In each Gp we define three operations such thatGp is an algebra, that is, a vector space and a ring, the operations are define as [(U, f

dim(M ) = dim(T

M )

Carlos Henrique Silva Alcantara E-mail: carlos.henalcant@gmail.com

This text is an exercise of manifold theory and we will showdim(M) =dim(T_pM), that is, the dimension of a differential manifolds (M,A), sometimes justM - topological manifolds with a maximal atlas of compatible charts - is the same of tangent space, but what is tangent space?

In this textM is always a differential manifold.

Definition 1. Letp∈M, define a set

Γ_p={(U, f) :U open neighborhood of p andf :U →Rsmooth function}

in Γ_p set the relation: (U, f) ∼ (V, g) ⇐⇒ there is a open neighborhood W of p such that f|_W =g|_W, we denote Gp:= Γ_p\ ∼and an element of Gp is calledgerm inp.

In each G_p we define three operations such thatG_p is an algebra, that is, a vector space and a ring, the operations are define as [(U, f)] + [(V, g)] = [(U ∩V, f +g)], λ[(U, f)] = [(U, λf)] (for vector space) and [(U, f)][(V, g)] = [(U ∩V, f g)] (for ring), where the [·] is the class of (·,·), it is easy verify the well-definition of these operations and the axioms of vector space and ring.

When the domain U of some f is clear or not important we denote just [f] for the class of f in Gp.

Definition 2. A derivation at p ∈ M is a R-linear map v :G_p → R, such that the Leibniz’s rule is satisfied:

v([f][g]) =v([f])g(p) +f(p)v([g])

the set of all derivation in p is called tangent space and denoted TpM. If we define sum and product by scalar pointwise, that is, (v+λw)([f]) =v[f] +λw([f]), then T_pM is a real vector space.

Remark. Now we will show dim(M) = dim_RTpM. The sketch is first show an isomorphism betweenT_pM and a abstract, but useful, vector space, then show the dimension of this space is finite and equal to dimM.

Definition 3. Let p ∈ M, define Fp := {X ∈Gp : f(p) = 0,∀f ∈X}, of course f(p) = 0 for some f ∈X so h(p) = 0 for all h ∈X. In terms of Fp we defineF_p² := {

n

X

i=1

[fi][gi] : [fi],[gi]∈ F_p and nis any positive integer}.

Remark. It easy to see Fp is an ideal and subspace of Gp, analogous for F_p², that is, F_p² is an ideal and a subspace ofF_p, it makes sense in the vector space and ring contextF_P/F_p², i.e., this set is both subspace and sub ring with induced operations.

Theorem 1. Let p∈M, then Fp/F_p^2∗ is isomorphic to TpM.

Proof. Set the function

ϕ:F_p/F_p^2∗ −→T_pM

α7−→ϕ(α) :G_p −→R

[f] 7−→ϕ(α)([f]) =α(([f]−[f(p)]) +F_p²)

where f(p) : M → R, f(p)(q) = f(p), ∀q ∈ M and [·] +F_p² is the class of [f] in Fp/F_p². Because ϕ,ϕ(α) are clearly R-linear, [f]−[f(p)] = [(U, f)]−[(M, f(p))] = [(U, f −f(p))] and (f −f(p))(p) = 0 we have ϕwell-defined ifϕ(α) is a derivation.

In fact it is, but we need of a trivial lemma before the checking.

Lemma 2. Let f, g, h, t:X →R functions from set a X to R, then f g−ht=h(g−t) + (f − h)t+ (f −h)(g−t)

Proof.

h(g−t) + (f−h)t+ (f−h)(g−t) =hg−ht+f t−ht+f g−f t−hg+ht

= (hg−hg) + (−ht+ht) + (f t−f t) + (−ht+f g)

=f g−ht

Now, let [f],[g]∈G_p, then

ϕ(α)([f][g]) =ϕ(α)([f g])

=α([f g]−[f(p)g(p)] +F_p²)

=α([f g−f(p)g(p)] +F_p²) by the last lemma

ϕ(α)([f g]) =α([f(p)(g−g(p)) + (f−f(p))g(p) + (f −f(p))(g−g(p)) +F_p²) but (f −f(p))(g−g(p))∈F_p², so

ϕ(α)([f g]) =α([f(p)(g−g(p)) + (f −f(p))g(p)] +F_p²) note that [f(p)][h] =f(p)[h], for all [h] inG_p, therefore

α([f(p)(g−g(p)) + (f−f(p))g(p)] +F_p²) =f(p)α([g−g(p)] +F_p²) +α([f−f(p)] +F_p²)g(p) that is,ϕ(α)([f g]) =ϕ(α)([f])g(p) +f(p)ϕ(α)([g]).

Set the function

ψ:T_pM −→F_p/F_p^2∗

v7−→ψ(v) :Fp/F_p²−→R

[f] +F_p² 7−→ψ(v)([f] +F_p²) =v([f])

we haveψ(v) well-defined, let is see, if [f] and [g] satisfy [f] +F_p²= [g] +F_p², then [f]−[g] is in F_p², hence v([f]−[g]) = 0 once v(X) = 0 for everyX ∈F_p² by the Leibniz’s rule. Clearly ψ(v) isR-linear, so ψ is well-defined too.

We complete the proof if we show ϕ=ψ⁻¹. Checking,

(ψ◦ϕ)(α)([f] +F_p²) =ψ(ϕ(α))([f] +F_p²)

=ϕ(α)([f])

=α(([f]−[f(p)]) +F_p²),[f]∈Fp

=α([f] +F_p²) On the other hand,

(ϕ◦ψ)(v)([f]) =ϕ(ψ(v))([f])

=ψ(v)(([f]−[f(p)]) +F_p²)

=v([f]) Thereforeϕ=ψ⁻¹.

Lemma 3. For every smooth function f :Rⁿ→R, we can write f(x) =f(0) +∇f(0)·x+

n

X

i=1 n

X

j=1

xixi

Z 1 0

Z 1 0

∂²f

∂x_i∂x_j(tsx)tdsdx.

Proof. Start with

f(x)−f(0) = Z 1

0

(f(tx))⁰dt= Z 1

0

∇f(tx)·xdt then (1)f(x) =f(0) +

n

X

i=1

x_i Z 1

0

∂f

∂x_i(tx)dt , denoteg_i(x) = Z 1

0

∂f

∂x_idt, the same argument show

gi(x) =gi(0) +

n

X

j=1

xj

Z 1 0

Z 1 0

∂²f

∂x_i∂x_j(stx)tdsdt butgi(0) =

Z 1 0

∂f

∂x_i(0)dt= ∂f

∂x_i(0) plugging in (1) f(x) =f(0) +

n

X

i=1

xi

Z 1 0

∂f

∂xi

(tx)dt

=f(0) +

n

X

i=1

x_ig_i(x)

=f(0) +

n

X

i=1

x_i(∂f

∂xi

(0) +

n

X

j=1

x_j Z 1

0

Z 1 0

∂²f

∂xi∂xj

(tsx)tdsdt)

=f(0) +∇f(0)·x+

n

X

i=1 n

X

j=1

x_ix_j ∂²f

∂x_i∂x_i(tsx)tdsdt

Proposition 4. Letp∈M, thenF_p/F_p²is a finite dimensional vector space anddimF_p/F_p²= dim(M).

Proof. Consider a chart (U, ξ) around p, we may assume ξ(U) = Rⁿ and ξ(p) = 0. Denote ξ(q) = (u_i(q),· · ·, u_n(q)), n= dim(M), u_i :U →R coordinates functions, let X ∈G_p, choose f :W =int(W)⊂U → Rsuch that [f] =X, by the previous lemma and the fact ξ(p) = 0 we have

(f◦ξ⁻¹)(ξ(q)) =f(ξ⁻¹(ξ(p)) +∇(f◦ξ⁻¹)(ξ(p))·ξ(q) +

n

X

i=1 n

X

j=1

ui(q)uj(q) Z 1

0

Z 1 0

∂²(f ◦ξ⁻¹)

∂xi∂xj

(tsξ(q))tdsdt

=f(p) +

n

X

i=1

u_i(q)∂(f ◦ξ⁻¹)

∂xi

(ξ(p))+

+

n

X

i=1 n

X

j=1

ui(q)uj(q) Z 1

0

Z 1 0

∂²(f ◦ξ⁻¹)

∂xi∂xj

(tsξ(q))tdsdt

Denoteh_i= ^∂(f◦ξ_∂x⁻¹⁾

i (ξ(p)) andg_ij(q) =^R₀^1R₀^1∂2_∂x^(f◦ξ−1)

i∂xj (tsξ(q))tdsdt, then f(q) =f(ξ⁻¹(ξ(q))) =f(p) +

n

Xui(q)hi+

n

X

n

Xui(q)uj(q)gij(q)

for each q∈W. If [f]∈F_p², we obtain [f] +F_p²=

n

X

i=1

hi[ui] +

n

X

i=1

[ui][vi] +F_p²

where vi(q) = ^Pn_j=1uj(q)gij(q), note that ξ(p) = 0 =⇒ uk(p) = 0,∀k, then each [u_i]∈ Fp, so each u_ig_ij ∈F_p, hencev_i ∈F_p, therefore

n

X

i=1

[u_i][v_j]∈F_p², that is, [f] +F_p²=h_i[u_i] +F_p² In other words, [f]∈span{[u₁] +F_p²,· · · ,[u_n] +F_p²}.

Claim. {[u₁] +F_p²,· · ·,[u_n] +F_p²} is linearly independent.

Consider the derivation _∂u^∂

i([f]) = ^∂(f◦ξ_∂x⁻¹⁾

i (ξ(p)), every element of [f] agree with f in a open neighborhood of p, so the value _∂u^∂

i([f]) is well-defined. Now, suppose α1,· · · , αn∈Rsuch that α1[u1] +F_p² +· · ·+αn[un] +F_p² ∈ F_p², then α1[u1] +· · ·αn[un] ∈ F_p², on one hand we have

∂

∂ui(α₁[u₁] +· · ·α_n[u_n]) = 0, because _∂u^∂

i is a derivation evaluated in an element of F_p², on the other hand

∂

∂ui

(α₁[u₁] +· · ·α_n[u_n]) =^X

j=1

α_jδ_ij

since _∂u^∂

i([u_j]) = ^∂(uj_∂x^◦ξ−1)

i (ξ(q)) = ^∂x_∂x^j

i(ξ(q)) = δ_ij, we conclude 0 = ^P_j=1α_jδ_ij = α_i, ∀i ∈ {1,· · ·, n}, that is, {[u₁] +F_p²,· · · ,[un] +F_p²} is linearly independent. Follow (Fp/F_p²) finite dimensional and dim(F_p/F_p²) =n=dim(M).

Corollary 5. dim(M) =dim(TpM).

Proof.

dim(M) =dim(F_p/F_p²), finite dimensional

=dim(F_p/F_p²)^∗, Theorem 1

=dim(TpM).

Corollary 6. In the previous context, p∈M, ∂

∂u₁,· · ·, ∂

∂u_n

is base ofTpM.

Proof. By the proof of Proposition 4 we have{[u₁] +F_p²,· · · ,[u_n] +F_p²}, as ∂

∂u_i([u_j]) =δ_ij follow ψ

∂

∂u_i

([uj] +F_p²) =δij, that is,

ψ ∂

∂u₁

,· · · , ψ ∂

∂u_n

is base of (F_p/F_p²)^∗ ϕisomorphism, then

ϕ

ψ

∂

∂u1

,· · ·, ϕ

ψ ∂

∂un

= ∂

∂u1

,· · · , ∂

∂un

is base of TpM.

Definition 4. Let γ : I ⊂ R → M, I connected, we say γ is smooth at t ∈ int(I) if exist (f ◦γ)⁰(t) for every smooth function f defined in a neighborhood ofγ(t). It value is denoted γ(t). That is,˙

γ˙ :R−→T M

t7−→γ(t) :˙ G_γ(t) −→R

[f] 7−→(f◦γ)⁰(t)

Proposition 7. For each p ∈ M and v ∈ TpM there is a smooth curve γ : R→ M such that γ(0) =p and γ(0) =˙ v

Proof. Let (ϕ, U) chart around p such that ϕ(U) = Rⁿ, ⁿ_∂u^∂

1,· · · ,_∂u^∂

n

o

is base of TpM, so v=

n

X

i=1

α_i ∂

∂u_i, forα_i∈R, then define γ(t) =ϕ⁻¹(ϕ(p) +t(α₁,· · ·, α_n)),γ(0) =ϕ⁻¹(ϕ(p)) =p.

Checking the ˙γ(0) =v,

γ˙(0)([f]) = (f◦γ)⁰(0)

= (f◦ϕ⁻¹◦α)⁰(0), whereα(t) =ϕ(p) +t(α₁,· · ·, α_n)

=∇(f ◦ϕ⁻¹)(ϕ(p))·(α1,· · · , αn)

=

n

X

i=1

∂

∂xi

(f◦ϕ⁻¹)(ϕ(p))αi

=

n

X

i=1

αi

∂

∂u_i([f])

=v([f]) therefore ˙γ(0) =v.

References

[L] Lee, J.,Introduction to Smooth Manifolds. Graduate Texts in Mathematics, Vol. 218, Springer, 2002.