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/sci/ - Science & Math

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4792911 No.4792911 [Reply] [Original]

Does anybody here know dick about group cohomology?

About the definition as a derived functor: if G is a group and A a ZG module (ZG = integral group ring) apparently group cohomology are the derived functors of passing from A to the submodule of G-invariants of A.

So to get group cohomology shouldn't we take a resolution of A, switch over to G-invariants (i.e. apply our functor to the resolution) and then compute homology of our new chain complex?

Because in the wiki-article (in the paragraph about ext) we take a resolution of Z, not A, and then apply hom(-,A) to some projective resolution of Z in the category of ZG modules (with Z as a trivial ZG module).

Now apparently hom(Z,A) (in the category of ZG modules) is isomorphic to the submodule of G-invariant elements of A, but that's all the connection I could find.

So can somebody shed some light on this?

>> No.4792935
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4792935

Britney Spears bump.
Picture related, it's Britney Spears.

>> No.4792961
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4792961

More Britney Spears.

>> No.4792981
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4792981

In fact in Lang's Algebra it says: (E is the standard resolution of Z as in the wiki-article)
"Show that if <span class="math">H^Q(G,A)[/spoiler] denotes the q-th homology of the complex Hom(E,A), then <span class="math">H^0(G,A) = A^G[/spoiler] (this is the submodule of G-invariants). Thus the left derived functors of <span class="math">A \mapsto A^G[/spoiler] are the homology groups of the complex Hom(E,A)."

Could someone explain the "thus"-part to me?

>> No.4793005
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4793005

>>4792981
that's supposed to read <span class="math">H^q(G,A)[/spoiler].

it's from page 826, 3rd edition. Picture related, it's Serge Lang.

>> No.4793052
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4793052

>>4793005
the book has 800 pages??
I only know cohomology in a differential geometric setting and while I guess there the group is an uncomplicated one (addition of forms) I don't see where the group actions come into plan and when I try to read the wikipedia article on derived functors I see that they don't even try to give a direct definition.
I could ask if you like.