5.1 Low dimensional examples
5.1.2 Euclidean plane
= − ∗
− sin uf (u) + (R + r cos u) r f0(u)
du ∧ dv
= sin u
r(R + r cos u)f (u) − 1 r2f0(u),
and on the other, if we regard f (u)du as a 1-form on T2/S1 = S1(r) we get (using the notation ¯∗ := ∗S1(r))
(dS1(r)+ d†S1(r))(f (u)du) = − ¯∗d¯∗f (u)du = −¯∗d 1 rf (u)
= −1 r2f0(u).
Thus, in view of (5.2), we see how Theorem 4.22 is verified for ω.
Finally, the operator D of Theorem 4.42 in this case is just D0 =D = DS1(r)+1
2
r sin u R + r cos u
c(du)ε,
where DS1(r) denotes the Hodge-de Rham operator of S1(r) with respect to the metric d¯s2 = r2du2. As the zero order term of D is bounded then, by Remark 4.47 and the Gauß-Bonnet theorem, we obtain the index formula
ind(Dev) = ind(DSev1(r)) = χ(S1(r)) = 0.
5.1.2 Euclidean plane
Let us consider now the real plane M = R2 equipped with the usual orientation and with the Euclidean metric, which can be written in polar coordinates as gT R2 = dr2+ r2dθ2 for r ≥ 0, θ ∈ [0, 2π]. Even tough this manifold is not compact, the Hodge-de Rham operator is essentially self-adjoint since the metric is complete. We study the action of S1 on R2 by counter-clockwise rotations around the origin. This action is semi-free and the fixed point set consists only of the origin. Consequently, the principal orbit is M0 = R2− {(0, 0)}. It is clear that M0/S1 can be identified with the open interval R> := (0, ∞) and the quotient metric is simply gT R> = dr2. In particular, note that (R>, gT R>) is not complete.
We start by computing the main geometric quantities. The generating vector field of the action is V = ∂θ. Its associated unit vector field is X = r−1∂θ and the corresponding characteristic form is χ = rdθ. The mean curvature vector field is computed from the Levi-Civita connection as
H = ∇XX = 1
r2∇∂θ∂θ= −1 r∂r,
and the its associated mean curvature 1-form is κ = X[= −dr/r. In this case we have, as in the example above,
ϕ0= dχ + κ ∧ χ = dr ∧ dθ +
−dr r
∧ rdθ = 0.
These computations show that the operator D0 of Theorem 4.42 is D0=D = DR>− 1
2rc(dr)ε, where DR> = dR>+ d†
R> is the Hodge-de Rham operator of (R>, gT R>).
Remark 5.1 (DR> is not essentially self-adjoint). With respect to the decomposition of forms by degree (0-forms and 1-forms) we can express
DR> =
0 −∂r
∂r 0
. Observe now the following facts:
• e−r ∈ L2(R>).
−1 as an eigenvalue, i.e.
0 −∂r
This shows that the deficiency indices are not zero and therefore DR> is not essentially self-adjoint when defined on the core Ωc(R>).
With respect to degree decomposition, as in the remark above, we can express D = then we can write this operator as
D = γ ∂
Here we explicitly see that we can write D as a regular singular first order differential operator in the sense of Br¨uning and Seeley (see Appendix A). In particular, since the cone coefficient satisfies
we verify that D is indeed essentially self-adjoint by Theorem A.3 (we will go deeper into this argument in the next chapters). This result of course follows directly from (4.17) and Corollary 4.51.
Remark 5.2 (Deficiency indices). For λ ∈ C and ¯ω = f0(r) + f1(r)dr consider the eigenvalue problem for the operator D,
−f10 − 1
2rf1 = λf0, f00 − 1
2rf0 = λf1.
We can uncouple this system to obtain two independent second order differential In order to compute the deficiency indices we need to study these equations for λ = ±i.
From [60, Lemma 4.2] it follows that in this case there exist no solutions for (5.4).
Hence, the corresponding deficiency indices for D are zero. This also shows that the operator D is essentially self-adjoint by [69, Proposition 3.7].
Remark 5.3. It is easy to verify that the functions f±(r) = r±1/2 satisfy the relation f±0 (r) = ± 1
2rf±(r),
and therefore D(f++ f−dr) = 0. However, this does not show that the kernel of D is zero since f±∈ L/ 2(R>).
To study the kernel of D we begin recalling a fundamental inequality.
Lemma 5.4 (Hardy’s inequality, [49]). Suppose 1 < p < ∞, f ∈ Lp(R>) and
Moreover, the equality holds if, and only if, f = 0 almost everywhere.
Lemma 5.5. The operator Dev satisfies ker(Dev) = {0}. and using that fn has compact support we arrive to the expression
kDevfnk2L2(R>)=
Finally, using Lemma 5.4, we conclude that fn−→ 0.
Corollary 5.6. We have ind(Dev) = 0.
Proof. First observe thatDodd is Fredholm by Theorem A.1. Hence, it suffices to show that ker(Dodd) = 0. Using the notation of the previous proof and using Lemma 5.4, we can estimate for fn∈ C∞(R>)
If we assume that Doddfn −→ 0, then these two integrals should also converge to zero (as they are both positive). In particular, from the convergence to zero of the first integral we see, again from Lemma 5.4, that fn−→ 0.
On the other hand, observe that the Euler characteristic χ(R>) = 0 because R> is contractible. Hence, we see that ind(Dev) = χ(R>) = 0.
We conclude this example by verifying Lemma 4.48. First, a straightforward computation shows that On the other hand we compute,
∇R>
Hence, the right hand side of Lemma 4.48 applied to a smooth function f0 is
and similarly for a 1-form f1dr we have
Thus, the desired statement is verified. In particular observe that we can express D2 = ∆R>+ 1
The aim of this example is to illustrate the whole procedure to obtain the operators of Theorem 4.31 and Theorem 4.42. This is intended to get a better understatement of the theory. We are going to consider the semi-free circle action on the unit 2-sphere M = S2 ⊂ R3 by rotations along the z-axis. The fixed point set is MS1 = {N , S}, where N and S denote the north and south pole respectively. On its complement M0= S2− {N , S} the action is free. We equip S2 with the induced metric coming from the Euclidean inner product of R3. As this metric is rotational invariant we see that S1 acts on S2 by isometries. Let us consider the local parametrization of S2 by spherical coordinates
x = sin θ cos φ, y = sin θ sin φ, x = cos θ,
where 0 < θ < π and the 0 < φ < 2π. With respect to this parametrization the metric on S2 takes the form
gT S2 = dθ2+ sin2θdφ2. (5.5) From this it follows that the induced inner products on 1-forms are
hdθ, dθi = 1, hdθ, dφi = 0, hdφ, dφi = (sin θ)−2.
If we choose the orientation on S2 using the outward normal vector then the associated Riemannian volume element is volS2 = sin θdθ ∧ dφ.
Now let us describe the S1-action concretely. An element eiϕ ∈ S1 acts on a point represented by the pair (θ, φ) by eiϕ(θ, φ) 7−→ (θ, φ + ϕ). In particular we see, in view of (5.5), that S1 acts on S2 effectively by orientation preserving isometries. The quotient manifold M0/S1 can be identified with the open interval I := (0, π), which we equip with the flat metric gT I = dθ2 so that the orbit map πS1 : M0 −→ M0/S1 becomes a Riemannian submersion.
(θ, φ) N
S x
z
Figure 5.2: Local chart of S2 defined by the polar angle 0 < θ < π and the azimuthal angle 0 < φ < 2π.
The generating vector field of the action is clearly V = ∂φ and its associated unit vector field is
X = 1 sin θ∂φ.
From (5.5) we see that the corresponding characteristic form is χ = sin θdφ. The mean curvature vector field is by definition
H = 1
sin2θ∇∂φ∂φ= 1
sin2θ(− sin θ cos θ∂θ) = − cot θ∂θ,
and its associated dual 1-form is then κ = − cot θdθ. This 1-form can be also computed using Proposition 4.7(2)
κ = −d log(kV k) = −d log(sin θ) = −cos θdθ
sin θ = − cot θdθ.
Note in particular that it satisfies dκ = 0 as expected. Again using (4.3) we find that ϕ0= 0 since
dχ + κ ∧ χ = d(sin θdφ) − cot θdθ ∧ sin θdφ = 0.
Having computed the relevant geometric quantities we are going to start by verifying Theorem 4.22. From Proposition 4.13 we know that any S1-invariant differential form on M0 can be written as
Now we will explicitly show that this operator is symmetric in L2(F, h). As we saw in Section 4.2.2, the vector bundle F −→ I is given by F = ∧CT∗I ⊕ ∧CT∗I and
This formula was expected since the orbits are circles of radius sin θ. As a consequence, the L2(F, h)-norm of a pair of differential forms with compact support f0(θ) + f1(θ)dθ
Using this relation we calculate the first component of the L2(F, h)-inner product Similarly for the other component of T (S2),
((DI+ cot θdθ∧)(f0+ f1dθ),g0+ g1dθ)L2(∧T∗I,h) These relations illustrate that the operator T (S2) is indeed symmetric with respect to the L2(F, h)-metric.
Now we want to conjugate this operator with the unitary transformation (4.12), which in this example just given by
U : Ωc(I) //Ωc(I)
¯
ω //(2π sin θ)−1/2ω.¯ First we verify Lemma 4.30,
d Theorem 4.31 takes the form
T (Db S2) := U−1T (DS2)U =
DI −12cot θbc(dθ) 0
0 DI+ 12cot θbc(dθ)
. Finally note that the operator D0 of Theorem 4.42 is
D0=D = DI −1
2cot θc(dθ)ε.
If we write a form ω0= f0(θ) + f1(θ)dθ ∈ Ωc(I) in column-vector notation as ω0 =
f0 f1
then when can express D as D =
0 −∂θ−12cot θ
∂θ−12cot θ 0
. (5.7)
The zero order part of the operator D is proportional to ±12cot θ. This potential blows up when θ −→ 0 and θ −→ π, i.e. when approaching the singular stratum. When θ −→ 0 the Taylor expansion of cot θ is
Note that, up to O(θ)-terms, this operator coincides with (5.3) in the example discussed above (up to a sign this also holds for θ −→ π). Hence, in view of Theorem A.3, we verify that the operator D with core Ωc(I) is essentially self-adjoint.
We want to end this example by verifying Lemma 4.48. The purpose of this is to give a feeling of how concrete computations can be done. It is easy to see that
∇IH¯(f0+ f1dθ) = − f00cot θ − f10cot θdθ, d†I(− cot θdθ) = − csc2θ,
k¯κ]k2= cot2θ.
Hence, for a smooth function f0 we have
and for a 1-form f1dθ we compute similarly
Therefore, with respect to the degree decomposition, we can write D2 =
Remark 5.7. The examples of the 2-torus and the 2-sphere can be easily generalized to general surfaces of revolution.