4

Let $G$ be any abelian group, and let $\triangle_G = \{(g,g)\mid g\in G\}\subset G\times G.$

Is there any significance in studying the quotient group ${(G\times G)}\left/{\triangle_G}\right.?$ If so, where does the study such an object naturally occur?

Motivation for such a question: Let $F$ be a field not of characteristic 2. Let $F^\ast=F\setminus\{0\},$ be the multiplicative group of units of $F.$ Define $G=F^\ast/(F^\ast)^2.$

The motivation for such a question comes from the study of maps $q:G\times G \longrightarrow Br(F),$ where $Br(F)$ denotes the Brauer group of $F.$ If $-1$ is a square in $F,$ that is if $-1\in (F^\ast)^2,$ then $q(x,x)=0\in Br(F)$ for all $x\in G.$ Since $q(x,x)=0,$ this motivates why I'd like to mod out by $\triangle_G.$

Example: In the event that $F=\Bbb{Q}_2[\sqrt{-1}]$ we have that $Br(F)\cong \Bbb{Z}/2\Bbb{Z},$ and we can view $G$ as a 4-dimensional vector space over $\Bbb{Z}/2\Bbb{Z}.$ One such map $q:G\times G \longrightarrow \Bbb{Z}/2\Bbb{Z}$ is the symplectic bilinear form. Another is the constant map $q := 0\in Br(F),$ although this one is not very interesting.

Chickenmancer
  • 4,470

1 Answers1

4

The group $H={(G\times G)}\left/{\triangle_G}\right.$ is just canonically isomorphic to $G$ itself. Indeed, the homomorphism $f:G\times G\to G$ defined by $f(g,h)=g-h$ is surjective and has $\triangle_G$ as its kernel, and thus induces an isomorphism from $H$ to $G$.

Eric Wofsey
  • 342,377
  • Would you happen to know under exactly what circumstance one might study this object, though? The map which carries $(g,h)\mapsto g-h$ would more generally be the assignment $(g,h)\mapsto gh^{-1}.$ But where might $g-h$ come up naturally? – Chickenmancer May 14 '18 at 22:13
  • Um, any time you want to study subtraction on a group? That arises naturally whenever you want to study whether two group elements are equal, by subtracting them and seeing if you get $0$. – Eric Wofsey May 14 '18 at 23:49
  • Could you be more specific? Is there something in terms of a trace map, possibly? I'm looking for context, rather than simply measuring when elements are equal. – Chickenmancer May 15 '18 at 18:40
  • I think you're vastly overthinking this. The map $(g,h)\mapsto g-h$ literally comes up ALL the time, whenever you compare two group elements. People just don't usually talk about the quotient $(G\times G)/\triangle_G$ because they immediately identify it with $G$ itself. – Eric Wofsey May 15 '18 at 18:46
  • To put it another way, if you encountered this quotient in some context and are looking for other contexts which it might be connected to, I think that is an unwise line of inquiry, since there's nothing unusual or interesting about this quotient. – Eric Wofsey May 15 '18 at 18:49
  • I'm not really familiar with the specific context you mentioned in the question, but I'm not sure $(G\times G)/\triangle_G$ is even relevant to it: it sounds like $q$ would be a bilinear map, not a linear map. (So the relevant quotient would instead be $G\otimes G$ modulo a certain "diagonal" subgroup.) – Eric Wofsey May 15 '18 at 18:52
  • There are contexts where the map $q$ is bilinear, and yes, there is a construction of such a map where one takes $G\otimes G$ and cuts away the symmetric tensors (this might fit the "diagonal" condition). However, my interest is much more general, and I was hoping to see if there was a necessity for such a study. I understand that this group is isomorphic to $G;$ however, sometimes simply looking at what it is isomorphic to may hide some information, and I had hoped to see the where one might look at this type of map, in hopes of uncovering techniques I had not considered. – Chickenmancer May 15 '18 at 19:25
  • @EricWofsey: What if $G$ is non-abelian? Does this quotient still isomorphic to $G$? – Bumblebee Jul 04 '22 at 12:45
  • 1
    @Bumblebee: If $G$ is nonabelian then $\triangle_G$ is not normal in $G\times G$ so the quotient does not make sense. – Eric Wofsey Jul 04 '22 at 12:50