Definition of Tensors with Low Regular Metrics

with a^lk := g^lk ∈C², and f =T₁(g). Since f =T₁(g) depends smoothly on the metric and its first derivatives,f =T1(g)∈C¹⊂C^1−ε, and Ricij ∈C^k,α by hipothesis. This is a linear elliptic operator acting on g_ij. Ifk+α≤1−ε, by elliptic regularityg∈C^k+2,α according to Theorem 7.

Ifk+α >1−εwe see thatg_ij ∈C^3−εfor allε >0. Repeating the argument, nowf =T₁(g)∈C^2−ε and a^lk∈C^3−ε. Ifk+α≤2−ε, we are done.

If k+α > 2−ε, we see g_ij ∈ C^4−ε, so f = T₁(g) ∈ C^3−ε. If k+α ≤ 3−ε we are done, and otherwise we keep going.

Repeating the argument k+ 1 times we obtain that both a^kl, T1(g)∈C^k+1−ε⊂C^k+α inεis small enough, sog_ij ∈C^k+2,α as desired.

Let us see (2). As both theg and RicareC^k,α fork≥2 andα >0 in some coordinates, we know by Proposition 24 that both of them are C^k,α in harmonic coordinates. Now we use (1) to see that g∈C^k+2,α. This yields (2) and proves the Theorem.

From this Theorem we see that if in harmonic coordinates φ the Ricci tensor is smooth and the metric is C^2,α, it follows that g is also smooth in the φ coordinates. In arbitrary coordinates the Theorem only says that we gain two derivatives on the metric when we express it in harmonic coordinates, but with respect to the regularity of the metric in the original coordinates, not to the regularity of the Ricci tensor. The following example ilustrates this.

Example 6.

(a) Consider (Rⁿ, g₀) beingg₀ =Idthe flat usual metric. Let ϕ be a diffeomorphism of classC³, so that the pull-back metric g1 := (ϕ⁻¹)∗g0 isC². As the Ricci tensor is invariant under isometries, Ric(g1) = 0, but the metric g1 is not smooth, not even C³. This happens because the coordinates given byϕ are not harmonic.

In particular, if n= 2 we see that g₁(x, y) = D(ϕ⁻¹)^t(x, y)D(ϕ⁻¹)(x, y) can not be the identity matrix multiplied by a function unless ϕ∈C^∞, by Proposition 25. This is a curious conclusion.

(b) Suppose that in harmonic coordinates the metric g is C², and the curvature tensor r is zero.

Then in these coordinates Ric= 0, sog is smooth.

Remark 41. If we suppose thatu, v∈Ethen from this it follows thatuv∈E, since the distributional derivative keeps being the same. The same applies for v⁻¹ under the additional assumption of non- vanishing.

Proof. First suppose that v ∈ C¹(Ω). Let ϕ ∈C₀¹(Ω) be a text function, and K := Supp(ϕ). Note thatϕxiv∈C₀¹(Ω) is also a test function, for which the usual product rule applies. Therefore

Z

Ω

uvϕ_xidx= Z

Ω

u[(vϕ)_xi−v_xiϕ]dx=− Z

Ω

(u_xiv+uv_xi)ϕdx.

For general v ∈ E(Ω), take vⁿ ∈ C¹(Ω) such that vⁿ → v in W^1,2(K). By the previous case and Holder’s inequality we have

| Z

Ω

uvϕxidx+ Z

Ω

(uxiv+uvxi)ϕdx|

=| Z

Ω

(v−vⁿ)uϕxidx+ Z

Ω

uvⁿϕxidx+ Z

Ω

(uxiv+uvxi)ϕdx|

=| Z

Ω

(v−vⁿ)uϕxidx+ Z

Ω

{u_xi(v−vⁿ) +u(vxi−vⁿ_xi)}ϕdx|

≤ kv−vⁿk_L2(K)

kuϕ_xik_L2(Ω)+ku_xiϕk_L2(Ω)

+kv_xi−v_xⁿ_ik_L2(K)kkuϕk_L2(Ω)→0

as n → ∞. This proves that the distributional derivative of uv is ∇(uv) := v∇u+u∇v (note that this is true only supposing that u, v ∈ W_loc^1,2(Ω)). Now, as u, v ∈ L^∞_loc(Ω), then ∇(uv) ∈ L²_loc(Ω), so uv∈E(Ω).

Now suppose that v(x)≥c >0 a.e. x∈Ω, and let vⁿ∈C¹(Ω) such thatvⁿ→v in W^1,2(K) and vⁿ(x)→ v(x) a.e. x∈K. Since almost everywhere subsets are dense, by continuity|vⁿ(x)| ≥c and

|vⁿ(x)| ≤ kvk_L^∞_(K) for allx∈K, so the usual derivative of (vⁿ)⁻¹ is −(vⁿ)⁻²∇vⁿ . Now we compute

Z

Ω

[v⁻¹ϕxi+v⁻²vxiϕ]dx

= Z

Ω

[v⁻¹−(vⁿ)⁻¹]ϕxidx+ Z

Ω

(vⁿ)⁻¹ϕxidx+ Z

Ω

v⁻²vxiϕdx

= Z

Ω

[v⁻¹−(vⁿ)⁻¹]ϕ_xidx+ Z

Ω

[v⁻²v_xi−(vⁿ)⁻²v_xⁿ_i]ϕdx

≤ Z

Ω

vⁿ−v vvⁿ ϕ_xidx

+ Z

Ω

(vⁿ)²vxi−v²vⁿ_xi v²(vⁿ)² ϕdx

≤

vⁿ−v vvⁿ

L²(K)

kϕ_xik_L2(Ω)+

(vⁿ)²vxi−v²v_xⁿ_i v²(vⁿ)²

_L2(K)

kϕk_L2(Ω) :=A_n+B_n Now note that both sequencesf_n:=k[vⁿ−v](vvⁿ)⁻¹kandg_n:=

[(vⁿ)²v_xi −v²v_xⁿ_i](vvⁿ)⁻²

converge almost everywhere to 0 inK, and moreover fnand gnare uniformly bounded functions inK, so they are dominated by an L²(K) function. By the Dominated Convergence Theorem we see that both An

and B_n tend to 0, as desired. This proves the Lemma.

Definition 44. Given Ω ⊂ Rⁿ an open let and 1 ≤ p ≤ ∞ and p⁰ its conjugate. We denote W^−1,p0(Ω) :=W^1,p(Ω)⁰ for the dual space ofW₀^1,p(Ω).

Using the isometric embbeding ι : W₀^1,p(Ω) → F := L^p(Ω)ⁿ⁺¹ : u → (u,∇u) we can have an idea of how acts an element T of W^−1,p0(Ω). Indeed, we see that T :W₀^1,p(Ω)→ R induces a linear

continuous map ˆT := T ◦ι⁻¹ on its image ι(W₀^1,p(Ω)) ⊂ F. By the Hanh-Banach Theorem we can extend ˆT to a continuous functionalT^∗ on F, i.e, T^∗ ∈F⁰ =L^p0(Ω)ⁿ⁺¹. Therefore, when we restrict T^∗ toι(W₀^1,p(Ω)), we see that there exists functions v₀, . . . , v_n∈L^p0(Ω) such that

T(u) = Z

Ω

uv0dx+

n

X

i=1

Z

Ω

uxividx

Note that the functions v_i are not unique, as neither is the extension T^∗ of ˆT.

Remark 42. Givenf ∈L^p0(Ω) = W^0,p0(Ω), we can consider f as a distribution, i.e, f ∈(L^p(Ω))⁰ ⊂ D⁰(Ω). Then, as D(Ω) =C_c^∞(Ω) is dense in W₀^1,p(Ω) it is obvious that its distributional derivatives f_xi ∈D⁰(Ω) can be uniquely extended to an element to W^−1,p0 and this element is of course the one that acts by (f_xi, u) :=−(f, u_xi) fro any u∈W₀^1,p(Ω). So we see that derivatives ofL^p0 functions fall inW^−1,p0. This should explain a little bit the notation.

We prove now a couple of technical propositions that will be useful later.

Proposition 27. Let Ω⊂Rⁿ be an open bounded set, and letE =W^1,2∩L^∞(Ω) as before. Given f, g∈E, we have

∂_l[f g] =f ∂_lg+g∂_lf ∈L¹

∂k[f ∂lg] =∂kf ∂lg+f ∂klg as elements ofW^−1,q0 for any q > n. (155) Proof. The first equality follows from Proposition 33. Let us see the second. First note that, as commented above, ∂_kf ∂_lg+f ∂_klg ∈ L¹(Ω) +W^−1,q0(Ω) ⊂ W^−1,q0(Ω) for any q > n, , being q⁰ the Holder conjugate of q. On the other hand,f ∂lg∈L², so∂k[f ∂lg]∈W^−1,2(Ω)⊂W^−1,q0(Ω). We shall see that both are equal inW^−1,q0(Ω). Indeed, givenϕ∈W₀^1,q(Ω), on one hand we have

(∂_k[f ∂_lg], ϕ) =− Z

Ω

f ∂_lg∂_kϕdx

On the other hand, recall that as q > n≥2 we haveW₀^1,q(Ω)⊂W₀^1,2(Ω) so by Lemma 33 we see that the pointwise productf ϕ∈E and ∇(f ϕ) =ϕ∇f+f∇ϕ∈L², so we have

(∂kf ∂lg+f ∂klg, ϕ) = Z

Ω

ϕ∂kf ∂lgdx− Z

Ω

∂lg∂k[f ϕ]dx

= Z

Ω

ϕ∂kf ∂lgdx− Z

Ω

ϕ∂lg∂kf dx− Z

Ω

f ∂lg∂kϕdx=− Z

Ω

f ∂lg∂kϕdx and this proves the Proposition.

Proposition 28. Let Ω ⊂ Rⁿ, n ≥ 2 be open and bounded. Given f ∈ E(Ω) and g = g₁+g₂ ∈ W^−1,2(Ω) +L¹(Ω) we can define the product f g in such a way that f g ∈ W^−1,q0(Ω) +L¹(Ω) and besidesf g is the pointwise product of functions wheng is aL^p(Ω) function for somep≥2.

Remark 43. If h∈W^−1,q0(Ω) can be expressed as h=eh⁰ for some e∈E(Ω) andh⁰ ∈W^−1,2(Ω) + L¹(Ω), then for f ∈E we can still define f h:= (f e)h⁰.

Proof. Let f ∈ E(Ω), g =g₁+g₂ ∈ W^−1,2(Ω) +L¹(Ω). It is clear that we can define the pointwise product f g2 ∈L¹(Ω). It remains to see how we define the product f g1. This will be similar to what it is done for distributions. Let q > n ≥2. Note that, by the Sobolev Embeding Theorem and the definition of norms, we know that W₀^1,q ⊂C^0,δ ⊂L^∞ for some 0< δ <1, with continuous inclusions.

Taking duals we haveL^p(Ω)⊂L¹(Ω)⊂W^−1,q0(Ω) forp≥1. Also note that W^1,q(Ω)⊂E(Ω).

That said, we claim now that we have a well defined map

·f :W₀^1,q(Ω)→W₀^1,2(Ω) :g7→f g.

First note that, asg is continuous and has zero trace, theng is zero restricted to the boundary of the coordinate patch Ω.

Let us see that the map above is well defined. As g∈W₀^1,q(Ω)⊂E(Ω), by Lemma 33 we see that

∇(f g) = f∇g+g∇f ∈L²(Ω), since f, g ∈L^∞(Ω). From this we see that f g ∈W^1,2(Ω). Note that this fails forq < nas we are not sure that g∇f ∈L²(Ω).

Finally let us see that T r(f g) = 0. Let φ_n → f in W^1,2(Ω), φ_n ∈ C(Ω). We can suppose φn∈L^∞(Ω) since we can changeφn by

φ⁰_n=φ_nχ_{{|φn|≤kfkL}∞(Ω)}+sign(φ_n)kfk_L∞(Ω)χ_{|φn|>kfkL∞(Ω)}

which satisfies φ⁰_n ∈C(Ω), kφ⁰_nk_L^∞_(Ω) ≤ kfk_L^∞_(Ω), and |φ⁰_n(x)−f(x)| ≤ |φ_n(x)−f(x)|so φ⁰_n are a better approximation off. So we directly supposeφn∈E(Ω). Recall that this impliesφng∈W^1,2(Ω).

We claim in addition that φ_ng → f g in W^1,2(Ω). To see this first note that kφ_ng−f gk_L2(Ω) ≤ kφ_n−fk_L2(Ω)kgk_L2(Ω)→0 as n→ ∞. Also we know that the product rule applies so

k∇(φ_ng)− ∇(f g)k_L2(Ω)≤ kgk_L2(Ω)k∇φ_n− ∇fk_L2(Ω)+k∇gk_L2(Ω)kφ_n−fk_L2(Ω).

So, as T r :W^1,2(Ω)→ L²(∂Ω) is continuous and T r(φng) = (φng)|_∂Ω = 0, we see that T r(f g) = 0, sof g∈W₀^1,2(Ω) as we wanted. This proves that ·f is well defined as claimed. Therefore we have the transpose map (·f)⁰ :W^−1,2(Ω)→W^−1,q0(Ω).

Suppose now that for some function g we have thatIg∈W^−1,2(Ω), (whereIg acts onW₀^1,2(Ω) by integrating against g), and suppose we have also that I_{f g} ∈ W^−1,q0(Ω). For example this is true if g∈L^p(Ω) for somep≥2. We might wonder whether (·f)⁰I_g =I_{f g}, and we claim this is true. Indeed, given h∈W₀^1,q(Ω) we have

((·f)⁰I_g, h) = (I_g, f h) = Z

Ω

gf hdx= (I_{f g}, h)

So, as (·f)⁰ coincides with·f when that makes sense, we denote simply (·f)⁰=·f, and the proposition is proved.

Now we shall see how to define tensors that involve two derivatives of the metric if we only suppose g∈E, and tensors that involve three derivatives supposing g∈E⁰.

First we write explicitly the expressions of the (3,1) curvature tensor R, the Ricci tensor Ric, the Schouten tensor s, the Weyl tensor w and the (3,1)-Weyl tensor W. In Einstein notation the tensors obtained by lowering and raising indexes are denoted with the same letter, so now we denote all tensors with capital letters. Moreover the tensors obtained taking traces are denoted also with the same letter. In particular note that the Ricci tensor is denotedR_ij since it is obtained by taking the trace from the curvature, Rij =Raija=Raijlg^al. In analogy, the scalar curvature is obtained as the trace of the Ricci tensor, so it is denoted byR=R^a_a=R_abg^ab. That said, according to Definition 10, we have

R_abc^d=∂_aΓ^d_bc−∂_bΓ^d_ac+ Γ^m_bcΓ^d_am−Γ^m_acΓ^d_bm S_ab= 1

2−n

R_ab+ 1

2(1−n)g^rsR_rsg_ab

Wabcd =Rabcd−gacSbd+gadSbc−Sacgbd+Sadgbc

(156)

W_abc^l=W_abcdg^ld =R_abc^l−g_acS_bdg^ld+δ_a^lS_bc−S_acδ^l_b+S_adg^ldg_bc

=Rabcl+ 1 2−n

h

−g_acRbdg^ld+δ_a^lRbc−δ_b^lRac+Radg^ldgbc

i

+ 1

2(1−n)(2−n)g^rsRrs

h

−g_acg_bdg^ld+δ^l_ag_bc−δ^l_bgac+g_adg^ldg_bc i

=R_abc^l+ 1 2−n

h

−g_acR_bdg^ld+δ_a^lR_bc−δ_b^lRac+R_adg^ldg_bc i

+ 1

(1−n)(2−n)g^rsRrs

h

δ^l_ag_bc−δ_b^lgac

i

(157)

The next proposition shows how to define the Weyl tensor if the metric has just one derivative.

Proposition 29. Given a Riemmannian manifold (M, g), and a system of coordinates (ϕ,Ω), if the metric g_ϕ ∈ E(Ω) = W^1,2(Ω)∩L^∞(Ω), then we can make sense of the components of the Riemann curvature, Weyl, and Ricci tensors in these coordinates (as distributions), and moreover if λ∈E(Ω) is a positive function, and we define g⁰ = λg ∈ E(Ω), then the Weyl tensors w and w⁰ of g and g⁰ satisfy the same relation than in the case where g andg⁰ are C², i.e,w⁰ =λw.

Proof. With the formulae above in mind, fix a coordinate patch Ω⊂Rⁿ. In the following discussion all the spaces are considered in Ω unless explicit mention. Suppose thatg_ij ∈E. Then|g|and|g|⁻¹∈E by Lemma 33, so its inverse matrix g^ij ∈E. Therefore we have

Γ^c_ab = 1

2g^cl[∂ag_bl+∂_bg_al−∂_lg_ab]∈L²

sinceL^∞L^p ⊂L^p for all p≥1. Also, if we consider Γ^c_ab ∈ D⁰, then looking at (156) we see that both R_abc^d andR_bc∈W^−1,2+L¹.

In order to make sense of Sabforn≥3, we would want to make sense ofg^rsRrs. Looking at (156), we haveg^rsg_ab ∈E and Rrs∈W^−1,2+L¹ so by Proposition 28 we have g^rsg_abRrs∈W^−1,q0+L¹ and R_abcd =R_abc^lg_ld∈W^−1,q0+L¹. AsW^−1,2⊂W^−1,q0, and we see that bothS_ab, W_abc^d can be regarded inW^−1,q0+L¹.

Also, we see thatW_abcd=W_abc^lg_ld ∈W^−1,q0+L¹, becauseW_abc^d∈E(W^−1,2+L¹) so we can still multiply W_abc^d by functions on E (see Remark 43). It is important to note that with this definition the curvature Rabcd has all the symmetries of the Riemann curvature tensor. To see this, look at the expression of R_abc^d in (156), and note that we can expand, multiply and differentiate the expression of the Christoffel symbols in terms of the metric as in the regular case, i.e,

Γ^c_abΓ^f_de = ¹₄g^clg^{f l}∂ag_bl∂_dg_el+¹₄g^clg^{f l}∂ag_bl∂eg_dl+· · · ∈L¹

∂_kΓ^c_ab =∂_k[¹₂g^cl][∂_ag_bl+∂_bg_al−∂_lg_ab] +¹₂g^cl[∂_kag_bl+∂_kbg_al−∂_klg_ab]∈L¹+W^−1,q0

Note that after expanding these expressions and inserting them in the formula forR_abc^d, we have lost a bit a regularity, since nowRabcd∈W^−1,q0+L¹, but this is irrelevant. The simmetries ofRabcdderive from its formal expression in terms of the metric, so they are the same as in the smooth case. Note also that, regardingR_abc^d ∈W^−1,q +L¹, we can still make sense of R_abcd =g_ldR_abc^l because R_abc^l is a sum of things that lie inE(W^−1,2+L¹), see Remark 43.

ThereforeR_abcdhas the same expression in terms ofg_ij that in the smooth case, so necessarily it has the same simmetries. Exactly the same argument applies to see that W_abcd has the same simmetries as in the smooth case, so in particular its Ricci contraction is zero, i.e, Ric(W)bc=Wabca= 0.

Finally note that L¹ ⊂ (L^∞)⁰ ⊂ W^−1,q0, since W^1,q ⊂ L^∞ with continuous inclusion by the Sobolev Embeding. We see that all the tensors in (156) and (157) can be regarded as elements of W^−1,q0+L¹⊂W^−1,q0.

Not let λ∈E, and define the conformal metric g⁰ :=λg ∈E beingλ≥k >0 for some constant k. We claim that the conformal behaviour of the weekly defined Weyl tensor remains the same, i.e, W_abcd⁰ =λW_abcd. The proof of this boils down to check that all the computations we did in 3 and 1 to compareW_abcd⁰ and W_abcd in the regular case (i.e, λ, g∈C²) are valid also in this less regular setting.

Note that in the regular case the equality W_abcd⁰ = λWabcd holds, and it is derived from the fact that, asg⁰_ij =λg_ij, we can apply Leibnitz’s rule when taking derivatives up to order 2, and this gives us a relation between both metrics. We will use Proposition 27 to show that we can do the same thing for metrics in E.

In first place note that ifg∈E, it is still true that the Riemann curvature tensorradmits a unique descompositionr=w+s?gbeingwa curvature tensor with Ricci contraction 0. This descomposition was proved in 1 using the simmetries ofr and taking traces twice. We have seen that these simmetries are the same here and we can take traces, so nothing changes.

Given this, it suffices to see that R_abc^dand R⁰_abc^d are related in the same way than they are in the regular case, because in that case we are done by Proposition 3 and the Corollary 1 below, since the same arguments apply. To see that R_abc^d and R⁰_abc^d are related as if they were regular, it is enough to see that we can apply the Leibnitz rule when computing

R⁰_abc^d=∂_aΓ^0d_bc−∂_bΓ^0d_ac+ Γ^0m_bcΓ^0d_am−Γ^0m_acΓ^0d_bm. To see this note that

Γ^0m_bc = 1

2(λ)⁻¹g^ml{∂_b[λg_cl] +∂_c[λg_bl]−∂_l[λg_bc]}= 1

2(λ)⁻¹g^mlg_cl∂_bλ+1

2(λ)⁻¹g^mlλ∂_bg_cl+. . . (158) since by Proposition 27 we can apply Leibnitz’s rule to each of the sumands and also we can apply the distributive law. Note that each of the summands is in L². So the products Γ^0m_bcΓ^0d_am can be manipulated as if the metric were smooth.

Also, by the algebra properties of E in Lemma 33 and Proposition 27, we can apply Leibnitz’s rule to each of the summands of the expanded expression of Γ^0m_bc in (158) as if these summands were smooth so it follows that

∂_aΓ^0m_bc = 1

2∂_a[(λ)⁻¹g^mlg_cl∂_bλ] +· · ·= (λ)⁻¹g^mlg_cl∂_abλ−λ⁻²g^mlg_cl∂_aλ∂_bλ+· · · ∈W^−1,q0 Note that after expanding the expression of Γ^0m_bc we lose a bit of information because we know (by another argument) that∂_aΓ^0d_bc actually lies inW^−1,2.

We conclude from this that R_abc^d and R_abc^{0 d} are related in the same way as in the smooth case, this time laying both inW^−1,q0 (not in W^−1,2+L¹ as before). From this we contract with the metric and we see that also R_abcd =g_ldR_abc^l and R⁰_abcd =λg_ldR⁰_abc^l are related as if they were smooth. We must see that this contraction with the metric is valid. Although we a priori cannot make sense ofgldh for general h ∈W^−1,q0, if h has the special form h=eh⁰ with e∈E and h⁰ ∈W^−1,2, then we define g_ldh:= (g_lde)h⁰ and everything works fine, by Remark 43. Note that by the computations above, both Rabcd and R⁰_abc^d are sums of functions hi = eih⁰_i with this special form, so we can contract with the metric without any problem.

Thus, as in the regular setting, R⁰_abcd = λ[R_abcd + (b_u ?g)_abcd], being u := ¹₂log(λ), and b_u = Hess(u)−du⊗du+¹₂|grad(u)|²_gg. Note thatu∈E and ∂au= ¹₂λ⁻¹∂a[λ] by the Chain rule for weak derivatives. Now we apply the argument given in Corollary 1 and we see thatW_abcd⁰ =λW_abcd. This completes the proof of the proposition.

Remark 44. One might wonder if it is really necessary to ask g∈E to define the Weyl tensor in the weak sense. Of course the problem is not on defining the derivatives of g, since we can differentiate g as many times as we want regardingg as a distribution. The problem lies on defining the products

Γ^c_abΓ^k_ij which appear in R_abc^d, since we cannot multiply distributions in a satisfactory way (we do multiply distributions with functions, but not with proper distributions).

Note for example that the tensorial product of distributionsu1, u2 ∈ D⁰(Ω) given by (u1⊗u₂, ϕ) :=

(u₁, ϕ)(u₂, ϕ) for ϕ ∈ D(Ω) does not work for two reasons. First it does not extend the product on functions, i.e, (I_f⊗I_g, ϕ) =R

f ϕdxR

gϕdx6=R

f gϕdx= (I_{f g}, ϕ). The second reason, is thatu₁⊗u₂ is not linear, it is bilinear, so it does not belong to D⁰(Ω).

In fact, it is known that we cannot define a product on distributions that extend the pointwise product of functions (see [18]). It is enough to note that x⁻¹²χ_(−1,1) is a distribution in Ω = (−1,1), but its square is x⁻¹χ_(−1,1) is not integrable near 0.

So we are not able to generalize the Riemann curvature tensor (and thus any of the above tensors) unless the Christoffel symbols are regarded as functions, and this demands on g to have at least one weak derivative.

Remark 45. Suppose now that the metricgisC_∗^rfor somer >1 in some local coordinates. Theng^jk areC_∗^rby the algebra properties of Zigmund spaces, and thus it is clear that Γ^c_abareC_∗^r−1. This implies that R_abc^d and R_ab areC_∗^r−2, since differentiation (in distributional sense) subtracts one exponent in the Zigmund spaces (see Remark after 9). Moreover, we can define the product g^jkRab∈C_∗^r−2 by the Corollary after Lemma 15, since r > 1 implies r > |r−2|. Therefore P_ab ∈ C_∗^r−2 and for the same reasonW_abc^d and W_abcd∈C_∗^r−2.

Now we see that if we demand one more derivative on the metric, the Cotton tensor can also be defined in the weak sense in a satisfactory manner. This time we need some properties of the space E⁰ =W^2,2∩W^1,∞in some open bounded set Ω⊂Rⁿ, which the next propositions summarize.

Proposition 30. Let Ω⊂Rⁿ. Letf, g∈E⁰(Ω) =W^2,2(Ω)∩W^1,∞(Ω). Thenf g∈E⁰(Ω). Therefore E⁰(Ω) is an algebra.

Moreover, if we suppose that f(x)≥k >0 a.e. x∈Ω for some constant kthenf⁻¹ ∈E⁰(Ω).

Proof. AsE⁰(Ω)⊂E(Ω) =W^1,2(Ω)∩L^∞(Ω), we use Proposition 33 to see that (f g)xi =gfxi+fxig.

Now we note thatf_xi, g_xi ∈E(Ω), so (f g)_xi ∈E, and this gives f g∈E⁰(Ω).

If now we suppose f(x) ≥k > 0, by Proposition 33 we know that f⁻¹ ∈ E and that (f⁻¹)_xi =

−f⁻²fxi. From here we see that (f⁻¹)xi ∈E, since both f⁻² =f⁻¹f⁻¹ and fxi lie inE, which is an algebra.

Proposition 31. Let Ω⊂Rⁿ be an open and bounded set, let f ∈E⁰(Ω) = (W^2,2∩W^1,∞)(Ω) and g∈L²(Ω). Then f g∈L² and moreover (f g)xi =gfxi +f gxi as elements in W^−1,2(Ω) (in particular we are saying that f gxi ∈W^−1,2(Ω)).

Proof. First note that the multiplication by f satisfies

·f :W₀^1,2(Ω)→W₀^1,2(Ω) :ϕ7→f ϕ (159) Recall Proposition 28, where we only supposed that f ∈ E(Ω) = (W^1,2 ∩L^∞)(Ω) and then the multiplication by f was worse behaved since it mapped the smaller space W₀^1,q(Ω) (for q > n) into W₀^1,2(Ω) . However, iff ∈E⁰(Ω),ϕ∈W₀^1,2(Ω), then by Lemma 155 we know that the distributional derivative of f ϕ is ∇(f ϕ) = ϕ∇f +f∇ϕ. As f ∈ L^∞(Ω),∇f ∈ L^∞(Ω;Rⁿ), we see that ∇(f ϕ) ∈ L²(Ω;Rⁿ), sof ϕ∈W^1,2(Ω).

Now let us see that T r(f ϕ) = 0 ∈ L²(∂Ω). As ϕ ∈ W₀^1,2(Ω) we can find φn ∈ D(Ω) such that φ_n → ϕ in W^1,2(Ω). By the Sobolev Embeding, as f ∈ W^1,∞(Ω), f is continuous (in fact Lipschitz), so f φn ∈ Cc(Ω) and therefore T r(f φn) = 0. If we see that f φn → f ϕ in W^1,2(Ω), by

the continuity of the trace we will conclude that T r(ϕf) = 0 as we want. To see this first note that kf φ_n−f ϕk_L2(Ω)≤ kfk_L2(Ω)kφ_n−ϕk_L2(Ω) →0 asn→ ∞. Also, aplaying the product rule, we have

k∇(f φ_n)− ∇(f ϕ)k_L2(Ω) ≤ k∇fk_L2(Ω)kφ_n−ϕk_L2(Ω)+kfk_L2(Ω)k∇φ_n− ∇ϕk_L2(Ω) →0 asn→ ∞ So we have seen the mapping property (159). Now we can consider the transpose map (·f)⁰ of

·f, which we shall denote also by (·f)⁰ := ·f, since it is an extension of the original ·f regarding W₀^1,2(Ω)⊂W^−1,2(Ω) as usual. Therefore we have a well defined map

·f :W^−1,2(Ω)→W^−1,2(Ω) :T 7→f T (160) This shows in particular that for g ∈L²(Ω) we can define f g_xi. Now let us see (f g)_xi =gf_xi+f g_xi en W^−1,2(Ω). Take ϕ∈W₀^1,2(Ω), and note that the product rule applies tof ϕ. It follows that

((f g)_xi, ϕ) =−(f g, ϕ_xi) =− Z

Ω

f gϕ_xidx (f_xig+g_xif, ϕ) =

Z

Ω

f_xigϕdx− Z

Ω

g(f ϕ)_xidx=− Z

Ω

f gϕ_xidx and this completes the Proposition.

Now let us write the expression of the Cotton tensor C_abc in local coordinates.

(∇_∂aS)(∂_b, ∂_c) =∂_aS_bc−S(Γ^l_ab∂_l, ∂_c)−S(∂_b,Γ^l_ac∂_l) =∂_aS_bc−Γ^l_abS_lc−Γ^l_acS_bl Cabc= (∇_∂aS)(∂b, ∂c)−(∇_∂bS)(∂a, ∂c) =∂aSbc−∂bSac+ Γ^l_bcSal−Γ^l_acSbl

(161) Proposition 32. Given a Riemmannian manifold (M, g), and a system of coordinates (ϕ,Ω), if the metric gϕ ∈ E⁰(Ω) = (W^2,2 ∩W^1,∞)(Ω), then we can make sense of the components of the Cotton tensor in these coordinates (as distributions), and moreover if λ ∈E⁰(Ω) is a positive function, and we defineg⁰ =λg ∈E⁰(Ω), then the Cotton tensors cand c⁰ ofg andg⁰ satisfy the same relation than in the case where g and g⁰ are C³, i.e, c⁰ =c−w(·,·,·, grad(u)), with u := ¹₂log(λ). In particular if n= 3 we have c⁰ =c.

Proof. All spaces will be considered in Ω unless explicit mention. First note that by Proposition 30 the inverse g^ij ∈E⁰ and therefore Γ^c_ab = ¹₂g^cl[∂ag_bl+∂_bg_al−∂_lg_ab]∈E =W^1,2∩L^∞. Also, from the formula of Rabcd given in (156) we see that Rabcd ∈ L²+E ⊂ L², so also Rbc ∈ L², and Sab ∈ L². From the formula 161 we readily see thatC_abc∈W^−1,2+L² ⊂W^−1,2.

It remains to see the transformation of the Cotton tensor under conformal change. Put g⁰ = λg withλ∈E⁰. We shall see that we can apply the Leibnitz rule for the derivatives of the metric up to order 3. If we see this, then the components C_abc⁰ and C_abcof the Cotton tensors associated to g⁰ and g will be related by the same formal expression (in terms of the metricg and λ) than in the smooth case (this time making sense as distributions), and this will give the claim.

Recall that we already saw in Proposition 29 that we can make operations with the components of tensors that involve ≤2 derivatives of the metric as if the metric were smooth, and it remains to see that in this case we can do it with components of tensors that involve up to three derivatives of the metric.

More concisely, to analize how the Cotton tensor change we are interested in the components S_ab⁰ of the Schouten tensor, which depend on the second derivatives of the metric via the Ricci tensor R⁰_ab and the scalar curvatureR⁰_rsg^0rsg_ab⁰ (so via the Ricci tensor). The higher order terms (i.e, less regular,

In document Master Thesis (página 110-118)