Heat supertraces and local index density
Pair positive chiral spectra and distinguish a constant integrated supertrace from a generally varying pointwise heat kernel.
On this page
The question for today
What you’ll understand
- Derive the McKean–Singer identity from positive-eigenvalue pairing.
- State the local spin Dirac index limit with a compatible connection.
- Check exact finite-dimensional cancellation and identify an unmatched spectral-cutoff artifact.
Heat flow reveals the index by suppressing positive eigenvalues, but a stronger cancellation makes the answer independent of heat time from the beginning. Positive modes in the two chiralities occur in matched pairs; only unpaired zero modes survive in a supertrace. The small-time heat kernel then expresses the same global integer using a local curvature density. This bridge requires analytic control of kernels and asymptotics, not merely the observation that exponential factors decay.
Positive eigenvalues cancel between chiralities
Let D be a self-adjoint odd Dirac operator on a closed even-dimensional manifold, with D⁻ the adjoint of D⁺. The nonnegative operators D⁻D⁺ and D⁺D⁻ have discrete spectra, and their heat operators are trace class at every positive time. If a positive eigenvalue λ occurs on the plus side, applying D⁺ gives a nonzero minus eigenvector with the same λ. The inverse on this eigenspace is λ⁻¹D⁻, so multiplicities agree exactly.
Subtracting the two heat traces therefore cancels every positive eigenvalue contribution. At eigenvalue zero the two dimensions need not agree; their difference is the index. This proves the integrated supertrace identity for every positive time, once trace-class convergence and spectral pairing are justified. A trace of a single chirality has no such constancy and usually diverges as time approaches zero. The parity sign is indispensable to the cancellation.
The local theorem identifies what survives
For the spin Dirac operator twisted by a Hermitian bundle with compatible unitary connection, the diagonal supertrace heat density tends to the top-degree  times Chern-character form as time tends to zero. The chirality, curvature normalization and oriented density are those fixed earlier. Individual diagonal traces have singular small-time expansions, but Clifford-supertrace cancellations remove their lower terms. Getzler rescaling isolates a curvature-dependent model that computes the surviving coefficient.
The pointwise heat supertrace density need not be independent of time even though its integral is. Likewise, the global McKean–Singer identity alone does not identify a local polynomial: different pointwise functions can have the same integral. The local index theorem is the stronger geometric statement supplying that identification. It is being stated here under its spin Dirac hypotheses; an arbitrary elliptic differential operator need not have the same pointwise limit or the same characteristic density.
Regularization must preserve the pairing
A finite-dimensional graded map exhibits the cancellation without any convergence issues. Its two squared singular-value lists agree away from zero, so the two ordinary traces may vary while their difference remains constant. In an infinite-dimensional calculation, separately truncating the two spectra at incompatible modes can introduce a false remainder. A valid approximation must preserve the paired spaces or prove that the unmatched error vanishes in the limit under consideration.
Large time and small time serve different purposes. Large time projects toward zero modes and makes the analytic index visible. Small time localizes the heat kernel and exposes curvature. Exchanging these limits with a trace, integral or cutoff requires the relevant estimates, not a formal substitution of zero into a divergent series. On manifolds with boundary a spectral boundary condition changes the heat analysis and produces a boundary correction; the closed formula cannot simply be reused there.
Work it through
Two changing traces, one fixed difference
Let A:C³→C² be the matrix with rows (1,0,0) and (0,2,0). Regard A as the positive-to-negative block of an odd operator. Compute its index and both heat traces.
The kernel of A is spanned by the third coordinate, while A is surjective. Thus its index is one. Direct multiplication gives A*A=diag(1,4,0) and AA*=diag(1,4).
The plus heat trace is exp(−t)+exp(−4t)+1 and the minus heat trace is exp(−t)+exp(−4t). Their positive-mode factors are identical even though each trace changes with t.
Subtracting leaves one at every t>0, equal to the index. Replacing the entry two by three changes the paired factors exp(−4t) to exp(−9t) in both traces and leaves the same answer.
Pairing cancels all nonzero squared singular values. The surviving constant counts an unpaired positive-chirality zero mode.
The living notebook
Explore the model
Interactive lab · heat
Inspect individual heat factors
Pause and Reset, then set mode n = 1 and diffusivity κ = 1. After two Steps, t=1, the amplitude is e^(−1). Reset and set both n = 2 and κ = 1, then take two Steps: it is e^(−4). These match two factors in the worked example. The instrument displays one scalar mode at a time, not two graded spectra, a heat trace or a curvature density; the supertrace cancellation must be checked algebraically.
- Current amplitude
- 1
- Spatial L² norm
- 1.772
- Decay rate · κn²
- 3.6 /s
The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forDecay factors explain individual modes, while a supertrace requires an additional parity pairing.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the de Rham operator on the circle of length 2π, compare the complete heat traces on functions and one-forms. Then truncate functions to |n|≤N and one-forms to |n|≤N−1, for N≥1. Compute the false finite difference and explain its origin.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Both complete Laplacians have eigenvalues n² indexed by all integers. In the mismatched truncation exactly two function modes remain unpaired.
Compare with the solution
The full function and one-form heat traces are each Σ_(n in Z)exp(−tn²), convergent for positive t, so their difference is zero, agreeing with b₀−b₁=1−1. The stated finite truncation leaves the two function modes n=±N and gives 2exp(−tN²). That is not an index: it depends on t and comes solely from a cutoff that breaks the positive spectral pairing. For fixed positive t it tends to zero as N grows, but for fixed N its small-time limit is two. Thus the order of cutoff and heat-time limits can matter, and the unmatched truncation does not supply a local index theorem.
Keep the idea
Take it with you
- Positive chiral spectra pair, making the integrated supertrace independent of heat time.
- The local index theorem computes a curvature density beyond global cancellation.
- Spectral cutoffs must preserve pairing or come with an explicit error and limit argument.
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