Why is Milnor K-theory not ad hoc? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T09:38:20Z http://mathoverflow.net/feeds/question/4246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4246/why-is-milnor-k-theory-not-ad-hoc Why is Milnor K-theory not ad hoc? S1 2009-11-05T11:31:50Z 2010-11-03T12:09:39Z <p>When Milnor introduced in "Algebraic K-Theory and Quadratic Forms" the Milnor K-groups he said that his definition is motivated by Matsumoto's presentation of algebraic <img src="http://latex.mathoverflow.net/png?K%5F2%28k%29" alt="K_2(k)" title="" /> for a field <img src="http://latex.mathoverflow.net/png?k" alt="k" title="" /> but is in the end purely ad hoc for <img src="http://latex.mathoverflow.net/png?n%20%5Cgeq%203" alt="n \geq 3" title="" />. My questions are:</p> <ol> <li>What exactly could Milnor prove with these <img src="http://latex.mathoverflow.net/png?K" alt="K" title="" />-groups? What was his motivation except for Matsumoto's theorem?</li> <li>Why did this ad hoc definition become so important? Why is it so natural? </li> </ol> http://mathoverflow.net/questions/4246/why-is-milnor-k-theory-not-ad-hoc/4258#4258 Answer by Denis-Charles Cisinski for Why is Milnor K-theory not ad hoc? Denis-Charles Cisinski 2009-11-05T14:53:43Z 2009-11-05T14:53:43Z <p>Milnor K-theory gives a way to compute étale cohomology of fields (i.e. Galois cohomology): if E is a field of characteristic different from a prime l, there is a residue map from the nth Milnor K-group of E mod l to the nth étale cohomology group of E with coefficients in the sheaf of lth roots of unity to the n (i.e. tensored with itself n times). There is the Bloch-Kato conjecture, which predicts that these residue maps are bijectvive. It happens that the case l=2 was conjectured by Milnor (up to a reformulation I guess). The Milnor conjecture has been proved by Voevodsky (and it was the first great achievements of homotopy theory of schemes, which he initiated with Morel during the 90's), and he got his Fields medal in 2002 for this. Now Rost and Voevodsky claimed they have a proof of the full Bloch-Kato conjecture for any prime l (which should appear some day, thanks to the work of quite a few people, among which Charles Weibel is not the least). Note also that the Bloch-Kato conjecture makes sense for l=p=char(E), but then, you have to replace étale cohomology by de Rham-Witt cohomology (and this has also been proved by Bloch and Kato). Suslin and Voevodsky also proved that the Bloch-Kato conjecture implies the Beilinson-Lichtenbaum conjecture, which predicts the precise relationship between torsion motivic cohomology of varieties with torsion étale cohomology.</p> <p>Milnor K-theory is related to motivic cohomology (i.e. higher Chow groups) in degree n and weight n H^n(X,Z(n)): for X=Spec(E), H^n(X,Z(n)) is the nth Milnor K-group. This is how homotopy theory of schemes enters in the picture (one of the main feature introduced by Voevodsky to study motivic cohomology with finite coefficients is the theory of motivic Steenrod operations). On the other hand, Rost studied Milnor K-theory for itself: among a lot of other things, he proved that, if you consider it as a functor from the category of fields, with all its extra structures (residue maps interacting well), you can reconstruct higher Chow groups of schemes (over a field), via some Gersten complex.</p> <p>Milnor K-theory is also a crucial ingredient in Kato's higher class field theory.</p> http://mathoverflow.net/questions/4246/why-is-milnor-k-theory-not-ad-hoc/5625#5625 Answer by Benjamin Antieau for Why is Milnor K-theory not ad hoc? Benjamin Antieau 2009-11-15T16:21:41Z 2009-11-15T18:50:27Z <p>To help answer Question 1, Milnor proved a local-global theorem for Witt rings of global fields. Recall that The Grothendieck-Witt ring $\widehat{W}(k)$ of a field $k$ is the ring obtained by starting with the free abelian group on isomorphism classes of quadratic modules and moding out by the ideal generated by symbols of the form $[M]+[N]-[M']-[N']$, whenever $[M]\oplus[N]\simeq [M']+[N']$. The multiplication comes from tensor product of quadratic modules. There is a special quadratic module $H$ given by $x^2-y^2=0$. This is the hyperbolic module. The Witt ring $W(k)$ of a field $k$ is the quotient of $\widehat{W}(k)$ by the ideal generated by $[H]$.</p> <p>Now, the main theorem of Milnor's paper is that there is a split exact sequence $$0\rightarrow W(k)\rightarrow W(k(t))\rightarrow \oplus_\pi W(\overline{k(t)}_\pi)\rightarrow 0,$$ where $\pi$ runs over all irreducible monic polynomials in $k[t]$, and $\overline{k(t)}_\pi$ denotes the residue field of the completion of $k(t)$ at $\pi$.</p> <p>The morphisms $W(k(t))\rightarrow W(\overline{k(t)}_\pi)$ come from first the map $W(k(t))\rightarrow W(k(t)_\pi)$. Then, there is a map $W(k(t)_\pi)\rightarrow W(\overline(k(t))_\pi)$ that sends the quadratic module $u\pi x^2=0$ to $ux^2=0$, where $u$ is any unit of the local field.</p> <p>Interestingly, Milnor $K$-theory is not used in the proof. However, the proof for Witt rings closely models the proof of a similar fact for Milnor $K$-theory: the sequence $$0\rightarrow K_n^M(k)\rightarrow K_n^M(k(t))\rightarrow\oplus_\pi K_{n-1}^M(\overline{k(t)}_\pi)\rightarrow 0.$$</p> <p>The important new perspective is the formal symbolic perspective, which was already existent for lower $K$-groups, but is very fruitful for studying the Witt ring as well.</p> http://mathoverflow.net/questions/4246/why-is-milnor-k-theory-not-ad-hoc/10137#10137 Answer by Chandan Singh Dalawat for Why is Milnor K-theory not ad hoc? Chandan Singh Dalawat 2009-12-30T13:05:40Z 2010-11-03T10:42:27Z <p>To see a few places where $K_2$ shows up, consult <a href="http://arxiv.org/abs/math/0311099" rel="nofollow">arXiv:math/0311099v4</a>.</p> http://mathoverflow.net/questions/4246/why-is-milnor-k-theory-not-ad-hoc/12176#12176 Answer by norondion for Why is Milnor K-theory not ad hoc? norondion 2010-01-18T07:31:13Z 2010-01-18T07:39:42Z <p>Another application of Milnor K-groups:</p> <ol> <li>The following are equivalent:</li> </ol> <p>${a_1, \ldots, a_n} = 0 \in K^M_n(K)/2$</p> <p>$\langle\kern-0.2em\langle{a_1, \ldots, a_n}\rangle\kern-0.2em\rangle$ (Pfister form) is totally hyperbolic</p> <p>$\langle\kern-0.2em\langle{a_1, \ldots, a_n}\rangle\kern-0.2em\rangle$ is isotropic</p> <p>$a_n$ is represented by $\langle\kern-0.2em\langle{a_1, \ldots, a_{n-1}\rangle\kern-0.2em}\rangle$</p> <ol> <li>higher local class field theory: The class formation of an $n$-dimensional local field is $K^M_n(K)$. <a href="http://www.emis.de/journals/GT/ftp/main/m3/" rel="nofollow">http://www.emis.de/journals/GT/ftp/main/m3/</a></li> </ol> http://mathoverflow.net/questions/4246/why-is-milnor-k-theory-not-ad-hoc/14609#14609 Answer by olli_jvn for Why is Milnor K-theory not ad hoc? olli_jvn 2010-02-08T06:44:05Z 2010-02-08T06:44:05Z <p>As already mentioned above by Denis-Charles Cisinski, Rost has shown (see "Chow Groups with Coefficients") that some version of higher Chow groups can be constructed via Milnor K groups.</p> <p>In fact, Gillet in his survey "K Theory and Intersection Theory" (googleable, I believe originally in the K-Theory Handbook) explains on page 24 and most importantly page 25 (middle) how one may even <strong>motivate</strong> the defining relations of Milnor K (i.e. the Steinberg relation) by intersection-theoric ideas. Whether you find this explanation natural or not is your free choice, but there is some beauty in it.</p> http://mathoverflow.net/questions/4246/why-is-milnor-k-theory-not-ad-hoc/44670#44670 Answer by Baptiste Calmès for Why is Milnor K-theory not ad hoc? Baptiste Calmès 2010-11-03T12:09:39Z 2010-11-03T12:09:39Z <p>Also, about the motivations of Milnor, it is quite natural to try to understand the Witt ring of a field, classifying quadratic forms over this field (in char not 2). This ring has a natural filtration by the fundamental ideal, and it is natural to try to understand the associated graded ring, which is simpler than the Witt ring. One approach is to understand it by generators and relations. The relations defining Milnor's K-theory are elementary ones obviously satisfied in the graded Witt ring, and there are very few of them. Milnor's conjecture (now a theorem) says that Milnor K-theory mod 2 is isomorphic to that graded Witt ring. It is equivalent to the formulation with étale cohomology, but probably, an important part of Milnor's original motivation was about quadratic forms, as one can see in his original paper.</p> <p>It is quite surprising that, a posteriori, this simple K-theory appears as such a fundamental object in intersection theory. There is a nice (and seminal) paper by Totaro explaining in rather elementary and geometric terms the simplest case of the connexion between Milnor K-theory and higher Chow groups, mentioned by Denis-Charles Cisinski. </p> <p>Milnor K-theory is the Simplest Part of Algebraic K-Theory, K-theory 6, 177-189, 1992</p>