Timeline for Where is the Steenrod Realization problem at?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 10 at 14:11 | comment | added | Mark Grant | In answer to your last comment, any $z\in H_5(M^n;\mathbb{Z})$ is Poincare dual to a class in $x\in H^{n-5}(M^n;\mathbb{Z})$. The primary obstruction to embeddability is $St^5_3(x)\in H^n(M^n;\mathbb{Z})$ which vanishes for the reason I gave above. The higher obstructions will live in cohomology groups in degrees larger than $n$. | |
| Sep 10 at 14:08 | comment | added | Mark Grant | @CrashBandicoot The Steenrod power $St^5_3$, aka $\beta P^1_3$, is definitely to do with embeddings and not just Steenrod realizability. It's not until Chapitre 3 that Thom tackles Steenrod's problem by passing to an associated manifold (regular negihbourhood of an embedding in some Euclidean space) and the duals Poincare duals of the Steenrod powers come into play. | |
| Sep 10 at 11:47 | comment | added | Mark Grant | @CrashBandicoot: Yes every class in $H_5(M^8;\mathbb{Z})$ is represented as an embedding, because for the dual class $x\in H^3(M^8;\mathbb{Z})$ the image $St^5_3(x)\in H^8(M^8;\mathbb{Z})\cong\mathbb{Z}$ is $3$-torsion, and all the higher obstructions vanish for dimension reasons. | |
| Sep 10 at 3:06 | comment | added | Crash Bandicoot | Thank you Mark, I am asking here since I noticed you worked a lot on the difference between embedded, immersed Thom criteria | |
| May 16 at 21:07 | history | edited | LSpice | CC BY-SA 4.0 |
Name of preprint
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| May 16 at 16:48 | comment | added | Ryan Budney | Thanks Mark. This looks helpful. | |
| May 16 at 16:39 | history | answered | Mark Grant | CC BY-SA 4.0 |