User jason polak - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T10:23:15Z http://mathoverflow.net/feeds/user/1437 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130270/what-is-a-good-reference-preferably-thorough-for-the-derived-category-of-a-sche/130282#130282 Answer by Jason Polak for What is a good reference (preferably thorough) for the Derived Category of a scheme/orbifold/stack? Jason Polak 2013-05-10T17:52:06Z 2013-05-10T17:52:06Z <p>I'll just mention some sources that are not already in the other question indicated by Franz's answer, in case you need to become more comfortable with the derived machinery itself. (The book by Huybrechts on Fourier-Mukai Transforms in the other answer will probably be the best, depending on your interests, and there is also "Fourier-Mukai and Nahm Transforms in Geometry and Mathematical Physics", by Bartocci, Bruzzo, and Ruipérez that contains additional information and good references).</p> <p>So here are the other references that might help with derived categories themselves:</p> <p>I think most of the texts or papers treating derived categories applied to algebraic geometry are a bit too terse when it comes to derived categories themselves. There is a nice book called "Interactions Between Homotopy Theory and Algebra", which is from a summer school that was held in Chicago, and in this book there are two nice lectures by Henning Krause on derivated categories. His second lecture actually consists of about 40 exercises, which might help you with understanding the machinery of derived categories without having to worry about how they work in algebraic geometry yet. These two chapters are also available on the arXiv (<a href="http://arxiv.org/abs/math/0511047" rel="nofollow">Lecture</a>, <a href="http://arxiv.org/abs/math/0609479" rel="nofollow">Exercises</a>), though I recommend the whole book too.</p> <p>Yet another book is the edited volume "Handbook of Tilting Theory" (LMS 332), which is a series of longer (most introductory papers) on tilting theory, so you'll see plenty of applications to module categories and representation theory, but there is also a chapter on the Fourier-Mukai transform there. </p> <p>In the book "Derived Equivalences for Group Rings" (K&ouml;nig, Zimmerman, et al.), there are several chapters that include introductions to aspects of derived categories including unbounded derived categories and many examples, which might also be useful.</p> http://mathoverflow.net/questions/129762/what-arithmetic-information-is-contained-in-the-algebraic-k-theory-of-the-integer/129786#129786 Answer by Jason Polak for What arithmetic information is contained in the algebraic K-theory of the integers Jason Polak 2013-05-05T22:16:03Z 2013-05-05T22:16:03Z <p>Let $p$ is an odd prime and $C$ the $p$-Sylow of the class group of $\mathbb{Q}(\zeta_p)$. If $C^\sigma$ denotes the group fixed by complex conjugation then Vandiver's conjecture is that $C^\sigma = 0$. Both Kurihara and Soulé have made some partial progress towards this conjecture, and their methods rely on knowledge of the torsion piece of the groups $H_i(\mathbb{Z})$. A good introduction is Soulé's 14-page paper on the matter entitled "Perfect forms and the Vandiver conjecture". Kurihara's paper "Some remarks on conjectures about cyclotomic fields and $K$-groups of $\mathbb{Z}$" is another very readable source, which points out many arithmetic applications.</p> http://mathoverflow.net/questions/128872/reference-for-rationality-in-algebraic-groups-in-the-language-of-schemes Reference for Rationality in Algebraic Groups in the Language of Schemes? Jason Polak 2013-04-26T22:57:57Z 2013-04-26T23:04:23Z <p>There are a few standard sources for learning about linear algebraic groups, such as Humphreys' book "Linear Algebraic Groups" and Borel's book by the same title. Both are not written in the language of group schemes. Humphreys does everything over an algebraically closed field, whereas Borel already introduces fields of definition and studies rationality properties through the text. This makes Borel's book more comprehensive but unfortunately the older language of algebraic geometry and in particular $k$-structures is a bit disconcerting especially if one is used to and prefers the language of schemes and functor of points.</p> <p>Hence this reasoning led me to choose Humphreys as a source to learn this material, who only works over an algebraically closed field, which has the advantage that the language difference isn't really an issue.</p> <p>I'd read through parts of Borel on things that I need, but now I feel like I'd like to have a deeper understanding about rationality, $\overline{k}/k$ forms of groups over some base field, etc. without having to go through Borel since the language is awkward from my point of view. So, I'm looking for references:</p> <blockquote> <p>An ideal reference would cover the following material, via the language of schemes and group schemes: everything that Borel (or Springer, in his book) does on rationality and fields of definition, and perhaps some more recent results as well, seasoned with plenty of Galois cohomology. Some elegant papers using the group scheme language could also be a substitute.</p> </blockquote> <p>In other words:</p> <blockquote> <p>Are there any good references on the basic rationality properties (such as the existence of maximal tori and Borel subgroups defined over the base field) for someone who as already read the basic material covered in a book like Humphreys, that uses the language of schemes? </p> </blockquote> <p>I don't have too high hopes for a book, but perhaps there is some survey paper I've missed? I find the seeming lack of sources on this material in a modern language very frustrating.</p> <hr> <p>Note that <a href="http://mathoverflow.net/questions/17662/books-on-reductive-groups-using-scheme-theory" rel="nofollow">Question 17662</a> is different in that I'm not looking for a book that covers all of the classification of reductive groups; for this, Humphrey's book is fine together with <a href="http://math.stanford.edu/~conrad/papers/luminysga3.pdf" rel="nofollow">Conrad's fine notes</a>.</p> http://mathoverflow.net/questions/128740/homotopy-theoretic-measure-of-operations-on-sheaves-failing-to-be-sheaves Homotopy-theoretic measure of operations on sheaves failing to be sheaves Jason Polak 2013-04-25T18:02:16Z 2013-04-25T18:02:16Z <p>Here's something I've been wondering about for a few weeks:</p> <p>Consider a topological space $X$ and a sheaf of rings $\mathscr O_X$ on $X$. Suppose $\mathscr{F}$ and $\mathscr{G}$ are $\mathscr O_X$ modules. </p> <p>We can define a presheaf on $X$ via $U\mapsto \mathscr{F}(U)\otimes_{\mathscr O(U)}\mathscr{G}(U)$. This is usually not a sheaf. (For instance one can take $X = \mathbb{P}^1_k$ over some algebraically closed $k$ and ${\mathscr O}_X$ the structure sheaf, $\mathscr{F} = \mathscr O_X(n)$ and $\mathscr{G} = \mathscr O_X(m)$, where these are the $\mathscr O_{X}$ modules of invertible sheaves attached to a divisor of degree $-n,-m$ respectively, and $n,m > 0$.) We write $\mathscr{F}\otimes_{\mathscr O_X}\mathscr{G}$ for the sheaf associated to this presheaf.</p> <p>On the other hand, if $X=\mathrm{Spec}(A)$ is an affine scheme and $\mathscr{F},\mathscr{G}$ are quasicoherent then the above presheaf is a sheaf.</p> <hr> <p>Here is the general question:</p> <blockquote> <p>Is there some kind of "obstruction theory" or homotopy theoretic-gadget that will explain when various colimit constructions of sheaves in the category of presheaves fail to be sheaves? </p> </blockquote> <p>However, this being rather vague, here are two examples of specific questions that I would consider special cases of the above:</p> <blockquote> <p>Let $X$ be a topological space (perhaps with some "nice" properties, like locally compact Hausdorff), and $\mathscr O_X$ be the ring of complex-valued continuous functions on $X$. Are there homotopy-theoretic conditions we can place on $X$ to ensure that for any two "nice" (perhaps finite-type or coherent) sheaves $\mathscr{F},\mathscr{G}$ of $\mathscr O_X$-modules, the tensor presheaf is a sheaf? (One could also ask this question for arbitrary direct sum presheaves, etc.)</p> </blockquote> <p>And:</p> <blockquote> <p>Let $X$ be a scheme, $\mathscr{F},\mathscr{G}$ two quasicoherent $\mathscr O_X$ modules. Are there conditions on $X$ such that $U\mapsto \mathscr{F}(U)\otimes_{\mathscr O_X(U)}\mathscr{G}(U)$ is already a sheaf? (I'd be interested to see an example also where $X$ is not affine and yet $U\mapsto \mathscr{F}(U)\otimes_{\mathscr{O}(U)}\mathscr{G}(U)$ is also a sheaf, if this is possible at all).</p> </blockquote> http://mathoverflow.net/questions/118081/nice-algebraic-statements-independent-from-zf-vl-constructibility Nice Algebraic Statements Independent from ZF + V=L (constructibility) Jason Polak 2013-01-04T19:45:16Z 2013-02-08T23:28:23Z <h2>Background and Motivation</h2> <p>I've always been fascinated about algebraic statements independent from ZFC set theory. One such fascinating example comes from considering $\rm{Ext}^1_\mathbb{Z}(A,\mathbb{Z})$. If $A$ is free then this abelian group is trivial. Is the converse true? The converse is known as the Whitehead problem.</p> <p>Now the Whitehead problem was shown independent of ZFC by Shelah. This is somewhat unsatisfying, but it can be proved assuming the axiom of constructibility ($V=L$). In fact, there are other reasonable sounding and simple statements in analysis and topology that are also independent from ZFC but become theorems once $V=L$ is assumed. Another statement is that the global (a.k.a. homological) dimension of the ring $\prod_{i=1}^\infty \mathbb{C}$ is two if and only if the continuum hypothesis holds, which is implied by adding $V=L$ again. Thus I feel warmly about the axiom of constructibility.</p> <p>Of course, the dissatisfaction remains, because there are other statements that are independent from $ZFC + V=L$ (well, perhaps I should write $ZF + V=L$ to save space). <a href="http://mathoverflow.net/questions/18058/" rel="nofollow">Question 18058</a> and <a href="http://mathoverflow.net/questions/11480" rel="nofollow">Question 11480</a> are examples.</p> <h2>Question, Loosely Stated</h2> <blockquote> <p>Now, I am curious if there are any known <strong>algebraic</strong> (see postscript) statements, reasonably naturally sounding (use judgement), that are independent from $ZF + V=L$? Or perhaps independent from $ZF + A$ where $A$ is your favourite set-theoretic axiom independent from $ZFC$? Perhaps some easy low-hanging fruit for this search would be in the area of homological dimension theory? Has anyone done work on this type of thing?</p> </blockquote> <p>I am sure there must be some statements of <em>some</em> kind. In the proof of the Whitehead problem under adding $V=L$, one can first deduce some combinatorial statement that requires little set-theoretic machinery and then use it to prove the Whitehead problem. So perhaps adding other axioms, one can also deduce various combinatorial gadgets and them use them to get new algebraic statements that are independent from the original $ZFC$? I would even like to hear about statements implied by additional axioms, but whose independence is not proved. (One of Devlin's books explains this).</p> <p>(Remark: Although with enough brute force, one should be able to churn out such things no matter how many new axioms one adds, I would be interested in finding enough axioms of set theory so that the remaining independent statements would be so bizarre sounding that they would be essentially be uninteresting for all of time. Presumably as one adds more and more axioms to set theory, this would happen, no?)</p> <p><em>Since I am not a set theorist, I would appreciate answers that are understandable to someone who knows the basics of set theory (say a typical first grad course) but knows very little about forcing.</em></p> <hr> <h2>Postscript</h2> <p>By "algebraic" I mean roughly some statement in the language of groups, rings, ideals, modules, fields, etc., somewhat natural sounding, that does not itself refer to the additional axiom (e.g. using some set whose cardinal is inaccessible, or something along these lines). Thus I am not looking for statements about real numbers or set theory, although those are interesting too.</p> http://mathoverflow.net/questions/121228/injective-modules-over-group-rings/121233#121233 Answer by Jason Polak for Injective Modules over Group Rings Jason Polak 2013-02-08T19:32:41Z 2013-02-08T19:32:41Z <p>The notion of Frobenius algebra is still useful in the general case, but then group rings that are Frobenius algebras aren't necessarily quasi-Frobenius rings, as your example notes. However if $M$ is a noetherian Frobenius $A$-algebra where $A$ is a commutative and self-injective ring then $M$ itself is a quasifrobenius ring so the notions of projective and injective coincide again. This is the content of Corollary 19 in [1].</p> <p>If $A$ is not self-injective then your method of constructing a counterexample for group rings shows that the conclusion cannot hold in general for $A[G]$, since $A$ itself is not quasifrobenius.</p> <p>[1].Eilenberg and Nakayama. "On the dimension of modules and algebras. II". Nagoya Mathematical Journal, 9. pp 1-16, 1955</p> http://mathoverflow.net/questions/111916/confusing-point-in-proof-semisimple-automorphism-fixes-torus Confusing Point in Proof: Semisimple Automorphism Fixes Torus Jason Polak 2012-11-09T17:19:36Z 2012-11-11T17:09:47Z <p>I am reading a proof on p.51 of Robert Steinberg in his book "Endomorphisms of Algebraic Groups" and I am having a bit of difficulty understanding one point in the proof.</p> <p>The setting is as follows. Consider an algebraic group $G$ over an algebraically closed field of arbitrary characteristic. We say that an automorphism $\sigma:G\to G$ is <em>semisimple</em> if there exists an embedding $G\hookrightarrow G'$ such that $\sigma$ is realised by conjugation in $G'$ by a semisimple element. The theorem is:</p> <blockquote> <p>If $G$ is solvable and $\sigma:G\to G$ is a semisimple automorphism then there exists a maximal torus $T\subseteq G$ such that $\sigma(T) = T$.</p> </blockquote> <p>I shall paraphrase the proof on pp.51-52 of Steinberg. Assume $G$ is connected. We choose at first an arbitrary maximal torus $T\subseteq G$, and let $U$ be the normal subgroup of $G$ of all unipotents; then we have $G = T\ltimes U$. We embed $G$ in a larger group so that $\sigma:G\to G$ is realised by conjugation by a semisimple element $s$. Then $sTs^{-1}$ and $T$ are maximal tori of $G$, so we can write $T = usTs^{-1}u^{-1}$ where $u\in G$ and by our decomposition, we can assume $u\in U$.</p> <p>It then suffices to show that $us$ is actually conjugate to $s$: i.e. $us = xsx^{-1}$ for some $x\in G$, for then $\sigma$ will fix $x^{-1}Tx$. </p> <p>Then Steinberg says: <em>"and on replacing $us$ by its semisimple part we may assume it to be semisimple."</em> After this, he proceeds to show that if $us$ is semisimple, then it indeed is conjugate to $s$, which completes the proof.</p> <blockquote> <p>Question: why can we assume that $us$ is semisimple? This <em>would</em> follow if the semisimple part could be written as $u's$ where $u' \in U$, but I cannot see why this is true at the moment.</p> </blockquote> <p>Thanks!</p> http://mathoverflow.net/questions/104223/justifying-explaining-math-research-in-a-public-address/104241#104241 Answer by Jason Polak for Justifying/Explaining math research in a public address Jason Polak 2012-08-07T22:29:46Z 2012-08-07T22:29:46Z <p>I‘m all for talking about the applications of math and your research. I think such things can be very interesting. However, let me add one more point. I feel you should also express to the audience why we really study mathematics; that is, it can be a source of immense abstract and intellectual beauty. Of course, some people do like to solve real world problems, but even there the most proximal reward is the joy of seeing an entire solution come together.</p> <p>I feel that to omit such a point removes the human component of mathematics and places it into the realm of austere practicality. </p> <p>I suggest you take your favourite ideas, make them accessible by considerable simplification, and attempt to explain why YOU care and like the mathematics, not why the audience should care.</p> <p>I have found that most people never fail to respond to the genuine enthusiasm and well-communicated passion of another person.</p> <p>If practical applications interest you, talk about those. If you‘re much more interested in the wonders and effectiveness of group character theory, speak about that provided you can give at least some illuminating examples. I think this is the only way to be honest.</p> http://mathoverflow.net/questions/103784/geometric-invariant-theory-for-geometers/103808#103808 Answer by Jason Polak for Geometric invariant theory for geometers Jason Polak 2012-08-02T18:43:31Z 2012-08-02T18:43:31Z <p>If you just want to get a feeling for invariant theory, here are some books that aren't necessarily comprehensive but nevertheless are enlightening at a more leisurely pace as compared to GIT, which would be useful for someone who isn't as familiar with algebraic groups and algebraic geometry:</p> <ul> <li><p>Santos and Rittatore - Actions and Invariants of Algebraic Groups: Minimal prerequisites. A very gentle introduction to some aspects of invariant theory, including some motivation via Hilbert's 14th problem. This book also contains most of the required theory of linear algebraic groups.</p></li> <li><p>Dolgachev - Lectures on Invariant Theory: This takes a more geometric viewpoint and might be something you are interested in. This only requires some basic knowledge of algebraic geometry.</p></li> <li><p>Schmitt - Geometric Invariant Theory and Decorated Principal Bundles: this might also be interesting if you are interested in the geometric applications and the related geometry, though I haven't looked into this book very much, but Part 1 does contain a fairly leisurely-looking introduction to GIT</p></li> </ul> <p>There is also Popov's and Vinberg's treatise "Invariant Theory" in the Ecyclopedia of Mathematical Sciences Volume 55 (Springer) which contains a good summary of the classical results in characteristic zero.</p> http://mathoverflow.net/questions/89598/higher-cartan-eilenberg-resolutions Higher "Cartan-Eilenberg" Resolutions Jason Polak 2012-02-26T19:15:04Z 2012-05-22T18:37:08Z <p>I am a number theory graduate student learning a bit of homological algebra, and I am curious about higher complexes in abelian categories. I apologize if my post is slightly vague as I am not an expert in this area. I will use capital roman letters to denote objects or complexes, but the usage is clearly stated.</p> <h2>Motivation</h2> <p>To motivate my question, let us start from a single object $M$ in an Abelian category $\mathcal{C}$. If $\mathcal{C}$ has enough projectives, we can form a projective resolution $P\to M$ of $M$, and apply a right exact additive functor $F:\mathcal{C}\to\mathcal{D}$ to $P$, and calculuate homology. Here $\mathcal{D}$ is just some other abelian category. This will give us the derived functors of $F$, and is a standard and well-known construction.</p> <p>Of course, it doesn't stop there. In the same abelian category $\mathcal{C}$ with enough projectives, any <em>chain complex</em> $M$ also has a (left) Cartan-Eilenberg resolution $P\to M$. Recall $P$ is an upper half plane double complex and the map $P\to M$ is just a chain map $P_{\bullet,0}\to M_\bullet$. Finally, $P$ is required to satisfy some axioms making it into a sort of 2-dimensional version of a projective resolution. I won't go into detail because this is also fairly standard.</p> <p>The point is that we can also apply a right-exact functor $F$ to this double complex $P$ and take the the homology of the total direct-sum complex of $P$ (if it exists!); that is $H_i(Tot^\oplus(FP))$, to get the <em>hyperderived</em> functors of $F$.</p> <h2>The Question</h2> <p>It seems as though there is a natural generalization. One can easily define an $n$-complex in an analogous fashion to $2$-complexes. Higher dimensional complexes don't really show up much as far as I can tell, although I believe in Cartan-Eilenberg a $4$-complex is used somewhere (sorry, I don't have the book with me!).</p> <p>So I suppose my question is:</p> <blockquote> <p>Suppose $\mathcal{C}$ is an abelian category with enough projectives. Is it true that for any $n$, an $n$-complex $M$ has some appropriate higher Cartan-Eilenberg resolution (which would be an $n+1$-complex)?</p> </blockquote> <p>Appropriate means that if $P\to M$ is this hypothetical higher Cartan-Eilenberg resolution, then applying a right exact additive functor $F$ to $P$ and taking the homology of the total direct-sum (if it exists) complex gives the "correct" notion of $n$-hyperderived functors.</p> <h2>Comments</h2> <p>I have searched the literature for this concept but I could not find anything relevant. I am thinking that there are two possibilities (a) yes, higher Cartan-Eilenberg resolutions exists and are interesting, or (b) yes, higher Cartan-Eilenberg resolutions exist but don't capture any new information and so are not that interesting. I'd be a bit surprised if they <em>don't</em> exist but I do not have enough experience in homological algebra to understand the bigger picture here.</p> <p>Also, we could have phrased this question in terms of injectives and (right) Cartan-Eilenberg resolutions.</p> <p>Thanks</p> http://mathoverflow.net/questions/93716/cov-right-exact-additive-functors-that-dont-commute-with-direct-sums Cov. right-exact additive functors that don't commute with direct sums? Jason Polak 2012-04-10T23:37:21Z 2012-05-10T18:26:25Z <h2>Background</h2> <p>Recently, I have been writing up some notes on derived functors and I came across the Eilenberg-Watts theorems [1], which essentially explain why it is hard to find derived functors besides Ext and Tor. Before asking my question, allow me to briefly state these theorems.</p> <p>Let $R$ and $S$ be rings and $M$ and $S-R$ bimodule. A basic property of the functor $M\otimes_R-:R-\mathrm{Mod}\to S-\mathrm{Mod}$ is that it is an additive covariant right-exact functor. In fact, it also commutes with direct sums. Curiously these properties are enough to characterize it: any covariant additive $T:R-\mathrm{Mod}\to S-\mathrm{Mod}$ that is right-exact and commutes with direct sums is in fact naturally equivalent to some $M\otimes_R -$ for some $S-R$ bimodule $M$. This is the statement of the Eilenberg-Watts theorem for tensor functors.</p> <p>For completeness, I should state the other version for Hom. If $T:R-\mathrm{Mod}\to S-\mathrm{Mod}$ is an additive left-exact contravariant functor which converts direct sums into direct products (i.e. $T(\oplus M_i) \cong \prod T(M_i)$) then there is an $R-S$ bimodule $M$ such that $T$ is naturally equivalent to $\mathrm{Hom}_{R}(-,M)$.</p> <p>Finally, if $T:R-\mathrm{Mod}\to \mathbb{Z}-\mathrm{Mod}$ is left-exact covariant that commutes with inverse limits then there is a left $R$-module $M$ such that $T$ is naturally equivalent to $\mathrm{Hom}_{R}(M,-)$.</p> <h2>The Prompting for the Question</h2> <p>From the above, any functor satisfying the hypotheses of these Eilenberg-Watts type theorems are going to be naturally equivalent to either a tensor or a Hom, and thus its derived functors will just be Tor or Ext respectively. For instance, if $G$ is a group then $G-\mathrm{Mod}\to \mathbb{Z}-\mathrm{Mod}$ given by $A\mapsto A_G$, where $A_G$ is the quotient of $A$ by the submodule generated by $ga - a$ for all $g\in G$ and $a\in A$ is just the usual coinvariant functor, whose left derived functors are the homology groups $H^i(G,A)$. By Eilenberg-Watts, $-_G$ must be equivalent to some tensor functor, and in fact it is easy to prove that $A\mapsto A_G$ is naturally equivalent to $A\mapsto \mathbb{Z}\otimes_{\mathbb{Z}G}A$</p> <p>Incidentally, the proof given by C.E. Watts is explicit enough so that the above natural equivalence is apparent.</p> <hr> <h2>The Question</h2> <p>Notice that in each of these the hypothesis of playing nice with limits is required. I am actually interested in functors which do <em>not</em> play nicely with limits. For instance,</p> <blockquote> <p>What are some examples of covariant right-exact functors $T:R-\mathrm{Mod}\to S-\mathrm{Mod}$ that do <em>not</em> commute with all direct sums? [Edit: $T$ also should not be left exact in this case.]</p> </blockquote> <p>Such a $T$ of course cannot be a left-adjoint for otherwise it would commute with direct sums. Such a $T$ could be interesting because its left-derived functors may not be "like" the Tor functor. The question also goes for dropping the playing-nice-with-limit hypotheses in the other forms of the theorem. I tried a Google search but could not seem to find anything relevant. </p> <p>Since I am asking for a list of examples, I have made this a community wiki. Thanks!</p> <p>[1] Watts, "Intrinsic Characterizations of Some Additive Functors". Proceedings of the American Mathematical Society, Vol. 11, No. 1 (Feb., 1960), pp. 5-8</p> <h2>Addendum (edit)</h2> <p>Thanks everyone for their answers; I think I should have been more precise and asked a question more along the lines of:</p> <blockquote> <p>What are some derived functors that are not Ext or Tor?</p> </blockquote> <p>Which I believe some of the existing answers are. In essence I wanted examples that were neither tensors nor Homs in disguise...</p> http://mathoverflow.net/questions/94383/first-group-homology-with-general-coefficients/94457#94457 Answer by Jason Polak for First group homology with general coefficients Jason Polak 2012-04-18T20:19:25Z 2012-04-19T20:21:43Z <p>In the trivial case the abelianization comes from the short exact sequence</p> <p>$0\to J\to \mathbb{Z}G\to \mathbb{Z}\to 0$</p> <p>where $J$ is the augmentation ideal. The homology $H_*(G,-)$ are just derived functors and give a long exact sequence in homology, which since $H_1(\mathbb{Z}G,\mathbb{Z})$ is always trivial, gives a four term exact sequence which looks like</p> <p>$0\to H_1(G,\mathbb{Z})\to J_G\to (\mathbb{Z}G)_G\to\mathbb{Z}\to 0$</p> <p>Here the subscript $-_G$ just means the coinvariant functor from which the homology is derived: $M_G$ is the quotient of $M$ by the submodule generated by $gm - m$ for all $m\in M$. It is not hard to verify that $(\mathbb{Z}G)_G\to\mathbb{Z}$ is an isomorphism so that $H_1(G,\mathbb{Z})$ is isormorphic to $J_G$, which is in turn the abelianization. All of this is classical and in Brown, Weibel, or probably even Mac Lane, which you can consult for more details.</p> <p>Alternatively, topologically it isn't hard to prove that $H_1(G,\mathbb{Z})$ is just the homology of a topological classifying space with $G$ as its fundamental group and whose higher homotopy groups are trivial, and from algebraic topology we know that $H_1$ of that space is $G/[G,G]$.</p> <p>If $M$ is not a trivial $G$ module then you probably won't get as nice a description as for cohomology because the tensor product is not as nice as this case as the Hom set. Andy's comment on derivations/principal derivations comes from (or at least morally) from the bar resolution, which for any group $G$ is a free $\mathbb{Z}G$ resolution of $\mathbb{Z}$. This is great because homology is just the derived functor of $\mathbb{Z}\otimes_{\mathbb{Z} G}-$, which makes sense because $-_G$ is right exact and commutes with arbitrary direct sums.</p> <p>So you can still work out a kind of analogue to the description for cohomology. If you actually use the bar resolution, you get that $H_1(G,M)$ for an arbitrary right $G$-module $M$ is just the homology $\mathrm{ker} (d)/\mathrm{im} (d)$ (abuse of notation, using the same $d$) of the complex</p> <p>$M\otimes B_2\xrightarrow{1\otimes d} M\otimes B_1\xrightarrow{1\otimes d} M\otimes \mathbb{Z}G$</p> <p>where the tensor is taken over $\mathbb{Z}G$. The module $B_1$ is the free $G$ module on the symbols $[g_1]$ for $g_1\in G$ and $g_1\not=1$. If we let $[]\in\mathbb{Z}G$ be the identity, then $d([g_1]) = g_1[] - []$. Similarly, $B_2$ is the free $G$ module on $[g_1|g_2]$ where neither are the identity, and $d([g_1|g_2]) = g_1[g_2] - [g_1g_2] + [g_1]$. Here $[g_1g_2] = 0$ if $g_1g_2 = 1$ in the group. You can do the same for left-$G$ modules by forming the analogous right free $G$-module resolution of $\mathbb{Z}$.</p> <p>We could also have used the unnormalized bar resolution; regardless, it is now true that this is just a complex with each term a direct sum of copies of $M$, but the difficulty is determining what exactly the boundary formulas turn out to be. For a specific $M$ often it is not too hard to make a computation. For cyclic groups you can use the usual period resolution of $\mathbb{Z}$ to get that</p> <p>$H_1(\mathbb{Z}/k,M)\cong M^{\mathbb{Z}/k}/NM$</p> <p>where $N$ is the norm element, defined for a general (finite!) group to be $\sum_{g\in G}g$.</p> http://mathoverflow.net/questions/85643/nonstandard-methods-or-model-theory-and-arithmetic-geometry/94237#94237 Answer by Jason Polak for Nonstandard Methods ( or Model Theory ) and Arithmetic Geometry Jason Polak 2012-04-16T18:19:18Z 2012-04-16T18:19:18Z <p>Since you also mention algebraic geometry, you may want to take a look at "Model Theory and Algebraic Goemetry" edited by Elizabeth Bouscaren (Springer LNM 1696). Its primary purpose is to introduce Hrushovki's proof of Mordell-Lang for function fields. Except for some elementary model theory, it is self-contained, and shows that deep results in model theory can be used to prove nontrivial statements. Although this might be more geometric than arithmetic, perhaps some of the techniques could still be useful to you.</p> http://mathoverflow.net/questions/16994/linear-algebra-texts/17001#17001 Answer by Jason Polak for Linear Algebra Texts? Jason Polak 2010-03-03T20:00:06Z 2010-05-22T19:10:53Z <p>There's also Nicholson's Elementary Linear Algebra or the slightly more advanced Linear Algebra: With Applications. If your students react negatively to the intro of abstract vector spaces, I don't think Hoffman and Kunze's book would be good for them. While I love that book myself it might be a little too daunting for your class. Also I think that if you want to introduce abstract vector spaces from the start there's no reason you can't cover the chapter on abstract vector spaces first.</p> http://mathoverflow.net/questions/20551/sources-for-bibtex-entries/20585#20585 Answer by Jason Polak for Sources for Bibtex entries Jason Polak 2010-04-07T02:51:06Z 2010-04-07T02:51:06Z <p>There's also <a href="http://lead.to/amazon/en/" rel="nofollow">Lead2Amazon</a> which can export in a variety of formats.</p> http://mathoverflow.net/questions/19046/open-source-mathematical-software/19057#19057 Answer by Jason Polak for Open source mathematical software. Jason Polak 2010-03-22T19:45:29Z 2010-03-22T19:45:29Z <p><a href="http://www-fourier.ujf-grenoble.fr/~sergerar/Kenzo/" rel="nofollow">Kenzo</a> and <a href="http://chomp.rutgers.edu/" rel="nofollow">Chomp</a> are for computing homology. Kenzo for instance can take an arbitrary abstract simplicial complex and compute the simplicial homology groups, and it has various spaces already built in. You can compute the homology of products and other neat things with it.</p> http://mathoverflow.net/questions/18271/what-out-of-print-books-would-you-like-to-see-re-printed/18303#18303 Answer by Jason Polak for What out-of-print books would you like to see re-printed? Jason Polak 2010-03-15T20:08:35Z 2010-03-15T20:08:35Z <p>"Essays In Group Theory" edited by S.M. Gersten, which in particular contains Gromov's paper "Hyperbolic Groups".</p> http://mathoverflow.net/questions/17410/criteria-for-accepting-an-invitation-to-become-an-editor-of-a-scientific-journal/17443#17443 Answer by Jason Polak for Criteria for accepting an invitation to become an editor of a scientific journal Jason Polak 2010-03-08T01:57:22Z 2010-03-08T01:57:22Z <p>What strikes me most is that this letter is generic and not directed at you specifically, in the sense that there is no explained reason why they contacted you and not someone else. And "highest publication quality possible"? Please.</p> http://mathoverflow.net/questions/12638/taking-lecture-notes-in-lectures/12641#12641 Answer by Jason Polak for Taking lecture notes in lectures Jason Polak 2010-01-22T15:30:53Z 2010-01-22T15:30:53Z <p>Yes. It's essential for me to take notes or else I learn almost nothing unless it's something I've been thinking about for a while. I've discovered that my best learning technique is seeing symbols on paper. I often see something that many people find intuitively obvious but I don't feel sure about it until I do the computations myself.</p> <p>There are two things I've found useful in lectures: a general overview or intuition from the professor's perspective on a topic and the professor going through the proof in detail. However, if I had to choose one I'd choose the latter, because I benefit more from seeing the syntax than hearing the intuition or meaning, and I can remember that part anyway. So when I do write down notes I pretty much just take down the syntax and listen to the prof when I'm writing.</p> <p>Note that this doesn't apply to lectures for which I already know the topic well, but in this case I still have paper to do extra calculations, since the symbols on paper are inspiring.</p> http://mathoverflow.net/questions/12342/reductio-ad-absurdum-or-the-contrapositive/12355#12355 Answer by Jason Polak for Reductio ad absurdum or the contrapositive? Jason Polak 2010-01-20T00:21:14Z 2010-01-20T00:21:14Z <p>There does seem to be a difference, although I don't know how to distinguish them formally, but here are two cases which distinguish the two:</p> <p>*Let V be a vector space. A linear map f on V is injective if its kernel is 0. (yes I know it's an iff). The contrapositive is, if f is a linear map and ker f is not 0, then f is not injective. Proof: $v\in \ker f$ nonzero means $f(v) = f(0) = 0$. We don't have to assume in the proof that f is injective first and get a contradiction.</p> <p>Here we are arguing about a class of objects with one property and we are trying to prove another of that class.</p> <p>--</p> <p>*sqrt(2) is irrational. We can word this (artificially) as x = sqrt(2) implies there does not exist a,b rational such that x = a/b. However if we try the contrapositive, we get, there exists a,b rational such that x = a/b implies x is not sqrt(2). But to prove this we still need to assume that a/b = sqrt(2) and derive a contradiction.</p> <p>Here we are trying to prove that a single object does not possess a certain characteristic. There must be a more formal treatment of these differences, however...</p> http://mathoverflow.net/questions/10419/depressed-graduate-student/10491#10491 Answer by Jason Polak for Depressed graduate student. Jason Polak 2010-01-02T15:55:51Z 2010-01-02T15:55:51Z <p>There are several different approaches you can use. As mentioned, health is important and you should go to the gym or exercise in some way, preferably with someone. Also reading about a variety of topics is also good. In that domain I recommend you read a history of mathematics, and in particular biographies of mathematicians. Some of them have lived very interesting lives and can be an inspiration. Norbert Wiener's "I Am A Mathematician" and Halmos's "I Want To Be A Mathematician" are two examples.</p> <p>However, when it comes down to doing something for a long time, and especially something so intense like mathematics, you have to keep it fun. You can't take it too serious. Once you do that, you'll just poison yourself with self-destructive thoughts. "I'm not good enough" will only bring you down. The mind is capable of amazing things but you'll never realise your potential unless you start practising mathematics with a light heart. That can be particularly frustrating if you've just entered graduate school and you start to interact with people around you that are often faster and more knowledgeable than you. However, that does not negate the fact that there is enough mathematics for everyone, and that if you work hard enough and long enough in your own field, chances are you'll be able to do some interesting mathematics too.</p> <p>Math is also a social activity. It's not something to be practised by one's lonesome in a dark room, and although isolation can help when working through problems, you should always return and talk about it with others. Pick the right people as well. Find the right professor or the right group of students who will bring you up and not enervate your soul. Whatever you learn you should release. It's not good to keep your math trapped inside you. Let it out to others and help others. Volunteer in the undergrad math help room(s) if such a thing exists.</p> <p>Math is a field of logic but it's also an art. Don't always focus on mere logic and correctness. Rigour and proof is fundamental in mathematics, but once you have that, ask yourself, what is <em>beautiful</em> about this? Make the mathematics you do an art to be admired by yourself and by others. </p> http://mathoverflow.net/questions/3739/your-experience-of-computer-science-programming-in-mathematics-education/9848#9848 Answer by Jason Polak for Your experience of Computer Science/Programming in Mathematics Education? Jason Polak 2009-12-27T03:00:46Z 2009-12-27T03:00:46Z <p>My undergrad was B.Sc. Math, and my program required four courses when I entered: two intro courses using Java, a programming lab which just was practise in a particular language (either C++, Prolog, or something else I can't remember), and a data structures course.</p> <p>The latter two were dropped from the requirements since the entire system was redesigned, but I took two extra courses anyway: advanced C++ and formal languages.</p> <p>I can't comment on any struggles, because I was bunched up with engineers and CS majors. Great courses though, and if anything they helped because experimental mathematics is just really fun.</p> http://mathoverflow.net/questions/9786/use-of-the-word-data-not-in-the-statistical-sense Use of the word "data" not in the statistical sense Jason Polak 2009-12-26T07:57:29Z 2009-12-26T15:35:33Z <p>Occasionally I see the use of the word "data" in definitions. For instance, one definition of an exact sequence starts off by saying, "An exact sequence of abelian groups (or modules or vector spaces) is given by the data of two homomorphisms [...]" (Perrin, Algebraic Geometry).</p> <p>I've heard this term used in class as well once. In these instances data of course does not refer to data as used in statistics, as in data from an experiment.</p> <p>What is the purpose of using such a strange word in abstract mathematics? Has anyone noticed this word?</p> http://mathoverflow.net/questions/6675/periods-and-commas-in-mathematical-writing/6699#6699 Answer by Jason Polak for Periods and commas in mathematical writing Jason Polak 2009-11-24T15:38:10Z 2009-11-24T15:38:10Z <p>This is something I've never paid attention to until graduate school, but virtually every book uses the convention that formulae in display mode are part of the text. Every Springer text for instance uses these conventions.</p> <p>If we define the function $f:\mathbb{R}\rightarrow\mathbb{R}$ by</p> <p>$$f(x) = e^x,$$</p> <p>then we can place a comma after the definition to indicate a pause one might take if speaking such a sentence. We could also have defined the function by</p> <p>$$f(x) = \sin(x).$$</p> <p>As this last definition was the end of a sentence, it ought to have a period. Finally we could also have</p> <p>$$|f(x) - f(x_0)| &lt; \varepsilon$$</p> <p>whenever $|x - x_0| &lt; \delta$. Here, no punctuation was needed.</p> <p>There are exceptions: Spanier's Algebraic Topology doesn't follow these conventions, but Hardy does, and all modern books that I've read do. Unless I pay attention, I don't even notice the punctuation.</p> http://mathoverflow.net/questions/5532/the-importance-of-zf/5569#5569 Answer by Jason Polak for The Importance of ZF Jason Polak 2009-11-14T20:42:51Z 2009-11-14T20:42:51Z <p>People use whatever is most useful. ZFC just happens to be a fairly simple formalization of the way people think of sets such that we can eliminate imprecision sufficiently to do good mathematics. There are no "theorems" independent of ZFC in ZFC. CH is not a theorem in ZFC. Choice isn't either, but on the other hand choice is useful so everyone uses it. If you don't accept the axiom of choice then you can't have things like arbitrary products, and for some strange reason studying infinite products is actually very useful to practicing mathematicians. </p> <p>That's not to say that studying things like CH isn't useful. I think it is, but right now it's just not pertinent to most mathematicians.</p> http://mathoverflow.net/questions/3044/tools-for-collaborative-paper-writing/5101#5101 Answer by Jason Polak for Tools for collaborative paper-writing Jason Polak 2009-11-11T20:38:50Z 2009-11-11T20:38:50Z <p>For joint collaboration I've used subversion. Having a central repository I think is a really good idea for a paper, although you can set up central repositories in bzr and git. </p> <p>I would stay away from git for two reasons, however. First, Windows support doesn't seem to me that great. You have to either use something like Cygwin or the port, which is hardly optimal. Unfortunately many people use Windows. The second is that it's a bit harder to use git. I love git myself but subversion is straightforward: there's commit (and conflict), whereas with git you have a local repository you can commit to and then you can also push to a remote repository.</p> <p>In the end I like subversion for collaborative efforts. The only downside is if you really like to branch your work, then subversion is horrible compared to git or bzr. Bzr and svn also have nice version numbers ;)</p> http://mathoverflow.net/questions/5095/resources-for-getting-maths-on-to-the-web/5099#5099 Answer by Jason Polak for Resources for getting maths on to the web. Jason Polak 2009-11-11T20:28:04Z 2009-11-11T20:28:04Z <p>You can also install MediaWiki, the software used for Wikipedia, with LaTeX support, on your own server (say, your own domain or a department server);</p> <p><a href="http://www.mediawiki.org/wiki/MediaWiki" rel="nofollow">http://www.mediawiki.org/wiki/MediaWiki</a></p> <p>Possibly useful for things like a study group for problem sessions or a textbook where multiple people work on the same problems, which not only helps to spread solutions but also to practice writing mathematics where others will see it. If you need to copy the code for some reason then you can just view the page wiki code.</p> <p>This may also be useful for departments to keep track of their various seminars. Any attendee of the seminar could post their notes on a wiki. A collaborative, published effort may increase interest, especially amongst the undergraduates and graduate students.</p> http://mathoverflow.net/questions/3973/what-should-be-offered-in-undergraduate-mathematics-thats-currently-not-or-isn/4007#4007 Answer by Jason Polak for What should be offered in undergraduate mathematics that's currently not (or isn't usually)? Jason Polak 2009-11-03T22:18:15Z 2009-11-03T22:18:15Z <p>Inequalities!</p> <p>I don't think I've ever seen a course on inequalities, and there's certainly enough elementary material to cover in a one-semester course. Very few undergrads know much about inequalities.</p> http://mathoverflow.net/questions/3951/memorizing-theorems/3954#3954 Answer by Jason Polak for Memorizing theorems Jason Polak 2009-11-03T16:44:26Z 2009-11-03T16:44:26Z <p>Try coming up with counterexamples when you remove hypotheses. Play with the mathematics. The best way to know a theorem is to prove it. Try coming up with a different proof. Explain the theorem to someone else. Ask yourself where the theorem is used later. Rewrite the statement of the theorem. Does it generalise? How does it fit into the theory? For every theorem you ought to come up with a few examples that illustrate it, or at least understand the ones explained in class or the book.</p> <p>If you have to know a bunch of random proofs for a course, which isn't uncommon, then get together with some friends and take turns going over the proofs. Trying to memorise just the theorem statement without any idea of why it's useful or where it fits in won't get you very far, and won't be very motivating.</p> http://mathoverflow.net/questions/129762/what-arithmetic-information-is-contained-in-the-algebraic-k-theory-of-the-integer/129786#129786 Comment by Jason Polak Jason Polak 2013-05-05T22:57:42Z 2013-05-05T22:57:42Z That's ok, I was in a hurry typing this and admittedly did not read the question as thoroughly as I should have. http://mathoverflow.net/questions/128872/reference-for-rationality-in-algebraic-groups-in-the-language-of-schemes Comment by Jason Polak Jason Polak 2013-04-27T17:32:39Z 2013-04-27T17:32:39Z @ayanta: thank you for the references, I'll take a look. @Kidwell: I have taken a look at B. Conrad's notes from his course now and these are essentially the kind of thing I want. If you post your comment as an answer, I shall accept it. @Putman: Thank you also for the link; the notes look promising as well, and I'll keep an eye out for their progress. http://mathoverflow.net/questions/128872/reference-for-rationality-in-algebraic-groups-in-the-language-of-schemes Comment by Jason Polak Jason Polak 2013-04-27T16:19:17Z 2013-04-27T16:19:17Z @ayanta: yes sorry, I was being sloppy. I am aware that not all groups are split or quasisplit. The result on the existence of split and quasisplit forms is one thing for example that I‘d like to see. @Kidwell: thanks, I will take a look at these notes! http://mathoverflow.net/questions/128740/homotopy-theoretic-measure-of-operations-on-sheaves-failing-to-be-sheaves Comment by Jason Polak Jason Polak 2013-04-25T18:11:08Z 2013-04-25T18:11:08Z @Jacob: Yes for qc, this is essentially the third paragraph. http://mathoverflow.net/questions/118081/nice-algebraic-statements-independent-from-zf-vl-constructibility/118084#118084 Comment by Jason Polak Jason Polak 2013-01-04T20:44:28Z 2013-01-04T20:44:28Z (I meant to say $\mathrm{Ext}^1(A,\mathbb{Z}) = 0$) http://mathoverflow.net/questions/118081/nice-algebraic-statements-independent-from-zf-vl-constructibility/118084#118084 Comment by Jason Polak Jason Polak 2013-01-04T20:43:26Z 2013-01-04T20:43:26Z Thank you for the detailed answer. Since forcing extensions cannot satisfy $ZFC + V=L$, what about considering some theory that does not satisfy $V=L$. For instance, in some models, nonfree abelian groups $A$ with $\mathrm{Ext}^1(A,\mathbb{Z})$ do exist, so would it be easier to find further algebraic statements in such models that are independent? http://mathoverflow.net/questions/118081/nice-algebraic-statements-independent-from-zf-vl-constructibility Comment by Jason Polak Jason Polak 2013-01-04T20:04:08Z 2013-01-04T20:04:08Z Your question, also interesting, looks for open problems that were proven in some definition of &quot;countable case&quot; and not necessarily independent or implied by stronger set theoretic axioms, so I am hoping that the different criteria and wording might prompt some answers. http://mathoverflow.net/questions/116309/the-non-traveling-mathematician-problem Comment by Jason Polak Jason Polak 2012-12-13T21:17:23Z 2012-12-13T21:17:23Z Even if you go to four conferences a year lasting a week each, which to me seems like a lot, that's four weeks a year....is it too much for you to be away from your family for that long??? http://mathoverflow.net/questions/112424/space-filling-curve-to-simplify-vector-addition/112427#112427 Comment by Jason Polak Jason Polak 2012-11-14T23:58:16Z 2012-11-14T23:58:16Z @Benjamin: you still need 2. The curve is the other piece of information. http://mathoverflow.net/questions/111916/confusing-point-in-proof-semisimple-automorphism-fixes-torus/111988#111988 Comment by Jason Polak Jason Polak 2012-11-12T15:46:36Z 2012-11-12T15:46:36Z @Jim: Yes, what I was doing was too complicated indeed. I think my misunderstanding was with format of the semisimple part of $us$. Thank you very much for your answer and comments, again. http://mathoverflow.net/questions/111916/confusing-point-in-proof-semisimple-automorphism-fixes-torus/111988#111988 Comment by Jason Polak Jason Polak 2012-11-11T14:55:29Z 2012-11-11T14:55:29Z Thank you very much for your answer. There is one point here that is still not clear to me; replacing $us$ by its semisimple part is essentially conjugating by the inverse of the unipotent part, so we get (us)_u^(-1)usTs^-1u^-1(us)_u = T. If we let u‘=(us)_u^(-1)us then we have u‘s semisimple, but the next part of the theorem uses that u‘ should be in U. Does this follow? this seems to be the crucial part of the proof. http://mathoverflow.net/questions/111917/is-it-necessary-that-gcd-1-of-an-infinite-set Comment by Jason Polak Jason Polak 2012-11-09T17:42:17Z 2012-11-09T17:42:17Z You can consider a filtration $A_1\subset A_2\subset A_3 ...$ of your set of natural numbers $S$ such that $\cup A_i = S$ and every $A_i$ is finite. Then take the $\mathrm{gcd}$. http://mathoverflow.net/questions/107902/finite-subgroups-of-sl-2r/107903#107903 Comment by Jason Polak Jason Polak 2012-09-23T15:14:55Z 2012-09-23T15:14:55Z Or, choose a nontrivial matrix whose corresponding angle of rotation is minimal. http://mathoverflow.net/questions/105269/non-split-extension-of-the-rationals-by-the-integers/105368#105368 Comment by Jason Polak Jason Polak 2012-08-24T19:30:47Z 2012-08-24T19:30:47Z So this means that, $\mathbb{R}\cong (\prod_p \mathbb{Z}_p)/\mathbb{Z}$? A $\mathrm{lim}^1$ exact sequence in Weibel's book shows that $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z})$ is isomorphic to the latter. http://mathoverflow.net/questions/104932/kostants-theorem-on-invariant-polynomials-in-positive-characteristic Comment by Jason Polak Jason Polak 2012-08-18T01:46:48Z 2012-08-18T01:46:48Z @Jim: Hey thanks! Good to know :)