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Nov 15, 2022 at 12:36 history edited Martin Sleziak
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Dec 21, 2016 at 16:42 answer added Nicholas Kuhn timeline score: 6
Dec 20, 2016 at 16:46 vote accept user84144
Dec 20, 2016 at 7:50 comment added user84144 What I mean is that I believe I ``know'' which class I want to pull back to get $w_n$ (in response to your very reasonable complaint that one needs to pin down what one means by "generators"). For instance, I'm optimistic that I could write down an explicit representative by a Schubert cell, if necessary. (We wouldn't know that it corresponds to a symmetric polynomial without doing more work.) Now I'm not suggesting that this is an efficient definition, but in my mind it illustrates a rough conceptual divide between a "classifying space definition" and "Steenrod square definition".
Dec 19, 2016 at 23:46 comment added Oscar Randal-Williams @DenisNardin: Being homogeneous does not characterise generators; the cohomology of BO has many graded ring automorphisms.
Dec 19, 2016 at 23:42 comment added Oscar Randal-Williams @user84144: Is this really a definition? The only way I can think of showing that map i) is injective, and ii) hits the symmetric polynomials, needs me to already have a definition of Stiefel-Whitney classes and have proved the sum formula for them.
Dec 19, 2016 at 22:47 comment added user84144 Well, you give me too much credit. What I actually have in my head is a specific choice of generator labelled $w_n$, corresponding to a specific symmetric polynomial under the embedding $H^*(BO(n)) \hookrightarrow H^*(BO(1)^n)$. I lazily decided not to be more precise, thinking that nobody would care, but that seems to have been a wrong presumption.
Dec 19, 2016 at 22:32 comment added Denis Nardin @OscarRandal-Williams I think he meant the homogeneous polynomial generators, which I believe in this case is in fact a characterization of the elements of $H^*(BO;\mathbb{Z}/2)$ we are talking about
Dec 19, 2016 at 22:13 comment added user84144 Sorry, I purposefully did not say "the generators" but maybe should have said "certain generators", which I realize is not actually a definition either, but is probably enough indication of what I'm referring to.
Dec 19, 2016 at 22:04 answer added Carlo Beenakker timeline score: 7
Dec 19, 2016 at 21:58 answer added John Rognes timeline score: 17
Dec 19, 2016 at 21:52 comment added Oscar Randal-Williams That is not a definition at all, as ``the generators" does not define elements of a ring.
Dec 19, 2016 at 21:42 history asked user84144 CC BY-SA 3.0