User s1 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:54:26Z http://mathoverflow.net/feeds/user/717 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2610/neukirchs-class-field-axiom-and-cohomology-of-units-for-unramified-extension Neukirch's class field axiom and cohomology of units for unramified extension S1 2009-10-26T13:29:05Z 2012-01-30T22:37:07Z <p>This question may be too detailed but perhaps somebody knows the answer: Neukirch proofs in his algebraic number theory book in chapter IV, proposition 6.2, that his class field axiom implies that the tate cohomology groups H^n(G(L|K),U<sub>L</sub>) for n=0,-1 vanish for finite unramified extensions L|K, where U<sub>L</sub> is the group of units. He mentions in the proof that every element a \in A<sub>L</sub> can be written as a = \epsilon * \pi<sub>K</sub>^m, where \epsilon \in U<sub>L</sub> and \pi<sub>K</sub> is a prime element in A<sub>K</sub>. Why does this work? I absolutely understand this argument, when the image of the valuation just lies in \ZZ! But how does this work for a valuation whose image is \widehat{\ZZ}? Unless A is not a profinite module, I don't know what \pi<sub>K</sub>^m is for some general m \in \widehat{\ZZ}. Unfortunately this has to work in this generality for global class field theory. </p> <p>(\ZZ denotes the integers of course, sorry for my personal notation.)</p> http://mathoverflow.net/questions/4157/universal-definition-of-tangent-spaces-for-schemes-and-manifolds Universal definition of tangent spaces (for schemes and manifolds) S1 2009-11-04T22:47:03Z 2011-03-15T18:24:01Z <p>Both schemes and manifolds are local ringed spaces which are locally isomorphic to spaces in some full subcategory of local ringed spaces (local models). Now, there is the inherent notion of the Zariski tangent space in a point (dual of maximal ideal modulo its square) which is the "right" definition for schemes and for $C^\infty$-manifolds (over $\mathbb{R}$ and $\mathbb{C}$). But for $C^r$-manifolds over $\mathbb{R}$ with $r&lt;\infty$ this is not the correct definition. Here one has to take equivalence classes of $C^r$-curves through the point. Isn't there some general definition of tangent spaces which is always the right one?</p> <p>I am also not completely sure what "right" means. So far, I think that one wants the dimension of the tangent space to be equal to the dimension of the point. This is for example the problem with the Zariski tangent spaces for $C^r$-manifolds. Can this failure be explained geometrically?</p> http://mathoverflow.net/questions/3038/errata-database Errata database? S1 2009-10-28T12:41:40Z 2010-11-12T00:32:14Z <p>Some authors do a really great job by collecting errors and comments to their books and putting a list on their websites. I wonder if there is some (perhaps wiki-style) website where errata are collected. Does anybody know?</p> 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/3798/automorphisms-of-the-totally-ordered-group-zn-with-lexicographical-order Automorphisms of the totally ordered group Z^n with lexicographical order S1 2009-11-02T14:28:06Z 2010-10-21T21:02:35Z <p>It is easy to see that the totally ordered group Z (the integers) with the natural order has no non-trivial automorphisms. Is this also true for Z^n with the lexicographical order?</p> http://mathoverflow.net/questions/4547/definitions-of-hecke-algebras Definitions of Hecke algebras S1 2009-11-07T18:45:32Z 2010-10-04T11:16:46Z <p>There is a definition of Iwahori-Hecke algebras for Coxeter groups in terms of generators and relations and there is a definition of Hecke algebras involving functions on locally compact groups. Are these two concepts somehow related? I think I read somewhere that the Hecke algebra with functions includes Iwahori-Hecke algebras, is that correct? Is there a good motivation for introducing and studying either type of algebras? How much is known about their representations?</p> http://mathoverflow.net/questions/20153/is-every-homogeneous-g-variety-of-the-form-g-h Is every homogeneous G-variety of the form G/H? S1 2010-04-02T09:43:53Z 2010-08-25T15:51:16Z <p>Let $G$ be an algebraic group over an algebraically closed field $k$. Then G/H is a quasi-projective homogeneous G-variety for any closed subgroup $H$. Now, several times I have seen something like "Let $X$ be a homogeneous $G$-variety, i.e. $X = G/H$ for a closed subgroup $H$ of $G$" and I wonder if this "i.e." is correct. This would imply that any homogeneous $G$-variety is already quasi-projective. I think this is true, when $\mathrm{char}(k) = 0$, because then the canonical <em>abstract</em> isomorphism $\pi:X \rightarrow G/G _x$ is separable and thus an isomorphism of varieties for any $x \in X$ (is this correct?). But what about $\mathrm{char}(k) > 0$? Are there counter-examples or is any (quasi-projective) homogeneous $G$-variety up to isomorphism of the form $G/H$?</p> http://mathoverflow.net/questions/33662/how-canonical-is-the-triangular-decomposition-of-a-rational-cherednik-algebra How canonical is the triangular decomposition of a rational Cherednik algebra? S1 2010-07-28T14:55:31Z 2010-07-29T17:35:34Z <p>Introduction: Let $V$ be a finite-dimensional $\mathbb{C}$-vector space, let $G \leq \mathrm{GL}(V)$ be a finite subgroup and let $\kappa:V \times V \rightarrow \mathbb{C}G$ be an alternating bilinear map. Let $I$ be the ideal in $\mathrm{T}(V) \sharp \mathbb{C}G$ generated by the elements $vw-wv-\kappa(v,w)$ with $v,w \in V$. ($\sharp$ denotes the smash product). If we let elements of $V$ have degree $1$ and elements of $\mathbb{C}G$ have degree 0, we get a grading on $\mathrm{T}(V) \sharp \mathbb{C}G$ and so a filtration on $A := \mathrm{T}(V) \sharp \mathbb{C} G/I$. As $\mathrm{gr}(A)$ is commutative in degree 1, the quotient morphism $\mathrm{T}(V) \sharp \mathbb{C} G$ induces a surjective graded algebra morphism $\xi: \mathrm{S}(V) \sharp \mathbb{C} G \rightarrow \mathrm{gr}(A)$.</p> <p>Now, $A$ is called a rational Cherednik algebra (or perhaps in this more general setting it's called a graded Hecke algebra or Drinfeld-Hecke algebra) if $\xi$ is an isomorphism.</p> <p>If I did not make a mistake, the condition that $\xi$ is an isomorphism is equivalent to the following: for any linear section $s$ of the quotient map $\mathrm{T}(V) \rightarrow \mathrm{S}(V)$, the vector space morphism $\theta_s:\mathrm{S}(V) \otimes \mathbb{C} G \rightarrow A$, $x \otimes g \mapsto q(s(x) \otimes g)$, is an isomorphism, where $q:\mathrm{T}(V) \otimes \mathbb{C} G \rightarrow A$ is the quotient map. </p> <p>Hence, a rational Cherednik algebra $A$ is $\mathrm{S}(V) \otimes \mathbb{C}G$ as a vector space, but only up to the choice of a linear section. Now, is there a <em>canonical</em> way to identify $A$ with $\mathrm{S}(V) \otimes \mathbb{C}G$? </p> <p>This might be a little pedantic, but nobody ever mentions a choice of a section and I fear that I miss an important point here. </p> <p>In an earlier version of my question I also asked if there is also a canonical way to view $S(V)$ as a subalgebra of $A$ but I see that this can't be true in general.</p> <p><br></p> <p>An additional (not totally unrelated) question: If the PBW morphism $\xi$ is an isomorphism, the algebra $A$ is already pretty nice because several properties of $\mathrm{S}(V) \sharp \mathbb{C}G$ (noetherian, prime, finite homolgical dimension, Cohen-Macaulay) are transported to $A$. But $\xi$ would in the same way as above already exist if $\kappa$ would be a map to $V \cdot \mathbb{C} G \subseteq \mathrm{T}(V) \sharp \mathbb{C}G$, i.e. if it would involve degree 1 elements. Why aren't those algebras interesting? (Perhaps because they don't have a triangular decomposition or they aren't deformations of $\mathrm{S}(V) \sharp \mathbb{C}G$?) I'm just looking for a natural reason to look at the $A_\kappa$ (I remember very well my question and the answers about reasons for studying rational Cherednik algebras but let's try it this way...)</p> http://mathoverflow.net/questions/32524/dualizing-sheaf-on-a-cohen-macaulay-variety Dualizing sheaf on a Cohen-Macaulay variety S1 2010-07-19T19:10:57Z 2010-07-19T21:43:44Z <p>Let $k$ be an algebraically closed field and let $X$ be a Cohen-Macaulay variety over $k$, i.e. all local rings are Cohen-Macaulay (perhaps this can later be generalized). What is the dualizing sheaf on $X$, what is its central/characterizing property and can one write it down explicitly?</p> <p>I'm asking this in such a naive way (knowing that there is a vast literature on this) because I have the following problem: I'm (obviously) not an expert in algebraic geometry and came across a definition stating that the dualizing sheaf of $X$ is $\omega_X := j_* \omega_{X_s}$, where $j:X_s \rightarrow X$ is the inclusion of the smooth locus $X_s$ of $X$ and $\omega_{X_s} := \mathrm{det}(\Omega_{X_s/k}^1)$. I can take this of course as a definition (it is pretty easy, and I also know something about dualizing sheaves on smooth varieties) but I don't know what the central properties of this sheaf in this setting are. I know that there is Grothendieck-Verdier-Neeman(-more names) duality and while browsing through Hartshorne's book <em>Residues and Duality</em> I tried to deduce this definition from the abstract "nonsense" but I failed. I know that there exists this very general duality and that there happens something in the Cohen-Macaulay case but this was over my head!</p> <p>Amnon Neeman defines in his article <em>Derived categories and Grothendieck duality</em> a <em>dualizing complex</em> of a noetherian and separated scheme $X$ to be an object $\mathcal{J} \in \mathbf{D}^b( \mathrm{Coh}(X))$ such that $\mathbb{R}\mathcal{H}om(-,\mathcal{J}):\mathbf{D}^b(\mathrm{Coh}(X))^{op} \rightarrow \mathbf{D}^b(\mathrm{Coh}(X))$ is an (triangulated) equivalence. Is it correct that if $X$ is a (separated?) noetherian Cohen-Macaulay scheme the sheaf defined above is a dualizing complex in this sense and that this is the characterizing property I was asking for? This is the only idea I have so far...</p> http://mathoverflow.net/questions/32353/confusing-definitions-in-lius-algebraic-geometry-and-arithmetic-curves Confusing definitions in Liu's Algebraic geometry and arithmetic curves? S1 2010-07-18T12:13:19Z 2010-07-18T14:21:38Z <p>In Qing Liu's book <em>Algebraic geometry and arithmetic curves</em> I came across several confusing definitions. Several times he defines a notion only for a subclass of schemes/morphisms but later he is never <em>explicitly</em> mentioning these extra conditions again. Here are some examples:</p> <ul> <li><p>Let $X$ be a locally Noetherian scheme, and let $x \in X$ be a point. We say that $X$ is <em>regular</em> at $x$ if [...]. We say that $X$ is <em>regular</em> if it is regular at all of its points. <b>Question:</b> If he later says "Let $X$ be a regular scheme", then is it implicit that $X$ is locally Noetherian? If so, then why doesn't he say "A scheme is called regular if it is locally Noetherian and [...]"?</p></li> <li><p>Let $X$ be a reduced Noetherian scheme. Let $\xi_1,\ldots,\xi_n$ be the generic points of $X$. We say that a morphism of finite type $f:Z \rightarrow X$ is a <em>birational</em> morphism if [...]. <b>Question</b>: If he later says that a morphism $f:Z \rightarrow X$ of (arbitrary) schemes is birational, then is it implicit that $X$ is reduced Noetherian and that $f$ is of finite type? If so, then why doesn't he say "A morphism $f$ is called birational if it is of finite type, if $X$ is reduced Noetherian and if [...]"?</p></li> <li><p>Now it gets really confusing: Let $X$ be a reduced locally Noetherian scheme. A proper birational morphism $\pi:Z \rightarrow X$ with $Z$ regular is called a <em>desingularization</em> of $X$. <b>Question:</b> He defined <em>birational</em> only for reduced Noetherian schemes. What is <em>birational</em> for reduced locally Noetherian schemes? Is his <em>desingularization</em> now automatically of finite type?</p></li> </ul> <p>Edit:</p> <ol> <li><p>In Liu's book I found the following definition now: We say that a morphism $f:X \rightarrow Y$ is <em>proper</em> if it is of finite type, separated and universally closed. So, first of all, I think that this definition is now given in the non-confusing style, and second, this implies that the desingularizations above are of finite type (although it doesn't answer the locally Noetherian/Noetherian question).</p></li> <li><p>I was asking "...then why doesn't he say that..." because I wasn't sure (and I'm still not sure) if there is some "higher truth" in this style of definition. Of course nobody except for Liu himself can answer this but perhaps someone else has more experience than I have and can give an explanation for this...</p></li> </ol> http://mathoverflow.net/questions/14440/how-can-i-conclude-that-i-live-in-a-solar-system How can I conclude that I live in a solar system? S1 2010-02-06T22:39:42Z 2010-07-11T00:09:57Z <p>Well, this is an awkward question and I don't know if it is mathematical enough for MO (I'm sorry if not) but I'll try it: What observations in the coordinate system centered in my fixed position on earth are necessary to conclude that the earth (and the planets) move (approximately) in ellipses around the sun and that earth is rotating around itself? </p> http://mathoverflow.net/questions/29060/classification-of-finite-complex-reflection-groups Classification of finite complex reflection groups S1 2010-06-22T08:38:08Z 2010-06-22T12:18:38Z <p><b>Background:</b></p> <p>Let $K$ be a field and let $V$ be a finite-dimensional $K$-vector space. A <em>pseudoreflection</em> (or usually imprecisely just <em>reflection</em>) in $V$ is an element $1 \neq s \in \mathrm{GL}(V)$ fixing a hyperplane. A <em>reflection representation</em> of a group $W$ over $K$ is a $K$-linear representation $\rho:W \rightarrow \mathrm{GL}(V)$, such that $\rho(W)$ is generated by reflections. A group $W$ is called a <em>reflection group</em> over $K$ if it admits a reflection representation over $K$.</p> <p>Shephard-Todd classified (see below) the finite irreducible reflection groups over $\mathbb{C}$ (i.e. those finite groups admitting an irreducible reflection representation over $\mathbb{C}$). </p> <p><b>Question:</b></p> <p>Is there also a classification of the finite irreducible reflection representations over $\mathbb{C}$? </p> <p>Edit: This question is very imprecise as indicated in the comments below. I should say what "classification of representations" means, and I have to admit: I don't know. A few ideas in this direction are:</p> <ul> <li><p>determine the isomorphism classes of finite irreducible reflection representations over $\mathbb{C}$, where an <em>isomorphism</em> between two reflection representations $\rho:W \rightarrow \mathrm{GL}(V)$, $\rho':W' \rightarrow \mathrm{GL}(V')$ is a vector space isomorphism $f:V \rightarrow V'$ such that $f \rho(G) f^{-1} = \rho'(G)$. (I think the Shephard-Todd classification is a classification relative to this notion!?)</p></li> <li><p>the same as above but an isomorphism is a vector space isomorphism $f:V \rightarrow V'$ and a group isomorphism $\varphi:W \rightarrow W'$ such that $f \rho(g) f^{-1} = \rho'( \varphi(g) )$ for all $g \in W$.</p></li> <li><p>consider pairs $(W,T)$ consisting of a finite irreducible reflection group over $\mathbb{C}$ and a subset $T$ which are generating reflections of some irreducible reflection representation of $W$ and then determine isomorphism classes of such pairs.</p></li> <li><p>[Insert your idea here].</p></li> </ul> <p>My motivation for this question is something like this: A Cherednik-Algebra is defined for any finite irreducible reflection representation over $\mathbb{C}$. In what sense does the algebra depend on the group alone and not on the choice of a particular reflection representation? </p> http://mathoverflow.net/questions/22957/what-are-the-motivations-for-studying-cherednik-symplectic-reflection-graded-he What are the motivations for studying Cherednik (symplectic reflection, graded Hecke) algebras? S1 2010-04-29T09:36:41Z 2010-06-14T14:07:56Z <p>Several times I have come across these algebras and I wonder why any of these are interesting; I'm very sure they are, but I could not find an answer in the literature. </p> <p>For example (the very general version of) a graded Hecke algebra for a finite group $G$ acting on a finite dimensional $\mathbb{C}$-vector space $V$ is defined as an algebra $A_\kappa := (T(V) \sharp \mathbb{C}G)/\langle vw - wv - \kappa(v,w) \mid v,w \in V \rangle$ satisfying the PBW-property $S(V) \sharp \mathbb{C}G \cong \mathrm{gr}(A_\kappa)$, where $\kappa:V \times V \rightarrow \mathbb{C}G$ is an alternating bilinear form. For these algebras I've seen an explanation like "we want to study the geometry of the action of $G$ on $V$, but the commutative algebraic geometry, i.e. $S(V)^G$, is bad, so we better study $S(V) \sharp \mathbb{C}G$"; but then I don't know why I'm interested in those $A_\kappa$. Can anybody explain this and make this precise? </p> <p>I was also told that graded Hecke algebras (for Weyl groups) were introduced by Lusztig in order to study affine Hecke algebras which in turn are important in the representation theory of semisimple split $p$-adic groups. Do the general graded Hecke algebras above have a similar use?</p> <p>The Cherednik and symplectic reflection algebras are special cases of graded Hecke algebras. If I would have a nice motivation for graded Hecke algebras, I would also have one for them; but I am pretty sure that there are different motivations for them!?</p> <p>I am thankful for any explanations and hints to literature.</p> http://mathoverflow.net/questions/15271/primary-decomposition-of-zero-dimensional-modules Primary decomposition of zero-dimensional modules S1 2010-02-14T16:57:31Z 2010-05-29T04:40:54Z <p>(I removed my motivation because it may be misleading :) )</p> <p>Let $A$ be a noetherian commutative ring and let $M \neq 0$ be a finitely generated zero-dimensional (i.e. $\mathrm{dim} \ \mathrm{Supp}(M) = 0$) $A$-module. Then the submodule $0 &lt; M$ has primary decomposition $0 = \bigcap_{\mathfrak{p} \in \mathrm{Supp}(M)} M(\mathfrak{p})$, where $M(\mathfrak{p})$ is the $\mathfrak{p}$-primary component of $M$, i.e. the kernel of the canonical morphism $M \rightarrow M_{\mathfrak{p}}$. I have proven (it's hopefully correct) that the canonical morphism $\mathbf{j}:M \rightarrow \bigoplus_{\mathfrak{p} \in \mathrm{Supp}(M)} M/M(\mathfrak{p})$ ist an isomorphism and that $\mathbf{j}(M \lbrack \mathfrak{q}^\infty \rbrack) \leq M/M(\mathfrak{q}) \leq \bigoplus_{\mathfrak{p} \in \mathrm{Supp}(M)} M/M(\mathfrak{p})$, where $M \lbrack \mathfrak{q}^\infty \rbrack$ is the $\mathfrak{q}$-torsion part of $M$ ($0$-th local cohomology with support $\mathfrak{q}$). Now, my question is if in general $\mathbf{j}(M \lbrack \mathfrak{q}^\infty \rbrack) = M/M(\mathfrak{q})$ so that $M = \bigoplus_{\mathfrak{p} \in \mathrm{Supp}(M)} M \lbrack \mathfrak{p}^\infty \rbrack$, or do I need to assume that $A$ is a Dedekind ring (and perhaps also that $M$ is a torsion module) and if so, how can I prove this?</p> http://mathoverflow.net/questions/21892/formal-deformations-of-algebras-over-not-necessarily-commutative-rings Formal deformations of algebras over not necessarily commutative rings S1 2010-04-19T21:04:55Z 2010-04-19T21:26:58Z <p>In Iain Gordon's survery article <a href="http://www.maths.ed.ac.uk/~igordon/pubs/ICRAsurvey_dec9.pdf" rel="nofollow">"Symplectic reflection algebras"</a> the concept of formal deformations of algebras over semisimple artinian (not necessarily commutative) rings is summarized (chapter 2). Unfortunately, deformations are needed in this generality and there are a few general things I don't understand: </p> <ol> <li><p>An algebra over a semisimple artinian $\mathbb{C}$-algebra $k$ is "defined" as a $k$-bimodule $A$ with a $k$-bimodule morphism $A \otimes_k A \rightarrow A$. Is this a standard definition and is it correct that there is no associativity or unity assumption? I could not find a single book defining an algebra over a not necessarily commutative ring.</p></li> <li><p>In the definition of a formal deformation given in the survey there also seems to be no associativity or unity assumption. Is this a standard definition and does Hochschild cohomology also work in this setting with the same interpretation? My problem is that when I don't restrict my deformations to be associative or have a unit, I might get a lot more deformations.</p></li> </ol> <p>Is there some literature discussing this in more detail?</p> http://mathoverflow.net/questions/10937/proof-of-steinbergs-tensor-product-theorem Proof of Steinberg's tensor product theorem S1 2010-01-06T17:24:15Z 2010-03-23T17:35:18Z <p>Let $G$ be a simply connected semi-simple algebraic group over an algebraically closed field of positive characteristic. The Steinberg tensor product theorem gives a tensor product decomposition of an irreducible rational $G$-module $S(\lambda)$ with heighest weight $\lambda$ according to the $p$-adic expansion of $\lambda$. </p> <p>I am trying to understand the proof of this theorem as given by Cline, Parshall, Scott in Journal of Algebra 63, 264-267 (1980). I have two major problems with this proof:</p> <ol> <li><p>In theorem 1 where it is proven that any irreducible $\mathcal{U}$-module over the restricted universal enveloping algebra extends uniquely to a rational $G$-module, how do I get this representation $\rho:G \rightarrow PGL(V)$? I was able to see that I have a morphism $G \rightarrow \mathrm{Aut}_{_kAlg}(\mathrm{End}_k(V))$. But what next? It seems to be well-known that the group of algebra-automorphisms of the endomorphism ring is just $PGL(V)$, but why? Moreover, why do I get a morphism of <em>algebraic</em> groups?</p></li> <li><p>In the proof of theorem 2, why acts the Lie algebra $L(G)$ trivially on $\mathrm{Hom}_{L(G)}(S _1, S)$? (Here $S$ is some irreducible $G$-module, $S _1$ is an irreducible $L(G)$-submodule of $S$ and $S _1$ also denotes the unique extension to a rational $G$-module). This is not explicitly mentioned, but I think it is used in b). This must have something to do with theorem 1!?</p></li> </ol> <p>Any hints and ideas are welcome.</p> http://mathoverflow.net/questions/18271/what-out-of-print-books-would-you-like-to-see-re-printed/18295#18295 Answer by S1 for What out-of-print books would you like to see re-printed? S1 2010-03-15T18:36:26Z 2010-03-15T18:36:26Z <p><em>Methods of representation theory</em> (Vol 1+2) by Curtis and Reiner. It's a shame that this is out of print!</p> http://mathoverflow.net/questions/17594/realizations-and-pinnings-epinglages-of-reductive-groups Realizations and pinnings (épinglages) of reductive groups S1 2010-03-09T10:28:27Z 2010-03-10T03:46:52Z <p>Let $G$ be a reductive group over an (say, algebraically closed) field $k$. Springer (in his book on algebraic groups) calls for a chosen maximal torus $T$ in $G$ a family $(u_\alpha) _{\alpha \in \Phi(G,T)}$ of immersions $u _\alpha:\mathbf{G}_a \rightarrow G$ such that</p> <p>(i) $t u_\alpha(c) t^{-1} = u_\alpha( \alpha(t) c)$ for all $c \in \mathbf{G}_a$ and $t \in T$,</p> <p>(ii) $n_\alpha := u_\alpha(1) u_{-\alpha}(-1) u_\alpha(1)$ lies in $\mathrm{N}_G(T) \setminus T$,</p> <p>(iii) $u_\alpha(x) u_{-\alpha}(-x^{-1}) u_\alpha(x) = \alpha^\vee(x) n_\alpha$ for all $x \in k^\times$,</p> <p>a <em>realization</em> of (G,T) (or $\Phi(G,T)$) in $G$. We then have $\mathrm{Im}(u_\alpha) = U_\alpha$.</p> <p>In the book of Conrad-Gabber-Prasad on pseudo-reductive groups a <em>pinning</em> of $G$ is defined as a tuple $(T,\Phi^+,(\varphi_\alpha)_ {\alpha \in \Delta})$ where $T$ is a maximal torus, $\Phi^+$ is a positive system for $\Phi(G,T)$, $\Delta$ is the corresponding basis and $\varphi _{\alpha}: (\mathrm{SL} _2, \mathrm{SL} _2 \cap \mathrm{D} _2) \rightarrow (G _\alpha, G _\alpha \cap T)$ are central isogenies such that $\varphi _\alpha( \mathrm{diag}(x,x^{-1}) ) = \alpha^\vee(x)$ for all $x \in k^\times$, where $G _\alpha = \langle U _\alpha,U _{-\alpha} \rangle$.</p> <p>My question is: are these two notions somehow equivalent? If a pinning is given, then by defining </p> <p>$u_\alpha(x) = \varphi_\alpha\begin{pmatrix} 1 &amp; x \\ 1 &amp; 0 \end{pmatrix}$, $u_{-\alpha}(x) = \varphi_\alpha\begin{pmatrix} 1 &amp; 0 \\ x &amp; 1 \end{pmatrix}$</p> <p>I get closed immersions satisfying the properties above, but unfortunately, as I have $\varphi_\alpha$ only for $\alpha \in \Delta$ this does not yet define a realization. How can I define the $u_\alpha$ for $\alpha \notin \Delta \cup -\Delta$? What about the other direction?</p> <p>Moreover (as C-G-P also mentions) in SGA3, exposé XXIII, there is defined the notion of <em>épinglages</em> and Conrad mentions that these carry the same information as the pinnings above. Can somebody make this precise? Moreover in SGA, it is mentioned that an épinglage induces monomorphisms $p_\alpha: \mathbf{G}_a \rightarrow G$ for $\alpha \in \Delta \cup -\Delta$. I suspect that these are the morphisms I defined above, but again, can I get a realization from this?</p> <p>A further problem is the following: For a given realization and a total order on $\Phi(G,T)$ Springer defines <em>structure constants</em> which appear in the expression of the commutator $\lbrack u_\alpha(x), u_\beta(y) \rbrack$ in terms of $u_\gamma$ for linearly independent $\alpha, \beta \in \Phi$. Springer shows that for root systems NOT of type $G_2$ a realization with integral structure constants exist. Demazure also calculates these commutators in SGA3, exposé XXII, for the $p_\alpha$ mentioned above in case of rank 2 root systems. Here, I was surprised that the structure constants seem to be <em>independent</em> of the pinning chosen. Is this now a rank 2 phenomenon that is also true for realizations or does this mean that pinnings/épinglages are more restrictive than realizations? </p> <p>I hope, somebody can help me here.</p> http://mathoverflow.net/questions/17233/if-the-morphism-of-root-data-induced-by-an-isogeny-of-a-reductive-group-is-a-frob If the morphism of root data induced by an isogeny of a reductive group is a Frobenius, is then the isogeny itself a Frobenius? S1 2010-03-05T22:17:51Z 2010-03-06T19:09:01Z <p>Let $G$ be a reductive (or just semisimple) algebraic group over an algebraically closed field $k$ of characteristic $p > 0$, let $T$ be a maximal Torus and let $f:G \rightarrow G$ be an isogeny. Suppose the induced morphism of root data $f^\star: \Psi(G,T) \rightarrow \Psi(G,T)$ is a Frobenius morphism multiplying roots by $q = p^r$ (confer 9.6.3 in Springer's book on algebraic groups). Is it true that $f$ is the geometric Frobenius of an $\mathbb{F}_{q}$-structure on $G$?</p> <p>Equivalent formulation: Is the morphism $F:G \rightarrow G$ defined by $F( u_\alpha(c) ) = u_\alpha(c^q)$ for all roots $\alpha$ and $F(t) = t^q$ for all $t \in T$ the geometric Frobenius of an $\mathbb{F}_{q}$-structure on $G$?</p> <p>I'm pretty sure this is true (perhaps under some additional conditions). I know that there is an easy characterization of Frobenius morphisms in terms of the comorphism of the coordinate rings but I was not able to verify this.</p> http://mathoverflow.net/questions/16382/is-the-cartan-matrix-a-complete-invariant-of-a-kac-moody-algebra Is the Cartan matrix a complete invariant of a Kac-Moody algebra? S1 2010-02-25T10:29:27Z 2010-02-25T10:29:27Z <p>In chapter 1 of Kac's book "Infinite dimensional Lie algebras" it is mentioned that two Kac-Moody algebras are isomorphic if and only if their Cartan matrices are isomorphic (i.e. they are the same up to permutation). A reference to the article "Infinite flag varieties and conjugacy theorems" by Peterson and Kac is given, where it is proven that Cartan subalgebras are conjugate (which implies this result). But in this article it is assumed that the Cartan matrix is symmetrizable! Now, is this result true for any Cartan matrix or just for symmetrizable ones? </p> <p>At the end of the article it is mentioned that all results which do not use the bilinear form also hold for a general Cartan matrix but I don't want to go through this whole article.</p> http://mathoverflow.net/questions/15271/primary-decomposition-of-zero-dimensional-modules/15281#15281 Answer by S1 for Primary decomposition of zero-dimensional modules S1 2010-02-14T19:59:23Z 2010-02-14T20:04:35Z <p>I could answer this now myself (so, sorry for asking!): From a general result (presented e.g. in Eisenbud's commutative algebra book) it follows that $M \lbrack \mathfrak{q}^\infty \rbrack = \bigcap_{\mathfrak{p} \in \mathrm{Supp}(M), \mathfrak{p} \neq \mathfrak{q}} M(\mathfrak{p})$ and then it follows that $\mathbf{j}(M \lbrack \mathfrak{q}^\infty \rbrack) = M/M(\mathfrak{q})$.</p> http://mathoverflow.net/questions/15199/infinite-products-of-topological-groups/15201#15201 Answer by S1 for Infinite products of topological groups S1 2010-02-13T18:20:24Z 2010-02-13T23:12:57Z <p>You can define the product of an arbitrary family $(G_i)_{i \in I}$ of topological groups $G_i$ by equipping the group-theoretic product $G = \prod_{i \in I} G_i$ with the product topology; the product topology is indeed compatible with the group structure (confer Bourbaki, General topology, III.2.9, but it's pretty obvious actually).</p> <p>Perhaps your problem is the product topology? Note that a basis for the product topology are the sets $(U_i)_{i \in I}$ where $U_i \subseteq G_i$ is open and $U_i = G_i$ for all but finitely many $i \in I$. (confer <a href="http://en.wikipedia.org/wiki/Product%5Ftopology" rel="nofollow">wiki</a> for the product topology).</p> http://mathoverflow.net/questions/15107/algebra-unital-associative-algebra-better-terminology Algebra / unital associative algebra: better terminology? S1 2010-02-12T14:28:53Z 2010-02-12T16:50:12Z <p>In Bourbaki an <em>algebra</em> over a commutative ring $k$ is defined to be a $k$-module $A$ together with a $k$-bilinear map $A \times A \rightarrow A$. We then have the obvious notion of morphisms of $k$-algebras. This terminology is nice, because e.g. Lie algebras are then a special kind of algebras and so on. But in 95% of my work I am using unital associative $k$-algebras with unital morphisms. Now, is there any better (in particular shorter) terminology available to distinguish these two cases? I don't want to add this "unital associative" and "unital morphisms" all the time. Is perhaps something like <em>prealgebra</em> for the first case or another short word used in the literature?</p> http://mathoverflow.net/questions/14877/how-much-of-differential-geometry-can-be-developed-entirely-without-atlases/14929#14929 Answer by S1 for How much of differential geometry can be developed entirely without atlases? S1 2010-02-10T18:12:15Z 2010-02-10T18:12:15Z <p>There is something called <em>abstract differential geometry</em> "developed" by Mallios et al (<a href="http://en.wikipedia.org/wiki/Abstract%5Fdifferential%5Fgeometry" rel="nofollow">wikipedia</a>). Perhaps you should skim through the first chapter of his book "Modern Differential Geometry in Gauge Theories: Maxwell fields". It looks pretty nice (despite the physical title). But I don't know how readable or how awesome all this is, I did not read it (there are too many nice explicit things to figure out...).</p> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme How can I really motivate the Zariski topology on a scheme? S1 2009-12-08T16:26:04Z 2010-02-06T17:18:53Z <p>First of all, I am aware of the questions about the Zariski topology asked here and I am also aware of the discussion at the <a href="http://sbseminar.wordpress.com/2009/08/06/algebraic-geometry-without-prime-ideals/" rel="nofollow">Secret Blogging Seminar</a>. But I could not find an answer to a question that bugged me right from my first steps in algebraic geometry: how can I <i>really</i> motivate the Zariski topology on a scheme?</p> <p>For example in classical algebraic geometry over an algebraically closed field I can define the Zariski topology as the coarsest $T_1$-topology such that all polynomial functions are continuous. I think that this is a great definition when I say that I am working with polynomials and want to make my algebraic set into a local ringed space. But what can I say in the general case of an affine scheme? </p> <p>Of course I can say that I want to have a fully faithful functor from rings into local ringed spaces and this construction works, but this is not a motivation. </p> <p>For example for the prime spectrum itself, all motivations I came across so far are as follows: well, over an algebraically closed field we can identify the points with maximal ideals, but in general inverse images of maximal ideals are not maximal ideals, so let's just take prime ideals and...wow, it works. But now that I know that one gets the prime spectrum from the corresponding functor (one can of course also start with a functor) by imposing an equivalence relation on geometric points (which I find very geometric!), I finally found a great motivation for this. What is left is the Zariski topology, and so far I just came across similar strange motivations as above...</p> http://mathoverflow.net/questions/13882/closed-vs-rational-points-on-schemes/13886#13886 Answer by S1 for Closed vs Rational Points on Schemes S1 2010-02-02T23:03:31Z 2010-02-02T23:13:46Z <p>If $k$ is algebraically closed and $X$ is a $k$-scheme locally of finite type, then the $k$-rational points are precisely the closed points. (See EGA 1971, Ch. I, Corollaire 6.5.3).</p> <p>More generally: if $k$ is a field and $X$ is a $k$-scheme locally of finite type, then $X$ is a Jacobson scheme (i.e. it is quasi-isomorphic to its underlying ultrascheme) and the closed points are precisely the points $x \in X$ such that $\kappa(x)|k$ is a finite extension.</p> <p>You should also confer the appendix of EGA 1971. There it is shown that for any field $k$ the category of $k$-schemes locally of finite type with morphisms locally of finite type is equivalent to the category of $k$-ultraschemes (a $k$-ultrascheme is locally the maximal spectrum of a $k$-algebra). </p> http://mathoverflow.net/questions/12095/f-q-structures-on-schemes F_q-structures on schemes S1 2010-01-17T12:30:12Z 2010-01-20T18:27:46Z <p>Let $k|\mathbb{F}_q$ be a field extension. An $\mathbb{F}_q$-structure on a $k$-algebra $A$ is an $\mathbb{F}_q$-subalgebra $A _0$ of $A$ such that $A _0 \otimes _{\mathbb{F}_q} k \cong A$ via the <em>canonical</em> morphism $a \otimes \lambda \mapsto a \lambda$.</p> <p>Now, my question is if this notion can be properly globalized to $k$-schemes? I saw a definition like: an $\mathbb{F}_q$-structure on a $k$-scheme $X$ is an $\mathbb{F}_q$-scheme $X _0$ such that $X \cong X _0 \times _{\mathrm{Spec}(\mathbb{F}_q)} \mathrm{Spec}(k)$ as $k$-schemes (see for example "Representations of finite groups of Lie type" by Digne and Michel, where $\cong$ is even replaced by $=$). But my problem is that here the particular choice of the canonical morphism as above does not appear so that on affines this definition is not the same as above. Is this a problem?</p> <p>(The reason why I care about this is that I want to defined the (geometric) Frobenius on a $k$-Scheme with $\mathbb{F}_q$-structure as the "base change" of the canonical Frobenius (raising to the $q$-th power) on the $\mathbb{F}_q$-structure $X _0$.)</p> http://mathoverflow.net/questions/11327/preschemes-and-schemes/11337#11337 Answer by S1 for Preschemes and schemes S1 2010-01-10T15:44:05Z 2010-01-10T15:44:05Z <p>In the 1971 edition of EGA (this is a revised version of the original 1960 EGA) you can find the following remark in the introduction:</p> <p><em>Signalons enfin, par rapport à la première édition, un changement important de terminologie: le mot «schéma» désigne maintenant ce qui était appelé «préschéma» dans la première édition, et les mots «schéma séparé» ce qui était appelé «schéma».</em></p> <p>The 1971 terminology should be standard today.</p> http://mathoverflow.net/questions/9611/eigenvalue-characters "Eigenvalue characters" S1 2009-12-23T11:18:53Z 2009-12-24T00:02:51Z <p>This question is an addition to my <a href="http://mathoverflow.net/questions/9541/simultaneous-diagonalization" rel="nofollow">question</a> on simultaneous diagonalization from yesterday and it is probably also obvious but I just don't know this: Let $G$ be a commutative affine algebraic group over an algebraically closed field $k$. Let $G_s$ be the semisimple part of $G$. Let $\rho:G \rightarrow GL_n(V)$ be an embedding. Then $\rho(G_S)$ is a set of commuting diagonalizable endomorphisms and I know from yesterday that I have unique morphisms of algebraic groups $\chi_i: \rho(G_s) \rightarrow \mathbb{G}_m$, $1 \leq i \leq r$, and a decomposition $V = \bigoplus _{i=1}^r E _{\chi_i}$, where $E_{\chi_i} = \lbrace v \in V \mid fv = \chi_i(f)v \ \forall f \in \rho(G_s) \rbrace$. Now, my question is: are the morphisms $\chi_i$ independent of $\rho$ so that I get well-defined morphisms $\chi_i:G_s \rightarrow \mathbb{G}_m$? </p> <p>If somebody knows what I'm talking about, then please change the title appropriately! :)</p> http://mathoverflow.net/questions/9541/simultaneous-diagonalization Simultaneous diagonalization S1 2009-12-22T14:38:38Z 2009-12-23T15:06:31Z <p>I'm pretty sure that the following (if true) is a standard result in linear algebra but unfortunately I could not find it anywhere and even worse I'm too dumb to prove it: Let $k$ be a field, let $V$ be a finite-dimensional $k$-vector space and let $S \subseteq \mathrm{End}_k(V)$ be a subset of pairwise commuting (i.e. $\lbrack S, S \rbrack = 0$) endomorphisms. Then the following holds:</p> <ol> <li><p>If all $f \in S$ are diagonalizable, then there exist maps $\chi_i:S \rightarrow k$, $i=1,\ldots,r$, such that $V = \bigoplus_{i=1}^r E_{\chi_i}(S)$, where $E_\chi(S) := \lbrace v \in V \mid fv = \chi(f)v \ \forall \ f \in S \rbrace$. </p></li> <li><p>The maps $\chi_i$ in 1 are unique.</p></li> <li><p>1 is equivalent to the existence of a basis $\mathcal{B}$ of $V$ such that for each $f \in S$ the matrix $M_{\mathcal{B}}(f)$ of $f$ with respect to $\mathcal{B}$ is diagonal. (I believe that this might not be true)</p></li> <li><p>If all $f \in S$ are trigonalizable, then there exists a basis $\mathcal{B}$ of $V$ such that for each $f \in S$ the matrix $M_{\mathcal{B}}(f)$ of $f$ with respect to $\mathcal{B}$ is upper triangular and for each diagonalizable $f \in S$ the matrix $M_{\mathcal{B}}(f)$ is diagonal.</p></li> </ol> <p>I know that a set of commuting diagonalizable endomorphisms can be simultaneously diagonalized in the sense of 3 but I don't know how to prove 1 (my problem is the "glueing" of the $\chi$-maps when I try to prove this by induction on $\mathrm{dim}V$). Also, I know that the first part of 4, the simultaneous trigonalization, holds but I don't know how to show that there exists a basis which then also diagonalizes all diagonalizable endomorphisms. This should follow from 1, I think.</p> <p>Perhaps, because all this is probably standard stuff, I should mention that this is not a homework problem :)</p> <p>One additional question: Suppose that $k$ is algebraically closed and that $G$ is an affine commutative algebraic group over $k$ which coincides with its semisimple part, embedded as a closed subgroup in some $GL(V)$. Are the maps $\chi_i:G \rightarrow \mathbb{G}_{m}$ morphisms of algebraic groups?</p> http://mathoverflow.net/questions/33662/how-canonical-is-the-triangular-decomposition-of-a-rational-cherednik-algebra/33815#33815 Comment by S1 S1 2010-07-29T20:02:08Z 2010-07-29T20:02:08Z So, I think my first comment is just exactly what you explained in general (thanks for this). http://mathoverflow.net/questions/33662/how-canonical-is-the-triangular-decomposition-of-a-rational-cherednik-algebra/33815#33815 Comment by S1 S1 2010-07-29T20:00:27Z 2010-07-29T20:00:27Z My problem was that I was searching for an embedding of $S(V)$ into a general graded Hecke algebra and this cannot work. In case of rational Cherednik algebras I'm only embedding 'half' of this algebra and I didn't realize this. Anyways, the problem remains if I have to choose a section as above to get my vector space decomposition or if this can be done canonically (it's still very pedantic but I'm still not sure if I miss a point here). http://mathoverflow.net/questions/33662/how-canonical-is-the-triangular-decomposition-of-a-rational-cherednik-algebra/33815#33815 Comment by S1 S1 2010-07-29T19:54:41Z 2010-07-29T19:54:41Z Well, I started too general and this generality was source of my confusion. Above I mentioned (if it's not wrong!) that for any section I have a vector space isomorphism $S(V) \otimes \mathbb{C}G \rightarrow A_\kappa$ (equivalent to PBW for $A_\kappa$). Now, in the case of a rational Cherednik algebra I have as vector space $V \oplus V^*$ and the above gives a vector space isomorphism $S(V) \otimes \mathbb{C}G \otimes S(V^*) \rightarrow A_\kappa$ (the triangular decomposition!?). AND, this also gives an algebra embedding of $S(V)$ and $S(V^*)$ into $A_\kappa$ because $[V,V^*]=0$ in $A_\kappa$ http://mathoverflow.net/questions/33662/how-canonical-is-the-triangular-decomposition-of-a-rational-cherednik-algebra Comment by S1 S1 2010-07-29T10:55:29Z 2010-07-29T10:55:29Z At this point I wasn't referring to your answer but to the nonsense I was writing (the isomorphism). As I didn't want to leave something wrong there, I removed this. But you're right; I will rewrite this... http://mathoverflow.net/questions/33662/how-canonical-is-the-triangular-decomposition-of-a-rational-cherednik-algebra/33664#33664 Comment by S1 S1 2010-07-28T15:19:07Z 2010-07-28T15:19:07Z I missed a point in my first comment: Is $\xi$ in the case of the Weyl algebra still an isomorphism? If so, then is the restriction to the particular $\kappa$ just something to make life simpler when considering deformations of $S(V) \sharp \mathbb{C} G$? As for the pedantic point: Perhaps I could have just summarized this in the question &quot;what precisely is the vector space isomorphism $\mathrm{S}(V) \otimes \mathbb{C}G \rightarrow A$ the PBW-property induces&quot;? Is it unique? http://mathoverflow.net/questions/33540/existence-of-a-minimal-generating-set-of-a-module/33542#33542 Comment by S1 S1 2010-07-27T17:33:39Z 2010-07-27T17:33:39Z Your remark about perfect rings is interesting! Thanks. http://mathoverflow.net/questions/33250/references-general-idea-of-kervaire-invariant-problem Comment by S1 S1 2010-07-25T07:39:27Z 2010-07-25T07:39:27Z Hopkins gave a talk on this at Atiyah's birthday conference in 2009. Perhaps you meant this talk or you already know its contents, but anyways, you can find a video and the slides of this talk here: <a href="http://www.maths.ed.ac.uk/~aar/atiyah80.htm" rel="nofollow">maths.ed.ac.uk/~aar/atiyah80.htm</a> (the title is &quot;Applications of algebra to a problem in topology&quot;) http://mathoverflow.net/questions/32524/dualizing-sheaf-on-a-cohen-macaulay-variety Comment by S1 S1 2010-07-19T22:06:16Z 2010-07-19T22:06:16Z @Donu, Karl: Okay, as I don't know exactly what I want (the reason for asking this question), I am not qualified to comment on that. At least in Liu Quing's book the canonical sheaf on a <i>smooth</i> variety $X$ is defined as $\mathrm{det}\Omega_X^1$ (6.4.2). But I'll try to put things together... http://mathoverflow.net/questions/32524/dualizing-sheaf-on-a-cohen-macaulay-variety/32539#32539 Comment by S1 S1 2010-07-19T21:28:17Z 2010-07-19T21:28:17Z Okay, that's something I found while browsing through the literature. But is the sheaf $\omega_X$ I defined above for a (proper, normal?) CM-variety now a dualizing sheaf in this sense. Moreover, this definition excludes affine (CM-) varieties which I am particularly interested in! http://mathoverflow.net/questions/32524/dualizing-sheaf-on-a-cohen-macaulay-variety Comment by S1 S1 2010-07-19T21:23:49Z 2010-07-19T21:23:49Z @t3suji: 'semi-separated' was perhaps what I was looking for. But except for separated schemes I don't know of any semi-separated ones. Is there some condition that for a morphism $f:X \rightarrow Y$ of varieties into a semi-separated/separated/affine variety ensures that $X$ is also semi-separated? http://mathoverflow.net/questions/32524/dualizing-sheaf-on-a-cohen-macaulay-variety Comment by S1 S1 2010-07-19T21:19:02Z 2010-07-19T21:19:02Z @t3suji: Of course! I was somewhere else... @Donu: I think it is $j_*\Omega_X^{\dim X}$ as long as $X$ is irreducible!? @all others: I will see if I can understand your comments... http://mathoverflow.net/questions/32353/confusing-definitions-in-lius-algebraic-geometry-and-arithmetic-curves Comment by S1 S1 2010-07-19T07:57:54Z 2010-07-19T07:57:54Z @Qing: I was pretty sure about that :) I hope you don't take it personal that I keep mixing up your first and last name :) http://mathoverflow.net/questions/32450/chevalleyshephardtodd-theorem Comment by S1 S1 2010-07-19T07:52:11Z 2010-07-19T07:52:11Z An additional question: Is it correct that Chevalley only gave a proof for reflection groups (reflection = pseudo-reflection of order 2) in this paper and that Serre later realized that the proof also works for pseudo-reflection groups? http://mathoverflow.net/questions/32353/confusing-definitions-in-lius-algebraic-geometry-and-arithmetic-curves Comment by S1 S1 2010-07-18T15:36:21Z 2010-07-18T15:36:21Z choice. So the style of definition in Quing's book (which gives the definition in a restricted situation which is probably easier) indeed makes sense as long as one mentions the extra conditions all along. http://mathoverflow.net/questions/32353/confusing-definitions-in-lius-algebraic-geometry-and-arithmetic-curves Comment by S1 S1 2010-07-18T15:34:03Z 2010-07-18T15:34:03Z @Georges: Moreover, thanks to your link, I think I understand, that my proposal for an alternative definition isn't the right way to do it. Take for example the definition of a smooth morphism in Quing's book: Let $Y$ be locally noetherian and let $f:X \rightarrow Y$ be a morphism of finite type. [...]. We say that $f$ is <i>smooth</i> if [...]. Now, I would have changed that to &quot;a morphism $f:X \rightarrow Y$ of schemes is called smooth if it is of finite type, if $Y$ is locally noetherian and if [...]&quot;. But due to the more general definition given in the stacks project, this wouldn't be a good