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## Why does the Grothendieck group $K_0(R)$ of a ring not depend on our choice of using left modules instead of right modules?

I am under the impression that in the definition of the Grothendieck group $K_0(R)$ of a (non-commutative) ring it doesn't matter whether we apply the usual $K_0$ construction to the exact category of all finitely generated projective left $R$-modules, or if we apply the construction to the category of all finitely generated projective right $R$-modules.

Is this obvious? To be honest my first reaction is disbelief but given that certain other notions for rings (such as semisimplicity of a ring or Morita equivalence for rings) are left/right-independent, I guess I can believe it. But it is not clear to me why this is true in the case of $K_0$.

I've taken a look in the standard references for classical algebraic $K$-theory but none of them seem to mention this point and either deal just with left $R$-modules throughout, or make no mention at all of the handedness of their modules.

This circle of ideas leads to a more general and vague follow up question: Does anyone have any intuition for why certain notions for non-commutative rings (such as the examples mentioned above: semisimplicity, Morita equivalence, $K_0$) do not depend on whether we look at properties of the collection of left modules over the ring or whether we look at properties of the collection of right modules over the ring.

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First of all, right $R$-modules are the same as left $R^{op}$-modules. Hence you are asking whether the $K$-theory changes if you pass form $R$ to $R^{op}$. The answer is: It does not change.

The reason for the independence is that one has a very concrete picture.

$$K_0(R) = Gr(V(R))$$

where $Gr$ denotes the Grothendieck group, and $V(R)$ denotes the monoid of conjugacy classes of idempotents in $\cup_{n \in \mathbb N} M_n(R)$. Clearly, neither the notion of idempotent nor conjugacy class depends on the order in which your multiplication is performed. Hence, the $K$-theories of $R$ and $R^{op}$ are canonically isomorphic.

The argument for $K_1$ is even easier, since $K_1$ is defined as the abelianization of $GL_{\infty}(R)$.

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 Hm, this does not show that $K_0(R)$ and $K_0(R^{op})$ are isomorphic, but rather anti-isomorphic, right? – Martin Brandenburg Nov 19 2010 at 7:42 And since every group is anti-isomorphic to itsself via $g \mapsto g^{-1}$, we get an isomorphism ... – Martin Brandenburg Nov 19 2010 at 7:43 $K_0(R)$ is abelien, so I do not see the point in considering anti-isomorphisms etc. However, maybe you want to look at my post mathoverflow.net/questions/46599/… which is related to your thoughts. – Andreas Thom Nov 19 2010 at 7:46 ... also the addition on $K_0(R)$ comes from the monoid structure on the set of idempotents. This monoid structure does not depend on the multiplication but only on the linear structure of $\cup_{n} M_n(R)$. – Andreas Thom Nov 19 2010 at 7:50 Alright, thanks. – Martin Brandenburg Nov 19 2010 at 8:16
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Here is an alternative to Andreas proof (which if you unfold it is not so different): We have a functor $M\mapsto \mathrm{Hom}_R(M,R)=:M^\ast$ which gives both a contravariant functor from left $R$-modules to right $R$-modules and vice versa. We also have a naturanl transformation $M\to M^{\ast\ast}$ given by $m\mapsto(f\mapsto f(m))$. Now, as $R^\ast=R$ and $(-)^\ast$ is additive it takes f.g. projective modules to f.g. projective modules and the map $P\mapsto P^{\ast\ast}$ is an isomorphism. This shows that the category of left f.g. right projective modules is anti-equivalent to the category of f.g. left projective modules. However, $K_0(-)$ induces an isomorphism not only for equivalences but also for anti-equivalences.

Unless I am mistaken the same argument works also for Quillen's higher K-groups.

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 Thanks Torsten. – Beren Sanders Nov 19 2010 at 17:55