Explicit equation of Dickson invariant / quasideterminant / special orthogonal group over the integers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:47:57Z http://mathoverflow.net/feeds/question/114253 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114253/explicit-equation-of-dickson-invariant-quasideterminant-special-orthogonal-gr Explicit equation of Dickson invariant / quasideterminant / special orthogonal group over the integers Gro-Tsen 2012-11-23T15:47:29Z 2013-03-19T14:13:40Z <p>Consider $2n$ coordinates $x_1,\ldots,x_n,y_1,\ldots,y_n$ and the quadratic form $q = \sum_{i=1}^n x_i y_i$. Now call $O(q,A)$ (orthogonal group of $q$) the group of $(2n)\times(2n)$ matrices, with coefficients in a commutative ring $A$, which preserve $q$. (This is an algebraic group over $\mathop{\mathrm{Spec}}\mathbb{Z}$.) When $A$ is a field of characteristic $\neq 2$, the determinant restricted to $O(q,A)$ takes its values in ${\pm 1}$, its kernel $O(q,A) \cap SL(2n,A)$ defines a subgroup $SO(q,A)$ of index $2$. When $A$ is a field of characteristic $2$, the determinant is identically $1$ on $O(q,A)$ but there is still a subgroup of index $2$, which one might still denote $SO(q,A)$, defined by the so-called "<a href="http://en.wikipedia.org/wiki/Orthogonal_group#The_Dickson_invariant" rel="nofollow">Dickson invariant</a>", also known as quasideterminant or pseudodeterminant, and it is relatively straightforward to give an explicit polynomial in the coefficients of $A$ which defines an equation of $SO(q,A)$ inside $O(q,A)$ (see, Dickson's book, <a href="http://archive.org/details/lineargroupswith00dickuoft" rel="nofollow"><em>Linear Groups</em></a>, theorem 205 on page 206).</p> <p>Now for an arbitrary ring $A$, there is still a subgroup $SO(q,A)$ of $O(q,A)$, which is the kernel of a morphism $\deg$ of $O(q,A)$ to the group $(\mathbb{Z}/2\mathbb{Z})(A)$ of idempotents of $A$ (so, when $A$ is connected, $SO(q,A)$ is a subgroup of order $2$) natural in $A$ and which coincides with $\frac{1}{2}(1-\det)$ when $2$ is invertible in $A$ and with Dickson's invariant when $2$ is zero in $A$. This is due to H. Bass ("Commutative Algebras and Spinor Norms over a Commutative Rings", Amer. J. Math. <strong>96</strong> (1974), 156–206).</p> <p>Concretely, this means that there exists a polynomial $\deg$ in $4n^2$ variables over $\mathbb{Z}$ such that, modulo the ideal $I$ of relations defining $O(q,A)$, we have $\deg^2 = \deg$ and $\det = 1-2\deg$. And $I_0 := I+(\deg)$ is the ideal defining $SO(q,A)$.</p> <p>My question is: can one give an <em>explicit</em> expression of $\deg$ (as a polynomial in $4n^2$ variables), or perhaps an explicit set of equations of $SO(q,A)$ (e.g., Gröbner basis of $I_0$ for some term order)? (At least for the particular quadratic form $q = \sum_{i=1}^n x_i y_i$ if not in general.)</p> http://mathoverflow.net/questions/114253/explicit-equation-of-dickson-invariant-quasideterminant-special-orthogonal-gr/114324#114324 Answer by Matthieu Romagny for Explicit equation of Dickson invariant / quasideterminant / special orthogonal group over the integers Matthieu Romagny 2012-11-24T10:46:00Z 2012-11-24T10:46:00Z <p>In fact, in order to define <code>$\deg=\frac{1}{2}(1-\det)$</code> it is enough that <code>$2$</code> be a nonzerodivisor in <code>$A$</code>. In particular, since the orthogonal groups are defined over <code>$\mathbb{Z}$</code> i.e. come from the orthogonal groups over <code>$\mathbb{Z}$</code> by the base change <code>$\mathbb{Z}\to A$</code>, it is ok to define <code>$\deg$</code> when <code>$A=\mathbb{Z}$</code>.</p> <p>This may also be phrased, more cleanly in my opinion, in terms of 'dilatations'. You may have a look at section 4.1.2 in the paper <a href="http://perso.univ-rennes1.fr/matthieu.romagny/articles/adjoint_quotient.pdf" rel="nofollow">On the adjoint quotient of Chevalley groups over arbitrary base schemes</a> joint with Pierre-Emmanuel Chaput.</p> http://mathoverflow.net/questions/114253/explicit-equation-of-dickson-invariant-quasideterminant-special-orthogonal-gr/124963#124963 Answer by Olivier Benoist for Explicit equation of Dickson invariant / quasideterminant / special orthogonal group over the integers Olivier Benoist 2013-03-19T14:13:40Z 2013-03-19T14:13:40Z <p>Hi Gro-Tsen ! The following is the outcome of a discussion with Olivier Taïbi.</p> <p>Let V be the free module of rank $2n$ over $\mathbb{Z}$ with basis $f_1,\dots,f_n,g_1,\dots,g_n$ and equipped with the split quadratic form $q=\sum_{i=1}^nx_iy_i$. Let $C(V,q)$ be the Clifford algebra of $(V,q)$ and $C^+(V,q)$ be its even part. The orthogonal group $O(V,q)$ acts on both. Moreover, $C^+(V,q)$ is the product of two matrix algebra, hence an element of the orthogonal group either preserves these factors or exchanges them. This gives a morphism $\deg:O(V,q)\to \mathbb{Z}/2\mathbb{Z}$ that is precisely the Dickson invariant we want to compute. [For all these facts, see for instance Conrad's notes <a href="http://math.stanford.edu/~conrad/252Page/handouts/O(q).pdf" rel="nofollow">http://math.stanford.edu/~conrad/252Page/handouts/O(q).pdf</a>, Lemma 1.4 and around.]</p> <p>Now, $C^+(V,q)$ has two non-trivial projectors (one on each factor), so that an element of the orthogonal group will have Dickson invariant $0$ (resp. $1$) if and only if it preserves these projectors (resp. it exchanges them). When $n=1$, it is not difficult to compute these projectors: one is given by $e=f_1g_1\in C^+(V,q)$ (and the other is $1-e$). It is then possible to compute inductively these projectors, using for instance the formula [SGA7 II Exp. XII (1.10.1)]. When $n=2$, one of the projectors is given by $e=f_1g_1+f_2g_2+2f_1f_2g_1g_2$ (and the other is $1-e$). By induction, it is possible to give a closed formula as follows. If <code>$I=\{i_1&lt;\dots &lt;i_k\}$</code>, let <code>$f_I:=f_{i_1}\dots f_{i_k}$</code> and <code>$g_I:=g_{i_1}\dots g_{i_k}$</code>. Then one of the projectors is given by: <code>$$e=-\sum_{k\geq 1}(-1)^{\frac{k(k+1)}{2}}2^{k-1}\sum_{I\subset\{1,\dots,n\},|I|=k}f_Ig_I.$$</code></p> <p>To obtain a formula for the Dickson invariant of $u\in O(V,q)$, compute $u(e)$ in $C(V,q)$. By the above, it is equal to $e$ if $\deg(q)=0$ and $1-e$ if $\deg(q)=1$. Hence, it suffices to write down $u(e)$ in the basis $(f_Ig_J)$ of $C(V,q)$: the coefficient of <code>$Id=f_{\varnothing}g_{\varnothing}$</code> will be $\deg(u)$. This gives an explicit polynomial expression for $\deg(u)$ in the matrix coefficients of $u$. </p> <p>In characteristic $2$, the fact that $e=\sum_{i=1}^nf_ig_i$ allows to recover the very simple expression of the Dickson invariant in characteristic $2$. When $n=2$, it is not difficult to use this to write down by hand an expression for the Dickson invariant over $\mathbb{Z}$. If $a,b,c,\dots, n,o,p$ are the matrix coefficients of $u$ from left to right then up to bottom (sorry for the poor notations), the Dickson invariant $\deg(u)$ is given by: $$ic+mg+jd+nh+2[(ib+mf)(kd+oh)+(id+mh)(jc+ng)-(jd+nh)(ic+mg)].$$</p> <p>It is also possible to obtain a (probably not very useful) closed formula from the procedure above by understanding how the $f_Ig_I$ term in the formula above contributes to the coefficient of $Id$ in $u(e)$. To write it down, let me introduce the following notation. If <code>$I\subset\{1,\dots, n\}$</code> has $k$ elements, <code>$\tilde{I}=I\cup(n+I)$</code> and $P_I$ is the set of partitions of <code>$\tilde{I}$</code> into $k$ pairs. To every such partition $P\in P_I$, it is possible to associate a natural sign $\varepsilon(P)$. Then: <code>$$\deg(u)=-\sum_{k\geq 1}(-1)^{\frac{k(k+1)}{2}}2^{k-1}\sum_{I\subset\{1,\dots,n\},|I|=k}\bigg(\sum_{P\in P_I}\varepsilon(P)\prod_{\{i&lt;j\}\in P}\sum_{k=1}^n u_{n+k,i}u_{k,j}\bigg).$$</code></p>