User evgeny shinder - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:44:28Z http://mathoverflow.net/feeds/user/2260 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51798/generalizing-eichler-shimura-to-higher-dimension-again Generalizing Eichler-Shimura to higher dimension, again Evgeny Shinder 2011-01-12T00:09:13Z 2011-01-12T03:16:22Z <p>This question is related to </p> <p><a href="http://mathoverflow.net/questions/19390/intuition-behind-the-eichler-shimura-relation" rel="nofollow">http://mathoverflow.net/questions/19390/intuition-behind-the-eichler-shimura-relation</a></p> <p>and</p> <p><a href="http://mathoverflow.net/questions/50004/l-functions-and-higher-dimensional-eichler-shimura-relation" rel="nofollow">http://mathoverflow.net/questions/50004/l-functions-and-higher-dimensional-eichler-shimura-relation</a></p> <p>Answering the first question above, Matt Emerton gives a sketch of a proof of the Eichler-Shimura congruence relation which tells that the Hecke correspondence mod p is a sum of the graph of the Frobeniusm and it's transpose.</p> <p>I am wondering how the statement and the proof can be generalized to moduli of higher dimensional abelian varieties with level structure (and maybe some more structure, like PEL).</p> <p>It seems like the reason for having only two isogenies $E \to E'$ in Matt's answer is that p-isogenies correspond to subgroups of order p in E[p] as a scheme, and E[p] (if we assume E to be ordinary) in char p is a product of $Z/p$ and the dual group, $\mu_p$, and apparently these two groups: $Z/p$ and $\mu_p$ are the only nontrivial subgroups in E[p], so taking quotients we come up with the Frobenius and Verschiebung (dual isogeny).</p> <p>Now let's say A is an abelian variety of dimension g in char p, which has maximal p-rank, i.e. $A[p] = Z/p^g \times \mu_{p^g}$. What are the subgroups of order $p^g$ of such a group? I am not familiar with how local groups behave, but I can assume there will be g+1 isogenies $A \to A'$, each one having the Kernel of the kind $H_1 \times H_2$, where $H_1$ and $H_2$ are subgroups in $Z/p^g$ and in $\mu_{p^g}$ respectively.</p> <p>Also one probably needs a statement that abelian varieties with maximal p-rank are Zarisky open and dense in the moduli space, which is true in dimension 1.</p> <p>Then the reduction mod p of the Hecke correspondence $T_p$, appropriately defined will be a sum of these g+1 cycles? Is that making any sense?</p> <p>Thanks.</p> http://mathoverflow.net/questions/11621/obstructions-to-descend-galois-invariant-cycles Obstructions to descend Galois invariant cycles Evgeny Shinder 2010-01-13T04:10:54Z 2010-12-23T19:20:50Z <p>Let $X$ be a smooth projective variety over $F$, and $E/F$ - finite Galois extension. There is an extension of scalars map $CH^*(X) \to CH^*(X_E)$. The image lands in the Galois invariant part of $CH^*(X_E)$, and in the case of rational coefficients, all Galois-invariant cycles are in the image (EDIT: this follows from taking the trace argument).</p> <p>With integer coefficients Galois-invariant classes don't have to descend. For example, for $CH^1(X) = Pic(X)$ there is an exact sequence: $$0 \to Pic(X) \to Pic(X_E)^{Gal(E/F)} \to Br(F),$$ so we can say that the obstruction to descend a cycle lies in a Brauer group.</p> <p>Are there any known obstructions to descend elements of higher groups $CH^i$ with integer coefficients?</p> <p>In my case I have a cycle in $CH^*(X_E)^{Gal(E/F)}$ and I want to find out whether or not it's coming from $CH^*(X)$. </p> <p>(The actual cycle is described in <a href="http://mathoverflow.net/questions/9200/geometry-of-the-multilagrangian-grassmannian" rel="nofollow">here</a>.)</p> http://mathoverflow.net/questions/50004/l-functions-and-higher-dimensional-eichler-shimura-relation L-functions and higher-dimensional Eichler-Shimura relation Evgeny Shinder 2010-12-20T22:17:28Z 2010-12-21T00:36:00Z <p>From what I have been reading I understand that it is a part of the Langlands program to express Zeta-function of a Shimura variety associated to the group G in terms of L-functions attached to G. I am not sure to what extent this already has been proven.</p> <p>In the cases I "know", which are the modular curves and Shimura curves - the essential step is to use the Eichler-Shimura congruence relation, which gives a connection between the Frobenius action and the Hecke correspondences.</p> <p>Now, in the higher-dimensional case, say that of Shimura varieties of PEL-type, is there a generalization of the Eichler-Shimura relation, and does this generalization play the same role in the proof of the statement?</p> <p>Maybe someone could give a rough overview of what is used in the proof in the known cases (as far as I understand, PEL case is known)?</p> <p>Thank you.</p> http://mathoverflow.net/questions/40221/an-alternative-description-of-k-nml An alternative description of K^*/Nm(L^*) Evgeny Shinder 2010-09-27T21:19:52Z 2010-09-28T15:11:44Z <p>Is there a nice explicit description for the group $K^*/Nm_{L/K}(L^*)$ for a finite field extension $L/K$? </p> <p>What if for example, $L$ is obtained from $K$ by ajoining an n-th root of some $\alpha \in K$ (and assuming that $K$ contains the n-th root of unity)? I don't see a nice answer even for the case $n=2$.</p> <p>Thank you. </p> <p>EDIT: Thanks for the answers! Is that correct that in the case of a cyclic extension this group is isomorphic to $Br(L/K)$, since both these groups are identified with $H^2(Gal(L/K), L^*)$?</p> http://mathoverflow.net/questions/30527/delignes-proof-of-ramanujans-conjecture Deligne's proof of Ramanujan's conjecture Evgeny Shinder 2010-07-04T16:22:40Z 2010-07-11T20:55:55Z <p>I am trying to understand Deligne's proof of the Ramanujan conjecture and more generally how one associates geometric objects (ultimately, motives) to modular forms.</p> <p>As the first step, which I understand more or less, one identifies the space of cusp forms $S_k(\Gamma)$ with the first cohomology group $W$ of $X(\Gamma)$ with coefficients in some sheaf (Deligne calls W parabolic cohomology). If we assume the weight is equal to 2, then this parabolic cohomology would probably just become $H^1(X(\Gamma), \mathbb C)$.</p> <p>After that point I unfortunately understand practically nothing in Deligne's paper (modular forms and l-adic cohomology), so maybe someone could give an informal explanation:</p> <ol> <li><p>What are the next steps in order to identify the Dirichlet series corresponding to a cusp form $f$ with a Hasse-Weil series of a motive (is this what we are doing?)</p></li> <li><p>Also, what role does the Hecke action play here (in particular, how is some adelic gadget that Deligne calls the Hecke action related to the usual one)? </p></li> <li><p>And what role does the field $\mathbb Q_f$ constructed by adjoining the coefficients of $f$ play?</p></li> </ol> <p>Thanks a lot.</p> http://mathoverflow.net/questions/30302/geometric-vs-arithmetic-frobenius Geometric vs Arithmetic Frobenius Evgeny Shinder 2010-07-02T12:51:08Z 2010-07-04T15:05:04Z <p>If an algebraic variety $X$ over a field characteristic p is given by equations $f_i(x_1,...,x_k) = 0$, we can consider the variety $X^{(p)}$ obtained by applying p-th powers to all the coefficients of all $f_i$'s. Frobenius morphism, as I understand it, is a morphism $X \to X^{(p)}$, given on points as raising all coordinates to p-th power. </p> <p>Can anyone please explain me, what is the geometric Frobenius, as opposed to the arithmetic one?</p> <p>EDIT: Thanks to Florian and George for the answers. I understand the difference now. I accepted Florian's answer because he was first and also because I found the last link <a href="http://www.math.mcgill.ca/goren/SeminarOnCohomology/Frobenius.pdf" rel="nofollow">http://www.math.mcgill.ca/goren/SeminarOnCohomology/Frobenius.pdf</a> he provided especially helpful.</p> http://mathoverflow.net/questions/25971/meaning-of-the-mobius-transformations-video Meaning of the Mobius transformations video Evgeny Shinder 2010-05-26T05:16:00Z 2010-05-26T13:05:17Z <p>What is this video trying to tell us? <a href="http://www.youtube.com/watch?v=JX3VmDgiFnY" rel="nofollow">http://www.youtube.com/watch?v=JX3VmDgiFnY</a></p> <p>The statement that fractional linear transformations correspond to rotations of the sphere under the stereographic projection is wrong (since for example some fractional linear transformations have only one fixed point, which is impossible for the rotation).</p> http://mathoverflow.net/questions/22985/natural-embedding-gl-nc-cn2-0-induces-zero-on-cohomology Natural embedding GL_n(C) -> C^{n^2} \ {0} induces zero on cohomology Evgeny Shinder 2010-04-29T14:39:56Z 2010-04-29T19:58:11Z <p>The question looks like an exercise in elementary algebraic topology, but I didn't manage to solve it. I am considering this question because it is a toy example in a problem I'm thinking about.</p> <p>Let's consider the natural embedding $GL_n(\mathbb C) \to \mathbb C^{n^2} \backslash {0}$. As was discussed <a href="http://mathoverflow.net/questions/18677/cohomology-rings-of-gl-nc-sl-nc" rel="nofollow">in this question</a>, cohomology with rational coefficients of $GL_n(\mathbb C)$ is an exterior algebra on generators in degrees 1, 3, ..., 2n-1 (one generator in each degree), whereas $\mathbb C^{n^2} \backslash {0}$ is homotopy equivalent to a sphere $S^{n^2-1}$.</p> <p>I'd like to prove that the map induced on cohomology of degree $n^2 - 1$ is a zero map. Any ideas?</p> <p>Thanks.</p> http://mathoverflow.net/questions/22907/confusion-about-how-the-first-cohomology-classifies-torsors/22925#22925 Answer by Evgeny Shinder for Confusion about how the first cohomology classifies torsors Evgeny Shinder 2010-04-29T01:54:32Z 2010-04-29T01:59:59Z <p>I think the right thing to look at is $H^1(X, \mathbb C^*)$. This classifies line bundles with a flat connection, or equivalently, line bundles with locally constant transition functions.</p> <p>Now the natural embedding $\mathbb C^* \to \mathbb O_X^*$ induces on a map on cohomology $H^1(X, \mathbb C^*) \to H^1(X, \mathbb O_X^*)$ which is forgetting the flat connection.</p> http://mathoverflow.net/questions/18677/cohomology-rings-of-gl-nc-sl-nc Cohomology rings of GL_n(C), SL_n(C) Evgeny Shinder 2010-03-18T22:53:46Z 2010-03-19T09:14:54Z <p>Can anyone provide me with the reference for the following fact (idea of the proof will be appreciated too):</p> <p>Cohomology ring with $\mathbb Q$-coefficients of a group $G$ (I don't know precisely what the assumptions are: reductive complex algebraic group or maybe complex Lie group G with some restrictions. The cases I'm interested in are $GL_n(\mathbb C)$ and $SL_n(\mathbb C)$) is the exterior algebra on the generators of odd degrees, with the number of generators equal to the rank of $G$.</p> <p>This fact is attributed to H.Hopf, but I wasn't able to find a reference. </p> <p>Thanks.</p> http://mathoverflow.net/questions/16621/what-do-you-lose-when-passing-to-the-motive/16658#16658 Answer by Evgeny Shinder for What do you lose when passing to the motive? Evgeny Shinder 2010-02-28T05:09:25Z 2010-02-28T05:09:25Z <p>Both examples you consider have the property that the additive structure of the Chow groups are the same, but the multiplicative structures are different. In the first case multiplication depends on the Chern classes of the bundle, and in the second case intersection forms on $CH^1$ are $x^2 - y^2$ and $2uv$ which are different integrally. So if we capture multiplicative structure of the motive (which is $\Delta: M(X) \to M(X) \otimes M(X)$ I think) we'll be able to do better.</p> <p>For your example 2, both varieties are cellular (glued from affine spaces), and the numbers of cells in each dimension are the same. Such varieties always have isomorphic (Tate) motives.</p> <p>In general, I think there's no better answer to the question "What can we recover about the variety X from its motive?" rather than the trivial one: "We can recover all the reasonable cohomology theories evaluated on X". I am very curious what other people will say, though.</p> <p>As an aside note, I remember reading somewhere that it is expected that the integral motive of a quadric determines the quadractic form itself.</p> http://mathoverflow.net/questions/14587/understanding-the-definition-of-the-lefschetz-pure-effective-motive/14711#14711 Answer by Evgeny Shinder for Understanding the definition of the Lefschetz (pure effective) motive Evgeny Shinder 2010-02-08T23:32:35Z 2010-02-08T23:32:35Z <p>Let me make a comment on the question number 3: why do we have to invert $\mathbb L$ in order to get a rigid category?</p> <p>Note that the category formed by direct sums of powers of $\mathbb L$ is equivalent to the category of nonnegatively graded vector spaces over $\mathbb Q$ (if we work with $\mathbb Q$ - coefficients), with $\mathbb L$ corresponding to $\mathbf Q[1]$ a 1-dimensional vector space in degree 1. The equivalence is provided by $CH^*$ (Chow groups) functor.</p> <p>The fact that it is an equivalence follows from the fact that $Hom(\mathbb L^i, \mathbb L^j) = \mathbb Q$ if $i=j$ and 0 otherwise.</p> <p>Now to make a rigid category out of nonegatively graded vector spaces we must include all $\mathbb Z$-graded spaces, so that $\mathbb Q[1]$ is dual to $\mathbb Q[-1]$. This corresponds to inverting $\mathbb L$.</p> <p>And once we invert $\mathbb L$ we indeed get a rigid category, as Matt Emerton explained: the dual to $M(X)$ is $M(X)(-dim X)$ (I believe that whether you have a minus sign or not depends on definitions).</p> <p>As for the question how do we think of $\mathbb L$, the picture is that the decomposition $M(\mathbb P^1) = 1 \oplus \mathbb L$ corresponds to cellular decomposition of $\mathbb P^1$ into disjoint union of a point and a line. This generalizes to any variety admitting cellular decomposition (Grassmannians and quadrics for example): their motives are sums of Lefschetz motives, a summand $L^d$ stands for each cell of dimensions d.</p> http://mathoverflow.net/questions/6834/kunneth-formula-for-motivic-cohomology/12948#12948 Answer by Evgeny Shinder for Kunneth formula for motivic cohomology Evgeny Shinder 2010-01-25T17:52:38Z 2010-01-25T18:13:36Z <p>I now remember a nice argument, why there's no Kunneth formula for Chow groups of $X \times X$ unless $X$ has a Tate motive. Let $X$ be smooth projective of dimension $d$. We start with a decomposition of a diagonal: $$[\Delta] = \sum_{i,j} \alpha^i_j \beta^{d-i}_j \in \oplus_i CH^i(X) \otimes CH^{d-i}(X)$$ We can assume $\alpha^i_j$ are linearly independent. In this case we can show that $\alpha^i_j$ form a basis of Chow groups and $\beta^{d-i}_j$ is the dual basis.</p> <p>Indeed, as a correspondence $[\Delta]$ acts as identity on Chow groups, so for any class c, $$c = [\Delta]c = \sum_{i,j} \alpha^i_j deg(\beta^{d-i}_j \cup c),$$ and the claim follows if we substitute $c = \alpha^i_j$.</p> <p>Now $CH_i(X) = Hom(\mathbb Z(i)[2i], M(X))$ and we can consider the set of $\alpha^i_j$ as a morphism of motives $$\oplus_{i,j}\mathbb Z(i)[2i] \to M(X).$$ A simple computation shows that it is an isomorphism with the inverse given by $\beta^i_j$.</p> <p>And of course, on the other hand, if $X$ has a Tate motive, then Kunneth formula for Chow groups follows (it doesn't answer the question, since I only consider smooth projective varieties).</p> http://mathoverflow.net/questions/11614/grassmannian-bundle-theorem Grassmannian bundle theorem Evgeny Shinder 2010-01-13T03:32:03Z 2010-01-13T23:46:41Z <p>Let's consider a vector bundle $E$ of rank $n$ over a compact manifold $X$. Consider the associated Grassmannian bundle $G$ for some $k &lt; n$, obtained by replacing each fiber $E_x$ by $Gr(k,E_x)$.</p> <p>Let's suppose that there is a full flag of subbunldes $F_1 \subset F_2 \dots \subset F_n \subset E$. I think that in this case we are able to define relative Schubert cycles on G which restrict to usual Schubert cycles on each fiber so that we can apply Leray-Hirsh theorem to deduce that $H^*(G) = H^*(X) \otimes H^*(Gr(k,n))$.</p> <ol> <li>Is the reasoning above correct?</li> <li>Can we still compute $H^*(G)$ in the case when the full flag of subbundles doesn't exist?</li> </ol> <p>EDIT: I meant complex vector bundles and complex Grassmannians. Also the bundle can be assumed holomorphic or algebraic if it makes a difference.</p> <p>EDIT: Ben in his answer mentions Serre's spectral sequence that can be used in this case. Is there a reason why it will degenerate to leave $H^*(X) \otimes H^*(Gr(k,n))$ as a result?</p> http://mathoverflow.net/questions/10820/is-the-scalar-extension-functor-for-chow-motives-conservative Is the scalar extension functor for Chow motives conservative? Evgeny Shinder 2010-01-05T17:09:04Z 2010-01-13T03:34:14Z <p>Denote $CHM(F)$ to be <a href="http://en.wikipedia.org/wiki/Chow%5Fmotive#Definition%5Fof%5Fpure%5Fmotives" rel="nofollow">the category of Chow motives</a> over a field $F$.</p> <p>Let's consider an algebraic exension $E/F$, then there is a natural extension of scalars functor $CHM(F) \to CHM(E)$.</p> <p>I was wondering if this functor is conservative, i.e. if a morphism $f: M \to N$ becomes an isomorphism after a field extension, does it imply $f$ is an isomorphism itself?</p> <p>A related question is: if a motive $M$ becomes zero after a field extension, does it imply that $M = 0$? I believe this question is weaker, than that of being conservative.</p> <p><a href="http://www.math.ucla.edu/~merkurev/papers/nilpotence9.dvi" rel="nofollow">Merkurjev-Gille-Chernousov</a> (Corollary 8.4) prove this for motives of homogenous spaces for algebraic group actions (so-called Rost Nilpotence theorem, since it was originally prove by Rost for quadrics). </p> <p>Do people believe that this holds in general? Is it related to some standard motivic conjectures?</p> <p>Thanks.</p> http://mathoverflow.net/questions/9961/colimits-of-schemes/10007#10007 Answer by Evgeny Shinder for Colimits of schemes Evgeny Shinder 2009-12-29T05:58:13Z 2009-12-29T05:58:13Z <p>Direct limits include in particular quotients.</p> <p>I think that if $Z \subset X$ is a closed subscheme, then $X/Z$ usually doesn't exist.</p> <p>Let's take for $Z$ a divisor on $X$, such that there exist a rational function $f: X \to \mathbb P^1$ with poles only on $Z$. Then according to the universal property of quotients such a function descends to a function $f \bar: X/Z \to \mathbb P^1$, with the pole consisting of one point (I'm little uncertain, why it's a point, but perhaps it must be a point). But that's impossible unless $dim(X)=1$.</p> http://mathoverflow.net/questions/9492/how-does-one-find-vanishing-algebraic-cycles How does one find vanishing algebraic cycles? Evgeny Shinder 2009-12-21T17:12:18Z 2009-12-29T05:18:09Z <p>I have a question, related to what I <a href="http://mathoverflow.net/questions/9200/geometry-of-the-multilagrangian-grassmannian" rel="nofollow">asked before</a>. Let's consider a smooth hyperplane section $X$ of a smooth projective variety $Y$ over $\mathbb C$. According to Weak Lefschetz theorem, cohomology groups of $X$ coincide with those of $Y$ in all dimensions except for the middle one. In the middle dimensions the pull-back $i^*: H^d(Y) \to H^d(X)$ is injective, but not surjective, and the "new" cycles on $X$ are called vanishing cycles. (The reason for such a name is that these "new" cycles will vanish when we approach singular fibers on the Lefschetz pencil.) Vanishing cycles also can be described as the ones that live in the kernel of $i_*: H^d(X) \to H^{d+2}(Y)$. </p> <p>Let's consider the case when $X$ is even-dimensional, so that we can hope that the vanishing cycles are algebraic.</p> <p>For example, for a smooth even-dimensional quadric in $\mathbb {CP}^n$ there exist one vanishing cycle - it is a difference $[E_1] - [E_2]$ of two maximal linear subspaces from different classes. </p> <p>Another example I thought about is a smooth cubic surface in $\mathbb {CP}^3$, vanishing cycles here are generated by differences of pairs of lines $[l_1] - [l_2]$ lying on the cubic.</p> <p>Now I wanted to ask, what are other examples people have in mind? I'm interested in the case, when vanishing cycles actually are algebraic.</p> <p>Is there a general method to describe such cycles in concrete situations (like <a href="http://mathoverflow.net/questions/9200/geometry-of-the-multilagrangian-grassmannian" rel="nofollow">MG(3,6)</a>)?</p> <p>Thanks</p> <p>EDIT: Probably winter break is not a best time to start a bounty...</p> http://mathoverflow.net/questions/9200/geometry-of-the-multilagrangian-grassmannian Geometry of the multilagrangian Grassmannian Evgeny Shinder 2009-12-17T20:54:21Z 2009-12-22T20:24:35Z <p>Let's introduce the following variety $MG(3,6)$, which is a "multisymplectic" analog of a Lagrangian Grassmannian $LG(3,6)$.</p> <p>Consider a 3-form $\omega = dx1 \wedge dx2 \wedge dx^3 - dx4 \wedge dx5 \wedge dx^6$ on a 6-dimensional vector space $V$ over an algebraically closed field $K$, $char(K) = 0$ (I feel that it's safe to assume $K = \mathbb C$). We can consider a subvariety $MG(3,6)$ of Grassmannian $Gr(3,6)$ consisting of 3-planes $E$ satisfying $\omega(E)=0$.</p> <p>$MG(3,6)$ turns out to be an 8-dimensional smooth hyperplane section in $Gr(3,6)$ w.r.t the Plucker embedding. I believe that $MG(3,6)$ is NOT a homogenous space for a group action.</p> <p>Now the question is: What can we say about geometry of $MG(3,6)$? More precisely, I'd like to compute the cohomology $H^*(MG(3,6))$ in terms of some "canonical" cycles, like Chern classes of something etc. </p> <p>According to the Weak Lefschetz theorem, there's no problem in small codimensions: the map $H^{2k}(Gr(3,6)) \to H^{2k}(MG(3,6))$ is an isomorphism for $k &lt; 4$, and therefore lower codimensional cohomology groups are generated by Chern classes of canonical bundle coming from Grassmannian.</p> <p>So the question really is: how do I find a nice description of cycles in the middle dimension? I know for sure, that there is one cycle which doesn't pull-back from Grassmannian.</p> <p><em>Remark</em>. The reason I'm looking for a "canonical" representation of cycles is that in fact I have some twisted form $X/F$ of $MG(3,6)/\bar F$. What I'm really looking at is cohomology (Chow groups, actually) of $X$, so I hope to find a basis of cycles which would descend to the field of definition.</p> <p>Thanks.</p> http://mathoverflow.net/questions/9492/how-does-one-find-vanishing-algebraic-cycles/9552#9552 Answer by Evgeny Shinder for How does one find vanishing algebraic cycles? Evgeny Shinder 2009-12-22T17:43:47Z 2009-12-22T18:02:44Z <p>Here's the method I had in mind looking at the examples I had. Let $X \subset Y$ be a smooth hyperplane section. Our goal is to detect cycles on $X$ that are not complete intersections of $X$ with cycles on $Y$. </p> <p>Let's consider a cycle $Z$ on $Y$ such that $Z \cap X$ is reducible, say $$Z \cap X = A + B$$ for some cycles $A$, $B$ on $X$. Then we may hope $A - B$ or some similar combination can be a vanishing.</p> <p>For quadrics and cubics above we take $Z$ to be tangent linear subspace of appropriate dimension.</p> <p>Has anybody seen something like that applied in other cases?</p> <p>The sad thing is that I <em>sort of</em> can make it work for $MG(3,6)$ to describe a vanishing cycle, but it's unclear from the description I get whether the cycle is rational or not.</p> http://mathoverflow.net/questions/9321/does-milnor-k-theory-arise-from-waldhausen-k-theory/9545#9545 Answer by Evgeny Shinder for Does Milnor K-Theory arise from Waldhausen K-Theory Evgeny Shinder 2009-12-22T16:22:02Z 2009-12-22T16:22:02Z <p>I don't know if there any evidence for this to be true. Note that Quillen K-groups <em>are defined</em> as homotopy groups of some space (+-construction, Q-construction, Waldhausen construction etc), whereas Milnor K-groups were defined in terms of generators and relations, which generalize generators and relations for classical K_2.</p> <p>More invariantly Milnor K-groups can be constructed using homology of GL_n (paper of Suslin and Nesterenko) or as certain motivic cohomology groups of a field (Suslin-Voevodsky). However, these constructions are unrelated to any homotopy groups.</p> <p>Also, I'm not sure how you define Milnor K-theory for a general ring R? (I was interpreting your question with "ring R" replaced by "field F".)</p> http://mathoverflow.net/questions/8244/what-is-the-name-for-the-following-categorical-property/8310#8310 Answer by Evgeny Shinder for What is the name for the following categorical property? Evgeny Shinder 2009-12-09T04:10:00Z 2009-12-09T16:24:18Z <p>As other people commented, the language of categories is richer than the language of sets with structures (Bourbaki structures). There are many categories, where objects don't have an underlying set.</p> <p>However, one can restate the property you formulate as follows: the category <strong>C</strong> admits a <a href="http://ncatlab.org/nlab/show/faithful+functor" rel="nofollow">faithful</a> <a href="http://ncatlab.org/nlab/show/conservative+functor" rel="nofollow">conservative</a> functor to <strong>Sets</strong>. Then we can interpret the fiber of this functor over a given set S as the set of structures on S and call the functor a forgetful functor. By faithfulness the homs in <strong>C</strong> will be subsets of those in <strong>Sets</strong>, and we can say that the homs in <strong>C</strong> preserve the structure.</p> http://mathoverflow.net/questions/7715/what-is-restriction-of-scalars-for-a-torus/7755#7755 Answer by Evgeny Shinder for What is "restriction of scalars" for a torus? Evgeny Shinder 2009-12-04T06:15:07Z 2009-12-04T06:15:07Z <p>It's not that hard at all. Here is an example. Let $k = \mathbf R$ and $K = \mathbf C$. Consider a 1-dimensional torus $G_m$ over $\mathbf C$. It basically the group $\mathbf C^*$ over $\mathbf C$. </p> <p>Now $G = Res_{\mathbf C/\mathbf R} G_m$ is the same group $\mathbf C^*$ considered as group over $\mathbf R$.</p> http://mathoverflow.net/questions/7604/indexing-the-line-bundles-over-a-grassmannian/7614#7614 Answer by Evgeny Shinder for Indexing the line bundles over a Grassmannian. Evgeny Shinder 2009-12-02T21:01:41Z 2009-12-03T17:55:10Z <p>Algebraic line bundles on a smooth variety $X$ are classified by the Picard group $Pic(X) = H^1(X, \mathbf O_X^*)$. This is an exercise in Hartshorne's book, basically every line bundle is mapped to it's gluing cocycle. The group $Pic(X)$ is also equal to the group $CH^1(X)$ of divisors modulo rational equivalence. The map sends a line bundle to it's first Chern class.</p> <p>Now for a projective space Picard group is $\mathbf Z$ generated by $\mathbf O(1)$. Picard group of a Grassmannian is also $\mathbf Z$. I believe that the generator is a pullback of $\mathbf O(1)$ for a Plucker embedding $Gr(n,k) \subset \mathbf P^N$.</p> <p>One can prove it as follows: both $\mathbf P^n$ and $Gr(n,k)$ are algebraically cellular, meaning that they consist of pieces isomorphic to affine spaces: for $\mathbf P^n$ this is obvious, and for $Gr(n,k)$ the cells are Schubert cells.</p> <p>For such varieties Chow groups coincide with cohomology groups and are generated by cells (it's like computing cohomology of CW complex without cells of odd dimension). Since there is only one cell of complex codimension 1 for projective spaces and Grassmannians, we get $Pic(X) = H^2(X, \mathbf Z) = \mathbf Z$.</p> <p>EDIT: </p> <ul> <li><p>Actually, a projective space IS a Grassmannian. :)</p></li> <li><p>As other people commented, the generator of the $Pic(Gr(n,k))$ is the n'th wedge power of the canonical vector bundle of rank n.</p></li> </ul> http://mathoverflow.net/questions/7664/how-do-we-study-the-theory-of-reductive-groups/7691#7691 Answer by Evgeny Shinder for How do we study the theory of reductive groups? Evgeny Shinder 2009-12-03T17:50:57Z 2009-12-03T17:50:57Z <p>I found the following introductory book very useful: Waterhouse, Introduction to Affine Group Schemes.</p> <p>It is short, clear and fun to read. The book doesn't assume the previous knowledge of algebraic geometry, and depending on your background it can be either advantage or a disadvantage.</p> <p>However, it's not going deep into theory of reductive groups. More advanced standard textbooks on algebraic groups are Springer, Borel and Humphreys.</p> http://mathoverflow.net/questions/51798/generalizing-eichler-shimura-to-higher-dimension-again/51816#51816 Comment by Evgeny Shinder Evgeny Shinder 2011-01-12T03:32:25Z 2011-01-12T03:32:25Z Matt, thanks for the reference! The sum on page 25 in that paper looks more or less like I was thinking: in the case of Siegel modular varieties, T_p mod p is a sum of g+1 cycles (with multiplicities), each one with the prescribed type of isogeny's kernel. http://mathoverflow.net/questions/11621/obstructions-to-descend-galois-invariant-cycles/50252#50252 Comment by Evgeny Shinder Evgeny Shinder 2010-12-24T19:05:39Z 2010-12-24T19:05:39Z Alena, thank you for reply. When I was asking the question I was keeping in mind the vanishing cycle in the Chow group of the middle dimension which was CH^4 in my case. However, by now I realize that my question was not a very good one - the technique I was looking for clearly does not exist in general. Your paper seems very interesting to me and I will take a look at it when I have time. http://mathoverflow.net/questions/30527/delignes-proof-of-ramanujans-conjecture/31462#31462 Comment by Evgeny Shinder Evgeny Shinder 2010-07-12T17:06:05Z 2010-07-12T17:06:05Z Brian, thank you for your advise. However, first of all I am not at Northwestern right now. Moreover, I feel more comfortable asking questions here, since this way I am not wasting anybody's time - only the people who are willing to answer will do so. Also, in this particular case since I gave up 500 points of my reputation for this question, I thought Thanos wouldn't mind claryfing his answer. http://mathoverflow.net/questions/30527/delignes-proof-of-ramanujans-conjecture/31462#31462 Comment by Evgeny Shinder Evgeny Shinder 2010-07-12T15:03:55Z 2010-07-12T15:03:55Z 5. After the Dirichlet series is identified with the Artin L-series of the Galois representation W, what makes it possible to use Weil conjectures? Will this L-series be a factor in the Hasse-Weil zeta function of M, because W is a &quot;piece&quot; of cohomology of M? http://mathoverflow.net/questions/30527/delignes-proof-of-ramanujans-conjecture/31462#31462 Comment by Evgeny Shinder Evgeny Shinder 2010-07-12T14:53:25Z 2010-07-12T14:53:25Z 4. &quot;The first theorem of Eichler and Shimura tells you that this isotypic component has dimension 2&quot; Is it because the space of weight 12 forms is 1-dimensional? http://mathoverflow.net/questions/30527/delignes-proof-of-ramanujans-conjecture/31462#31462 Comment by Evgeny Shinder Evgeny Shinder 2010-07-12T14:51:38Z 2010-07-12T14:51:38Z 3. What is &quot;f-isotypic&quot;? http://mathoverflow.net/questions/30527/delignes-proof-of-ramanujans-conjecture/31462#31462 Comment by Evgeny Shinder Evgeny Shinder 2010-07-12T14:50:50Z 2010-07-12T14:50:50Z 2. Also, what does &quot;Big Hecke algebra&quot; mean? http://mathoverflow.net/questions/30527/delignes-proof-of-ramanujans-conjecture/31462#31462 Comment by Evgeny Shinder Evgeny Shinder 2010-07-12T14:49:50Z 2010-07-12T14:49:50Z 1. Is there an easy way to see why GL(A)-action must be related to Hecke-correspondeces action? http://mathoverflow.net/questions/30527/delignes-proof-of-ramanujans-conjecture/31462#31462 Comment by Evgeny Shinder Evgeny Shinder 2010-07-12T14:47:03Z 2010-07-12T14:47:03Z Thanos, thank you for your answer. http://mathoverflow.net/questions/30302/geometric-vs-arithmetic-frobenius/30309#30309 Comment by Evgeny Shinder Evgeny Shinder 2010-07-02T15:25:52Z 2010-07-02T15:25:52Z Florian, thank you for the answer. I'll think about all that. Actually the Deligne's 68/69 Bourbaki seminar is one of the papers I'm struggling with right now. http://mathoverflow.net/questions/25971/meaning-of-the-mobius-transformations-video/26014#26014 Comment by Evgeny Shinder Evgeny Shinder 2010-05-26T16:18:10Z 2010-05-26T16:18:10Z Thanks. I am glad that there is some mathematical content in this video. http://mathoverflow.net/questions/22985/natural-embedding-gl-nc-cn2-0-induces-zero-on-cohomology/22988#22988 Comment by Evgeny Shinder Evgeny Shinder 2010-04-29T19:59:31Z 2010-04-29T19:59:31Z This matches with n^2=1+2+...+(2n-1), the degree of a top cohomology class in the exterior algebra H^*(GL_n(C)). http://mathoverflow.net/questions/22985/natural-embedding-gl-nc-cn2-0-induces-zero-on-cohomology/22988#22988 Comment by Evgeny Shinder Evgeny Shinder 2010-04-29T14:57:55Z 2010-04-29T14:57:55Z That was quick. Thanks! http://mathoverflow.net/questions/18677/cohomology-rings-of-gl-nc-sl-nc/18721#18721 Comment by Evgeny Shinder Evgeny Shinder 2010-03-23T19:52:01Z 2010-03-23T19:52:01Z Thanks! Can you please give me a reference for the statement that &quot;H(G;Q) is the exterior algebra on the generators shifted down one degree&quot;. http://mathoverflow.net/questions/16621/what-do-you-lose-when-passing-to-the-motive/16653#16653 Comment by Evgeny Shinder Evgeny Shinder 2010-02-28T05:13:50Z 2010-02-28T05:13:50Z Over Q Severi-Brauer varieties and quadrics all have the same motives indeed, but integrally they are different.