Double orthogonal complement of a finite module - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:20:20Z http://mathoverflow.net/feeds/question/75262 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75262/double-orthogonal-complement-of-a-finite-module Double orthogonal complement of a finite module Carl 2011-09-13T00:30:38Z 2011-09-13T02:40:41Z <p>Crossposted from <a href="http://math.stackexchange.com/questions/64016/periodicity-of-the-nullspace-of-a-module" rel="nofollow">math.stackexchange</a> since I'm not getting any answer.</p> <p>Let $W$ be the finite $\mathbb{Z}$-module obtained from $\mathbb{Z}_q^n$ with addition componentwise where $\mathbb{Z}_q$ is the integers mod $q$. Let $V$ be a submodule of $W$. Let $V^{\perp} = \{w \in W \, : \, \forall v \in V \quad v.w = 0 \}$ where "." is the dot product. Is it true that ${(V^{\perp})}^{\perp} = V$ for all $q \geq 2$? If not, when is it the case?</p> http://mathoverflow.net/questions/75262/double-orthogonal-complement-of-a-finite-module/75266#75266 Answer by Pete L. Clark for Double orthogonal complement of a finite module Pete L. Clark 2011-09-13T01:16:12Z 2011-09-13T01:16:12Z <p>Your question does not specify what $q$ is. But if $q$ is an odd prime -- so that you are talking about quadratic spaces over the field $\mathbb{F}_q$ of odd characteristic -- then the answer is <strong>yes</strong>. </p> <p>In this case your inner product $\cdot$ is the bilinear form associated to the quadratic form $q(x_1,\ldots,x_n) = x_1^2 + \ldots + x_n^2$. This quadratic form is nondegenerate, so the result you want is Proposition 7 in <a href="http://math.uga.edu/~pete/quadraticforms.pdf" rel="nofollow">these notes</a>. (They are nothing so special: any sufficiently basic text on quadratic forms will contain this material.) </p> <p>I am not really used to thinking about quadratic forms either in characteristic $2$ or over rings which are not domains, so if you're really interested in the case of $q$ not necessarily an odd prime, please say so, so that someone else can give a more complete answer. (But I will guess that the result is also true when $q$ is an odd prime power, for instance.)</p> http://mathoverflow.net/questions/75262/double-orthogonal-complement-of-a-finite-module/75268#75268 Answer by Chris Godsil for Double orthogonal complement of a finite module Chris Godsil 2011-09-13T01:19:57Z 2011-09-13T01:19:57Z <p>The answer is yes. The easiest way for me is to appeal to character theory. If $\zeta$ is a complex $q$-th root of 1 then the map from $\mathbb{Z}_q^n$ to $\mathbb{C}$ given by $$x \mapsto \zeta^{a^Tx},\qquad (a\in\mathbb{Z}_q^n)$$ is a character of the abelian group $W=\mathbb{Z}_q^n$. The set of characters obtained as $a$ varies over the elements of $W$ is the character group <code>$W^*$ of $W$</code>. If $V$ is a subgroup of $W$, then <code>$V^\perp=V^*$</code> is isomorphic to the subgroup <code>$(W/V)^*$</code> of $W^*$.</p> <p>A convenient source for the relevant character theory is from our own KConrad: <code>http://www.math.uconn.edu/~kconrad/blurbs/</code> (under characters of finite abelian groups).</p>