Timeline for What do Atiyah and Segal mean by $K_G^*(X)$?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 25, 2018 at 1:54 | comment | added | John Klein | Yes, I believe I've explained all that already. They are making use of the Thom isomorphism and Bott periodicity. Look again at what I wrote above. | |
| Mar 23, 2018 at 11:17 | comment | added | theinvariant | Maybe I'm being slow here, but that still doesn't explain to me how there's a non-shifted $K$-term appearing on the right of $K_G(X\times S^{2n+1})$: i understand there's a injection $K/\xi\cdot K\rightarrow K_G(X\times S^{2n+1})$ but surely without appealing to the 6-term exact sequence involving $K^0$ and $K^1$, we don't have any information about what one can put after $K_G(X\times S^{2n+1})$ in the sequence? Are they not relying on a Thom isomorphism for $K^1$ to say $K_G^0(X\times S^{2n+1})$ maps into the kernel of multiplication by $\xi$ of a $K^1$-ring? | |
| Mar 23, 2018 at 0:16 | comment | added | John Klein | I can explain: I t's enough to show that the term $K/\xi\cdot K$ that you are refering to is just the image of $K_G(X\times D^n) \to K_G(X\times S^{n-1})$. Note first that $K_G(X\times D^n) =K$. The point then that this map is not injective: there's a kernel given by $\xi K$. Why? That is precisely explained in the paragraph above the exact sequence (**) on page 6. | |
| Mar 22, 2018 at 22:51 | comment | added | theinvariant | That makes sense to me. My only question is then on the notation they choose: when they write $_{\xi}K:=\{x\in K:\xi\cdot x=0\}$ they're talking about a submodule of the $R(\mathbb{T})$-module $K^{\ast+1}_G(X)$, but if I understand correctly, when they write $K/\xi\cdot K$ at the start of this exact sequence, by $K$ they mean $K^{\ast}_G(X)$. Is this a mistake on their part or is there something I'm not getting? | |
| Mar 22, 2018 at 1:46 | history | edited | John Klein | CC BY-SA 3.0 |
added 44 characters in body
|
| Mar 21, 2018 at 20:25 | comment | added | John Klein | The long exact sequence of the pair has a connecting homomorphism $K^\ast_G(X\times S^{n-1}) \to K^{\ast+1}_G(X\times D^n,X\times S^{n-1})$. The Thom isomorphism identifies the target of this with $K^{\ast-n+1}_G(X)$. In their case $n$ is even so the latter group is identified with $K^{\ast+1}_G(X)$ by periodicity. The image of the connecting map with respect to these identifications is what they are denoting by ${}_\xi K$. | |
| Mar 21, 2018 at 18:19 | comment | added | theinvariant | I see. This generates the following question for me: I understand the six-term exact sequence that Bott periodicity generates, but when they state on page 6 "Consider the exact sequence for the pair $(X\times D^n,X\times S^{n+1})$" and then use the Thom isomorphism, which as I understand it is defined in terms of the $0$th $K$-ring of an equivariant bundle over X, could someone tell me what comes after $K^*(X\times S^{n+1})$ in the sequence they point to? The paper I'm talking about can be found here projecteuclid.org/download/pdf_1/euclid.jdg/1214428815 | |
| Mar 21, 2018 at 18:03 | history | edited | John Klein | CC BY-SA 3.0 |
deleted 4 characters in body
|
| Mar 21, 2018 at 17:58 | history | edited | John Klein | CC BY-SA 3.0 |
deleted 4 characters in body
|
| Mar 21, 2018 at 17:56 | comment | added | LSpice | It stands for (depending on context) "the collection of groups $K_G^q(X)$" or "the same collection, assembled in some way into a single group" (e.g., as $\bigoplus_q K_G^q(X)$, the associated graded group). | |
| Mar 21, 2018 at 17:56 | history | edited | John Klein | CC BY-SA 3.0 |
added 214 characters in body
|
| Mar 21, 2018 at 17:54 | comment | added | theinvariant | Thank you, I suspect you're right. In this case what does $K^*_G(X)$ signify? | |
| Mar 21, 2018 at 17:46 | history | answered | John Klein | CC BY-SA 3.0 |