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May 18 at 13:39 history edited hyyyyy CC BY-SA 4.0
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May 17 at 17:37 vote accept hyyyyy
May 17 at 11:09 answer added Dave Benson timeline score: 5
May 16 at 19:33 history edited hyyyyy CC BY-SA 4.0
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May 16 at 19:07 comment added hyyyyy @DaveBenson I see. I missed the term coming from lower homology. For the $p=0$ term I think it's still 0 since it will be $\bigwedge^{i}(\mathbb{C}) \otimes_{\mathbb{Z}[\mathbb{C}^{*}]} \mathbb{Z}$ with the multiplication action but the $q=0$ term will not be zero.
May 16 at 19:02 history edited hyyyyy CC BY-SA 4.0
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May 16 at 17:57 comment added Dave Benson I don't see why that makes $E_{2,0}$ and $E_{0,2}$ equal to zero. For example, isn't $H_2(\mathbb{C},\mathbb{Q})$ equal to $\Lambda^2_{\mathbb{Q}}H_1(\mathbb{C},\mathbb{Q})$, hence infinite dimensional? So with $\mathbb{Z}$ coefficients it won't be zero.
May 16 at 17:52 comment added hyyyyy @DaveBenson $G$ can be represented as $\mathbb{C} \rtimes \mathbb{C}^{*}$ where the action of adjunction is given by multiplication. $\mathbb{C}$ is some direct product of $\mathbb{Q}$ and $\mathbb{C}^{*}$ is some copy of $\mathbb{Q}$ and one copy of $\mathbb{Q}/\mathbb{Z}$, so $E_{2,0}=E_{0,2}=0$ and $E_{3,0}=\mathbb{Q}/\mathbb{Z}$, so $H_{2}(G)$ is $E_{1,1}$.
May 16 at 17:02 comment added Dave Benson Then I don't understand your computation of $H_2(G,\mathbb{Z})$. Could you please explain.
May 16 at 13:59 comment added hyyyyy I edited the question for clarity.
May 16 at 13:57 history edited hyyyyy CC BY-SA 4.0
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May 16 at 13:32 comment added hyyyyy @DaveBenson They are discrete groups. I think otherwise they will not be called Schur multiplier?
May 16 at 7:28 comment added Dave Benson Are you computing homology as a discrete group or as a Lie group via the classifying space? They're very different. For example, what do you think $H_2(\mathbb{R},\mathbb{Z})$ is?
May 16 at 6:52 history edited hyyyyy
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May 16 at 6:45 history asked hyyyyy CC BY-SA 4.0