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§3. Connection with Confocal Quadrics 151
The normal to U
These are the desired eigenfunctions, if λ
orthogonal to y, the eigenfunction for λ =0,wehaveP
, ξ = ξ
λ = λ
j
j
at ξjis given by
λ
j
ϕ =(λ
− A)−1ξj.
j
=0. Indeed, since they are
j
ϕ = ϕ; hence for
y
(λ − L)ϕ = Py(λ − A + x ⊗ x)ϕ = Py(ξ + xx, ϕ)=Py(ξ + xQ(x, ξ)),
where we abbreviate Q
by Q.Sincefors = s
λ
j
j
Q(x, ξ)=Q(x + sy, ξ)=Q(ξ)=−1,
we conclude
(λ − L)ϕ = P
(ξ −x)=Py(sy)=0.
y
proving the statement.
This implies that the n−1 normals ϕ
=(λj−A)−1ξj(j =2, ..., n) and y
j
form an orthogonal frame. This is a well-known theorem of elementary geometry,
due to Chasles (Salmon and Fiedler [17]): The normals of two confocal quadrics
, U
U
z
theorem corresponds to the orthogonality of the frame Φ
− A)
at the points of contact with a common tangent are perpendicular. This
z
1
2
−1
ξj. Actually the identity {Φ
, Φ
} =0can be deduced from this
z
z
1
2
= y, Φj=(λj−
1
geometrical fact (see Moser [12]). Another consequence of this geometrical
fact is that the common tangents of n − 1 confocal quadrics U
, ..., U
z
1
z
n−1
generically form a normal congruence, i. e. can be viewed as the normals of
(n −1)-dimensional surfaces (Bianchi [2]). These surfaces are the level surfaces
of a function S = S(x, y) which we will determine below (see Section 4).
d. Joachimsthal’s Integral
In the classical literature one finds for the geodesic flow in the 3 axial
ellipsoid the integral of Joachimsthal. It is given as the product p ·d where for
a given tangent to the ellipsoid d is the diameter of the ellipsoid in the direction
parallel to the tangent and p is the distance from the origin to the tangent plane
of the ellipsoid containing the given tangent. We express this integral in terms
of the integrals G
, G2, G3defined by
1
Φ
=
z
n
j=1
G
j
. (3.12)
z − α
j
152 Geometry of Quadrics and Spectral Theory
One finds
2
pd
= −
Indeed, at a point ξ ∈ U
−1
by A
ξ, x =1and its distance p from x =0by
2
,i.e.,A−1ξ, ξ =1, the tangent plane is given
0
p = |A
j=1
−1ξ|−1
For a tangent vector ξ + η at ξ we have A
2
d
2
so that
2
pd
=
2
By comparing the coefficients of z
(d/dz)Φ
at z =0, one finds for x = ξ, y = η,
z
η, η
=
A−1η, η
A−1η, ηA−2ξ, ξ
−1
in the expansion of (3.12), and taking
η, η =
−1
A
η, ηA−2ξ, ξ = −
3
G
j
j=1
3
−2
α
G
j
.
−1
ξ, η =0and
η, η
3
j=1
3
j=1
j
,
Gj,
.
.
−2
α
Gj,
j
which gives the stated identity. Clearly this integral has the same interpretation
for higher dimensions, but one cannot expect such a geometrical meaning for all
integrals.
§ 4. The Hyperelliptic Curve
a. The Isospectral Manifold M(λ)
We investigate the manifold M = M(λ
which the matrices L = L(x, y) of the form (2.4) have the fixed eigenvalues (λ
, ..., λn). We will assume that the 2n numbers λj, αkare distinct.
1,λ2
, ..., λn) of x, y ∈ R2nfor
1,λ2
§4. The Hyperelliptic Curve 153
In this section we will allow all quantities to be complex and will not go into reality considerations. If the vector fields X
the tangent space of M. Furthermore, if we consider M as being embedded in
the symplectic space (R
2n
,ω) where ω =
are linearly independent, they span
G
j
n
dyj∧dxj,thenM is a Lagrange
j=1
manifold, since
,X
ω(X
G
)=−{Gj,Gk} =0.
G
j
k
Since M can also be characterized by the equations
)
l(α
k
G
= ck,ck= −
k
a(αk)
,
by Sard’s lemma we can assume that the λ
hence X
are linearly independent on M, provided (a, b + c, d) =(0, 0, 0).
G
j
This is a restriction, since on the linear space x
and thus linear dependence of the dG
independence of the dG
,sothatM is a manifold without singularities. It is an
j
can be so chosen that the dGjand
j
∗
∗
= y
k
. We impose the restriction of linear
j
=0one has dG
k
∗
=0,
k
algebraic manifold. In this section we will study M and relate it to the Jacobi
variety of some hyperelliptic curve of genus n − 1.
We first note that there are two trivial symplectic group actions on (R
2n
,ω)
leaving M invariant — which we will factor out.
The first group is discrete and is generated by the n reflections
τ
:(x, y) → (Tkx, Tky)
k
where T
and therefore the spectrum of L(x, y) is invariant under τ
under τ
: xj→ (1 − 2δjk)xj. Clearly
k
x, Tky)=T
L(T
k
.
k
The second group action g
−1
L(x, y)Tk,
k
,i.e.,M is invariant
k
t
is one-dimensional and is generated by the
Hamiltonian
n
1
G =
2
j=1
Gj(x, y)=
The corresponding vector field X
x
= sx + dy, y= −ax −sy,
1
(ax, x +2sx, y + dy, y).
2
is given by the linear differential equations
G
√
with the characteristic exponents ±
−θ,whereθ = ad −s2.
154 Geometry of Quadrics and Spectral Theory
Let Γ be the Abelian group generated by the τk(k =1, 2, ..., n) and
t
the g
(t ∈ C). We are interested in the set of Γ-orbits M/Γ on M.Ifwe
exclude the points on a quadric cone K, this quotient will be a manifold. For
this purpose we construct a cross section which is crossed transversally by the
2
G-orbits. For θ = ad − s
=0, i.e. case (i), one can find a quadratic form
F = a
x, x +2sx, y+ dy, y (4.1)
such that
F = {F, G} =2√−θF.
X
G
This is clear because the map F →{F, G} has the eigenvalues 0, 2α, −2α
where α =
√
−θ. Thus
F = F
t=0
2αt
e
,
and all orbits which do not lie on the cone F =0cross the section F =1
transversally.
In the case (ii) where θ =0but a, s, d are not all zero, one can construct a
quadratic function (4.1) satisfying
X
F = {F, G} = G,
G
so that
F = F
t=0
+ tG.
Thus for orbits outside the cone G =0we have such a cross section in F =0.
Finally, in case (iii), if a = s = d =0we take G =
1
x, x and F = x, y
2
and have a cross section in F =0for all orbits outside the cone G =0.
Thus if we define the cone K by the equation F =0, G =0, x, x =0in
case (i), (ii), (iii) respectively, then,
(M − M ∩ K)/Γ=M
is a manifold of dimension n − 1. Indeed, any orbit gt(x, y) intersects the
orbit τg
manifolds x
trivial bundle over M
t
(x, y) (where τ =idis a product of some τ1, τ2, ..., τn) on the linear
∗
∗
= y
k
=0which do not intersect M. Thus M − M ∩ K is a
k
with complex lines as fibers in cases (ii) and (iii). In
case (i) the fibers are cylinders, i. e. complex lines identified by the periods
−1
t = iπα
.
§4. The Hyperelliptic Curve 155
Incidentally, in case (ii) we can choose the eigenvalues λ
that 2G =
M
= M/Γ.
To take account of the group generated by the τ
u
= x
k
n
Gk=
k=1
2
, vk= xkyk, wk= y
k
n
k=1
ck=0,sothatK ∩ M = ∅. In that case we have
2
, which are invariant under this group and also
k
j
characterize equivalence classes. Moreover, all functions G
polynomials in the u
have the n quadratic relations u
the intersection of 2n quadrics in the hyperplane C
, vk, wk, while F and G are linear functions. Since we
k
kwk
2
= v
, the manifold Mcan be viewed as
k
3n−1
, λ2, ..., λnso
1
we use the functions
(x, y) are quadratic
j
given by F =const.
The object of this section is to prove the following
Theorem 4. If Δ=ad − bc =0and the above assumptions hold, then
the quotient M
curve
=(M − K ∩ M)/Γ is the Jacobi variety g of the hyperelliptic
2
= P(z)=a(z)(r2a(z) − Δl(z)), (4.2)
w
with l(z)=det(zI − L), a(z)=det(zI − A), of genus n −1. Moreover, if
μ
k
n
k=2
defines the Jacobi map of the divisor classes (p
on the hyperelliptic curve into C
z
#
n−j
P (z)
k=2
dz
n−1
n
c
,j=2, 3, ..., n, (4.3)
= s
j
, then the vector fields X
∂
(mod XG)
jk
∂s
k
, ...,pn), pj=(μj,#P (μj)),
2
are given by
G
j
with constant coefficients c
2
Finally, x
, xkyk, y
k
;i.e.,theGj-flow is linear in the sk.
jk
2
(k =1, 2, ..., n) restricted to Mare abelian
k
functions on g and can b e expressed as ratios of θ-functions on g.
The main point of this result is that the linear structure on g given by the X
G
agrees with the linear structure as it is given by Abel’s theorem. The solutions of
any Hamiltonian ϕ(G
θ-functions plus an exponential e
trivial part X
.
G
, ..., Gn)=H can be expressed in terms of Jacobi
1,G2
±αt
or a linear function t coming from the
An alternate form of the polynomial P(z) in (4.2) is
P (z)=a(z)
2
Δdet(sym(C−1) − Q),
j
156 Geometry of Quadrics and Spectral Theory
where sym ( ) denotes the symmetric part of a matrix and
C =
ab
cd
,Q=
Q
(x) Qz(x, y)
z
Q
(x, y) Qz(y)
z
.
We will give the proof of the above statement in the three cases of Section 2:
(i) θ = ad − s
2
=0. We begin with case (i) in detail and then indicate the necessary
= r
2
=0, (ii) θ =0but d =0, and (iii) a = s = d =0but Δ=
modifications for the other cases.
b. An Inverse Spectral Problem
We assume θ = ad − s
2
=0. Without loss of generality we can take
a = d =0and therefore consider
L(x, y)=A + bx ⊗ y + cy ⊗x (4.4)
with bc = −Δ =0, b + c =0,asθ =0. In this case our formula become
l(z)
a(z)
Φ
(x, y)=(b + c)Qz(x, y)+bc(Qz(x)Qz(y) − Q
z
, αjare the eigenvalues of L, A respectively, we have
If λ
j
l(z)=
det(zI − L)
=
det(zI − A)
n
(z −λj),a(z)=
j=1
=1−Φ
n
j=1
(x, y) (4.5)
z
2
(x, y)). (4.6)
z
(z −αj).
To parametrize points on M we use the eigenvalues of another matrix,
namely
, (4.7)
yAPy
where P
= I −
y
M = M(y)=P
y ⊗ y
is — for y, y =0— the projection into the orthogonal
y, y
complement of y. Thus M(y) has the eigenvalue 0 with the eigenvector y and
n − 1 other eigenvalues μ
m(z)=y
, μ3, ..., μn.Ifweset
2
n
2
(z −μj), y2= y, y =
j=2
n
k=1
2
y
k
, (4.8)
we have the identity
m(z)
a(z)
§4. The Hyperelliptic Curve 157
n
2
y
=
det(zI − M )
z
det(zI − A)
= Q
z
(y)=
j=1
2
y
j
z − α
, (4.9)
j
which defines the μ
, μ3, ..., μnas the zeros of Qz(y) provided y2=0.
2
The length y of y is irrelevant for M (y),andweset
μ
= y2.
1
We consider the following «inverse spectral problem»: Given λ
, μ2, ..., μnconstruct the L = L(x, y), M = M(y) so that λjare the
μ
1
eigenvalues of L,sothatμ
μ
= y2.Theλj, μjrepresent 2n variables, which turn out to be sufficient to
1
determine the 2n variables x
, ..., μnand 0 are the eigenvalues of M ,andsothat
2
, ykexcept for branchings and singularities, which
k
, λ2, ..., λn,
1
will not be investigated here.
To carry out the construction we require some standard formulae for elliptic
coordinates. Since μ
, ..., μndepend only on the yk, we will determine the
1
latter first. From (4.9) one reads off that
)
m(α
2
y
=
j
In the real case when α
<μ2<α2<... <μn<αnand μ1> 0, the right
1
side of (4.10) is positive and allows for 2
, μ3, ..., μncan be viewed as coordinates on the sphere S
μ
2
orthogonal coordinate system
1
. In the following we will derive these properties
j
. (4.10)
a(αj)
n
real vectors y.Forμ1= y2=1the
n−1
which form an
but ignore the reality conditions.
By logarithmic differentiation of (4.10) one finds
⎧
1
⎪
⎪
y for k =1,
⎨
∂y
∂μ
k
2μ
⎪
⎪
⎩
1
1
(μ
− A)−1y for k 2,
k
2
=
(4.11)
which implies
1
E. Rosochatius refers to them as «elliptische Kugelkoordinaten»: see [16], in particular p. 29.
&
∂y
∂μ
k
'
∂y
,
∂μ
= g
j
jδjk
,
158 Geometry of Quadrics and Spectral Theory
with
We verify (4.12) only for distinct μ
⎧
1
⎪
⎪
⎪
⎨
4μ
⎪
⎪
⎪
⎩
1
(μj)
m
−
4a(μj)
, μ3, ..., μnand for μ1=0,usingthe
2
=
g
j
for j =1,
(4.12)
for j 2.
resolvent identity
− A)−1(μj− A)−1=
(μ
k
− 1
μj− μ
((μj− A)−1− (μk− A)−1).
k
From (4.11) we find for j = k, j, k 2,
(
∂y
∂μ
k
'
∂y
,
∂μ
j
=
4(μj− μk)
− 1
(Q
(y) − Q
μ
j
(y)) = 0,
μ
k
and for j = k 2 we compute
&
'
2
∂y
∂μ
j
1
(μ
=
− A)−2y, y = −
j
4
1
d
4
dz
Q
z
(y)
z=μ
,
j
which by (4.9) gives the desired result for j 2.Forj 2 we have by (4.11)
&
∂y
∂μ
1
&
,
∂y
∂μ
∂y
∂μ
'
j
'
2
1
= −
=
4μ
1
4μ
2
1
1
Q
1
y2=
(y)=0,
μ
j
4μ
1
,
1
which establishes (4.12). Thus we have
n
k=1
dy
n
2
=
k
j=1
gjdμ
2
.
j
Having determined y, we turn to the construction of x, which we represent
in the orthogonal frame ∂y/∂μ
as
j
x =
n
j=1
X
∂y
. (4.13)
j
∂μ
j
§4. The Hyperelliptic Curve 159
Taking the inner product with ∂y/∂μ
, we find for the coefficients Xj,us-
k
ing (4.11), (4.12),
⎧
1
⎪
⎪
x, y for k =1,
⎨
g
kXk
=&x,
∂y
∂μ
'
k
2μ
⎪
⎪
⎩
1
1
Q
(x, y) for k 2.
μ
k
2
(4.14)
=
The terms on the right-hand side can be expressed in terms of λ, μ, thus determining x as a function of λ, μ. To show this we compare the coefficients of z
−1
as z →∞in (4.6):
n
tr(L − A)=
so that
To compute Q
(x, y) for j 2,wesetz = μjin (4.6), (4.5), using Q
μ
j
(λj− αj)=(b + c)x, y =2sx, y, (4.15)
j=1
n
x, y =
1
2s
(λj− αj).
j=1
(y)=
μ
j
=0:
l(z)
for z = μ
a(z)
=(b + c)Q
1 −
, ..., μn. This is a quadratic equation for Q
2,μ3
(x, y) −bcQ
z
Δ=−bc, 2r = b − c, we obtain for z = μ
Q
(x, y) −
z
P (z)=(a(z)r
b + c
2bc
2
=
− l(z)Δ)a(z).
, j 2,
j
1
a(z)Δ
#
2
(x, y)
z
P (z),
(x, y).Ifweset
μ
j
(4.16)
Thus by (4.13), (4.14), (4.15), (4.16) we can express x, y in terms of λ, μ,
solving the inverse spectral problem.
To construct (M − K ∩M)/Γ, we construct the orbits of G =
= sx, y,i.e.of
The function F of (4.1) can be chosen as F = y
x
= sx, y= −sy.
t=0
e
F = F
−2st
2
and
,
1
Gj=
2
j
160 Geometry of Quadrics and Spectral Theory
and the cross section is given by
2
F = y
=1, or μ1=1.
We can take account of the discrete group generated by τ
with uj= x
2
, vj= xjyj, wj= y
j
2
. We observe that these 3n functions are
j
by parametrizing M
k
rational symmetric functions of the divisor
(p
, ..., pn),
2,p3
where p
by (4.10) w
=(μj,#P (μj)) is a point on the Riemann surface of (4.2). Indeed,
j
j
2
= y
is a symmetric polynomial of μ2, μ3, ..., μn. By (4.11),
j
(4.12), (4.13), and (4.14) we have
n
= xjyj=
v
j
k=1
−1
x, yy
= μ
1
n
Using x, y =(1/2s)
a rational symmetric function of the divisor p
for u
j
= v
−1
2
w
j
j
j=1
and therefore for any rational function of uj, vj, wj.
∂y
j
X
2
j
yj=
k
∂μ
k
n
a(μk)
−
m(μk)
k=2
Q
(x, y)(μk− αj)−1y
μ
k
2
.
j
(λj−αj), (4.16), and (4.10), it is clear that vjis also
, ..., pn. Finally, the same is true
2
It is well known (see Neumann [13], Siegel [18]) that all rational symmetric
functions of p
Jacobi variety and thus also u
rise to a map of g → M
λ
, ..., λnthis map is given by (4.10), (4.13). Since uj, vj, wjare sufficient to
2
distinguish points on M
, p3, ..., pncan be represented in terms of θ-functions on the
2
, vj, wjcan all be represented in this way, giving
j
. For nonspecial divisors (p2,p3, ..., pn) and fixed λ1,
, this mapping is an isomorphism, showing that M∼ g.
c. The Symplectic Structure
For the proof of Theorem 4 in case (i), it remains to show that the vector
fields X
For this purpose we express the symplectic form ω =
the variables λ
have constant coefficients with respect to the variables sjof (4.3).
G
j
, μk. Since the λkas well as the μk, being functions of y alone,
k
n
dyj∧ dxjin terms of
j=1
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