J.T. Lyczak, March 2017
Grothendieck spectral sequences
These are the notes for the seminar on obstructions to local-global principles at Leiden University in
2017. The website for this seminar can be found at http://pub.math.leidenuniv.nl/~vissehd/
BMseminar/. The reference for everything in these notes is the book by Weibel [Wei94], where
you can find many more details and examples.
Using the results of last week we will derive the Grothendieck spectral sequence, which is
actually the main example of a spectral sequence we are interested in. Here is the statement.
Theorem 1. Let A, B and C be abelian categories, such that A and B have enough injectives. Let
G : A → B and F : B → C be two left exact additive functors, such that G maps injective objects
in A to a F -acyclic object in B. Then for any object A ∈ A there exists a cohomological spectral
sequence starting with
E2pq = Rp F ◦ Rq G(A)
converging to
Rp+q (F ◦ G)(A).
First we will need the following lemma.
Lemma 2. Consider a cochain complex J • in an abelian category B of injective objects
J0 → J1 → J2 → J3 → . . .
such that all cocylces Z n , coboundaries B n and cohomology groups H n are injective as well. Let
F : B → C be a left exact functor. Then F commutes with taking cohomology, i.e.
F H n (J • ) ∼
= H n (F J • ).
Proof. Let us first fix some notation. We will write Z p := ker(J p → J p+1 ) and B p := Im(J p−1 →
J p ) for the cocycles and the coboundaries in J p . The corresponding cohomology groups will be
p
denoted by H p := Z p /B p . For these objects of the complex F J • we will write Z := ker(J p →
p
p
p
p
p+1
p−1
p
J
), B := Im(J
→ J ) and H := Z /B .
p
By definition we have exact sequences 0 → Z p → J p → J p+1 and 0 → Z → F J p → F J p+1 .
If we apply the functor F to the first sequence we get two exact sequences (since F is left-exact)
in C, which fit together as follows
0
F Zp
0
Z
p
F Jp
F J p+1
F Jp
F J p+1
p
By the universal property of kernels we see that F Z p = Z .
Since Z p is injective we know that the short exact sequence 0 → Z p−1 → J p−1 → B p → 0
splits and again we find two compatible exact sequences in C:
0
0
F Z p−1
Z
p−1
F J p−1
F Bp
F J p−1
B
p
So F Z p = Z .
1
p
0
0
We now apply the same trick to the following short exact sequence 0 → B p → Z p → H p → 0
and we find
0
F Bp
F Zp
F Hp
0
0
B
p
Z
p
H
p
0
So we get natural isomorphisms
p
H p (F J • ) = H = F H p = F H p (J • ).
Now we can construct the Grothendieck spectral sequence.
Proof of the theorem. Fix an injective resolution A → I • of A in A and a Cartan-Eilenberg resolution J •,• of the cochain complex G(I • ) in B:
J 02
J 12
J 22
J 01
J 11
J 21
J 00
J 10
J 20
G(I 0 )
G(I 1 )
G(I 2 )
0
0
0
We will use both spectral sequences of the double cochain complex F J •,• . Let T n be the limit of
these two spectral sequences.
First we will show that of the spectral sequences associated to this double cochain complex
has the correct second page. So, let us consider first consider the spectral sequence which flips the
double cochain complex:
II
E0 =
F J 20
F J 21
F J 22
F J 10
F J 11
F J 12
F J 00
F J 01
F J 02
which gives
II
E1 =
H 2 (F J •,0 )
H 2 (F J •,1 )
H 2 (F J •,2 )
H 1 (F J •,0 )
H 1 (F J •,1 )
H 1 (F J •,2 )
H 0 (F J •,0 )
H 0 (F J •,1 )
H 0 (F J •,2 )
=
2
F H 2 (J •,0 )
F H 2 (J •,1 )
F H 2 (J •,2 )
F H 1 (J •,0 )
F H 1 (J •,1 )
F H 1 (J •,2 )
F H 0 (J •,0 )
F H 0 (J •,1 )
F H 0 (J •,2 )
REFERENCES
REFERENCES
where we used the previous lemma on each object.
But H q (J •,p ) is an injective resolution of H q (GI • ), so
E2 = Rp F (H q (GI • )).
II
Also, I • is an injective resolution of A, so H q (GI • ) = Rq G(A) and we find a converging spectral
sequence where the objects on the second page are
Rp F ◦ Rq G(A).
Now we consider the other spectral sequence of the double cochain complex associated to
F J •,• . We have
I
E0 =
F J 02
F J 12
F J 22
F J 01
F J 11
F J 21
F J 00
F J 10
F J 20
Since J p,• is an injective resoltion of GI p , we see that the cohomology of these columns are the
higher derived functors of F applied to GI p :
I
E1 = Rq F (GI p ) =
R2 F (GI 0 )
R2 F (GI 1 )
R2 F (GI 2 ))
R1 F (GI 0 )
R1 F (GI 1 )
R1 F (GI 2 )
R0 F (GI 0 )
R0 F (GI 1 )
R0 F (GI 2 )
=
0
0
0
0
0
0
F G(I 0 )
F G(I 1 )
F G(I 2 )
since G maps injective objects to F -acyclic objects and R0 F = F .
Now we use that I • is an injective resolution of A, so by definition we find
E2p0 = Rp (F G)(A).
I
This means that T n = Rn (F G)(A).
References
[Wei94] Charles A. Weibel. An introduction to homological algebra, volume 38 of Cambridge
Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1994.
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