Timeline for When and why are Adams operations "non-negative"?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 25, 2019 at 18:56 | answer | added | John Greenwood | timeline score: 5 | |
| Nov 23, 2019 at 1:59 | comment | added | Will Sawin | @JohnGreenwood Good point. Maybe a rank-2 tautological bundle on a Grassmanian $G(2,n)$ for $n$ sufficiently large will do the trick instead, as these approximate the classifying space of $GL_2$ and the standard rep of $GL_2$ gives a counterexample. | |
| Nov 22, 2019 at 23:48 | comment | added | John Greenwood | @WillSawin The tangent bundle to $\mathbb{P}^{n}$ is stably a sum of line bundles, so I don' t think that will work. More precisely, if $T$ is the tangent bundle then there's a line bundle $L$ with $[T]=(n+1)[L] - 1$. Applying $\psi^{k}$ we get $(n+1)[L^{k}]-1$. But the first term is a rank $n+1$ bundle with zero top chern class, so it has a nonvanishing section. That means "subtracting 1" actually makes sense at the level of bundles. I think a similar argument will get you non-negativity of the Adams operations on anything that's stably a difference of line bundles. | |
| Nov 11, 2019 at 6:55 | comment | added | John Baez | Right, I get it now. The splitting principle is good for reducing identities between characteristic classes to the case where a vector bundle is a sum of line bundles, but of no use for what I was trying to use it for. | |
| Nov 3, 2019 at 9:05 | comment | added | Will Sawin | Just for clarity I think the splitting principle argument is also wrong. The map being objective on $K$-theory implies almost nothing about what it does on vector bundle classes. Probably the tangent bundle to $\mathbb P^2$ provides a counterexample. | |
| Nov 3, 2019 at 1:37 | comment | added | John Baez | Given the mistake that Will Sawin caught, I'm gonna delete the stuff about representation rings in my original post. Nonetheless I'm happy to hear about which lambda-operations are nonnegative in which representation rings. | |
| Nov 3, 2019 at 1:32 | history | edited | John Baez | CC BY-SA 4.0 |
deleted stuff about representation rings
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| Nov 3, 2019 at 1:29 | comment | added | John Baez | "For $\rho$ a representation, $\rho(g^k)$ is not a representation unless $G$ is abelian." Oh, duh. I guess this somehow analogous to the splitting principle then: people study Adams operations on $\mathrm{Rep}(G)$ with $G$ a compact Lie group by looking at what they do to $\mathrm{Rep}(T)$ when $T$ is a maximal torus, where the formula I gave actually makes sense! | |
| Nov 3, 2019 at 1:24 | comment | added | Will Sawin | There might be many examples of operations that are only nonnegative for symmetric monoidal abelian categories of a certain type. Simple Lie groups with no automorphisms of their Dynkin diagram have only self-dual representations, making $\operatorname{Sym}^2 V + \wedge^2 V -1$ nonnegative, at least when $V$ is nonzero. | |
| Nov 3, 2019 at 1:19 | comment | added | Will Sawin | For $\rho$ a representation, $\rho(g^k)$ is not a representation unless $G$ is a abelian. No operation preserves non negativity for all groups unless it is a nonnegative combination of Schur functors, as you can see by plugging in the standard representation of $GL_n$. An operation preserves non negativity in the representation ring of all abelian groups if and only if the associated symmetric polynomial has nonnegative coefficients. | |
| Nov 3, 2019 at 1:06 | history | asked | John Baez | CC BY-SA 4.0 |