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Home , Basis (linear algebra), Hodge star operator, Manifold, Riemannian manifold, Tensor, Tensor field

THE HODGE STAR, POINCARE´ , AND

SCOTT MORRISON

Abstract. Riemannian and pseudo- Riemannian are objects with rich geometric structure and considerable interest in . In this essay we consider the behaviour of differential forms on pseudo- Riemannian manifolds. In particular an inves- tigation is made of the properties of the , an between k differential forms and rank n-k dif- ferential forms on n dimensional manifolds. Two main applica- tions of this operator are then explored. Poincar´eDuality can be proved in a great variety of settings—for example in accordance with Poincar´e’soriginal ideas, in the setting of simplicial mani- folds, or in a much more general case as a corollary of . With the aid of the Hodge Decomposition Theorem, we will see that the Hodge star in fact induces Poincar´eDuality for Riemannian manifolds. Finally, we use the Hodge star to express Maxwell’s of electromagnetism in a simple and general form, and exhibit a short proof of Lorentz invariance.

1. pseudo- Riemannian manifolds We define pseudo- Riemannian manifolds in very much the same way as Riemannian manifolds are defined. If M is a smooth , a pseudo- Riemannian is a smooth field g : C∞(TM) ⊗ C∞(TM) C∞(M), where C∞(TM) denotes the of vector fields −→ on M, and C∞(M) denotes the set of smooth functions on M, such that for all p M, gp : TpM TpM R defined by ∈ ⊗ −→ gp :(Xp,Yp) g(X,Y )(p) 7→ is symmetric and nondegenerate. The pair (M, g) is then called a pseudo- . Riemannian manifolds are thus a spe- cialisation of pseudo- Riemannian manifolds, for which we demand that at every p M gp is positive in the sense that gp(Xp,Xp) 0 Xp ∈ ≥ ∀ ∈ TpM, that is, gp is an inner on TpM. 2. The Hodge Star Operator 2.1. . We at first restrict our attention to the oriented vector E = Rn, the model for the space at a on an abstract orientable manifold. Let g : E E R be a symmetric and nondegenerate 2-tensor. A result from× linear−→ algebra Date: November 8, 1999. 1 2 SCOTT MORRISON allows a choice of positively oriented vectors e1, e2, . . . , en E such that ∈ n g = c ei ei, X i i=1 ⊗ i where e E∗ denotes the dual vector to ei, and with ci = 1. Denote by index(g)∈ the number of coefficients equal to -1. Such a set± of vectors we call a g-orthonormal . 1 n Choose a g- e1, . . . , en, and define µ = e e . ∧· · ·∧ Then for any other g-orthonormal basis f1, . . . , fn, µ(f1, . . . , fn) = 1, and so µ is independent of the particular g-orthonormal basis chosen, depending only on g. Call µ the g-. Using g, we can define a symmetric bilinear on the space of alternating , g :Λk(E) Λk(E) R, by its values on the k × −→ for Λ (E). If σ1 < < σk and τ1 < < τk then ··· ···

cσ cσ if σ1 = τ1, , σk = τk g(eσ1 eσk , eτ1 eτk ) =  1 ··· k ··· ∧ · · · ∧ ∧ · · · ∧ 0 otherwise We say that this basis for Λk(E) is g-orthonormal.

2.2. Existence and uniqueness. We now define the Hodge star op- erator.

k n k Lemma 2.1. There is a unique isomorphism ? :Λ (E) Λ − (E) satisfying −→ α ?β = g(α, β)µ α, β Λk(E). ∧ ∀ ∈ Proof. We first prove uniqueness. Suppose we have such a ? and let σ1 σk τ1 τk β = e e and α = e e where σ, τ Sn and ∧ · · · ∧ ∧ · · · ∧ ∈ τ1 < < τk. Then α ?β = 0 unless σ1 = τ1, . . . , σk = τk. Thus ?β = ···aeσk+1 eσn for∧ some a R. Using this, we calculate β ?β = σ ∧· · ·∧ ∈ σ ∧ a( 1) µ = g(β, β)µ, and g(β, β) = cσ1 cσn , so a = ( 1) cσ1 cσn . Therefore− ? is uniquely determined, since··· it is uniquely− determined··· on the basis of Λk(E), by

σ1 σk σ σk+1 σn ?e e = ( 1) cσ . . . cσ e e , ∧ · · · ∧ − 1 n ∧ · · · ∧ where σ1 < < σk and σk+1 < < σn. Because µ depends only on g, and not the··· g-orthonormal basis··· chosen, ? also depends only on g. Using this as our definition of ?, it is clear that it is an isomorphism, since it the g-orthonormal basis of Λk(E) to the g-orthonormal n k basis of Λ − (E).

2.3. Properties of the Hodge star operator. The Hodge star op- erator has the following easily verified properties. THE HODGE STAR, POINCARE´ DUALITY, AND ELECTROMAGNETISM 3 Lemma 2.2. For all α, β Λk(E), ∈ α ?β = β ?α ∧ ∧ ?1 = µ ?µ = ( 1)index(g) − index(g)+k(n k) ? ? α = ( 1) − α − 2.4. The Hodge star operator on pseudo- Riemannian mani- folds. Let (M, g) be an n-dimensional pseudo- Riemannian manifold. Firstly, note than index(g) is constant. There exists a unique volume n element µ Ω (M) such that µ(X1, ,Xn) = 1 for all positively oriented g-orthonormal∈ bases on the tangent··· to M. Using this, k n k we define the Hodge star operator, ? :Ω (M) Ω − (M) pointwise, by −→ (?α)(p) = ?(α(p)). It is clear that the properties proved in Subsection 2.3 carry across. In particular, index(g)+k(n k) ? ? α = ( 1) − α, k− n k and so ? is an isomorphism Ω (M) ∼= Ω − (M). If M is compact, we can define g :Ωk(M) Ωk(M) R by × −→ g(α, β) = Z α ?β = Z g(α, β)µ. M ∧ M If M is in fact Riemannian, that is, if index(g) = 0, then since at each point p M gp(α, α) 0, g(α, α) 0 also, so this g is an inner product on Ωk(M∈). ≥ ≥ 2.5. The codifferential operator. Define the codifferential k 1 k δ :Ω − (M) Ω (M) −→ by δω = 0 if ω Ω0(M), and otherwise ∈ δβ = ( 1)nk+1+index(g) ? d ?. − index(g)+k(n k) Note that since ? ? α = ( 1) − α, − index(g) k(n k) δδ = ?d ??d? = ( 1) ( 1) − ? dd? = 0. − − Lemma 2.3.

g(dα, β) = Z dα ?β = Z α ?δβ = g(α, δβ) M ∧ M ∧ Proof. Using δβ = ( 1)nk+1+index(g) ? d ? β, − dα ?β α ?δβ = dα ?β + ( 1)nk+index(g)α ?? d ? β ∧ − ∧ ∧ − nk+2index(g)+∧k(n k) = dα ?β + ( 1) − α d ? β ∧ − ∧ = dα ?β + ( 1)kα d ? β ∧ − ∧ = d(α ?β), ∧ 4 SCOTT MORRISON where we have used the fact that k2 + k is even. Integrating both sides over M, and applying Stokes’ theorem,

Z dα ?β Z α ?δβ = Z d(α ?β) M ∧ − M ∧ M ∧

= Z α ?β ∂M ∧

= Z α ?β 0 ∧ = 0, we obtain the desired result.

3. Poincare´ Duality In this we restrict our attention to compact Riemannian man- ifolds. Thus g :Ωk(M) Ωk(M) R is an inner product, and, in accordance with Lemma× 2.3, d and−→δ are adjoints with respect to g.

3.1. The Laplace-de Rham operator, and the Hodge Decom- Theorem. Define ∆ : Ωk(M) Ωk(M) by ∆ = dδ + δd. If ∆α = 0 we say α is harmonic, and write−→k(M) for the of harmonic forms on M. H Lemma 3.1. ∆α = 0 is equivalent to dα = δα = 0. Proof. If dα = δα = 0, then clearly ∆α = 0. If ∆α = 0, then g(∆α, α) = g(dδα, α) + g(δdα, α) = g(δα, δα) + g(dα, dα) = 0. Since g is positive, this implies dα = δα = 0.

We now state without proof the following important theorem. Theorem 3.2 (Hodge Decomposition). Let ω Ωk(M). Then there k 1 k+1 ∈ k exists a unique α Ω − (M), β Ω and γ (M) such that ω = dα + δβ + γ. ∈ ∈ ∈ H 3.2. Poincar´eDuality. Since δδ = 0, we can define associated ho- mology groups,

k k 1 ker(δ :Ω (M) Ω − (M)) H0k = −→ . (δ :Ωk+1(M) Ωk(M)) −→ Lemma 3.3.

k ? : H (M) = H0n k ∼ − k n k Proof. We rely on the fact that ? :Ω (M) Ω − (M) is an isomor- phism. Firstly, if dα = β, then δ ? α = ( 1)−→p ? dα = ( 1)p ? β for some − − THE HODGE STAR, POINCARE´ DUALITY, AND ELECTROMAGNETISM 5 integer p. Similarly if δα = β, then d ? α = ( 1)q ? β for some integer q also. Thus − k k+1 n k n k 1 ? : ker(d :Ω (M) Ω (M)) ∼= ker(δ :Ω − (M) Ω − − (M)) k 1 → k n k+1 → n k ? : img(d :Ω − (M) Ω (M)) = img(δ :Ω − (M) Ω − (M)) → ∼ → and so taking quotients we have the required result. k k k Lemma 3.4. The maps , H (M) and , H0k(M) given by H → H → k inclusion followed by are , and so H (M) ∼= H0k(M). Proof. The proof for each is very similar, so we only give the argument for Hk(M). If γ k, ∆γ = dγ = 0. Thus [γ] Hk(M). We first show that the map is∈ injective.H Suppose [γ] = 0. Thus∈ γ = dβ k 1 for some β Ω − (M). δγ = 0, so g(γ, γ) = g(γ, dβ) = g(δγ, β) = g(0, β) = 0,∈ and therefore γ = 0. To prove the map is surjective, suppose [ω] Hk(M). Using the Hodge Theorem, there exist α k 1 ∈ k+1 k ∈ Ω − (M), β Ω (M) and γ so ω = dα + δβ + γ. Now dω = 0, and so dδβ ∈= 0. Thus g(δβ, δβ∈) H = g(β, dδβ) = 0, and δβ = 0. Then ω = dα + γ, and [ω] = [γ]. Therefore the map is an isomorphism. Combining the two previous lemmas we have proved the following result. k n k Theorem 3.5 (Poincar´eDuality). H (M) ∼= H − (M) 4. Electromagnetism In this section, we show that Maxwell’s on R4 with the Lorentz metric can be written succinctly in terms of differential forms and the Hodge star operator. We use this formulation to extend Maxwell’s equations to any Lorentz manifold. A Lorentz manifold is a pseudo- Riemannian 4-manifold (M, g) with index(g) = 1, for exam- ple the manifolds of . In this general setting, we prove the Lorentz invariance of Maxwell’s equations. The invariance is unsur- prising - since our formulation uses no geometry beyond the differen- tial structure of the underlying manifold, and the pseudo- Riemannian metric, a , which by definition preserves these objects, must also preserve solutions of the electromagnetic equations. 4.1. Maxwell’s Equations. On R4, with coordinates x, y, z, t, write F = E1dx dt + E2dy dt + E3dz dt + ∧ ∧ ∧ +B1dy dz + B2dz dx + B3dx dy, ∧ ∧ ∧ and let E = (E1,E2,E3) and B = (B1,B2,B3). Also let j = J 1dx + 2 3 1 2 3 J dy+J dz ρdt, and J = (J ,J ,J ). Let e1, e2, e3, e4 be the standard basis of R4,− and define g : R4 R4 R by g = e1 e1 + e2 e2 + e3 e4 e4 e4. Thus index(g×) = 1.−→ ⊗ ⊗ ⊗ − ⊗ 6 SCOTT MORRISON Theorem 4.1. Maxwell’s equations, divE = ρ divB = 0 ∂B curlE + = 0 ∂t ∂E curlB = J − ∂t are equivalent to the equations dF = 0 and δF = j In this form, the equations for electromagnetism make no reference to the structure of R4 beyond its diffeomorphism class and Lorentz struc- ture, and so are a good starting point for generalisation to arbitrary Lorentz manifolds. Given a Lorentz manifold (M, g), define the set of Lorentz transfor- mations as

gp(X,Y ) = gψ(p)(ψ∗X, ψ∗Y ) = ψ C∞(M,M)  L ∈ p M and X,Y TpM ∀ ∈ ∈ k Lemma 4.2. If ψ , and α Ω (M), then ?ψ∗α = ψ∗ ? α. ∈ L ∈ Proof. Fix p M. If e1, . . . , en is a g-orthonormal basis for gp, ∈ { } then define ai = ψ ei Tψ(p)M, 1 i n. Since gψ(p)(ai, aj) = ∗ ∈ ≤ ≤ gp(ei, ej) = ciδij, a1, . . . , an is a g-orthonormal basis for gψ(p). We { } k prove the result by showing it holds on the basis of Ω (M). Let σ Sn, σ1 σk ∈ σ1 < < σk and σk+1 < < σn. If α = e e , then we can calculate··· ··· ∧ · · · ∧

σ σk+1 σn ?α = ( 1) cσ cσ e e − 1 ··· k ∧ · · · ∧ σ σk+1 σn ψ∗ ? α = ( 1) cσ cσ a a − 1 ··· k ∧ · · · ∧ σ1 σk ψ∗α = a a ∧ · · · ∧ σ σk+1 σn ?ψ∗α = ( 1) cσ cσ a a − 1 ··· k ∧ · · · ∧ So ?ψ∗α = ψ∗ ? α. We are now in a position to prove the Lorentz invariance of Maxwell’s equations. Theorem 4.3. Let (M,g) be a Lorentz manifold, and F Ω2(M), j 1 ∈ ∈ Ω (M) such that dF = 0 and δF = j. If ψ , then dψ∗F = 0 and ∈ L δψ∗F = ψ j. ∗ k Proof. dψ∗F = ψ∗dF = 0. Since ?ψ∗α = ψ∗ ? α α Ω (M), 8+1+1 ∀ ∈ δψ∗F = ( 1) ? d ? ψ∗F = ψ∗ ? d ?F = ψ∗δF . Thus δψ∗F = ψ∗j, as required.− THE HODGE STAR, POINCARE´ DUALITY, AND ELECTROMAGNETISM 7 References 1. R. Abraham, J. E. Marsden, and T.Ratiu, Manifolds, tensor analysis and appli- cations, second ed., Springer-Verlag, 1988. 2. William M. Boothby, An introduction to differentiable manifolds and , second ed., Pure and , Academic Press, 1986. 3. Klaus J¨anich, , Undergraduate Texts in Mathematics, Springer-Verlag, 1984, Translated by Silvio Levy. 4. Mikio Nakahara, Geometry, topology and physics, Adam Hilger, 1990. E-mail address: [email protected], student number 2231252