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Grothendieck Group

  1. Sep 13, 2014 #1
    In Lang's book,page 39-40, he factorizes ##F_{ab}(M)## with respect to the subgroup generated by all elements of type ##[x+y]-[x]-[y]##. I don't quite understand why he does this. I know that he is trying to create inverse elements, but I don't see why that factorization necessarily satisfies the universal property.
     
  2. jcsd
  3. Sep 13, 2014 #2
    Are we basically factoring through the commutator of the free group, ##F_{ab}(M)##.
     
  4. Sep 15, 2014 #3
    No, [itex] F_{ab}(M) [/itex] is the free abelian group so it has trivial commutator. We want to make the monoid [itex] M [/itex] into an abelian group in a way which is universal, but only with respect to monoid homomorphisms. The simplest choice is to take the free abelian group generated by [itex] M [/itex] itself however this is universal with respect to all set maps from M to an abelian group, not just monoid homomorphisms into an abelian group. The characteristic difference is that for a monoid homomorphism, [itex] f(x+y)=f(x)+f(y) [/itex] or equivalently [itex] f(x+y)-f(x)-f(y)=0 [/itex]. So when we quotient out [itex] F_{ab}(M) [/itex] by the subgroup generated by [itex] [x+y]-[x]-[y] [/itex] to get the Grothendieck group [itex] K(M)[/itex], every induced homomorphism from the free group that came from a monoid homomorphism will vanish on this subgroup since [itex] \bar f([x+y]-[x]-[y])=f(x+y)-f(x)-f(y)=0 [/itex] and so yields a well-defined map [itex] f_*:K(M)\to A [/itex]. On the other hand, any set map which is not a monoid homomorphism will not vanish on this subgroup and hence won't be well defined on the quotient. This shows that the quotient is now universal only with respect to the monoid homomorphisms out of [itex] M [/itex] rather than all set maps. So you can think of taking the quotient by this subgroup as throwing away all the induced maps from the free group which came from set maps [itex] g:M\to A [/itex] which were not monoid homomorphisms.
     
  5. Sep 15, 2014 #4
    Why do we write it as [x+y]-[x]-[y], shouldn't it be [x+y]-[y]-[x] since your adding inverses to the right? Or are we assuming the fact that the representatives are abelian, so we can write [x+y]=[x]+[y]=[y]+[x].
     
    Last edited: Sep 15, 2014
  6. Sep 15, 2014 #5
    Everything in this example is abelian. The monoid Lang starts out with is commutative, you construct the free abelian group and then take a quotient which is again abelian so everything in sight commutes.

    I'm not sure I understand quite what you mean by adding inverses though. Taking the quotient of [itex] F_{ab}(M) [/itex] isn't what creates inverses since [itex] F_{ab} (M) [/itex] is already a group by definition. So every desired inverse is already here before taking a quotient. The reason to take a quotient is that [itex] F_{ab}(M) [/itex] is too big to satisfy the universal property as I explained in my previous post so we take a quotient of it to get rid of all the extra bits we don't need and force it to have the universal property we want.
     
  7. Sep 15, 2014 #6
    I was just stating that if ##[x+y]=[x]+[y] \rightarrow [x+y]-[y]=[x]+[y]-[y]=[x]\rightarrow [x+y]-[y]-[x]=0##. But this isn't important since everything were working with is abelian.

    I understood the rest of your explanation Terandol. Thanks.
     
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