Is Higher K-functor the derived functor of K0? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:14:35Z http://mathoverflow.net/feeds/question/11421 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11421/is-higher-k-functor-the-derived-functor-of-k0 Is Higher K-functor the derived functor of K0? Peter Lee 2010-01-11T11:10:48Z 2012-04-01T09:18:34Z <p>It might be a stupid question. I wonder whether the derived functor of functor K0 is Quillen Higher K-functor?</p> <p>If not, is there any relationship between derived functor of K0(or satellites of K0-functor) and Quillen Higher K-functor?</p> <p>The motivation to ask this question is if this statement holds, then Quillen higher K-functor is universal. </p> http://mathoverflow.net/questions/11421/is-higher-k-functor-the-derived-functor-of-k0/11434#11434 Answer by Clark Barwick for Is Higher K-functor the derived functor of K0? Clark Barwick 2010-01-11T14:57:43Z 2010-01-11T14:57:43Z <p>I don't think it's stupid, but I guess it depends what you mean by "derived functor." This is true in the weak sense that K-theory is naturally a space- or spectrum-valued functor, and the $K_i$ is the i-th homotopy of this functor. But it seems not to be the case that $K$-theory is a derived functor in the sense of Cartan-Eilenberg.</p> <p>Let me discuss the question of the universality of $K$-theory:</p> <p>I'll abuse terminology and refer to "categories" when I mean categories of a suitable kind, with appropriate added structure --- e.g., exact categories if you want to do Quillen K-theory, Waldhausen categories if you want to do Waldhausen K-theory, Waldhausen $\infty$-categories if you want to do K-theory with them, etc. ...</p> <p>Now if one translates the sense in which $K_0$ is universal as an abelian-group-valued functor on "categories" into the language of stable homotopy theory, one arrives at the universal property satisfied by K-theory as a spectrum-valued functor on "categories."</p> <p>More precisely, we have additive $K_0$, denoted $K_0^{\oplus}$, which is simply the functor that assigns to any "category" $\mathcal{C}$ the group completion of the abelian monoid whose elements are isomorphism (or equivalence) classes of objects of $\mathcal{C}$, where the sum is $\oplus$. This functor is "inadequate" in the sense that there might be some exact (or fiber) sequences of $\mathcal{C}$ that $K_0^{\oplus}$ cannot see.</p> <p>To address this, for any "category" $\mathcal{C}$, we can build a new "category" $\mathcal{E}(\mathcal{C})$ whose objects are exact sequences. This "category" admits two functors to $\mathcal{C}$ that send an exact sequence $[0\to A'\to A\to A''\to 0]$ to either $A'$ or $A''$. For any functor $F$ from categories to abelian groups, we get an induced homomorphism $F\mathcal{E}(\mathcal{C})\to F\mathcal{C}\oplus F\mathcal{C}$. Let's say that $F$ <em>splits the exact sequences of</em> $\mathcal{C}$ if this morphism is an isomorphism, and let's say that $F$ is <em>additive</em> if $F$ splits the exact sequences of every "category."</p> <p>Now $K_0$ has the following pleasant universal property. It is the initial object in the category of additive functors receiving a natural transformation from $K_0^{\oplus}$.</p> <p>Now to translate all this into stable homotopy. We have additive K-theory, denoted $K^{\oplus}$, which is simply the functor that assigns to any "category" $\mathcal{C}$ the <em>spectrum</em> corresponding to the group completion of the $E_{\infty}$ space given by the (nerve of the) subcategory of $\mathcal{C}$ comprised of the isomorphisms (or weak equivalences), where the sum is $\oplus$. This functor is again "inadequate" in the sense that there might be some exact (or fiber) sequences of $\mathcal{C}$ that $K^{\oplus}$ cannot see.</p> <p>Now for any functor $F$ from categories to <em>spectra</em>, we get an induced homomorphism $F\mathcal{E}(\mathcal{C})\to F\mathcal{C}\vee F\mathcal{C}$. Let's say that $F$ <em>splits the exact sequences of</em> $\mathcal{C}$ if this morphism is an equivalence, and let's say that $F$ is <em>additive</em> if $F$ splits the exact sequences of every "category."</p> <p>Now $K$ has the following homotopy-universal property. It is the homotopy-initial object in the category of additive functors receiving a natural transformation from $K^{\oplus}$.</p> <p>So the universality of K-theory arises not from thinking of the disembodied K-groups, but rather from interpreting K-theory as a spectrum, and rewriting the universal property of $K_0$ in suitably homotopical language.</p> <p>(References: Gonçalo Tabuada has a <a href="http://www.math.uiuc.edu/K-theory/0855/" rel="nofollow">paper</a> in which he characterizes K-theory by a similar universal property, and John Rognes and I have begun a similar paper in the context of Waldhausen $\infty$-categories, an incomplete draft of which is on my webpage.)</p> http://mathoverflow.net/questions/11421/is-higher-k-functor-the-derived-functor-of-k0/15743#15743 Answer by Zoran Škoda for Is Higher K-functor the derived functor of K0? Zoran Škoda 2010-02-18T19:18:52Z 2010-02-18T19:18:52Z <p>There is a definition due Sasha Rosenberg, of a variant of the algebraic K-theory which is universal practically by definition and in the setting much more general than Quillen exact categories. By universal I mean something along Cartan-Eilenberg and Tohoku. He first defines the notion of a right exact structure on a category, what is just a Grothendieck pretopology whose covers are singletons which are strict epimorphisms. Now the collection of all small categories with right exact structure has a left exact structure, what is a dual notion (Grothendieck precotopology which...). Now he extends the formalism of Tohoku to define universal delta functors or delta star functors (I never know which one is which) for right or left exact structures. So, one can define K-zero by hand, on the collection of all categories with right exact structures and try to extend it to a universal delta functor. And there is such. In particular, Quillen exact categories have a canonical right exact structure. Now this variant of K-theory has all other standard properties of Quillen K-theory, like resolution by devissage, exactness and so on. But it is not clear if it is equal or not to Quillen K-theory yet (for this one should evaluate Quillen recipe on the injective resolution by the categories with right exact structures; if one gets zero, voila!). The article of Rosenberg is at</p> <p>A. Rosenberg, <a href="http://www.mpim-bonn.mpg.de/preprints/send?bid=3623" rel="nofollow">Homological algebra of noncommutative 'spaces' I (pdf)</a>, 199 pages, preprint Max Planck, Bonn: MPIM2008-91.</p> http://mathoverflow.net/questions/11421/is-higher-k-functor-the-derived-functor-of-k0/92799#92799 Answer by M Mueger for Is Higher K-functor the derived functor of K0? M Mueger 2012-04-01T08:49:12Z 2012-04-01T09:18:34Z <p>The algebraic $K$-groups of a commutative unital ring can indeed be defined as derived functors, but one needs to work in the context of non-abelian homological algebra in the sense of A. Dold, D. Puppe, Homologie nicht-additiver Funktoren, Ann. Inst. Fourier 11 (1961), and M. Tierney, W. Vogel: Simplicial resolutions and derived functors, Math. Zeit. 111 (1969). A useful introduction is given in the book `Non-abelian homological algebra and its applications' by Hvedri Inassaridze (Kluwer, 1997). </p> <p>For a group $G$, define $\displaystyle Z_\infty (G)=\lim_{\leftarrow} G/\Gamma_i(G)$, where ${\Gamma_i(G)}$ is the lower central series of $G$. This defines a functor $Z_\infty: Gr\rightarrow Gr$. Now Theorem 5.1 in the cited book roughly reads as follows:</p> <p>Let $L_*Z_\infty$ be the left derived functors of the functor $Z_\infty$. Then $L_i Z_\infty(GL(R))$ is isomorphic to Quillen's $K_i(R)$.</p>