Given a group $G$ an a subgroup $H<G$, a representation of $H$ on $V$ is a pair $(\rho, V)$ where $\rho \colon H \to \mathrm{GL}(V)$ where $V$ is a vector space over a field $K$. We can the construct an induced representation of $G$ by $$\mathrm{Ind}_H^G \, \rho = K[G] \otimes_{K[H]} V \, .$$

Is there a way to generalize this for an arbitrary group homomorphism $f \colon H \to G$ such that when applying the construction for subgroups and the inclusion map $i \colon H \to G$ one recovers the original induced representation? That is, if instead of $H$ being a subgroup of $G$ we consider only a group homomorphism between $H$ and G.

At first I thought that, at least for the algebraic construction for finite groups, one could just change $H$ for $\mathrm{im}\, f$ and get done with it, but seems a too naive approach to work in general.

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    $\begingroup$ Arnaud is right. Google "change of rings" for the general setting. $\endgroup$ – Matthew Towers Nov 4 '18 at 20:39
  • $\begingroup$ @ArnaudD. Thank you for the comment. So a group homomorphism $f \colon H \to G$ would induce a ring homomorphism $K[G] \to K[H]$ by precomposing composition with $f$ and then to make $K[G]$ a $K[H]$-module I use extension of scalars. Does this make sense? $\endgroup$ – user314159 Nov 5 '18 at 9:56
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    $\begingroup$ No, $f$ induces a ring homomorphism $K[H]\to K[G]$, and then you can make $K[G]$ a $K[H]$-module by restriction of scalars. $\endgroup$ – Arnaud D. Nov 5 '18 at 10:51
  • $\begingroup$ @ArnaudD. Haha, just the opposite of what I wrote. Thank you very much! Maybe you can write that as an answer so the question is no longer unanswered. $\endgroup$ – user314159 Nov 5 '18 at 13:10
  • $\begingroup$ @user314159 I've written an answer based on my comments (I've corrected a mistake in one of them by the way). $\endgroup$ – Arnaud D. Nov 5 '18 at 14:00

Extensions of scalars works for general ring homomorphisms, so the induced representations must work for general group homomorphisms.

In order for the tensor product $K[G]\otimes_{K[H]} V$ to define a $K[G]$-module, all you need is that $K[G]$ is a $(K[G],K[H])$-bimodule. For this, you can use any group homomorphism $f:H\to G$ to define an action of $H$ on $G$ on the right by $g\cdot h=gf(h)$, and this will be compatible with the translation action of $G$ on itself on the left. Then it suffices to extend this bilinearly to get a bimodule, and then the tensor product $K[G]\otimes_{K[H]} V$ (where $g\otimes \rho(h)v=gf(h)\otimes v$) is a left $K[G]$-module, pretty much as in the case where $f$ is the inclusion of a subgroup.

Moreover, we still have a natural correspondance between maps $\mathrm{Ind}_H^G (V,\rho)\to (W,\sigma)$ and maps $(V,\rho) \to \mathrm{Res}_H^G(W,\sigma)=(W,\sigma \circ f)$, for all representations $(W,\sigma) $ of $G$.

You mention that you tried to take the same construction as in the case where $H$ is a subgroup but with $H$ replaced by $im(f)$. That is indeed a bit too simple to work, but it's not that far from correct either : if you factorize $f$ as a $H\to \frac{H}{\ker(f)}\to G$, then since $im(f)\cong \frac{H}{\ker(f)}$ your suggestion actually does half of the job! So it is enough to find the representation induced by the quotient map $H\to \frac{H}{\ker(f)}$, i.e. the $K\left[\frac{H}{\ker(f)}\right]$-module $K\left[\frac{H}{\ker(f)}\right]\otimes_{K[H]}V$. This is actually isomorphic to the module of coinvariants $V_{\ker(f)}$, which is defined as the quotient of $V$ by the ideal generated by terms of the form $v-\rho(k)v$, where $v\in V$ and $k\in \ker(f)$.

Thus you can construct $K[G]\otimes_{K[H]}V$ as $$\bigoplus_{g\operatorname{im}(f)\in G/\operatorname{im}(f)} V_{\ker(f)},$$ with the action defined as in the case where $H$ is just a subgroup.

  • $\begingroup$ Amazing! Thank you very much! $\endgroup$ – user314159 Nov 5 '18 at 19:28

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