Let $A$ be an finite abelian group and $B$ be a subgroup of $A$. Then we defined the orthogonal of $B$ : $$B^{\perp} = \{f:(A,+) \to (\mathbb{Q}/\mathbb{Z},+) \mid \forall b \in B ,f(b) = 0 \}$$

I understand the definition, but I don't understand what it really represents. In linear algebra, the orthogonal of a sub-vector space has a very nice geometric meaning. Here I don't understand what it represents and why it's useful?

Hence here I am really seeking for intuition. For example, if I take famous groups like the dihedral group of the symmetric group and take a subgroup of one of these group. Then what the orthogonal will represent?

I we think of a group as a set of transformations that can act on some set $X$ maybe the orthogonal has a more geometric meaning?

Finally, why do we take morphism that go to the group : $\mathbb{Q}/ \mathbb{Z}$ ? I know this has a link with representation theory but isn't there a simple explanation to take this group ?

Thank you.


Note that you can't really take the dihedral or symmetric groups, as this is about abelian groups.

Here's an interesting analogy : for a vector space $V$, you have a dual vector space $V^*:= \hom(V,k)$, and if $W\leq V$ is a subvector space, then $W^\bot\leq V^*$ is a subvector space. Here, for an abelian group $A$ you have a dual abelian group $\check{A} := \hom (A, \mathbb{Q/Z})$, and given $B\leq A$ a subgroup you have $B^\bot \leq \check{A}$.

Essentially it will be as useful as $V^*,W^\bot$ are in the context of vector spaces.

The reason why $\mathbb{Q/Z}$ is that, just as $k$ is a cogenerator in the category of vector spaces, $\mathbb{Q/Z}$ is one in the category of abelian groups. What does it mean ? Here, essentially, it reduces to the fact that given $x\in A\setminus\{0\}$, there is a map $A\to \mathbb{Q/Z}$ such that $f(x) \neq 0$, this is because $\mathbb{Q/Z}$ is injective, i.e. given a subgroup $B\leq A$ and a map $f:B\to \mathbb{Q/Z}$, there is a map $g:A\to \mathbb{Q/Z}$ extending $f$.

Then for $x\neq 0 \in A$, either $x$ has a finite order $n$, in which case $x\mapsto \frac{1}{n}$ mod $\mathbb{Z}$ gives a map $\langle x\rangle \to \mathbb{Q/Z}$ which sends $x$ to something nonzero; either $x$ has infinite order, in which case you can send it to anything nonzero in $\mathbb{Q/Z}$; in either case there is a nonzero $f:\langle x\rangle \to \mathbb{Q/Z}$, which can then be extended to $A$.

So $\check{A}$ separates points of $A$, and so plays a similar role to $V^*$ in the context of $k$-vector spaces.

  • $\begingroup$ Thank you very much for your answer. You are completely right for the symmetric and dihedral group, that was a stupid question on my part. I think I understand better why we use the group $\mathbb{Q}/\mathbb{Z}$. Nevertheless, I don't understand how it gives a geometric/intuition about the orthogonal of a group. I already knew your analogy, but I don't understand how it helps understanding the orthogonal of a subgroup. Also I don't understand your last sentence : "... separates points of $A$". What separates points means and how did you conlude that ? (so there is an intuition about the dual?) $\endgroup$ – auhasard Dec 2 '18 at 18:46
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    $\begingroup$ It doesn't really give geometric intuition, unless you already have some concerning the orthogonal of a subvector space. If you do, then you can think of it as being similar, so it does give some geometric insight. $\check{A}$ separates points of $A$ means if $x\neq y$, then there is some $f\in\check{A}$ such that $f(x) \neq f(y)$, and it follows directly from what I mentioned. $\endgroup$ – Max Dec 2 '18 at 18:53
  • $\begingroup$ +1Thank you. The problem is that for me the orthogonal of a subvector space $W$ is:$\{x \in V \mid B(x,y) = 0 \forall y \in W\}$ where $B$ is a bilinear form hence it's composed of vectors not of functions whereas the orthogonal of a subgroup is a set of functions. Hence for me the orthogonal of a s.v.s $W$ is really just the space orthogonal to $W$ in $V$, so geometricaly this is very intuitive. For the orthogonal of a group it doesn't seem like I can relate the element of $B^\perp$ with the one of $B$ since there are different (one are morphisms whereas in $B$ we just have element of a g $G$ $\endgroup$ – auhasard Dec 2 '18 at 19:02
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    $\begingroup$ That's a version of the orthogonal when you are given a bilinear form $B$. Similarly if you were given a bilinear map $A\times A\to \mathbb{Q/Z}$, you could define the orthogonal subgroup as a subgroup of $A$. But for a general vector space $V$, with no given bilinear form, the orthogonal is indeed a subvector space of $V^*$. To relate the two, note that if you are given a bilinear form $B$, you have a clear map $V\to V^*$ (analogously $A\to \check{A}$) defined by $j: x\mapsto (y\mapsto B(x,y))$, and then the orthogonal subvector space of $W\leq V$ is $j^{-1}(W^\bot)$, where $W^\bot\leq V^*$ $\endgroup$ – Max Dec 2 '18 at 19:10
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    $\begingroup$ (and similarly $j^{-1}(B^\bot)$, where $B^\bot \leq \check{A}$). Here in some sense it's more general, it's a "universal" orthogonal because you can find any orthogonal defined by a bilinear form thanks to it $\endgroup$ – Max Dec 2 '18 at 19:11

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