User dan grayson - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:32:19Z http://mathoverflow.net/feeds/user/15247 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/128344/k-theory-of-monoidal-categories/128502#128502 Answer by Dan Grayson for K-theory of monoidal categories Dan Grayson 2013-04-23T15:35:40Z 2013-04-23T16:31:42Z <p>(I'm not sure the term <em>exact monoidal category</em> is a standard one in the literature, so I'll just assume I know what you mean by it.)</p> <p>Yes, $K_n$ commutes with products, as some have mentioned, but the tensor monoidal functor $\otimes : \mathcal C \times \mathcal C \to \mathcal C$ is not exact, so it doesn't induce a map $K_n \mathcal C \times K_n \mathcal C \to K_n \mathcal C$. Rather, the functor $\otimes : \mathcal C \times \mathcal C \to \mathcal C$ is <em>bi-exact</em>, in the sense that it is an exact functor in each of its two variables, separately. (That's what makes $K_0 \mathcal C$ into a ring.)</p> <p>The result is that the collection of abelian groups $K_n \mathcal C$, indexed by $n$, forms a graded ring, with products $K_m \mathcal C \otimes K_n \mathcal C \to K_{m+n} \mathcal C$. Assuming the tensor product is commutative up to natural isomorphism, the ring is commutative in the graded sense that $x y = (-1)^{mn} y x$.</p> <p>The actual construction is intricate, the main idea being to construct a map $K \mathcal C \wedge K \mathcal C \to K \mathcal C$ of spectra. See, for example, Waldhausen's approach in section 9 of Algebraic K-theory of generalized free products. I, II. Ann. of Math. (2) 108 (1978), no. 1, 135–204. It uses the Q-construction of Quillen, but could be simplified by using the S-construction of Segal that Waldhausen developed extensively.</p> <p>My paper with Gillet, <a href="http://www.math.illinois.edu/~dan/Papers/loopsQ.pdf" rel="nofollow">The loop space of the Q-construction</a>, Illinois Journal of Mathematics, 31 (1987) 574-597, gives another approach that avoids both spectra (i.e., delooping) and the phony multiplication trap that Steve mentioned.</p> <p>PS: My paper <a href="http://www.math.uiuc.edu/K-theory/0988/" rel="nofollow">Algebraic K-theory via binary complexes</a>, Journal of the American Mathematical Society, 25 (2012) 1149-1167, gives an algebraic description of the elements of the higher K-groups that converts the construction of the product into a simple exercise involving tensor product of chain complexes.</p> http://mathoverflow.net/questions/125969/homology-of-b-s-1-s-computation-in-the-proof-that-q/126000#126000 Answer by Dan Grayson for homology of $B S^{-1} S$ computation in the proof that $+ = Q$ Dan Grayson 2013-03-30T11:43:00Z 2013-03-30T11:43:00Z <p>If you want to apply Kunneth, then you should take into account that $H_0 X$ is the free abelian group on $\pi_0 X$, so you have to replace your use of $K_0R$ by the free abelian group on $K_0R$. Then the Tor vanishes, and we see that Srinivas' formula is incorrect.</p> <p>Simpler than applying Kunneth would be to think about how the homology groups of a space are related to the homology groups of its connected components: it's the direct sum. So if a space $X$ has all its components homotopy equivalent to each other, as in this case, then $H_p X$ is $\coprod_{\pi_0 X} H_p X^0$, where $X^0$ is one of the components.</p> <p>One further issue to think about is naturality of the isomorphism, since the homotopy equivalence between two components of $X$ depends on a choice.</p> http://mathoverflow.net/questions/125125/categorical-description-of-the-second-k-group/125136#125136 Answer by Dan Grayson for Categorical description of the second K-group Dan Grayson 2013-03-21T08:19:42Z 2013-03-21T08:19:42Z <p>Algebraic generators and relations for Quillen's K-group $K_n(P)$ are given in this paper: "<a href="http://dx.doi.org/10.1090/S0894-0347-2012-00743-7" rel="nofollow">Algebraic K-theory via binary complexes</a>". Those you mention don't give the Quillen K-group $K_1(P)$ in general, but just for Quillen-exact categories where every short exact sequence splits; the group they give is known as Bass' $K_1$.</p> http://mathoverflow.net/questions/99713/classify-matrices-up-to-similarity-over-arbitrary-commutative-ring/99848#99848 Answer by Dan Grayson for Classify matrices up to similarity over arbitrary (commutative) ring. Dan Grayson 2012-06-17T17:20:28Z 2012-06-17T17:20:28Z <p>The category $P(R,\mathbb G_m)$ is defined in such a way that a morphism $(R^m,A) \to (R^n,B)$ is a matrix C with CA=BC. The category (groupoid) $iP(R,\mathbb G_m)$ of isomorphisms consists of those morphisms where C is invertible, i.e., where C is a similarity between A and B, and for good rings, such as commutative nonzero rings, that will force m=n. Perhaps that will help with the original "headache".</p> <p>The automorphisms are those isomorphisms where m=n and A=B.</p> <p>To write the category (groupoid) $iP(R,\mathbb G_m)$ of automorphisms as equivalent to a disjoint union of groupoids, each with just one object, one chooses an object in each isomorphism class and discards the others, which is what is what prompted the original question about classification results.</p> <p>Some papers of mine discuss the K-theory of that category, but without investigating the original question, so whether they are relevant depends on the unstated goal. In case they are relevant, here they are:</p> <p><a href="http://www.math.uiuc.edu/~dan/cv.html#endo" rel="nofollow">http://www.math.uiuc.edu/~dan/cv.html#endo</a></p> <p><a href="http://www.math.uiuc.edu/~dan/cv.html#witt" rel="nofollow">http://www.math.uiuc.edu/~dan/cv.html#witt</a></p> <p><a href="http://www.math.uiuc.edu/~dan/cv.html#auto" rel="nofollow">http://www.math.uiuc.edu/~dan/cv.html#auto</a></p> <p><a href="http://www.math.uiuc.edu/~dan/cv.html#wfilt2" rel="nofollow">http://www.math.uiuc.edu/~dan/cv.html#wfilt2</a></p> <p>The approach chosen by the original poster, using $S^{-1} S$, will address only the direct sum K-theory of the category, not the (usual) exact sequence K-theory of the category. The difference between the two is relevant to the last paper on the list, and is treated by Suslin in </p> <p><a href="http://www.math.uiuc.edu/K-theory/0588/" rel="nofollow">http://www.math.uiuc.edu/K-theory/0588/</a></p> http://mathoverflow.net/questions/82390/k-theory-and-regular-rings/82393#82393 Answer by Dan Grayson for K-theory and regular rings Dan Grayson 2011-12-01T16:36:04Z 2011-12-01T16:36:04Z <p>I don't know how to answer that question, but suppose we strengthen the hypothesis by making it apply also to all polynomial rings $R[t_1,\dots,t_n]$ over $R$, for $n \ge 0$. Then because $K_i(R-Mod) \to K_i(R[t]-mod)$ is an isomorphism (Quillen, Theorem 8, Higher Algebraic K-theory:I) for all noetherian rings $R$, it follows that $K_i(R) \to K_i(R[t_1,\dots,t_n])$ is an isomorphism for all $n$ --- that is called $K_i$-regularity of $R$. The paper "$K$-regularity, cdh-fibrant Hochschild homology, and a conjecture of Vorst", J. Amer. Math. Soc. 21 (2008), no. 2, 547–561, by Cortiñas, Haesemeyer, and Weibel available at <a href="http://www.math.uiuc.edu/K-theory/0783/chwpost.pdf" rel="nofollow">http://www.math.uiuc.edu/K-theory/0783/chwpost.pdf</a> shows that $K_i$-regularity of $R$ for all $i$ implies regularity of $R$ if $R$ is an algebra essentially of finite type over a field of characteristic 0.</p> http://mathoverflow.net/questions/68630/gersten-for-homotopy-invariant-k-theory-of-non-singular-varieties/79805#79805 Answer by Dan Grayson for Gersten for homotopy invariant K-theory of non-singular varieties. Dan Grayson 2011-11-02T06:02:16Z 2011-11-02T06:02:16Z <p>Your question is an interesting one. But smoothness is essential to Quillen's proof of Gersten's conjecture (in his paper, Higher Algebraic K-theory : I), which, I'm assuming, you intend to use to deduce the consequence you state.</p> http://mathoverflow.net/questions/9321/does-milnor-k-theory-arise-from-waldhausen-k-theory/65685#65685 Answer by Dan Grayson for Does Milnor K-Theory arise from Waldhausen K-Theory Dan Grayson 2011-05-22T00:02:50Z 2011-05-22T00:02:50Z <p>The original question seems not to have been answered yet. One answer might be that it would be unnatural to expect all the Milnor K-groups of a field R to arise as the homotopy groups of a single space $K(F(R))$, because the natural way they currently arise is as homotopy groups of separate spaces, or better, of separate spectra. The spectra are the Eilenberg-MacLane spectra $\mathbb Z(n)$ associated to the chain complexes that compute motivic cohomology of $R$, namely, $K_n^M R = \pi_{-n} \mathbb Z(n)$.</p> http://mathoverflow.net/questions/65635/rationalised-k-theory-of-number-fields/65650#65650 Answer by Dan Grayson for Rationalised K-theory of number fields Dan Grayson 2011-05-21T15:22:27Z 2011-05-21T20:41:13Z <p>The answer is yes, but only after tensoring with $\mathbb R$.</p> <p>Thinking of the Beilinson regulator map with values in Deligne cohomology is simpler than thinking about the Borel regulator map; it's been proved that they agree with each other. </p> <p>The topological Chern character map $ch_n : kU_{i} \to H^{2n-i}(pt,\mathbb Q)$ is an isomorphism $kU_{2n} \otimes \mathbb Q \ \xrightarrow \cong \ \mathbb Q$ when $i=2n$.</p> <p>The corresponding algebraic Chern character map with values on Deligne cohomology is $ch_n : K_{i}\mathbb C \to H^{2n-i}_{\cal D}(pt,\mathbb Q(n))$. Here $\mathbb Q(n)$ (or $\mathbb Z(n)$) denotes a certain cochain complex of sheaves for the analytic topology on a complex manifold $X$. It starts in degree 0 with $\mathbb Q$ (resp., $\mathbb Z$), in cohomological degree 1 it has $\mathbb C$, and the differential map $d^0 : \mathbb Q \to \mathbb C$ is multiplication by $(2 \pi i)^ n$. The term in degree $i+1$ is the sheaf of holomorphic differentials $\Omega^i$ if $i &lt; n$ and is $0$ if $i \ge n$. The exponential map $\mathbb C \to \mathbb C ^ \times$ given by $z \mapsto e^z$ gives a quasi-isomorphism $\mathbb Z (1) \to \mathbb C ^ \times [-1]$; the degree shift there answers your second question, partially; another way of saying that is that there is a degree shift in the boundary map $c_1 : H^1(X,\mathbb C^\times) \to H^2(X,\mathbb Z)$. I say "partially", because one must know also that the regulator map involves no further degree shift; in degree 1 it's because the map $\mathbb C ^ \times \to \mathbb R$ given by $z \mapsto {\rm log} |z|$ involves no degree shift.</p> <p>Now consider the projection $\mathbb C \to \mathbb R$ that sends $(2 \pi i)^n$ to $0$ and $i^{n-1}$ to $1$; perhaps there is a better normalization for this map, such as choosing to send $(2 \pi i)^{n-1}$ to $1$. It induces a map of cochain complexes $\mathbb Q(n) \to \mathbb R[-1]$; the map it induces on Deligne cohomology, composed with the Chern character map above, is the Beilinson regulator map $$ch_n : K_{i}\mathbb C \to H^{2n-i}(pt,\mathbb R[-1]) = H^{2n-i-1}(pt,\mathbb R),$$whose only nonzero possibility is the map $ch_{n} : K_{2n-1}\mathbb C \to H^{0}(pt,\mathbb R) = \mathbb R$. For the ring of integers $A$, we get a map $K_{2n-1} A \to K_{2n-1}( A \otimes \mathbb C ) \to H^{0}(Spec(A \otimes \mathbb C),\mathbb R) = \mathbb R^{s+2t}$. Borel's theorem is recast as saying that for $n > 1$ the map induces an isomorphism of $(K_{2n-1} A) \otimes \mathbb R$ with the appropriate eigenspace for the action of $G = Gal(\mathbb C/\mathbb R)$ on $$H^0(Spec(A \otimes \mathbb C), \mathbb R (n-1)) = \mathbb R^{s+2t},$$where now I use $\mathbb R (n-1)$ to remind us how $G$ acts on this real vector space of dimension 1.</p> <p>(<strong>Added later</strong>: actually, it may be more natural to replace the $\Sigma$ in the question by $\Omega$ and to use the anti-invariants (or invariants, depending on the parity of n) under $G$ acting on $K^{top}(A \otimes \mathbb C) \otimes \mathbb C$ instead of the invariants. Thus the degree shift can be viewed as $-1 = 1 - 2$)</p> http://mathoverflow.net/questions/99713/classify-matrices-up-to-similarity-over-arbitrary-commutative-ring/99848#99848 Comment by Dan Grayson Dan Grayson 2012-06-18T19:58:07Z 2012-06-18T19:58:07Z Yes............