# Why do characters fail to characterize non-abelian LCH-groups?

For any locally compact Hausdorff abelian group (LCA group) $$A$$, a character $$\xi\colon A\mapsto\mathbb{T}$$ is definined as a continuous group homomorphism to the unit circle $$\mathbb{T}\subseteq\mathbb{C}$$. The space of characters is called the dual space of $$A$$ and is denoted $$\hat{A}$$. To my knowledge, characters are vital for Fourier analysis on abelian groups, e.g. establishing the homeomorphism $$\hat{A}\rightarrow \mathrm{sp}L^1(A),\,\xi\mapsto m_\xi$$ (where $$\mathrm{sp}$$ denotes the Gelfand spectrum and $$m_\xi(f):=\hat{f}(\xi)=\int_A f(x)\,\overline{\xi(x)}\,\mathrm{d}x$$) and the Pontryagin duality $$A\rightarrow\hat{\hat{A}},\,x\mapsto\delta_x$$ (where $$\delta_x(\xi):=\xi(x)$$).

For non abelian locally compact Hausdorff abelian group it seems to be more complicated, as representations (which I do not understand yet) are being used instead of characters. Are there clear examples why characters fail in the non abelian case?

• There are groups for which all nontrivial elements are conjugate to each other. Thus the character group is trivial, so it does not provide much information on the structure of the group. Sep 13 at 6:32

The short answer is that if $$G$$ is a group and $$A$$ an abelian group (like $$\mathbb T$$), then every group homomorphism $$\phi\colon G\to A$$ factors through the abelianization $$G/[G,G]$$. Here $$[G,G]$$ is the subgroup generated by commutators, i.e. elements of the form $$ghg^{-1}h^{-1}$$ with $$g,h\in G$$. Thus continuous group homomorphisms from $$G$$ to $$\mathbb T$$ can really only tell you things about the abelianization, which is not enough to understand the group itself. As an example, the free group $$\mathbb F_d$$ and the free abelian group $$\mathbb Z^d$$ have the same abelianization and thus the same group homomorphisms to $$\mathbb T$$.