Why not _co_free modules? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T16:00:27Z http://mathoverflow.net/feeds/question/38085 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38085/why-not-co-free-modules Why not _co_free modules? Theo Johnson-Freyd 2010-09-08T18:36:47Z 2010-09-09T00:24:46Z <p>Let $R$ be a ring, and $R\text{-Mod}$ its category of all left modules. There is a "forgetful" functor $\operatorname{Forget}: R\text{-Mod} \to \text{AbGp}$, which is additive, continuous, and cocontinuous (in particular, exact). Since $R\text{-Mod}$ is both complete and cocomplete, $\operatorname{Forget}$ has both a left adjoint $\operatorname{Free}: \text{AbGp} \to R\text{-Mod}$ and a right adjoint $\operatorname{Cofree}: \text{AbGp} \to R\text{-Mod}$.</p> <p>You can see what these functors are explicitly. Let me write <code>$_R R_{\mathbb Z}$</code> for "$R$ as a left module" and <code>$_{\mathbb Z} R _R$</code> for "$R$ as a right module". The $\operatorname{Forget}$ functor is (isomorphic to) the functor <code>$\operatorname{Hom}_R({_R R_{\mathbb Z}},-)$</code> &mdash; this description makes it clearly continuous, and its left adjoint is <code>$\operatorname{Free} \cong {_R R_{\mathbb Z}} \otimes_\mathbb Z (-)$</code>. But we also have <code>$\operatorname{Forget} \cong {_{\mathbb Z} R _R}\otimes_R (-)$</code>, whence its right adjoint is <code>$\operatorname{Cofree} \cong \operatorname{Hom}_{\mathbb Z}({_{\mathbb Z} R _R},-)$</code>.</p> <p>I feel like I have some positive amount of experience with free modules. (I would say, given the above, that the correct definition of "free module" is "object in the essential image of $\operatorname{Free}$", although what's actually used is "object of the form $\operatorname{Free}(\mathbb Z^{\oplus \kappa})$ for some cardinal $\kappa$.) But I hardly ever come across the essential image of $\operatorname{Cofree}$, or indeed the cofree functor at all. (Again, maybe the "standard" definition of "cofree module" is "module isomorphic to $\operatorname{Cofree}((\mathbb Q/\mathbb Z)^{\times \kappa})$," or something.) The functors are not the same: when $R = \mathbb Z/2$, then $\operatorname{Free}(\mathbb Z) = \mathbb Z/2$, whereas $\operatorname{Cofree}(\mathbb Z) = 0$. If you would rather replace $\mathbb Z$ by a field throughout, then they are still not the same when $R$ is infinite-dimensional (for example).</p> <p>So: Do people use cofree modules? If so, how? If not, why not? Are free modules just a lot nicer than cofree ones, and if so, how?</p> http://mathoverflow.net/questions/38085/why-not-co-free-modules/38091#38091 Answer by Emerton for Why not _co_free modules? Emerton 2010-09-08T19:10:06Z 2010-09-09T00:24:46Z <p>This construction is used frequently (at least, I use it frequently in my work). For example, it appears in the usual proof that module categories have enough injectives. (In this case one studies $Cofree(\mathbb Q/\mathbb Z)$, as you anticipated.)</p> <p>If we generalize slightly, and replace $\mathbb Z$ by the group ring $k[H]$ and $R$ by the group ring $k[G]$ (with $H$ being a subgroup of $G$), then $Hom_{k[H]}(k[G],\text{--})$ is precisely the functor of induction from $H$-representations to $G$-representations, and the adjointness you note is a form of Frobenius reciprocity.</p> <p>If $R$ is a Hecke algebra (over $\mathbb Z$) on a space of weight $k$-cuspforms of some level, then $Cofree(\mathbb Z)$ is the space of modular forms of weight $k$ with coefficients in $\mathbb Z$. (This technical relationship between Hecke operators and the space of modular forms on which they operate is used frequently by number theorists working on the arithmetic of modular forms.)</p> <p>There are lots of other contexts in which this functor (and its variants, replacing $\mathbb Z$ by other rings) appear, but maybe I've said enough for now.</p>