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DeRham cohomology

  1. Apr 15, 2004 #1
    I am pretty familiar with deRham cohomology. for me, deRham cohomology is synonymous with cohomology in general.

    but I often run into that cocycle condition, you know, where c(g,h)+c(gh,k)=c(g,hk)+c(h,k) you need to look at conditions of this type when deciding whether a projective representation of a group can straightened into a linear rep, or a rep of a central extension.

    I know this is a question of cohomology, and the language (cocycle) is very suggestive of stuff I know from deRham cohomology, but beyond that, I don't know much.

    so does someone want to give me a 50 cent sketch of the details? what is the "d" operator in this cohomology? does the group have to be a manifold? there is such a thing as cohomology of discrete groups, isn't there? and Lie algebras? is there a cup product? what is the name of the cohomology where this cocycle condition lives?
    Last edited: Apr 20, 2004
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  3. Apr 16, 2004 #2

    matt grime

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    De Rham cohomolgy is one of the more esoteric cohomologies and certainly isn't synonymous with cohomology in my mind, but then I'm biased. It uses real coeffs which make it easy to work out. The truly hard case stuff uses Z.

    In order to have a cohomology theory one needs a cochain complex, that is a set of objects X* indexed by Z, with maps indexed by Z, denoted d*, the differential, where d^n :X^n---->X^{n+1} (it increases the degree) and the composition of two consecutive differentials is zero

    The cohomology groups are then H* indexed by Z with H^n = kerd^n/Imd^{n-1}

    So what are these things for when the underlying object is a group? Well, there are several ways of doing this, one is to take the geometric version, which I don't understand. The other is to take the trivial module, and take a projective resolution of it, or simpler, a free resolution: give me some module, and I can find a free module surjecting onto it. I take the kernel of that map and I can get a surjection from a free module onto that and so on, so we get:

    T the trivial module
    Then if we apply Hom( T,?) to this complex we get another complex, the cohomology of which is callled the cohomology of G. Of course I'm missing some details - what kinds of modules and so on. Lots of people want ZG modules and triv is then Z, I would tend to use some algebrtaically closed field. Not of char zero because then the cohomology is trivial. Moreover, we don't have to limit to using triv, we can resolve any module Y and take Hom(X,?) with X any module too, and the results are teh Ext groups - the extensions of one by the other.

    The cohomological information nowadays is studied as the derived category, and the ext groups measure the failure of Hom(X,?) to be an exact functor - if X is projective the functor is exact and the Ext groups vanish.

    So that's my 50 cent on group cohomology. There is a lot more than that. Try cyclic homology, sheaf cohomology, simplicial, elliptic... they all have this complex and differential though.
  4. Apr 16, 2004 #3


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    Lethe, is the product in your cocycle condition the native cup product of cohomology or some other product? Are the cocycles from the same spaces or different ones (for example base, fiber and top space of a fibration)? There are a lot of techniques in cohomology but the underlying topology determines what you use.
  5. Apr 16, 2004 #4
    i think maybe in physics, de Rham is very common.

    right. i think this is a paraphrase of the Steenrod axioms, right?

    trivial module... is that a module made up of just a 0?
    i don't know what a resolution is. is what follows the definition?

    where does my group fit in all this? where does the cocycle condition come from in all this?
  6. Apr 16, 2004 #5
    its just addition of real numbers. (note: i had the formula wrong last night. i have updated it)

    they are just real numbers.
    the arguments of those functions are group elements, which underlying topology are you talking about? i think cohomology of groups is purely algabraic. what does topology have to do with it?

    the cocycle condition is often used in physics to decide when a group admits projective representations. this is important in quantum theory, and i would just like to know the math underlying this stuff
  7. Apr 16, 2004 #6

    matt grime

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    The trivial module for the group ring RG is the one where all elements of G ect by 1. Judging by your questions I think I would need to go and discuss a lot of basics which arent' really necessary, as we'll hopefully see.

    A resolution of a module is an acyclic chain complex of projective modules, with the exception of the degree -1 term where it is the module you start with. The group comes into it because these are all modules for the group. The free module is RG, and using direct sums of RG it is easy to write down a resolution (And any two will produce the same answer) but it grows exponentially in dimension.

    I think I needed Hom(?,X) for a contravariant functor to make a cochain complex as it happens but you didnt' pick up on that one.

    What I gave isn't a paraphrase of the steenrod rod axioms, it is the definition of the cohomology groups of a cochain complex (when it makes sense to talk of these things). It satisfies the steenrod axioms when we use the case of cochain complexes of group modules and maps in its homotopy category.

    If you want to interpet low degree group homology then you get the kind of results you were talking about.

    Calculating any of these things is very difficult, even if it can be done at all. Cyclic groups can have their cohomology calculated.

    There are ways of calculating group cohomolgy using topological spaces too.
  8. Apr 19, 2004 #7
    too many things i don't know. it sounds to me like i simply don't have enough training in algebra to learn chomology of groups. ho hum
  9. Apr 19, 2004 #8

    matt grime

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    I don't know about that. It's just that these are things no one knows enough about to be able to teach courses on them at a basic level. We just can't calculate these things. There is an alternative method using simplicial complexes (or do I mean singular) where the complexes/cells are treated as free modules for the group ring.

    Do you know what an exact sequence is? There is another interpretation of the n'th cohomology group ext(M,N) as the set of isomorphism classes of exact sequences with n+2 entries.
  10. Apr 19, 2004 #9


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    In simplicial cohomology the basic units are elementary simplices of a triangulation of the manifold. Singular cohomolgy is based on continuous maps of spheres into the manifold. Where all these diffenernt cohomologies can be defined together, they coincide. Evaluation of expressions is indeed based on exact sequences, and also on known cohomology operations like the Steenrod operations. Outside of the classic manifolds (spheres, tori) it's like pulling teeth and every little advance is greeted with joy. Nash and Sen's fairly straightforward book or Nakahara's harder and comprehensive one might help you. It would be equivalent to a semester course.

    Nash & Sen, Topology and Geometry for Physicists, Academic Press.

    Nakahara, Geometry, Topology, and Physics, Institute of Physics Publishing, Graduate Student Series in Physics.
  11. Apr 19, 2004 #10
    simplicial complexes can be used to make homology groups. i am familiar with this. i did not know that you could make cohomology groups this way.

    as i said before, the only cohomology groups that i know are the de Rham cohomology groups. these are the groups of closed forms mod exact forms on a smooth manifold.

    well, the only homology groups i know are the simplicial homology groups. the group of boundaryless chains mod boundary chains.

    i wanted to learn more about projective reps and group cohomology and where that cocycle condition fits in. I have been looking at a book on homological algebra by Rotman, but the book is probably meant for algebraists and for me, the learning curve for that book is very steep, and so i was hoping to get something more dumbed down on the forum. but your explanations are also hard for me.

    i know what an exact sequence is. i don't know what ext(M,N) means though. is it a notation for exact sequences?
  12. Apr 19, 2004 #11
    i am pretty sure you mean simplicial homology

    are you addressing this comment to me? this seems rather strange. how would those books help me learn group cohomology? I have read both of those books, and i can assure you, they don't breath a word on the subject. they both have a nonrigorous colloquial and incomplete introduction to de Rham cohomology, simplicial homology, and the homotopy groups Map(S^n,X).
  13. Apr 19, 2004 #12

    matt grime

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    homology an cohomology are essentially the same thing, the only difference is that in homology the d sends degree n to n-1 and in cohomology it sends degree n to n+1. If the complexes are unbounded in both directions this is the same, otherwise it is different. See Ext versus Tor.

    Rotman is fairly basic for a *topologist*. It doesn't treat group cohomology very well at all. But then no book does really, not unless you're at least a 3rd year postgraduate. It is not an easy topic, there is not an easy way to deal with it. Sorry. It takes at least two years to even be happy with what projective means in my opinion, and then to use it.

    I would suggest readin Alperin's Local Representation Theory as background (omitting the last two chapters) before reading Benson's books on group cohomology. For a general overview of homological algebra then try Wiebel's book on it.

    Remember that any chain complex is turned into a cochain complex by applying a contravariant functor.
    Last edited: Apr 19, 2004
  14. Apr 20, 2004 #13
    yeah, and they are dual to one another. or at least, simplicial homology (or is it singular? did we decide what the difference is between those two?) is dual to de Rham cohomology.

    well, i saw Rotman cited as a reference in the bibliography of a physics book for exactly this topic.

    what does projective mean, anyway?

    here is what i think about a projective representation: it is a representation of a central extension of a group.

    there is another usage of projective that i am familiar with, in the context of the projective spaces RP^n, CP^n and HP^n. these are, respectively, the space of rays in R^n, C^n, and H^n. more generally, you can consider the projective space of rays through any vector space.

    the two notions are related by the fact that a projective representation of a group on a vector space is the same as a representation on a projective vector space.

    also, with regards to projective linear groups, like PGL_n. i understand PGL_n as the space of lines through the origin in the vector space M_n (nxn matrices) intersect GL_n

    a colleague of mine in the math department said that a definition of the projective group associated to any group G is G/Z(G). i have never seen this definition in the books, but it agrees with my definition of PGL_n, and others.

    do you know this definition for a projective group?

    but i don't know what it means for a module to be projective.

    thanks for the references. i will take a look when i get a chance

    what is the point of this statement? to show me how a homology group is related to a cohomolgy group? so what?
  15. Apr 20, 2004 #14
    i have found a nice book: "Lie groups, Lie algebras, cohomolgoy and some applications in physics" by Azcárraga and Izquierdo

    it is a mathematical physics book, i think. that is, it is a book on mathematical topics but mostly motivated by physical applications. so, closer to my needs than a pure math book, probably.

    they define group cohomology groups in chapter 5

    given a group G, an abelian group A, and a realization [itex]\sigma[/itex] of G on the group of (outer) automorphisms of A (i.e. a homomorphism from G --> Aut(A))

    we have n-cochains [itex]C^n(G,A)[/itex] the set of mappings [itex]\alpha_n\colon G\times\dotsb\times G\rightarrow A[/itex]

    \alpha_n\colon (g_1,\dotsc,g_n)\mapsto \alpha_n(g_1,\dotsc,g_n)\inA

    and the coboundary operator [itex]\delta_n\colon C^n\rightarrow C^{n+1}[/itex]


    this operator satisfies the usual [itex]\delta_n\circ\delta_{n-1}=0[/itex] and so we can define the cohomology groups


    then the zeroth cohomology group is the subgroup of A of fixed invariant points under the action of G.

    the first cohomology group is the group of crossed homomorphisms from G to A mod by principle homomorphisms (what the **** is a crossed homomorphism?)

    the second cohomology group is the one we use in physics. it characterizes the way that we can centrally extend G by A. in other words, it tells us the possible projective representations of G

    the third cohomology group tells us the possible nonassociative representations.

    this seems to be the sort of stuff that i was looking for. this book is a little difficult, but more or less aimed at a guy at my level. i don't understand everything, but so far, i am learning something from this book.
    Last edited: Apr 20, 2004
  16. Apr 20, 2004 #15
    one interesting thing about these cohomology groups i notice is that they take their coefficients in A (not in the real line). matt grime said before that the fact that de Rham cohomology groups are over the reals makes them easier.

    now, how about the cup product? i didn't see any mention of this topic in the book. my understanding is that any cohomology group should have a cup product. is there one here?
    Last edited: Apr 20, 2004
  17. Apr 20, 2004 #16
    one of the upshots of the chapter on group cohomology is the result that nonrelativistic quantum mechanics of a neutral spin-0 particle has to use a complex Hilbert space, whereas relativistic quantum mechanics of a neutral spin-0 particle can use a real valued Hilbert space.
  18. Apr 20, 2004 #17

    matt grime

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    1) Projective means a summand of a free module, or that Hom(P,?) is an exact functor. Over a field of char zero everything thing is projective, so the theory there is trivial.

    2) cannot for the life of me remember why i made that parting comment, seems like one of my better non sequiturs.

    3) The resolution in the book is called the free resolution. It uses G's (well, kG's tensored really) and it it semi-useful because you can easily say what it is, but the resolution grows exponentially so it's impractical.

    4) There is a cup product. You can interpret the cohomology as 'n fold extensions of k by k' where k means the trivial module of the group and the n fold extenions means an exact sequence of modules starting and ending with k and with n terms in between. Thus H^1(G,k) is the set of short exact sequences starting and ending in k. It also has the other interpretations you mention you've found. If you use this idea you can splice together these exact sequences into longer ones. This is the Yoneda splice. You may also realize it as the total tensor product of the two exact sequences.
  19. Apr 20, 2004 #18
    hmm... that definition seems opaque to me. can you make contact between this usage of the word "projective" and any of the definitions i gave above?
  20. Apr 20, 2004 #19

    matt grime

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    A module P is projective if any short exact sequence 0-->X-->Y--->P--->0 is split exact. Ie there are no non-trivial extensions, or H^1 is zero.

    it is a quite opaque definition, but they are the cohomologically trivial objects, since the projective resolution is just non zero in one degree only. they do take a while to get used to. The main thing is that things which are true for free modules tend to be true fo projective modules as they are summands of free modules.

    They have important lifting properties that ensure eveything you want to be true about cohomology being independent of a choice of resolution is true: if P* is a complex of projectives with homology in degree zero only where it is X say, and Q* is the same, then P* and Q* are homotopic.
  21. Apr 20, 2004 #20
    I am still chewing on things. a few things, firstly, why doesn't the cohomology group that i showed above have a name? is its name simply the group cohomology group? that sounds stupid.

    in what sense does a cohomology group over a char zero become trivial? i mean, de Rham cohomology groups contain nontrivial topological information about the smooth manifolds they are defined on.
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