Timeline for Decomposition of direct image of a smooth morphism, Deligne's theorem, motives
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 7, 2022 at 10:26 | vote | accept | Geordie Williamson | ||
| Mar 3, 2022 at 9:06 | answer | added | Chris H | timeline score: 2 | |
| Oct 13, 2021 at 14:09 | comment | added | rvk | Ok, I see why even dim > 1, now. Canonical fundamental class. Slow and rusty. | |
| Oct 13, 2021 at 13:58 | comment | added | rvk | @GeordieWilliamson : Doh! You are too kind. That was a completely dumb idea on my part. You at least need the fibers to have dim > 0 (I dont immediately see right now why you need dim > 1) otherwise your hope holds rather trivially. This is what I get for not thinking a bit carefully. I am tempted to delete to hide my shame! | |
| Oct 13, 2021 at 4:34 | comment | added | Geordie Williamson | @rvk: thanks a lot. I'll look at this, but I suspect that the example you give (approximations to EZ/Z -> BZ/2) won't work as any example should have fibres of dimension $> 1$. (Btw, I basically convinced myself that Anonymous' proposal works. There is one very natural looking identification that I couldn't convince myself of, but I'm hoping to still do so and post the answer.) Welcome back by the way! | |
| Oct 7, 2021 at 20:43 | comment | added | rvk | Geordie: I dont know a counterexample off the top of my head. However, the first thing that came to my mind was that this would get you into trouble with Steenrod operations (over $\mathbb{F}_2$ say). So perhaps look at approximations to the map $E\mathbf{Z}/2 \to B\mathbf{Z}/2$. However, it has been a long time since I thought about these things and my current intuition might be way off. | |
| Sep 8, 2021 at 1:26 | comment | added | Geordie Williamson | @Anonymous: This looks like a very promising counter-example, thank you. I'll try to work through the details. | |
| Sep 7, 2021 at 12:06 | comment | added | Anonymous | Here's a candidate example: If $f:X \to Y$ is a smooth proper genus $0$ curve, then I think $R^i f_* k$ looks the same as that for $\mathbf{P}^1_Y \to Y$ (as $\mathrm{PGL}_2$ is connected). So the obstruction to splitting with $\mathbf{Z}/2$-coefficients is an element of $H^2(Y, \mathbf{Z}/2(1))$ that I think is the Brauer class of $f$. So it will be non-trivial in general. | |
| Sep 7, 2021 at 5:58 | history | asked | Geordie Williamson | CC BY-SA 4.0 |