Fuetterer, Schwab - Spectral Sequences

8182019 Fuetterer Schwab - Spectral Sequences

httpslidepdfcomreaderfullfuetterer-schwab-spectral-sequences 119

Spectral Sequences

Miriam Schwab Michael Fuumlerer

ᵗʰ September

We explain the notion of a spectral sequence and various ways of construct-

ing them before illustrating the concept by some examples We do not treat

spectral sequences in the greatest possible generality but tried to keep the text

accessible

is note grew out of the seminar ldquoStable cohomology of the mapping classgrouprdquo by JProf Dr Gabriela Weitze-Schmithuumlsen and Dipl-Inform Tobias

Columbus at the Karlsruhe Institute of Technology in summer term We

would like to thank both of them and especially Tobias for explaining us a lot

about spectral sequences

Parts of this text follow [Wei] rather closely

Contents

1 Spectral Sequences and Convergence 2

2 Construction of Spectral Sequences 6

3 Double Complexes 10

4 Examples 12

e five lemma

e snake lemma

Balancing Tor Base change for Tor

e universal coefficients theorem

References 19



1 Spectral Sequences and Convergence

Let C be an abelian category

Definition (Spectral sequences) A (homologically graded) spectral sequence is a family

of objects E r pq

of C for all p q isin Z and r ge a (with a fixed a isin Z) together with

differentials d r pq E r pq

E r p minusr q +r minus1 which satisfy d r d r = 0 Furthermore we require

that there are isomorphisms

E r +1 pq H (E

r pq ) = ker(d r pq )

1048623im(d r p +r q minusr +1)

By n = p + q we denote the total degree of E r pq

e collections (E r pq ) p q isinZ for fixed r are called the sheets or pages of the spectral sequence

By the isomorphisms required above we imagine that we get from one sheet to the next

one (ldquoturing a page aroundrdquo) by taking homology

To get an imagination of how the differentials look like we first regard the sheet of thespectral sequence where r = 0

E 003 E 013 E 023 E 033

E 002 E 012 E 022 E 032

E 001 E 011 E 021 E 031

E 000 E 010 E 020 E 030

On the first sheet the differentials go from the right to the le as is shown here

E 103 E 113 E 123 E 133

E 1

02 E 1

12 E 1

22 E 1

32

E 101 E 111 E 121 E 131

E 100 E 110 E 120 E 130



In the following picture we see as an example the 2-sheet with differentials d 2 pq E 2 p q

E 2 p minus2q +1

E 203 E 213 E 223 E 233

E 202

E 212

E 222

E 232

E 201 E 211 E 221 E 231

E 200 E 210 E 220 E 230

Example (First quadrant spectral sequence) A spectral sequence with E r pq = 0 for p lt

0 or q lt 0 is called a first quadrant spectral sequence We observe that for r large enough

the differential with codomain E r pq has domain 0 and the differential with domain E r pq has

codomain 0 We get

E r +1 pq = H (E r pq ) = ker(d )

983087im(d ) = E r pq

9830870 = E r pq

e stable value E r pq = E k pq for k ge such r is named E infin pq

We restrict ourselves to first quadrant spectral sequences which makes our life much easier

Definition (Convergence) Let H n be a family of objects of C

We say a spectral sequence

(a) weakly converges to H lowast if there exists a filtration

sube F p minus1H n sube F p H n sube F p +1H n sube sube H n

for each n isin Z and furthermore isomorphisms

β pq E infin pq

F p H p +q 983087F p minus1H p +q

(b) approaches H lowast if it weakly converges to H lowast andsup1

H n =991786

F p H n and991785

F p H n = 0

(c) converges to H lowast if it approaches H lowast and H n = limlarrminusminus

983080H nF p H n

983081sup2

We denote convergence by E r pq rArr H p +q

Remark Let a spectral sequence converging to H lowast have E 2 pq = 0 unless p = 0 1 en

for each n there is a short exact sequence

0 E 20n H n E 21nminus1

0

sup1 e first of the two conditions is called exhaustive and the second one Hausdorff sup2 A filtration which satisfies foralln existt F t H n = H n fulfills this additional condition



Proof First we draw a picture of the 2-sheet of the spectral sequence

0

0

E 203

E 213

0

0

0

0

E 2

02

E 2

12

0

0

0 0 E 201 E 211 0 0

0

0 E 200 E 210

0

0

We see that E 2 pq = E infin pq and since we know that the spectral sequence converges we know

that there exists a filtration F p H n which fulfills E infin pq F p H p +q F p minus1H p +q

If p 0 1 we get 0 = E 2 pq = F p H p +q F p minusq H p +q which tells us F p H p +q = F p minus1H p +q and

therefore the filtration looks like

F minus2H n = F minus1H n sube F 0H n sube F 1H n = F 2H n =

Moreover we know thatcap

F p H n = 0 so F minus1H n = F minus2H n = = 0 and sincecup

F p H n =H n we get F 1H n = F 2H n = H n

Now let p = 0 We notice that

E 20n = E infin0n F 0H n

983087F minus1H n = F 0H n

For p = 1 we get

E 21nminus1 = E infin1nminus1 F 1H n 983087F 0H n = H n 983087F 0H n

and obviously the short exact sequence

0 F 0H n H n H n983087F 0H n

0

turns into the short exact sequence


0

Remark Let a spectral sequence converging to H lowast have E 2 pq = 0 unless q = 0 1 en

there is a long exact sequence

H p +1 E 2 p +10

d E 2 p minus11

H p E 2 p 0

d E 2 p minus21

H p minus1



Proof Again we first draw a picture

0

0

0

0

E 2minus11

E 2

01

E 2

11

E 2

21

E 2minus10 E 200 E 210 E 220

0

0

0

0

and we see that the objects at the 3-sheet are the objects E infin pq For q 1 0 we get

F p H p +q 983087F p minus1H p +q E

infin pq = E 3 pq = H (E 2 pq ) = ker(d 2 pq )

1048623im(d 2 p +2q minus1) = 0

and therefore F p H p +q = F p minus1H p +q

Letrsquos have a look at the filtration of H n

= F nminus3H n = F nminus2H n sube F nminus1H n sube F nH n = F n+1H n =

capF p H n = 0 implies F r H n = 0 for r le p +q minus 2 and

cupF p H n = H n implies F r H n = H n for

r ge n = p + q

Let q = 0 We get

H p 983087F p minus1H p = F p H p

983087F p minus1H p E

infin p 0 = E 3 p 0 = ker(d 2 p 0)

1048623im(d 2 p +2minus1) = ker(d 2 p 0)

And q = 1 leads to

F p H p +1 = F p H p +1983087F p minus1H p +1 E

infin p 1 = E 3 p 1

= H (E 2 p 1) = ker (d 2 p 1)1048623

im(d 2 p +20) = E 2 p 1

1048623im(d 2 p +20)



We can now put all this together to a long exact sequence

0

0

E 2 p +10 kerd 2 p +10

H p F p minus1H p

H p +1 E 2 p +10 E 2 p minus11 H p

kerd 2 p +10 F p minus1H p

0 0 0 0

2 Construction of Spectral Sequences

In this section we introduce the important notions of exact couples and filtered complexes

Every exact couple yields a spectral sequence and every filtered complex yields an exact

couple Finally we prove a convergence theorem for spectral sequences obtained in this

way

Construction (Exact couples and their derivations) An exact couple is a pair of objects

D and E of C and morphisms i j and k such that the diagram

D

D

E

i

j k

is exact at each vertex Since d = jk complies with d 2 = ( jk )2 = j (k j )k = j 0k = 0 we

can apply homology by seing H (E ) = ker(d )im(d ) and get the derived exact couple

i (D ) i (D )

H (E )

i |i (D )

j (2)k (2)

with j (2)(i (x )) = [ j (x )] and k (2)([e ]) = k (e ) x isin D and [e ] isin H (E ) It is an easy computa-

tion that j (2) and k (2) are well-defined and that the derived couple is exact It suggests itself

to iterate this process and set E 1 = E and E r = H (E r minus1) d 1 = d = jk and d r = j (r )k (r )



e (r + 1)-th exact couple looks like

i r (D ) i r (D )

E r +1

i |i r (D )

j (r +1)k (r +1)

Remark Notice that the cycles can be described as Z r = k minus1(i r (D )) and the boundaries

as B r = j (ker(i r ))

Proof One just calculates

Z r = ker(d r ) = ker( j (r )k (r )) = ker( ji minusr +1k )

= k minus1(ker( ji minusr +1)) = k minus1(i r minus1(ker j ))

= k minus1(i r minus1(im i )) = k minus1(i r (D ))

B r = im(d r ) = im( j (r )k (r )) = im( ji minusr +1k )

= j (i minusr +1k (E pq )) = j (i minusr +1 im(k ))

= j (i minusr +1(ker(i )) = j (i minusr (0)) = j (ker(i r ))

We now regardD =oplus

D pq and E =oplus

E pq with the morphismsi j and k having bidegree

(1 minus1) (0 0)sup3 and (minus1 0) respectively (ie i (D pq ) sube D p +1q minus1 and so on) e bidegrees

of i and k do not change by passage to the derived couple while the bidegree of j (r ) in the

r -th couple is deg( j (r )) = deg( j ) minus (r minus 1) deg(i ) = (0 0) minus (r minus 1)(1 minus1) = (minus(r minus 1) r minus 1)since j (r )(i r minus1(x )) = [ j (x )] and therefore the bidegree of the differential d r = j (r )k (r ) is

deg(d r ) = (minusr r minus 1) just as claimed in the definition of spectral sequences

In this way the exact couple yields a spectral sequence with objects E

r

pq Example (Exact couple of a filtered complex) Let C lowast be a complex with a filtration

F p C lowast sube F p +1C lowast sube sube C lowast

such that there are integers s lt t for eachn with F s C n = 0 and F t C n = C n (Such a filtration

is called bounded Particularly this means F k C n = 0 for all k le s and F k C n = C n for all

k ge t We will always consider canonically bounded filtrations ie s = minus1 and F nC n = C nfor all n ndash what leads to a first quadrant spectral sequence)

We get short exact sequences

0 F p minus1C lowast

i

F p C lowast

π p

F p C lowast983087F p minus1C lowast

0

and by applying homology long exact secquences

H p minus1q +1(F p minus1C lowast)

i H p +q (F p C lowast)

j H p +q

983080F p C lowast

983087F p minus1C lowast

983081

δ H p minus1+q (F p C lowast)

sup3 To avoid confusion because of all the indices and variables we assume a = 0 ndash but j could also have bidegree(minusa a ) a 0 would not be more difficult but we would have to take care of it



where i and j are the maps induces by i and π p above and δ is the map delivered by the

snake lemma

Now we can roll up those long exact sequences into an exact triangle

oplusH p +q (F p C lowast) oplusH p +q (F p C lowast)

oplusH p +q (F p C lowastF p minus1C lowast)

i

j

k

which gives us a spectral sequence E r pq

From now on we concentrate on spectral sequences arising from an exact couple which

complies the following two conditions we require D pq = 0 if p lt 0 and for q lt 0 we want

i D p minus1q +1 D pq to be an isomorphism

Remark ose two facts guarantee that the spectral sequence lives in the first quadrant

Proof First let p lt 0 Since k E pq D p minus1q = 0 has im(k ) = 0 we get d = jk = 0

Furthermore 0 = D pq j

E pq says that

E pq = ker(d )983087

im(d ) = ker(d ) = im( j ) = 0

Now we look at the case q lt 0 We have d E pq D p minus1q

j

D p minus1q and we know

that i D p minus1q sim D p q minus1 is an isomorphism But since im(k ) = ker(i ) = 0 this leads to

d = jk = 0 Moreover we have D p minus1q +1

i

sim D pq j

E pq and since ker( j ) = im(i ) =D p minus1q +1 we get

E pq = ker(d ) = ker(k ) = im( j ) = 0

Let Z r pq = ker(d r pq ) and B r pq = im(d r p +r q minusr +1) be the cycles and boundaries given by the

fact that the r -sheet of the spectral sequence contains chain complexes

Furthermore we set H n = limminusminusrarr

D p nminus p (the injective limit along the maps i D pq

D p +1q minus1 ie the disjoint union of the D p nminus p where two elements are said to be equal if

they are equal under compositions of i ) and F p H n = im(D pq H p +q ) by what we get

a filtration

F p minus1H n sube F p H n sube

since i (D p minus1nminus p minus1) sube D p nminus p is filtration is even canonically bounded D p nminus p = 0for p lt 0 causes F p H n = 0 If p gt n (which is equal to q = n minus p lt 0) we know that

D p n minus p D p +1nminus( p +1) and therefore H p +q = im(D p nminus p H p +q ) = F p H n

In particular the filtration satisfiescup

p F p H n = H n = limlarrminusminus

(H n F p H n) andcap

p F p H n = 0

So if a spectral sequence weakly converges to H lowast it even converges to H lowast

Proposition ere is a natural inclusion of F p H n F p minus1H n in E infin p nminus p e spectral se-

quence E r pq weakly converges to H lowast if and only if

Z infin =991785r

k minus1(i r D ) equals k minus1(0) = j (D )



Proof First we define K pq to be the kernel of D pq H p +q Beeing an element of K pq is

equivalent to lying in the kernel of i r for an integer r so K pq =cup

r ker(i r ) It follows that

j (K pq ) =cup

r j (ker(i r )) =cup

r B r pq = B infin pq as one can easily see that the infinite boundaries

are the union of the B r pq (and similarly Z infin pq =cap

r Z r pq ) We look at the following diagram

0

0 K p minus1q +1 D p minusq q +1 F p minus1H n

0

0 K pq D pq F p H n 0

coker(i |K p minus1q +1)

coker(i ) F p H nF p minus1H n

i |K p minus1q +1 i

with short exact sequences as rows

Before applying the snake lemma we regard the cokernels

D pq 983087i (D p minus1q +1) = D pq

983087im(i ) = D pq

983087ker( j )

tells us that coker(i ) = j (D pq ) Furthermore we have im(i |K p minus1q +1) = ker( j |K pq ) () and

therefore coker(i |K p minus1q +1) j (K pq ) = B infin pq e snake lemma now says that

0 B infin pq j (D pq ) F p H n

983087F p minus1H n

0

is a short exact sequence

For all r isin Z we have

j (D pq ) = im( j ) = ker(k ) = k minus1(0) sube k minus1(i r D p minusr minus1q +r ) = Z r pq

what means j (D pq ) subecap

r Z r pq = Z infin pq We get a natural inclusion

F p H n983087F p minusq H n

j (D pq )1048623B infin pq sube Z infin pq

1048623B infin pq = E infin pq

and in particular we know now that the spectral sequences weakly converges to H lowast ie

F p H n F p minus1H n = E infin pq if and only if j (D ) = Z infin

Corollary (Convergence eorem) A spectral sequence with the requirements mentioned

above converges to H lowast = limminusminusrarr D E r pq rArr H p +q

If a spectral sequence arises from an exact couple of a complex C lowast with canonically bounded

filtration it converges to limminusminusrarr

D = H lowast which is the homology of the filtered complex C lowast

Proof A remark above tells us that the spectral sequence converges to H lowast if it weakly con-

verges to H lowast All we have to show is that Z infin = k minus1(0) if the spectral sequence complies

the two facts that D pq = 0 for p lt 0 and i D p minus1q +1

sim D pq for q lt 0



Let p and q be arbitrary integers ere exists an integer r such that p minus r minus 1 lt 0 which

implies i r (D p minusr minus1q +r ) = i r (0) = 0 We now see that Z r pq = k minus1(i r (D p minusr minus1q +r ) =

k minus1(0) en Z infin pq =cap

r prime Z r prime

pq = k minus1(0) and therefore Z infin = k minus1(0)

at means E r pq rArr H p +q

Let F p minus1C lowast sube F p C lowast sube sube C lowast be a canonically bounded filtration of a complex C lowastSince F p C n = 0 for all p lt 0 we get D pq = H p +q (F p C lowast) = 0 for those p For q lt 0 we have

p = n minusq gt n and F p C n = C n is leads to D pq = H n(F p C lowast) = H n(C lowast) = H n(F p +1C lowast) =D p +1q minus1 = D p +1q minus2 Alltogether it follows that a spectral sequence arising from a

canonically bounded complex converges to H lowast

It remains to show hat H lowast is the homology of C lowast

Let p lt n we then know that i is an isomorphism and therefore we get

H p +q im(D p nminus p H p +q ) D p nminus p = H p +q (F p C lowast) = H p +q (C lowast)

3 Double ComplexesIn Example we explained that a filtered complex yields an exact couple which in turn

yields a spectral sequence We will now go one step further and introduce double complexes

which yield filtered complexes and finally spectral sequences

Definition (Double complex) A double complex C lowastlowast is a family (C pq ) p q isinZ of objects

of an abelian category together with maps

d h pq C pq C p minus1q d v pq C pq C p q minus1

such that for each q resp p the families (C lowastq d hlowastq ) resp (C p lowast d v p lowast) are chain complexes

and additionally in every square the relation

d v p minus1q d h pq = plusmn d h p q minus1 d v pq

is satisfied In case of a ldquo+rdquo in this relation we call the double complex commutative and

in case of a ldquominusrdquo we call it anticommutative We say that a double complex lives in the first

quadrant if C pq = 0 whenever p lt 0 or q lt 0

We can picture a double complex like this (here we have a first quadrant double complex)

C 02

C 12

C 22

C 01 C 11 C 21

C 00 C 10 C 20



Definition Let C lowastlowast be an anticommutative double complex en we define the total

complex aached to C lowastlowast as the chain complex Totlowast(C lowastlowast) with entries⁴

Totn(C lowastlowast) =991893

p +q =n

C pq

and differentials d = d v + d h

A straightforward calculation using anticommutativity shows d 2 = 0 so that the total

complex is indeed a chain complex

We can turn the total complex of a given double complex into a filtered chain complex by

considering the following two obvious filtrations

ldquoFiltration by rowsrdquo

0 0 0

lowast lowast lowast


ldquoFiltration by columnsrdquo

lowast lowast 0lowast lowast 0lowast lowast 0

More precisely we set

F (R) p Totn(C lowastlowast) =

991893i le p

C i nminusi F (C)q Totn(C lowastlowast) =

991893 j leq

C nminus j j

In both cases we get

983080F p Totlowast(C lowastlowast)

983087F p minus1 Totlowast(C lowastlowast)

983081n

= C pq

for the n-th entry of the filtration quotients (where p + q = n) Note that when the double

complex lives in the first quadrant then both filtrations are canonically bounded ereforewe will assume from now on that every double complex lives in the first quadrant

e machinery developped in the previous sections now yields two spectral sequences

which we call (R)E r lowastlowast and (C)E r lowastlowast respectively By Corollary both spectral sequences

converge to the homology of the total complex Totlowast(C lowastlowast) Oen one is not really inter-

ested in this homology but one can play these sequences off against one another to obtain

information about the entries of the spectral sequence

Can we write down more explicitly how these spectral sequences look like We have to

apply the technique of exact couples to our situation Doing so we obtain

Proposition e E 1 page of (C)E r lowastlowast is obtained from the initial double complex C lowastlowast by

taking homology in vertical direction and the differentials d 1 on that page are the maps in-

duced by the horizontal differentials d h of C lowastlowast on this homology Dually the E 1 page of (R)E r lowastlowastis obtained by ldquotransposingrdquo this procedure

⁴ It is also common to consider a slightly differently defined total complex in which the direct sum is replacedby a direct product As we will mostly consider double complexes living in the first quadrant this does notmake a difference for us



Idea of proof To prove this we would have to go into the details of how the long homology

sequence coming from the short exact sequence of chain complexes

0 F p minus1 Totlowast(C lowastlowast) F p Totlowast(C lowastlowast) F p Totlowast(C lowastlowast)1048623F p minus1 Totlowast(C lowastlowast) 0

is constructed which in the end boils down to investigating the definition of the connectinghomomorphism coming from the snake lemma As this is a rather tedious calculation we

will omit this here

is proposition suggests to view the initial double complex C lowastlowast as the E 0 page of either

spectral sequence (possibly ldquotransposedrdquo)

Before giving some examples of applications of this construction we record the following

useful result

Lemma If we have a converging spectral sequence

E r pq rArr H p +q

and the E 2 page consists of 0 everywhere except in the boom row then we have

E 2 p 0 = H p

Proof is follows directly from Remark

Finally we note that we can also apply this whole technique to a commutative double com-

plex We can turn any such double complex into an anticommutative one by inserting some

minus1rsquos for example at every second horizontal or vertical arrow in this way in every square

exactly one map is multiplied by minus1 is does not affect the remaining properties of being

a double complex and does also not change homology

4 Examples

We conclude this note with a few examples that demonstrate the power of our techniques

e first examples are very simple whereas the later ones are a bit more interesting

41 The five lemma

As a first example we want to prove the five lemma Given have a commutative diagram

A B C D E

Aprime B prime C prime D prime E prime

f

д

f prime

д prime

α β γ δ ε

with exact rows we want to prove

(a) When β δ are monomorphisms and α is an epimorphism then γ is a monomorphism



(b) When β δ are epimorphisms and ε is a monomorphism then γ is an epimorphism

To get a double complex we flip this diagram insert kernels and cokernels on the le and

right and fill everything else with 0 to obtain

0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0

coker дprime E prime D prime C prime B prime Aprime ker f prime 0

In the following we will omit the 0rsquos again and draw only the non-zero part

From this we see immediately that the E 1 page of (R)E r lowastlowast consists only of 0 because to get

this E 1 page we have to take homology in horizontal direction but as the rows are exactthis vanishes So we conclude that (R)E r lowastlowast converges to 0 and our theory then tells us that

also (C)E r lowastlowast must converge to 0 What does this imply

Taking homology in vertical direction we obtain the E 1 page of (C)E r lowastlowast

lowast ker ε ker δ kerγ ker β ker α lowast

lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast

Here we have wrien ldquolowastrdquo to indicate any entry we are not interested in

We now use the assumption from (a) to get ker δ = ker β = cokerα = 0 By taking

homology again we finally look at the E 2 page

lowast

lowast

lowast

kerγ

lowast

lowast

lowast

lowast lowast lowast lowast lowast 0 lowast

As the spectral sequence converges to 0 we know that on the E infin page no entry can be

le But this means that kerγ has to vanish since otherwise it could never disappear on

the following pages is is just what we wanted to show in (a) e claim in (b) is shown

similarly



42 The snake lemma

e next application of our theory will be a proof of the snake lemma which proceeds quite

similar to the one of the five lemma We start again with a commutative diagram

A

B

C

0

0 Aprime B prime C prime

f

д

f prime

д prime

α β γ

with exact rows and as before we insert kernels cokernels and zeros to obtain the E 0 page

of our two spectral sequences

0 0 0 0 0 0 0

0 0 C B A ker f 0

0

coker дprime C prime B prime Aprime

0

0

As the rows are exact horizontal homology vanishes and again both spectral sequences

converge to 0 Now taking homology in vertical direction yields the E 1 page of (C)E r lowastlowast

0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ

Here we see part of the snake lemma sequence in the two rows and standard arguments

show that these are exact Finally we take homology again to look at the E 2 page

0 cokerφ 0 0 0

0

0

0

kerψ

0

As on the E infin page everything must have vanished this one remaining map must be an

isomorphism By inverting this isomorphism we can get a connecting homomorphism andobtain the snake lemma sequence

0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim

By simple arguments one can prove that this is indeed exact



43 Balancing Tor

e next application we want to consider is balancing Tor So let R be a ring A a right

R -module and B a le R -module Denoting by Torlowast(A minus) the derived functors of A otimesR minus

and by Torlowast(minus B ) the derived functors of minus otimesR B we want to show

Torlowast(A minus)(B ) = Torlowast(minus B )(A)

using spectral sequences

To get our machinery started we need a double complex We choose projective resolutions

P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0

of A and B respectively and define the double complex C lowastlowast by

C pq = P p otimes Q q

where the maps are the induced ones coming from the maps in the projective resolutions

Since projective modules are flat the rows and columns of this double complex are indeed

complexes and the squares in the double complex are commutative

Writing down the E 0 page of the column-filtered spectral sequence we have

P 0 otimes Q 2 P 1 otimes Q 2 P 2 otimes Q 2

P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1


Now since every P p is projective (hence flat) the sequence

P p otimes Q 1 P p otimes Q 0 P p otimes B 0

is exact so we have

coker(P p otimes Q 1 P p otimes Q 0) = P p otimes B

erefore the E 1 page looks like this

0

0

0

P 0 otimes B P 1 otimes B P 2 otimes B



What we have here in the boom row is exactly the complex used to calculate the derived

functors of minus otimes B So taking homology of this yields as E 2 page of (C)E r lowastlowast

0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)

In a totally analogous way one can see that the E 2 page of (R)E r lowastlowast is

0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )

Now in both spectral sequences we have an E 2 page whose only nonzero entries are in the

boom row In this situation we can apply Lemma which tells us that

(C)E r pq rArr Tor p +q (minus B )(A) and (R)E r pq rArr Tor p +q (A minus)(B )

But as both spectral sequences converge to the same thing we are done now

44 Base change for Tor

We now consider the following situation let R S be rings f R S a ring homomor-

phism A an R -module and B an S -module By f we can also consider B as an R -module

As in the previous example we consider again projective resolutions

P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0

of A as an R -module and B as an S -module respectively and form the double complex C lowastlowastgiven by

C pq = P p otimesR Q q

Here Q q might not be projective as an R -module But P p is so P p otimesR minus is exact and when

we consider the spectral sequence (C)E r lowastlowast filtered by columns we see by the same argument

as an the previous example that it converges to TorR p +q (A B ) So we know that also (C)E r lowastlowastconverges to the same thing but how does this spectral sequence look like explicitly

e E 0 page looks like this

P 2 otimesR Q 0 P 2 otimesR Q 1 P 2 otimesR Q 2





We now use

P p otimesR Q q = (P p otimesR S ) otimesS Q q

As (P p otimesR S ) p is just the complex used to calculate the derived functors of minus otimesR S and every

Q q is projective hence minus otimesS Q q is exact we obtain the following E 1 page

TorR 2 (A S ) otimesS Q 0



TorR 1 (A S ) otimesS Q 0 TorR 1 (A S ) otimesS Q 1 TorR 1 (A S ) otimesS Q 2




Now the complexes in these rows are the ones used to calculate the derived functors of

minus otimesS B and this means that the pq -th entry on the E 2 page is TorS p (TorR q (A S ) B ) So we

have proved

Proposition ere exists a spectral sequence

E 2 pq = TorS p (TorR q (A S ) B ) rArr TorR p +q (A B )

Why is this useful Let us consider the special case that S is flat as an R -module en

TorR i (A S ) vanishes fori gt 0 hence on the E 2 page of the spectral sequence we constructed

only the boom row contains nonzero entries erefore we can again use Lemma

which gives us

TorS i (A otimesR S B ) = TorR i (A B )

e requirement that S is a flat R -module is specifically fulfilled when G is a group H is a

subgroup and we take R = Z[H ] and S = Z[G ] In this case S is even R -free one can use

any system of coset representatives as a basis In terms of group homology (seing B = Z)

the statement then reads

H i (G indG H A) = H i (H A)

is is just Shapirorsquos lemma

45 The universal coefficients theorem

As a last application we want to prove a universal coefficients theorem Let (C lowast d lowast) =(C q d q )q be a chain complex consisting of free abelian groups and A be any abelian group

In this situation what can we say about the relation between H lowast(C lowast) and H lowast(C lowast otimes A)

We choose a projective resolution

P 2 P 1 P 0 A 0

of A and consider the double complex (P p otimes C q ) pq



We first look at the spectral sequence coming from the filtration by columns Its E 0 page is

P 0 otimes C 2 P 1 otimes C 2 P 2 otimes C 2



Since P p otimes minus is exact we get as E 1 page

P 0 otimes H 2(C ) P 1 otimes H 2(C ) P 2 otimes H 2(C )



Now the rows here are the complexes used to calculate the derived functors of A otimes minus so

the pq -th entry on the E 2 page is TorZq (A H p (C )) We want to examine this more closely

Applying A otimes minus to the short exact sequence of chain complexes

0 imd lowast kerd lowast H lowast(C lowast) 0

and deriving yields a long exact sequence Tor2(A imd nminus1) Tor2(A kerd n) Tor2(A H n(C ))

part Tor1(A imd nminus1) Tor1(A kerd n) Tor1(A H n(C ))

part Tor0(A imd nminus1)

for every n isin N0 But as im d nminus1 and kerd n are subgroups of the free abelian group C n

they are both free themselves and therefore the higher Tor groups vanish By the long

exact sequence this means that also Tori (A H n(C )) vanishes for i ge 2 Hence our E 2 page

looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0



We now have a look at the other spectral sequence coming from the filtration by rows

From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2



we as pq -th entry of the E 1 page Torq (AC p ) For q gt 0 this vanishes because every C p is

free so the E 1 page looks like this

0 0 0

A otimes C 0 A otimes C 1 A otimes C 2

Hence on the E 2 page we have again that everything except the boom row is 0 and in the

boom row we have

H 0(A otimes C lowast) H 1(A otimes C lowast) H 2(A otimes C lowast)

So by Lemma this (and hence both) spectral sequences converge to H lowast(A otimes C lowast) We

now apply Remark to the first spectral sequence For every n isin N0 we have (C)E 20n =

Tor0(A H n(C lowast)) = H n(C lowast) otimes A and (C)E 21nminus1 = Tor1(H nminus1(C lowast) A) So what we get is

the following universal coefficients theorem that describes the relation between H lowast(C lowast) and

H lowast(C lowast otimes A) in terms of the laer Tor group

Proposition ere is a short exact sequence

0 H n(C lowast) otimes A H n(C lowast otimes A) Tor1(H nminus1(C ) A) 0

References

[Wei] CA Weibel An Introduction to Homological Algebra Cambridge Studies in Ad-

vanced Mathematics Cambridge Cambridge University Press



1 Spectral Sequences and Convergence

Let C be an abelian category

Definition (Spectral sequences) A (homologically graded) spectral sequence is a family

of objects E r pq

of C for all p q isin Z and r ge a (with a fixed a isin Z) together with

differentials d r pq E r pq

E r p minusr q +r minus1 which satisfy d r d r = 0 Furthermore we require

that there are isomorphisms

E r +1 pq H (E

r pq ) = ker(d r pq )

1048623im(d r p +r q minusr +1)

By n = p + q we denote the total degree of E r pq

e collections (E r pq ) p q isinZ for fixed r are called the sheets or pages of the spectral sequence

By the isomorphisms required above we imagine that we get from one sheet to the next

one (ldquoturing a page aroundrdquo) by taking homology

To get an imagination of how the differentials look like we first regard the sheet of thespectral sequence where r = 0

E 003 E 013 E 023 E 033

E 002 E 012 E 022 E 032

E 001 E 011 E 021 E 031

E 000 E 010 E 020 E 030

On the first sheet the differentials go from the right to the le as is shown here

E 103 E 113 E 123 E 133

E 1

02 E 1

12 E 1

22 E 1

32

E 101 E 111 E 121 E 131

E 100 E 110 E 120 E 130




E 2 p minus2q +1

E 203 E 213 E 223 E 233

E 202

E 212

E 222

E 232

E 201 E 211 E 221 E 231

E 200 E 210 E 220 E 230




codomain 0 We get

E r +1 pq = H (E r pq ) = ker(d )

983087im(d ) = E r pq

9830870 = E r pq








β pq E infin pq



H n =991786

F p H n and991785

F p H n = 0


983080H nF p H n

983081sup2





0





0

0

E 203

E 213

0

0

0

0

E 2

02

E 2

12

0

0

0 0 E 201 E 211 0 0

0

0 E 200 E 210

0

0












For p = 1 we get



0 F 0H n H n H n983087F 0H n

0



0



H p +1 E 2 p +10

d E 2 p minus11

H p E 2 p 0

d E 2 p minus21

H p minus1




0

0

0

0

E 2minus11

E 2

01

E 2

11

E 2

21

E 2minus10 E 200 E 210 E 220

0

0

0

0




1048623im(d 2 p +2q minus1) = 0






r ge n = p + q

Let q = 0 We get



infin p 0 = E 3 p 0 = ker(d 2 p 0)

1048623im(d 2 p +2minus1) = ker(d 2 p 0)

And q = 1 leads to


infin p 1 = E 3 p 1

= H (E 2 p 1) = ker (d 2 p 1)1048623

im(d 2 p +20) = E 2 p 1

1048623im(d 2 p +20)




0

0

E 2 p +10 kerd 2 p +10

H p F p minus1H p

H p +1 E 2 p +10 E 2 p minus11 H p


0 0 0 0





way



D

D

E

i

j k



i (D ) i (D )

H (E )

i |i (D )

j (2)k (2)







i r (D ) i r (D )

E r +1

i |i r (D )

j (r +1)k (r +1)











D pq and E =oplus







r








i

F p C lowast

π p


0




j H p +q

983080F p C lowast


983081






snake lemma




i

j

k








E pq says that

E pq = ker(d )983087

im(d ) = ker(d ) = im( j ) = 0


j




i

sim D pq j


E pq = ker(d ) = ker(k ) = im( j ) = 0







a filtration







p F p H n = 0




Z infin =991785r







j (K pq ) =cup





0


0




i |K p minus1q +1 i




983087im(i ) = D pq

983087ker( j )




983087F p minus1H n

0



















sim D pq for q lt 0






r prime Z r prime






















C 02

C 12

C 22

C 01 C 11 C 21

C 00 C 10 C 20






p +q =n

C pq







0 0 0







991893i le p


991893 j leq

C nminus j j




983081n

= C pq




















will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References






E 2 p minus2q +1

E 203 E 213 E 223 E 233

E 202

E 212

E 222

E 232

E 201 E 211 E 221 E 231

E 200 E 210 E 220 E 230




codomain 0 We get

E r +1 pq = H (E r pq ) = ker(d )

983087im(d ) = E r pq

9830870 = E r pq








β pq E infin pq



H n =991786

F p H n and991785

F p H n = 0


983080H nF p H n

983081sup2





0





0

0

E 203

E 213

0

0

0

0

E 2

02

E 2

12

0

0

0 0 E 201 E 211 0 0

0

0 E 200 E 210

0

0












For p = 1 we get



0 F 0H n H n H n983087F 0H n

0



0



H p +1 E 2 p +10

d E 2 p minus11

H p E 2 p 0

d E 2 p minus21

H p minus1




0

0

0

0

E 2minus11

E 2

01

E 2

11

E 2

21

E 2minus10 E 200 E 210 E 220

0

0

0

0




1048623im(d 2 p +2q minus1) = 0






r ge n = p + q

Let q = 0 We get



infin p 0 = E 3 p 0 = ker(d 2 p 0)

1048623im(d 2 p +2minus1) = ker(d 2 p 0)

And q = 1 leads to


infin p 1 = E 3 p 1

= H (E 2 p 1) = ker (d 2 p 1)1048623

im(d 2 p +20) = E 2 p 1

1048623im(d 2 p +20)




0

0

E 2 p +10 kerd 2 p +10

H p F p minus1H p

H p +1 E 2 p +10 E 2 p minus11 H p


0 0 0 0





way



D

D

E

i

j k



i (D ) i (D )

H (E )

i |i (D )

j (2)k (2)







i r (D ) i r (D )

E r +1

i |i r (D )

j (r +1)k (r +1)











D pq and E =oplus







r








i

F p C lowast

π p


0




j H p +q

983080F p C lowast


983081






snake lemma




i

j

k








E pq says that

E pq = ker(d )983087

im(d ) = ker(d ) = im( j ) = 0


j




i

sim D pq j


E pq = ker(d ) = ker(k ) = im( j ) = 0







a filtration







p F p H n = 0




Z infin =991785r







j (K pq ) =cup





0


0




i |K p minus1q +1 i




983087im(i ) = D pq

983087ker( j )




983087F p minus1H n

0



















sim D pq for q lt 0






r prime Z r prime






















C 02

C 12

C 22

C 01 C 11 C 21

C 00 C 10 C 20






p +q =n

C pq







0 0 0







991893i le p


991893 j leq

C nminus j j




983081n

= C pq




















will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References






0

0

E 203

E 213

0

0

0

0

E 2

02

E 2

12

0

0

0 0 E 201 E 211 0 0

0

0 E 200 E 210

0

0












For p = 1 we get



0 F 0H n H n H n983087F 0H n

0



0



H p +1 E 2 p +10

d E 2 p minus11

H p E 2 p 0

d E 2 p minus21

H p minus1




0

0

0

0

E 2minus11

E 2

01

E 2

11

E 2

21

E 2minus10 E 200 E 210 E 220

0

0

0

0




1048623im(d 2 p +2q minus1) = 0






r ge n = p + q

Let q = 0 We get



infin p 0 = E 3 p 0 = ker(d 2 p 0)

1048623im(d 2 p +2minus1) = ker(d 2 p 0)

And q = 1 leads to


infin p 1 = E 3 p 1

= H (E 2 p 1) = ker (d 2 p 1)1048623

im(d 2 p +20) = E 2 p 1

1048623im(d 2 p +20)




0

0

E 2 p +10 kerd 2 p +10

H p F p minus1H p

H p +1 E 2 p +10 E 2 p minus11 H p


0 0 0 0





way



D

D

E

i

j k



i (D ) i (D )

H (E )

i |i (D )

j (2)k (2)







i r (D ) i r (D )

E r +1

i |i r (D )

j (r +1)k (r +1)











D pq and E =oplus







r








i

F p C lowast

π p


0




j H p +q

983080F p C lowast


983081






snake lemma




i

j

k








E pq says that

E pq = ker(d )983087

im(d ) = ker(d ) = im( j ) = 0


j




i

sim D pq j


E pq = ker(d ) = ker(k ) = im( j ) = 0







a filtration







p F p H n = 0




Z infin =991785r







j (K pq ) =cup





0


0




i |K p minus1q +1 i




983087im(i ) = D pq

983087ker( j )




983087F p minus1H n

0



















sim D pq for q lt 0






r prime Z r prime






















C 02

C 12

C 22

C 01 C 11 C 21

C 00 C 10 C 20






p +q =n

C pq







0 0 0







991893i le p


991893 j leq

C nminus j j




983081n

= C pq




















will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References






0

0

0

0

E 2minus11

E 2

01

E 2

11

E 2

21

E 2minus10 E 200 E 210 E 220

0

0

0

0




1048623im(d 2 p +2q minus1) = 0






r ge n = p + q

Let q = 0 We get



infin p 0 = E 3 p 0 = ker(d 2 p 0)

1048623im(d 2 p +2minus1) = ker(d 2 p 0)

And q = 1 leads to


infin p 1 = E 3 p 1

= H (E 2 p 1) = ker (d 2 p 1)1048623

im(d 2 p +20) = E 2 p 1

1048623im(d 2 p +20)




0

0

E 2 p +10 kerd 2 p +10

H p F p minus1H p

H p +1 E 2 p +10 E 2 p minus11 H p


0 0 0 0





way



D

D

E

i

j k



i (D ) i (D )

H (E )

i |i (D )

j (2)k (2)







i r (D ) i r (D )

E r +1

i |i r (D )

j (r +1)k (r +1)











D pq and E =oplus







r








i

F p C lowast

π p


0




j H p +q

983080F p C lowast


983081






snake lemma




i

j

k








E pq says that

E pq = ker(d )983087

im(d ) = ker(d ) = im( j ) = 0


j




i

sim D pq j


E pq = ker(d ) = ker(k ) = im( j ) = 0







a filtration







p F p H n = 0




Z infin =991785r







j (K pq ) =cup





0


0




i |K p minus1q +1 i




983087im(i ) = D pq

983087ker( j )




983087F p minus1H n

0



















sim D pq for q lt 0






r prime Z r prime






















C 02

C 12

C 22

C 01 C 11 C 21

C 00 C 10 C 20






p +q =n

C pq







0 0 0







991893i le p


991893 j leq

C nminus j j




983081n

= C pq




















will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References






0

0

E 2 p +10 kerd 2 p +10

H p F p minus1H p

H p +1 E 2 p +10 E 2 p minus11 H p


0 0 0 0





way



D

D

E

i

j k



i (D ) i (D )

H (E )

i |i (D )

j (2)k (2)







i r (D ) i r (D )

E r +1

i |i r (D )

j (r +1)k (r +1)











D pq and E =oplus







r








i

F p C lowast

π p


0




j H p +q

983080F p C lowast


983081






snake lemma




i

j

k








E pq says that

E pq = ker(d )983087

im(d ) = ker(d ) = im( j ) = 0


j




i

sim D pq j


E pq = ker(d ) = ker(k ) = im( j ) = 0







a filtration







p F p H n = 0




Z infin =991785r







j (K pq ) =cup





0


0




i |K p minus1q +1 i




983087im(i ) = D pq

983087ker( j )




983087F p minus1H n

0



















sim D pq for q lt 0






r prime Z r prime






















C 02

C 12

C 22

C 01 C 11 C 21

C 00 C 10 C 20






p +q =n

C pq







0 0 0







991893i le p


991893 j leq

C nminus j j




983081n

= C pq




















will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References






i r (D ) i r (D )

E r +1

i |i r (D )

j (r +1)k (r +1)











D pq and E =oplus







r








i

F p C lowast

π p


0




j H p +q

983080F p C lowast


983081






snake lemma




i

j

k








E pq says that

E pq = ker(d )983087

im(d ) = ker(d ) = im( j ) = 0


j




i

sim D pq j


E pq = ker(d ) = ker(k ) = im( j ) = 0







a filtration







p F p H n = 0




Z infin =991785r







j (K pq ) =cup





0


0




i |K p minus1q +1 i




983087im(i ) = D pq

983087ker( j )




983087F p minus1H n

0



















sim D pq for q lt 0






r prime Z r prime






















C 02

C 12

C 22

C 01 C 11 C 21

C 00 C 10 C 20






p +q =n

C pq







0 0 0







991893i le p


991893 j leq

C nminus j j




983081n

= C pq




















will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References






snake lemma




i

j

k








E pq says that

E pq = ker(d )983087

im(d ) = ker(d ) = im( j ) = 0


j




i

sim D pq j


E pq = ker(d ) = ker(k ) = im( j ) = 0







a filtration







p F p H n = 0




Z infin =991785r







j (K pq ) =cup





0


0




i |K p minus1q +1 i




983087im(i ) = D pq

983087ker( j )




983087F p minus1H n

0



















sim D pq for q lt 0






r prime Z r prime






















C 02

C 12

C 22

C 01 C 11 C 21

C 00 C 10 C 20






p +q =n

C pq







0 0 0







991893i le p


991893 j leq

C nminus j j




983081n

= C pq




















will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References








j (K pq ) =cup





0


0




i |K p minus1q +1 i




983087im(i ) = D pq

983087ker( j )




983087F p minus1H n

0



















sim D pq for q lt 0






r prime Z r prime






















C 02

C 12

C 22

C 01 C 11 C 21

C 00 C 10 C 20






p +q =n

C pq







0 0 0







991893i le p


991893 j leq

C nminus j j




983081n

= C pq




















will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References








r prime Z r prime






















C 02

C 12

C 22

C 01 C 11 C 21

C 00 C 10 C 20






p +q =n

C pq







0 0 0







991893i le p


991893 j leq

C nminus j j




983081n

= C pq




















will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References








p +q =n

C pq







0 0 0







991893i le p


991893 j leq

C nminus j j




983081n

= C pq




















will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References









will omit this here




useful result


E r pq rArr H p +q


E 2 p 0 = H p







4 Examples



41 The five lemma


A B C D E


f

д

f prime

д prime

α β γ δ ε








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References








0 0 0 0 0 0 0 0

coker д E D C B A

ker f

0








lowast

coker ε

coker δ

cokerγ

coker β

coker α

lowast




lowast

lowast

lowast

kerγ

lowast

lowast

lowast





similarly



42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References





42 The snake lemma



A

B

C

0


f

д

f prime

д prime

α β γ



0 0 0 0 0 0 0

0 0 C B A ker f 0

0


0

0



0

kerγ

ker β

ker α

ker f

0

0

cokerдprime

cokerγ

coker β

cokerα

0

φ

ψ



0 cokerφ 0 0 0

0

0

0

kerψ

0



0

ker f

kerα

ker β

kerγ

cokerα

coker β

cokerγ

coker дprime 0

cokerφ kerψ

φ

ψ

sim




43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References





43 Balancing Tor







P 2 P 1 P 0 A 0

Q 2 Q 1 Q 0 B 0








P 0 otimes Q 1

P 1 otimes Q 1

P 2 otimes Q 1




is exact so we have



0

0

0






0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References







0 0 0 0 0

Tor0(minus B )(A)

Tor1(minus B )(A)

Tor2(minus B )(A)

Tor3(minus B )(A)

Tor4(minus B )(A)


0 0 0 0 0

Tor0(A minus)(B )

Tor1(A minus)(B )

Tor2(A minus)(B )

Tor3(A minus)(B )

Tor4(A minus)(B )









P 2 P 1 P 0 A

0

Q 2

Q 1

Q 0

B 0












We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References





We now use













have proved






which gives us












P 2 P 1 P 0 A 0






















looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References























looks like this

Tor0(A H 2(C )) Tor1(A H 2(C )) 0 0

Tor0(A H 1(C ))

Tor1(A H 1(C ))

0

0

Tor0(A H 0(C )) Tor1(A H 0(C )) 0 0




From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References






From the E 0 page

P 2 otimes C 0

P 2 otimes C 1

P 2 otimes C 2





0 0 0



boom row we have









References



Fuetterer, Schwab - Spectral Sequences

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