Cohomologies?

lethe

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?

matt grime

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.

lethe

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.

also
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.

matt grime

real coeffs is easier because there is no torsion to worry about. the universal coeffs theorem tells you if you work out Z ceoffs you can work out all, so that is the harder one to figure out.

Group cohomology over C, say, is trivial because there is *no* cohomology. every module is projective, all the resolutions are trivial, and therefore when i say trivial i mean it in a technical sense: it is the trivial group.

you have the cohomology ring (that's the usual thing) of the group with coeffs in something. they are called the Ext groups, or the right/left derived functors of Hom or tensor.

mathwonk

homology and cohomology are basically just measures of the failure of some maps to be surjectuve or injective.

if we want to measure the structure of some object we start by mapping a trivial object ionto ti and then ask how far the map is from being an isomorpohism. that tells us how far the object is from beign trivial.

say we have an abelkian group. then we just map a direct sum of copies of Z onto it and ask whetehr it is an isomorpohism. if not we map another direct sum of copies of Z onto the kernel. then the theorem is that this last map can alwauys be made injective. so we can represent any abelian group as a cokernel of a map of free abelian groups. this is called the fundamenatl theorem of abelian groups.

If the group is finitely genmerated we can diagonal;ize the matrix of this map and represent the group as a quotient of a free abelian group by a "diagonal" subgroup, thus proving all finitely generated abelian groups are direct sums of cyclic ones.


de raham cohomology is based on the fact that the curl of a gradient is zero. i.e. the subspace of differential forms with crul equal to zero, contains thoise which are gradients. so to emasure the esxtrent to which curl = 0 is equiavlent to being a gradient we define the derham cohomology group {curl = 0}/ {gradients}.

this turns our to be the group whose generators are the holes in our space.

in ant setting we try to find some necessary condition and mod out by the sufficient condition. this cohomology group then measures the failure of the necesdsary condition to also be sufficient.


the examples are endless. for group cohomology i recommend reading a book, like oops ai am having a seniopr moment and cannot recall the name of the famous book on homolgy by the prod=fessor from chiocago, who cowrote with arggh!!! yes!!! MacLane!