Laplacian commutes with Local Isometries

The fact that the laplacian on a Riemannian manifolds commutes with isometries is a well known fact and several proofs are available. The fact that the laplacian commutes with local isometries and hence with covering map, even if well known, is not threated in detail anywhere I could find. These notes are an attempt to give an explicit proof of this fact.

Preliminaries

Given two Riemannian manifolds (M1,g1),(M2,g2)(M_1,g_1),(M_2,g_2) we say that g1g_1 is a pullback metric of g2g_2 if g1=f∗g2g_1 = f^\ast g_2 for some smooth map ff.

An isometry is a diffeomorphism f:M1→M2f: M_1 \to M_2 such that g1=f∗g2g_1 = f^\ast g_2.
A local isometry is defined in the obvious way, i.e. by requiring that each point in M1M_1 has an open neighbourhood UU such that f∣U:U→f(U)f|_U: U \to f(U) is an isometry.

Remark

Actually a smooth map f:M1→M2f: M_1 \to M_2 is a local isometry if and only if g1=f∗g2g_1 = f^\ast g_2 holds. One implication is trivial since g1=f∗g2g_1 = f^\ast g_2 is a local statement. Viceversa, if g1=f∗g2g_1 = f^\ast g_2 holds, for each point p∈M1p \in M_1 (f∗)p:TpM→Tf(p)N{(f_\ast)}_p: T_pM \to T_{f(p)}N is an isometry of vector spaces, in particular it is an isomorphism. By inverse function theorem, ff is locally a diffeomorphism.

It is well known that

Proposition(Laplacian commutes with isometries)

Given an isometry f:M1→M2f: M_1 \to M_2:

f∗Δ2=Δ1f∗f^\ast \Delta^2 = \Delta^1 f^\ast

where Δi\Delta^i denotes the laplacian on MiM_i and f∗:h↦h∘ff^\ast: h \mapsto h \circ f is the pull-back.

Proof.

By definition Δ=−divgrad\Delta = - \mathrm{div} \mathrm{grad}. We have f∗gradf∗=gradf_\ast \mathrm{grad} f^\ast = \mathrm{grad}. Infact

⟨f∗gradf∗ϕ,X⟩=⟨f∗gradf∗ϕ,f∗(f∗)−1X⟩=⟨gradf∗ϕ,(f∗)−1X⟩=\langle f_\ast \mathrm{grad} f^\ast\phi, X \rangle = \langle f_\ast \mathrm{grad} f^\ast\phi, f_\ast (f_\ast)^{-1}X \rangle = \langle \mathrm{grad} f^\ast\phi, (f_\ast)^{-1}X \rangle ==((f∗)−1X)(f∗ϕ)=X(ϕ)=⟨gradϕ,X⟩=((f_\ast)^{-1}X)(f^\ast\phi) = X(\phi) = \langle \mathrm{grad} \phi , X\rangle

We have f∗div(f∗X)=divXf^\ast \mathrm{div} (f_\ast X)= \mathrm{div} X.
Infact from the definition of divergence

(divX)vol=LX(vol)=d(ιXvol)( \mathrm{div} X) \mathrm{vol} = L_X( \mathrm{vol}) = d(\iota_X \mathrm{vol})

so that

(f∗div(f∗X))vol1=f∗(div(f∗X)vol2)=f∗d(ιf∗Xvol2)(f^\ast \mathrm{div} (f_\ast X)) \mathrm{vol}_1 = f^\ast ( \mathrm{div}(f_\ast X) \mathrm{vol}_2) = f^\ast d(\iota_{f_\ast X} \mathrm{vol}_2)

Since the pullback commutes with the differential

=d(f∗ιf∗Xvol2)=d(ιX(f∗vol2))=d(ιX(vol1)=div(X)vol1=d(f^\ast\iota_{f_\ast X} \mathrm{vol}_2) = d(\iota_X(f^\ast \mathrm{vol}_2)) = d(\iota_X( \mathrm{vol}_1) = \mathrm{div}(X) \mathrm{vol}_1

Putting these together

Δf∗=−divgradf∗=−f∗f−1∗divf∗−1f∗gradf∗\Delta f^\ast = - \mathrm{div} \mathrm{grad} f^\ast = - f^\ast {f^{-1}}^\ast \mathrm{div} f_\ast^{-1} f_\ast \mathrm{grad} f^\ast=−f∗divgrad=f∗Δ= - f^\ast \mathrm{div} \mathrm{grad} = f^\ast \Delta

The proof crucially relies on the fact ff is an isometry and not just a local isometries. Infact we used the inverse map of ff.

The key fact needed to extend the result to local isometries is that differential operators (gradient, divergence, laplacian, …) are local operators and local operators can be restricted to open subsets.

Given vector bundles E1,E2→ME_1,E_2 \to M, an operator D:Γ(E1)→Γ(E2)D: \Gamma(E_1) \to \Gamma(E_2) (i.e. R\mathbb{R}-linear map) is called local if supp(Df)⊆supp(f)\text{supp}(Df) \subseteq \text{supp}(f).

Proposition

Differential operators are local

Proof.

See Nicolaescu, “Lectures on the Geometry of Manifolds”, Lemma 10.1.3

Local operators can be restricted to open subsets:

Proposition

Let D:Γ(M,E)→Γ(M,F)D: \Gamma(M,E) \to \Gamma(M,F) be a local operator. Then for each open U⊆MU \subseteq M exists DU:Γ(U,E)→Γ(U,F)D_U : \Gamma(U,E) \to \Gamma(U,F) such that:

DU(σ∣U)=D(σ)∣U∀σ∈Γ(M,E)D_U(\sigma|_U) = D(\sigma)|_U \quad \forall \sigma \in \Gamma(M,E)
Proof.

See Tu, “Differential Geometry”, Theorem 7.20.

So on a Riemannian manifold (M,g)(M,g) for each open U⊆MU \subseteq M there exist a restricted laplacian ΔU:C∞(U)→C∞(U)\Delta_U: C^\infty(U) \to C^\infty(U) such that, for each ϕ∈C∞(M)\phi \in C^\infty(M): ΔU(ϕ∣U)=(Δϕ)∣U\Delta_U (\phi|_U) = (\Delta \phi)|_U

The Proof

Let f:M1→M2f: M_1 \to M_2 be a local isometry. We want to show:

f∗(Δ2ϕ)=Δ1(f∗ϕ)∀ϕ∈C∞(M2)f^\ast (\Delta^2 \phi) = \Delta^1 (f^\ast\phi) \quad \forall \phi \in C^\infty(M_2)

that is

(f∗(Δ2ϕ))(x)=(Δ1(f∗ϕ))(x)∀ϕ∈C∞(M2),∀x∈M1(f^\ast (\Delta^2 \phi))(x) = (\Delta^1 (f^\ast\phi))(x) \quad \forall \phi \in C^\infty(M_2), \forall x \in M_1

Given an open cover {Uα} \{ U_\alpha\} of M1M_1, will be enough to show that

(f∗(Δ2ϕ))∣Uα=(Δ1(f∗ϕ))∣Uα∀ϕ∈C∞(M2)(f^\ast (\Delta^2 \phi))|_{U_\alpha} = (\Delta^1 (f^\ast\phi))|_{U_\alpha} \quad \forall \phi \in C^\infty(M_2)

For each x∈Mx \in M there is an open neighbourhood UU such that f∣Uf|_U is an isometry onto its image U≅f(U)=:VU \cong f(U) =: V, call fU:U→f(U)f_U: U \to f(U) this isometry. These neighbourhood forms an open cover for M1M_1, so we must prove the relation for each such UU.
We will use the elementary identities:

f∗ϕ∣U=(ϕ∘f)∣U=ϕ∘f∣U=ϕ∣V∘fU=fU∗(ϕV)f^\ast\phi|_U = (\phi \circ f)|_U = \phi \circ f|_U = \phi|_V \circ f_U = f_U^\ast (\phi_V)

By locality and the identity above

(Δ1f∗ϕ)∣U=ΔU1(f∗ϕ)∣U=ΔU1(fU∗(ϕ∣V))(\Delta^1 f^\ast\phi)|_U = \Delta^1_U (f^\ast\phi)|_U = \Delta^1_U (f_U^\ast(\phi|_V))

Since fUf_U is an isometry

=fU∗(ΔV2ϕ∣V)=fU∗((Δ2ϕ)∣V)=(f∗Δ2ϕ)∣U = f_U^\ast(\Delta^2_V \phi|_V) = f_U^\ast ((\Delta^2\phi)|_V) = (f^\ast\Delta^2 \phi)|_U

where we used again locality and the elementary identity. So we proved

(Δ1f∗ϕ)∣U=(f∗Δ2ϕ)∣U(\Delta^1 f^\ast\phi)|_U = (f^\ast\Delta^2 \phi)|_U

and hence we proved

Proposition(Laplacian commutes with local isometries)

Let f:M1→M2f: M_1 \to M_2 be a local isometry, then

Δ1f∗=f∗Δ2\Delta^1 f^\ast = f^\ast\Delta^2

Since riemannian coverings are local isometries we get

Corollary(Laplacian commutes with covering maps)

Given a riemannian covering map π:M→N\pi : M \to N we have:

ΔMπ∗=π∗ΔN\Delta_M \pi^\ast = \pi^\ast \Delta_N

Applications

In spectral geometry the following results are important.
Let Eλ(M)E_\lambda(M) denote the laplace eigenspace of MM related to the eigenvalue λ\lambda.

Proposition

Let f:M→Nf: M \to N be a local isometry. Then for each λ∈R\lambda \in \mathbb{R}

f∗(Eλ(N))⊆Eλ(M)f^\ast(E_\lambda(N)) \subseteq E_\lambda(M)
Proof.

If ϕ∈Eλ(N)\phi \in E_\lambda(N), then (ΔN−λ)ϕ=0(\Delta_N - \lambda)\phi = 0. By our result

(ΔM−λ)(f∗ϕ)=f∗((ΔN−λ)ϕ)=0(\Delta_M - \lambda)(f^\ast\phi) = f^\ast((\Delta_N - \lambda)\phi) = 0

Given a GG-action on a vector space VV denote by VGV^G the space of GG-invariant vectors.

Proposition

Given a GG-covering π:M→N\pi: M \to N we have an isomorphism π∗:Eλ(N)→EλG(M)\pi^\ast: E_\lambda(N) \to E^G_\lambda(M)

Proof.

The GG-action on MM extends to a R\mathbb{R}-linear GG-action on C∞(M)C^\infty(M) (or L2(M)L^2(M) if you prefer) by (g⋅f)(x)=(f∘g−1)(x)=f(g−1x)(g \cdot f)(x) = (f \circ g^{-1})(x) = f(g^{-1}x). Recall that N≅M/GN \cong M/G. By the universal property of quotient space π∗\pi^\ast gives an isomorphism C∞(N)≅C∞(M)GC^\infty(N) \cong C^\infty(M)^G. By the previous lemma π∗∣Eλ(N){\pi^\ast}|_{E_\lambda(N)} is an isomorphism Eλ(N)≅EλG(M)E_\lambda(N) \cong E^G_\lambda(M)

The last result is used in a beatiful theorem in Spectral Geometry, named after Sunada, which can be used to constructs a pair of manifolds which are isospectral but not isometric. See https://en.wikipedia.org/wiki/Hearing_the_shape_of_a_drum