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Friday, November 8, 2024

Algebraic Description of Tangent/Cotangent Space via Derivations and Ideals of Taylor Series

Cotangent Space

 

Cotangent Space

The definition

Let R[X1,...,Xn] be the polynomial ring and R[[X1,...,Xn]]​​ be the formal power series ring.

Let Ik:(X1α1...Xnαn),l(α)=α1+...+αn=k. Easy to see that Ik=I1k. This ideal will appear at different rings, so distinguish it according to the context.

Let C∞(M)p be the stalk at p and the chart (U,φ),p∈U​ induce a isomorphism

(1)C∞(U)≅C∞(φ(U)),f⟼f∘φ−1⟹C∞(M)p≅C∞(Rn)φ(p)

and consider the Taylor series τk:C∞(M)p→R[X1,...,Xn]/Ik, which is a R algbra homomorphism.

We denote the kernel of τk as Mk and kenrel of τ:C∞(M)p→R[[X1,...,Xn]] as M∞.

Notice that M∞ is nontrivial since C∞(M)p is not a intergal domain hence zero is not prime ideal but R[[X1,...,Xn]] is, hence M∞ is a prime ideal and M∞≠(0).

The inverse limit of

(2)...R[X1,...,Xn]/Ik+1→πkR[X1,...,Xn]/Ik...

is R[[X1,...,Xn]] and kenrel of πk is Ik. From the diagram we see that:

(3)τk=πk∘τk+1

Thus

(4)Mk=τk−1(0)=τk+1−1(πk−1(0))⊃τk+1−1(0)=Mk+1

Also, let pk be the canonical map from the limit, pk:R[[X1,...,Xn]]→R[X1,...,Xn]/Ik.

(5)τk=pk∘τ∞

Hence

(6)Mk=τk−1(0)=τ∞−1(pk−1(0))=τ∞−1(Ik)=τ∞−1(I1k)=(τ∞−1(I1))k=M1k=mpk

The equation

(7)τ∞−1(I1k)=(τ∞−1(I1))k

hold since I1⊂Im(τ∞)​, and contraction and extension of ideal is a pair of inverse when we consider surjective. Click here. Let e(I) be the extension of I from C∞(M)p to R[[X1,...,Xn]] and c(I) be the contraction.

(8)τ∞−1(I1k)=c(Ik)=c(e(c(I))k)=c(e((c(I))k))=(c(I))k=(τ∞−1(I1))k

From (3) we see that Mk⊃Mk+1, hence we have M∞=⋂i=1∞Mi=⋂i=1∞mpi.

Then we define the cotangent space Tp∗(M) as mp/mp2​​​​.

This consturction could be generalized to locally ringed space, for example, affine scheme. For px∈Spec(R) consider the stalk at x, Rpx whcih is a local ring as well, then consider px/px2, whcih is a Rpx/px vector space.

Cotangent space as a functor

Let Man∗ be the category of smooth manifold with base point, then we could define the cotangent space functor:

(9)Tp∗(−):Man∗op→R−Mod

Let g:Mp→Ng(p) be a smooth function, then we induce a ring homomorphism

(10)g∗:C∞(N)g(p)→C∞(M)p

And

(11)g∗:mg(p)→mp,mg(p)2→mp2⟹g∗:Tg(p)∗(N)→Tp∗(M)

Tangent Space

Proposition.

(12)DerR(C∞(M)p,−)≅HomR−Mod(mp/mp2,−)

Proof. Let DerR(C∞(M)p,V) be the set such that D is R linear and D(fg)=f(p)D(g)+D(f)g(p).

Let D:mp→V be a linear map and KerD⊇mp2. Then for f,g∈C∞(M)p, define D(f)=D(f−f(p))​.

Notice that

(13)fg=(f(p)+(f−f(p)))(g(p)+(g−g(p)))

Simplify it we get

(14)f(p)g(p)+f(p)(g−g(p))+g(p)(f−f(p))+(f−f(p))(g−g(p))

then we have

(15)D(fg)=D(f(p)g(p)+f(p)(g−g(p))+g(p)(f−f(p))+(f−f(p))(g−g(p)))

The last term in mp2, hence we have

(16)D(fg)=f(p)D(g)+D(f)g(p)

Conversely, let D be a derivation, then restrict D to mp we have

(17)D(fg)=f(p)D(g)+D(f)g(p)=0

Then KerD⊇mp2. Hence we have a bijection between derivation and linear map from mp to sth and KerD⊇mp2.

The following theorem establish the bijection

between linear map from mp to sth and KerD⊇mp2 to HomR−Mod(mp/mp2,−).

image-20241108101750457

The naturalness leaves to readers.

Corollary. DerR(C∞(M)p,R)≅HomR−Mod(mp/mp2,R). Hence Tp(M)≅DerR(C∞(M)p,R).

The basis of mp/mp2 is x1−x1(p),...,xn−xn(p) since τ2−1(0)=mp2, τ2:mp→Im(τ2)≅mp/mp2.

Easy to see that the dual basis of xi−xi(p) is ∂∂xi|p.

Usually we denote the elements of mp/mp2 as dx1,...,dxn, dxi(∂∂xj|p)=∂∂xj|pxi=δi,j.

Tangent Space Functor

As we have seen, Tp(−) should be defined as HomR−Mod(Tp∗(−),R).

Let (x1,...,xm) be the coordinate around p and (y1,...,yn) be the coordinate around g(p).

For a smooth function g:Mp→Ng(p),

(18)Tp(g)(∂∂xj)=(g∗)∗(∂∂xj)=∂∂xj(g∗(−))=∂∂xj(−∘g)=∑i=1n∂gi∂xj∂∂yi.

 

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