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Friday, November 8, 2024

Topics in Differential Geometry

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

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),pU​ induce a isomorphism

(1)C(U)C(φ(U)),ffφ1C(M)pC(Rn)φ(p)

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

We denote the kernel of τk as Mk and kenrel of τ:C(M)pR[[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+1πkR[X1,...,Xn]/Ik...

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

(3)τk=πkτk+1

Thus

(4)Mk=τk1(0)=τk+11(πk1(0))τk+11(0)=Mk+1

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

(5)τk=pkτ

Hence

(6)Mk=τk1(0)=τ1(pk1(0))=τ1(Ik)=τ1(I1k)=(τ1(I1))k=M1k=mpk

The equation

(7)τ1(I1k)=(τ1(I1))k

hold since I1Im(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)τ1(I1k)=c(Ik)=c(e(c(I))k)=c(e((c(I))k))=(c(I))k=(τ1(I1))k

From (3) we see that MkMk+1, hence we have M=i=1Mi=i=1mpi.

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 pxSpec(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():ManopRMod

Let g:MpNg(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)2mp2g:Tg(p)(N)Tp(M)

Tangent Space

Proposition.

(12)DerR(C(M)p,)HomRMod(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:mpV be a linear map and KerDmp2. Then for f,gC(M)p, define D(f)=D(ff(p))​.

Notice that

(13)fg=(f(p)+(ff(p)))(g(p)+(gg(p)))

Simplify it we get

(14)f(p)g(p)+f(p)(gg(p))+g(p)(ff(p))+(ff(p))(gg(p))

then we have

(15)D(fg)=D(f(p)g(p)+f(p)(gg(p))+g(p)(ff(p))+(ff(p))(gg(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 KerDmp2. Hence we have a bijection between derivation and linear map from mp to sth and KerDmp2.

The following theorem establish the bijection

between linear map from mp to sth and KerDmp2 to HomRMod(mp/mp2,).

image-20241108101750457

The naturalness leaves to readers.

Corollary. DerR(C(M)p,R)HomRMod(mp/mp2,R). Hence Tp(M)DerR(C(M)p,R).

The basis of mp/mp2 is x1x1(p),...,xnxn(p) since τ21(0)=mp2, τ2:mpIm(τ2)mp/mp2.

Easy to see that the dual basis of xixi(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 HomRMod(Tp(),R).

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

For a smooth function g:MpNg(p),

(18)Tp(g)(xj)=(g)(xj)=xj(g())=xj(g)=i=1ngixjyi.

 

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