I've taken courses previously on representation theory and Harmonic Analysis, and I'm working on some research which has taken me to studying Haar measures and integration over general locally compact groups in order to do Harmonic analysis.

In general real analysis, when discussing dual spaces, we generally define the dual of a function space (for instance $\mathcal{S}$) to be the set of bounded linear functionals $\ell : \mathcal{S} \mapsto \mathbb{R}$ (In this case, the set of tempered distributions). When reading some notes on Harmonic analysis over locally compact groups, we encounter the idea of Pontryagin duality. In the notes I'm reading, this is defined to be set of characters on $G$. So, for instance, $\hat{\mathbb{R}} = \{ e^{2\pi rx} : r \in \mathbb{R} \} \cong \mathbb{R}$.

As an insight to this, I know that in the case of finite $G$, the set of irreducible characters is a basis for the space of class functions of $G$. The set of class functions $f : G \mapsto \mathbb{C}$ is in fact a set of functionals on $G$. I'm unsure if it is the set of bounded linear functions, however. I'm also unsure if the same property of irreducible characters holds for locally compact $G$, but it seems to me that there is some similarity between these ideas of dual spaces.

My question is what is the relationship between these things? Are these the same dual? Can I use irreducible representations/characters (if I know them) on a locally compact group to identify the dual space in this sense?

  • 1
    $\begingroup$ For functions $f:G\to\Bbb C$ (btw, not "$\mapsto$"), "bounded linear" means nothing. $\endgroup$ Mar 13 at 14:46
  • $\begingroup$ @AnneBauval Yes, sorry, I'm not being precise. I feel like I have trouble putting these two concepts together in my head, so I'm not sure how to write it. $\endgroup$
    – beeselmane
    Mar 13 at 14:48
  • $\begingroup$ If this did not answer your question, please edit your post to clarify what is your question actually. $\endgroup$ Mar 13 at 14:49
  • $\begingroup$ @AnneBauval My question is are these two notions of dual spaces the same. I'm aware of the result for class functions only on finite groups. Is the same true for locally compact groups? $\endgroup$
    – beeselmane
    Mar 13 at 14:52
  • 1
    $\begingroup$ By my first comment, they cannot be "the same" (even for finite groups). $\endgroup$ Mar 13 at 14:55

1 Answer 1


Here's the situation for complex representations of finite groups:

$$ \begin{array}{l|l|l} \textrm{space} & \textrm{basis} & \textrm{dual basis} \\ \hline \textrm{all functions} & \textrm{group elements} & \textrm{matrix coefficients} \\ \textrm{class functions} & \textrm{conjugacy classes} & \textrm{irreducible characters} \end{array} $$

Any function $f:G\to\mathbb{C}$ can be decomposed uniquely as a linear combination $f(x)=\sum f(g)\delta_g(x)$ of Kronecker delta functions indexed by group elements. Given (WLOG) a unitary basis for the underlying inner product space, the operators $\rho(g)$ can be represented as matrices whose individual entries may be interpreted as functions of $g$; all functions are uniquely expressible as a linear combination of these matrix coefficient functions drawn from all unitary irreps (one basis chosen per irrep).

With the conjugation action of $G$ on itself extended to the space of all functions, the subspace of invariants is the comprised of class functions. One basis is the indicator function of conjugacy classes, typically normalized by $1/|G|$. Another basis for class functions is the basis of irreducible characters.

The group element basis for the full function space induces an inner product on the function space (often normalized), which the subspace of class functions inherits, and then we can speak of both the bases being unitary. The dual bases, suitably normalized if necessary, are also unitary, per Schur Orthogonality. The change-of-basis procedure from the conjugacy class basis to the irreducible character basis is none other than the Fourier transform.

[ By endowing the function space with pointwise addition and multiplication, and possibly using the group element basis to identify the function space with the group algebra (which has its own addition and multiplication, the latter corresponding to convolution in the function space), we can treat $\mathbb{C}[G]$ as a Hopf algebra (aka quantum group). The restriction to class functions corresponds to the center of the group algebra, and the corresponding conjugacy class basis and irreducible character basis become more than just vector space bases, but also idempotent elements with respect to the respective algebra and coalgebra structures. ]

Here I am using the general definition of "character" as the trace of a representation. On this general definition, characters are (non-canonically) in bijection with conjugacy classes. A narrower definition you'll sometimes see is as a homomorphism $G\to\mathbb{C}^\times$. On this narrower definition, characters form a "character group" under pointwise multiplication which is (non-canonically) in bijection with the abelianization $G^{\mathrm{ab}}=G/[G,G]$.

Generalizing to locally compact groups $G$ (note finite groups are discrete and compact), not everything generalizes. Firstly, the appropriate generalization of the function space is not all functions but all $L^2$-integrable functions (using any Haar measure). This means group elements no longer index a basis for $L^2(G)$; the diract delta distribution centered at the identity element for example would be an "approximate identity" (and the Kronecker delta function is equivalent to the zero function since they agree on all but a set of measure zero). Irreducible characters still form a basis for the class functions, though.

Things are nice if $G$ is finite abelian: the two rows of my table above become the same. The character group, or dual group $G^\wedge$ since $G$ is abelian, is (non-canonically) isomorphic $G$. Moreover, linearizing the evaluation map $G^\wedge\times G\to\mathbb{C}$ yields $\mathbb{C}G^\wedge\otimes\mathbb{C}G\to\mathbb{C}$ which is the appropriate dual pairing between the function space (viewed as the free vector space $\mathbb{C}G$ on $G$) and its dual.

If $G$ is locally compact abelian (LCA), things are still nice, the irreps are still 1D and the irreducible characters form a dual group $G^\wedge$, but $G^\wedge$ will generally not be isomorphic to $G$, even non-canonically, not to mention $G$ is no longer a basis (or even corresponds to elements) of the appropriate function space $L^2(G)$. For example, while $G=\mathbb{R}$ is self-dual ($\mathbb{R}^\wedge=\mathbb{R}$), for $G=\mathbb{Z}$ the dual is $\mathbb{Z}^\wedge=S^1$ (or vice-versa: $(S^1)^\wedge=\mathbb{Z}$). Still, even though $\mathbb{Z}$ and $S^1$ aren't even the same cardinality, the function spaces $\ell^2(\mathbb{Z})$ and $L^2(S^1)$ are still isomorphic via the Fourier transform. The fact the Fourier transform is a unitary operator between function spaces on $\mathbb{R}$ is the Plancheral theorem, or on $\mathbb{Z}$ and $S^1$ is Parseval's theorem.

For Banach spaces (of which Hilbert spaces are a subcategory), the dual is defined as the space of all continuous linear functions. Famously, the dual of the Hilbert space $L^p(X)$ is $L^q(X)$ where $1/p+1/q=1$, so in particular $L^2(G)$ is self-dual. And just as for the finite case, by "linearly extending" the evaluation map $G^\wedge\times G\to\mathbb{C}$ (with integration) we can identify the dual space as $L^2(G)^\wedge\cong L^2(G^\wedge)$, even if $G\not\cong G^\wedge$ for LCA $G$, generalizing the situation for finite abelian $G$.


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