Fredholm and Dirac Index Theorems and Zero-Mode Counting

How can analytic zero-mode data be related to topological characteristic data? On a closed manifold, an elliptic operator becomes Fredholm after its domain and target are completed in the appropriate Sobolev norms. Its analytic index is then a stable difference of two finite-dimensional zero-mode spaces. The Atiyah–Singer theorem identifies that integer with a topological index determined by the principal symbol.

For a closed, oriented, Riemannian spin manifold $M^{2m}$, a Hermitian bundle $W$ with unitary connection, and the index-theorem grading $\Gamma_{\mathrm{ind}}$ defined below, this identification becomes

$$\operatorname{ind}\mathcal D_{R,W}^{+} = \dim\ker\mathcal D_{R,W}^{+} - \dim\ker\mathcal D_{R,W}^{-} = \int_M \left[ \widehat A(TM)\wedge\operatorname{ch}(W) \right]_{\dim M}.$$

This is a net chiral zero-mode count. It does not usually determine the two kernel dimensions separately, and it does not apply unchanged on a manifold with boundary, on a noncompact space, or to a Lorentzian Dirac operator. Those qualifications are part of the theorem, not technical afterthoughts.

Required background. Spectra, Resolvents, Spectral Measures, and Functional Calculus supplies closed operators, adjoints, kernels, and ranges; Characteristic Classes and Chern–Weil Theory supplies $\operatorname{ch}(W)$, $p_1(TM)$, and the gauge-curvature normalization; and Spin Structures and Dirac Operators constructs the Euclidean twisted Dirac operator and establishes its ellipticity.

Fredholm and Dirac index hypotheses

The main setting on this page is therefore:

• $M$ is a smooth, closed manifold, meaning compact and without boundary;
• all bundles are finite-rank complex vector bundles with smooth Hermitian structures;
• the Dirac specialization assumes that $M^{2m}$ is oriented, Riemannian, and spin, and that $W$ has a unitary connection;
• the chirality grading and orientation are stated explicitly.

The word closed has two different uses below. A closed manifold is compact and boundaryless. A closed operator is one whose graph is closed. Context will distinguish them.

Fredholm operators isolate a stable defect

Let

$$T:\mathcal D(T)\subset H_0\longrightarrow H_1$$

be a densely defined closed operator between Hilbert spaces. Its domain is a Hilbert space in the graph norm

$$\|u\|_T^2=\|u\|_{H_0}^2+\|Tu\|_{H_1}^2,$$

and

$$T:\mathcal D(T)_{\mathrm{graph}}\longrightarrow H_1$$

is bounded. The operator $T$ is Fredholm when

$$\dim\ker T<\infty, \qquad \operatorname{ran}T\ \text{is closed}, \qquad \dim\operatorname{coker}T<\infty,$$

where

$$\operatorname{coker}T = H_1/\operatorname{ran}T.$$

Its analytic index is

$$\operatorname{ind}T = \dim\ker T-\dim\operatorname{coker}T.$$

Closed range gives the orthogonal decomposition

$$H_1 = \operatorname{ran}T \oplus \ker T^\dagger.$$

Consequently the orthogonal projection identifies $\operatorname{coker}T$ with $\ker T^\dagger$, and

$$\boxed{ \operatorname{ind}T = \dim\ker T-\dim\ker T^\dagger }.$$

The formula resembles finite-dimensional rank–nullity, but its force is different. A square finite-dimensional matrix always has index zero. An infinite-dimensional Fredholm operator can have a nonzero index because its kernel and cokernel need not have the same dimension.

The smallest nonzero example: the unilateral shift

On $\ell^2(\mathbb N_0)$, define

$$S(x_0,x_1,x_2,\ldots) = (0,x_0,x_1,\ldots).$$

It is injective, and its range is the closed codimension-one subspace orthogonal to

$$e_0=(1,0,0,\ldots).$$

Thus

$$\ker S=0, \qquad \operatorname{coker}S\simeq\mathbb C e_0, \qquad \operatorname{ind}S=-1.$$

Its adjoint is the left shift,

$$S^\dagger(x_0,x_1,x_2,\ldots) =(x_1,x_2,\ldots),$$

which is surjective and has kernel $\mathbb C e_0$. Therefore

$$\operatorname{ind}S^\dagger=+1.$$

This example already displays the two analytic quantities that the Dirac index will compare: a kernel of an operator and a kernel of its adjoint.

Why closed range matters

Consider multiplication by the coordinate on $L^2([-1,1])$:

$$(M_xf)(x)=x f(x).$$

Both $M_x$ and $M_x^\dagger=M_x$ have zero kernel because the single point $x=0$ has measure zero. The range is dense, but it is not closed. To see this, set

$$f_n(x)= \begin{cases} 1/x,& |x|\ge 1/n,\\ 0,& |x|<1/n. \end{cases}$$

Each $f_n$ lies in $L^2([-1,1])$, while

$$M_xf_n=\mathbf1_{\{|x|\ge1/n\}} \longrightarrow 1 \quad\text{in }L^2.$$

The constant function $1$ is not in the range, since $1/x\notin L^2$. Thus $M_x$ is not Fredholm. Merely subtracting the two kernel dimensions would give $0-0$, but that number would not carry Fredholm stability.

Stability is the reason to take the difference

The Fredholm operators form an open subset of the bounded operators, and the index is locally constant on that subset. If $T$ is Fredholm and $K$ is compact, then

$$T+K\ \text{is Fredholm}, \qquad \operatorname{ind}(T+K)=\operatorname{ind}T.$$

For a closed unbounded operator, the corresponding statement treats $K:\mathcal D(T)_{\mathrm{graph}}\to H_1$ as compact. Along a norm-continuous path of bounded Fredholm realizations, the individual dimensions of $\ker T$ and $\ker T^\dagger$ can jump, but their difference cannot. A kernel mode can therefore appear or disappear together with an adjoint-kernel mode without changing the index.

These statements follow from the parametrix characterization of Fredholm operators; Dai 2015, § 3.1, pp. 35–39, PDF gives the analytic construction and its stability consequences.

Ellipticity produces Fredholm operators on closed manifolds

Let

$$P:C^\infty(M,E)\longrightarrow C^\infty(M,F)$$

be a differential operator of order $k$. Its principal symbol

$$\sigma_k(P)(x,\xi):E_x\longrightarrow F_x$$

is obtained from the highest-derivative part. The operator is elliptic when $\sigma_k(P)(x,\xi)$ is invertible for every nonzero covector $\xi\in T_x^*M$.

This local symbol condition is not itself a Fredholm statement: $C^\infty(M,E)$ is not the Hilbert-space domain used in the definition above. On a closed manifold, however, $P$ extends for every real $s$ to a bounded map

$$P_s: H^s(M,E) \longrightarrow H^{s-k}(M,F).$$

The elliptic Fredholm theorem states:

If $M$ is closed and $P$ is elliptic, then every Sobolev extension $P_s$ is Fredholm. Its kernel consists of smooth sections, and its index is independent of $s$.

The hypotheses do distinct work. Ellipticity supplies a local inverse at high momentum. Compactness of $M$, absence of a boundary, and the Sobolev completion turn that local information into finite-dimensional kernels and a closed range.

Proof idea: a parametrix modulo compact errors

Elliptic symbol calculus constructs a pseudodifferential operator $Q$ of order $-k$ such that

$$\begin{aligned} QP&=\mathbf1-K_E,\\ PQ&=\mathbf1-K_F, \end{aligned}$$

where $K_E$ and $K_F$ are smoothing operators. On a compact manifold, smoothing gains derivatives, and the Rellich compactness theorem makes these remainders compact on the relevant Sobolev spaces. Thus $P_s$ is invertible modulo compact operators. Atkinson’s theorem then implies that $P_s$ is Fredholm.

Elliptic regularity supplies the second important step. If a Sobolev section lies in the kernel of $P_s$, then it is smooth. Under the natural Sobolev dual pairing, the cokernel is represented by distributional solutions of the formal adjoint $P^*$; elliptic regularity makes those solutions smooth as well. Thus both defect spaces are the same geometric solution spaces for every Sobolev exponent, and the index does not depend on $s$.

This is a proof of the elliptic-to-Fredholm step at the level of its main mechanism. Constructing the full pseudodifferential calculus and proving its estimates is beyond this page.

The general Atiyah–Singer statement

Ellipticity also gives a topological object. Because $\sigma_k(P)(x,\xi)$ is invertible away from the zero section, it defines a compactly supported $K$-theory class

$$[\sigma_k(P)]\in K_c^0(T^*M).$$

The analytic index is

$$\operatorname{ind}_{\mathrm a}(P) = \dim\ker P_s-\dim\operatorname{coker}P_s.$$

Topology constructs another homomorphism

$$\operatorname{ind}_{\mathrm t}: K_c^0(T^*M)\longrightarrow\mathbb Z$$

using a $K$-theory pushforward, equivalently an embedding, the Thom isomorphism, and Bott periodicity. The Atiyah–Singer index theorem says:

For an elliptic differential operator $P$ between complex vector bundles over a closed smooth manifold,

$$\operatorname{ind}_{\mathrm a}(P) = \operatorname{ind}_{\mathrm t} \bigl([\sigma_k(P)]\bigr).$$

Thus the stable analytic defect is determined by the homotopy class of the principal symbol, even though the two kernel dimensions depend on all coefficients of the operator.

This is the theorem statement, not a full proof. The parametrix explains why the analytic side is defined. The topological construction and the deformation argument needed to identify the two indices are the deep part of the original theorem. The original paper is Atiyah and Singer, “The Index of Elliptic Operators I,” Annals of Mathematics 87 (1968), pp. 484–530. A physics-oriented route from analytic index to symbol topology appears in Nakahara, 2003 second edition, Chapter 12, §§ 12.1–12.2, pp. 453–460.

The chiral Dirac block

Now let $M^{2m}$ be a closed, oriented, Riemannian spin manifold, and let $W\to M$ be a Hermitian bundle with unitary connection. The spin page used Hermitian gamma matrices and the analytic Clifford action

$$\gamma(v)^2=g(v,v), \qquad \mathbf c(v)=i\gamma(v), \qquad \mathbf c(v)^2=-g(v,v).$$

The distinction between these two Clifford conventions affects the label attached to chirality. For an oriented orthonormal frame, define the index-theorem grading by

$$\Gamma_{\mathrm{ind}} = i^m\, \mathbf c(e_1)\cdots\mathbf c(e_{2m}).$$

The earlier spin page defined

$$\Gamma_E = i^m\gamma^1\cdots\gamma^{2m}.$$

Since $\mathbf c=i\gamma$,

$$\Gamma_{\mathrm{ind}} = (-1)^m\Gamma_E .$$

The two labels agree in dimensions divisible by four and exchange in dimensions congruent to two modulo four. This is only a naming conversion, but it controls the sign of a chiral index.

Let $\Sigma_{\mathrm{ind}}^\pm M$ be the $\pm1$ eigenbundles of $\Gamma_{\mathrm{ind}}$. Clifford multiplication reverses the grading, so the twisted Riemannian Dirac operator decomposes as

$$\mathcal D_{R,W} = \begin{pmatrix} 0&\mathcal D_{R,W}^{-}\\ \mathcal D_{R,W}^{+}&0 \end{pmatrix}$$

on $\Sigma_{\mathrm{ind}}^+\otimes W\oplus \Sigma_{\mathrm{ind}}^-\otimes W$.

Regard the positive block as an unbounded operator on $L^2$:

$$\begin{aligned} \mathcal D(\mathcal D_{R,W}^{+}) &= H^1(M,\Sigma_{\mathrm{ind}}^+\otimes W), \\ \mathcal D_{R,W}^{+}&: \mathcal D(\mathcal D_{R,W}^{+}) \subset L^2(M,\Sigma_{\mathrm{ind}}^+\otimes W) \\ &\longrightarrow L^2(M,\Sigma_{\mathrm{ind}}^-\otimes W). \end{aligned}$$

Its unbounded $L^2$ adjoint is

$$\left(\mathcal D_{R,W}^{+}\right)^\dagger = \mathcal D_{R,W}^{-}, \qquad \mathcal D(\mathcal D_{R,W}^{-}) = H^1(M,\Sigma_{\mathrm{ind}}^-\otimes W).$$

Ellipticity and closedness of $M$ make both blocks Fredholm. If

$$n_\pm=\dim\ker\mathcal D_{R,W}^{\pm},$$

then

$$\operatorname{ind}\mathcal D_{R,W}^{+} = n_+-n_-.$$

The full operator $\mathcal D_{R,W}$, with domain $H^1$ in $L^2$, is self-adjoint in this closed-manifold setting. Its kernel and cokernel have the same dimension, so

$$\operatorname{ind}\mathcal D_{R,W}=0.$$

Only the chiral block can carry the nonzero index used for zero-mode counting.

The twisted Dirac index formula

With the index-theorem grading just fixed, Atiyah–Singer specializes to

$$\boxed{ \operatorname{ind}\mathcal D_{R,W}^{+} = \left\langle \widehat A(TM)\wedge\operatorname{ch}(W), [M] \right\rangle = \int_M \left[ \widehat A(TM)\wedge\operatorname{ch}(W) \right]_{2m} }.$$

The characteristic series begin

$$\widehat A(TM) = 1-\frac{p_1(TM)}{24} +\frac{7p_1(TM)^2-4p_2(TM)}{5760} +\cdots$$

and

$$\operatorname{ch}(W) = \operatorname{rank}W +c_1(W) +\operatorname{ch}_2(W) +\cdots.$$

For the anti-Hermitian curvature matrix $\mathbf F_W$ used on the characteristic-class page,

$$\operatorname{ch}(W) = \operatorname{tr}_W \exp\left(\frac{i\mathbf F_W}{2\pi}\right).$$

The right-hand side of the index formula is written using rational characteristic forms. The theorem implies that its integral is nevertheless an integer. That integrality is not visible from de Rham closedness alone.

If one keeps the $\Gamma_E$ labels from the spin page instead of converting to $\Gamma_{\mathrm{ind}}$, then the $+$ and $-$ bundles are exchanged when $m$ is odd. The corresponding formula is

$$\operatorname{ind} \mathcal D_{\Gamma_E,W}^{+} = (-1)^m \int_M \left[ \widehat A(TM)\wedge\operatorname{ch}(W) \right]_{2m}.$$

Reversing the orientation also exchanges the chiral labels and reverses the index. Multiplying the differential operator by a nonzero scalar does not change either kernel and therefore does not change the index.

Dai 2015, § 3.3, pp. 50–52, PDF gives the chiral decomposition and twisted Dirac formula in the $\mathbf c(v)^2=-g(v,v)$ convention used for $\Gamma_{\mathrm{ind}}$.

Why a local density counts global zero modes

The McKean–Singer formula gives a useful proof map. On a closed manifold the heat operators are trace class, and for every $t>0$,

$$\begin{aligned} \operatorname{Str}_{\mathrm{ind}}\, e^{-t\mathcal D_{R,W}^2} &= \operatorname{Tr}_{\Sigma_{\mathrm{ind}}^+} e^{-t\mathcal D_{R,W}^{-}\mathcal D_{R,W}^{+}} \\ &\quad - \operatorname{Tr}_{\Sigma_{\mathrm{ind}}^-} e^{-t\mathcal D_{R,W}^{+}\mathcal D_{R,W}^{-}} \\ &= \operatorname{ind}\mathcal D_{R,W}^{+}. \end{aligned}$$

Every positive eigenvalue of $\mathcal D_{R,W}^{-}\mathcal D_{R,W}^{+}$ is paired with the same eigenvalue of $\mathcal D_{R,W}^{+}\mathcal D_{R,W}^{-}$, so the two traces cancel away from zero. Only the two chiral kernels survive in the supertrace.

The same expression can be evaluated as $t\to0^+$. The short-time heat kernel is local, and its supertrace density has top-degree limit

$$\left[ \widehat A(TM)\wedge\operatorname{ch}(W) \right]_{2m}.$$

Integrating that local density produces the global index. This explains the bridge:

$$\text{chiral zero modes} \longleftrightarrow \text{heat-kernel supertrace} \longleftrightarrow \text{characteristic density}.$$

For the untwisted spin Dirac operator, Dai 2015, § 3.6, pp. 63–71, PDF derives the local $\widehat A$ density by heat-kernel asymptotics and Getzler rescaling. Dai 2015, § 3.4, pp. 53–54, PDF gives the general Clifford-module formula and identifies its twisting factor with $\operatorname{ch}(W)$ when the module is $\Sigma M\otimes W$.

Deriving the heat-kernel coefficients and proving their cancellation is not reproduced here; this is the architecture of an analytic proof, not a claim of a full proof.

A two-dimensional sign and flux check

Let $\Sigma$ be a closed oriented spin surface and let $L_n\to\Sigma$ be a Hermitian line bundle with

$$\int_\Sigma c_1(L_n)=n\in\mathbb Z.$$

There is no degree-two term in $\widehat A(T\Sigma)$. With the index-theorem grading,

$$\operatorname{ind}\mathcal D_{R,L_n}^{+} = \int_\Sigma c_1(L_n) =n.$$

In this dimension $m=1$, so the earlier $\Gamma_E$ labels give the opposite answer:

$$\operatorname{ind}\mathcal D_{\Gamma_E,L_n}^{+}=-n.$$

The conversion can be checked with Pauli matrices:

$$\gamma^1=\sigma_1, \qquad \gamma^2=\sigma_2,$$

for which

$$\Gamma_E=i\sigma_1\sigma_2=-\sigma_3, \qquad \Gamma_{\mathrm{ind}} = i(i\sigma_1)(i\sigma_2) =\sigma_3.$$

The theorem fixes only the difference $n_+-n_-=n$. Determining, for example, that all zero modes on one particular surface and connection have one chirality requires an additional vanishing theorem or an explicit solution. Nakahara 2003, § 12.6.2, pp. 471–472 develops related two- and four-dimensional examples; its chirality and curvature conventions must be translated by the displayed formulas above.

Four dimensions and a controlled gauge-background example

Let $X$ be a closed oriented Riemannian spin four-manifold. Keeping only degree four in $\widehat A(TX)\operatorname{ch}(W)$ gives

$$\operatorname{ind}\mathcal D_{R,W}^{+} = \int_X \left[ \operatorname{ch}_2(W) - \frac{\operatorname{rank}W}{24}p_1(TX) \right] .$$

Now take an $SU(N)$ background and a representation $\rho$. The site uses Hermitian gauge generators and

$$D=\mathrm d-i g_{\mathrm{YM}}A, \qquad \mathbf F_\rho=-i g_{\mathrm{YM}}F^aT_\rho^a.$$

Normalize the quadratic trace by

$$\operatorname{tr}_\rho(T_\rho^aT_\rho^b) = T(\rho)\delta^{ab}, \qquad T(F)=\frac12$$

for the fundamental representation $F$. Then

$$\operatorname{ch}_2(W_\rho) = \frac{g_{\mathrm{YM}}^2}{8\pi^2} \operatorname{tr}_\rho(F\wedge F).$$

The characteristic-class page fixed

$$\begin{aligned} Q &= \int_X\operatorname{ch}_2(W_F) \\ &= \frac{g_{\mathrm{YM}}^2}{8\pi^2} \int_X\operatorname{tr}_F(F\wedge F) \\ &= -\left\langle c_2(W_F),[X]\right\rangle \in\mathbb Z. \end{aligned}$$

Since quadratic traces differ by their Dynkin indices,

$$\int_X\operatorname{ch}_2(W_\rho) = \frac{T(\rho)}{T(F)}Q = 2T(\rho)Q.$$

Therefore

$$\boxed{ \operatorname{ind}\mathcal D_{R,\rho}^{+} = 2T(\rho)Q - \frac{\dim\rho}{24} \int_Xp_1(TX) }.$$

For $X=S^4$, the tangent bundle is stably trivial, so $\int_{S^4}p_1(TS^4)=0$. In the fundamental representation,

$$\operatorname{ind}\mathcal D_{R,F}^{+}=Q.$$

A sector with $Q=1$ therefore has

$$n_+-n_-=1,$$

and hence at least one zero mode. The theorem alone does not say that $n_+=1$ and $n_-=0$; that stronger statement needs a vanishing result. This is the controlled QFT-facing use of the theorem: topological charge fixes a net Euclidean chirality count before any physical consequence is drawn.

The coefficient $2T(\rho)$ is quadratic representation data. It is not the cubic representation coefficient governing a four-dimensional perturbative gauge anomaly. Jacobians, anomaly polynomials, descent, Ward identities, and cancellation conditions belong to Perturbative Chiral and Gauge Anomalies.

What the index does and does not determine

Let

$$I=\operatorname{ind}\mathcal D_{R,W}^{+}=n_+-n_-.$$

Then

$$n_++n_-\ge |I|.$$

A nonzero index forces zero modes. It gives a lower bound on their total number and fixes their net chirality. It does not generally give the total number $n_++n_-$ or the separate values of $n_+$ and $n_-$.

The flat two-torus gives the simplest warning. Choose the periodic spin structure and the trivial twisting bundle. Constant spinors supply one zero mode of each chirality:

$$n_+=n_-=1, \qquad \operatorname{ind}\mathcal D_R^+=0.$$

Index zero therefore does not mean that there are no zero modes. Under a deformation, such an opposite-chirality pair may move away from zero while the index remains unchanged.

The index also does not convert a mathematical kernel directly into a physical amplitude. In a fermionic path integral, normalizability, Grassmann integration, operator insertions, collective coordinates, and the dynamics of a chosen background must still be analyzed. That later chain is developed on Fermion Zero Modes, Index Data, and Selection Rules.

Where the closed-manifold theorem stops

Each omitted hypothesis changes the analytic problem.

Boundary. Integration by parts produces a boundary pairing, and the differential expression does not determine a Fredholm operator until an elliptic domain or boundary condition is chosen. Global spectral boundary conditions lead to additional boundary corrections. The closed formula above must not be applied unchanged.

Noncompact manifold. Ellipticity is local and does not control escape to infinity. One must specify an $L^2$ domain and asymptotic conditions and then prove Fredholmness, perhaps using a coercive or Callias-type argument. An instanton written on $\mathbb R^4$ cannot be inserted into the closed formula without a justified compactification or a separate noncompact index theorem.

Lorentzian signature. A nonzero null covector makes the Lorentzian Dirac symbol singular, so the operator is not elliptic. Hyperbolic propagation, not the elliptic Fredholm theorem above, is the relevant starting point.

Odd dimension. There is no ordinary chirality splitting $\Sigma^+\oplus\Sigma^-$ for the complex spin representation in odd dimension, so there is no ordinary chiral index of this form. Spectral flow and eta-type invariants answer related but different questions.

Non-Fredholm deformation. The index is constant along a continuous path in the Fredholm topology—operator norm for the bounded Sobolev realizations used here. If the range ceases to be closed or essential spectrum reaches zero, the protection argument has left its domain of validity.

Common pitfalls

Assigning the chiral index to the full Dirac operator. The full self-adjoint operator has equal kernel and cokernel dimensions and hence index zero. The nonzero invariant belongs to $\mathcal D_{R,W}^{+}$.

Calling ellipticity alone Fredholmness. Ellipticity is a symbol condition. Closedness of the manifold, the absence of a boundary, and the Sobolev domain are what make the operator here Fredholm.

Reading an index as a total count. The theorem fixes $n_+-n_-$, not $n_++n_-$. Exact one-chirality counts require a separate vanishing theorem or explicit analysis.

Suppressing the chirality convention. With $\mathbf c=i\gamma$, the standard index grading is $\Gamma_{\mathrm{ind}}=(-1)^m\Gamma_E$. Forgetting this conversion reverses two-dimensional signs.

Importing a gauge normalization. The site uses $\mathbf F=-ig_{\mathrm{YM}}F$ and $Q=\int\operatorname{ch}_2=-\langle c_2,[X]\rangle$. Omitting the coupling, trace representation, or orientation can change the displayed integer.

Ignoring the domain on a boundary or at infinity. The same local differential expression can define different closed operators, or no Fredholm operator, on different domains.

Confusing quadratic index data with an anomaly coefficient. $2T(\rho)Q$ is the gauge contribution to this four-dimensional Dirac index. A perturbative chiral gauge anomaly uses different representation data and additional field-theoretic reasoning.

Exercises

1. The shift and its adjoint

For the unilateral shift $S$, compute the kernel, range, cokernel, and index of both $S$ and $S^\dagger$. Why does this not contradict finite-dimensional rank–nullity?

Solution

The shift $S$ is injective and has range

$$\operatorname{ran}S = \{x\in\ell^2:x_0=0\} = (\mathbb C e_0)^\perp.$$

Therefore

$$\ker S=0, \qquad \operatorname{coker}S\simeq\mathbb C e_0, \qquad \operatorname{ind}S=-1.$$

The left shift $S^\dagger$ is surjective and $\ker S^\dagger=\mathbb C e_0$, so

$$\operatorname{ind}S^\dagger=1.$$

A square finite-dimensional matrix maps spaces of the same finite dimension, so rank–nullity forces kernel and cokernel to have equal dimension. Infinite-dimensional Fredholm operators need not satisfy that equality.

2. Translate the chirality label

In dimension $2m$, derive $\Gamma_{\mathrm{ind}}=(-1)^m\Gamma_E$. What happens in dimensions two and four?

Solution

Since $\mathbf c(e_a)=i\gamma^a$,

$$\begin{aligned} \Gamma_{\mathrm{ind}} &= i^m \prod_{a=1}^{2m}\mathbf c(e_a) \\ &= i^m i^{2m} \prod_{a=1}^{2m}\gamma^a \\ &= (-1)^m\Gamma_E. \end{aligned}$$

For $m=1$, the two-dimensional labels are exchanged, so the two definitions of the $+$ block have opposite indices. For $m=2$, the four-dimensional labels agree.

3. Extract the four-form

Starting from

$$\widehat A(TX)=1-\frac{p_1(TX)}{24}+\cdots$$

and

$$\operatorname{ch}(W) = \operatorname{rank}W+c_1(W)+\operatorname{ch}_2(W)+\cdots,$$

derive the twisted Dirac index in four dimensions.

Solution

The $\widehat A$-class has no degree-two term. The degree-four part of the product is therefore

$$\left[ \widehat A(TX)\operatorname{ch}(W) \right]_4 = \operatorname{ch}_2(W) - \frac{\operatorname{rank}W}{24}p_1(TX).$$

Integrating gives

$$\operatorname{ind}\mathcal D_{R,W}^{+} = \int_X \left[ \operatorname{ch}_2(W) - \frac{\operatorname{rank}W}{24}p_1(TX) \right].$$

The $c_1(W)$ term has degree two and has nothing of degree two in $\widehat A(TX)$ with which to combine.

4. QFT transfer without overclaim

Let $X=S^4$, let the gauge group be $SU(N)$, and let a fermion transform in a representation $\rho$. If the background has charge $Q$, what does the theorem establish? Name two further conclusions it does not establish.

Solution

Since $\int_{S^4}p_1(TS^4)=0$,

$$\operatorname{ind}\mathcal D_{R,\rho}^{+} = n_+-n_- = 2T(\rho)Q.$$

The theorem fixes the net Euclidean chirality and implies at least $\lvert2T(\rho)Q\rvert$ zero modes when this integer is nonzero. It does not determine $n_+$ and $n_-$ separately without a vanishing theorem. It also does not by itself compute a path-integral selection rule or a perturbative gauge-anomaly coefficient.

Synthesis and continuations

The analytic-to-topological bridge has four steps. A closed operator is Fredholm when its kernel and cokernel are finite and its range is closed. Ellipticity on a closed manifold supplies that Fredholm property after Sobolev completion. The principal symbol defines a $K$-theory class, and Atiyah–Singer equates its topological index with the analytic index. For a twisted chiral Dirac operator, the resulting characteristic number is $\int\widehat A\,\operatorname{ch}$, which fixes a stable net chirality of zero modes.

Continue according to the question:

• Characteristic Classes and Chern–Weil Theory supplies the characteristic forms and the gauge normalization used in the index density;
• Perturbative Chiral and Gauge Anomalies develops the field-theoretic anomaly calculation and its distinct representation coefficients;
• Fermion Zero Modes, Index Data, and Selection Rules turns normalizable zero modes into Grassmann-saturation conditions and physical selection rules.

References

• Michael F. Atiyah and Isadore M. Singer, “The Index of Elliptic Operators I”, Annals of Mathematics 87 (1968), 484–530. This is the original paper proving the equality of the analytic and topological indices.
• Xianzhe Dai, Lectures on Dirac Operators and Index Theory — Open PDF, lecture notes, 2015, § 3.1, pp. 35–39, §§ 3.3–3.4, pp. 50–54, and § 3.6, pp. 63–71. These sections support the Fredholm and parametrix discussion, stability, the chiral decomposition, the McKean–Singer supertrace, the twisted Dirac index formula, and the local heat-kernel proof. Dai uses Clifford multiplication squaring to $-g$, which is the page’s $\mathbf c=i\gamma$ convention.
• Mikio Nakahara, Geometry, Topology and Physics, second edition, Institute of Physics Publishing, 2003, Chapter 12, §§ 12.1–12.2, pp. 453–460, and § 12.6, pp. 468–472. This provides a physics-oriented account of analytic and topological indices, Euclidean chirality, the twisted spin index formula, and two- and four-dimensional gauge examples. The body explicitly translates its chirality, curvature, trace, and coupling conventions to those fixed on this site.