Timeline for Number of elementary abelian subgroups of extraspecial $2$-groups
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13 events
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| Oct 2, 2018 at 15:12 | comment | added | Pierre | This is going to sound pedantic, but you may enjoy looking into the following remark. The cohomology class of the extension of $G/Z(G)$ by $Z(G) = \mathbb{F}_2$ is an element of $H^2(G/Z(G), \mathbb{F}_2)$. Since $G/Z(G)$ is 2-elementary abelian, its cohomology ring is a polynomial ring; in particular, elements of $H^2(G/Z(G), \mathbb{F}_2)$ are homogeneous polynomials of degree 2. That's another way of saying "quadratic form"! and yes, it is in fact the same form in a different guise. Quillen's paper on extraspecial groups and spin groups is a wonder to read. | |
| Oct 1, 2018 at 13:24 | comment | added | benblumsmith | ... I see. Then $\beta(\bar u, \bar v) = -q(u)+q(u+v) - q(v) = u^{-2}uvuvv^{-2} = u^{-1}vuv^{-1} = [u,v^{-1}]$, and this is the same as $[u,v]$ because the characteristic is $2$. | |
| Oct 1, 2018 at 13:01 | comment | added | Derek Holt | Yes $q(vZ(G)) = v^2 \in Z(G)$. | |
| Oct 1, 2018 at 12:50 | comment | added | benblumsmith | @DerekHolt - Thank you, this is so clarifying! Do you mean $q(\bar v) = v^2$ for $v\in G, \bar v\in G/Z(G)$? (Taking you literally with $q(v) = v^2$ for $v\in G/Z(G)$ would yield a trivial form.) | |
| Oct 1, 2018 at 12:38 | comment | added | Derek Holt | @benblumsmith I am using this definition. A quadratic form of vector space $V$ over field $F$ is a map $q:V \to F$ such that $q(av)=a^2q(v)\,\forall a \in F,\,v \in V$; and the map $\beta:V \times V \to F$ defined by $\beta(u,v) = Q(u+v)-Q(u)-Q(v)$ is a bilinear form. In this case, with $F = {\mathbb F}_2$, we define $q(v) = v^2$ for $v \in G/Z(G)$, and the associated bilinear form $\beta$ is the commutator map. So if $W/Z(G)$ is a totally singular subspace, then $q(w)=0$ for all $w \in W$ translates to $w^2=1$, and also $[v,w]=1$ for all $v,w \in W$, so $W$ is elementary abelian. | |
| Oct 1, 2018 at 12:20 | comment | added | benblumsmith | @DerekHolt - I trust you completely, but I am confused by both claims. You and Pierre must be referring to a different quadratic form than I am thinking of? I assumed Pierre meant the commutator form $\mathbb{F}_2^n=G/Z(G)\rightarrow [G,G]=Z(G) = \mathbb{F}_2$ given by $\langle \bar a,\bar b\rangle =[a,b]$, where $a,b\in G$ represent $\bar a, \bar b\in G/Z(G)$. This is a skew form on a field of characteristic 2. All vectors satisfy $q(v)=0$, but many do not satisfy $\beta(v,w)=0$. | |
| Sep 30, 2018 at 11:06 | comment | added | Derek Holt | @benblumsmith a subspace $W$ of a space $V$ with a quadratic form $q$ is totally singular if $q(w)=0$ for all $w \in W$. That implies that it is also totally isotropic i.e. $\beta(v,w)=0\, \forall v,w \in W$, where $\beta$ is the associated bilinear form. As Pierre said, elementary abelian subgroups of $G$ that contain $Z(G)$ correspond exactly to totally isotropic subspaces of $G/Z(G)$. | |
| Sep 29, 2018 at 21:40 | comment | added | benblumsmith | (Aside: it seems to me the def. of "totally isotropic" given in the linked question differs from the Wikipedia def: en.wikipedia.org/wiki/Isotropic_quadratic_form. Does it mean all the vectors in the subsp. are orth. to each other, or just themselves?) | |
| Sep 29, 2018 at 21:38 | comment | added | benblumsmith | @Pierre - is this clear? If $H\subset G$ is a subgroup on which the commutator form vanishes, that just makes $H$ abelian, not nec. elem. abelian. E.g. the subgroup of $Q_8$ generated by $i$ corresponds to a subspace of $Q_8/Z(Q_8)$ with triv. commutator form, but it is not elem. ab. | |
| Sep 29, 2018 at 12:10 | comment | added | Pierre | It's the same thing as the number of totally isotropic subspaces of $G/Z(G)$, for the quadratic form defining the extension (= the cohomology class). For this, start with mathoverflow.net/questions/102615/… and follow the references. | |
| Sep 29, 2018 at 0:11 | history | edited | Marius Tarnauceanu | CC BY-SA 4.0 |
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| Sep 29, 2018 at 0:00 | review | First posts | |||
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| Sep 28, 2018 at 23:55 | history | asked | Marius Tarnauceanu | CC BY-SA 4.0 |