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Getzler rescaling combines the change $x\mapsto \sqrt{t}x$ in geodesic normal coordinates with the Clifford algebra filtration, assigning degree one to a derivative and to a Clifford generator and degree minus one to a coordinate. The associated graded Clifford algebra is the exterior algebra. This is useful because the spinor supertrace annihilates every Clifford monomial except the top-degree one. Applied to the Lichnerowicz formula for a twisted Dirac operator, it retains a quadratic curvature model, whose heat kernel is computed by a Gaussian ansatz; all terms of smaller rescaling degree disappear from the local index limit. Thus the cancellations in a supersymmetric heat trace can be computed before solving the complete curved heat equation.

Use the round unit sphere with its complex orientation and unique spin structure. Its chiral spinor bundles are $S^{+}=\mathcal{O}(-1)$ and $S^{-}={\Lambda }^{0,1}\otimes \mathcal{O}(-1)$, the latter a smooth line bundle of degree $+1$. The Dirac operator isHere Clifford multiplication has $c(v{)}^2=-|v{|}^2$, and $D$ acts self-adjointly on the L2 spinors with first-order Sobolev domain. The geometric spin connection is invariant under the lifted $SU(2)$ action, hence $D$ is equivariant.

We justify the spectral properties needed for its phase. The Peter-Weyl theorem and Frobenius reciprocity for compact groups give the decomposition of sections of an isotropy-weight-$q$ line bundle: $E_n$ occurs once exactly when $n\ge |q|$ and $n\equiv q(\mathrm{mod}2)$, because its torus weights are $n,n-2,\dots ,-n$. The spinor isotropy weights are $+1,-1$, so each chirality contains $E_1,E_3,E_5,\dots$, each once. The horizontal connection Laplacian is the total Casimir minus the vertical generator square, ${\nabla }^{\ast }\nabla =C-1/4$ on these weight-one bundles. Expanding the Dirac square in a normal orthonormal frame gives the diagonal connection-Laplacian terms and the off-diagonal curvature contraction. The Clifford relations contract spin curvature to $\mathrm{Scal}/4$, the Lichnerowicz formula for a twisted Dirac operator with no twisting bundle. Since the scalar curvature here is two,Thus there is no kernel, the eigenvalues of $D$ are $\pm 1,\pm 2,\dots$ with finite multiplicities, and the resolvent is compact by spectral truncation. These round-sphere spectral facts agree with the explicit calculation in Abrikosov's paper.

Define the chiral phase of the round-sphere Dirac operatorIt is unitary because both chiral kernels vanish; equivariance follows from that of $D$. The full phase is $F=D|D{|}^{-1}=\left(\begin{array}{cc}0& V^{\ast }\\ V& 0\end{array}\right)$.

For a smooth scalar $f$, the Leibniz rule gives $[D,M_f]=c(df)$, a bounded bundle endomorphism. The spectral integral for the sign operator gives, in the strong topology,Write $R_{\pm }=(D\mp it{)}^{-1}$. Since $2D(D^2+t^2{)}^{-1}=R_{+}+R_{-}$ and $[R_{\pm },M_f]=-R_{\pm }[D,M_f]R_{\pm }$, the commutator integral is norm convergent: its integrands are compact and bounded by $\Vert [D,M_f]\Vert /(1+t^2)$. A norm limit of their integrals is compact. Thus $[F,M_f]$ is compact, and its off-diagonal blocks say $V{M}_f^{+}-M_f^{-}V$ is compact. Uniform approximation of continuous $f$ by smooth functions extends this to all $f\in C(S^2)$, using the bound $2\Vert f-g{\Vert }_{\infty }$ for the change of commutator.

The chiral phase is an equivariant unitary which intertwines the two multiplication representations modulo compacts. This specifies the grading implicit in the subsequent index question; treating the full self-adjoint phase as an endomorphism would yield a different, zero, compressed index.

Let $M$ be a closed manifold of even dimension $n=2m$, with a Riemannian metric and a spin structure. Let $S=S^{+}\oplus S^{-}$ be its spinor bundle, and let $W$ be a Hermitian vector bundle with a unitary connection of vector-bundle curvature $F^W=({\nabla }^W{)}^2$. We use Clifford multiplication satisfying $c(v)c(w)+c(w)c(v)=-2⟨v,w⟩$, so the Twisted Dirac operatoris self-adjoint and $D_W^{-}=(D_W^{+}{)}^{\ast }$. Here $(e_i)$ is an oriented local orthonormal basis and $(e^i)$ its dual. The Atiyah-Singer index theorem in this setting isThe second factor is the Chern character; the first is the A-hat formThe square root has constant term one. All functions of the vector-bundle curvature matrices mean power series truncated above the dimension of $M$. The brackets select the top-degree differential form. The Bianchi identity shows that these forms are closed, and Chern-Weil connection transgression shows that their de Rham cohomology classes do not depend on the chosen connections on a vector bundle.

The global step is supersymmetric quantum mechanics. On the Hilbert space $L^2(S\otimes W)$, take the grading operator $\Gamma$, the odd supercharge $D_W$, and the nonnegative Hamiltonian $H=D_W^2$. They obey $\Gamma D_W=-D_W\Gamma$ and $D_W^2=H$. The principal symbol of $D_W$ is invertible away from the zero covector, so elliptic regularity and Rellich-Kondrachov compactness theorem give a smooth discrete orthonormal eigenbasis for $H$, finite-dimensional eigenspaces, and a trace-class operator $e^{-tH}$ for $t>0$. Define its supertrace by $\mathrm{Str}A=\mathrm{Tr}(\Gamma A)$.

For every positive eigenvalue $\lambda$ of $H$, the map ${\lambda }^{-1/2}{D}_W$ is an isomorphism between its positive and negative grading eigenspaces, with inverse itself. Their contributions to the supertrace cancel. The zero eigenspaces are $\ker D_W^{+}$ and $\ker D_W^{-}$. ConsequentlyThis also proves independence of $t$ without differentiating an unbounded operator inside a trace. If $K_t$ is the heat kernel, the heat kernel trace formula givesThe remaining task is to compute this local integrand as $t\downarrow 0$.

Fix $p\in M$, use geodesic normal coordinates $x$, and trivialize $S\otimes W$ by radial parallel transport. Expanding $D_W^2$ at $p$ gives the Lichnerowicz formula for a twisted Dirac operator:Indeed the terms with identical indices give the rough Laplacian. The other terms are ${\sum }_{i<j}c(e^i)c(e^j)[{\nabla }_i,{\nabla }_j]$. Decompose this vector-bundle curvature into spin curvature and $F^W$; substituting the spin curvature into the Clifford algebra relations and the algebraic Bianchi identity leaves $\mathrm{Scal}/4$. Thus the calculation retains the curvature of the twisting bundle explicitly.

There is an especially strong cancellation in the spinor supertrace. With $\Gamma =i^{m}c(e^1)\cdots c(e^{2m})$, every ordered product of distinct Clifford generators of degree less than $2m$ has zero supertrace, whileFor an even product of lower degree, conjugate $\Gamma$ times that product by a missing Clifford generator: it changes sign, so its matrix trace is zero. An odd product exchanges the grading summands and also has zero supertrace. For the full product, $(c(e^1)\cdots c(e^{2m}){)}^2=(-1{)}^m$, and $\dim S=2^m$, giving the displayed constant.

To exploit this, identify the Clifford algebra as a filtered vector space with the exterior algebra by sending an ordered Clifford product to the corresponding wedge product of differential forms. Implement Getzler rescaling: put $x=\epsilon y$, $t={\epsilon }^{2}s$, multiply the degree-$k$ exterior component of the kernel by ${\epsilon }^{-k}$, and multiply the whole kernel by ${\epsilon }^{2m}$. A derivative and a Clifford generator have rescaling degree one; a coordinate has degree minus one. Clifford contractions lower the degree by two relative to the wedge product of differential forms, so the leading product becomes exterior multiplication.

In the radially parallel frame, the linear Taylor term of the connection is determined by its vector-bundle curvature: if $A_i$ is its matrix, then $A_i(x)=\frac{1}{2}{x}^j{F}_{ji}(p)+O(|x{|}^2)$. To check the coefficient, radial gauge means $x^i{A}_i=0$, and integration of $x^j{F}_{ji}$ along the radial segment gives $A_i(x)={\int }_0^{1}s{x}^j{F}_{ji}(sx)ds$. Combining this formula with the spin representation, the symmetries of Riemann curvature tensor, and the Lichnerowicz formula for a twisted Dirac operator, the rescaled square has leading modelHere $\mathcal{R}$ is the skew matrix of tangent-bundle curvature two-forms at $p$; changing the curvature-matrix sign convention changes the sign inside the covariant derivatives but not the determinant below. The coefficients act in $\Lambda T_p^{\ast }M\otimes \mathrm{End}(W_p)$. The scalar-curvature term has degree zero and disappears after multiplication by ${\epsilon }^2$; the linear spin-connection term and the Clifford action of $F^W$ have degree two and survive. Higher Taylor terms have smaller degree and disappear.

For clarity, the analytic meaning of this limiting calculation is the following. Construct the local heat kernel expansion to any fixed order and insert the Taylor expansion of the rescaled coefficients into its transport equations. At each exterior degree, the leading equation is the heat equation of ${\mathcal{H}}_p$; the remaining terms acquire positive powers of $\epsilon$. Equivalently, successive substitutions ingive the same expansion. Polynomial spatial factors are integrable against the local Gaussian kernels, and only finitely many exterior degrees exist. A cutoff outside a fixed normal neighbourhood contributes $O(e^{-c/t})$. Choosing the initial expansion order arbitrarily large bounds its differentiated remainder to any required order. This proves convergence of the rescaled diagonal kernel to the model diagonal kernel, uniformly in $p$ on the compact manifold. Its degree-$2m$ component has no net power of $\epsilon$, precisely the component detected by the spinor supertrace.

We can compute the model kernel rather than invoke a local index formula. Set $F=F_p^W$ and seek its kernel from the origin in the formThe scalar even exterior coefficients of $\mathcal{R}$ commute. On this ansatz the rotational first-order term vanishes because $Q$ is a function of ${\mathcal{R}}^2$. Expanding the model operator gives $-\sum {\partial }_i^2+\frac{1}{2}(\mathcal{R}y)\cdot \nabla +\frac{1}{16}{y}^T{\mathcal{R}}^{2}y+F$, so substitution into the heat equation givesThe Euclidean initial condition $Q(s)\sim s^{-1}I$, $a(s)\sim (4\pi s{)}^{-m}$ selectsThese expressions have well-defined power series at $\mathcal{R}=0$. This is the curvature version of the Gaussian calculation underlying the Mehler formula for Hermite functions.

At $s=1$, $y=0$, multiply the top exterior coefficient by the spinor supertrace constant. Since $(4\pi {)}^{-m}(-2i{)}^m=(2\pi i{)}^{-m}$, and every curvature factor is a two-form, the result isIntegration and the constant supertrace established above prove the boxed Atiyah-Singer index theorem. The mechanism is that odd symmetry pairs the nonzero spectrum globally, while the Clifford filtration makes the surviving local contribution an explicitly solvable curvature oscillator. Odd-dimensional spin Dirac operators have no canonical chiral index of this type; their self-adjoint Fredholm index is zero. For comparison of normalization conventions, Raphaël Ponge's local index calculation writes the same unnormalized curvature model and spinor supertrace constant.