Note: Laplacian Constant Curvature

Finding Fundamental Solutions in Euclidean, Spherical and Hyperbolic Geometry

Steve Trettel

April 3, 2023 |

Here’s a quick computation of the fundamental solution to the laplacian $Δ f = δ$ in the constant curvature geometries. This is well known but I always have to hunt a bit when I need them or end up re-deriving so figured I’d write it down.

(Laplacian in Constant Curvature) If

X^{n}

is an

n

-dimensional constant curvature geometry,

O \in X

a point and

r = dist (-, O)

is the Riemannian distance from

O

, then the fundamental solution for the Laplacian based at

O

F (r) = \frac{1}{ω_{n - 1}} \int \frac{d r}{k (r)^{n - 1}}

For

ω_{n - 1}

the volume of the unit

n - 1

sphere, and

k (r)

is equal to

\sin (r)

r

\sinh (r)

depending on if

X

is spherical, Euclidean or hyperbolic respectively.

And here’s some specific ones in low dimensions

The fundamental solutions to the laplacian in constant curvature are proportional to the following functions, where $r$ is the geodesic distance from the chosen origin.

Two Dimensions:

$E^{2}$ : $F (r) = \frac{1}{2 π} \log (r)$
$S^{2}$ : $F (r) = \frac{1}{2 π} \log | \tan \frac{r}{2} |$
$H^{2}$ : $F (r) = \frac{1}{2 π} \log | \tanh \frac{r}{2} |$

Three Dimensions:

$E^{3}$ : $F (r) = \frac{- 1}{4 π} \frac{1}{r}$
$S^{3}$ : $F (r) = \frac{- 1}{4 π} \frac{1}{\tan r}$
$H^{3}$ : $F (r) = \frac{- 1}{4 π} \frac{1}{\tanh r}$

The Trick

In Euclidean, hyperbolic and spherical geometry we work in polar coordinates, where the laplacian factors additively into two components, one differentiating along the radius and the other along $n - 1$ spheres, the level sets of constant $r$ .

$Δ_{E^{n}} f = \frac{1}{r^{n - 1}} \frac{\partial}{\partial r} (r^{n - 1} \frac{\partial f}{\partial r}) + \frac{1}{r^{2}} Δ_{S^{n - 1}}$

$Δ_{H^{n}} f = \frac{1}{\sinh^{n - 1} (r)} \frac{\partial}{\partial r} (\sinh^{n - 1} (r) \frac{\partial f}{\partial r}) + \frac{1}{\sinh^{2} (r)} Δ_{S^{n - 1}}$

$Δ_{S^{n}} f = \frac{1}{\sin^{n - 1} (r)} \frac{\partial}{\partial r} (\sin^{n - 1} (r) \frac{\partial f}{\partial r}) + \frac{1}{\sin^{2} (r)} Δ_{S^{n - 1}}$

Looking at these coordinate expressions we take advantage of a certain similarity - they are all the same up to a choice of $\sin (r)$ , $r$ or $\sinh (r)$ . Thus, we can give a uniform expression abstracting these differences into a function $k (r)$ :

$Δ f = \frac{1}{k (r)^{n - 1}} \frac{\partial}{\partial r} (k (r)^{n - 1} \frac{\partial f}{\partial r}) + \frac{1}{k (r)^{2}} Δ_{S^{n - 1}}$

This still looks a bit scary, but because everything is symmetric under rotations we might think to look for solutions sharing the same symmetry group. We’ll call such solutions radial as they depend only on $r$ . In this case things simplify drastically

(Radial Solutions) In any of the constant curvature geometries, if

f

is a radial solution to

Δ f = 0

then

f \equiv f (r)

is proporitional to the following integral:

f (r) \propto \int \frac{1}{k (r)^{n - 1}}, d r

If $f = f (r)$ then it is constant along $n - 1$ spheres, and so its restriction to the unit sphere satisfies $Δ_{S^{n - 1}} f = 0$ . But if this term is zero, then the entire expression for the Laplacian collapses to

$Δ f = \frac{1}{k (r)^{n - 1}} \frac{\partial}{\partial r} (k (r)^{n - 1} \frac{\partial f}{\partial r}) = 0$

Cancelling the prefactor of $1 / k^{n - 1}$ we see

$\frac{\partial}{\partial r} (k (r)^{n - 1} \frac{\partial f}{\partial r}) = 0$

and as this derivative is identically zero, the quantity differentiated must be constant. That is, for some $C \in R$ we have

$k (r)^{n - 1} \frac{\partial f}{\partial r} = C$

And dividing through by $k^{n - 1}$ produces a differential equation for $f (r)$ , which upon integrating yields our formula

$f (r) = \int \frac{C}{k (r)^{n - 1}}, d r$

Our goal is to promote a radial solution to a fundamental solution, which means finding the correct multiplying factor $C$ . In practice, the easiest way for me to do this has been to find a function with suitably normalized derivative and then scale by the volume of the unit $n - 1$ sphere.

(Radial to Fundamental) If

\tilde{F} (r)

any radial solution to

Δ \tilde{F} = 0

, then

F (r) = \frac{1}{vol (S_{X} (1))} \frac{\tilde{F} (r)}{{\tilde{F}}^{'} (1)}

is a fundamental solution to the laplacian, where

S_{X} (1)

is the sphere of radius

1

X

centered at

r = 0

Let $X$ be one of the constant curvature geometries, and $d V$ be its Riemannian volume form. If $Δ F$ is to act like the delta distribution under integration, then for any region $Ω \subset X$ we must have

$\int_{Ω} Δ F d vol = {\begin{cases} 0 & O \notin Ω 1 & O \in Ω \end{cases}$

Where $O$ is the point $δ$ is based at ( $r = 0$ in our coordinate system). On any region not containing $O$ this is trivial as our radial solution $\tilde{F}$ has constantly zero Laplacian everywhere it is defined. So, we only need to concern ourselves with regions containing $O$ . For specificity, we can consider the unit ball $Ω = B_{X} (1) \subset X$ . Here we compute using the divergence theorem and set the result equal to 1 (since $O \in Ω$ ):

$\begin{aligned} \int_{B_{X} (1)} Δ F, d V & = \int_{B_{X} (1)} div \cdot grad F, d V & = \int_{\partial B_{X} (1)} ⟨ grad F, \hat{n} ⟩ d S & = \int_{S_{X} (1)} ⟨ grad F, \partial_{r} ⟩ d S & = \int_{S_{X} (1)} \frac{\partial F}{\partial r}, d S \end{aligned}$

Where $d S$ is the Riemannian $n - 1$ volume form induced on the sphere $S_{X} (1)$ by the overall volume $d vol$ . Since $F$ (and hence $F^{'}$ ) are functions of $r$ alone they are constant on the sphere $S_{X} (1)$ , and restricting to the sphere is the same as plugging in $r = 1$ . This allows the one term above to be pulled out of the integral, giving

$= \frac{\partial F}{\partial r} (1) \int_{S_{X} (1)} d S = \frac{\partial F}{\partial r} (1) vol (S_{X} (1))$

This quantity is supposed to be $1$ . If it is not, we can simply divide the current definition of $F$ by it, and notice this constant multiple carries all the way through the calculation unscathed.

In fact, we can even do a little better than this since we know the functions $k (r)$ that we are integrating. By the fundamental theorem of calculus

$F^{'} (r) = \frac{\partial}{\partial r} F (r) = \frac{\partial}{\partial r} \int \frac{1}{k (r)^{n - 1}}, d r = \frac{1}{k (r)^{n - 1}}$

Thus $F^{'} (1)$ is none other than $1 / k (1)^{n - 1}$ , and we may as well incorporate this directly:

X

is a constant curvature geometry, then a fundamental solution

F

to the laplacian is given by

F (r) = \frac{k (1)^{n - 1}}{vol (S_{X} (1))} \int \frac{d r}{k (r)^{n - 1}} d r

$F$ is a scalar multiple of $\int \frac{d r}{k (r)^{n - 1}}$ which we know to be harmonic on $X ∖ O$ . And, as we saw before integrating the radial derivative over the unit sphere gives $\begin{aligned} \int_{S_{X} (1)} \frac{\partial F}{\partial r} d S & = F^{'} (1) \int_{S_{X} (1)} d S & = F^{'} (1) vol (S_{1} (X)) \end{aligned}$ But now computing $F^{'} (1)$ we see we have correctly arranged things: $\begin{aligned} \frac{\partial}{\partial r} |_{r = 1} F (r) & = \frac{\partial}{\partial r} |_{r = 1} \frac{k (1)^{n - 1}}{vol (S_{X} (1))} \int \frac{d r}{k (r)^{n - 1}}, d r & = \frac{k (1)^{n - 1}}{vol (S_{X} (1))} \frac{\partial}{\partial r} |_{r = 1} \int \frac{d r}{k (r)^{n - 1}}, d r & = \frac{k (1)^{n - 1}}{vol (S_{X} (1))} \frac{1}{k (1)^{n - 1}} & = \frac{1}{vol S_{X} (1)} \end{aligned}$

Plugging this back into the original yields $\begin{aligned} \int_{S_{X} (1)} \frac{\partial F}{\partial r} d S & = F^{'} (1) vol (S_{1} (X)) & = \frac{1}{vol S_{X} (1)} vol (S_{1} (X)) & = 1 \end{aligned}$

But its even a little better than this! Because we are in the very special situation of constant curvature, we can actually compute the “volume” (ie surface area) of these spheres directly. For each $n$ let $ω_{n}$ be the volume of the unit $n$ -sphere in $E^{n}$ . Then for all radii $R$

$$\mathbb{E}^n:\hspace{1cm} \mathrm{vol}{\mathbb{S}{\mathbb{E}}(R)}=\omega{n-1} R^{n-1}=\omega_{n-1} k(r)^{n-1}\mathbb{S}^n:\hspace{1cm} \mathrm{vol}{\mathbb{S}_{\mathbb{S}}(R)}=\omega_{n-1} \sin(R)^{n-1}=\omega_{n-1} k(r)^{n-1}\mathbb{H}^n:\hspace{1cm} \mathrm{vol}{\mathbb{S}_{\mathbb{H}}(R)}=\omega_{n-1} \sinh(R)^{n-1}=\omega_{n-1} k(r)^{n-1}$$

Thus uniform across all geometries we have $\mathrm{vol}(\mathbb{S}X(1))=\omega{n-1}k(1)^{n-1} $a n d p l u g g i n g t h i s i n t o o u r c o r o l l a r y w e s e e t h a t t h e f a c t o r s o f$ k(1)^{n-1}$ always cancel out. This gives our main theorem:

(Laplacian in Constant Curvature) If

X^{n}

is an

n

-dimensional constant curvature geometry,

O \in X

a point and

r = dist (-, O)

is the Riemannian distance from

O

, then the fundamental solution for the Laplacian based at

O

F (r) = \frac{1}{ω_{n - 1}} \int \frac{d r}{k (r)^{n - 1}}

For

ω_{n - 1}

the volume of the unit

n - 1

sphere, and

k (r)

is equal to

\sin (r)

r

\sinh (r)

depending on if

X

is spherical, Euclidean or hyperbolic respectively.

Euclidean Space

Here are the explicit solutions for Euclidean spaces of low dimension, with $k (r) = r$ .

(2D)

F (r) = \frac{1}{2 π} \log r

First we compute the integral

\int \frac{d r}{k (r)} = \int \frac{1}{r}; d r = \log (r)

And then the normalizing factor: the “volume” of 1 dimensional sphere is just the cirumference of the unit circle, or

2 π

. Thus

F (r) = \frac{1}{2 π} \log (r)

(3D)

F (r) = \frac{- 1}{4 π} \frac{1}{r}

We first compute the integral to find a radial solution:

\int \frac{d r}{k (r)^{2}} = \int \frac{1}{r^{2}}, d r = \frac{- 1}{r}

And then normalize by

4 π

, the surface area of the unit sphere

F (r) = \frac{- 1}{4 π} \frac{1}{r}

The same pattern continues in higher dimensions,

(n-Dimensions)

Spherical Space

Here are the explicit solutions for spherical spaces of low dimension, with $k (r) = \sin (r)$ .

(2D)

F (r) = \frac{1}{2 π} \log | \tan \frac{r}{2} |

To find a radial solution we must compute $\int \frac{d r}{\sin r} = \int \frac{1}{\sin (r)}, d r = \int \csc (r), d r$ This is one of those annoying integrals you learn in calculus and then forget, because deriving it needs a trick. Anyway, $\int \csc (r), d r = \ln | \csc (r) - \cot (r) |$ There are many trig identites one could use to “simplify” here: for example this is equal to $- \ln | \csc r + \cot r |$ (which I didn’t know until wolfram alpha told me so). For us, its nicest to simplify as $\csc r - \cot r = \frac{1}{\sin r} - \frac{\cos r}{\sin r} = \frac{1 - \cos r}{\sin r} = \tan \frac{r}{2}$ where the last equality is the tangent half angle formula.

We are in 2 dimensions so the normalizing factor is the unit 1-sphere’s volume or $2 π$ : $F (r) = \frac{1}{2 π} \log | \tan \frac{r}{2} |$

(3D)

F (r) = \frac{1}{4 π} \frac{1}{\tan (r)}

Computing a radial solution to start, we recognize an easier integral from calculus involving the cotangent

\int \frac{d r}{k (r)^{2}} = \int \frac{1}{\sin^{2} (r)}; d r = \int \csc^{2} (r), d r = - \cot (r) = \frac{- 1}{\tan (r)}

For the normalizing constant in dimension 3 we look to the unit 2-sphere, whose area is

4 π

F (r) = \frac{- 1}{4 π} \frac{1}{\tan r}

Unlike in Euclidean space the calculations don’t stabilize in difficulty beyond $n = 3$ , and instead get more challenging as powers of $k$ grow in the integrand.

(4D)

F (r) = \frac{- 1}{4 π^{2}} (\frac{\cos r}{\sin^{2} r} - \log | \tan \frac{r}{2} |)

The radial solution is $\int \frac{d r}{k (r)^{3}} = \int \frac{1}{\sin^{3} (r)}; d r = \int \csc^{3} (r), d r$ This is an integral you might see in calculus if you’re looking for really tricky examples at the end of the chapter (perhaps the more commonly seen case is its sibling $\sec^{3} (x)$ ):

$\int \csc^{3} (r), d r = \frac{- \csc r \cot r + \log | \csc r - \cot r |}{2}$

We can simplify this a bit by (i) factoring out the constant $- 1 / 2$ , (ii) expanding cosecant and cotangent in terms of sines and cosines, and treating the last term with the half-angle identity, as we previously did. $= \frac{- 1}{2} (\frac{\cos r}{\sin^{2} r} - \log | \tan \frac{r}{2} |)$

Here the normalizing constant is the surface area of the unit 3-sphere. We can find this if we know the volume of $4 -$ balls grows as $\frac{1}{2} π^{2} R^{4}$ by differentiation, $2 π^{2} R^{3}$ , so the unit sphere has volume $2 π^{2}$ . Dividing by this gives the answer

$\frac{- 1}{4 π^{2}} (\frac{\cos r}{\sin^{2} r} - \log | \tan \frac{r}{2} |)$

In higher dimensions, we continue having to confront the integral of $1 / \sin^{N} (r)$ for various $N$ . Such integrals are doable (though not fun), with the tricks basically sorting by even and odd dimensions.

Hyperbolic Space

Here are the explicit solutions for hyperbolic spaces of low dimension, with $k (r) = \sinh (r)$ .

(2D)

F (r) = \frac{1}{2 π} \log | \tan \frac{r}{2} |

Very analogous to the spherical case, we have the integral

\int \frac{d r}{k (r)} = \int \frac{1}{\sinh r}, d r = \int csch r, d r

This integral can be accomplished by substitution and partial fractions unpacking the definition of

\sinh

in terms of the exponential…or just remembered in analogy with the spherical trigonometric case:

= \log | csch r - \coth r |

Writing in terms of

\sinh

and

\cosh

lets us recognize the hyperbolic tangent half angle identity:

\coth r - csch = \frac{\cosh r - 1}{\sinh r} = \tanh \frac{r}{2}

Where we have switched the order of the terms from the previous lines, as they occur inside an absolute value so it does not affect anything. Finally dividing by the normalizing constant of 2D, or

2 π

yields the answer

F (r) = \frac{1}{2 π} \log | \tan \frac{r}{2} |

(3D)

F (r) = \frac{- 1}{4 π} \frac{1}{\tanh r}

The radial solution is calculable as

\int \frac{d r}{k (r)^{2}} = \int \frac{1}{\sinh^{2} r}, d r = \int {csch}^{2} r, d r = - \coth r = \frac{- 1}{\tanh r}

The normalizing constant in three dimensions is

4 π

, so

F (r) = \frac{- 1}{4 π} \frac{1}{\tanh r}

(4D)

Higher Dimensions:

◀ Note: Laplacian Chain Rule

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