Timeline for Novikov conjecture
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 19, 2019 at 20:56 | comment | added | Yasha | Sounds, good. Leave it as an answer, and add some references please. I looked at Novikov's paper and he says something similar but also without references ("it is well known kinda thing"). I guess all this is indeed well known to experts but they are vanishing, so making this explicit would be of great help to some people. | |
| Jun 17, 2019 at 21:13 | comment | added | Ben Wieland | The situation simplifies in dim not a multiple of 4 by not having the signature obstruction, which is why I gave $S^2\times S^4$. In the case of $S^4\times S^4$, the mflds are, roughly, parameterized by $p_1\in H^4=Z^2$, with $p_2$ determined by the signature formula. Exercise: consider 4d vector bundles over $S^4$, ie, $O(4)$ bundles. parameterize them. Take the space of unit vectors to get $S^3$ bundles over $S^4$. Infinitely many have the homology of $S^3\times S^4$. Compute $p_1$ and see that anything is possible. Harder: show that infinitely many are homotopy equivalent to $S^3\times S^4$ | |
| Jun 16, 2019 at 19:29 | comment | added | Yasha | @BenWieland, Do you have a reference please? Also do you mean $S^4 \times S^4$? The signature formula is in dim 4n. | |
| Jun 15, 2019 at 4:30 | review | Close votes | |||
| Jun 16, 2019 at 5:42 | |||||
| Jun 15, 2019 at 3:34 | comment | added | Aleksandar Milivojević | @BenWieland Besides calculating the signature correctly, the Pontryagin numbers formed from the chosen Pontryagin classes have to satisfy a set of congruence relations as well, as given by the Hattori-Stong theorem. Let me know if I've misunderstood your claim. | |
| Jun 15, 2019 at 2:29 | comment | added | Ben Wieland | That is very false. Roughly speaking, surgery theory says if $M$ is a simply connected manifold that you can make a new manifold $N$ homotopy equivalent to $M$ with new Pontrjagin classes, whatever you want them to be (as long as it satisfies the Hirzebruch signature formula). (eg, an infinite family equivalent to $S^2\times S^4$ parameterized by $p_1$ and another infinite family for $\mathbb C\mathbb P^4$, parameterized by either $p_1$ or $p_2$, which are related by Hirzebruch.) The Novikov conjecture measures the extent to which this fails in the presence of fundamental group. | |
| Jun 15, 2019 at 0:31 | history | asked | Yasha | CC BY-SA 4.0 |