Timeline for How to write K-theory Conner-Floyd Chern classes in terms of Adams operations?
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24 events
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| Aug 14, 2019 at 5:39 | history | edited | YCor |
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| Oct 15, 2018 at 3:01 | comment | added | xir | okay i see what my problem is now. of course you're right, $\gamma(-1,t)$ is not what i thought. (just replying this so any future google parvenu/es can see) | |
| Oct 11, 2018 at 23:38 | comment | added | xir | in fact your very own $\gamma$-ring formalism seems to suggest that if, say, $L$ is a line bundle, so that $\lambda_t(L) = 1+ Lt$, then $\gamma_t(L)=\gamma_t(L-1)\gamma_t(1)=\gamma_t(L-1)=1+(L-1)t$ so that $c_1(L)=L-1$. okay i'm satisfied with what's going on now | |
| Oct 11, 2018 at 23:33 | comment | added | xir | see here for example - math.harvard.edu/~lurie/252xnotes/Lecture4.pdf - the canonical coordinate on K(BU(1)) is given by L-1, and on K(BU(n)), the conner-floyd chern classes should be elementary symmetric polynomials of these coordinates under the pullback BU(1)^n -> BU(n) classifying direct sum. | |
| Oct 11, 2018 at 22:53 | history | edited | xir | CC BY-SA 4.0 |
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| Oct 11, 2018 at 22:47 | history | edited | xir | CC BY-SA 4.0 |
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| Oct 11, 2018 at 22:17 | comment | added | xir | why do you say that changing virtual dimension would pose a problem? the classes $c_d$ under your suggested orientation $c_d(V)=\wedge^d V$ don't converge to zero in the $I$-adic topology ($I$ the aug. ideal) which seems morally wrong to me as well. | |
| Oct 11, 2018 at 22:13 | comment | added | xir | i'm still not really convinced though, since of course the orientation $c_1(L)=L-1$ gives the same thing. and this orientation is necessary also to make the formal group law associated to tensor product of line bundles $\hat{\mathbb{G}_m}$. | |
| Oct 11, 2018 at 22:08 | comment | added | xir | ah, that's the source of my confusion then. right, because maps into BU have to classify virtual dimension zero. so that's why my hirzebruch calculation gives the wrong result | |
| Oct 11, 2018 at 22:04 | comment | added | William Balderrama | $BU(1)\rightarrow BU$ is classifying $L-1$, so $c_1(L-1)=L-1$. | |
| Oct 11, 2018 at 21:55 | comment | added | xir | in your answer, don't you say that $c_1$ should be the pullback of $l-1$ with $l$ the tautological line bundle? i've never seen any orientation of K-theory besides $c_1(L)=L-1$ (and minor variants like 1-L, etc.) | |
| Oct 11, 2018 at 21:51 | comment | added | William Balderrama | It's possible my $c_i$'s are different from your $c_i$'s, although I don't think it's possible for $c_1$ to change virtual dimension and still have $KU(BU) = \mathbb{Z}[[c_1,\ldots]]$. | |
| Oct 11, 2018 at 21:46 | comment | added | xir | hm i'm quite confused now. it seems to me that it should be true that $\psi^1([L])=[L]$ while $c_1([L])=[L]-1$. | |
| Oct 11, 2018 at 21:37 | comment | added | William Balderrama | @xir As $ku(BU)\cong \mathrm{Op}(ku)$, there is an action of $ku(BU)$ on $ku(X)$ for any $X$. This is a set-map $a\colon ku(BU)\times ku(X)\rightarrow ku(X)$ satisfying among other things left-linearity and $a(f\smile g,x) = a(f,x)\smile a(g,x)$, but not e.g. right-linearity. In this sense, $\psi^k\in ku(BU)$ can act on $ku(X)$, and we generally only have $\psi^k = (\psi^1)^k$ for lines as $a((\psi^1)^k,x)=x^x$ and $\psi^k(x):=a(\psi^k,x) = x^k$ generally only for lines $x$. | |
| Oct 11, 2018 at 21:27 | vote | accept | xir | ||
| Oct 11, 2018 at 21:26 | comment | added | xir | well here i'm considering this idea of $\psi^k$ not as an operator, but as an element living in a certain K-theory ring containing the K-theory operations, where the usual ring structure does not correspond to the ring structure on the additive operations (which is given by composition rather than cup product). does that make sense? | |
| Oct 11, 2018 at 21:22 | comment | added | მამუკა ჯიბლაძე | Hm maybe I am confusing things but I thought $\psi^k$ of a line bundle is its $k$th tensor power, and this then extends to other vector bundles via e. g. the splitting principle, no? | |
| Oct 11, 2018 at 21:19 | comment | added | xir | i'm not sure what you mean by that. how are you proposing to "act" by $\psi^k$ on a line bundle? the only answer i can think of is by acting on $K^0$ of the relevant space via the action of the (additive) K-theory operations algebra, and that action is via the compositional algebra structure, not via the cup-product one, like i said. | |
| Oct 11, 2018 at 21:17 | comment | added | მამუკა ჯიბლაძე | But does not $\psi^k$ equal $(\psi_1)^k$ only on line bundles? | |
| Oct 11, 2018 at 21:06 | comment | added | xir | actually seeing the answer now i think perhaps one wants to restrict to additive operations to make things like "algebra action" make sense. | |
| Oct 11, 2018 at 21:01 | comment | added | xir | it does, in the usual cup-product multiplication on $K^0(BU)$. but $K^0(BU)$ has the "extra" multiplicative structure of composition, which gives it a second, different algebra structure, and it's via this algebra structure that it acts on other K-theory rings. | |
| Oct 11, 2018 at 19:46 | comment | added | მამუკა ჯიბლაძე | Why does not your last statement imply that $\psi^k$ is $(\psi_1)^k$? | |
| Oct 11, 2018 at 19:12 | answer | added | William Balderrama | timeline score: 10 | |
| Oct 9, 2018 at 3:51 | history | asked | xir | CC BY-SA 4.0 |