Timeline for Does $\overline{\mathbb{C}P^2 \setminus B^4}$ embed (smoothly/topologically) in $\mathbb R^5$?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| May 29, 2023 at 14:17 | comment | added | Martin Tancer | Thank you again. Hopefully, now I understand reasonably well thus I have accepted the answer. | |
| May 29, 2023 at 14:17 | vote | accept | Martin Tancer | ||
| May 26, 2023 at 13:29 | comment | added | David E Speyer | I don't know about the PL/topological questions. | |
| May 26, 2023 at 13:29 | comment | added | David E Speyer | See math.stackexchange.com/questions/1923402/… and ncatlab.org/nlab/show/basic+complex+line+bundle+on+the+2-sphere for some possibly useful background. | |
| May 26, 2023 at 13:27 | comment | added | David E Speyer | The K theory here is pretty trivial: Rank $r$ vector bundles on $S^2$ are classified by homotopy classes of maps $S^{1} \to GL(r) \simeq O(r)$, in other words, by $\pi_1(SO(r))$, which is trivial for $r=1$, $\mathbb{Z}$ for $r = 2$ and $\mathbb{Z}/2 \mathbb{Z}$ for $r>2$. Taking direct sum of vector bundles adds the classes of the summands, reducing modulo $2$ if necessary. So the claim is that $TS^2$ is trivial, $\underline{\mathbb{R}}$ is trivial but that $\mathcal{L} \cong \mathcal{O}(1)$, which is nontrivial mod $2$. | |
| May 26, 2023 at 13:19 | comment | added | Martin Tancer | Continued: This I do not know, but could it be the case that the Stiefel-Whitney classes depend on whole cohomology ring? | |
| May 26, 2023 at 13:14 | comment | added | Martin Tancer | Thank you for the answer and the comments. Now, I am trying to digest it (whether I am able to fully verify). I am not too familiar with K-theory, thus I am perhaps closer to verifying the comment on Stiefel-Whitney classes. Some double checks: The answer above provides no in "smooth" case but with Stiefel-Whitney classes, this seems to give "no" also in piecewise linear case? The isomorphism on second cohomology is meant of $T\mathbb{C}P^2$ and $TU$? If so, shouldn't there be still said something about the relation to $T\mathbb{C}P^1$? Is the second cohomology sufficient? | |
| May 26, 2023 at 2:57 | comment | added | Connor Malin | Phrased a little differently, the 2nd Stiefel-Whitney class of $\mathbb{C}P^2$ is nontrivial and since the inclusion of $\mathbb{C}P^1$ induces an isomorphism on second cohomology, we can deduce that the tangent bundle of the punctured $\mathbb{C}P^2$ is stably nontrivial, so does not admit a codimension 1 embedding into Euclidean space. | |
| May 25, 2023 at 18:49 | comment | added | David E Speyer | @MichaelAlbanese Fixed, thanks. | |
| May 25, 2023 at 18:47 | history | edited | David E Speyer | CC BY-SA 4.0 |
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| May 25, 2023 at 18:18 | comment | added | Michael Albanese | It's worth noting that $\mathcal{L} = \mathcal{O}(1)$ and $U$ is its total space. In your last line, did you mean to say $KO(S^2)$? | |
| May 25, 2023 at 17:39 | history | edited | David E Speyer | CC BY-SA 4.0 |
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| May 25, 2023 at 17:13 | history | answered | David E Speyer | CC BY-SA 4.0 |