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Can cohomologies always be defined as the quotient of the kernel and the image of a certain operation?

+ 2 like - 0 dislike
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For example, concerning de Rham cohomology as explained on the lower part of page 5 here, one considers the vector space of all forms on a manifold $M$, $A_{DR}(M)$ in the notation of the paper (?).

The $p$-th de Rham cohomology is then defined as the quotient of the kernel and the image of the exterior derivative $d$ acting on the space of $p$ and $p-1$ forms respectively as

\[H_{DR}^p(M) = \frac{Ker(d: A_{DR}^{p}(M)\rightarrow A_{DR}^{p+1}(M))} {Im(d: A_{DR}^{p-1}(M)\rightarrow A_{DR}^{p}(M))}\]

Can cohomology always be defined as the kernel divided by the image of a certain operation?

What does cohomology mean or "measure" in as simple as possible intuitive terms in the de Rham case but also more generally?

PS: I tried to read wikipedia of course, but it was (not yet?) very enlightening for me ...

asked Jan 21 in Mathematics by Dilaton (4,295 points) [ revision history ]

The $n$th cohomology group is always the quotient of a vector space of $n$-cocycles by the corresponding vector space of $n$-coboundaries (which are special $n$-cocycles constructed from arbitrary $(n-1)$-cocycles. They are a way to measure how nontrivial the topology is from which they are constructed. (They are topological invariants.)

Some of the low order cohomology groups measure in some applications the possibilities for extensions of an object, or the obstructions for possible deformations. in the latter case the nonvanishing is called an anomaly.

1 Answer

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As in the math.stackexchange answer linked by an anonymous user in a comment to your question, certain (topological) cohomology theories like K theory do not come from cohomology of a chain complex functor.

These generalized cohomology theories always come from a spectrum, however, which is a sequence of spaces analogous to the Eilenberg-Maclane spaces K(G,n) which represent ordinary cohomology in the sense that for any CW complex X, the homotopy classes of maps X -> K(G,n) form a group (!) isomorphic to $H^n(X,G)$.

The spectrum of K complex theory is 2-periodic, alternating between $BU \times \mathbb{Z}$ and its delooping.

answered Jan 26 by Ryan Thorngren (1,605 points) [ revision history ]
edited Jan 26 by Ryan Thorngren

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