The definition I know of for a cuspidal automorphic representation of, say, $G=\mathrm{GL}\_2$ over a number field $F$ (relative to a choice of compact open subgroup $K\_f$ of $G(\mathbf{A}\_F^\infty)$ and a maximal compact subgroup $K\_\infty$ of $G(F\otimes\_\mathbf{Q}\mathbf{R})$) is: an irreducible subquotient (equivalently subspace) of the space of cuspidal automorphic forms for $G$ over $F$. Since the full group $G(\mathbf{A}\_F)$ does not preserve the property of $K=K\_fK\_\infty$-finiteness, "subrepresentation" here really means a subspace which is simultaneously a $G(\mathbf{A}\_F^\infty)$-submodule and a $(\mathfrak{g}\_\infty,K\_\infty)$-submodule (and this can be phrased more concisely as a submodule for the appropriate global Hecke algebra). Now, I gather that there is a more ``analytic" theory, where one actually has a representation of $G(\mathbf{A}\_F)$ on a Hilbert space. My first question is: 


**Are these theories equivalent in some sense?** 


I should make this more precise. I know that any cuspidal automorphic representation in the first sense is unitarizable. If I were to take the completion, would I get a unitary (admissible) Hilbert space representation of $G(\mathbf{A}^\infty)$ whose space of $K$-finite vectors was isomorphic to the representation with which I started?


I also know that two irreducible unitary admissible representations of, say, $\mathrm{GL}\_2(\mathbf{R})$, are unitarily equivalent if and only if they are infinitesimally equivalent in the sense of having isomorphic associated $(\mathfrak{gl}\_2(\mathbf{R}),\mathrm{O}\_2(\mathbf{R}))$-modules (the spaces of $K$-finite vectors). My second question is:


**Does this extend to unitary cuspidal automorphic representations (which I guess would be defined as irreducible admissible subrepresentations of the appropriately defined space of cusp forms in $L^2(G(F)\setminus G(\mathbf{A}\_F),\omega)$, $\omega$ being the unitary central character)?**


That is, are two unitary cuspidal automorphic representations isomorphic if and only if their spaces of $K$-finite vectors are isomorphic as modules for the Hecke algebras? And, out of curiosity:


**If the answer to my second question is 'yes,' is the same true with $G$ an arbitrary connected reductive group of a number field?**


The reason I ask is because, in some of the literature, it seems like both viewpoints are being used, sometimes at the same time, sometimes implicitly. So I would like to know if I can freely pass back and forth between them. 


Thanks!



Cuspidal automorphic representations as the space of $K$-finite vectors in a unitary cuspidal automorphic representation.
**Yes.** All these viewpoints are equivalent. We write "rep" for irreducible representation.


Edit on request:


An analytic cuspidal rep is subrepresentation $\pi$ of $L\_0^2(G(F) \backslash G(A), \omega)$ for a central character $\omega$.


An algebraic cuspidal rep is an irreducible Harish-Chandra module and an irreducble $C\_c^\infty(G(A\_f))$-module. By Schur's lemma they automatically have a central character $\omega$. Moreover, you need appropiate growth conditions guaranteeing square-integrability. This guarantees also that the vectors are in the $L\_0^2(G(F) \backslash G(A), \omega)$ vector space. Hence every algebraic cuspidal rep is a dense subspace of an analytic cuspidal rep. Let us now discuss the opposite inclusion.


**Fact 1:** Every analytic cuspidal rep $\pi$ of $G(A)$ factors into a tensor product $\otimes\_v \pi\_v$ over the places $v$ of $F$ of unitary reps of $G(F\_v)$. We write $\pi\_\infty \otimes \pi\_f$. See Flath's article in the Corvallis Proceedings Theorem 4.


**Fact 2:** Two distinct unitary reps have distinct $tr \; \pi\_v : C\_c^\infty(G(F\_v)) \rightarrow \mathbb{C}$ functional, and according to Fact 1, this holds for $tr \; \pi : C\_c^\infty(G(A)) \rightarrow \mathbb{C}$ and $\pi$ cuspidal analytic as well. See this related question: [Character determines the representation?](https://mathoverflow.net/questions/88542/character-determines-the-representation). Hence the $C\_c^\infty(G(A))$-module structure determines uniquely the analytic cuspidal representation.


**Fact 3:** Every unitary representation of a real reductive Lie group has a unique Harish-Chandra Module associated to it. So the $C\_c^\infty(G(A\_f)) \otimes C\_c^\infty(G(A\_\infty))$-module structure is encoded uniquely in the Harish-Chandra module $\times C\_c^\infty(G(A\_f))$-module structure. This can be found on the related Wikipedia page, but you seem to believe this already.


So the missing point was fact 2, which allows an algebraic classification of unitary representations. This can be expressed in various languages, but I prefer the statement in terms of character distributions. 


