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A Hermitian metric on a holomorphic vector bundle $E\to X$ is a smoothly varying positive-definite Hermitian form $h_x$ on every complex fiber $E_x$.

Such a metric makes the underlying smooth bundle a Hermitian vector bundle.

Fix the input-index conventionExtend the connection on a vector bundle by the graded Leibniz rule, $D(\alpha \otimes s)=d\alpha \otimes s+(-1{)}^{\mathrm{deg}\alpha }\alpha \wedge Ds$. Applying it twice givessoThis is the Cartan curvature equation with input indices. In the more usual coefficient-column convention $A={\theta }^t$, and $(A\wedge A{)}^t=-\theta \wedge \theta$ because the entries have degree one. Thus $F=dA+A\wedge A={\Theta }^t$; the sign difference is entirely conventional.

The Chern connection of a Hermitian vector bundle with holomorphic structure is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. To construct it, use a holomorphic local frame and let $h_{ij}=⟨e_i,e_j⟩$, with the Hermitian inner product linear in the first variable. The second defining property makes $\theta$ type $(1,0)$. The first givesand therefore the only possible matrix isConversely, this formula has type $(1,0)$ and satisfies both type components of the metric identity, because $h={\overline{h}}^t$. Under a holomorphic frame change $e^{\prime }=ge$, $h^{\prime }=gh{\overline{g}}^t$ and direct differentiation givesThis is precisely the transformation obtained by differentiating $e^{\prime }=ge$. The locally constructed connections glue, proving existence, and the forced formula proves uniqueness of the Chern connection.

The derivative of $h^{-1}$ gives $\partial \theta =\theta \wedge \theta$. Consequently the Chern curvature isThe second assertion is ${\bar{\partial }}^2\theta =0$ in a holomorphic local frame; holomorphic transition functions make it intrinsic. Thus the curvature Dolbeault class is $\Psi =[\Theta ]\in H^1(M,{\Omega }_M^1\otimes \mathrm{End}E)$ by the Dolbeault theorem. Here ${\Omega }^1(\mathrm{End}E)$ denotes this tensor product. We use the identification with bundle-valued Dolbeault forms in which the antiholomorphic exterior factor is placed first, so $\bar{\partial }\theta$ has the stated positive descent sign.

For the stated transitions $e^{(\beta )}=g_{\alpha \beta }{e}^{(\alpha )}$, differentiating the frame relation yieldsTransport the alpha matrix to the beta frame before subtracting it. The difference is the holomorphic sectionOn a triple intersection, $g_{\alpha \gamma }=g_{\beta \gamma }{g}_{\alpha \beta }$, soThis is exactly the Čech cocycle condition after transporting all representatives to the gamma frame.

To identify its class, let $D_{\alpha }$ be the local holomorphic connection making the alpha frame parallel. Then $t_{\alpha }=D-D_{\alpha }$ is a genuine local section of $A^{1,0}(\mathrm{End}E)$, represented by ${\theta }_{\alpha }$, with $\bar{\partial }{t}_{\alpha }=\Theta$. Its Čech coboundary, using $t_{\beta }-t_{\alpha }$, is the intrinsic section represented by ${\sigma }_{\alpha \beta }$ in the beta frame. This local-primitive construction is the Čech–Dolbeault identification, and proves that $({\sigma }_{\alpha \beta })$ maps to $\Psi$. For an arbitrary trivializing cover use its natural map to sheaf cohomology; a Stein refinement gives the usual Čech computation. The resulting object is the Čech cocycle for the curvature Dolbeault class.

If the metric changes, the difference $B=D^{\prime }-D$ is a global smooth $(1,0)$ form with values in $\mathrm{End}E$. The local curvature formulas give ${\Theta }^{\prime }-\Theta =\bar{\partial }B$, so $\Psi$ is metric independent. If holomorphic frames change by $e_{\alpha }^{\prime }=u_{\alpha }{e}_{\alpha }$, the corresponding local tensors $t_{\alpha }^{\prime }-t_{\alpha }$, expressed in the old frame, are $B_{\alpha }=u_{\alpha }^{-1}\partial u_{\alpha }$. They are holomorphic, and their differences change the cocycle by $\delta B$. Hence $\Psi$ depends only on the holomorphic bundle, not on its metric or trivializations.

Apply exterior product and endomorphism composition to curvature to defineThese are closed for the Dolbeault operator by the product rule. Their independence can also be verified without commuting matrices. If ${\Theta }_1-{\Theta }_0=\bar{\partial }B$, thenThis follows by the noncommutative telescoping identity and the even total degree of curvature. The matrix products are interpreted in the fixed frame convention; powers of the degree-two curvature transpose to the corresponding usual curvature powers. The trace contraction is invariant under frame changes, so its class is well defined. This proves metric independence of curvature Dolbeault powers; for $k>\dim M$ the forms and groups in question vanish.

Finally take a smooth unitary frame. Metric compatibility gives ${\theta }^{†}=-\theta$, and the Cartan curvature equation with input indices gives ${\Theta }^{†}=-\Theta$. ThusThe second expression is therefore a real differential form. It is also $d$-closed: differentiating Cartan's equation gives $d\Theta =\theta \wedge \Theta -\Theta \wedge \theta$, and cyclicity of trace cancels the resulting commutator in $d\mathrm{tr}({\Theta }^k)$. On a compact Kähler manifold, the Hodge decomposition theorem for compact Kähler manifolds identifies $H^k(M,{\Omega }_M^k)$ with the $(k,k)$ component of complex de Rham cohomology. HenceIt is $k!$ times the degree-$2k$ component of the Chern character. Reality does not assert integrality of these individual power-sum classes.

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.

If $S$ is a holomorphic subbundle of a Hermitian vector bundle $E$, its induced Chern connection is $D_S={\pi }_S{D}_E$ on sections of $S$. Orthogonal projection preserves metric compatibility, and holomorphicity of $S$ preserves the required $(0,1)$ operator. Uniqueness of the Chern connection therefore identifies the projected connection.

A smooth self-adjoint projection $p:M\to M_N(\mathbb{C})$ of locally constant rank defines the Hermitian vector bundle $\mathrm{im}p$ inside the trivial bundle. Projecting the flat derivative gives the unitary connection $\nabla s=pds$ for $ps=s$. Differentiating $p^2=p$ gives $pdp=dp(1-p)$ and $pdpp=0$. For a section $s$, ${\nabla }^{2}s=pdp\wedge ds=p(dp{)}^{2}s$, proving the curvature formula. Relative to $\mathrm{im}p\oplus \ker p$, $dp$ is off-diagonal and $(dp{)}^2$ is diagonal. Therefore $d\mathrm{tr}(p(dp{)}^{2k})=\mathrm{tr}(dp{)}^{2k+1}=0$. These are explicit closed representatives of the Chern character.

For a smooth subbundle $S$ of a Hermitian vector bundle $E$, orthogonal projection identifies $E/S$ smoothly with $S^{\perp }$. The quotient metric is $h_Q([v],[w])=h_E((I-{\pi }_S)v,(I-{\pi }_S)w)$. It is independent of representatives and positive definite. For a holomorphic subbundle the resulting smooth quotient metric is defined on the holomorphic quotient, but the orthogonal identification itself need not be holomorphic.

On a spin manifold, tensor the spinor bundle $S$ with a Hermitian vector bundle $W$ carrying a unitary connection. Composing the induced tensor-product connection on a vector bundle with Clifford multiplication defines the twisted Dirac operator. In even dimension it maps $S^{+}\otimes W$ to $S^{-}\otimes W$ and conversely. On a closed manifold its chiral restriction $D_W^{+}:H^1(S^{+}\otimes W)\to L^2(S^{-}\otimes W)$ is a Fredholm operator.