Editing Character group (section)

== Definition ==
If ''G'' is an abelian group, then the set of characters ''f<sub>k</sub>'' forms an abelian group under pointwise multiplication. That is, the product of characters <math>f_j</math> and <math>f_k</math> is defined by <math>(f_j f_k)(g) = f_j(g) f_k(g)</math> for all <math>g \in G</math>.  This group is the '''character group of ''G''''' and is sometimes denoted as <math>\hat{G}</math>.  The identity element of <math>\hat{G}</math> is the principal character ''f''<sub>1</sub>, and the inverse of a character ''f<sub>k</sub>'' is its reciprocal 1/''f<sub>k</sub>''.  If <math>G</math> is finite of order ''n'', then <math>\hat{G}</math> is also of order ''n''.  In this case, since <math>|f_k(g)| = 1</math> for all <math>g \in G</math>, the inverse of a character is equal to the [[complex conjugate]].

=== Alternative definition ===
There is another definition of character group<ref>{{Cite book|last=Birkenhake|first=Christina |title=Complex Abelian varieties|date=2004|publisher=Springer|author2=H. Lange |isbn=3-540-20488-1|edition=2nd, augmented |location=Berlin|oclc=54475368}}</ref><sup>pg 29</sup> which uses <math>U(1) = \{z \in \mathbb{C}^*: |z|=1 \}</math> as the target instead of just <math>\mathbb{C}^*</math>. This is useful when studying [[complex torus|complex tori]] because the character group of the lattice in a complex torus <math>V/\Lambda</math> is [[canonical isomorphism|canonically isomorphic]] to the dual torus via the [[Appell–Humbert theorem]]. That is,<blockquote><math>\text{Hom}(\Lambda, U(1)) \cong V^\vee\!/\Lambda^\vee = X^\vee</math></blockquote>We can express explicit elements in the character group as follows: recall that elements in <math>U(1)</math> can be expressed as<blockquote><math>e^{2\pi i x}</math></blockquote>for <math>x \in \mathbb{R}</math>. If we consider the lattice as a [[subgroup]] of the underlying [[real number|real]] [[vector space]] of <math>V</math>, then a homomorphism<blockquote><math>\phi: \Lambda \to U(1)</math></blockquote>can be factored as a map<blockquote><math>\phi : \Lambda \to \mathbb{R} \xrightarrow{\exp({2\pi i \cdot })} U(1)</math></blockquote>This follows from elementary properties of homomorphisms. Note that<blockquote><math>\begin{align}
\phi(x+y) &= \exp({2\pi i }f(x+y)) \\
&= \phi(x) + \phi(y) \\
&= \exp(2\pi i f(x))\exp(2\pi i f(y))
\end{align}</math></blockquote>giving us the desired factorization. As the group<blockquote><math>\text{Hom}(\Lambda,\mathbb{R}) \cong \text{Hom}(\mathbb{Z}^{2n},\mathbb{R})</math></blockquote>we have the isomorphism of the character group, as a group, with the group of homomorphisms of <math>\mathbb{Z}^{2n}</math> to <math>\mathbb{R}</math>. Since <math>\text{Hom}(\mathbb{Z},G)\cong G</math> for any abelian group <math>G</math>, we have<blockquote><math>\text{Hom}(\mathbb{Z}^{2n}, \mathbb{R}) \cong \mathbb{R}^{2n}</math></blockquote>after composing with the complex exponential, we find that<blockquote><math>\text{Hom}(\mathbb{Z}^{2n}, U(1)) \cong \mathbb{R}^{2n}/\mathbb{Z}^{2n}</math></blockquote>which is the expected result.