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Update July 29, 2013.

I have still not found a good textbook for this topic, if you point one to me I will be grateful :) I have accepted BS's answer anyway, since their explanation was useful to me and gave me a good starting point for further research. I ended up finding a very helpful resource for this topic: the online notes of the course Introduction to Topological Groups, by Dikran Dikranjan, University of Udine. I recommend these notes to anyone interested on this topic, or on the Pontryagin-Van Kampen duality. The content of these notes has been partially published as An elementary approach to Haar integration and Pontryagin duality in locally compact abelian groups, Dikranjan and Stoyanov, Topology and its Applications, Volume 158, issue 15, 2011), p. 1942-1961, DOI: 10.1016/j.topol.2011.06.037, MR: 2825348, Elsevier Science.

A popular concept in quantum computation, used extensively to design algorithms for finite-abelian-groups, are the so-called orthogonal subgroups

Let $G=\mathbb{Z}_{d_1}\times\ldots\times\mathbb{Z}_{d_m}$ be a finite abelian group, the orthogonal subgroup $H^{\perp}$ of $H$ a subgroup of $G$ is defined as:

$$H^\perp:=\lbrace g\in H : \chi_g(h)=1 \quad\text{for all } h \in H\rbrace,$$

where $\chi_g$ are the characters of $G$: $$ \chi_g(h) = \exp{\left(2\pi \sum_{i=1}^{m} g_i h_i/d_i \right)} \quad \text{for all } \quad g, h \in G $$

Given two subgroups $H$ and $K$, basic Character Theory allows one to quickly derive

\begin{matrix} (1)\ H^{\perp\,\perp} = H & (2)\ |H^{\perp}| = |G|/|H| \\ (3)\ H\subset K \iff K^{\perp}\subset H^{\perp} & (4)\ (H\cap K)^{\perp} = \langle H^{\perp} , K^{\perp} \rangle \end{matrix}

Question.

This structure is extensively used in some important quantum algorithms and appears in quite a bunch of relatively-recent research papers. Yet, and though it looks pretty basic, I can not find some standard textbook where this is defined and that includes proofs of propositions (1-4). I would like to find such a reference since I often discuss these concepts with people not fluent with Character or Group theory. Also, I would like to know if the name "orthogonal-subgroup" is used by mathematicians.

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  • $\begingroup$ What is $\chi_g$? Also, you can write $+$ for the join, since the groups are abelian and thus the join is just the sum of $\mathbb Z$-submodules. $\endgroup$
    – darij grinberg
    Commented Feb 27, 2012 at 14:09
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    $\begingroup$ These are quite easy facts which anyone who knows about the rudiments of character theory of finite groups could prove in a few minutes. I coud not say when they were first explictly noted, but it is just a question of writing down the definitions, so I would be surprised if much older references do not exist. $\endgroup$
    – Geoff Robinson
    Commented Feb 27, 2012 at 15:44
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    $\begingroup$ Perhaps the key word is not orthogonal subgroups, but lattices and dual lattices. $\endgroup$
    – Marc Palm
    Commented Feb 27, 2012 at 18:21
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    $\begingroup$ Dear Juan, Here is the standard way to deduce (4): Consider the homomorphism $H \times K \to G$ given by $(h,k) \mapsto h - k$. The kernel is precisely $H \cap K$ (embedded diagonally inside $H \times K$). Passing to duals, obtain a right exact sequence $$\hat{G} \to \hat{H} \times \hat{K} \to \widehat{H\cap K} \to 0,$$ which we may rewrite as $$\hat{G} \to \hat{G}/H^{\perp} \times \hat{G}/K^{\perp} \to \hat{G}/(H\cap K)^{\perp} \to 0.$$ This gives $(H\cap K)^{\perp} = \langle H^{\perp}, K^{\perp}\rangle $. Many texts on character theory for abelian groups will contain the basic exactness $\endgroup$
    – Emerton
    Commented Jul 30, 2013 at 4:45
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    $\begingroup$ ... properties that I used here (in particular, the exact sequence $$0 \to H\cap K \to H \times K \to G$$ converted to an exact sequence in the other direction once I took duals), and property (4) is a basic consequence of them. Unfortunately, texts don't always list all the basic consequences of the properties that they state ... . Regards, Matthew $\endgroup$
    – Emerton
    Commented Jul 30, 2013 at 4:48

3 Answers 3

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These results belong to what is called duality in finite abelian groups, a theory that has been generalized by Pontryagin and others in the 30's to locally compact abelian groups.

Another keyword here is "Discrete Fourier Transform", although it is mainly used for cyclic groups (of order $2^N$ for FFT) or $GF(2)$ vector spaces (analysis of boolean functions).

It is also a chapter in the representation theory of finite groups, founded by Frobenius in the end of 19th century, namely the case of finite abelian groups, where the irreducible characters are all homomorphisms to $S^1$. This case was already known to Gauss and Dirichlet.

By the way, the orthogonal you define is more naturally seen as a subgroup of the dual group $\hat{G}=Hom(G,S^1)$, which in (multiplicative) duality with $G$ via the evaluation $G\times \hat{G}\to S^1$. It is isomorphic to $G$, but not naturally so.

I would bet that almost any book on either Pontryagin duality or representation theory of finite groups has the results you want to cite, but I have only been able to find this link to an online encyclopedy, which easily implies your claims, since 3 and 4 follow quickly from 1 and 2.

Hope this helps.

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  • $\begingroup$ Thanks for your answer. I have spent the last months learning about LCA groups and I have found some references for most of the above properties (except 4, curiously). I am still looking for a good textbook though: Morris, "Pontryagin duality and the structure of locally compact Abelian groups" is a good reference for this topic, yet somewhat old; I studied this topic mostly by reading the online lecture notes ``Introduction to Topological Groups" by Dikran Dikranjan, which have been only partially published. $\endgroup$
    – Juan Bermejo Vega
    Commented Jul 29, 2013 at 9:30
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In today's updates you ask specifically about published textbook treatments of duality for finite abelian groups, including in particular a proof of (4). That may be found in B. Huppert's books,

Endliche Gruppen. I (1967), §V.6 "Charaktere abelscher Gruppen" (pp. 487-490);

Character theory of finite groups (1998), §5 "Characters of abelian groups" (pp. 58-65).

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If $g_i$ with $i=1\ldots n$ is a (minimal) set of generators of $G$ and $\chi_g(h)=:\langle g,h\rangle$ is considered as a scalar product $G\times G\rightarrow k^\times$ (see BS's answer about Pontryagin-Duality!), you get a natural braiding $\tau_{ij}(x_i\otimes x_j)=q_{ij}(x_j\otimes x_i)$ with braiding matrix $q_{ij}:=\chi_{g_i}(g_j)$.

This turn the vector space $X$ with basis $x_1\ldots x_n$ into a braided vectorspace, the same as induced by the $G$-grading $x_i\mapsto g_i\otimes g_j$ and $G$-action $g_i\otimes x_j\rightarrow q_{ij}x_j$ (Yetter-Drinfel'd module)

This braiding can correspond via $\tau^2=q_{ij}q_{ji}\stackrel{!}{=}q_{ii}^{A_{ij}}$ to a Cartan matrix / Dynkin diagram $A_{ij}$ of a semisimple Lie algebra (or something more exotic, see (Nichols Algebra, that gets finite in these cases!)

Your case in question $\tau^2|_{V\otimes W}=id$ for some $U,V\subset X$ corresponds to

  • $U\perp V$ in the scalar product $\langle x_i,x_j\rangle=q_{ij}$ in the beginning.

  • two subspaces with a braiding $\tau^2=1$ being a symmetry (antisymmetry...)

  • a decomposition of the Dynkin diagram into mutuall disconnected components

If you have plain bosons or fermions with $q_{ij}=q_{ji}=\pm 1$, all points are disconnected! ($A_{i\neq j}=0$), but for anyons in quantumcomputing there is more going on ;-)

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