Timeline for RO(G) grading of Mackey functors
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 28, 2014 at 1:49 | comment | added | Peter May | I think you are still missing the point: many inequivalent representations have homotopy equivalent S^V: that is, as I said in the first place, the map from RO(G) to Pic is neither injective or surjective. This is not something you expect to calculate naively. | |
| Jul 28, 2014 at 1:41 | comment | added | KHBG | Thanks Peter. I didn't mean to invoke the ring structure on RO(G). Like you say, the additive group RO(G) has an action on the category of G-spectra. The full subcategory of rational G-spectra is equivalent to the reasonably concrete category of Mackey functors, and seems to be preserved by this action of (the group) RO(G). I would like an explicit formula for this action. That is, if E is a rational G-spectrum, and F is the corresponding Mackey functor, which Mackey functor corresponds to S^V E? For instance, what is it's value on G/H; can you write it in terms of F and V? | |
| Jul 28, 2014 at 1:19 | comment | added | Peter May | What it is is implicit in my answer: smashing with S^V gives you an action of the abelian group RO(G) on HoGS that factors through Pic(HoGS). You do not see the multiplication of RO(G) that way, so it is in no sensible sense an action of RO(G). | |
| Jul 28, 2014 at 0:41 | comment | added | KHBG | Peter, you seem to be saying I fell for something, but I cannot see what it is. Please tell me. If $E$ is a rational $G$-spectrum, and $V$ is a real representation of $G$, is $S^V E$ another rational $G$-spectrum or not? | |
| Jul 28, 2014 at 0:26 | comment | added | Peter May | Not an operation of RO(G). You've got to learn the math. | |
| Jul 27, 2014 at 19:39 | comment | added | KHBG | If E is a rational G-spectrum, is its smash product with a representation sphere not also a rational G-spectrum? If that's a meaningful operation, I would like to know a formula for it in terms of Mackey functors. | |
| Jul 27, 2014 at 19:03 | comment | added | Peter May | Delete the first use of ``how''. No meaning, no formula. | |
| Jul 27, 2014 at 2:20 | comment | added | KHBG | Thanks Peter, sorry for the malapropism. I think my question still makes sense: how does RO(G) act on the category of chain complexes of rational-vector-space-valued Mackey functors? How to give a formula for (S^V F)(G/H) in terms of F? | |
| Jul 26, 2014 at 20:12 | history | answered | Peter May | CC BY-SA 3.0 |