User najdorf - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:44:34Z http://mathoverflow.net/feeds/user/8811 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math Cardinal Arithmetic, foundations and constructive math Najdorf 2012-11-04T01:55:51Z 2012-11-04T11:01:02Z <p>This is not my area but a question occurred to me that I can not find the answer to. There is a very strong <strong>axiom of constructibility</strong> which <em>ironically</em> gives us highly non-constructive math (<strong>GCH</strong> is one of its implications). What would be an equally strong axiom in the opposite direction? And I mean direction in a philosophical sense, so what would be the strongest axiom that constructivists/intuitionists would approve of?</p> <p>My first idea was to find the largest $\kappa$ such that $2^{\aleph_0} = \aleph_{\kappa}$ is consistent with <strong>ZF</strong> but this set is unbounded ($\kappa$ can be any finite number) and $2^{\aleph_0} &lt; \aleph_{\omega}$. Which brings up the question, how much fundamental difference are there between <strong>CH</strong> and $2^{\aleph_0} = \aleph_{118}$ for example?</p> http://mathoverflow.net/questions/103273/a-possible-mistake-in-kacs-infinite-dimensional-lie-algebras A possible mistake in Kac's "Infinite Dimensional Lie Algebras" Najdorf 2012-07-27T06:18:46Z 2012-07-29T21:55:07Z <p>I have a paperback 3rd edition and on page 65 you can find Proposition 5.8. My question is about part (c):</p> <blockquote> <p>If $A$ is of indefinite type, then $$\overline{X} = \{ h \in { \frak h_{\mathbb{R}} }| \left \langle \alpha, h \right \rangle \geq 0 \text{ for all } \alpha \in \Delta_+^{im} \},$$ where $\overline{X}$ denotes the closure of $X$ in the metric topology of $\frak h_{\mathbb{R}}$.</p> </blockquote> <p>Some context: $A$ is a generalized Cartan matrix, $\frak h_{\mathbb{R}}$ is a real form of the Cartan subalgebra in the Kac-Moody algebra associated to $A$ and finally $X$ is the Tits cone. </p> <p>Question: if one reads the proof given by Kac, it is written for $X$ and not $\overline{X}$. Is there a mistake or am I missing something here?</p> http://mathoverflow.net/questions/55311/subsystems-of-peano-arithmetic-and-incompleteness-theorem Subsystems of Peano arithmetic and incompleteness theorem Najdorf 2011-02-13T12:43:05Z 2012-02-20T12:19:25Z <p>I think everyone is familiar with Goedel's incompleteness theorems. In particular they imply that PA (Peano arithmetic) can not prove its own consistency. Now my question is what is the largest subsystem of PA that "can" prove its own consistency? You definitely can't have the induction axiom but is that enough? </p> http://mathoverflow.net/questions/41493/explicit-isomorphism-between-distributions-and-universal-enveloping-algebra Explicit isomorphism between distributions and universal enveloping algebra Najdorf 2010-10-08T09:04:46Z 2012-01-31T07:42:45Z <p>The universal enveloping algebra of a Lie algebra $\mathfrak{g}$ is isomorphic to the algebra of distributions on the Lie group $G$ with support at the identity. The proof I have of this fact uses the universal property of the universal enveloping algebra and hence it is not constructive. I was wondering if there is an explicit map? For example what would happen for $\mathfrak{g} = \mathfrak{sl}(2, \mathbf{C})$ and $G = SL(2, \mathbf{C})$? For an explicit monomial in $\mathscr{U} \ \left ( \mathfrak{sl}(2, \mathbf{C}) \right )$ can we write the corresponding distribution and how it acts on functions on $SL(2, \mathbf{C})$?</p> http://mathoverflow.net/questions/61743/non-trivial-integral-forms-of-algebras Non-trivial integral forms of algebras Najdorf 2011-04-14T20:34:03Z 2011-07-24T01:59:43Z <p>Suppose $\mathcal{A}$ is a $\mathbf{C}$-algebra then an integral form would be a subring $\mathcal{B} \subset \mathcal{A}$ such that the canonical map $\mathcal{B} \otimes_{\mathbf{Z}} \mathbf{C} \rightarrow \mathcal{A}$ is a bijection. </p> <p>For some algebras there is an obvious integral form in the following sense: there is a preferred $\mathbf{C}$-basis for $\mathcal{A}$ and the $\mathbf{Z}$-span of that basis is $\mathcal{B}$. Now my question is do we have examples where $\mathcal{B}$ is non-obvious? In other words the basis coming from $\mathcal{B}$ would look very strange for those who only work with $\mathcal{A}$. Is there such an example where $\mathcal{A}$ is commutative?</p> http://mathoverflow.net/questions/70866/question-on-linear-operators Question on Linear Operators Najdorf 2011-07-21T01:15:29Z 2011-07-21T20:37:07Z <p>Let $V$ be a normed infinite dimensional vector space. Let $L: V \longrightarrow V$ be a bounded linear operator. Moreover assume that $L$ is 'locally nilpotent' that is: $$\forall v \in V \quad \exists n \in \mathbf{N}: L^n (v) = 0.$$ Now my question is if the linear operator: $$\exp (L) = \sum_{n=0}^{\infty} \frac{L^n}{n!}$$ is bounded or not.</p> http://mathoverflow.net/questions/52607/embedding-of-algebraic-surfaces Embedding of algebraic surfaces Najdorf 2011-01-20T12:11:39Z 2011-06-28T13:09:19Z <p>If I am not mistaken there is a theorem that says any curve $C$ can be embedded in $\mathbf{P}^3$. What can be said about surfaces? Do we have a theorem like:</p> <blockquote> <p>All surfaces can be embedded in $\mathbf{P}^{N}$ for some fixed $N$. </p> </blockquote> <p>And if not what is the simplest counterexample? A counterexample would be an infinite sequence of surfaces: $\{S_1, S_2, \cdots \}$ and infinite sequence of strictly increasing positive integers: $\{n_1, n_2, \cdots \}$ such that $S_i$ can not be embedded in $\mathbf{P}^{n_i}$. </p> http://mathoverflow.net/questions/32566/careers-advice-for-ph-d-s-without-current-postdocs-or-university-jobs/60031#60031 Answer by Najdorf for Careers advice for Ph.D.s without current postdocs or university jobs Najdorf 2011-03-30T03:07:26Z 2011-03-30T03:07:26Z <p>I am more or less in the same situation as you are. After careful consideration I decided that I should abandon academia for good. Because as far as I have seen your career in academia is very dependent on where you start out. If you do a postdoc in <em>University of Nowhere</em> the chances of getting a decent job after that will be even lower than it is now. There is a temptation to think that by hard work you can compensate for the bad start in your career, resist that temptation.</p> <p>As it stands now I will be applying for finance jobs later in summer. Consider applying to big traditional banks (J P Morgan, Chase Manhattan, Wells Fargo, ...) investment banks (Goldman-Sachs, Morgan Stanley) and big hedge funds (Renaissance Technologies, D E Shaw, Citadel). If you want more info about 2.5 years ago there was a small article in Notices on the transition from academia (specifically math) to finance. The author had a Ph.D. in number theory and went on working for D E Shaw (PDF): <a href="http://www.ams.org/notices/200806/tx080600700p.pdf" rel="nofollow">http://www.ams.org/notices/200806/tx080600700p.pdf</a> </p> <p>Also don't worry about the state of financial firms. They have fully recovered from the effects of the great recession of 2007-2009: <a href="http://motherjones.com/kevin-drum/2011/03/chart-day-finance-back" rel="nofollow">http://motherjones.com/kevin-drum/2011/03/chart-day-finance-back</a> </p> http://mathoverflow.net/questions/58838/resume-de-cours-by-j-tits "Resume de cours" by J. Tits Najdorf 2011-03-18T13:51:20Z 2011-03-20T22:10:14Z <p>I have been reading a number of papers by J. Tits (mostly written in the second half of 1980s) and in them he frequently refers to following publications of his:</p> <ol> <li>Resume de cours, Annuaire du college de France, 81e annee (1980-1981), 75-86.</li> <li>Resume de cours, Annuaire du college de France, 82e annee (1981-1982), 91-105.</li> </ol> <p>Unfortunately I have not been able to locate these two papers (internet, library, faculty). Any help with finding these two would be immensely appreciated. </p> http://mathoverflow.net/questions/55512/classification-of-generalized-cartan-matrices-gcms Classification of generalized Cartan matrices (GCMs) Najdorf 2011-02-15T12:24:17Z 2011-02-15T14:00:29Z <p>A GCM is square matrix $A = (a_{ij})$ satisfying: (1) $a_{ij} \in \mathbf{Z}$ (2) $a_{ii} = 2$ for all $i$. (3) $a_{ij} \leq 0$ for $i \neq j$. (4) $a_{ij} = 0$ iff $a_{ji} = 0$. There is a standard notion of irreducibility among GCMs and the common term for it is "indecomposable". Now if you look at most books there is a standard basic classification of GCMs: An indecomposable GCM is of 3 kinds: finite, affine and another category which can be described as all-other-GCMs-that-I-know-little-about-so-I-will-bundle-them-together. There is also another classification based on whether the GCM is "symmetrizable" or not (the finite and affine GCMs are <strike>symmetric and hence</strike> symmetrizable). </p> <p>What I am looking for is a smart classification of the GCMs that are not finite or affine. For example here is a number of questions: </p> <ul> <b>[1]</b> Is the number of indefinite GCMs (not finite, not affine but symmetrizable) finite? As far as I know indefinite GCMs is a larger set than hyperbolic GCMs which are only finitely many.</ul> <ul> <b>[2]</b> Are all indefinite GCMs invertible? </ul> <ul> <b>[3]</b> If $A$ is indefinite and of size $N$ what is the biggest matrix of finite type that can appear as a principal minor in $A$? </ul> <p></p> <p><strong>EDIT:</strong> corrected a mathematical mistake pointed out by Jim Humphreys, also improved the format as to make things more visible. Again suggested by Jim Humphreys. </p> http://mathoverflow.net/questions/55308/what-a-geometer-should-know What a geometer should know ... Najdorf 2011-02-13T12:29:45Z 2011-02-14T23:27:22Z <p>I am wondering what are the prerequisites for being a modern geometer? It seems that the amount you have to know is just huge: differential geometry, differential topology, algebraic topology, algebraic geometry, symplectic geometry, ... and then there is all kinds of overlap. So my question is the following: what should you know (from the topics I mentioned and the topics that I forgot) to call yourself a geometer (lets say by the time you get your Ph.D.)? And where are the best books/articles covering all that? </p> http://mathoverflow.net/questions/53069/quiver-varieties-and-the-affine-grassmannian Quiver varieties and the affine Grassmannian Najdorf 2011-01-24T15:05:53Z 2011-01-24T16:53:58Z <p>Recently I was watching a talk: <a href="http://media.cit.utexas.edu/math-grasp/Ben_Webster.html" rel="nofollow">http://media.cit.utexas.edu/math-grasp/Ben_Webster.html</a> and at the end the lecturer gave a correspondence (I was having trouble with subscripts so changed the notation a bit):</p> <p>$$\mathfrak{Q} (\lambda, \mu) \Leftrightarrow \mathfrak{M} (\lambda, \mu)$$</p> <p>Between Quiver varieties and slices of affine Grassmannian. I would like to learn how this exactly works (like precise definition of both objects is a start!). If you have the time and the patience you can post a lengthy response here or you could just point me towards the right place to start reading on my own. Thank you!</p> http://mathoverflow.net/questions/52898/a-question-about-the-affine-grassmanian A question about the affine Grassmanian Najdorf 2011-01-23T02:46:26Z 2011-01-23T04:28:00Z <p>For $SL(2, \mathbf{C} ((t)))$ the affine Grassmanian is defined as: $$SL(2, \mathbf{C}((t))) / SL(2, \mathbf{C} [[t]])$$ Now that is fine but $SL(2, \mathbf{C} ((t)))$ has smaller parabolic subgroups. Define: $$B = \begin{pmatrix} \mathcal{O}^{\times} &amp; \mathcal{O} \\ t\mathcal{O} &amp; \mathcal{O}^{\times} \end{pmatrix}$$ where $\mathcal{O} = \mathbf{C}[[t]]$. Then: $$SL(2, \mathbf{C} [[t]]) = B \left \{ 1, \begin{pmatrix} &amp; 1 \\ -1 &amp; \end{pmatrix} \right \} B$$</p> <p><strong>Question</strong>: Why are people so much interested in the affine Grassmanian when it actually sits in a larger, equally well behaved projective ind-scheme (in our example above $SL(2, \mathbf{C}((t)))/ B$)? Both are flag varieties for the loop group so it seems natural to go with the <strong>full flag variety</strong>.</p> http://mathoverflow.net/questions/38544/quotients-of-unipotent-groups Quotients of unipotent groups Najdorf 2010-09-13T06:55:49Z 2010-09-13T17:12:46Z <p>Let $U (\mathbf{R})$ be the standard unipotent subgroup of $SL(3, \mathbf{R})$. So $U(\mathbf{R})$ is the group of 3 by 3 upper triangular matrices with 1s on the diagonal. I am interested in the quotient space $U (\mathbf{R})/ U (\mathbf{Z})$. I think it is just the cube, $[0,1]^3$, but I am having difficulty writing the actual map. The only non-obvious part is the (1,3)-entry where the addition is not straightforward. Also is the generalization that one gets the $\frac{1}{2}(n^2 - n)$ dimensional cube for the standard unipotent subgroup of $SL(n, \mathbf{R})$ correct?</p> <hr> <p><strong>@ Bill</strong> </p> <p>Your answer helped clear a lot of confusion but it also created a number of new questions:</p> <ol> <li><p>So basically $U(\mathbf{R})$ has three copies of $\mathbf{R}$ however they way they are knit together is not the simple direct sum. Is that right? </p></li> <li><p>If I pick an element $u \in U$ is the group generated by $u$ isomorphic to $\mathbf{G}_a$ ?</p></li> <li><p>Is the space $U (\mathbf{R})/ U (\mathbf{Z})$ compact? The Haar measure on $SL(3, \mathbf{R})$ gives a measure on $U(\mathbf{R})$. Then presumably I get a measure on $U (\mathbf{R})/ U (\mathbf{Z})$ what is the volume of the quotient with respect to this measure? Is there a place I can find these calculations?</p></li> </ol> <p>Ultimately I am interested in the arithmetic quotient $SO(3)\backslash SL(3, \mathbf{R}) / SL(3, \mathbf{Z})$ and $U (\mathbf{R})/ U (\mathbf{Z})$ is supposed to be its 'base' and the end result should be an algebraic surface (hence a 4-dimensional real manifold I think). </p> http://mathoverflow.net/questions/38376/nilpotent-lie-algebras-and-unipotent-lie-groups Nilpotent Lie algebras and unipotent Lie groups Najdorf 2010-09-11T01:35:02Z 2010-09-11T05:11:49Z <p>$\mathbf{n}$ is nilpotent Lie algebra with $N$ being the corresponding <em>algebraic</em> Lie group. Now one neat feature of this setting is that you can take the exponential map to be identity. In other words you can define a group structure on $\mathbf{n}$ using the Campbell-Hausdorff formula. I have the following questions (they might be very easy; I just don't know):</p> <ol> <li>If $\pi: \mathbf{n} \rightarrow \text{End}(V)$ is a representation of the nilpotent Lie algebra then does $x \in N$ acts as $\exp (\pi (x))$ on $V$ ?</li> <li>If $\mathbf{m}$ is a subalgebra of $\mathbf{n}$ then you get a corresponding group $M \subset N$. Do we have: $N/M = \mathbf{n}/\mathbf{m}$ as sets?</li> </ol> <p><strong>EDIT:</strong> Victor's comments are to the point so in order to clear up the confusion I added that $N$ is algebraic.</p> http://mathoverflow.net/questions/36821/explicit-equations-for-schubert-varieties Explicit equations for Schubert varieties Najdorf 2010-08-26T23:49:12Z 2010-09-09T11:05:08Z <p>How can one compute the Schubert variety (by compute I mean having actual polynomials that define it) for SL(n)? If this is well known forgive my ignorance and just point me to the right book/paper.</p> <p><strong>EDIT:</strong> Sorry I did not return here for quite some time. It is kind of amusing that the way I learned about Schubert varieties is not even mentioned. Here is how I learned it:</p> <blockquote> <ol> <li>$G$ an algebraic group with Lie algebra $\mathbf{g}$.</li> <li>$L(\Lambda)$ is an <em>integrable</em> highest weight module for $\mathbf{g}$.</li> <li>For $w$, an element of the Weyl group, consider the 1-dimensional root space $L(\Lambda )_{w \cdot \Lambda}$. </li> <li>Denote the vector space $U(\mathbf{b}) \bullet L(\Lambda)_{w \cdot \Lambda}$ by $E_w(\Lambda)$ (you take the 1-dimensional root space and act on it by all the raising operators plus the Cartan). Then $E_w(\Lambda) \subset L(\Lambda)$. </li> </ol> <p>Now we are ready: since $L(\Lambda )_{w \cdot \Lambda}$ is 1-dimensional it becomes a single point in $\mathbf{P} \left (E_w(\Lambda) \right)$. We look at the orbit $B \bullet L(\Lambda)_{w \cdot \Lambda} \subset \mathbf{P} \left (E_w(\Lambda) \right)$. We call its closure the <strong><em>Schubert variety associated to $w$ and $\Lambda$</em></strong> and denote it by $S_{w, \Lambda}$.</p> </blockquote> <p>I don't know if this is a good way of computing things but in principle it should give you any Schubert variety you need.</p> http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math/111425#111425 Comment by Najdorf Najdorf 2012-11-05T15:09:27Z 2012-11-05T15:09:27Z Also above I said that my idea of constructivism is that it should be practically computable (I mentioned MP in addition to the law of excluded middle). I have no idea if this is even a topic of discussion in constructive/intuitionist circles but if you look at it from a programming point of view it is the most natural definition. http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math/111425#111425 Comment by Najdorf Najdorf 2012-11-05T15:07:09Z 2012-11-05T15:07:09Z Alright that example clarified things a bit. But I am fine with unprovabally finite subsets of a finite set. My initial thought after reading your statement was that, there is a freakish model where a finite set had a non-finite subset (that would is stupid but from my experience you can't just call something obviously stupid and move one hence my confusion). http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math Comment by Najdorf Najdorf 2012-11-05T09:17:31Z 2012-11-05T09:17:31Z Law of excluded middle is non-constructive. I have doubts about MP as well, if $A$ and $A \to B$ but it took a trillion years to get from $A$ to $B$ would that be constructive? I have realized that I use constructive in a much stronger form than anyone else here. For example, I have a class of graphs with $\chi = 5$ but the proof is classical, then I get a 5-coloring algorithm of order $O(|G|^{|G|})$. To my mind the algorithm is progress but still not constructive. In short being constructive without being computable in some sensible sense of the word does not make sense to me ... http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math/111425#111425 Comment by Najdorf Najdorf 2012-11-05T08:57:44Z 2012-11-05T08:57:44Z @Andrej: I am very sorry for the lack of clarity, I have not thought about these things and strangely enough I think I use constructive in a stronger form than anyone else here which just worsens the matters. The subset/Random Variable is not a <i>real subset</i>, so if the non-finite subsets of a finite set are only of that type then I am fine with the original statement. The next question is whether what I call <i>real subset</i> can be formalized in any sense or not ... http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math Comment by Najdorf Najdorf 2012-11-04T23:05:32Z 2012-11-04T23:05:32Z @all: I call an axiom <b>AX</b> non-constructive if it has non-constructive consequences in everyday classical mathematics we use. I thought that was the normal understanding of the term. http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math Comment by Najdorf Najdorf 2012-11-04T23:03:06Z 2012-11-04T23:03:06Z @Blass I think you are right in your first comment. http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math/111425#111425 Comment by Najdorf Najdorf 2012-11-04T23:01:58Z 2012-11-04T23:01:58Z @Andrej, @Guillaume: In those cases I would take issue with the way you define the subsets which depend on an odd black box. What you have is not a fixed subset, it can be different things in the future, it is essentially a random variable. http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math Comment by Najdorf Najdorf 2012-11-04T11:07:08Z 2012-11-04T11:07:08Z @Noah, my understanding is that the model for <b>V=L</b> done by Godel is &quot;constructive&quot; but the theory itself (which is always much bigger than any single model ...) is not. http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math Comment by Najdorf Najdorf 2012-11-04T11:02:56Z 2012-11-04T11:02:56Z @Ricky, well <b>GCH</b> implies <b>AC</b> ... http://mathoverflow.net/questions/111421/cardinal-arithmetic-foundations-and-constructive-math/111425#111425 Comment by Najdorf Najdorf 2012-11-04T11:01:31Z 2012-11-04T11:01:31Z Alright that was helpful but I still have some issues: (1) When you say there are models for your first list of statements, what is the base system of constructive math you are working with? (2) Throughout the whole discussion I am assuming a degree of reasonableness, an axiom stating the fact that a finite set can have non-finite subsets or that we have an embedding $\mathbf{R} \to \mathbf{N}$ are not resonable (3) I might be mistaken but I thought that intuitionists did not like Church's thesis ... http://mathoverflow.net/questions/103273/a-possible-mistake-in-kacs-infinite-dimensional-lie-algebras/103455#103455 Comment by Najdorf Najdorf 2012-08-14T07:54:27Z 2012-08-14T07:54:27Z Does Kac answer email? Actually I found what seems to be a full proof in &quot;Introduction to Kac-Moody Algebra&quot; by Zhexian Wan (Proposition 5.6, page 98). http://mathoverflow.net/questions/103273/a-possible-mistake-in-kacs-infinite-dimensional-lie-algebras/103455#103455 Comment by Najdorf Najdorf 2012-07-31T18:11:55Z 2012-07-31T18:11:55Z Why are all integral points of X^{\prime} (defined as points where all simple roots give us integers) dense in the metric topology? I think it would be far from dense and actually discrete. http://mathoverflow.net/questions/103273/a-possible-mistake-in-kacs-infinite-dimensional-lie-algebras Comment by Najdorf Najdorf 2012-07-27T13:45:16Z 2012-07-27T13:45:16Z I used the word &quot;mistake&quot; because I think the statement is false and you should replace the closure with the cone. Of course I might be mistaken myself, hence &quot;possible mistake&quot;. http://mathoverflow.net/questions/103273/a-possible-mistake-in-kacs-infinite-dimensional-lie-algebras Comment by Najdorf Najdorf 2012-07-27T13:43:10Z 2012-07-27T13:43:10Z But if you read the proof he says let X^{\prime} be the right hand side and then shows that it is equal to X itself! The closure in metric topology is not mentioned even once! http://mathoverflow.net/questions/64905/which-book-would-you-like-to-see-texified/64910#64910 Comment by Najdorf Najdorf 2011-05-14T21:35:17Z 2011-05-14T21:35:17Z Can you give a link? I can't find it.