Timeline for The space of homotopy classes of maps of products of spheres
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11 events
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| Mar 27, 2016 at 11:23 | answer | added | Mark Grant | timeline score: 7 | |
| Mar 24, 2016 at 2:12 | comment | added | Dylan Wilson | Since you only care about parallelizable spheres, you should be able to carry out Ryan's proposal (surely we know the first 20 or so homotopy groups, and definitely when k is large since then you're in the stable range where this becomes a question about reduced K-theory) | |
| Mar 24, 2016 at 2:01 | comment | added | Bilateral | @j.c. Indeed I meant parallelizable. | |
| Mar 24, 2016 at 1:58 | history | edited | Bilateral | CC BY-SA 3.0 |
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| Mar 24, 2016 at 1:23 | comment | added | j.c. | There's a discussion related to Dylan Wilson's comments in Hatcher's algebraic topology book section 4.A. Note that the generalization of the result you are quoting from Bott and Tu only reduces the problem of computing the homotopy classes of maps between two spaces $K\rightarrow X$ to two things you need to know: (1) the set of homotopy classes of maps fixing a basepoint in the two spaces ("pointed" maps) and (2) the action of the fundamental group of $X$ on this set. When $K=S^q$, (1) is just the definition of $\pi_q(X)$, but for your $K$ it may be quite hard to compute. | |
| Mar 23, 2016 at 20:43 | comment | added | Ryan Budney | You can describe the pointed maps $S^q \times S^p \to X$ in terms of homotopy groups, whitehead products and some obstruction theory. Unfortunately in your case this would require knowledge of the homotopy groups of your matrix groups. Often these can be difficult to compute, but you have enough information to do rational computations. | |
| Mar 23, 2016 at 20:21 | comment | added | Dylan Wilson | The collection of choices, however, is acted on by $\pi_1$ and the quotient gets rid of the ambiguity. | |
| Mar 23, 2016 at 20:21 | comment | added | Dylan Wilson | Focusing on your specific situation, here's what's happening: Given a pointed map $K \rightarrow X$ and a path $\gamma$ in $X$, we can try to imagine `shifting' this map along the path (think monodromy) and ending up with a new map where the basepoint of $K$ is sent to the endpoint of the path. Any map is basepoint-presrving for some basepoint of $X$ (trivially). Choosing a path from that one to $x$ gives a way of taking an arbitrary map and making it basepoint preserving. But there's ambiguity because there are different homotopy classes of paths between these two points. | |
| Mar 23, 2016 at 20:10 | comment | added | Bilateral | @DylanWilson: Thank you very much for the answer. Could you please elaborate a little bit more or point out some references? I am far from an expert in Algebraic Topology. Thanks. | |
| Mar 23, 2016 at 19:56 | comment | added | Dylan Wilson | If $K$ is well-pointed then there is a fibration $\text{map}(K,X) \rightarrow X$ given by evaluation at the basepoint and its fiber over a point in $X$ is the space of pointed maps. The result you're after is a special case of the corresponding `exact sequence' $\pi_1X \rightarrow [K,X]_* \rightarrow [K,X]$ which is (by definition) the statement that $\pi_1$ acts on the path components of the fiber and the quotient is given by the path components of the total space (assuming that $X$ is connected). | |
| Mar 23, 2016 at 19:49 | history | asked | Bilateral | CC BY-SA 3.0 |