Confused by a theorem in Milnor-Stasheff

In summary, the k-th Pontrjagin class of a real vector bundle is defined as the 2k-Chern class of the complexified bundle and lives in cohomology with integer coefficients. Theorem 15.9 states that if the coefficient ring is a PID containing 1/2, then the singular cohomology ring of a certain space G is the polynomial ring over the PID in the Pontrjagin classes. A natural map exists from H^*(G,Z) to H^*(G,PID), making cohomology contravariant in spaces and covariant in coefficient modules. This map is also used in the Milnor-Stasheff theorems.
  • #1
quasar987
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The k-th Pontrjagin class of a real vector bundle is defined as the 2k-Chern class of the complexified bundle. Therefor, a Pontrjagin class lives in cohomology with integer coefficients. But then the statement of Theorem 15.9 is that if the coefficient ring is taken to be a PID containing 1/2 (ex: Z[1/2] or Q, R, C), then the singular cohomology ring of a certain space G is the polynomial ring over in the Pontrjagin classes. But what is meant by a Pontrjagin class as an element of ?? Is there a natural map that allows such an identification!?
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  • #2
Looking at it more closely it seems to me that the definition of the Euler class of an oriented bundle as an integer cohomology class would work just as well by replacing Z with any PID (we need a unit). And since the Chern classes are defined only in terms of the Euler class, then those as well are defined with any PID as coefficients, and so finally also for the Pontrjagin classes.
 
  • #3
There is such a map as asked for in the last sentence of post #1. In general, cohomology is contravariant in the spaces and covariant in the coefficient modules, e.g. Thm. 3, page 237, of Algebraic Topology, by E. Spanier. The reason is that the Hom functor, used to turn chains into cochains, is covariant in the second variable, i.e. a map from A-->B, yields a map from maps X-->A to maps X-->B. And of course there is a unique ring map from the integers Z into to any ring with identity element.
 
  • #4
mathwonk said:
There is such a map as asked for in the last sentence of post #1. In general, cohomology is contravariant in the spaces and covariant in the coefficient modules, e.g. Thm. 3, page 237, of Algebraic Topology, by E. Spanier. The reason is that the Hom functor, used to turn chains into cochains, is covariant in the second variable, i.e. a map from A-->B, yields a map from maps X-->A to maps X-->B. And of course there is a unique ring map from the integers Z into to any ring with identity element.
Ahh, yep. Plus I knew of that construction from the all the theorems in Milnor-Stasheff of the form "The natural map H^*(X,Z)-->H^*(X,Z_2) sends such-and-such characteristic class to such-and-such Stiefel-Whitney class" but I didn't think of it. Instead I was trying to find the map in the Universal Coefficient Theorem for cohomology but couldn't find it there. Anyhow, thanks for the reply!
 
  • #5
you are welcome. i also looked for it in the univ coefficient section, then kept looking when i didn't find!
 

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