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Given a monoidal category $\mathcal{C}$ one can define the Green ring (or representation ring) $r(\mathcal{C})$ as the abelian group generated by the isomorphism classes $[V]$ of $\mathcal{C}$ modulo the relations $[M\oplus V]=[M]+ [V]$, and the multiplication is given by the tensor product.

Not all rings can be realized as the representation ring of some monoidal category $\mathcal{C}$. However I cannot seem to find some nice examples of this fact in literature. I'd very much like to see an example and the reasoning why this is true. Any references are very welcome.

Thank you.

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    $\begingroup$ Hmm, I think $\mathbb{Z}[i]$ is an example, but I need to think some more about it (the idea is that anything arising as such a decategorification will have a natural positive basis, and I don't think this ring has one). $\endgroup$ – Tobias Kildetoft Apr 19 '16 at 7:45
  • $\begingroup$ The answer should be fairly sensitive to exactly what other conditions you require on $C$. For example, is it abelian or something like it? (If not, what does direct sum mean?) Does the monoidal operation distribute over direct sums? How about finite colimits? Is $C$ semisimple? Does every object have finite length? In another direction, how about using short exact sequences to define the Grothendieck group? $\endgroup$ – Qiaochu Yuan Apr 19 '16 at 8:35
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    $\begingroup$ @QiaochuYuan Well, using short exact sequences would then require the category to be abelian, and even then we could still just consider the larger split Grothendieck group as in the question. But you are definitely right that what qualifies as an example depends heavily on what sort of categories one allows. $\endgroup$ – Tobias Kildetoft Apr 19 '16 at 8:38
  • $\begingroup$ Yes, I should have provided more conditions. Obviously I want $\mathcal{C}$ to be abelian. I also want the tensor operation to distribute over direct sums. It would be nice if the category were Krull-Schmidt as well with finitely many indecomposable objects. If possible, I do not want it to be a linear category. $\endgroup$ – Mathematician 42 Apr 19 '16 at 8:44
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    $\begingroup$ At least with the added conditions of being Krull-Schmidt with finitely many indecomposables will mean that the indecomposables will give you a positive basis. Without that, I have no idea what could possibly happen. $\endgroup$ – Tobias Kildetoft Apr 19 '16 at 10:03
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If $\mathcal{C}$ is a monoidal Krull-Schmidt category with finitely many indecomposables, then this translates to $r(\mathcal{C})$ being a free $\mathbb{Z}$-module of finite rank, so clearly an interesting example will be of this form.

Take the ring to be $\mathbb{Z}[i]$ and note that if this arises as $r(\mathcal{C})$ for some category as above then this category will have precisely two indecomposables, and one of these must be the neutral element for the tensor product in $\mathcal{C}$, which must decategorify to the unit in the ring. Let $X$ be the other indecomposable and assume that $X$ decategorifies to some $a + bi$ in $\mathbb{Z}[i]$ (we can clearly assume that $b\neq 0$).

Now we also note that in general, the $\mathbb{Z}$-basis of $r(\mathcal{C})$ given by the decategorifications of the indecomposables will be positive, since clearly there is no such thing as minus in the category. But it is a basic exercise in linear algebra to check that $(a+bi)^2 = (-a^2 - b^2)\cdot 1 + 2a\cdot (a + bi)$ so the structure constants will never be non-negative no matter which element we try. Thus, $\mathbb{Z}[i]$ cannot be the representation ring of such a category.

The existence of a positive basis as mentioned above is one of the very strong features of categorification, though there are still plenty of things that can happen for positive based algebras that do not correspond to anything coming from the categorical picture (though the examples I am mainly familiar with are ones where the algebra is indeed the decategorication of some $2$-category but where some representation of the algebra cannot come from a $2$-representation of the $2$-category).

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  • $\begingroup$ Great answer, and indeed, the category being Krull-Schmidt is strong condition. Thank you for this answer, I never thought of showing that $\mathbb{Z}[i]$ cannot be a representation ring due to the structure constants being negative. $\endgroup$ – Mathematician 42 Apr 24 '16 at 19:50
  • $\begingroup$ @Mathematician42 Thank you. Krull-Schmidt is a strong condition, but also a fairly reasonable one (at least there is plenty to do even if one restricts to studying these). Having a positive basis is indeed one of the nicest properties of these sorts of rings (I happen to have done a bit of research on positively based algebras so I quite like those). $\endgroup$ – Tobias Kildetoft Apr 25 '16 at 7:03
  • $\begingroup$ One small remark, we do not necessarily have that the unit object for the tensor product is indecomposable. Even if we assume that there are finitely many indecomposable objects, we still need that the tensor product of two indecomposable objects is non-zero, right? Or I'm missing an easy argument. $\endgroup$ – Mathematician 42 Apr 28 '16 at 13:05
  • $\begingroup$ @Mathe Hmm, I suppose I am just used to having the assumption of the unit being indecomposable, so I have not thought about what happens otherwise. $\endgroup$ – Tobias Kildetoft Apr 28 '16 at 15:04
  • $\begingroup$ Nevermind, for $\mathbb{Z}[i]$ you automatically have that the unit is indecomposable, I found some argument for that. But it nicer to assume the unit is indecomposable anyway. $\endgroup$ – Mathematician 42 Apr 28 '16 at 17:19

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