# Seminar on Homological Algebra

This seminar was held in #mathematics on EFnet on ** 16th of May 2004 ** at ** 19:00 EDT ** by ** nerdy2 and dublisk**.

It covered group cohomology, Galois cohomology, Brauer groups, abelian categories, delta and derived functors, derived categories.

## Seminar log

`
[13:05] (Tohoku) Allright, so today's topic is homological algebra and the speakers are dublisk and nerdy2
[13:06] (Tohoku) dublisk will go first and nerdy2 will follow
[13:06] (dublisk) Hi everybody, thanks Tohoku
[13:06] (dublisk) Ok, today I'll be talking about group cohomology
[13:07] (dublisk) I won't do things in the most generality, because
nerdy2 will cover some of that. The background material I will assume is
familiarity with notions of free-modules, group actions, etc. not too
much
[13:07] (dublisk) So I'll start by a construction that is very similar to homology of simplicial complexes
[13:08] (dublisk) Let G be a group, and let G^{q+1} denote the cartesian product of G with itself q+1 times
[13:08] (dublisk) Let C_q be the free Z-module with basis elements of
G^{q+1}. Then C_q is a free ZG-module where ZG is the group ring, with
basis orbits of the action
[13:09] (dublisk) Since (g_0,g_1,...,g_q) = g_0^-1 (1,g_0 g_1^-1, ... ,
g_0 g_q^-1), C_q has basis {(1,x_1,...,x_q) : x_i \in G} as a
ZG-module
[13:11] (dublisk) We can write (1,x_1, ... , x_q) = (1,g_1,g_1*g_2, ....
,g_1*g2*...*g_q) by setting x_1 = g_1 g_2 = x_1^-1*x_2 etc. Denote this
element by [g_1 | g_2 | ... | g_q]
[13:11] (dublisk) I will be constructing what is known as the Eilenberg-MacLane Bar resolution
[13:12] (dublisk) I define a chain map d : C_q -> c_{q-1} by
d(g_0, ..., g_q) = sum_i (-1)^i (g_0, ..., g_{i-1}, g_{i+1} , ... g_q)
[13:13] (dublisk) note that in algebraic topology, if I was defining the
boundary operator on the simplicial complex with vertices g_0 , ... ,
g_q it would have the same formula, this gives some motivation for the
formula
[13:13] (dublisk) the key point here is that d_q o d_{q-1} = 0 map
[13:13] (dublisk) d_q being C_q -> C_{q-1}
[13:14] (dublisk) thus, we have a bunch of maps -> C_q -> C_{q-1} -> C_{q-2} -> etc.
[13:14] (dublisk) such that whenever you move two arrows, you get 0
[13:14] (dedekind) d_{q-1} o d_q = 0?
[13:15] (dublisk) erm, yeah
[13:15] (dublisk) whenever you have such a sequence of maps, for example of R-modules, it is called a _chain complex_
[13:16] (dublisk) we can define the k^th homology of the complex by taking ker d_q / im d_{q+1}
[13:16] (dublisk) we are actually working towards cohomology though, so we aren't done yet
[13:18] (dublisk) In terms of the bar notation d_q [ g_1 | ... | g_n] =
g_1[g_2 | ... |g_q] - [g_1*g_2 | g_3 | ... | g_q] + ... + (-1)^q [g_1 |
g_2 | ... | g_{q-1}]
[13:19] <Syzygy-> !
[13:19] (dublisk) so we have a chain complex -> C_2 -> C_1 ->
C_0 -> Z -> 0 where the last map is induced by sending [] -> 1
[13:19] (dublisk) Syzygy ?
[13:19] <Syzygy-> Maybe I'm just tired, but what would the third
term in bar notation be? I think I'll see it clearly if I see that.
[13:20] (dublisk) it ends jup just having a shifting
[13:20] (dublisk) so for example for q = 2
[13:20] <Syzygy-> You stated the two first and the last term in a
development for any q. I don't see straight off what the third term
would be.... It's just that q.
[13:21] (dublisk) d [ g_1 | g_2] = (g1, g1*g2) - (1 , g1*g_2) + (1,g_1)
in our initial Z-module format formula, which we re-write as g_1(1,g_2) -
(1,g_1*g_2) + (1,g_1) , and in bar notation g_1[g_2] - [g_1 * g_2] +
[g_1]
[13:22] (dublisk) erm the first equality came from [g_1,g_2] = (1,g_1,g1*g_2)
[13:22] (dublisk) sorry if this is confusing
[13:22] <Syzygy-> NP. Go on.
[13:22] (dublisk) [g_1 | g_2 ] = (1,g_1,g_1*g_2) rather
[13:23] <Syzygy-> Ah. Now it's obvious. Thank you.
[13:23] (dublisk) ok, now there are some important properties of the chain complex we constructed
[13:23] (dublisk) firstly, each C_q is a _free_ ZG-module
[13:24] (dublisk) secondly, it is actually exact, meaning the kernel of one boundary map is the image of the previous
[13:24] (dublisk) I haven't shown the second property yet, but these are the two important qualities
[13:24] (Tohoku) !
[13:24] (dublisk) Tohoku?
[13:24] (Tohoku) a boundary map is d_q?
[13:24] (dublisk) yes
[13:24] (Tohoku) thanks
[13:25] (dublisk) In fact, as nerdy2 will probably go into, I could
replace my construction by any such construction with the above two
properties, and furthermore we could replace free with projective
[13:25] (dublisk) To see that it is exact, we can construct what is known as a chain homotopy
[13:26] (dublisk) given two chain complexes {C_q} and {B_q}, a chain map
f is a map from each C_q -> D_q commuting with all boundary maps
[13:28] (dublisk) a chain homotopy between two chain maps f and g is a
sequence of maps h_i from C_i to B_{i+1} such that d_{i+1} o h_i +
h_{i-1} o d_i = f - g here I am using d_i for the boundary maps of both
complexes just for notation
[13:29] <Syzygy-> !
[13:29] <Syzygy-> NM.
[13:29] (dublisk) note that a chain map induces a map on homology because of how it commutes
[13:29] (dublisk) any two chain maps that are homotopic will induce the same map on homology
[13:30] dioid (prj@flash.visit.se) joined #mathematics.
[13:30] (dublisk) So in our case to show our sequence is exact, we can
apply the trick of showing that the identity map mapping our chain
complex to itself is homotopic to the 0-map
[13:31] (dublisk) Note that having 0-homology means precisely that the sequence is exact
[13:32] (dublisk) You can check that the map h_i : C_q ->
C_{q+1} mapping (g_1, ... , g_i) -> (1,g_1, ... ,g_i) gives such a
chain homotopy
[13:32] (dublisk) Ok, so now we are almost done in making group cohomology, the only thing we have to do is get coefficients
[13:33] (dublisk) we do this by applying the functor Hom_ZG( _ , M) where M is a ZG-module to our complex
[13:33] (dublisk) then after taking the cohomology of this complex, it
is denoted H^q(G,M) and is called group cohomology. Here cohomology just
refers to the fact that our arrows got reversed when applying the Hom
functor, it is still simply the kernel modulo the image
[13:35] (dublisk) Now, Since C_q is a free-module with basis [g_1 | ...
|g_q], and we are taking a map of ZG-modules from C_q to M, we could
look at Hom_(sets) (G^q , M)
[13:35] esojo (esojo@200.79.144.162) left #mathematics.
[13:36] (dublisk) sometimes you'll just see things as such maps, the
reason why I didn't do that, is that I developed first a free-ZG complex
ending in Z
[13:36] (dublisk) the latter can be really generalized in terms of derived functors
[13:37] (dublisk) In the case that K/k is a (finite) galois extension,
H^q(Gal(K/k),K^*) is called the galois cohomology, where K* is the group
of units
[13:40] (dublisk) So, after identifying Hom_ZG(C_q,K^*) with
Hom_set(G^q,K^*), and using multiplicative notation as we do with K^*, a
cocycle is a map f : G -> K^* such that f(gh) = g*f(h)*f(g),
and for f to be a coboundary, there should be an element a \in K such
that f(g) = (g(a))*a^-1
[13:40] (dublisk) here cocycle means it goes to zero under a boundary map
[13:40] (dublisk) and coboundary means it is the image of a boundary map
[13:40] (dublisk) so sometimes you see cohomology described as cocycles module coboundaries
[13:41] (dublisk) A theorem (Hilbert's Theorem 90) states that the first galois group as described is always trivial
[13:42] (dublisk) I thing the additive group of K can also be added and it is still refered to as galois cohomology too
[13:42] (dublisk) There's also an additive version of Hilberts theorem, which says that this latter cohomology is also trivial
[13:44] (dublisk) To prove the Hilbert's theorem 90, it's not too hard,
given a cycle f : G -> K^*, there is by linear independence of
characters, an element a \in K such that b = sum_g f(g)g(a) is not zero.
Applying some group element h to both sides and using the cocycle
condition, we can see that h(b) = f(h)^-1*b for all h, so that f is a
coboundary, I'll let you fill in the details
[13:45] (dublisk) I guess I'll concretely work out H^1 some more to clarify things
[13:45] (dublisk) well firstly H^0(G,M) = {m \in M : for alll g, gm = m}
[13:46] (dublisk) and c is a cycle iff for all g_1, g_2, c([g_1g2]) = g_1*c([g_2] + c([g_1])
[13:47] (dublisk) and b is a boundary if b(g) = gm - m. If we view M
also as a trivial right ZG-module (it is a left ZG-module to start),
then H^1 corresponds to derivations G->M modulo inner derivations,
[13:48] (dublisk) The acse of H^2 is much more interesting
[13:49] (dublisk) It turns out it helps classify a certain extension
problem. Given say group, A, G with A a normal subgroup of G, we want to
see what E we can fit into a short exact sequence 1 -> A -> E
-> G -> 1, i.e. that E/A =~ G
[13:50] (dublisk) this is a bit too hard, but the H^2 questions solves a bit of an easier case
[13:50] (dublisk) Tohoku wants me to mention that you can do
H^q(Gal(K/k),E) where E is an elliptic curve, ask him for more
info :)
[13:51] <Syzygy-> !
[13:51] (dublisk) Syzygy?
[13:51] <Syzygy-> Why would there be an s.e.s. begining with a normal subgroup of the group at the end? Seems weird...
[13:52] (Poinky) !
[13:52] (dublisk) Poinky?
[13:52] (Poinky) you can do H^q(Gal(K/k), M) for any Gal(K/k)-module M (for example E(K), K-points on an elliptic curve over k)
[13:52] (dublisk) yeah that doesn't make sense either, typo in my notes
[13:53] (dublisk) yeah, H^q(G,M) here M was a ZG-module, this was needed for applying HomZG( _ , M) to make sense
[13:55] (Tohoku) actually, what I was trying to say is that "galois
cohomology" is a fairly generic name which applies to any case Poinky
pointed out. H^q(Gal(K/k),K^*) is the Galois cohomology of K^*. that's
all :)
[13:55] (dublisk) so given an exact sequence 1 -> A -> E -> G
-> 1, lets just think of A as a normal subgroup of E, then E acts on A
via conjugation, and from E -> Aut(A) factoring through E/A since A
acts trivially on itself by conjugation we get a map E/A = G ->
Aut(A).
[13:55] (dublisk) Ok, thanks Tohoku
[13:55] (dublisk) the action is g*a = h*a*h^-1 where h \in E maps to G
[13:56] (dublisk) i.e our action depends on a section s : G -> E
, so if we called our map E -> G by pi, pi o s = id, i.e. s gives
choices of coset reps, also known as a transversal
[13:57] (dublisk) I should point out that two extensions 1 -> A ->
E -> G -> 1 and 1 -> A -> E' -> G -> 1 are said to be
equivalent if there is a map E -> E' making the diagram commute with
identity maps on A,H. SUch a map is an isomorphism by the five-lemma
[13:59] (dublisk) Not for a fixed transveral s, we get a bijection E
with A x G sending (a,g) to as(g). WE multiply two elemts via
(a_1,g_1)*(a_2,g_2) = a_1 g_2*a_2 s(g_1)*s(g_2) under this bijection,
and g_2*a_2 is the action of G on A
[13:59] (dublisk) If is is a homomorphism, then the multplication shows that E is the semi-direct product of A and G
[14:00] (dublisk) So let c : G x G -> A be defined as c(g_1,g2) =
s(g_1)s(g_2)s(g_1*g_2)^-1, measuring the failure for s not to be a
homomorphism.
[14:02] (dublisk) We multiply (a_1,g_1)*(a_2,g_2) to (a_1 g_1*_a_2 *
c(g_1,g_2), g_1*g_2), and we ask what such functions c : GxG-> A
make this multipliactino on A x G into a group
[14:02] (dublisk) if it were a group, we could put this as E and get an extension
[14:03] (dublisk) it turns out that the condition for c to give associativity is exactly the condiciton that c be a 2-cocycle
[14:04] (dublisk) moreover, if you change transversals, say s' is
another transversal, define b : G -> A by b(g) = s'(g)s(g)^-1,
measuring the difference on our choice of transversal. A calculation
shows that if c' is the 2-cocycle associated to s', then c' = d(b)*c,
i.e that the cycles differ by abounday, hence they are equal in
cohomology.
[14:05] (dublisk) Thus we have a partial solution: Given a ZG-module A,
we have a bijection between extensions 1 -> A -> E -> G -> 1
realizing the a
[14:05] (dublisk) the G-action on A and element of H^2(G,A), taking an
extension to [c], c the cocycle above associated to some transversal
[14:06] (dublisk) So this is what H^2 does in general, but there is also
the question of what H^2 is in the case of galois cohomology
H^2(Gal(K/k),K^*)
[14:06] (dublisk) It turns out this is isomorphic to the Brauer group Br(K/k), I'll quickly sketch some ideas about this
[14:07] (dublisk) A central k-algebra is a k-algebra such that Z(A) = k.
It is called simple if A is simple as a ring (no proper-nonzero two
sided ideals)
[14:08] (dublisk) If A is finite dimensional, A =~ M_n(D) for some central division k-algebra by the Artin-Wedderburn Theorem.
[14:09] (dublisk) We call two central-simple k-algebras A,A' similar, if
A =~ M_n(D) A' =~ M_s(D), i.e they are matrix rings over the same
divison algebra
[14:10] (dublisk) The Brauer group Br(k) consists of the similarity
classes of (finite dimensional ) central simple k-algebras, where you
multiply via tensor product
[14:10] (dublisk) it takes some technical lemmas concerning centralizers to see that it is well defined and everything works
[14:10] (dublisk) Given an extension K/k, we get a homomorphism Br(k)
-> Br(K) via [A] -> [K (x) A], and Br(K/k) is the kernel of this
map
[14:11] (Tohoku) !
[14:11] (dublisk) I should also say that Br(k) hence classifies central
division k-algebras because of how a central simple k-algebra similarity
class is determined by the division algebra it is a matrix algebra over
[14:11] (dublisk) Tohoku?
[14:11] (Tohoku) could you give an example of what Br(k) is, for some k?
[14:11] (dublisk) Yeah
[14:12] (dublisk) So Br(R) = Z/2Z
[14:12] (dublisk) Since the quaternions and the reals are give central
division R-algebras, this shows that they are they _only_ ones. This is
called the frobenius theorem
[14:13] (dublisk) This can be proved with cohomology, since actually
Br(C/R) = Br(R), and so we only need to compute H^2(Gal(C/R),C^*), given
that we knew about this bijection bewteen the Brauer group and H^2
[14:13] (dublisk) So this is a pretty neat application
[14:14] (dublisk) I'm basically done now I guess
[14:14] (Tohoku) coolness
[14:14] (Poinky) ! any other examples?
[14:14] (dublisk) It would take me too afar to prove the brauer group
[14:14] (dublisk) But I should say that the bar-resolution is a bit too specific
[14:15] (dublisk) Br(k) = 0 fir a finite field
[14:15] (dublisk) If you wanted to compute that Br(C/R) group, you actually might not want to use the bar resolution
[14:15] (Tohoku) 1
[14:15] (Tohoku) er, !
[14:15] (dublisk) This is because I only gave one specific resolution
[14:15] (dublisk) yes Tohoku?
[14:15] (Tohoku) is Br(k) always abelian?
[14:16] (dublisk) I want to say yes :)
[14:16] (dublisk) I'm a bit rusty on this, and by no means an expert
[14:17] (Poinky) in the way you defined it, it is quite clear, since D (x) D' is isomorphic to D' (x) D.
[14:17] (dublisk) yeah
[14:17] (dublisk) right
[14:17] (dublisk) Anyway, you could replace my bar-resolution with ANY projective resolution C_q -
[14:18] (dublisk) C_q -> C_{q-1} -> ... -> Z
[14:18] (dublisk) for example, if you know the group G is cyclic, there
is a very simple reslution that makes things easier to calculate
[14:19] (dublisk) in general you can define an Ext functor, which is a
derived functor of Hom, and then group cohomology is Ext^q_ZG(Z,M).
Nerdy2 will probably talk about this
[14:19] (Poinky) !
[14:19] (dublisk) I hope some people got a feel for cohomology though
[14:19] (dublisk) yes Ponky?
[14:20] (dublisk) Poinky
[14:20] (Poinky) i just want to mention that things like Br Q can be
quite horrible things. However, for a local field Q_p, it is equal to
Q/Z, and Br Q fits into 0 -> Br Q -> sum(p 'prime') Br Q_p ->
Q/Z -> 0
[14:21] (Tohoku) !
[14:21] (dublisk) Tohoku
[14:21] (Poinky) (so it is a horrible, torsion abelian group)
[14:21] (Tohoku) does this work if we replace Z by any dedekind domain and Q by its field of fractions?
[14:21] (dublisk) Go ahead an discuss things freely now, I'm done
[14:21] Action: Syzygy- applauds!
[14:21] (Tohoku) cool, nice talk
[14:21] Action: Tohoku claps
There was a break.
[15:35] (Tohoku) so, we reconvene. nerdy2 will continue about homological algebra
[15:35] (nerdy2) btw, i got a shiny new bike ;)
[15:35] (Tohoku) please start :)
[15:35] (nerdy2) ok, so we want to generalize homology/cohomology
[15:36] (nerdy2) these are really just nice functors from a sufficiently nice category :)
[15:36] (nerdy2) by which i mean, we have a category (G-modules) where
'short exact sequences' makes sense, and these functors take them to
long exact sequences
[15:37] (nerdy2) so from here on out, we'll always want to work with
categories that are like abelian groups or R-modules or G-modules...
[15:37] (nerdy2) is jerold banned for a reason, he's asking me to unban him
[15:37] (Tohoku) let's discuss that in #math
[15:38] (nerdy2) so we define an abelian category
[15:38] (nerdy2) and i'll surely get it wrong... :)
[15:39] (nerdy2) but an abelian category is a category with a zero
object (initial and terminal), where the Hom-sets are abelian groups,
where there are finite products and coproducts (and these agree)
[15:39] (nerdy2) composition is bilinear
[15:39] (nerdy2) and kernels, cokernels, images, coimages all exist
[15:39] (nerdy2) and the ker coker f = f, and coim im f = f
[15:40] (Tohoku) !
[15:40] (nerdy2) yes
[15:40] (Tohoku) im(f) = ker(coker(f))?
[15:41] (nerdy2) ok, sorry, how about im = ker coker, ker = coim im
[15:42] (nerdy2) are those reasonable?
[15:42] (Tohoku) sorry, I dunno what coim is
[15:42] <Kit`> !
[15:42] (nerdy2) heh yea?
[15:43] (nerdy2) coim is the dual notion to im
[15:43] (nerdy2) coim:im::coker:ker
[15:43] Landgraf (Dimfrost@a204.lambo.student.liu.se) joined #mathematics.
[15:43] <Syzygy-> !
[15:43] (nerdy2) yes?
[15:44] <Syzygy-> How about exemplifying with familiar notions?
[15:44] <Kit`> If it's not too much of a digression, what exactly
is kernel in an arbitrary category? Something to do with a monomorphism
that when composed with the map gives zero, but I'm not clear on quite
how it works.
[15:44] <Syzygy-> im and ker are natural enough - but, for
instance with R-modules for a ring - what happens with coker and coim?
[15:45] <Kit`> (If it is too much of a digression I'll dig up an explanation from wiki. :)
[15:45] (Tohoku) fair enough. so coim in, say, R-modules, is M -> im(f) ?
[15:45] (nerdy2) kernel is the pullback of a -> b, 0 -> b
[15:46] (Tohoku) oh, coim = cok(ker(f)), gotcha. sorry for the stupidity :)
[15:46] (nerdy2) the right condition should be ker coker f = f for mono's f, and coker ker f = f for epi's f
[15:47] (nerdy2) which should be equivalent to something i wrote involving ker/cok/im/coim
[15:47] (nerdy2) and also every map can be factored into an epi followed by a mono
[15:47] <Kit`> I'm not entirely clear on what a pullback is, but I
think I've hijacked this enough. I'll just assume they mean what they
sound like for now. :)
[15:48] (nerdy2) anyways, the reason the def'n is less important than most is that they are what you think they are:
[15:48] (nerdy2) Thm: abelian categories are full subcategories of (R-modules) for rings R
[15:48] (nerdy2) (R may be noncommutative, but is associative with unit)
[15:50] (nerdy2) so we will be primarily concerned with functors F : A -> B between two abelian categories A and B
[15:51] (nerdy2) for example: (G-modules) -> (Ab.) (M -> M^G = { m : gm = m for all g })
[15:51] (nerdy2) now, since we are in an abelian category we shouldn't consider all functors....
[15:51] (nerdy2) we should consider ones which respect the structure somehow
[15:51] (nerdy2) so we always want additive functors, i.e. ones for which F induces /homomorphisms/ Hom(a,b) -> Hom(Fa,Fb)
[15:52] (nerdy2) but the invariant functor and ones like it are nicer
[15:52] (nerdy2) they are _almost_ exact
[15:52] (nerdy2) that is, if we have a short exact sequence 0 -> A
-> B -> C -> 0 in (G-mod), we get an exact seq 0 -> A^G
-> B^G -> C^G in (Ab)
[15:53] (nerdy2) that is, it is _left exact_
[15:53] (nerdy2) likewise, if SES's go to right sided seq's, we say the functor is right exact
[15:53] (nerdy2) (if the functor is contravariant, we still have these
names: if SES's go to left/right sided exact seq's, we say it is
left/right exact)
[15:54] (nerdy2) you will recall that A^G = H^0(G,A)
[15:54] (nerdy2) exact functors are nice, it would be nice if given such
an almost exact functor we could get an exact one out of it
[15:55] (nerdy2) we almost can
[15:55] (nerdy2) consider a sequence {T^i} of functors A -> B
[15:56] (nerdy2) we call them a delta-functor if for each short exact
seq 0 -> a -> b -> c -> 0, we have boundary maps d (or
delta) : T^i(c) -> T^{i+1}(a)
[15:56] (nerdy2) such that:
[15:56] (nerdy2) for every short exact seq 0 -> a -> b -> c
-> 0, the long exact seq 0 -> T^0(a) -> T^0(b) -> T^0(c)
-> T^1(a) -> .... is exact
[15:58] (nerdy2) and the delta's are natural, so if we have a map of
short exact seq's, from 0->a->b->c->0 into
0->a'->b'->c'->0, the square consisting of the maps: T^i(c)
-> T^{i+1}(a), T^{i+1}(a) -> T^{i+1}(a'), T^i(c) -> T^i(c'),
T^i(c') -> T^{i+1}(c') commutes
[15:58] (nerdy2) we call T = (T^i) universal if for any other T' =
(T'^i), and any map f^0 : T^0 -> T'^0, there is a _unique_ seq
of maps f^i : T^i -> T'^i, starting with f^0 and commuting with
the d's (boundary maps)
[15:59] (nerdy2) by this property a universal functor is uniquely determined by T^0
[15:59] (nerdy2) universal delta-functor rather
[15:59] (nerdy2) so we can think of it is the rightful extension of the functor T^0
[15:59] (nerdy2) so we start with a functor F and ask if we can complete it to a universal delta-functor {T^i} with T^0 = F
[16:01] (nerdy2) Call a functor T^i effaceable if for each a in A, there is a mono u : a -> m, such that T^i(u) = 0
[16:02] (nerdy2) Thm: If {T^i} is a delta functor, and T^i is effaceable for i > 0, then {T^i} is universal
[16:02] (nerdy2) so we can check pretty easily if something is universal this way
[16:02] (nerdy2) however it doesn't allow us to construct it
[16:02] (nerdy2) there's sort of an ideal way to find a mono into something though
[16:03] (nerdy2) call an object i of A injective if hom(-,i) is exact
[16:03] (nerdy2) sorry, hom(i,-)
[16:03] (nerdy2) sorry, hom(-,i) ;)
[16:03] (nerdy2) this time i'm done :)
[16:03] (nerdy2) ok
[16:04] (nerdy2) so we say A 'has enough injectives' if every object has a mono into an injective object
[16:04] (nerdy2) let's leave this for now, and assume that A has enough injectives
[16:04] (nerdy2) then for any a in A, we have a resolution 0 -> a -> i^0 -> i^1 -> ... built up inductively
[16:04] (nerdy2) where i^* are injective
[16:05] (nerdy2) and we can apply F to this, and we can get the complex F(i^*)
[16:05] (nerdy2) if we take the homology of this complex, we set H^i(F(i^*)) = R^iF(a)
[16:05] (nerdy2) and call it the i-th right derived functor of F (on a)
[16:06] (nerdy2) but what happens if i take another resolution?
[16:06] (nerdy2) i never guaranteed uniquenes
[16:06] (nerdy2) it turns out that any two are chain homotopic (and so
the resulting R^iF(a) are naturally isomorphic) - this is an easy
exercise
[16:07] (nerdy2) also, from the injectivity of the objects in the
resolutions, we get that R^iF is actually a functor (we can define it on
maps)
[16:07] (nerdy2) moreover, from construction we see that R^iF(j) = 0 if j is injective
[16:08] (nerdy2) and since a short exact sequence of complexes gives a
long exact sequence in homology, we have that R^i F is a delta functor
[16:08] (nerdy2) but it's a delta functor and vanishes on injectives, and everything has a mono into an injective
[16:08] (nerdy2) therefore it's effaceable, therefore(!) it's universal
[16:08] (nerdy2) so these are 'the' right guys
[16:09] (nerdy2) (if there are not enough injectives, existence and other things become harder....)
[16:09] (dublisk) !
[16:09] (nerdy2) this brings up the q, when are there enough injectives ?
[16:09] (nerdy2) yes?
[16:09] (dublisk) you are using the snake lemma to get this last long exact sequence?
[16:09] <Syzygy-> !
[16:10] (nerdy2) it's more complicated, but similar to the snake lemma
[16:10] (nerdy2) you wind around at each point
[16:10] (nerdy2) the boundary maps come from a snake lemma argument
[16:10] (nerdy2) the other maps are already there
[16:10] (nerdy2) syz, yes?
[16:10] <Syzygy-> Errrr. And why was it that R^iF(j)=0
[16:10] (nerdy2) because 0 -> j -> j -> 0 is an injective resolution of j
[16:11] (nerdy2) so R^iF(j) = 0 for i > 0
[16:11] <Syzygy-> Because any of them are chain homotopic, and
thus the homologies that make out R^iF(j) can be found from this
particular resolution?
[16:12] (nerdy2) yes
[16:12] <Syzygy-> Thanks.
[16:12] (nerdy2) make sure you ask plenty of q's
[16:12] (nerdy2) i'm going fast, and we haven't reached shore (concrete stuff) yet :)
[16:14] (nerdy2) so when does a category have enough injectives?
[16:15] (nerdy2) well there's an easy answer that is usually applicable:
[16:16] (nerdy2) a grothendieck abelian category A is one that (a)
satisfies (Ab5): that is it is cocomplete (has all colimits), and
filtered colimits are exact and (b) has a generator z (i.e. every object
of A has an epi from some coproduct of copies of z)
[16:16] (nerdy2) Thm: grothendieck abelian cat's have enough injectives
[16:17] (nerdy2) Cor: (R-modules) have enough injectives (the generator is R)
[16:17] (dublisk) !
[16:17] (nerdy2) yes?
[16:17] (dublisk) what does it mean for filtered colimits to be exact
[16:18] (nerdy2) it means that when you take a filtered system of SES's, and you take their colimit, you get a SES
[16:19] (nerdy2) landen, keeping up?
[16:19] (landen) sorry not at all
[16:20] (nerdy2) (filtered colimits are a little bit different than direct limits, but i think of them as about the same ...)
[16:21] (nerdy2) in (R-modules), filtered colimits are exact
[16:22] (nerdy2) (try this as an exercise - if you don't know 'filtered colimit' replace with 'direct limit')
[16:23] (nerdy2) anyways, so when we have a left exact covariant functor
F, on an abelian cat. with enough injectives, we can define their right
derived functors R^iF
[16:23] (nerdy2) note that, we can define the dual to injective, that is, projective objects
[16:24] (nerdy2) and we can take the left-derived functors L^iF of a right-exact covariant functor using projective res's
[16:25] (nerdy2) (likewise, contravariant left-exact, projective res
=> right derived, contravariant right-exact, injective => left
derived)
[16:25] (nerdy2) note that free implies projective
[16:27] (dublisk) !
[16:27] (nerdy2) so that H_q(G, M) = H_q(C_* tens._ZG M) is really the left-derived functor of H_0(G,M)
[16:27] (nerdy2) dub, yes?
[16:27] (dublisk) how do you define a free object in an abelian category
[16:28] (nerdy2) ok, i'm really thinking of it as a category of R-modules say or something like that
[16:28] (dublisk) ok
[16:28] (nerdy2) i really need a forgetful functor to sets
[16:28] (nerdy2) and then its left adjoint (the free object functor)
[16:28] (Tohoku) in a grothendieck category, it could be defined as a co-product of copies of z, right?
[16:28] (dublisk) right
[16:29] (nerdy2) tohoku, not quite, Z + (Z/2) is a generator in (Ab)
[16:29] (Tohoku) how is that a coproduct of copies of z = Z ?
[16:29] (Tohoku) oh, nm,
[16:29] (nerdy2) i'm saying take that for z :0
[16:29] (nerdy2) :)
[16:30] (Tohoku) yeah, sorry. please continue
[16:30] (nerdy2) anyways, by 'free' i mean assume you have an exact
functor U : (your cat) -> (Sets), and you are taking its left
adjoint
[16:30] (nerdy2) and then it's clear that hom(Fx,-) = hom(x,U-) is exact
[16:31] (nerdy2) (you need to define exact here :))
[16:31] (nerdy2) anyways, it's clear what it means in (R-modules) or such
[16:32] (nerdy2) the point was just to get that in (ZG-modules) =
(G-modules) or (R-modules) we can compute left derived functors of
covariant right exact functors by free res's
[16:33] (nerdy2) the attentive reader has many details to fill in ;)
[16:33] (nerdy2) so anyways, H_q(G,M), is the left-derived functor of H_0(G,M) = M/IM where I = <s - 1> subset ZG
[16:34] (nerdy2) and H^q(G,M) is the right derived functor of H^0(G,M) = M^G
[16:34] (nerdy2) but these functors are M (x)_ZG Z, and Hom_ZG(Z, M) respectively, so we'll see them soon in another light
[16:35] (nerdy2) in general, Hom is left exact (in either variable), and tensor is right exact
[16:35] (nerdy2) so you can take the derived functors and get Ext^q(M,N) and Tor_q(M,N)
[16:35] (nerdy2) so H^q(G,M) = Ext^q_ZG(Z,M), and H_q(G,M) = Tor^ZG_q(Z,M)
[16:36] (dublisk) !
[16:36] (nerdy2) yes?
[16:36] (dublisk) So it seems like Ext works in any abelian category since you have Hom, but can you get Tor?
[16:38] (nerdy2) i'm not sure
[16:38] (dublisk) Ok no problem
[16:39] (nerdy2) anyways, a real problem is that most ab. cat's don't have enough projectives
[16:39] (nerdy2) (oh, i didn't mention - that it doesn't matter what variable you derive Hom/tens in)
[16:39] (nerdy2) (you get the same no matter what, and what's more Tor^R_q(M,N) ~= Tor^R_q(N,M))
[16:40] (nerdy2) so i guess let's talk about the derived category a bit
[16:41] (nerdy2) (i'm in an ultra-abstract mode today :))
[16:41] (nerdy2) so for an abelian category we can form the category C(A) of chain complexes in A
[16:41] (nerdy2) (with differential increasing degree say)
[16:42] (nerdy2) and then we can consider instead of all maps,
equivalence classes of maps between complexes, where two are equivalent
if they are chain homotopic
[16:42] (nerdy2) and form a new category this way, call it K(A)
[16:43] (nerdy2) but in some sense, we don't just want homotopy
equivalent maps identitified, we want all quasi-isomorphisms to be
isomorphisms
[16:43] (nerdy2) a quasi-isomorphism is a homology isomorphism
[16:43] (nerdy2) so an injective resolution 0 -> a -> i^0 -> ..., is just a quasi-isomorphism a -> i^*
[16:44] (nerdy2) which gives you the reasoning behind all this (in part)
- we want to replace something by a 'good' resolution before applying
functors
[16:44] (nerdy2) so what we do is form a new category D(A) by inverting all the quasi-isomorphisms
[16:45] (nerdy2) there are a few problems here - it is not clear that
D(A) is even a category in an ordinary sense (to 'invert' maps we need
to consider Hom's which are equivalence classes from a /class/, it is
not clear that they form a set)
[16:45] (nerdy2) i think it works when A is a grothendieck ab. cat.
[16:46] (nerdy2) second of all, it's not clear that we have good resolutions
[16:46] (nerdy2) even if we have enough injectives, this gives us
quasi-iso's from bounded below complexes to bounded below complexes of
injectives
[16:46] (nerdy2) but if we want to consider unbounded things we are in trouble
[16:47] (nerdy2) so we consider special complexes: i^* is K-injective if
for any exact (acyclic) complex a^*, hom^*(a^*,i^*) is acyclic
[16:47] (dublisk) !
[16:47] (nerdy2) yes?
[16:47] (dublisk) bounded means that eventually the arrows terminate to 0's ?
[16:48] (nerdy2) eventually the objects are 0
[16:49] (nerdy2) so ... -> c^0 -> c^1 -> c^2 -> ... is bounded below if there is an N, so that c^n = 0 for n < N
[16:50] (nerdy2) hom^*(a^*,i^*) is the complex of hom's, so that hom^n(a^*,i^*) is chain maps a^* -> i^* of degree n
[16:50] (nerdy2) exercise - what are the differentials
[16:50] (nerdy2) if you do it right, you'll find H^0 hom^*(a^*,i^*) = homotopy classes of maps a^* -> i^*
[16:51] (nerdy2) Thm: for a groth. ab. cat., there are enough K-injectives
[16:51] (nerdy2) so every complex is quasi-isomorphic to a K-inj complex
[16:51] (nerdy2) so if you have a left exact cov. functor F : A
-> B and A is groth. you can define RF : D(A) -> D(B)
[16:52] (nerdy2) by choosing for any a^* in D(A) a quasi-isomorphic K-injective and applying F to that
[16:53] (nerdy2) in this form, suppose that F : A -> B, G :
B -> C are left exact cov., and that A is groth., as is B, then
RF,RG exist, and R(G o F) = RG o RF
[16:53] (nerdy2) under one condition:
[16:53] (nerdy2) that F takes injectives in A to G-acyclic objects in B
[16:54] (nerdy2) (actually i should check this, it might not be true, it
is true if you restrict to D^+ - bounded below complexes)
[16:54] (nerdy2) i'm getting tired now :)
[16:54] (nerdy2) and this is prolly a good place to stop
[16:55] (nerdy2) even though i've covered nothing
[16:55] Action: dublisk applauds
[16:55] Action: Syzygy- claps and hopes for the continuation to be held on similarily friendly times.
[16:55] Action: Tohoku applauds
[16:56] (dublisk) Is this all in Weibel?
[16:56] (nerdy2) some is
[16:56] (Tohoku) okay, so do we have your permission(s) to upload the channel log?
[16:56] (nerdy2) sure
[16:56] (nerdy2) he does mention the derived category at the end
[16:56] (Tohoku) cool, end we stop relaying then
`