities aresimply the integer lattice points satisfying

q ≥ 0, n ≥ 0, 0.25q + 0.05n ≤ 1, 000 .

The last equation can be rewritten as c := 20, 000 − 5q − n ≥ 0, so thenumber is the same as the number of non-negative integer lattice points(q, n, c) satisfying

5q + n+ c = 20, 000 = 5k .

Here k = 4, 000 represents the most quarters the slot machine will award. Ofcourse we can generalize the problem and consider different maximums.

In order to apply the index theorem, notice that each triple of latticepoints can be encoded as the coefficients of a monomial θqηnγc. These mono-mials will be a basis of the kernel of a differential operator that we willconstruct.

As a warm up consider the space of degree k polynomials defined over thecomplex numbers. We can identify this space of polynomials as the kernel of adifferential operator. Consider two examples: f(z) = zz is not a polynomial,and g(z) = z2 clearly is a polynomial. If we set ∂ := dz ∂

∂zwe will obtain

∂f = z dz and ∂g = 0. A standard result from complex analysis implies thatany function h(z) satisfying ∂h = 0 may be expressed as a convergent powerseries h(z) =

∑∞n=0 cnz

n.To restrict to degree k functions we need to further specify the geometry of

the domain of the operator. First recall that the version of the index theoremthat we stated is valid for closed manifolds. To use the index theorem in thissetting we add a point at infinity viewed as the ratio [1 : 0]. The space of allratios is complex projective space

CP 1 := {(z, w) ∈ C2 − {(0, 0)}}/ ∼ , (z, w) ∼ (λz, λw) ,

for λ ∈ C− {0}, and we denote the equivalence class of (z, w) (the ratio) by[z : w].

The degree k condition enters through the following definition:

Lk := {(z, w, f) ∈ (C2 − {(0, 0)})× C}/ ∼ , (z, w, f) ∼ (λz, λw, λkf) ,

for λ ∈ C − {0}. Notice that we have a projection π : Lk → CP 1. Theinverse image of each point has the structure of a vector space. This is anexample of something called a vector bundle.

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It is instructive to think about this using real numbers and k = 1. Withthe real numbers every ratio [z : w] with w 6= 0 is equivalent to one of the form[r : 1], and we may view [1 : 0] as infinity, so the analogous base RP 1 is justthe result of adding one infinity to the real numbers, so it is homeomorphic toa circle. Moving to the real version of L1, one sees that every point (z, w, f)will be equivalent to one with z2 + w2 = 1 via multiplication by a positivereal number. By multiplying by −1 if need be one may further assume thatw ≥ 0. Thus every point in the real L1 is equivalent to one of the form(cos θ, sin θ, f) with θ ∈ [−π, π]. There are still points left to identify. Thepoint (1, 0, f) is equivalent to (−1, 0,−f) via multiplication by −1. Thus thereal L1 is the result of identifying opposite edges of the [−π, π]× R strip inthe opposite direction – a Mobius band. This is a great example of a vectorbundle. Locally it looks just like the product of a space with a vector space,but globally it is twisted.

Figure 13: Mobius band vector bundle

The domain of an elliptic differential operator is a space of sections of avector bundle. A section of a vector bundle π : E → X is a map s : X → Eso that π ◦ s(x) = x for every x ∈ X. In the special case of a trivial vectorbundle E = X × V with π just projection on the second factor, and sectionmay be identified with a vector valued function and vice versa. Indeed, avector-valued function f : X → V induces the section sf : X → X×V givenby sf (x) = (x, f(x)). It is in this sense that sections of a vector bundle area generalization of vector-valued functions. The space of smooth sections ofa vector bundle E is denoted C∞(E).

To see how this works in the case of the bundle Lk, notice that any

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section s : CP 1 → L such that π ◦ s = idCP 1 is completely specified by s([z :w]) = [z, w, f(z, w)] where f(λz, λw) = λkf(z, w). The Cauchy-Riemannoperator may be defined on such sections by ∂

∂zf(z, 1). For s ∈ ker(∂) we

have ∂f∂w|(1,w) = 0, so f(1, w) =

∑∞n=0 bnw

n and f(z, 1) = zkf(1, z−1) =zk∑∞

n=0 bnz−n. The only way this latter expression can can be a power

series is if bn = 0 for n > k in which case it is a polynomial of degree k.It is possible to show that the cokernel of the Cauchy-Riemann operator is

trivial in this situation. Thus the index is just the dimension of the kernel ofthe Cauchy-Riemann operator, i.e. the dimension of the space of polynomialsof degree k. Whereas it is not beyond the level of an undergraduate tounderstand the details of the computation of the characteristic classes andthe integral in the index formula, it would take a bit more space that we havein this present article. These details will appear in the related paper [5].

Now return to the jackpot problem (counting the non-negative integersolutions to 5q + n+ c = 20, 000 = 5k) and consider the space

Lk ∼= {(θ, η, γ, f) ∈ (C3 − {0})× C}/ ∼ ,(θ, η, γ, f) ∼ (λ5θ, λη, λγ, λ5kf) , for λ ∈ C− {0} .

This projects toX := {(θ, η, γ) ∈ C3 − {0}}/ ∼ ,

(θ, η, γ) ∼ (λ5θ, λη, λγ) .

A section in C∞(Lk) is expressed as s([θ, η, γ]) = [θ, η, γ, f(θ, η, γ)] forsome smooth f satisfying f(λ5θ, λη, λγ) = λ5kf(θ, η, γ).

As in the toy problem of polynomials of degree k, good examples ofsuch smooth functions are given by polynomials. The monomial f(θ, η, γ) =θ6η2γ3 satisfies f(λ5θ, λη, λγ) = λ5·7f(θ, η, γ). We can think of it as a rep-resentation of a pay out of 6 quarters, 2 nickels, on a slot machine with amaximum pay out of 7 quarters. Thus it is 3 nickels short of the maximumpay out. Compare these numbers with the exponents of the specific mono-mial f . It should be clear that we can create a bijection between monicmonomials and lattice points in the lattice polytope describing all possiblejackpots.

In fact, a bit of thought should convince the reader that there is a bijectionfrom the lattice points in a lattice polytope and the lattice points of theintersection of an affine subspace with the non-negative cone in a higherdimensional lattice. Just arrange each inequality defining the polytope tobe of the form something ≥ 0 and then declare each something to be a new

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(slack) variable. The lattice points in this latter section can then be identifiedwith monic weighted-homogenous polynomials. If these polynomials can beidentified as a basis of the kernel of an elliptic differential operator, thenthe Atiyah-Singer Theorem can be used to compute the number of latticepoints in the lattice polytope. This is indeed the case. The operator is justa generalization of the Cauchy-Riemann operator, so the kernel consists ofconvergent power series which must be weighted polynomials in order to bewell defined as sections of the vector bundle. Details of the computation ofintegral in the index theorem in this example take more room than we havehere, but may be found in [5].

We can use the invariance of the Euler characteristic discussed at the startof the paper to complete the solution of the jackpot problem. To see howthis works, let ∆ be a lattice polytope in the plane, i.e. a union of triangleswith vertices living in the lattice Z2. It is known that the area of a latticetriangle that does not contain any interior lattice points is 1/2. Furthermore,any lattice polytope in the plane is a union of such simple triangles.

In any triangulation of a polytope, one may relate the number of vertices,number of boundary vertices, Euler characteristic and the number of bound-ary vertices. To do so we repeat the argument from the start of the paperused to find the invariant expression for the Euler characteristic. Put dotsin each triangle next to each edge as in the following figure.

Figure 14: Triangulated Polytope

We can count the number of dots in two different ways. On one hand

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the number of dots (D) is 3 number of faces = 3F . On the other hand thenumber of boundary edges is equal to the number of boundary vertices (B),if we added one dot for each boundary edge then each edge would have twodots. Thus the number of dots plus the number of boundary dots is equal totwice the number of edges:

2E = D +B = 3F +B .

Thus the Euler characteristic can be expressed in terms of the number ofinterior vertices (I), number of boundary vertices (B), and number of faces:

χ = V − E + F = I +B − 3

2F − 1

2B + F .

In the special case of a triangulation consisting entirely of triangles con-taining no other lattice points than their vertices, the area will be the sumof the area of the triangles, and this will just be one half of the number offaces. Thus,

A =1

2F = I +

1

2B − χ .

This result is known as Pick’s theorem [15]. We can rearrange this to get thenumber of lattice points inside the region:

number of lattice points = I +B = A+1

2B + χ .

Thus the number of lattice points in the generalized lattice polygon for thejackpot problem is

5

2k2 +

7

2k + 1 .

One can verify directly that this formula is correct for k = 0, 1, and 2. To-gether with the expected quadratic growth in k this gives an easier derivationof the specific formula.

It is clear that the Atiyah-Singer index theorem is not the easiest wayto solve the jackpot problem. The point is that the index theorem has alarge number of surprising and varied applications. It may often be usedto deduce unexpected connections between different types of mathematics,and often points to different generalizations of results. The discussion sofar, should make it clear that there are higher dimensional versions of Pick’stheorem that follow from the index theorem. Pommersheim wrote a nice

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dissertation in the mid 1990s about the application of the index theoremto lattice counting questions [17]. Also see [6, 7, 8]. We obtain the answerto our original jackpot question by substituting k = 4, 000 into the generalformula to see that there are 40, 014, 001 possible jackpots of $1, 000 or lesswith nickels and quarters. This is a large number, but it may be smaller thanthe number of possible applications of the index theorem.

Conclusion

Michael Atiyah and Isadore Singer announced the Index theorem 50 yearsago. This theorem connects deep topological properties of spaces to thestructure of elliptic differential operators. Since the two different sides of theequation come from different branches of mathematics, the theorem serves asa bridge that connects different mathematical results. The index theorem hasmany applications and many generalizations, and is one of the great math-ematical achievements of the last century. Atiyah and Singer were awardedthe Abel Prize in 2004, in part, for their index theorem.

When I was an undergraduate, Ed Ihrig showed me the index theorem inan undergraduate geometry class. (Thank Ed.) It inspired me to learn moreabout topology. I hope this exposition inspires others to explore and learnmore about the wonderful world of geometry.

References

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