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Let $G$ be a group, $k$ be a commutative unital ring.

Consider $\mathbf{Alg}_k^{inv}$ the category of unital $k$-algebras whose multiplicative semigroup is a group. Then there is a forgetful functor $$U:\mathbf{Alg}_k^{inv}\to \mathbf{Grp}$$ that assigns each algebra its multiplicative group. This functor has a right adjoint $$K:\mathbf{Grp}\to \mathbf{Alg}_k^{inv}$$ which assigns to each group its group algebra. One can see this by observing that the group homomorphism $G\to U(k[G])$ is initial in the comma category $G\to U(\mathbf{Alg}_k^{inv})$ and thus is a unit.

There is something wrong with it, because a $k$-representation of a group is supposed to be equivalent to a $k[G]$-module. But now it is weird as it corresponds to an algebra morphism $k[G]\to Aut_k(V)$. Please help me point out what is the correct way to understand group algebra.

OK, I realized why this is wrong. But what is the right way to think about it? It the group algebra construction a left adjoint to some functor?

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  • $\begingroup$ Can you explain how you are viewing $Aut_k(V)$ as an algebra? $\endgroup$ – user2055 Dec 17 '14 at 19:01
  • $\begingroup$ @JasonPolak As a subalgebra of $End_k(V)$. $\endgroup$ – mez Dec 17 '14 at 19:52
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    $\begingroup$ You can't add automorphisms. $k[G]$ maps into the endomorphism algebra of $V$ which is precisely the information that gives you a $k[G]$-module. $\endgroup$ – Siddharth Venkatesh Dec 17 '14 at 20:02
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The group algebra is a functor $\mathsf{Grp} \to \mathsf{Alg}_K$ which is left adjoint to the "group of units" functor $\mathsf{Alg}_K \to \mathsf{Grp}$, $A \mapsto A^\times$. It is defined for all unital $K$-algebras. No non-trivial ring has the property that its underlying semigroup is a group, because $0$ is not invertible. If $V$ is some $K$-module, then a $K[G]$-module structure on $V$ (which restricts to the given $K$-module structure) corresponds to a $K$-algebra homomorphism $K[G] \to \mathrm{End}_K(V)$, hence (by adjunction) to a group homomorphism $G \to \mathrm{Aut}_K(V)$. So this is precisely a representation of $G$ on $V$.

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  • $\begingroup$ Nice. Does the "group of units" functor also have aright adjoint? (I would guess something like the coordinate ring of the group seen as a discrete algebraic group, though that might require extra assumptions). $\endgroup$ – Tobias Kildetoft Dec 18 '14 at 8:58
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    $\begingroup$ The "group of units" functor from unital $K$-algebras to groups has no right adjoint because it is not cocontinuous. In fact, $K$ is the initial $K$-algebra, but $K^\times$ is (usually...) not the initial group. $\endgroup$ – Martin Brandenburg Dec 18 '14 at 9:00
  • $\begingroup$ Ahh, ok. Thanks. $\endgroup$ – Tobias Kildetoft Dec 18 '14 at 9:01
  • $\begingroup$ Yea I said multiplicative semigroup is a group, it excludes $0$. $\endgroup$ – mez Dec 18 '14 at 10:38
  • $\begingroup$ This means that you only consider division rings. But $K[G]$ is usually not a division ring. $\endgroup$ – Martin Brandenburg Dec 19 '14 at 16:03

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