Heat Kernel for Simply-Connected Riemann Surfaces
aa r X i v : . [ m a t h . DG ] J u l Heat Kernel for Simply-connected Riemann Surfaces
Trevor H. Jones Dan KucerovskyFebruary 27, 2018
Abstract
From the uniformization theorem, we know that every Riemann surface has a simply-connectedcovering space. Moreover, there are only three simply-connected Riemann surfaces: the sphere, theEuclidean plane, and the hyperbolic plane. In this paper, we collect the known heat kernels, or Green’sfunctions, for these three surfaces, and we derive the differential forms heat kernels as well. Then we givea method for using the heat kernels on the universal cover to generate the heat kernel on the underlyingsurface.
For the purposes of this paper, we will be concentrating on the three simply-connected Riemann surfaces:the sphere, the plane, and the hyperbolic plane. We will study the solution operator for the differential formsheat equations on these surfaces. For a Riemann surface, M , the heat equation can be written as,( ∂ t + ∆) ω ( x , t ) = 0 ω ( x ,
0) = ν ( x ) (1.1)where ω is a differential form depending on a point x ∈ M and on time, t , with a bounded, L initialcondition ν ( x ). The Laplace-Beltrami operator, ∆ = ( d + d ∗ ) = dd ∗ + d ∗ d , is a linear differential operatorwhich maps k -forms to k -forms. Thus we do not need to consider mixed degree forms. This means the heatequation on M is actually three separate equations, one for 0-forms, which are identified with functions on M , one for 2-forms, which is isomorphic to the 0-form case, and one for 1-forms, which yields two differentialequations, which are sometimes coupled. Let us begin with the heat kernel for differential forms in the Euclidean plane, with the usual metric ds = dx + dy . Let ω = A ( x , t ) + B ( x , t ) dx + C ( x , t ) dy + D ( x , t ) dx ∧ dy, be an arbitrary differential form withcoefficients having appropriate partial derivatives. Then the heat equation applied to ω has the followingform: (∆ + ∂ t ) ω = ( A t − A xx − A yy ) + ( B t − B xx − B yy ) dx (2.1)+ ( C t − C xx − C yy ) dy + ( D t − D xx − D yy ) dx ∧ dy = 0The only way for this to be true is for each component to be zero. This means each of the coefficient functionsof ω must satisfy u t = u xx + u yy . This has a well known Green’s function, or heat kernel, K ( x , x , t ) = 14 πt exp (cid:18) − | x − x | t (cid:19) , where x i = ( x i , y i ) and | x | is the usual norm on R . So the heat kernel for differential forms on the Cartesianplane is K ( x , x , t ) = 14 πt exp (cid:18) − | x − x | t (cid:19) [1 x ⊗ x + dx ⊗ dx + dy ⊗ dy + dx ∧ dy ⊗ dx ∧ dy ] . (2.2)1f we change to polar coordinates, x i = ( r i , θ i ) with the metric ds = dr + r dθ , the heat kernel becomes K ( x , x , t ) = 14 πt exp (cid:18) − | x − x | t (cid:19) [1 x ⊗ x + Ω + r dr ∧ dθ ⊗ r dr ∧ dθ ] (2.3)where | x − x | = r + r − r r cos( θ − θ ) andΩ = cos( θ − θ ) [ dr ⊗ dr + r dθ ⊗ r dθ ] + sin( θ − θ ) [ dr ⊗ r dθ − r dθ ⊗ dr ] . For the case of the hyperbolic plane, we will use the hyperboloid model, which we denote H , with the metric ds = dr + sinh rdθ . If we let ω = A ( x , t ) + B ( x , t ) dr + C ( x , t ) sinh rdθ + D ( x , t ) sinh rdr ∧ dθ, be anarbitrary differential form with coefficients having appropriate partial derivatives. Then the heat equationapplied to ω has the following form:( ∂ t + ∆) ω = (cid:18) A t − A rr − cosh r sinh r A r − r A θθ (cid:19) + (cid:18) B t − B rr − cosh r sinh r B r + 1sinh r B − r B θθ + 2 cosh r sinh r C θ (cid:19) dr + (cid:18) C t − C rr − cosh r sinh r C r + 1sinh r C − r C θθ − r sinh r B θ (cid:19) sinh rdθ + (cid:18) D t − D rr − cosh r sinh r D r − r D θθ (cid:19) sinh rdr ∧ dθ = 0 . (3.1)We chose sinh rdθ and sinh rdr ∧ dθ since these differential forms make the different parts of the heat equationsimilar, thus simplifying our problem. It is also worth noting that these forms have unit length under thelocal inner product.Notice that the dr and sinh rdθ portion of the equation are coupled with a derivative in θ . If we considerthe Laplacian applied only to a radially symmetric argument, then all terms involving a derivative in θ disappear, so the equations are no longer coupled. For forms independent of θ , the heat equation is( ∂ t + ∆) ω = (cid:18) A t − A rr − cosh r sinh r A r (cid:19) + (cid:18) B t − B rr − cosh r sinh r B r + 1sinh r B (cid:19) dr + (cid:18) C t − C rr − cosh r sinh r C r + 1sinh r C (cid:19) sinh rdθ + (cid:18) D t − D rr − cosh r sinh r D r (cid:19) sinh rdr ∧ dθ =0 . (3.2)The heat kernel for functions, or 0-forms, on H is given by McKean [9] and a derivation of the formulacan be found in Chavel [3]. The heat kernel for functions on H is K ( x , y , t ) = 12 π Z ∞ P − + iρ (cosh d H ( x , y )) ρ exp (cid:18) − (cid:18)
14 + ρ (cid:19) t (cid:19) tanh πρdρ. (3.3)After some simplification this agrees with the heat kernel given in [9]: K ( d H ( x , y ) , t ) = √ (cid:18) − t (cid:19) (4 πt ) − Z ∞ d H ( x , y ) s exp (cid:16) − s t (cid:17) (cosh s − cosh d H ( x , y )) ds. The approach we will use to arrive at the 1-form heat kernel is similar to the approach used by Chavel[3], that is, to find an integral transform for the radial heat equation, and then use translation to arrive atthe full heat kernel. To find the integral transform required for this, we will solve the following eigenfunctionproblem: − E rr − cosh r sinh r E r + 1sinh r E = λE = (cid:18)
14 + ρ (cid:19) E (3.4)2ince the Laplacian is a positive operator, there are no negative or complex eigenvalues. Since the continuousspectrum of the Laplacian on H is [ , ∞ ), as stated in [4, 5], we are justified in representing λ by + ρ .By changing variables, x = cosh r and + ρ = − v ( v + 1), (3.4) can be transformed into Legendre’s equationof order v and degree 1. Since we will be considering only bounded solutions, the solution to (3.4) is E ρ ( r ) = P − + iρ (cosh r ), with ρ ≥
0. We are able to disregard ρ < P − iρ ( x ) = P − + iρ ( x ).With these eigenfunctions, we use the integral transform given by b f ( ρ ) := h E ρ ( r y ) dr y , f ( r y ) dr y i = Z H E ρ ( r y ) dr y ∧ ∗ f ( r y ) dr y . Since f and E ρ are independent of θ y , the integral transform simplifies to b f ( ρ ) = 2 π Z ∞ E ρ ( r y ) f ( r y ) sinh r y dr y . (3.5)This is proportional to the Mehler-Fock transform of order −
1, [10], with inverse transform f ( r y ) dr y = − π Z ∞ b f ( ρ ) P − − + iρ (cosh r y ) ρ tanh πρdρ ⊗ dr y (3.6)= 12 π Z ∞ b f ( ρ ) ρ tanh πρ + ρ E ρ ( r y ) dρ ⊗ dr y Because the Laplacian is self-adjoint with respect to this inner product, when we apply the transform tothe dr portion of the heat equation (3.2), we have h E ρ ( r y ) dr y , ( ∂ t + ∆) B ( r y , t ) dr y i = b B t ( ρ, t ) + (cid:18)
14 + ρ (cid:19) b B ( ρ, t ) = 0 b B ( ρ,
0) = b f ( ρ )where B ( r y ,
0) = f ( r y ). Thus b B ( ρ, t ) = b f ( ρ ) exp (cid:18) − (cid:18)
14 + ρ (cid:19) t (cid:19) . So, to solve for B we apply the inversetransform (3.6), and we get B ( r x , t ) dr x = 12 π Z ∞ (cid:18) π Z ∞ E ρ ( r y ) f ( r y ) sinh r y dr y (cid:19) exp (cid:18) − (cid:18)
14 + ρ (cid:19) t (cid:19) ρ tanh πρ + ρ E ρ ( r x ) dρ ⊗ dr x = Z ∞ f ( r y ) sinh r y (cid:18)Z ∞ exp (cid:18) − (cid:18)
14 + ρ (cid:19) t (cid:19) ρ tanh πρ + ρ E ρ ( r y ) E ρ ( r x ) dρ (cid:19) dr y ⊗ dr x Since P − + iρ (cosh r ) = ddr P − + iρ (cosh r ) and P − + iρ (cosh r y ) P − + iρ (cosh r x ) = 12 π Z π P − + iρ (cosh r y cosh r x − sinh r y sinh r x cos α ) dα (3.7)(see formulas in [1, 10, 6]), we have B ( r x , t ) dr x = 12 π Z ∞ f ( r y ) sinh r y Z ∞ exp (cid:18) − (cid:18)
14 + ρ (cid:19) t (cid:19) ρ tanh πρ + ρ × ∂ ∂r y ∂r x Z π P − + iρ (cosh r cosh r x − sinh r y sinh r x cos α ) dαdρdr y ⊗ dr x . In (3.7), the variable α can be replaced by θ y − θ x , with the integration with respect to θ y , and keeping thesame bounds. This means the argument of the Legendre function is cosh d H ( x , y ), where x = ( r x , θ x ) and3 = ( r y , θ y ). Since the integral is over a full period, the result in independent of θ y as well as θ x . This meansthat we can replace the partial derivatives with exterior derivatives. In other words, B ( r x , t ) dr x = 12 π Z ∞ f ( r y ) sinh r y Z ∞ exp (cid:18) − (cid:18)
14 + ρ (cid:19) t (cid:19) ρ tanh πρ + ρ × Z π d x d y P − + iρ (cosh d H ( x , y )) dθ y dρ. (3.8)Using similar arguments, we can write C ( r x , t ) sinh r x dθ x = 12 π Z ∞ g ( r y ) Z ∞ exp (cid:18) − (cid:18)
14 + ρ (cid:19) t (cid:19) ρ tanh πρ + ρ × Z π ∗ x ∗ y d x d y P − + iρ (cosh d H ( x , y )) dρdr y . (3.9)Note that for equations (3.8) and (3.9), that the “missing” differentials dr y and dθ y come from the exteriorderivative d y .Now let G = 12 π d x d y Z ∞ exp (cid:18) − (cid:18)
14 + ρ (cid:19) t (cid:19) ρ tanh πρ + ρ P − + iρ (cosh d H ( x , y )) dρ. Then B ( r x , t ) dr x = Z π Z ∞ G ∧∗ y f ( r y ) dr y and C ( r x , t ) sinh r x dθ x = Z π Z ∞ ∗ x ∗ y G ∧∗ y g ( r y ) sinh r y dθ y .Since G ∧ ∗ y g ( r y ) sinh r y dθ y = 0 and ∗ x ∗ y G ∧ ∗ y f ( r y ) dr y = 0, we can combine this statement into B ( r x , t ) dr x + C ( r x , t ) sinh r x dθ x = Z π Z ∞ ( I + ∗ x ∗ y ) G ∧ ∗ y ( f ( r y ) dr y + g ( r y ) sinh r y dθ y ) . Thus it appears that ( I + ∗ x ∗ y ) G plays the role of the heat kernel for radially symmetric initial conditions.We will show that ( I + ∗ x ∗ y ) G is the heat kernel for 1-forms on the hyperbolic plane. Theorem 3.1
Consider the heat equation on the hyperbolic plane with metric ds = dr + sinh rdθ : (∆ + ∂ t ) ω ( x , t ) = 0 ω ( x ,
0) = ν ( x ) (3.10) where ω and ν have the form f dr + g sinh rdθ and ν is a bounded L K ( x , y , t ) isgiven by the formula K ( x , y , t ) = ( I + ∗ x ∗ y ) G where G = 12 π d x d y Z ∞ exp (cid:18) − (cid:18)
14 + ρ (cid:19) t (cid:19) ρ tanh πρ + ρ P − + iρ (cosh d H ( x , y )) dρ. Proof:
By existence and uniqueness of heat kernels, [2], (3.10) has a unique solution, ω . We will showthat ω = Z H ( I + ∗ x ∗ y ) G ∧ ∗ y ν ( y ). First, let us define an inner product on forms as h f, g i = Z H f ∧ ∗ g. It is known that d and d ∗ are adjoint under this inner product and that h∗ f , g i = − h f, ∗ g i . With thisnotation, we will show that ω = h ( I + ∗ x ∗ y ) G, ν ( y ) i . Recall the 0-form heat kernel from (3.3), and note that4 t G = − d x d y K ( x , y , t ). We will let G = Z ∞ t d x d y K ( x , y , τ ) dτ . h ( I + ∗ x ∗ y ) G, ν ( y ) i = h G, ν ( y ) i + ∗ x h∗ y G, ν ( y ) i = Z ∞ t [ d x h d y K ( x , y , τ ) , ν ( y ) i + ∗ x d x h∗ y d y K ( x , y , τ ) , ν ( y ) i ] dτ = Z ∞ t (cid:2) d x (cid:10) K ( x , y , τ ) , d ∗ y ν ( y ) (cid:11) − ∗ x d x (cid:10) K ( x , y , τ ) , ∗ y d ∗ y ν ( y ) (cid:11)(cid:3) dτ = Z ∞ t [ d x d ∗ x ω ( x , τ ) − ∗ x d x ∗ x d ∗ x ω ( x , τ )] dτ = Z ∞ t ∆ ω ( x , τ ) dτ = Z ∞ t − ∂ τ ω ( x , τ ) dτ = ω ( x , t )In this calculation, we used that fact that the Laplacian commutes with the Hodge star, the exterior deriva-tive, and the coderivative to obtain results like: 0 = d x (∆ + ∂ t ) ω = (∆ + ∂ t ) d x ω with initial conditions d x ν ( x ) = d x ω ( x , (cid:4) Using the notation from the proof of Theorem 3.1, we can write the 1-form heat kernel on H as K ( x , y , t ) = ( I + ∗ x ∗ y ) Z ∞ t d x d y K ( x , y , τ ) dτ . For the sphere, we use the unit sphere with the metric ds = dφ + sin φdθ with θ ∈ [0 , π ), φ ∈ [0 , π ]. Let ω = A ( x , t ) + B ( x , t ) dφ + C ( x , t ) sin φdθ + D ( x , t ) sin φdφ ∧ dθ. Here we choose sin φdθ and sin φdφ ∧ dθ forthe obvious reasons.( ∂ t + ∆) ω = (cid:18) A t − A φφ − cos φ sin φ A φ − φ A θθ (cid:19) + (cid:18) B t − B φφ − cos φ sin φ B φ + 1sin φ B − φ B θθ + 2 cos φ sin φ C θ (cid:19) dφ + (cid:18) C t − C φφ − cos φ sin φ C φ + 1sin φ C − φ C θθ − φ sin φ B θ (cid:19) sin φdθ + (cid:18) D t − D φφ − cos φ sin φ D φ − φ D θθ (cid:19) sin φdφ ∧ dθ = 0 . (4.1)If we have a radially symmetric solution, one that is independent of θ , then the heat equation simplifiesas before and we have( ∂ t + ∆) ω = (cid:18) A t − A φφ − cos φ sin φ A φ (cid:19) + (cid:18) B t − B φφ − cos φ sin φ B φ + 1sin φ B (cid:19) dφ + (cid:18) C t − C φφ − cos φ sin φ C φ + 1sin φ C (cid:19) sin φdθ + (cid:18) D t − D φφ − cos φ sin φ D φ (cid:19) sin φdφ ∧ dθ = 0 . (4.2)Notice a similarity between equations (3.2) and (4.2). By similar techniques, the eigenfunctions forthe 0-form case are E n ( φ ) = c n P n (cos φ ), which has eigenvalues n + n . The normalizing coefficient is5 n = r n + 14 π , using the inner product h f, g i = Z S f ∧ ∗ g . We require the eigenfunction to be boundedand L , so n is restricted to the positive integers.We can write the 0-form radially symmetric heat kernel on S as G ( φ x , φ y , t ) = 14 π ∞ X n =0 (2 n + 1) exp ( − n ( n + 1) t ) P n (cos φ x ) P n (cos φ y ) . We can simplify this expression by setting φ y = 0. This means P n (cos φ y ) = 1. We also note that φ x is justthe distance from the “north” pole to x , so we can recover the full heat kernel on S by replacing φ x by thedistance between the points x and y . Thus K ( x , y , t ) = 14 π ∞ X n =0 (2 n + 1) exp ( − n ( n + 1) t ) P n (cos d S ( x , y )) . (4.3)The situation is similar for the 1-form case: E n ( φ ) = c n P n (cos φ ), with c n = s n + 14 πn ( n + 1) , and eigen-value n + n . Note that n >
0, so there are no harmonic L dφ x ⊗ dφ y portion of the heat kernel we proceed in a manner similar to that of Section 3. Let G ( φ x , φ y , t ) = 14 π ∞ X n =1 n + 1 n ( n + 1) exp ( − n ( n + 1) t ) P n (cos φ x ) P n (cos φ y ) dφ x ⊗ dφ y . Using some identities of Legendre functions [6], we have P n (cos φ ) = ddφ P n (cos φ ) . This means we can replace P n (cos φ x ) dφ x with d x P n (cos φ x ), with a similar expression for the y term as well. So G φ ( φ x , φ y , t ) = ∞ X n =1 F ( n, t ) d x d y P n (cos φ x ) P n (cos φ y ) , where F ( n, t ) = 14 π n + 1 n ( n + 1) exp ( − n ( n + 1) t ). By similar arguments we find that G θ ( φ x , φ y , t ) = ∞ X n =1 F ( n, t ) ∗ x ∗ y d x d y P n (cos φ x ) P n (cos φ y ) . Therefore the radially symmetric 1-form heat kernel can be written as G ( φ x , φ y , t ) = ∞ X n =1 F ( n, t ) ( I + ∗ x ∗ y ) d x d y P n (cos φ x ) P n (cos φ y ) . To recover the full heat kernel, we will make use of the identity Z π P n (cos φ x ) P n (cos φ y ) dθ y = Z π P n (cos φ x cos φ y + sin φ x sin φ y cos( θ x − θ y )) dθ y which is derived from the “addition theorem” (see 8.814, [6]). The distance formula on the unit sphere isgiven by cos d S ( x , y ) = cos φ x cos φ y +sin φ x sin φ y cos( θ x − θ y ), so we will use this to simplify the expression.6e will consider the solution of the radial heat equation u ( φ x , t ) dφ x = Z S ∞ X n =1 F ( n, t ) d x d y P n (cos φ x ) P n (cos φ y ) ∧ ∗ y f ( φ y ) dφ y = Z S ∞ X n =1 F ( n, t ) ( d x d y P n (cos φ x ) P n (cos φ y )) ∧ f ( φ y ) sin φ y dθ y = Z π ∞ X n =1 F ( n, t ) f ( φ y ) sin φ y d x d y Z π P n (cos φ x ) P n (cos φ y ) dθ y = Z π ∞ X n =1 F ( n, t ) f ( φ y ) sin φ y d x d y Z π P n (cos d S ( x , y )) dθ y = Z S ∞ X n =1 F ( n, t ) d x d y P n (cos d S ( x , y )) ∧ ∗ y f ( φ y ) dφ y Following the same argument, we have u ( φ x , t ) sin φ x dθ x = Z S ∞ X n =1 F ( n, t ) ∗ x ∗ y d x d y P n (cos d S ( x , y )) ∧ ∗ y g ( φ y ) sin φ y dθ y In these two calculations, f ( φ x ) dφ x and g ( φ x ) sin φ x dθ x are initial conditions for the heat equation. Fromthese calculations we have a candidate for the full 1-form heat kernel: K ( x , y , t ) = 14 π ∞ X n =1 n + 1 n ( n + 1) exp ( − n ( n + 1) t ) ( I + ∗ x ∗ y ) d x d y P n (cos d S ( x , y )) (4.4)The proof that this is indeed the 1-form heat kernel follows exactly that of Theorem 3.1. As stated earlier, any Riemann surface will have one of the three simply-connected Riemann surfaces as itsuniversal covering space. In this section we will show how to generate the heat kernel for a Riemann surfacebased upon its universal cover and its covering group.Let M be an arbitrary Riemann surface, U its universal cover, and G its covering group. Then thereis a projection map π : U → M such that π ( g · x ) = π ( x ) for all x ∈ U and all g ∈ G . This map inducesa metric on M compatible with the metric on U , with distance d M ( x , y ) = min g ∈ G d M ( e x , g · e y ), where e x and e y are fixed pre-images of x , y ∈ M . Since the Laplacian is defined in terms of the metric, we have∆ U e u ( x ) = ∆ M u ( π ( x )) for x in a small open set in U , and e u is the lift of the function u to the universalcover. With this information we can state the following theorem: Theorem 5.1
Let M be an arbitrary Riemann surface, U its universal cover, and G its covering group. Let K U ( x , y , y ) be the heat kernel on U . Then the heat kernel on M is given by K M ( x , y , t ) = X g ∈ G K U ( e x , g · e y , t ) for fixed pre-images e x , e y under the projection map π . To prove this theorem we use the following definition and results which can be found in [7].
Definition 5.2
Let M be a manifold, and G a group which acts on the manifold. A function, f , on themanifold is said to be G -periodic if f ( g · x ) = f ( x ) for each g ∈ G and x ∈ M . roposition 5.3 Let M be a manifold, f M be a cover of M with covering group G , and let V be a vectorspace. Suppose we have a G -periodic function f : f M → V . Then there is a unique function b f : M → V suchthat f = b f ◦ π , where π is the covering map from f M to M . Proof:
We know that if y = π ( x ), then π − ( y ) = { g · x | g ∈ G } . Let f be G -periodic. We will define b f asfollows: b f ( y ) = f ( π − ( y )). The inverse image of the point y is a set of points in f M . However, as statedabove, each of those points is the image of one point under the action of the group G . Since f is G -periodic,the set π − ( y ) is mapped to a single point by f , making the function b f well-defined.To show that b f is unique, assume that f = b f ◦ π = b g ◦ π . Since π is surjective, π ( f M ) = M , hence b f ( x ) = b g ( x ) for all x ∈ M . (cid:4) Theorem 5.4
Let M be a Riemannian manifold and G be a group of isometries acting on M which preserveorientation. Then the solution of the heat equation with G -periodic initial conditions is G -periodic. Proof:
Recall that the heat kernel is invariant under the action of isometries, that means K ( g · x , g · y , t ) = K ( x , y , t ) for all isometries g . To show the solution is G -periodic, we will use a substitution y = g · z with d y = d z . ω ( g · x , t ) = Z M K ( g · x , y , t ) ∧ ∗ y f ( y ) d y = Z M K ( g · x , g · z , t ) ∧ ∗ z f ( g · z ) d z = Z M K ( g · x , g · z , t ) ∧ ∗ z f ( z ) d z = Z M K ( x , z , t ) ∧ ∗ z f ( z ) d z = ω ( x , t ) (cid:4) The proof of Theorem 5.1 involves lifting the initial conditions on the surface M to the universal cover U . This creates a G -periodic initial condition for the heat equation on U , which has, by Theorem 5.4, a G -periodic solution. Then by Proposition 5.3, this solution projects to a unique solution of the heat equationon M . We will conclude by summarizing the results. First, for the simply-connect surfaces, the 1-form heat kerneltakes the form: K ( x , y , t ) = ( I + ∗ x ∗ y ) Z ∞ t d x d y K ( x , y , τ ) dτ . A generalization to manifolds of higher dimension can be found in [8]. Also, the 2-form heat kernel is givenby K ( x , y , t ) = ∗ x ∗ y K ( x , y , t ) . K ( x , y , t ) = 14 πt exp (cid:18) − d R ( x , y ) t (cid:19) Euclidean plane K ( x , y , t ) = 12 π Z ∞ P − + iρ (cosh d H ( x , y )) ρ exp (cid:18) − (cid:18)
14 + ρ (cid:19) t (cid:19) tanh πρdρ hyperbolic plane K ( x , y , t ) = 14 π ∞ X n =0 (2 n + 1) exp ( − n ( n + 1) t ) P n (cos d S ( x , y )) sphereAlso, from Section 5, if we know the covering group for an arbitrary Riemann surface, M , then the heatkernel on M is given by K M ( x , y , t ) = X g ∈ G K U ( e x , g · e y , t ) , where U is the universal cover of M . References [1] Milton Abramowitz and Irene A. Stegun, editors.
Handbook of Mathematical Functions with Formulas,Graphs, and Mathematical Tables . Dover, 1972.[2] Ingolf Buttig and Jurgen Eichhorn. The heat kernel for p -forms on manifolds of bounded geometry. Acta Sci. Math. , 55:33–51, 1991.[3] Isaac Chavel.
Eigenvalues in Riemannian Geometry , volume 115 of
Pure and Applied Mathematics .Academic Press, 1984.[4] Jozef Dodziuk. Harmonic Forms on Complete Manifolds l . In Seminar on Differential Geometry , pages291–302. Princeton University Press, 1982.[5] Harold Donnelly. The Differential Form Spectrum of Hyperbolic Space. manuscripta mathematica ,33:365–85, 1981.[6] I. S. Gradshteyn and I. M. Ryzhik.
Table of Integrals, Series, and Products . Academic Press, 6thedition, 2000.[7] Trevor H. Jones.
The Heat Kernel on Noncompact Riemann Surfaces . PhD thesis, University of NewBrunswick, 2008.[8] Trevor H. Jones. Heat kernel for open manifolds.
Differential Geometry and its Applications , 28(5):518–522, October 2010.[9] H. P. McKean. An upper bound to the spectrum of ∆ on a manifold of negative curvature.
Journal ofDifferential Geometry , 4:359–366, 1970.[10] Fritz Oberhettinger and T. P. Higgins.