You cannot hear the shape of a hyperbolic drum
Ben Linowitz
Department of Mathematics
University of Michigan
Outline
Motivation: Can you hear the shape of a drum?
Vigneras’ construction
Embedding orders into quaternion algebras
Isospectral non-isometric hyperbolic surfaces of small volume
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
Let M be a compact Riemannian manifold.
The spectrum of M is the multiset of eigenvalues of the
Laplace-Beltrami operator of M.
Inverse spectral geometry asks for the extent to which the
spectrum of M determines its geometry and topology.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
Theorem (Weyl, 1911)
The spectrum of M determines the volume.
More precisely,
Vol(M)
#{λ : λ < x}
=
.
x→∞
x
2π
lim
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
Leon Green (1960) asked if the spectrum of M determines its
isometry class.
The spectrum of M is essentially the collection of frequencies
produced by a drumhead shaped like M.
Mark Kac (1966) popularized this question for
planar domains:
Can you hear the shape of a drum?
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
Milnor had already shown that shown that isometry class is not in
general a spectral invariant.
Theorem (Milnor, 1964)
There exist lattices Γ1 , Γ2 ⊂ R16 such
that the tori R16 /Γ1 and R16 /Γ2 are
isospectral but not isometric.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
Kac’s question was finally answered in 1992.
Theorem (Gordon, Webb, Wolpert, 1992)
One cannot hear the shape of a drum.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
What about vibrating surfaces? Specifically, compact Riemannian
manifolds of dimension 2 without boundary?
Such a surface X still has a Laplacian ∆ = −div grad and a
corresponding spectrum of eigenvalues, the frequencies at which
the surface will vibrate.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
The hyperbolic plane H2 is a simply connected surface with
constant curvature −1 and can be be modelled by the disc (Circle
Limit IV, by M.C. Escher):
The symmetries of this diagram, mapping one angel to any other
angel, form a group of isometries acting on H2 . A discrete
subgroup of orientation-preserving isometries of H2 is called a
Fuchsian group.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
Vigneras later constructed isospectral hyperbolic 2 and 3-manifolds.
Theorem (Vigneras, 1980)
There exist isospectral but not
isometric hyperbolic 2 and 3-manifolds.
Actually, Vigneras’ result is much stronger and proves the existence
of isospectral but not isometric Riemmanian n-manifolds for every
n ≥ 2.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
In 1994, Colin Maclachlan (Aberdeen, Scotland) and Gerhard
Rosenberger (Dortmund, Germany) claimed to have produced a
pair of orbifolds of genus zero using Vigneras’ method, but their
examples fail due to an arithmetic condition known as selectivity.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
Virtually all examples of isospectral but not isometric manifolds
rely on a general method due to Sunada.
Let G be a finite group with subgroups H1 and H2 .
H1 and H2 are almost conjugate if for every g ∈ G
#([g ] ∩ H1 ) = #([g ] ∩ H2 ),
where [g ] denotes the G -conjugacy class of g .
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
Theorem (Sunada, 1985)
Let M be a Riemannian manifold upon which a
finite group G acts by isometries. If H1 and H2
are almost conjugate subgroups of G then the
quotient manifolds H1 \M and H2 \M are
isospectral.
Sunada’s theorem has since been generalized in a number of ways;
for example, to Lie groups.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
Most of the ideas that we’ve seen thus far have their origins in
number theory.
Theorem (Perlis, 1977)
Let F /Q be a finite Galois extension containing
subfields F1 , F2 . Then ζF1 (s) = ζF2 (s) if and
only if Gal(F /F1 ) and Gal(F /F2 ) are almost
conjugate subgroups of Gal(F /Q).
Perlis does not use the term almost conjugate but rather
Gassmann-equivalent. Fritz Gassmann studied subgroups of this
type in 1926.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Vigneras’ construction
Vigneras’ examples stand out because they cannot be constructed
using Sunada’s method or its variants.
They are somewhat less understood by the differential geometry
community because of the background required to construct them.
In particular, they arise from orders in quaternion algebras.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Vigneras’ construction
Recall that Isom+ (H2 ) ∼
= PSL2 (R).
Every orientable hyperbolic 2-orbifold is of the form H2 /Γ for some
discrete subgroup Γ of PSL2 (R).
We want to generalize the following construction of PSL2 (Z):
M2 (Q) ⊃ M2 (Z) −→ SL2 (Z) −→ PSL2 (Z).
We will replace M2 (Q) with a quaternion algebra and M2 (Z) with
a quaternion order.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
A brief introduction to quaternion algebras and orders
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
In the 1830s and 1840s William Rowan Hamilton sought a number
system which would play a role in three-dimensional geometry
analogous to that of the complex numbers for two-dimensional
geometry.
“Every morning in the early part of the above-cited month [October
1843], on my coming down to breakfast, your (then) little brother
William Edwin, and yourself, used to ask me: Well, Papa, can you
multiply triplets? Whereto I was always obliged to reply, with a sad
shake of the head: No, I can only add and subtract them.”
– Hamilton (in a letter to his son)
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
Theorem (Hamilton, 1843)
The R-algebra H with basis {1, i, j, ij} and
defining relations
i 2 = −1
j 2 = −1
ij = −ji
is a four-dimensional division algebra.
Hamilton was so excited by this discovery that he carved these
relations into the stone of the Brougham Bridge.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
Let’s write (−1, −1, R) in place of H.
This notation suggests a number of ways to generalize H.
For instance, let (1, 1, R) be the R-algebra with basis {1, i, j, ij}
and defining relations
i2 = 1
j2 = 1
Then (1, 1, R) ∼
= M2 (R) via i 7→
Ben Linowitz
0 1
1 0
ij = −ji.
and j 7→
1 0
.
0 −1
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
More generally, we can define (a, b, R) to be the R-algebra with
basis {1, i, j, ij} and defining relations
i2 = a
j2 = b
ij = −ji
a, b ∈ R∗ .
It’s not too hard to show that
(a, b, R) ∼
= H if a, b < 0 and
(a, b, R) ∼
= M2 (R) otherwise.
Thus (a, b, R) is either a division algebra or else M2 (R).
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
There are other ways that we could have generalized (−1, −1, R).
If F is a field of characteristic zero and a, b ∈ F ∗ we can define the
generalized quaternion algebra (a, b, F ).
Let F = Q and consider the Q-algebra (−1, −1, Q).
As (−1, −1, Q) ( (−1, −1, R), we see that (−1, −1, Q) is a
division algebra.
As before we also see that (1, 1, Q) ∼
= M2 (Q).
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
Recall that if F = R, (a, b, F ) is a division algebra or else M2 (R).
Theorem (Wedderburn)
For any field F , if the F -algebra (a, b, F ) is
not a division algebra then
(a, b, F ) ∼
= M2 (F ).
Note that the case F = R is special in that in general there will
not be a unique quaternion division algebra over F .
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
A “complex” example: F = C.
Let A be a quaternion algebra over C.
By the fundamental theorem of algebra, A cannot be a division
algebra.
Then by Wedderburn’s theorem, A ∼
= M2 (C).
Thus M2 (C) is the only quaternion algebra over C.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
Extension of scalars:
Let F be a field and F 0 /F a field extension.
If A is a quaternion algebra over F then we can consider the
quaternion algebra A ⊗F F 0 over F 0 .
Concretely, if A = (a, b, F ) then A ⊗F F 0 = (a, b, F 0 ).
This is especially important in number theoretic applications.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
Let F be a number field, p a prime of F and Fp the p-adic
completion of F .
If A is a quaternion algebra over F then define Ap := A ⊗F Fp .
By Wedderburn’s theorem, Ap is either a matrix algebra or else a
division algebra.
If Ap is a division algebra then p ramifies in A.
If Ap is a matrix algebra then p splits in A.
We’ve already seen that all complex primes split in A.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Quaternion Algebras
Reduced norm:
Let A be a quaternion algebra over a number field F .
We’ve already seen that we have an embedding A ,→ M2 (C).
The reduced norm of A is the composite map
A ,→ M2 (C) →det C
For A = M2 (F ) the reduced norm coincides with the determinant.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Orders in quaternion algebras
Let F be a number field with ring of integers OF .
An order of an F -algebra is a subring which is also a finitely
generated OF -module containing an F -basis of the algebra.
Example 1: Z[i] is a quadratic order of the Q-algebra Q(i).
Example 2: M2 (Z) is a maximal order of M2 (Q).
Example 3: OF [i, j] is always an order of the F -algebra (a, b, F )
when a, b ∈ OF .
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
From orders to surfaces
Constructing surfaces from quaternion orders
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
From orders to surfaces
Let K be a totally real number field.
Let B/K be a quaternion algebra in which a unique real prime
splits.
Let O1 , O2 be non-conjugate maximal orders of B (if they exist!).
Consider the representation ρ : B → M2 (R).
Restricting ρ to Oi1 and projecting onto PSL2 (R) gives embeddings
ρ̄ : Oi1 → PSL2 (R)
Ben Linowitz
i = 1, 2.
You cannot hear the shape of a hyperbolic drum
From orders to surfaces
ρ̄(O11 ) and ρ̄(O21 ) are discrete subgroups of isometries with finite
covolume.
If B is a division algebra, then ρ̄(O11 ) and ρ̄(O21 ) are cocompact.
If ρ̄(O11 ) and ρ̄(O21 ) are torsion-free then ρ̄(O11 )\H2 and ρ̄(O21 )\H2
are hyperbolic 2-manifolds.
The latter condition amounts to saying that no cyclotomic field
embeds into B.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Vigneras’ construction
Theorem (Vigneras)
Suppose that O1 and O2 have the property that for every
quadratic field extension L of K and order Ω ⊂ L, there exists an
embedding of Ω into O1 if and only if there exists an embedding of
Ω into O2 . Then the surfaces ρ̄(O11 )\H2 and ρ̄(O21 )\H2 are
isospectral but not isometric.
ρ̄(O11 )\H2 and ρ̄(O21 )\H2 are not isometric:
An isometry would lift to an isometry of H2 , given as conjugation
by an element of PSL2 (R). This can be pulled back into B to
show that O11 and O21 are conjugate in B × . But this implies that
O1 and O2 are conjugate, a contradiction.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Vigneras’ construction
ρ̄(O11 )\H2 and ρ̄(O21 )\H2 are isospectral:
We use the fact that two surfaces are isospectral if they have the
same geodesic length spectra.
Let γ ∈ ρ̄(O11 ) be a hyperbolic element and `(γ) the corresponding
value in the spectrum of ρ̄(O11 )\H2 . Let u ∈ O1 be the element
whose image in PSL2 (R) is γ.
Set Ω := OK [u]. It is well-known that the multiplicity to which
`(γ) occurs in the spectrum of ρ̄(O11 )\H2 is equal to the number
Emb(Ω, O1 )/O11 of embeddings of Ω into O1 modulo the action of
O11 .
Working adelically, one can show that if Ω embeds into both O1
and O2 then Emb(Ω, O1 )/O11 = Emb(Ω, O2 )/O21 .
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Embedding orders into quaternion algebras
How can we tell if the hypothesis in Vigneras’ theorem is satisfied?
That is, if we only know that Ω ,→ O1 , how can we tell if Ω ,→ O2 ?
Before answering this question, let’s take a step back.
Let K be any number field and L/K a quadratic field extension.
Let B/K be a quaternion algebra.
Important problem:
Determine when there exists an embedding of K -algebras L ,→ B.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Embedding orders into quaternion algebras
Theorem (Albert-Brauer-Hasse-Noether, 1931)
There is an embedding of L into B if and only if no prime p of K
for which Bp = B ⊗K Kp is a division algebra splits in L/K .
The actual ABHN theorem is more general and concerns
embeddings of fields into central simple algebras of finite degree.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Embedding orders into quaternion algebras
Partial Proof:
Suppose that we have
K ⊂ L ⊂ B.
Extending scalars to Kp gives us
Kp ⊂ L ⊗K Kp ⊂ Bp .
If p ramifies in B but splits in L/K , then Bp is a division algebra
and L ⊗K Kp ∼
= Kp2 .
Contradiction!
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Embedding orders into quaternion algebras
It’s natural to ask for an integral analogue of the ABHN theorem.
Such an analogue would ask when an order Ω ⊂ L embeds into an
order O ⊂ B.
Work towards such a analogue began in the 1930s with the work of
Chevalley.
In what follows we’ll talk about some recent progress.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Can you hear the shape of a drum?
Theorem (Chevalley, 1936)
Let B = M2 (K ) and L/K be a quadratic field
extension. If Ω = OL then the proportion of
isomorphism classes of maximal orders whose
representatives admit an embedding of Ω is equal
to
[HK ∩ L : K ]−1 ,
where HK is the Hilbert class field of K .
Remarks:
1. Since [L : K ] = 2, the proportion in the theorem is
1
2
or 1.
2. If the above proportion is 1, then Ω embeds into every maximal
order of B and the hypothesis in Vigneras’ theorem on isospectral
surfaces is satisfied.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Embedding orders into quaternion algebras
Afterwards, partial generalizations of Chevalley’s result were
obtained by various authors (Vigneras, Brzezinski, etc).
In 1999 Chinburg and Friedman proved:
Theorem (Chinburg-Friedman, 1999)
Suppose that B is not a totally definite quaternion algebra and
L ⊂ B is a maximal subfield. If Ω ⊂ L is an order then Ω can be
embedded into either every maximal order of B or those
representing exactly one half of the isomorphism classes of
maximal orders.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Embedding orders into quaternion algebras
The full statement of Chinburg and Friedman’s theorem also
describes necessary and sufficient conditions for each proportion to
occur. For instance, a necessary condition for the latter proportion
is that both B and L/K be unramified at all finite primes of K .
Chinburg and Friedman’s theorem is the reason why the examples
of Maclachlan and Rosenberger failed to be isospectral. In their
situation the relevant proportion was 21 , so the hypothesis in
Vigneras’ theorem was not satisfied. Because of the explicitness of
Chinburg and Friedman’s theorem, it is actually possible to specify
a length which lies in the geodesic length spectrum of exactly one
of Maclachlan and Rosenberger’s surfaces.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Isospectral non-isometric hyperbolic surfaces of small
volume
Vigneras’ original example was as follows:
√
√
K = Q( 10), Ram(B) = {p∞ , (3), (7), (11 + 3 10)}.
The only values
cos(π/n) ∈ K are n = 2, 3, but (7)
√
√ of n for which
splits in K ( −1) and K ( −3) so neither embeds into B.
This algebra has type number 2, so we can find non-conjugate
maximal orders O1 and O2 .
By Chinburg and Friedman’s theorem, an order Ω embeds into O1
if and only if it embeds into O2 .
By Vigneras’ theorem, ρ̄(O11 )\H2 and ρ̄(O21 )\H2 are isospectral
but not isometric compact hyperbolic 2-manifolds.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Isospectral non-isometric hyperbolic surfaces of small
volume
Vigneras’ original example was as follows:
√
√
K = Q( 10), Ram(B) = {p∞ , (3), (7), (11 + 3 10)}.
The only values
cos(π/n) ∈ K are n = 2, 3, but (7)
√
√ of n for which
splits in K ( −1) and K ( −3) so neither embeds into B.
This algebra has type number 2, so we can find non-conjugate
maximal orders O1 and O2 .
By Chinburg and Friedman’s theorem, an order Ω embeds into O1
if and only if it embeds into O2 .
By Vigneras’ theorem, ρ̄(O11 )\H2 and ρ̄(O21 )\H2 are isospectral
but not isometric compact hyperbolic 2-manifolds.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Isospectral non-isometric hyperbolic surfaces of small
volume
Vigneras fixed her example in her book.
√
She did this by setting Ram(B) = {p∞ , (7), (11), (11 + 3 10)}.
We can calculate the example’s volume using a formula of Borel:
3/2
Vol(ρ̄(O11 )\H2 ) =
8π∆K ζK (2)
Q
p|∆(B) N(p)
(4π 2 )[K :Q]
−1
,
where ∆K (respectively ∆(B)) is the discriminant of K
(respectively B).
In this case Vol = 201600 ∗ 2π and g = 100801!
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Isospectral non-isometric hyperbolic surfaces of small
volume
Theorem (L., Voight)
Let Γ, Γ0 be maximal arithmetic Fuchsian groups such that the
orbifolds X = Γ\H and X 0 = Γ0 \H are isospectral but not
isometric.
Then Vol(X ) = Vol(X 0 ) ≥ 23π/6, and this bound is achieved by
exactly 6 distinct pairs of genus zero surfaces.
These 6 = 3 + 3 pairs of surfaces fall naturally into 2 Galois orbits.
We also have a result with a pair of manifolds of genus 6 and
volume 20π.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Isospectral non-isometric hyperbolic surfaces of small
volume
8
7
2
7
9
6
6 5
2
4
3
3
5
8
2
4
3
2
1
12
10
1
9
14
13
11
12
10
9
8
7
7
1
13
12
11
5
6
11
A pair of isospectral hyperbolic 2-orbifolds with genus 0.
Ben Linowitz
You cannot hear the shape of a hyperbolic drum
Thanks!
Ben Linowitz
You cannot hear the shape of a hyperbolic drum