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Grothendieck Groups and the Eilenberg Swindle July 12, 2009

Posted by Akhil Mathew in algebra, representation theory.
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The following topic came up in a discussion with my mentor recently. Since the material is somewhat general and well-known, but relevant to my project area, I decided to write this post partially to help myself understand it better.

Definition

Consider an abelian category {\mathbf{A}}. Then:

Definition 1 The Grothendieck group of {\mathbf{A}} is the abelian group {K(\mathbf{A})} defined via generators and relations as follows: {K(\mathbf{A})} is generated by symbols {[M]} for each {M \in \mathbf{A}}, and by relations {[M] - [M'] - [M'']} for each exact sequence

\displaystyle   0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0.\ \ \ \ \ (1)

Note here that if {M,N} are isomorphic, then {[M] = [N]} in {K(\mathbf{A})} by considering the exact sequence

\displaystyle  0 \rightarrow M \rightarrow N \rightarrow 0 \rightarrow 0.

The Grothendieck group has an important universal property:

Proposition 2 To give a function {\chi: \mathbf{A} \rightarrow X} for an abelian group {X} satisfying {\chi(M) = \chi(M')+\chi(M'')} for each exact sequence as in (1) (i.e. an Euler-Poincaré map, is equivalent to giving a group-homomorphism {K(\mathbf{A}) \rightarrow X}.

This property, which follows from the definition, determines the Grothendieck group up to isomorphism as the unique group making the above result valid.

The Grothendieck Group of Representations

Let {G} be a finite group. Then consider the category {Rep(G)} defined as follows: the objects of {G} are the finite-dimensional representations of {G}, and morphisms are {{\mathbb C}[G]}-module homomorphisms (also called intertwining operators).

Then:

Proposition 3 The Grothendieck group {K(Rep(G))} is the abelian group {F} generated by the irreducible characters.

Suppose the irreducible representations are {V_1, \dots, V_t}, corresponding to characters {\chi_1, \dots, \chi_t}. We have a group-homomorphism {F \rightarrow K(Rep(G))} sending a sum {n_1 \chi_1 + \dots + n_t \chi_t} to {\bigoplus_i n_i V_i} (this being well-defined as the characters are linearly independent). This is surjective since we can decompose a representation as a direct sum of irreducibles. We define the inverse {K(Rep(G)) \rightarrow F} by using the above universal property. First, define the Euler-Poincaré map {Rep(G) \rightarrow F} by sending {V= \bigoplus_i m_i V_i \rightarrow \sum m_i \chi_i}; this is valid by the previous post, since each object can be decomposed uniquely into irreducibles. One then gets a map {K(Rep(G)) \rightarrow F}. These two maps between {K(Rep(G))} and {F} (and vice versa) are checked to be inverse to each other.

This result can be generalized to semisimple abelian categories.

The Eilenberg Swindle

Suppose an abelian category {\mathbf{A}} admits infinite direct sums. Then I claim:

Theorem 4 {K(\mathbf{A})=0}.

This is proved using the Eilenberg swindle. Given {X \in \mathbf{A}}, we show that {[X] = 0}. But

\displaystyle  X \oplus \bigoplus_{i=1}^\infty X = \bigoplus_{i=1}^\infty X;

we thus have an exact sequence

\displaystyle 0 \rightarrow X \rightarrow \bigoplus_{i=1}^\infty X \rightarrow \bigoplus_{i=1}^\infty X \rightarrow 0,

which gives us the relation {[X]=0}.

The Eilenberg swindle can also be stated in a slightly different form.

The notion of Grothendieck group is useful in modular representation theory—which works in fields other than {{\mathbb C}}, having nonzero characteristic not relatively prime to {card \ G}. Then, it’s possible to say that certain maps between categories may not be surjective, but they are “mostly” surjective in that the induced map on Grothendieck groups is so. We’ll probably use Grothendieck groups more later.

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Comments»

1. Qiaochu Yuan - July 12, 2009

My understanding was that the Grothendieck group construction was most generally defined on a commutative semigroup. The construction you give can be broken down into two steps, both of which satisfy a universal property: turn an abelian category into a commutative semigroup (also a monoid), then turn the semigroup into an abelian group.

I mention this because the simplest example of the Grothendieck group construction is quite familiar: it’s the construction of the integers from the positive integers.

2. Harrison - July 12, 2009

Seconding what Qiaochu said — Grothendieck would be saddened that you’re not thinking in full generality. :P

I’ll have to double-check this when I get a chance, but I believe that the construction of the ring of virtual species from the semiring of species works along essentially the same lines as the (general) Grothendieck group construction. (Question: is the category of combinatorial species — equivalently, I guess, the functor category B^B, where B is the category of sets with bijections as morphisms — abelian? If so, is the Grothendieck group of it isomorphic to the additive group of virtual categories?)

3. Qiaochu Yuan - July 12, 2009

As far as I know, the coproduct and product are disjoint union and Cartesian product in the species sense, so no. (I would be very surprised if this wasn’t the case.) Nonetheless, disjoint union does get us a monoidal category which is commutative up to isomorphism, which is, I believe, essentially how the virtual species construction works.

4. Akhil Mathew - July 12, 2009

Thanks for the comments! I am learning quite a bit from these posts and discussions. However, how does one make the abelian category into a commutative semigroup? One could use direct sums, but then one wouldn’t get enough relations for the exact sequences as above (unless the category is semisimple).

5. Qiaochu Yuan - July 12, 2009

Hmm. If it’s not enough to quotient by natural isomorphism, then perhaps it’s best to regard the categorical Grothendieck group construction as a “categorification” of the semigroup one.

6. Pete L. Clark - February 13, 2011

Just happened to be passing by (I’m websurfing via “Eilenberg Swindle”). One point: you need some “smallness” condition on your abelian category in order for this construction to make sense: explicitly, you want the isomorphism classes of objects in your category form a set. For instance, you cannot take the Grothendieck group of the category of all abelian groups: set theoretic difficulties arise.

This is related to Qiaochu’s comment: for a “skeletally small” additive category (or whatever the term is…), you can define an associated commutative monoid with objects the isomorphism classes of objects in the category and with addition law the direct sum. Then you can take the Grothendieck group of this. However, this does not force every short exact sequence to split, so it is not in general the same as the construction you give. (For that, you need to impose suitable relations.) So far as I know, both constructions are useful and come up in nature.

Akhil Mathew - February 13, 2011

Dear Pete, thanks for the comment! I tend to forget such set-theoretic issues — in this case, I think perhaps the problem can be solved by fixing a Grothendieck universe once and for all.


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