Timeline for The space of homotopy classes of maps of products of spheres
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 22, 2018 at 7:44 | comment | added | Mark Grant | @PPR: For $p=q=1$ this is quite easy to see: suspension gives a group homomorphism $\pi_1(S^1\vee S^1)\to \pi_2(S^2\vee S^2)$ whose target is an abelian group, and the attaching map represents a commutator $aba^{-1}b^{-1}$. For the general case, see Whitehead's "Elements of homotopy theory", Theorem X.8.20. | |
| Aug 21, 2018 at 21:28 | comment | added | PPR | @MarkGrant, in relation to my question here math.stackexchange.com/questions/2885139/… would you mind explaining why the suspension of the attachment map is nullhomotopic? The last answer to the linked question seems to have quite a big argument to support that; what is the way to argue here? | |
| Aug 10, 2016 at 10:47 | comment | added | Mark Grant | @DLIN: Yes, I think that's right (although you would have to be careful with the group structure). In general, the elements of a semi-direct product $G\rtimes H$ (where $H$ is acting on $G$) are ordered pairs $(g,h)$ for $g\in G$ and $h\in H$. The group operation is not the same as in $G\times H$, however, and depends on the action. | |
| Aug 10, 2016 at 9:45 | comment | added | DLIN | Thank you for your comments. I still have a question, can we say that any element of $[S^p\times S^q,GL_N(\mathbb C)]$ can be represented as the product (matrix product)of three parts corresponding to $\pi_p(GL_N(\mathbb C))$, $\pi_q(GL_N(\mathbb C))$ and $\pi_{p+q}(GL_N(\mathbb C))$ | |
| Aug 9, 2016 at 8:20 | comment | added | Mark Grant | @DLIN: Well, I didn't claim that it was Abelian! This group structure is induced by the group structure on $X=Gl_N(\mathbb{C})$ given by matrix multiplication, which is clearly not Abelian, so if $[S^p\times S^q,X]$ were Abelian it would require some other argument. If it turns out not to be Abelian, this split exact sequence still describes it as a semi-direct product. | |
| Aug 9, 2016 at 7:51 | comment | added | DLIN | But I still do not understand the group structure of the homotopic group $[S^p\times S^q,X]$. Could you tell me why the $[S^p\times S^q,X]$ is Abelian, some reference is very welcome. | |
| Aug 9, 2016 at 7:12 | history | edited | Mark Grant | CC BY-SA 3.0 |
added special case in response to a comment
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| Aug 9, 2016 at 6:56 | comment | added | Mark Grant | @DLIN: Actually, it seems you are right (up to a shift in dimension), see my edit. | |
| Aug 9, 2016 at 2:06 | comment | added | DLIN | No, actually, I mean $[S^p\times S^q,X]=\pi_p(X)\oplus \pi_q(X)\oplus \pi_{p+q-1}(X)$. | |
| Aug 8, 2016 at 20:08 | comment | added | Mark Grant | @DLIN: Do you want to conclude that $\langle S^p\times S^q, X\rangle\cong\pi_p(X)\oplus\pi_q(X)$ when $X=GL_N(\mathbb{C})$? Since all Whitehead products are trivial in a path-connected H-space, the last map in my final exact sequence above will be trivial, but I'm not sure about the first map. | |
| Aug 8, 2016 at 11:30 | comment | added | DLIN | If $X=GL(N,\mathbb C)$, $p$ is even and $q$ is odd, can we say the homotopy group $[S^p\times S^q,X]$ splits? | |
| Mar 27, 2016 at 11:23 | history | answered | Mark Grant | CC BY-SA 3.0 |