## Finite, abelian, yet “fugitive” orthogonal subgroups

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 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 use 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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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. – darij grinberg Feb 27 2012 at 14:09
I clarified the notation for $\chi_g$, typically in our field one starts with a cyclic decomposition of $G$ and then the notation makes more sense. Is it clear now? Also, I removed the join entirely, since one can say the same thing using generating brackets, which are more common. – Juan Bermejo Vega Feb 27 2012 at 14:48
Actually, if I choose $G=\mathbf{Z}_d^m$, then it seems that the "orthogonal subgroup" is the syzygy of a module: ams.org/notices/200604/what-is.pdf – Juan Bermejo Vega Feb 27 2012 at 15:16
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. – Geoff Robinson Feb 27 2012 at 15:44
Perhaps the key word is not orthogonal subgroups, but lattices and dual lattices. – Marc Palm Feb 27 2012 at 18:21

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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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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