User lucas seco - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T12:24:17Z http://mathoverflow.net/feeds/user/12170 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58339/origin-of-terms-flag-flag-manifold-flag-variety/104157#104157 Answer by Lucas Seco for Origin of terms "flag", "flag manifold", "flag variety"? Lucas Seco 2012-08-06T22:25:44Z 2012-08-06T22:25:44Z <p>@Jim, I always asked myself the same question. You say that the "notion of flag variety or flag manifold evolved from the older and still somewhat mysterious use of the term flag in projective geometry, for instance to refer to an incident point-line pair." Once I saw a drawing in an expository article that solved that mistery for me: picture a point in the projective plane as the corresponding line in space, and a projective line as the corresponding plane in space, then an incident point-line becomes a line waving a plane in space, like a flag :)</p> http://mathoverflow.net/questions/60935/a-question-on-the-root-systems-of-simple-lie-algebras-in-the-90-degree-case/68028#68028 Answer by Lucas Seco for A question on the root systems of simple Lie algebras in the 90 degree case Lucas Seco 2011-06-17T04:29:19Z 2011-06-17T04:29:19Z <p>Ben, I also asked myself that same question and the notion that made it clear for me was that of an isomorphism between root systems: it is a linear map that sends all the roots of one system to all the roots of the other system, preserving the Cartan-Killing numbers of the corresponding roots (see Humphreys' book, Section 9.2)</p> <p>The beauty is that such an isomorphism does not need to be an isometry! So the the infinitely many two-dimensional root systems of the 90$^o$ case are all isomorphic between themselves, in particular all isomorphic to $A_1 \times A_1$.</p> <p>In terms of the corresponding Lie algebras, you have also infinitely many, all of them isomorphic to ${\frak sl}(2) \times {\frak sl}(2)$.</p> <p>PS - It can be shown that an isomorphism between two <strong><em>irreducible</em></strong> root systems must be conformal: it must scale the metrics by a constant factor. In particular, every automorphism of an irreducible root system is an isometry. (Ask me if you need a proof of this, it is very simple but it is not in Humphreys' book, I think.)<br> The irreducibility here is crucial since an $A_1 \times A_1$ root system with distinct root lengths admit an automorphism which is not an isometry.</p> http://mathoverflow.net/questions/52873/reference-request-probability-ergodic-theory-without-measure-spaces/52904#52904 Answer by Lucas Seco for Reference request: probability / ergodic theory without measure spaces Lucas Seco 2011-01-23T04:07:12Z 2011-01-23T04:07:12Z <p>I don't know if this is exactly what you are searching for, but maybe it is worth to take a look:</p> <p>Probability Theory: The Logic of Science by. E. T. Jaynes.</p> <p><a href="http://bayes.wustl.edu/etj/prob/book.pdf" rel="nofollow">http://bayes.wustl.edu/etj/prob/book.pdf</a></p> http://mathoverflow.net/questions/51685/how-did-bernoulli-prove-lhopitals-rule/51809#51809 Answer by Lucas Seco for How did Bernoulli prove L'Hôpital's rule? Lucas Seco 2011-01-12T01:54:15Z 2011-01-12T01:54:15Z <p>Regarding the above answers, it is important to state what is considered (see <a href="http://planetmath.org/encyclopedia/LHospitalsRule.html" rel="nofollow">http://planetmath.org/encyclopedia/LHospitalsRule.html</a>) to be L'Hôpital rule: $$\lim_{x\to a} f(x)/g(x) = \lim_{x\to a} f'(x)/g'(x)$$ whenever $f(a)=g(a)=0$ and the righmost limit make sense.</p> <p>Note that the weaker rule stated in the answer above $$\lim_{x\to a} f(x)/g(x) = f'(a)/g'(a)$$ is an easy consequence of the definition of the derivative, dividing both $f(x)$ and $g(x)$ by $x-a$ and taking limits. Despite the temptation to state and prove L'Hôpital in this weaker form, this form becomes useless whenever you have to use L'Hôpital rule twice to obtain an indefinite limit.</p> http://mathoverflow.net/questions/28569/is-there-a-morse-theory-proof-of-the-bruhat-decomposition/51800#51800 Answer by Lucas Seco for Is there a Morse theory proof of the Bruhat decomposition? Lucas Seco 2011-01-12T00:28:54Z 2011-01-12T00:28:54Z <p>Yes, there is!</p> <p>The best place I know where this is done are the initial sections of </p> <p>J. J. Duistermaat, J. A. C. Kolk, and V. S. Varadarajan. Functions, ﬂows and oscillatory integrals on ﬂag manifolds and conjugacy classes in real semisimple Lie groups. Compositio Math., 49(3):309–398, 1983</p> <p>I don't know much about complex groups (I work in the real context), but the above article also considers the complex case. Below I give you a picture of what happens for a real Lie semisimple noncompact Lie group $G$, hope this can help.</p> <p>The key is the action of a regular split-element of the Lie algebra on the maximal flag manifold $\mathbb{F} = G/P$ of $G$: this action is Morse, and the Morse function is beautifully simple: it is the height function of a natural embeeding of $\mathbb{F}$ in the Lie algebra of $G$ under an appropriate metric on $\mathbb{F}$. Its critical points are computed to be $wP$ and their stable manifolds are computed to be $PwP$. </p> <p>In case you are interested, they even prove the double-coset Bruhat decomposition $$G = \coprod_w P_\Theta w P_\Delta,$$ which is disjoint when $w$ runs trough the double coset $P_\Theta \backslash W / P_\Delta$. Here $P_\Theta, P_\Delta$ are the standart parabolic subgroups of type $\Theta$ and $\Delta$: they contain the minimal parabolic subgroup, which plays the role of the Borel subgroup in the real theory. The key here is the action of a (possibly not regular) split-elemet on $\mathbb{F}$: this action is Morse-Bott with the same beatifull Morse function! Its critical manifolds are computed to be the orbits of $wP_\Delta$ and their stable manifolds are computed to be the orbits of $P_\Theta w P_\Delta$ on the flag manifold. The hard part is to show that the above critical manifolds are indeed disjoint: to do this the above article appeals to algebraic constructions involving Tits buildings and other things I don't understand... This was disappointing to me since I expected a self-contained purely dynamical solution! </p> <p>After some stubborn tries I was able to do this step by purely dynamical arguments. It turns out that these critical manifolds are again flag manifolds: actually, flag manifolds of semisimple subgroups of $G$! This came as a nice surprise to me and my PhD advisor. Using this fact and the previous regular Bruhat decomposition, one can show by some simple dynamical arguments that these critical manifolds are disjoint. So the question can be settled by purely dynamical arguments and in a nice inductive manner. This is the content of my article</p> <p>Seco, L. . "A Note on the Bruhat Decomposition of Semisimple Lie Groups". Journal of Lie Theory, v. 18, p. 725-731, 2008.</p> <p>I would be very interested to know if the same method applies in the complex case to obtain an analogous double-coset Bruhat decomposition.</p> http://mathoverflow.net/questions/383/motivating-the-laplace-transform-definition/2141#2141 Comment by Lucas Seco Lucas Seco 2011-11-23T03:32:13Z 2011-11-23T03:32:13Z The link of the 2009 lecture is broken, but searching for &quot;differential equations Laplace transform&quot; on the OCW from MIT I found a link that is working as of today: <a href="http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-19-introduction-to-the-laplace-transform/" rel="nofollow">ocw.mit.edu/courses/mathematics/&hellip;</a> I don't know how OCW handles their links, so maybe in a near future the link will become broken again. Anyway, very nice lecture and very nice motivation! Thank you very much for sharing. http://mathoverflow.net/questions/28569/is-there-a-morse-theory-proof-of-the-bruhat-decomposition/51800#51800 Comment by Lucas Seco Lucas Seco 2011-06-24T02:22:44Z 2011-06-24T02:22:44Z @Dr Shello: not yet, but I would be very interested in exploring that! Just now I have seen Jim Humphreys' recomendation given above of the following reference MR628640 (83a:14035) 14L30 (14M17 20G05) Akyıldız, Ersan, Bruhat decomposition via Gm-action. (Russian summary) Bull. Acad. Polon. Sci. S&#180;er. Sci. Math. 28 (1980), no. 11-12, 541–547 (1981). It seems that it is hard to find, but I found a review paper of the same author <a href="http://www3.iam.metu.edu.tr/iam/images/8/81/Preprint13.pdf" rel="nofollow">www3.iam.metu.edu.tr/iam/images/8/81/&hellip;</a> which mentions this subject in the Remark of page 14, in ways which I don&#180;t understand yet. http://mathoverflow.net/questions/51685/how-did-bernoulli-prove-lhopitals-rule/51809#51809 Comment by Lucas Seco Lucas Seco 2011-01-30T23:47:53Z 2011-01-30T23:47:53Z @Andy: that's right, except that you are assuming that $G(x)$ has a derivative at $0$, that is, you are assuming that the limit $\lim_{x\to 0} f(x)/x^2$ exists. Do you know how to prove this? I've seen approximations to second derivatives that looks like this in numerical Calculus textbooks, but I don't know if they justify it rigorously. I really like your point of view about L'Hopital and would consider using it in my Calculus 1 classes if I am able to follow it throughly. http://mathoverflow.net/questions/51685/how-did-bernoulli-prove-lhopitals-rule/51809#51809 Comment by Lucas Seco Lucas Seco 2011-01-23T22:59:27Z 2011-01-23T22:59:27Z @Andy: exactly, but how do you prove that $G(0) = f''(0)$? http://mathoverflow.net/questions/51685/how-did-bernoulli-prove-lhopitals-rule/51809#51809 Comment by Lucas Seco Lucas Seco 2011-01-23T04:43:24Z 2011-01-23T04:43:24Z @Andy: Maybe you are right... But how does one gives a calculus 1 proof that $f(x) = x^k F(x)$ when the derivatives of order $&lt; k$ of $f$ vanishes? For $k=1$ it is an easy consequence of the definition of derivative, but for $k=2$ already I am not able to give a simple proof... Can you give me some indications? I imagine that an appropriate $k=2$ argument can be adapted to give a proof for any $k$ by induction.