For Theorem3.3 we did not need to test whetherdFµ(v)∈TF(µ)W(M) for all v∈TµW(N), since we only needed (F(µt), dFµt(vt)) to be an a.c. couple. But is it still true, in case (µt, vt) is a tangent couple?
Let us first properly assert that dFµ(v)∈L2(µ). In the rest of this section letF :W(M)→W(N) be as in Theorem3.3 anddFµ(v) as in formula (3.1), unless stated otherwise.
Lemma 3.5. For everyµ∈W(M)andv∈TµW(M),dFµ(v)∈L2(F(µ)).
Proof. As in the proof of Theorem3.3,kdFµ(v)kL2(F(µ))≤CkvkL2(µ), which is finite.
Now that we know that dFµ(v) is always an element of L2(F(µ)), we can consider formula (3.1) as the prescription for a map between TµW(M) and L2(F(µ)). It is useful to know that this map is always bounded.
Theorem 3.6(Boundedness ofdF). For eachµ∈W(M),dFµ:TµW(M)→ L2(F(µ))is bounded with
(3.2) |||dFµ||| ≤esssupµ
x∈M
|||dfx|||.
Here,|||·|||denotes the operator norm of the respective linear map andess supµx∈M the essential supremum with respect toµ.
The right-hand side of equation (3.2) is finite since we demanded supx∈M|||dfx|||
to be finite.
Proof. Taking similar steps as in the proof of Theorem 3.3, we have:
|||dFµ|||2 = sup
We have proven an inequality in Theorem 3.6. However, there are indeed functionsF such that equality holds for everyµ.
Example 3.7. Let g : M → M be a Riemannian isometry, i.e. g∗h = h, wherehis the Riemannian metric tensor on M. Then, forF =g# and for all µ∈W(M),|||dFµ|||= ess supµx∈M|||dfx|||= 1.
This is, because on the one hand, for allx∈M,|||dgx|||= supkvk=1kdgx(v)k= 1, sincedg is an isometry between the tangent spacesTxM and Tg(x)M and the norm is taken with respect to the Riemannian tensor. On the other hand,
|||dg#|||= sup
To come back to our question, whether dFµ(v) is always an element of TF(µ)W(M), we first want to study the following simple cases.
Lemma 3.8. Let µ = δx, for x ∈ M. Then dFµ(v) ∈ TF(µ)W(N) for all
Proof. For the Riemannian metric honM and for every vector fieldX h(∇(ϕ◦g−1), X) = d(ϕ◦g−1)(X) =dϕ(dg−1(X)) =h(∇ϕ, dg−1(X))
= h(dg(∇ϕ), X).
Since we know from Theorem 3.6that dg#µ is bounded and therefore con-tinuous for everyµ∈W(M), we can infer the following more general statement.
Corollary 3.10. Let g :M →M be a Riemannian isometry and TµW(M)3 v= limn→∞∇ϕn. Thendg(v) = limn→∞∇(ϕn◦g−1)∈TF(µ)W(M).
Example 3.11. A simple example for an isometry on R3 is f(x1, x2, x3) = (x1, x3, x2). For a vector field v(x) = (v1(x), v2(x), v3(x)) it is then, for every µ∈W(R3), dFµ(v) =dg(v) = (v1(g(x)), v3(g(x)), v2(g(x))). Furthermore, we can compute directly that∇(ϕ◦g−1) = (∂x1ϕ, ∂x3ϕ, ∂x2ϕ) =dg(∇ϕ).
OnR3, a vector fieldvis conservative, i.e. the gradient of a scalar function, if and only if its curl∇×vis zero. Recall that this can be reformulated in terms of 1-forms: On a general Riemannian manifold (M, h), there is a duality between vector fieldsvand 1-formsv[by the formulav[(·) :=h(v,·). The curl of a vector field and the exterior differentiald1of a 1-form on a three-dimensional manifold are related by the formula curl =] ? d1[, where]is the inverse procedure to[ and?the Hodge star operator. In particular, curlv= 0 if and only ifd1v[= 0.
For the Euclidean spaceR3 this implies∇ ×v= 0 if and only if d1v[= 0.
However, in our situation we are more generally interested in determining whenvis anL2-limit of conservative vector fields. So instead of looking at the de Rham complex, we want to make use of the so calledL2-de Rham complex onR3 (see for example [L¨uc02] for details on the following).
To be able to do this, let Ωpc(R3) be the set of smoothp-forms on R3 with compact support. On this set we consider theL2-inner product
hω, ξiL2(p,h):=
Z
R3
ω∧?ξ.
This inner product depends on the choice of the Riemannian tensorh on R3. LetL2Ωp(R3, h) denote the Hilbert space completion of Ωpc(R3).
Now let us look at the following linear operators from the L2-de Rham complex onR3:
d0: Ω0c(R3) −→ L2Ω1(R3, h) d1: Ω1c(R3) −→ L2Ω2(R3, h)
and their respective minimal closures ¯d0 and ¯d1. It is im( ¯d0)⊂ker( ¯d1). And sinceker( ¯d1) is closed (being the kernel of a closed operator), alsoim(d0)L
2(1,h)
⊂ ker( ¯d1).
The principal idea is then that finding an elementωin the domain of ¯d1 (an element which is smooth,L2 and ¯d1(ω) is again L2) which is not in its kernel, should correspond to finding an element which is not in some tangent space TF(µ)W(R3). However, theL2-structures in TF(µ)W(R3) ={∇ϕ}L
2((R3,h),F(µ))
and inL2Ωp(R3, h) (and thus, in particular in im(d0)L
2(1,h)
), or rather in its dual with respect to h, are a priori not related. In both cases, for the scalar product two vector fields are first inserted into the Riemannian tensor h, but whereas in the second case, this is integrated with respect to the metric volume form, in the first case, this is integrated with respect to a probability measure.
To adapt to this, we formulate the following lemma.
Lemma 3.12. Let η be the Euclidean metric tensor on R3, ν : R3 → R>0
a positive function on R3, h := ν(x)2η a new Riemannian tensor and dν = ν(x)2dvolη, wheredvolη denotes the measure induced by the metric volume form volη. Then
{∇ηϕ|ϕ∈ Cc∞}L
2((R3,η),ν)∼= {∇hϕ|ϕ∈ Cc∞}L
2((R3,h),volh)
as Hilbert spaces. Here, ∇η and ∇h denote the gradients that are taken with respect to the metricsη andh, respectively. In particular,
w= lim
It remains to check whether the scalar product is preserved. For this we note thatdvolh=ν3dvolη (see for example [Bes87] for validation). Then
Z 3
Now, this isometry can continuously be extended to the closure of{∇ηϕ}.
So, finding a vector field which is not a member of {∇hϕ}L
2((R3,h),volh) , corresponds to finding a vector field which is not a member of{∇ηϕ}L
2((R3,η),ν) . Based on this, to find elements that are notL2-limits of gradients, we look at theL2-de Rham complex on a Riemannian manifold of the form (R3, h=ν2η).
Proof. This follows from Z
R3
h(v, w)dvolh= Z
R3
w[∧?v[. (Consult for example [Fec06] for calculations on this.)
We conclude the following: If∇ ×w6= 0 holds true forw:=df(v),v=∇ϕ, ϕ ∈ Cc∞(R3) and f a smooth map, we know that also dw[ 6= 0, where w[ is the dual 1-form of w with respect to the Euclidean metric. But we need to know whether the dual 1-form of ν12w with respect to h is in the kernel of d or not. Luckily, both forms coincide, i.e. h(ν12w,·) = νν22η(w,·) =w[(·). So if w[ is smooth, L2 and ¯d1(ω) is againL2, w[ is in the domain of ¯d1, but not in its kernel, which means that it cannot be an element of im(d0)L
2(1,h)
. With Lemma 3.13, then, ν12w is not a member of {∇hϕ}L
2((R3,h),volh)
and Lemma 3.12tells us thatwis not part ofTνW(R3), in caseν is a probability measure.
We want to apply this consideration in the following corollary.
Corollary 3.14. Consider the mapf : R3 →R3, (x, y, z)7→ (x, y, x). Then there is a function ϕ ∈ Cc∞(R3) such that df(∇ϕ) is not an L2(ν)-limit of gradients, where ν is of the form dν = ν2dλ and ν : R3 → R>0 such that R
R31dν <∞.
Proof. Let us choose ϕ(x) := exp
1 kxk2−1
∈ C∞c (R3) for all kxk < 1, x = (x, y, z), andϕ(x) := 0 otherwise. Then∇ ×df(∇ϕ)6= 0. Forv=∇ϕit is
df(v)(f(x)) = (v1(f(x)), v2(f(x)), v1(f(x)))
= (ϕ,1(x, y, x), ϕ,2(x, y, x), ϕ,1(x, y, x)).
Here,ϕ,idenotes the partial derivative ofϕwith respect to thei-th component.
For validation of our statement, we only need to compute the third component of the curl ofdf(v):
(∇ ×df(v))3(f(x)) =df(v)2,1−df(v)1,2=v2(f(x)),1−v1(f(x)),2
= (ϕ,2◦f),1−(ϕ,1◦f),2
= 4xy 4x2+ 2y2−1
(2x2+y2−1)4exp 1 2x2+y2−1.
This last expression is unequal to zero for example forx=y =z = 1/2. The final conclusion follows from the remarks above.
The counterexample we have found in the proof of Corollary 3.14is not a counterexample forµ=δ, as we have seen in Lemma3.8, since all vector fields are regarded as elements of an L2-space. However, as soon as, for example, supp(ϕ)⊂supp(µ),dFµ(∇ϕ)6=∇ϕ.˜