I've been looking at some definitions of Grothendieck rings. However I've not found a good definition that I've understood. Any recommendations? I'm referring to the definition in tensor categories, more specifically I've discovered there are some structure coefficients in a Grothendieck Ring, I understand the mathematics from a physics perspective. I learned this first from Physics papers, but I want to understand more deeply the mathematical structure. http://www.math.sunysb.edu/~kirillov/tensor/tensor.html is the set of lecture notes I'm referring to, and I'd love if someone could help me tag this post more effectively.
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2$\begingroup$ Perhaps you could mention those at which you looked already: so that noone suggests those, and it is easier to infer what would be 'good'. $\endgroup$– user9072Commented Sep 3, 2011 at 11:53
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2$\begingroup$ I agree with quid's comment, but I'll make an attempt anyway. The first example you should contemplate is the Grothendieck ring of finite dimensional vector spaces. You have a symbol for each isomorphism class of spaces. Adding symbols corresponds to direct sum, and multiplying to tensor products. It seems a bit like the dimension doesn't it? $\endgroup$– Donu ArapuraCommented Sep 3, 2011 at 12:09
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5$\begingroup$ Also, I think the question is ambiguous, because there's more than one sense in which people use the term "Grothendieck ring". $\endgroup$– Todd TrimbleCommented Sep 3, 2011 at 12:11
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$\begingroup$ I'm referring to the Grothendieck Ring in Tensor Categories. I'm not sure if this is the same.... $\endgroup$– Peadar CoyleCommented Sep 3, 2011 at 14:37
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1$\begingroup$ Yes, it's the same as what I was thinking. Another example, perhaps closer to what you want, is the category of complex reps of compact Lie group (e.g. finite group). This a tensor category. Every object is determined up to iso. by its character $\chi_V$. Note $\chi_{V\oplus W}=\chi_V+\chi_W$ and $\chi_{V\otimes W}=\chi_V\chi_W$ which are the same relations as in the Grothendieck ring. So perhaps you can view the Groth. ring as a generalization of character theory. I'm sure someone will write more. $\endgroup$– Donu ArapuraCommented Sep 3, 2011 at 15:08
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2 Answers
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I'll expand my comments into an answer. Since I'm not quite sure what parts bother you, I'll assume it's everything! The Grothendieck construction is actually a family of related constructions, which is brilliant in its simplicity. Whenever, you have a collection of things (e.g. finite sets) that can split into parts, you can force it be an abelian group by requiring that the sum of parts correspond to addition in the group. If your things have more structure, then the Grothendieck group can be expected to inherit this as well.
To get closer to what you seem to be interested in, suppose that $G$ is a discrete or Lie group, and $C$ is the category of finite dimensional complex representations (as a Lie group). Then $C$ is a good example of a tensor category: It's abelian, so we can speak of exact sequences, there are also tensor products (usual product with $g(v\otimes w) =gv\otimes gw$), and various compatibilities hold.
Given $\rho:G\to V$, one can attach a character $\chi_V= g\mapsto trace(\rho(g))$ which sometimes determines $V$ when $G$ is compact, but not in general. Nevertheless, the standard relations
- $\chi_V = \chi_{U}+\chi_{W}$ when $V$ is extension of $W$ by $U$
- $\chi_{V\otimes W} = \chi_{V}\chi_{W}$
always hold, which makes this notion quite useful.
Now suppose that $C$ is a more general tensor category. What would be the analogue of the ring of characters on $G$? It would be the Grothendieck ring, generated by symbols $\chi_V$ where we simply impose the above relations.
answered Sep 4, 2011 at 0:35
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$\begingroup$ This is off-topic, but: Are you sure you want to say "Lie group" rather than "semisimple Lie group" or something like that? I am new to representation theory of Lie groups, but I would be somewhat surprised if the Grothendieck ring of the representations of every Lie group would be its character ring... $\endgroup$ Commented Sep 4, 2011 at 9:06
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$\begingroup$ You're probably right. I tried to convey the idea that $K_0(C)$ is formally similar to the ring of characters, and that it works fine even on a nonsemisimple category... $\endgroup$ Commented Sep 4, 2011 at 12:26
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$\begingroup$ It is a fact that continues to amaze me that there are any groups at all for which the Grothendieck ring is the ring of characters. But what Donu well-illustrates is that there is a map from the Grothendieck ring to the ring of characters. Since for many calculations it is enough to know that characters satisfy both 1 and 2 above, it is then natural to consider the "universal" ring with such properties. I like the idea that addition correspond to any notion of "breaking an object into parts". In the case of abelian categories, any exact sequence "breaks the middle into sub-plus-quotient". +1 $\endgroup$ Commented Sep 12, 2011 at 4:41
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Let $\mathrm{Var}_k$ denote the category of varieties over a field $k$. Then $K_0(\mathrm{Var}_k)$ is the free abelian group generated by symbols $[X]$ for $X\in \mathrm{Var}_k$, subject to the relations:
(i) $[X]=[Y]$ if $X \cong Y$;
(ii) $[X]=[X\setminus Y]+[Y]$ if $Y$ is a closed subscheme of $X$ (the so called scissor relation).
We can define a multiplication on $K_0(\mathrm{Var}_k)$ by $[X]\cdot [Y] := [X \times_k Y]$. The resulting ring is called the Grothendieck ring (of varieties).
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9$\begingroup$ In this case the Grothendieck ring is a universal Euler characteristic. The Euler characteristic $\chi:\text{Var}_k \to \mathbb{Z}$ satisfies the relations, $\chi(X) = \chi(Y)$ if $X \cong Y$, $\chi(X) = \chi(X\setminus Y) + \chi(Y)$ for $Y$ a closed subsceme of $X$, and $\chi(X \times Y) = \chi(X)\chi(Y)$. So $\chi$ and any map from $\text{Var}_k$ to a commutative ring $R$ satisfying those relations factors uniquely as a ring homomorphism on $K_0(\text{Var}_k)$. Other interesting examples of such maps are the maps sending $X \to \text{#}X(\mathbb{F}_q)$ (the number of $\mathbb{F}_q$ points). $\endgroup$ Commented Sep 4, 2011 at 0:58
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$\begingroup$ @DoriBejleri maybe the Grothendieck ring is the recipient of the universal Euler characteristic? $\endgroup$– user138661Commented Apr 23, 2019 at 12:16