Timeline for K-Theory as a special $\lambda$-ring
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 11, 2013 at 18:49 | comment | added | Will Sawin | If that's so, then the key issue is whether the $K$-theory ring is still generated by exterior powers. Seems like it should be. | |
| Mar 11, 2013 at 17:59 | comment | added | darij grinberg | 1. You mean that the exterior power might not be well-defined in $K_0$? I am pretty sure they are (when we're talking about finitely generated projective modules). | |
| Mar 11, 2013 at 6:34 | comment | added | Will Sawin | 1. Which exterior power? There's more than one. Are they obviously equivalent? I don't quite see it. 2. No, I'm referring to the category with one object whose morphisms are the elements of the group. The functors from this category to $Vect$ are trivially the representations of the group. | |
| Mar 11, 2013 at 6:20 | comment | added | darij grinberg | "But the category of $n$-dimensional vector spaces is equivalent to the group $GL_n$." You mean the category of representations of the group $GL_n$? Sorry, I don't think so. And if I restrict myself to dimension $n$, my isomorphisms will only be canonical w.r.t. isomorphism of representations, not w.r.t. homomorphism of representations... | |
| Mar 11, 2013 at 6:18 | comment | added | darij grinberg | I'm pretty sure $\lambda^k$ is defined in characteristic $p$ (as the exterior power). | |
| Mar 10, 2013 at 22:57 | comment | added | Will Sawin | 1. It's characteristic $0$. $\lambda^k$ is not even defined in characteristic $p$, I don't think. 2. Yes, you have to shuffle addends to the other side. All I am doing is interpreting a natural isomorphism as a natural transformation, Clearly the $\lambda$-operations must be interpreted in this context as functors $Vect \to Vect$. To study vector bundles we only need to consider vector bundles up to isomorphism, so we only need the source category to consist of $n$-dimensional vector spaces, say. But the category of $n$-dimensional vector spaces is equivalent to the group $GL_n$. | |
| Mar 10, 2013 at 22:38 | comment | added | darij grinberg | I also can't really say I understand the last paragraph, about how you get a natural isomorphism. (Notice that the polynomials $P_k$ and $P_{ik}$ don't always have nonnegative coefficients, so one has to shovel some of the addends onto the other side, potentially obscuring the combinatorics.) | |
| Mar 10, 2013 at 22:37 | comment | added | darij grinberg | What characteristic are we in? I know this argument (or at least one that sounds like this -- I don't recall it ever saying anything geometric) being used in characteristic $0$, but I'm not sure how well it applies beyond that case. | |
| Mar 10, 2013 at 21:35 | history | answered | Will Sawin | CC BY-SA 3.0 |