I've seen two different ways to define induced representation.

One is as in the book Introduction to representation theory: If $G$ is a group, $H$ is a subgroup of it, and $V$ is a representation of $H$, then the induced representation $Ind^G_H V$ is the representation of $G$ with $$Ind^G_H V=\{f:G\longrightarrow V|f(hx)=\rho_V(h)f(x)\forall x\in G, h\in H\}$$ and the action $g(f)(x)=f(xg)$, $\forall g\in G$.

If we choose a representative $x_{\sigma}$ from each right $H$-coset $\sigma$ of $G$, then any $f\in Ind^G_H V$ is uniquely determined by $\{f(x_\sigma)\}_{\sigma}$.

I've also seen another way to define induced representation. Let $G$ be a group, $H$ a subgroup of it, and $V$ a representation of $H$. The underlying vector space of $Ind^G_H V$ is the direct sum $$\bigoplus_{\tau\in G/H}\tau V$$ with $\tau$ going over all the left cosets in $G/H$. It multiplication operation is defined by choosing a set $\{g_\tau\}_{\tau\in G/H}$ of coset representatives and setting $$g(\tau v)=\beta (hv)$$ where $\tau v$ is an element in $\tau V$, $\beta$ is the unique left coset containing $gg_{\tau}$, and $gg_{\tau}=g_{\beta}h$ for some $h\in H$. It's easy to verify this definition of $Ind|^G_HV$ does not depend on the choice of the representatives $\{g_\tau\}_{\tau\in G/H}$.

Induced representation is the left adjoint to the restricted representation. So any definition of it should be unique up to unique isomorphism. But I cannot see why the two definitions above are equivalent. Can anybody show me where the problem is? Thanks.

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    $\begingroup$ A minor point: in the second definition, if you choose different coset representatives, you get an equivalent representation, though not necessarily exactly the same representation. $\endgroup$ – Geoff Robinson Feb 10 '15 at 18:12
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    $\begingroup$ This is just the difference between induction and coinduction right? In particular, whether you think of induction as a right or left adjoint to restriction. These are isomorphic functors for finite groups but for infinite groups I thought these could be different? Maybe this is relevant math.columbia.edu/~woit/LieGroups-2012/inducedreps.pdf . $\endgroup$ – Jay Taylor Feb 10 '15 at 18:36
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    $\begingroup$ The finite groups case is discussed in detail at this mathstackexchange post math.stackexchange.com/questions/225730/… . $\endgroup$ – Jay Taylor Feb 10 '15 at 18:42
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    $\begingroup$ @Jay: You make good points, but it's always been hard for me to distinguish easily between "induction" and "coinduction" in situations where they give some kind of dual constructions. I'm not sure how these terms originated, which might shed light on the usage. $\endgroup$ – Jim Humphreys Feb 10 '15 at 19:59
  • $\begingroup$ @Jim: I guess the isomorphism makes it implicitly difficult to distinguish them. However, it is also surprisingly useful to use the equivalence of both constructions in the case of finite groups. Maybe the definition by tensor products was preferable because of the universal property of the tensor product, or was just an artefact of who first defined induced representations (Frobenius?). $\endgroup$ – Jay Taylor Feb 12 '15 at 9:49

These two versions of the induced representation are not the same in general. You get isomorphic objects only if you add finiteness conditions. Indeed your second definition corresponds to the subspace of functions supported on a finite number of $H$-cosets.

The two definition agree if e.g. $H$ is of finite index in $G$, or if you add topological conditions. For instance if $G$ is totally disconnected and $H$ is open, then you have the notion of compactly induced representation ${\rm c-Ind}_H^G \, V$ which agrees with your second definition. This compactly induced representation is the subspace of ${\rm Ind}_H^G \, V$ formed of those function which have compact support modulo $H$. Note that in that case ${\rm Ind}_H^G$ is right-adjoint to restriction and ${\rm c-Ind}_H^G $ is left-adjoint ...

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    $\begingroup$ Do you know if there is some catch theorem classifying when the functors are isomorphic, or fail to be isomorphic, for infinite groups? Or do you just have to examine each situation on its own? $\endgroup$ – Jay Taylor Feb 12 '15 at 9:45
  • $\begingroup$ Hi Paul, Can you give a reference on compactly induced representation? Is it defined only for the case $G$ is totally disconnected and $H$ is open? Thanks. $\endgroup$ – Megan May 18 '15 at 22:16
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    $\begingroup$ @Megan. I only know the definition in the case of a totally disconnected group, which does not mean that this object does not exist more generally. For a reference you can read Bernstein's notes : math.ubc.ca/~cass/research/books.html, or Casselman's Introduction to admissible representations of p-adic groups : math.ubc.ca/~cass/research/publications.html . The subgroup H does not need to be open. $\endgroup$ – Paul Broussous May 21 '15 at 14:43

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