Timeline for Induced map on $H_4$ of Eilenberg–MacLane spaces
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 29, 2021 at 12:43 | vote | accept | Chris Schommer-Pries | ||
| Sep 29, 2021 at 11:45 | answer | added | Bad English | timeline score: 3 | |
| Sep 28, 2021 at 13:06 | comment | added | Chris Schommer-Pries | Thank you everyone! I don't want to leave this question open. @BadEnglish or someone else should make a brief answer, so I can accept it! | |
| Sep 28, 2021 at 13:04 | comment | added | Chris Schommer-Pries | @TylerLawson Thank you. I think I understand my confusion. I guessed this formula before the OP. But then I tried applying it to $a \wedge b \wedge a \wedge d$, i.e. $a=c$. I should get zero. But if you directly apply the formula you get $$\gamma(a \wedge b + a \wedge d) + \gamma(-a \wedge b + a \wedge d) - 2\gamma(a \wedge b) - 2 \gamma(a \wedge d)$$and that doesn't look at all like zero. However I now gather that actually it is zero after all. The key is to verify that $x \cdot (-y) = \gamma(x -y) - \gamma(x) - \gamma(-y) = -\gamma(x + y) + \gamma(x) + \gamma(y) = - (x \cdot y)$. | |
| Sep 28, 2021 at 12:56 | comment | added | Chris Schommer-Pries | @BadEnglish The answer to your first comment is yes. $H_4(K(B,2); \mathbb{Z}) = \Gamma(B)$ for all $B$. This is a Theorem of Whitehead from the 50's. However $\Gamma(\mathbb{Z}/p^a) = \mathbb{Z}/p^b$ with $b=a$ for any odd prime and $b= a +1$ for $p=2$. | |
| Sep 28, 2021 at 8:54 | comment | added | Bad English | @ChrisSchommer-Pries, the isomorphism with the second divided power is induced by $\gamma(x)\to \frac{x^2}{2}=\gamma_2(x)$, hence the product in divided powers is given by the formula in Tyler's comment. But the torsion case leaves me confused. Could you please reply on my first comment? I have some inconsistency with my current knowledge on homology of EM-spaces. I claim that $H_4(K(\mathbb{Z}/p^a,2),\mathbb{Z})=\mathbb{Z}/p^b$ with $b=a$ for a prime $p>2$ and with $b=a+1$ for $p=2$. This follows from BSS done by May. Contrary, the functorial answer predicts $\mathbb{Z}/p^{2a}$, right? | |
| Sep 28, 2021 at 3:44 | comment | added | Tyler Lawson | @ChrisSchommer-Pries If I understand the identification correctly, $x \cdot y = \gamma(x+y) - \gamma(x) - \gamma(y)$, and so $a \wedge b \wedge c \wedge d$ goes to: $$ \gamma(a \wedge b + c \wedge d) - \gamma(a \wedge c + b \wedge d) + \gamma(a \wedge d + b \wedge c) - \gamma(a \wedge b) - \gamma(c \wedge d) + \gamma(a \wedge c) + \gamma(b \wedge d) - \gamma(a \wedge d) - \gamma(b \wedge c) $$ I think the method suggested by DylanWilson and BadEnglish applies to the universal example when $A = \Bbb Z^4$ and $B = \Lambda^2 \Bbb Z^4$, enough to show this formula is correct in general. | |
| Sep 28, 2021 at 0:20 | comment | added | Chris Schommer-Pries | @BadEnglish Yes, this all holds for non-free B, and the map is functorial in A and B (A is required to be free, but the maps need not be free). How does your formula in your second comment translate into the $\gamma(b)$'s? | |
| Sep 27, 2021 at 23:29 | history | edited | LSpice | CC BY-SA 4.0 |
Name of paper; \wedge -> \bigwedge
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| Sep 27, 2021 at 21:59 | comment | added | Bad English | also, at least in the case $B$ is torsion-free the group $\Gamma(B)$ identifies with divided powers in degree 2, i.e., $\Gamma^2(B)$. In the universal example we can send $a\wedge b\wedge c \wedge d$ to $a\wedge b\cdot c\wedge d- a\wedge c\cdot b\wedge d+a\wedge d\cdot b\wedge c$ | |
| Sep 27, 2021 at 21:33 | comment | added | Dylan Wilson | In the universal case everything is torsion-free, so you could look at cohomology instead. But then the map is determined by what happens on H^2 since the source is a polynomial algebra on stuff in H^2. | |
| Sep 27, 2021 at 21:29 | comment | added | Bad English | Does this statement holds for non free $B$? (it is not clear from the paper you mentioned) Is it true that $\Gamma(\mathbb{Z}/p^a)=\mathbb{Z}/p^{2a}$? It seems to be bigger than what I expected of $H_4(K(\mathbb{Z}/p^a,2),\mathbb{Z})$... | |
| Sep 27, 2021 at 21:19 | comment | added | Chris Schommer-Pries | Note: One can consider the universal case that $B = \wedge^2 A$, and they we are asking for a natural map $\wedge^4(A) \to \Gamma(\wedge^2A)$. | |
| Sep 27, 2021 at 19:37 | history | asked | Chris Schommer-Pries | CC BY-SA 4.0 |