Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 7 (Atiyah Singer Index Theorem)

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.

Use ${\mathbb{T}}^n={\mathbb{R}}^n/(2\pi \mathbb{Z}{)}^n$ and the periodic Sobolev space conventionThe Cauchy-Schwarz inequality givesTo check the last sum, partition the lattice into $|k|<1$ and dyadic shells $2^j\le |k|<2^{j+1}$. A shell contains at most $C_n{2}^{jn}$ points and contributes at most $C_n{2}^{j(n-2s)}$. This is a convergent geometric series exactly when $s>n/2$.

Thus the Fourier series ${\sum }_k\hat{f}(k)e^{ik\cdot x}$ converges absolutely and uniformly on the torus. Its sum is a continuous function and has the original Fourier coefficients, so it represents the original periodic distribution. In particular the representative is unique andThis proves the requested inclusion, with its continuous Sobolev embedding estimate.

Let $V$ be a finite-dimensional real or complex inner product space, and let $A\subseteq \mathrm{End}(V)$ be a unital subalgebra stable under the adjoint operator. Define its commutant of an operator algebra by $A^{\prime }=\{T:Ta=aT\text{ for all }a\in A\}$. The double commutant theorem for star-algebras asserts $\boxed{A^{\prime \prime }=A}$. The inclusion $A\subseteq A^{\prime \prime }$ follows immediately from the definition; we prove the reverse.

Write $N=\dim V$, fix a basis $v_1,\dots ,v_N$, and act diagonally on $V^N$. The linear subspaceis invariant under every diagonal $a$. Since $A$ is closed under adjoints, its orthogonal complement $W^{\perp }$ is invariant too: $⟨aw,z⟩=⟨w,a^{\ast }z⟩=0$ for $w\in W^{\perp }$, $z\in W$. Hence the orthogonal projection $P$ onto $W$ commutes with every diagonal $a$.

Writing $P$ as an $N$ by $N$ block matrix $(P_{ij})$, this says $P_{ij}a=a{P}_{ij}$, so each $P_{ij}\in A^{\prime }$. If $b\in A^{\prime \prime }$, it commutes with every $P_{ij}$, and its diagonal action therefore commutes with $P$ and preserves $W$. Because $1\in A$, $(v_1,\dots ,v_N)\in W$. Therefore there is one $a\in A$ such thatEquality on the basis gives $b=a$, proving the theorem over either scalar field.

The identity hypothesis is necessary. For a possibly nonunital adjoint-stable subalgebra, adjoining the ambient identity does not change its commutant of an operator algebra, so the exact statement is $A^{\prime \prime }=A+\mathbb{F}I$. For example $A=\{0\}$ has $A^{\prime \prime }=\mathbb{F}I$ on a nonzero $V$. Thus the usual exam formulation takes the algebra to contain the identity.

At a fixed $t$, write $A=f(t)$ and $B=f^{\prime }(t)$. Differentiating the matrix exponential in the direction $B$ givesThe order of factors matters because matrix multiplication need not be commutative.

To prove the derivative of the matrix exponential formula, expand its power series. The coefficient linear in $h$ in $(A+hB{)}^k$ is ${\sum }_{j=0}^{k-1}{A}^{j}B{A}^{k-1-j}$. On bounded sets of matrices, the exponential series and its directional derivative series converge uniformly: the latter is dominated in operator norm by $\Vert B\Vert {\sum }_{k\ge 1}k\Vert A{\Vert }^{k-1}/k!$. Terms with two or more $hB$ factors have sum $O(h^2)$ locally. Hence the derivative isOn the other hand, the integral in the box expands absolutely asThe integral identity follows by repeated integration by parts. Grouping by $a+b=k-1$ gives the same derivative. Finally differentiability means $f(t+h)=A+hB+o(h)$; the locally bounded derivative of the exponential sends the $o(h)$ error to $o(h)$, proving the chain rule even when $f^{\prime }$ is not continuous. If $[f,f^{\prime }]=0$, the formula simplifies to $e^f{f}^{\prime }$.

Let $e(\omega )$ denote exterior multiplication, $e(\omega )\alpha =\omega \wedge \alpha$. For a $q$-form $\alpha$, evaluation of the proposed operator on vector fields $Y_0,\dots ,Y_q$ givessince ${\sum }_i{\omega }_i(Y_a)X_i=Y_a$. This expression is independent of the chosen local basis.

Expand each covariant derivative into the derivative of $\alpha (Y_0,\dots ,{\hat{Y}}_a,\dots ,Y_q)$ minus the terms obtained by differentiating each argument. For every pair $a<b$, those argument terms combine intoThe Levi-Civita connection is a torsion-free connection, so the vector in this expression is $[Y_a,Y_b]$. The result is exactly the defining evaluation formula for the exterior derivative:Therefore $\boxed{d=\sum _{i}e({\omega }_i){\nabla }_{X_i}}$. No orthonormality or compactness is needed for this identity; the dual-frame property and vanishing torsion tensor suffice.

Suppose first that the bounded Fredholm operator $T:H\to H$ has Fredholm index zero. Put $N=\ker T$ and $C=(\mathrm{ran}T{)}^{\perp }=\ker T^{\ast }$. These finite-dimensional linear subspaces have the same dimension. The restricted map $T:N^{\perp }\to \mathrm{ran}T$ is a bounded bijection, with bounded inverse by the bounded inverse theorem. Choose an isomorphism $J:N\to C$ and extend it by zero on $N^{\perp }$, writing the resulting finite-rank operator as $K$.

Relative to the orthogonal decompositions $H=N\oplus N^{\perp }$ and $H=C\oplus \mathrm{ran}T$, the operator $T+K$ is the direct sum of $J$ and the invertible restricted map. Thus $T+K$ is invertible, and $T=(T+K)-K$ is the sum of an invertible operator and a compact operator.

Conversely write $T=U+K$ with $U$ invertible and $K$ compact. It suffices to treat $I+C$ with $C=U^{-1}K$. A compact operator on a Hilbert space admits arbitrary operator norm approximation by finite-rank operators: take a finite epsilon-net of its unit-ball image and project onto the span of the net. Choose $F$ of finite rank with $\Vert C-F\Vert <1$, put $R=C-F$, and factorThe first factor is invertible by the Neumann series. Set $G=(I+R{)}^{-1}F$ and $E=\mathrm{ran}G$, a finite-dimensional linear subspace. On $E\oplus E^{\perp }$, the second factor has block formLeft multiplication by the invertible block matrix $\left(\begin{array}{cc}I& -B\\ 0& I\end{array}\right)$ reduces it to $A\oplus I$. Its range is closed, and its kernel and cokernel have equal dimension by finite-dimensional rank-nullity theorem for $A:E\to E$. Invertible factors preserve all these properties. Hence $T$ is a Fredholm operator of Fredholm index zero, proving both directions:

The Clifford algebra relation impliesIndeed move $C(v)$ past $C(b)$ and then past $C(a)$ using the two anticommutator identities; the three-generator products cancel.

Since all $C(w)$ are self-adjoint,Each scalar vanishes because $T$ is skew-adjoint. This proves $\boxed{\pi (T{)}^{\ast }=-\pi (T)}$.

For the requested commutator, the first identity givesThe first sum is $C(Tv)$ by the orthonormal basis expansion of $v$. For the second, the skew-adjoint property gives ${\sum }_i(T{e}_i,v)e_i=-Tv$. By real linearity of $C$, the difference is thereforeThis calculation explains the factor $1/4$: it makes infinitesimal rotations of $V$ act correctly on its Clifford generators.

Products below combine matrix multiplication with the wedge product of differential forms of differential forms. Differentiate the projection identity $p^2=p$ to getMultiplying by $p$ on both sides also gives $pdpp=0$; applying the same argument to $1-p$ gives $(1-p)dp(1-p)=0$. Thus at each point $dp$ is off-diagonal relative to $\mathrm{im}p\oplus \ker p$, whereas $(dp{)}^2$ is diagonal and commutes with $p$. For $\omega =p(dp{)}^2$,because $d(dp)=0$.

The graded Leibniz rule and cyclicity of the matrix trace give, since $\omega$ has even degree,Here $(d\omega ){\omega }^{k-1}$ is off-diagonal: $(dp{)}^3$ is off-diagonal and every power of $\omega$ is diagonal. Its matrix trace is zero. Hence $\boxed{d\mathrm{tr}({\omega }^k)=0}$ for every $k\ge 1$; for $k=0$ the trace is the constant matrix size.

For the determinant, work over the commutative algebra of even differential forms. If $f(0)=1$, write $f(\omega )=I+N$, where $N$ has strictly positive degree and is nilpotent for dimensional reasons. The finite power series identityholds over this algebra, as follows either from the formal determinant identity over characteristic-zero commutative rings or by differentiating $\det (I+sN)$ in the scalar variable $s$. Each $N^j$ is a linear combination of positive powers of $\omega$. The matrix trace of the logarithm is therefore a closed differential form, and its finite exponential is closed by the graded Leibniz rule. Thus $\boxed{d\det f(\omega )=0}$. The projection connection interprets $\omega$ as the curvature of $pd$ on $\mathrm{im}p$, but the proof above directly establishes both closedness claims.

The positive sign printed in the exponential uses the skew-adjoint convention for the Dirac operator: $D^{\ast }=-D$, so $D_{-}=-D_{+}^{\ast }$. Put $A=D_{+}$ and define the nonnegative self-adjoint operatorsFor $t>0$, the assumed compact elliptic spectral theorem for these elliptic differential operators gives discrete eigenvalues, finite multiplicities and trace-class operators $e^{-tP}$, $e^{-tQ}$.

If $Pu=\lambda u$ with $\lambda >0$, then $Q(Au)=A(Pu)=\lambda Au$ and $Au\ne 0$. Conversely $A^{\ast }$ sends the $\lambda$ eigenspace of $Q$ to that of $P$. On these eigenspaces the inverse of $A$ is ${\lambda }^{-1}{A}^{\ast }$. Thus the positive spectra agree with multiplicity, and absolute convergence of the operator traces permits cancellation term by term:The last expression is the Fredholm index of $A$, since its closed range has orthogonal complement $\ker A^{\ast }$. ConsequentlyThis is the supertrace cancellation used in Question 1. If the Dirac operator is instead defined to be self-adjoint, $D_{-}=(D_{+}{)}^{\ast }$ and the heat exponentials must have a minus sign. With that convention the positive exponentials in the printed formula are generally unbounded and have no finite operator trace.

Interpret $H_{-\infty }({\mathbb{T}}^n)={\bigcup }_{s\in \mathbb{R}}{H}^s({\mathbb{T}}^n)$, the periodic distributions of finite negative Sobolev space order. Smooth multiplication is bounded on every periodic Sobolev space. One direct proof uses the rapid decay of the Fourier coefficients of $V$: the inequalityreduces the weighted convolution formula for $\widehat{Vf}$ to an ${\ell }^1\ast {\ell }^2$ estimate. Thus $\Vert Vf{\Vert }_{H^s}\le C_{V,s}\Vert f{\Vert }_{H^s}$ for every real $s$.

Suppose initially $f\in H^s$. The equation gives $\Delta f=u-Vf\in H^s$, since a smooth function belongs to every periodic Sobolev space. For $k\ne 0$, the Fourier multiplier of $\Delta$ is $|k{|}^2$, andThe zero coefficient is already finite. Hence $f\in H^{s+2}$. Iterating the argument gives $f\in H^{s+2j}$ for every $j\ge 0$. Applying the Sobolev embedding theorem also to each distributional partial derivative produces a smooth representative. Therefore $\boxed{f\in C^{\infty }({\mathbb{T}}^n)}$. This is an explicit elliptic regularity bootstrap, valid for complex or real $V$ and without a sign assumption on it.

Every entire holomorphic function has a Taylor series $f(z)={\sum }_{n\ge 0}{a}_n{z}^n$, uniformly convergent on each circle. Angular orthogonality givesIntegrate in polar coordinates and use Tonelli theorem for the nonnegative sum:The last integral equals $n!/2$ by substituting $u=r^2$ and repeated integration by parts.

Conversely, for any $(b_n)\in {\ell }^2$, set $a_n=b_n/\sqrt{n!}$. On $|z|\le R$, the Cauchy-Schwarz inequality givesApplied to the tails, the same bound proves uniform convergence on every disk. The series therefore defines an entire holomorphic function, and the norm identity puts it in the Bargmann-Fock space $\mathcal{F}$. The coefficient map $f\mapsto (\sqrt{n!}{a}_n)$ is consequently an onto linear isometry $\mathcal{F}\to {\ell }^2$; polarization also identifies the inner products. Completeness of ${\ell }^2$ proves that $\mathcal{F}$ is a Hilbert space. Its standard coordinate orthonormal basis corresponds toThe norm identity shows both their unit norms and completeness; it also gives $|f(z)|\le e^{|z{|}^2/2}\Vert f{\Vert }_{\mathcal{F}}$.

To distinguish the two normalizations, denote the physicists' Hermite polynomial by $h_n(x)=(-1{)}^n{e}^{x^2}(d/dx{)}^n{e}^{-x^2}$. Define the normalized Hermite functions byIntroduce the differential operators $a=(x+d/dx)/\sqrt{2}$, $a^{\ast }=(x-d/dx)/\sqrt{2}$ on the Schwartz space. Direct multiplication gives $[a,a^{\ast }]=1$ andThe Gaussian $H_0={\pi }^{-1/4}{e}^{-x^2/2}$ satisfies $a{H}_0=0$, and repeated differentiation gives $H_n=(a^{\ast }{)}^n{H}_0/\sqrt{n!}$. The commutator identity implies $a{H}_n=\sqrt{n}{H}_{n-1}$ and $a^{\ast }{H}_n=\sqrt{n+1}{H}_{n+1}$. ThereforeFor normalization, integration by parts makes $a^{\ast }$ the adjoint operator of $a$ on these rapidly decreasing functions, so $\Vert (a^{\ast }{)}^n{H}_0{\Vert }^2=n!\Vert H_0{\Vert }^2=n!$. Distinct eigenvalues of the symmetric quantum harmonic oscillator give orthogonality.

The generating identity follows directly from the Rodrigues definition of the Hermite polynomials:The Taylor expansion of the entire Gaussian justifies the middle equality for every complex $w$. Substitute $w=z/\sqrt{2}$ and the Bargmann-Fock space basis $e_n(z)=z^n/\sqrt{n!}$ to obtainThe convergence is locally uniform in $z$ by the same entire generating series.

For $0<t<1$, $A_x(\sqrt{t}z)$ belongs to the Bargmann-Fock space: after writing $z=u+iv$, its squared modulus times $e^{-|z{|}^2}$ has strictly negative quadratic coefficients $-(1+t)u^2-(1-t)v^2$. Taking the Bargmann-Fock space inner product of the two generating functions and using the orthonormal basis proved in part (b) givesThe coefficient series converges absolutely by the Cauchy-Schwarz inequality. Substitution factors the right side intoUse the Gaussian integral $\int e^{-a{u}^2+bu}du=\sqrt{\pi /a}{e}^{b^2/(4a)}$ for $a>0$ and complex $b$; the complex case follows from the real case by analyticity. The resulting exponent isThus the Mehler formula for Hermite functions isAs a normalization check, its $t\downarrow 0$ limit is $H_0(x)H_0(y)$. Setting $t=e^{-2s}$ and multiplying by $e^{-s}$ yields the harmonic oscillator transition kernel with eigenvalues $2n+1$.

For a closed manifold $M$ with an oriented Riemannian metric, define $\delta$ to be the formal $L^2$ adjoint operator of the exterior derivative, equivalently $\delta =(-1{)}^{n(q+1)+1}\ast d\ast$ on $q$-forms. The Hodge Laplacian is ${\Delta }_q=d\delta +\delta d$. The Hodge decomposition theorem states that its space ${\mathcal{H}}^q=\ker {\Delta }_q$ of harmonic differential forms is finite-dimensional and smooth, and thatThe last isomorphism sends a harmonic differential form to its de Rham cohomology class: every class has exactly one harmonic representative. Out-of-range degree spaces are zero.

We first justify the analytic facts needed in the proof. In local coordinates and a smooth frame for the form bundle, the principal symbol of ${\Delta }_q$ is $g^{ij}{\xi }_i{\xi }_{j}I$. This follows from the exterior/contraction identities for $d$ and $\delta$. Freezing the leading coefficients, the Fourier transform estimate for the constant positive quadratic form gives, for each nonnegative integer $s$,Here is how to pass to variable coefficients. On a sufficiently small coordinate ball the difference of leading coefficients has arbitrarily small supremum norm. After differentiating $s$ times, its top-order contribution is bounded by $\epsilon \Vert u{\Vert }_{H^{s+2}}$; derivatives of the coefficients and first-order terms contribute $C_s\Vert u{\Vert }_{H^{s+1}}$. A partition of unity gives the same types of terms from commutators with the cutoffs. Absorb the epsilon term and use Sobolev interpolation to absorb $\Vert u{\Vert }_{H^{s+1}}$ into a small multiple of $\Vert u{\Vert }_{H^{s+2}}$ plus $C\Vert u{\Vert }_2$. This proves the estimate.

The first-order energy form $Q(u,u)=\Vert du{\Vert }_2^2+\Vert \delta u{\Vert }_2^2$ satisfies $Q(u,u)+c\Vert u{\Vert }_2^2\ge c^{\prime }\Vert u{\Vert }_{H^1}^2$ for a sufficiently large $c>0$. This follows from the same frozen-symbol argument at first order, or by integrating the principal part and absorbing the lower-order terms. The Lax-Milgram theorem thus constructs a bounded inverse $R=({\Delta }_q+c{)}^{-1}:L^2\to H^1$ in the weak sense. To obtain $H^2$ regularity of this weak solution, use a local difference quotient ${\delta }_{h}u$ and test the weak equation with $-{\delta }_{-h}({\chi }^2{\delta }_{h}u)$. Positivity of $g^{ij}$ bounds $\Vert \chi {\delta }_h\nabla u{\Vert }_2^2$; derivatives of $\chi$ and difference quotients of the smooth coefficients give terms bounded by $C\Vert u{\Vert }_{H^1}\Vert \chi {\delta }_h\nabla u{\Vert }_2$ and $C\Vert u{\Vert }_{H^1}^2$. The forcing contributes at most $C\Vert f{\Vert }_2(\Vert \chi {\delta }_h\nabla u{\Vert }_2+\Vert u{\Vert }_{H^1})$. Absorption gives a bound independent of $h$, so the difference-quotient criterion yields $u\in H^2$. Differentiating the equation and repeating gives the higher estimates and smoothness for smooth forcing. Thus the use of elliptic regularity here has been justified rather than assumed.

The inverse $R$ is compact as an $L^2$ operator by the Rellich-Kondrachov compactness theorem, and self-adjoint by symmetry of the weak energy form. It is positive and injective with dense range. The spectral theorem for compact self-adjoint operators now gives an orthonormal eigenbasis of ${\Delta }_q$, with nonnegative eigenvalues tending to infinity. Each eigenfunction is smooth by the estimates above. In particular ${\mathcal{H}}^q$ is finite-dimensional and there is a positive gap above its zero eigenvalue.