Timeline for Defining Transfers Algebraically
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2013 at 5:36 | comment | added | Steven Landsburg | Tor is commutative, so you can compute homology as either $Tor(−,{\mathbb Z})$ or $Tor({\mathbb Z},-)$. Unless I'm badly mistaken, I'm computing one of these and you're computing the other, and either yields a perfectly good argument. | |
| Mar 16, 2013 at 3:33 | comment | added | nerses | On homology level we get a map $\mathrm{H}_n(G;M)\rightarrow \mathrm{H}_n(H;M)$, since $P_\bullet$ is a projective resolution of $M$ as a left $\mathbb{Z}[H]$-module. | |
| Mar 16, 2013 at 3:30 | comment | added | nerses | Actually a variation of what you just said works. The map $\mathbb{Z}\rightarrow \mathbb{Z}\otimes_{\mathbb{Z}[H]}\mathbb{Z}[G]$ that sends $1$ to $1\otimes t$, where $t$ is a sum of coset representatives, is a right $\mathbb{Z}[G]$-homomorphism. Let $P_\bullet$ be a left projective resolution of $M$. Then we get a map of chain complexes: $\mathbb{Z}\otimes_{\mathbb{Z}[G]}P_\bullet\rightarrow \mathbb{Z}\otimes_{\mathbb{Z}[H]}\mathbb{Z}[G]\otimes_{\mathbb{Z}[G]} P_\bullet\simeq \mathbb{Z}\otimes_{\mathbb{Z}[H]} P_\bullet$. | |
| Mar 16, 2013 at 2:23 | comment | added | Steven Landsburg | Agh! I believe I misled you earlier. The map from $M$ to ${\mathbb Z}G\otimes_{{\mathbb Z}H}M$ is in fact $m\mapsto 1\otimes m$, just as you originally said, and not $m\mapsto t\otimes m$ as I said. I hope I didn't cause you to waste too much time on this. | |
| Mar 15, 2013 at 22:41 | comment | added | nerses | Suppose $H=\{1\}$ the trivial subgroup of $C_2$ ($\alpha$ is its generator) and let $M=\mathbb{Z}[C_2]$. Then the map $\mathbb{Z}[C_2]\rightarrow \mathbb{Z}[C_2]\otimes_{\mathbb{Z}}\mathbb{Z}[C_2]$, sends $1$ to $(1+\alpha)\otimes 1$ and $\alpha$ to $(1+\alpha)\otimes \alpha$. This is not a left $\mathbb{Z}[C_2]$-module map. | |
| Mar 13, 2013 at 17:04 | comment | added | Steven Landsburg | Apologies; an earlier version of this comment mixed up rights and lefts. I've repaired that and am deleting the original. I'm not sure what's confusing you. The G-module structure on ${\mathbb Z}G\otimes_{{\mathbb Z}H}M$ is given by multiplication on the left. (Were you maybe thinking it was on the right?) So ${\mathbb Z}G\otimes_{{\mathbb Z}H}M$ is a direct sum of $(G:H)$ copies of M, permuted by the elements of $G$. The element $t\otimes m$ corresponds to the element $(m,m,\ldots m)$. (This does assume $(G:H)$ is finite, but I don't think that's where you're having trouble.) | |
| Mar 12, 2013 at 17:29 | comment | added | nerses | Come to think of it, there is a bit of a problem: I don't understand why the map $m$ sending $t\otimes m$ is a $G$-module and how we get the map between homologies. It is if we have trivial $G$-module $M$, I think, we do recover the right transfer map in this case. I check out the case, $*\hookrightarrow C_2$ with $M=\mathbb{Z}[C_2]$ and it does not seem to work. | |
| Mar 12, 2013 at 17:17 | vote | accept | nerses | ||
| Mar 12, 2013 at 17:21 | |||||
| Mar 12, 2013 at 4:29 | vote | accept | nerses | ||
| Mar 12, 2013 at 17:14 | |||||
| Mar 12, 2013 at 4:27 | comment | added | Steven Landsburg | PS: Along with the map in 1), there is also an obvious surjection going the other way. Apply all of the same reasoning to that surjection, and you'll get the usual map $f:H(H,M)\rightarrow H(G,M)$. If you first do the obvious injection and then the obvious surjection, the induced map on $M$ is multiplication by $(G:H)$. This in turn tells you what happens when you compose the transfer with $f$. | |
| Mar 12, 2013 at 4:22 | comment | added | Steven Landsburg | nerses: I'm not sure what you mean by "1". The element $m$ maps to $t\otimes m$ where $t$ is a sum of coset representatives for $G$ over $H$. (It doesn't matter how you pick the representatives.) | |
| Mar 12, 2013 at 4:18 | comment | added | nerses | By obvious you mean $m$ goes to $1\otimes m$? | |
| Mar 12, 2013 at 4:13 | history | undeleted | Steven Landsburg | ||
| Mar 12, 2013 at 4:13 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
deleted 66 characters in body
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| Mar 12, 2013 at 3:51 | history | deleted | Steven Landsburg | ||
| Mar 12, 2013 at 3:38 | history | answered | Steven Landsburg | CC BY-SA 3.0 |