The Atiyah–Singer index formula
Identify analytic and topological indices and compute the appropriate characteristic pairing for spin and Dolbeault operators.
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The question for today
What you’ll understand
- State the closed-manifold index theorem without confusing its symbol and bundle inputs.
- Use normalized Chern–Weil classes in the twisted spin density.
- Compute a four-dimensional twisted index and a holomorphic line-bundle Euler characteristic.
The Atiyah–Singer theorem identifies the Fredholm index of an elliptic operator with the topological pushforward of its symbol class. A characteristic formula then makes that topological answer computable. The polynomial depends on the operator: the spin Dirac, de Rham and Dolbeault constructions do not all use the same characteristic density. A correct calculation therefore begins by naming the operator, its grading, the manifold's orientation and the geometric structure that makes the selected formula applicable.
The general theorem is a symbol theorem
For an elliptic differential operator between smooth complex bundles on a smooth compact manifold without boundary, the analytic index of its Sobolev realization equals the topological index of its class in K⁰_c(T*M). The topological map is the embedding, Thom and Bott construction of the previous lesson. No spin assumption is needed for this general statement. A spin assumption enters only when the operator is specialized to the spin Dirac construction and its density is simplified accordingly.
The theorem determines a difference of kernel dimensions, not the separate kernels or a complete eigenvalue list. It is unchanged by lower-order modifications that preserve the symbol class, although the actual solutions may change. Its proof requires substantial elliptic analysis and topology; computing a characteristic integral in one example verifies the consequences and normalization, rather than furnishing a general proof. Boundary or noncompact problems require additional hypotheses and correction terms that this closed-manifold version does not contain.
Normalize the characteristic forms
Use a unitary connection with skew-Hermitian curvature F_E and normalize the Chern character as tr exp(iF_E/(2π)). For the real tangent curvature matrix R, take p₁=−tr(R∧R)/(8π²). These choices give the  expansion starting with one minus p₁ divided by twenty-four. Only the top-degree component of the product is integrated. All lower-degree terms matter through multiplication, but terms above the manifold dimension make no contribution.
In real dimension four, the twisted spin integrand is one half of c₁(E)² minus c₂(E), minus rank(E) times p₁(TM)/24. Characteristic numbers may be supplied topologically or computed from normalized curvature forms. Changing the connection changes the representative by an exact form and leaves the integral unchanged on a closed manifold. The rational coefficients do not make the final answer an arbitrary rational number: the theorem imposes integrality for the stated geometric data.
Other operators have other densities
For the Dolbeault operator on a compact complex manifold, rolled from even to odd antiholomorphic degrees and twisted by a holomorphic bundle E, the index is the integral of Td(T^(1,0)M) times ch(E). This is the holomorphic Euler characteristic. A Kähler metric is convenient for many analytic identities but is not required for this Dolbeault index formula. On a complex curve the Todd class truncates to one plus half the first Chern class of the tangent line.
The de Rham parity operator instead pairs the Euler class with the fundamental class. A spin-c Dirac formula includes the exponential of half its determinant-line class along with Â. These variants explain why an oriented manifold with no spin lift can still have an index theorem for other elliptic operators. They also prevent a common mistake: replacing every occurrence of Todd or Euler by  simply because all three appear in the broader subject of index theory.
Work it through
A rank-two twist on a K3 surface
Take a K3 surface with its complex orientation, ∫p₁(TM)=−48, and a rank-two Hermitian bundle E with c₁(E)=0 and ∫c₂(E)=3. Compute the twisted spin Dirac index.
A K3 surface is spin; use the standard positive chirality from its complex orientation. The degree-four Chern character of E is (c₁²−2c₂)/2=−c₂, with integral minus three.
The rank contribution from  is −2p₁/24. Its integral is −2·(−48)/24=4. Higher characteristic terms have degree above four and do not contribute.
Adding the two contributions gives ind D_E⁺=−3+4=1. Thus the positive kernel dimension exceeds the negative one by one; the formula does not determine either dimension separately.
The twisted index is one. The untwisted  number would be two, so discarding the rank or twisting term would give the wrong result.
The living notebook
Explore the model
Interactive lab · function
Pair a density with an orientation
Pause and Reset, then set coefficient a = 1 and position x = 1: the signed area ∫₀¹ax²dx is 1/3. Reset and set both a = 1 and x = −1: it is −1/3. Predict the sign change from reversed integration limits. This is an ordinary one-dimensional integral illustrating oriented pairing, not a curvature form, a Chern number or a numerical index solver.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forA characteristic density needs the right normalization and orientation before its integral can be interpreted.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let L be a holomorphic line bundle of degree seven on a compact Riemann surface of genus three. Compute its Dolbeault index, then show that H¹(L)=0 and determine h⁰(L).
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The curve Todd contribution integrates to 1−g. For the separate cohomology dimensions, use Serre duality and the degree of K⊗L⁻¹.
Compare with the solution
In complex dimension one, the top product is c₁(L)+c₁(T)/2. Its integral is 7+(2−2·3)/2=5, so h⁰(L)−h¹(L)=5. The canonical line has degree 2g−2=4, hence K⊗L⁻¹ has degree −3. A nonzero holomorphic section of a line bundle would have an effective zero divisor whose degree equals the line degree; a negative degree makes that impossible. Serre duality therefore gives h¹(L)=h⁰(K⊗L⁻¹)=0, and h⁰(L)=5. The last equality uses this additional vanishing argument, not the index formula alone.
Keep the idea
Take it with you
- The general theorem identifies the analytic index with the topological symbol index.
- Spin, spin-c, de Rham and Dolbeault operators require their own characteristic densities.
- An index gives a difference; separate kernel counts require additional arguments.
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