Dieudonné module theory, part II: the classical theory...Dieudonné module theory, part II: the classical theory Alan Koch Agnes Scott College May 24, 2016 Alan Koch (Agnes Scott

Dieudonneacute module theory part IIthe classical theory

Alan Koch

Agnes Scott College

May 24 2016

Alan Koch (Agnes Scott College) 1 41

Outline

1 Overview

2 Witt Vectors

3 Dieudonneacute Modules

4 Some Examples

5 Duality

6 Want More


Pick a prime p gt 2

Recall it is unreasonable to think that every finite abelian Hopf algebraH of p-power rank over an Fp-algebra R can be classified withDieudonneacute modules

Last time we focused on Hopf algebras with a nice coalgebrastructure

This time we will make assumptions on both R and H

The assumptions on R are very restrictive and the ones on H aremore palatable


The situation du jour (et demain)

Throughout this talk let R = k be a perfect field of characterisic p

Any finite k -Hopf algebra H can be written as

H sim= Hr r otimes Hr ℓ otimes Hℓr otimes Hℓℓ

whereHr r and Hlowast

r r are both reduced k -algebrasHr ℓ is a reduced and Hlowast

r ℓ is a local k -algebraHℓr is a local and Hlowast

r ℓ is a reduced k -algebraHℓℓ and Hlowast

ℓℓ are both local k -algebras

If p ∤ dimk H then H = Hr r



Reduced k -Hopf algebras are ldquoclassified they correspond to finitegroups upon which Gal(kk) acts continuously

Then Hr r and Hr ℓ are classified and by duality so is Hℓr

Thus the most mysterious Hopf algebras are the ones of the form Hℓℓ

We will call these ldquolocal-localrdquo and assume that all Hopf algebras in thistalk are finite abelian and local-local over the perfect field k

If H is local-local then a result of Waterhouse is that H is a ldquotruncatedpolynomial algebrardquo ie

H sim= k [t1 tr ](tpn1

1 tpnr

r )

so the algebra structure isnrsquot terrible


Suggested readings

M Demazure and P Gabriel Groupes Algeacutebriques - Tome 1Good news partially translated into English as Introduction toAlgebraic Geometry and Algebraic GroupsBad news not the part you need (chapter 5)A Grothendieck Groupes de Barsotti-Tate et Cristaux deDieudonneacuteExplicit connection between Dieudonneacute modules and HopfalgebrasT Oda The first de Rham cohomology group and Dieudonneacutemodules The section on Dieudonneacute modules gave me my firstexplicit examplesR Pink Finite group schemeshttpspeoplemathethzch pinkftpFGSCompleteNotespdfGood intro to Dieudonneacute modules with proofs (group schemepoint of view)


The gist

What is a k -Hopf algebra

It is a k -algebra and a k -coalgebra with compatible structures

A Dieudonneacute module is a single module (over a ring TBD) whichencodes both structures

In Mondayrsquos talk the coalgebra structure was assumed and thealgebra structure was encoded by the action of F on Dlowast(H)


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


Witt polynomials

For each n isin Zge0 define Φn isin Z[Z0 Zn] by

Φn(Z0 Zn) = Z pn

0 + pZ pnminus1

1 + middot middot middot+ pnZn

Note Φn depends on the choice of prime p which we assume to beldquoourrdquo p

Define polynomials SnPn isin Z[X0 XnY0 Yn] implicitly by

Φn(S0 Sn) = Φn(X0 Xn) + Φn(Y0 Yn)

Φn(P0 Pn) = Φn(X0 Xn)Φn(Y0 Yn)

Fact Yes the coefficients are all integers


Φn(Z0 Zn) = Z pn

0 + pZ pnminus1




Example (low hanging fruit)

Φ0(Z0) = Z0

S0(X0Y0) = X0 + Y0

P0(X0Y0) = X0Y0

Exercise 1 Prove

S1((X0X1) (Y0Y1)) = X1 + Y1 minus1p

pminus1983142

i=1

983072pi

983073X i

0Y pminusi0

P1((X0X1) (Y0Y1)) = X p0 Y1 + X p

1 Y0 + pX1Y1


Let W (Z) = (w0w1w2 ) wi isin Z for all i Define addition andmultiplication on W (Z) by

(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )

(w0w1 ) middotW (x0 x1 ) = (P0(w0 x0)P1((w0 x1) (w0 x1)) )

These operations make W (Z) into a commutative ring

Actually W (minus) is a Z-ring scheme meaning that for all Z-algebras AW (A) is a ring

In particular set W = W (k) and call this the ring of Witt vectors withcoefficients in k (or ldquoWitt vectorsrdquo for short)


Properties of Witt vectors Show each of the following

Exercise 2 W is a ring of characteristic zero

Exercise 3 W is an integral domain

Exercise 4 (1 0 0 ) is the multiplicative identity of W

Exercise 5 W (Fp) sim= Zp

Exercise 6 W (Fpn) is the unramified extension of Zp of degree n

Exercise 7 The element (0 1 0 0 ) isin W acts as mult by p

Exercise 8 pnW is an ideal of W

Exercise 9 Let Wn = (w0w1 wnminus1) Then Wn is a ring withoperations induced from W

Exercise 10 WpnW sim= Wn In particular W0sim= k


Of particular importance will be two operators F V on W given by

F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )

F is called the Frobenius and V is called the VerschiebungShow each of the following

Exercise 11 FV = VF = p (multiplication by p)

Exercise 12 F V both act freely on W only F acts transitively

Exercise 13 If w isin pnW then Fw Vw isin pnW Thus F and V makesense on Wn as well

Exercise 14 Wn is annihilated by V n+1

Exercise 15 Any w = (w0w1w2 ) isin W decomposes as

(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


The Dieudonneacute ring

Now let wσ be the Frobenius on w isin W (invertible since k is perfect)

Let E = W [F V ] be the ring of polynomials with

FV = VF = p Fw = wσF wV = Vwσw isin W

We call E the Dieudonneacute ring

Note that E is commutative if and only if k = Fp


One more construction

Let us view Wn as a group scheme (as opposed to the ring Wn(k))

Let W mn be the mth Frobenius kernel of the group scheme Wn ie for

any k -algebra A

W mn (A) = (a0 a1 anminus1) ai isin A apm

i = 0 for all 0 le i le n minus 1

Some properties of W mn

Clearly F Wn rarr Wn restricts to F W mn rarr W m

n Also V Wn rarr Wn restricts to V W m

n rarr W mn

For each k -algebra A W (k) acts on W (A) through the algebrastructure map k rarr AEvery local-local Hopf algebra represents a subgroup scheme ofsome (W m

n )r


There are maps ι W mn rarr W m+1

n (inclusion) and ν W mn rarr W m

n+1(induced by V ) such that

W mn

ιminusminusminusminusrarr W m+1n983117983117983175ν ν

983117983117983175

W mn+1

ιminusminusminusminusrarr W m+1n+1

Then W mn is a direct system (where the partial ordering

(nm) le (nprimemprime) is given by the conditions n le nprime m le mprime)

Let983153W = limminusrarr

mnW m

n

(Not standard notation)


Let H be a local-local Hopf algebra Define

Dlowast(H) = Homk -gr983054983153W Spec(H)

983055

The actions of F and V on 983153W induces actions on Dlowast(H) as does theaction of W (k)

Thus Dlowast(H) is an E-module

Furthermore Dlowast(minus) is a covariant functor despite the fact this isreferred to as ldquocontravariant Dieudonneacute module theoryrdquo



983055

Properties of DlowastDlowast(H) is killed by some power of F and V Dlowast(H) is finite length as a W -modulelengthW Dlowast(H) = logp dimk HDlowast(H1 otimes H2) sim= Dlowast(H1)times Dlowast(H2)Note Spec(H1 otimes H2) sim= Spec(H1)times Spec(H2)

Theorem (Main result in Dieudonneacute module theory)Dlowast induces a categorical equivalence

983110983114

983112

p- power ranklocal-local

k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112

E-modulesof finite length over W (k)

killed by a power of F and V

983111983115

983113

We call E-modules satisfying the above Dieudonneacute modules


Legal disclaimer

983110983114

983112



983111983115

983113 = ldquoDieudonneacute modulesrdquo

This is not everyonersquos definition

Dieudonneacute modules can also be used to describep-power rank Hopf algebras (local separable)p-divisible groupsgraded Hopf algebrasprimitively generated Hopf algebrasa slew of other things


Another legal disclaimer

983110983114

983112



983111983115


Some literature will describe a Dieudonneacute module as a triple (MF V )where

M is a finite length W (k)-moduleF M rarr M is a nilpotent σ-semilinear mapV M rarr M is a nilpotent σminus1-semilinear map

This is just a different way to describe the same thing


A Hopf algebra interpretation


983055

Each W mn is an affine group scheme represented by

Hmn = k [t0 tnminus1](tpm

1 tpm

2 tpm

nminus1)

with comultiplications induced from Witt vector addition

∆(ti) = Si((t0 otimes 1 ti otimes 1) (1 otimes t0 1 otimes ti))

In theory we could possibly write something like

Dlowast(H) = Homk -Hopf(H limlarrminusmn

Hmn)

However I have never found this to be helpful


No wait come back

Suppose we are given a Dieudonneacute module What is thecorresponding Hopf algebra

Suppose V N+1M = 0 Let H = k [Tm m isin M] subject to therelations

TFm = (Tm)p

Tm1+m2 = SN((TV Nm1 TVm1 Tm1) (TV Nm2

TVm2 Tm2)

T(w0w1 )m = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)

mm1m2 isin M (w0w1 ) isin W

Define ∆ H rarr H otimes H by

∆(Tm) = SN((TV Nmotimes1TV Nminus1motimes1 Tmotimes1) (1otimesTV Nm 1otimesTm))

Then H is a local-local k -Hopf algebra


The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)

∆(Tm) = SN((TV Nm otimes 1 Tm otimes 1) (1 otimes TV Nm 1 otimes Tm))

The action of F describes the (interesting) algebra structure

The action of V describes the coalgebra structure

Exercise 16 Show that the N above need not be minimal

Exercise 17 Let M be a Dieudonneacute module Show that T0 = 0


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)


ExampleThe simplest possible (nontrivial) E-module is a k -vector space M ofdimension 1 with F and V acting triviallyIn other words M = EE(F V ) In particular N = 1 Let M havek -basis x and let H be the Hopf algebra with Dlowast(H) = M

Exercise 18 Prove H is generated as a k -algebra by t where t = Tx

Sincetp =

983054Tx

983055p= TFx = T0 = 0

we have H = k [t ](tp) Also by the formulas above t is primitive


Primitively generated flashback

Suppose H is primitively generated (and local-local) and letM = Dlowast(H)

Since


it follows that we may take N = 0

Thus VM = 0 hence M can be viewed as a module over k [F ] with

TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm

In this case itrsquos the same module as yesterday


ExampleLet M = E(F mV n) = Dlowast(H)

Exercise 19 What is pM

Exercise 20 Exhibit a k -basis for M

Exercise 21 Prove that

H = k [t1 tn](tpm

1 tpm

n )

Exercise 22 Write out the comultiplication (in terms of Witt vectoraddition)

Exercise 23 Show that Spec(H) = W mn


Let H = k [t ](tp5) with

∆(t) = t otimes 1 + 1 otimes t +pminus1983142

i=1

1i(p minus i)

tp3i otimes tp3(pminusi)1

What is Dlowast(H)Let M = Dlowast(H) and suppose there is an x isin M such that t = Tx Then

0 = tp5=

983054Tx

983055p5= TF 5x

so F 5x = 0 from which it follows that F 5M = 0 but F 4M ∕= 0 Also

∆(t) = S1983054((Tx)

p3 otimes 1Tx otimes 1) (1 otimes (Tx)p3 1 otimes Tx)

983055

= S1983054TF 3x otimes 1Tx otimes 1) (1 otimes TF 3x 1 otimes Tx)

Then N = 1 (ie V 2M = 0) and Vm = F 3m Thus

M sim= EE(F 5F 3 minus V V 2) = EE(F 5F 3 minus V )

This can be confirmed by working backwardsAlan Koch (Agnes Scott College) 29 41

Let H = k [t1 t2](tp2

1 tp2

2 ) with

∆(t1) = t1 otimes 1 + 1 otimes t1

∆(t1) = t2 otimes 1 + 1 otimes t2 +pminus1983142

i=1

1i(p minus i)

tpi1 otimes tp(pminusi)

1

What is Dlowast(H)Let M = Dlowast(H) and suppose x y isin M satisfy t1 = Tx t2 = Ty Clearly F 2x = F 2y = 0 and

∆(t1) = S0(Tx otimes 1 1 otimes Tx)

= S1((0 otimes 1Tx otimes 1) (1 otimes 0 1 otimes Tx))

∆(t2) = S1983054((Tx)

p otimes 1Ty otimes 1) (1 otimes (Tx)p 1 otimes Ty )

983055

= S1983054(TFx otimes 1Ty otimes 1) (1 otimes TFx 1 otimes Ty )

983055

Thus Vx = 0 and Vy = Fx So

M = Ex + Ey F 2M = V 2M = 0Vx = 0Fx = Vy


Exercises

Write out the Hopf algebra for each of the following Be as explicit asyou can

Exercise 24 M = E(F 2F minus V )

Exercise 25 M = EE(F 2V 2)

Exercise 26 M = EE(F 2 pV 2)

Exercise 27M = Ex + Ey F 4x = 0F 3y = 0Vx = F 2y V 2x = 0Vy = 0


More exercises

Find the Dieudonneacute module for each of the following

Exercise 28 H = k [t ](tp4) t primitive

Exercise 29 H = k [t1 t2](tp3

1 tp2

2 ) t1 primitive and


i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)

tp2i1 otimes tp2(pminusi)

1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


How does duality work

Very nicely with the proper definitions

For an E-module M we define its dual Mlowast to be

Mlowast = HomW (MW [pminus1]W )

This is an E-module where F and V act on φ M rarr W [pminus1]W asfollows

(Fφ)(m) = (φ(Vm))σ

(Vφ)(m) = (φ(Fm))σminus1

Note The σ σminus1 make Fφ Vφ into maps which are W -linear



PropertiesIf M is a Dieudonneacute module so is Mlowast(EE(F nV m))lowast = EE(F mV n)ldquoDuality interchanges the roles of F and V rdquoDlowast(Hlowast) = (Dlowast(H))lowast


An example

Let H = Fp[t ](tp3) with


i=1

1i(p minus i)

tp2i otimes tp2(pminusi)

The choice of k = Fp is only for ease of notation

Then M = EE(F 3F 2 minus V ) = kx oplus kFx oplus kF 2x as W -modules

We wish to compute HomW (MW [pminus1]W ) = HomZp(MQpZp)


HomZp(EE(F 3F 2 minus V )QpZp)

Note that any φ kx oplus kFx oplus kF 2x rarr QpZp is determined by theimages of x Fx and F 2x

Note also that V (Fx) = V (F 2x) = 0 which will be useful later

Since px = FVx = F (F 2x) = F 3x = 0

Mlowast sim= (w y z) isin (QpZp)3 pw py pz isin Zp

where φ isin Mlowast is given by

φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by

(Fφ)(x) = (φ(Vx))σ = (φ(F 2x))σ = zσ = z (Fφ)(y) = (Fφ(z)) = 0

and V acts on φ by

(Vφ)(x) = (φ(Fx))σminus1

= yσminus1= y

(Vφ)(Fx) = (φ(F 2x))σminus1

= zσminus1= z

(Vφ(z)) = 0

Thus F (w y z) = (z 0 0) V (w y z) = (y z 0)Clearly for all u = (w y z) we have V 3u = 0 and V 2u = FuIt follows that Mlowast = EE(V 3V 2 minus F )


H = Fp[t ](tp3) Mlowast = EE(V 3V 2 minus F )

Exercise 31 Using Mlowast give the algebra structure for Hlowast (Hint Itrequires two generators)

Exercise 32 Using Mlowast give the coalgebra structure for Hlowast

Exercise 33 What changes if k ∕= Fp


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


More is coming tomorrow

Thank you


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


Pick a prime p gt 2
























1 tpnr

r )



Suggested readings



The gist






Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


Witt polynomials


Φn(Z0 Zn) = Z pn

0 + pZ pnminus1








Φn(Z0 Zn) = Z pn

0 + pZ pnminus1





Φ0(Z0) = Z0

S0(X0Y0) = X0 + Y0

P0(X0Y0) = X0Y0

Exercise 1 Prove

S1((X0X1) (Y0Y1)) = X1 + Y1 minus1p

pminus1983142

i=1

983072pi

983073X i

0Y pminusi0

P1((X0X1) (Y0Y1)) = X p0 Y1 + X p

1 Y0 + pX1Y1



(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )


















F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Pick a prime p gt 2
























1 tpnr

r )



Suggested readings



The gist






Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


Witt polynomials


Φn(Z0 Zn) = Z pn

0 + pZ pnminus1








Φn(Z0 Zn) = Z pn

0 + pZ pnminus1





Φ0(Z0) = Z0

S0(X0Y0) = X0 + Y0

P0(X0Y0) = X0Y0

Exercise 1 Prove

S1((X0X1) (Y0Y1)) = X1 + Y1 minus1p

pminus1983142

i=1

983072pi

983073X i

0Y pminusi0

P1((X0X1) (Y0Y1)) = X p0 Y1 + X p

1 Y0 + pX1Y1



(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )


















F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you




















1 tpnr

r )



Suggested readings



The gist






Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


Witt polynomials


Φn(Z0 Zn) = Z pn

0 + pZ pnminus1








Φn(Z0 Zn) = Z pn

0 + pZ pnminus1





Φ0(Z0) = Z0

S0(X0Y0) = X0 + Y0

P0(X0Y0) = X0Y0

Exercise 1 Prove

S1((X0X1) (Y0Y1)) = X1 + Y1 minus1p

pminus1983142

i=1

983072pi

983073X i

0Y pminusi0

P1((X0X1) (Y0Y1)) = X p0 Y1 + X p

1 Y0 + pX1Y1



(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )


















F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you









1 tpnr

r )



Suggested readings



The gist






Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


Witt polynomials


Φn(Z0 Zn) = Z pn

0 + pZ pnminus1








Φn(Z0 Zn) = Z pn

0 + pZ pnminus1





Φ0(Z0) = Z0

S0(X0Y0) = X0 + Y0

P0(X0Y0) = X0Y0

Exercise 1 Prove

S1((X0X1) (Y0Y1)) = X1 + Y1 minus1p

pminus1983142

i=1

983072pi

983073X i

0Y pminusi0

P1((X0X1) (Y0Y1)) = X p0 Y1 + X p

1 Y0 + pX1Y1



(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )


















F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Suggested readings



The gist






Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


Witt polynomials


Φn(Z0 Zn) = Z pn

0 + pZ pnminus1








Φn(Z0 Zn) = Z pn

0 + pZ pnminus1





Φ0(Z0) = Z0

S0(X0Y0) = X0 + Y0

P0(X0Y0) = X0Y0

Exercise 1 Prove

S1((X0X1) (Y0Y1)) = X1 + Y1 minus1p

pminus1983142

i=1

983072pi

983073X i

0Y pminusi0

P1((X0X1) (Y0Y1)) = X p0 Y1 + X p

1 Y0 + pX1Y1



(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )


















F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


The gist






Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


Witt polynomials


Φn(Z0 Zn) = Z pn

0 + pZ pnminus1








Φn(Z0 Zn) = Z pn

0 + pZ pnminus1





Φ0(Z0) = Z0

S0(X0Y0) = X0 + Y0

P0(X0Y0) = X0Y0

Exercise 1 Prove

S1((X0X1) (Y0Y1)) = X1 + Y1 minus1p

pminus1983142

i=1

983072pi

983073X i

0Y pminusi0

P1((X0X1) (Y0Y1)) = X p0 Y1 + X p

1 Y0 + pX1Y1



(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )


















F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


Witt polynomials


Φn(Z0 Zn) = Z pn

0 + pZ pnminus1








Φn(Z0 Zn) = Z pn

0 + pZ pnminus1





Φ0(Z0) = Z0

S0(X0Y0) = X0 + Y0

P0(X0Y0) = X0Y0

Exercise 1 Prove

S1((X0X1) (Y0Y1)) = X1 + Y1 minus1p

pminus1983142

i=1

983072pi

983073X i

0Y pminusi0

P1((X0X1) (Y0Y1)) = X p0 Y1 + X p

1 Y0 + pX1Y1



(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )


















F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Witt polynomials


Φn(Z0 Zn) = Z pn

0 + pZ pnminus1








Φn(Z0 Zn) = Z pn

0 + pZ pnminus1





Φ0(Z0) = Z0

S0(X0Y0) = X0 + Y0

P0(X0Y0) = X0Y0

Exercise 1 Prove

S1((X0X1) (Y0Y1)) = X1 + Y1 minus1p

pminus1983142

i=1

983072pi

983073X i

0Y pminusi0

P1((X0X1) (Y0Y1)) = X p0 Y1 + X p

1 Y0 + pX1Y1



(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )


















F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Φn(Z0 Zn) = Z pn

0 + pZ pnminus1





Φ0(Z0) = Z0

S0(X0Y0) = X0 + Y0

P0(X0Y0) = X0Y0

Exercise 1 Prove

S1((X0X1) (Y0Y1)) = X1 + Y1 minus1p

pminus1983142

i=1

983072pi

983073X i

0Y pminusi0

P1((X0X1) (Y0Y1)) = X p0 Y1 + X p

1 Y0 + pX1Y1



(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )


















F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you



(w0w1 ) +W (x0 x1 ) = (S0(w0 x0)S1((w0w1) (x0 x1)) )


















F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you














F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you



F (w0w1w2 ) = (wp0 w

p1 w

p2 )

V (w0w1w2 ) = (0w0w1 )







(w0w1w2 ) =infin983142

i=0

pi(wpminusi

i 0 0 )


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you












any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you





any k -algebra A






n rarr W mn


n )r





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you





W mn


983117983117983175

W mn+1





mnW m

n





983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you




983055






983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you



983055



983110983114

983112


k-Hopf algebras

983111983115

983113 minusrarr

983110983114

983112



983111983115

983113



Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Legal disclaimer

983110983114

983112



983111983115






983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you



983110983114

983112



983111983115








983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you




983055



1 tpm

2 tpm

nminus1)





Hmn)



No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


No wait come back



TFm = (Tm)p


TVm2 Tm2)


0 wpminusN

N ) (TV Nm TVmTm)






The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


The gist part 2

TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


TFm = (Tm)p


TVm2 Tm2)

Twm = PN((wpminusN

0 wpminusN

N ) (TV Nm TVmTm)




Sincetp =

983054Tx

983055p= TFx = T0 = 0





Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you




Since




TFm = (Tm)p

Tm1+m2 = Tm1 + Tm2

Twm = w0Tm







H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you






H = k [t1 tn](tpm

1 tpm

n )






i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you




i=1

1i(p minus i)



0 = tp5=

983054Tx

983055p5= TF 5x


∆(t) = S1983054((Tx)


983055






1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you



1 tp2

2 ) with



i=1

1i(p minus i)


1




∆(t2) = S1983054((Tx)


983055


983055




Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Exercises







More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


More exercises




1 tp2



i=1

1i(p minus i)


1


1 tp2



i=1

1i(p minus i)


1


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you







(Fφ)(m) = (φ(Vm))σ







An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you





An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


An example



i=1

1i(p minus i)












φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you








φ(x) = wφ(Fx) = y

φ(F 2x) = z


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


φ(x) = w φ(Fx) = y φ(F 2x) = z


Now F acts on φ by


and V acts on φ by


= yσminus1= y


= zσminus1= z

(Vφ(z)) = 0








Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you







Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you


Outline

1 Overview

2 Witt Vectors


4 Some Examples

5 Duality

6 Want More



Thank you



Thank you


Documents

Dieudonné module theory, part II: the classical theory...Dieudonné module theory, part II: the classical theory Alan Koch Agnes Scott College May 24, 2016 Alan Koch (Agnes Scott