Timeline for Is there a four-manifold whose tangent bundle is an endomorphism bundle?
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| Apr 24, 2017 at 17:26 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Apr 24, 2017 at 17:20 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Apr 23, 2017 at 17:31 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Apr 23, 2017 at 17:11 | answer | added | Michael Albanese | timeline score: 3 | |
| Apr 21, 2017 at 23:25 | comment | added | Michael Albanese | @IgorBelegradek: Thank you so much for taking the time to explain this to me, I really appreciate it. I know that vector bundles don't have to be orientable to have a Pontryagin class, I was trying to describe why I wasn't able to proceed with my calculation (I obviously didn't explain myself very well). The entire problem was that I didn't think to use the (in retrospect obvious) fact that $(A\otimes_{\mathbb{R}}B)\otimes_{\mathbb{R}}\mathbb{C} = (A\otimes_{\mathbb{R}}\mathbb{C})\otimes_{\mathbb{C}}(B\otimes_{\mathbb{R}}\mathbb{C})$. Sorry for taking up so much of your time. | |
| Apr 21, 2017 at 20:04 | comment | added | Igor Belegradek | Correction: $E\otimes E^*$ and $E\otimes E$ are isomorphic real bundles, but after you complexify this breaks down. The complexification of $E\otimes E^*$ and the tensor product of $E\otimes\mathbb C$ and $E^*\otimes\mathbb C$, and the latter is the conjugate of the former. The even Charn classes of conjugate bundles agree and odd Chern classes differ by multiplication by $-1$. | |
| Apr 21, 2017 at 19:19 | comment | added | Igor Belegradek | In your case $End(E)\cong E^*\otimes E\cong E\otimes E$. The complexification of $E\otimes E$ is the tensor square of the complexification of $E$. The formulas for the SW-classes and Chern classes are obtained in the same way via the splitting principle. Your computation of $w_1$ and $w_2$ of $End(E)$ seems correct but I have not checked the higher SW-classes. Are you sure they all vanish? Similar formulas hold for Chern classes but I do not know them precisely. This will give you $c_2$ of the complexification of $End(E)$ and hence $p_1$. | |
| Apr 21, 2017 at 19:17 | comment | added | Igor Belegradek | I suggest read the relevant portion of Dold-Whitney, e.g. at see maths.ed.ac.uk/~aar/papers/doldwhit.pdf, and specifically, the corollary on p.674. The 4th cohomology of the base has no torsion because the manifold $M$ is orientable (as you claim in the question, i.e., you say that $TM=End(E)$ is orientable). Another thing I suggest is that you read the definition of Ponryagin classes, e.g. in Milnor-Stasheff. If you did that, you would know that $p_1(\xi)$ is defined as $-c_2(\xi\otimes\mathbb C)$. Orientability of $\xi$ is not needed. | |
| Apr 21, 2017 at 18:24 | comment | added | Michael Albanese | @IgorBelegradek: I still don't think $w_2$, $w_4$ and $p_1$ suffice (they are all stable classes, they can't tell the difference between stably trivial and trivial). The formulas you linked to above tell you how to obtain the Chern classes of a bundle formed out of complex vector bundles. Here $E$ is not assumed to be complex (it isn't even assumed to be orientable). In the computation of the Stiefel-Whitney classes I could split $E$ as a sum of real line bundles and get a corresponding splitting of $\operatorname{End}(E)$. What is the analogue of this here given that $E$ may not be complex? | |
| Apr 21, 2017 at 16:51 | comment | added | Igor Belegradek | You said above that $M$ must be orientable, which is why I thought that $w_2$. $w_4$, $p_1$ suffice. Dold-Whitney also include results for non-orientable bundles but then the answers are more complicated. As for $p_1$ why cannot you compute $c_2$ of the complexifcation in the same way you computed $w_2$, or alternatively, by the formulas I linked above? | |
| Apr 21, 2017 at 16:06 | comment | added | Michael Albanese | @IgorBelegradek: I believe Dold-Whitney is for oriented bundles, and the classes are $w_2$, $e$, and $p_1$ (otherwise $TS^4$ and $\varepsilon^4$ would be a counterexample). I haven't been able to make any progress with calculating $p_1$ in general using the splitting principle. If I assume $E$ is orientable, then I can calculate $p_1$, but not via the splitting principle. I will include what I have done later today. | |
| Apr 18, 2017 at 21:09 | comment | added | Igor Belegradek | Any rank 4 bundle over a closed oriented 4-manifold is uniquely determined by $w_2, w_4$, and $p_1$, see p.647 of [Dold-Whitney, Classification of Oriented Sphere Bundles Over A 4-Complex]. You can compute the Chern classes of the complexification in the same way as the SW classes, see e.g. math.stackexchange.com/questions/989147/…. This will give you $p_1$. I do not want to complete the computation because this is a good exercise, hence not posting this as an answer. Also $End(\xi)$ is isomorphic to $\xi\otimes\xi$. | |
| Apr 18, 2017 at 20:47 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Apr 18, 2017 at 20:17 | history | asked | Michael Albanese | CC BY-SA 3.0 |