Timeline for On the determination of a quadratic form from its isotropy group
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 8, 2012 at 23:23 | comment | added | Hugo Chapdelaine | It is possible though to save my idea about using unicity of $\rho_v^F$ by saying that is is the unique $F$-isometrie such that $s^2=1$, $s(w)=-w$ for all $w\in\mathbf{C}v:=L$ and such for all $w$ which are $F$-perp to $L$ one has $s(w)=w$. | |
| Feb 7, 2012 at 18:55 | comment | added | Hugo Chapdelaine | Ok I was being stupid, one simply unfolds the equality $\rho_v^F(w_1)\cdot_G\rho_v^F(w_2)=w_1\cdot_G w_2$ in order to get the equality. I finally got it! | |
| Feb 7, 2012 at 18:48 | comment | added | Hugo Chapdelaine | @Robert, you are right, it is not unique. Then how do you get the equality $\bigl(v\ \cdot_F\ v\bigr)\bigl(v\ \cdot_G\ w\bigr) = \bigl(v\ \cdot_G\ v\bigr)\bigl(v\ \cdot_F\ w\bigr).$ | |
| Feb 7, 2012 at 18:27 | comment | added | Robert Bryant | @Hugo: Umm,...no. I didn't use uniqueness at all (and it's false anyway, since, for example, the map $v\mapsto -v $ has all of these properties, too). What I used was that, for each $v\in V$ with $F(v)\not=0$, there is an explicit formula for an element $\rho^F_v$ in $O(F)$ and that assuming that all of those particular elements also belong to $O(G)$ implies the relation $F(v)G(w)=F(w)G(v)$ for all $v,w\in V$ purely by algebra. Of course, it then follows that $\rho^F_v = \rho^G_v$. I guess you could argue directly that $\rho^F_v\in O(G)$ implies $\rho^F_v = \rho^G_v$, but I didn't do that. | |
| Feb 7, 2012 at 17:22 | comment | added | Hugo Chapdelaine | So in other words, give a line $L$ and a quadratic form $F$, there exists a unique linear map $s:\mathbf{R}^n\rightarrow\RR^{n}$ such that $s(w)=-w$ for all $w\in L$, $s^2=1$ and $s$ preserves $F$-length. So this uniqueness result is important in your argument. | |
| Feb 7, 2012 at 17:08 | comment | added | Hugo Chapdelaine | Also a key observation in your argument is that the reflection $\rho_v^F$ only depends on the $\mathbf{C}$-line spanned by $v$, namely $\mathbf{C}v$. moreover there exists a unique line $L$ such that for all $w\in L$ one has that $\rho_v^F(w)=-w$ namely the line spanned by $v$. So using this obervation with the assumption that $O(F)=O(G)$ one may deduce that $\rho_v^F=\rho_v^G$. | |
| Feb 7, 2012 at 16:01 | comment | added | Robert Bryant | @Hugo: Thanks, but, even though I don't remember a reference for it off the top of my head, I'm sure that this argument is classical. Note that it works over any field of characteristic not equal to $2$ and it does not assume that the vector space has finite dimension. If you look up references to the Cartan-Dieudonné Theorem, you'll probably find it written up explicitly (and more elegantly, I expect) in one of those references. | |
| Feb 7, 2012 at 12:45 | vote | accept | Hugo Chapdelaine | ||
| Feb 7, 2012 at 4:21 | history | answered | Robert Bryant | CC BY-SA 3.0 |