User david mjc - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:44:00Z http://mathoverflow.net/feeds/user/10106 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/123438/finite-order-automorphisms-on-complex-simple-lie-algebras/128663#128663 Answer by David MJC for Finite Order Automorphisms on Complex Simple Lie Algebras David MJC 2013-04-24T21:45:19Z 2013-04-24T21:45:19Z <p>It may be worth adding (belatedly) that the subgroup of affine diagram automorphisms mapping to Inn(L) has a natural description. Namely, the (co)weight lattice acts by translation on the affine root system. Since the affine Weyl group contains the (co)root lattice acting by translations, the quotient group acts on the affine diagram. As is well known, this quotient is isomorphic to the center Z of the simply connected group of L. In this way, the automorphism group of the affine diagram is an extension of Out(L), the automorphism group of the Dynkin diagram, by Z.</p> http://mathoverflow.net/questions/58870/what-should-be-taught-in-a-1st-course-on-smooth-manifolds/59125#59125 Answer by David MJC for What should be taught in a 1st course on smooth manifolds? David MJC 2011-03-22T01:16:24Z 2011-03-22T01:16:24Z <p>Differential forms.</p> <p>Books by Darling (Differential forms and connections) and Madsen-Tornehave (From calculus to cohomology: de Rham cohomology and characteristic classes) may help.</p> http://mathoverflow.net/questions/50610/what-are-classical-groups/50689#50689 Answer by David MJC for What are "classical groups"? David MJC 2010-12-30T00:02:40Z 2010-12-30T00:02:40Z <p>A classical group is the isotropy subgroup of an open orbit in a representation of GL(n).</p> http://mathoverflow.net/questions/49181/what-is-a-convenient-shorthand-notation-for-a-category What is a convenient shorthand notation for a category David MJC 2010-12-12T22:52:58Z 2010-12-16T23:49:50Z <p>Set theory has a very convenient and well established curly brace notation to specify a set by its elements: <code>$\{2,3,4,6\}$</code> or <code>$\{\text{finite subgroups of }SU(2)\}$</code> are simple examples.</p> <p>There should be a similar convenient notation for specifying a category by its objects and morphisms. Such a notation should easily accommodate categorical constructions such as slice categories. For example a double slash notation to separate objects and morphisms would define a slice category by something like (I am making this up!) $$\mathcal{C}\downarrow X= [ Y\to^f X : [Y//f]\in \mathcal C\quad // \quad (Y\to^f X)\to^h (Z\to^g X): [Y,Z//h] \in \mathcal C, g\circ h=f]$$ (a commutative diagram in the second part of the specification would be more convenient here, but it should also be possible to typeset the notation inline).</p> <p>Do such notations already exist? Whether they do or not, what notations would contributors recommend or suggest?</p> <p><strong>Update</strong>. Many thanks for comments made here. So far I most like the observation that clearly describing the morphisms makes the objects implicit. Still, I think beginners need the objects too, and have been experimenting with a notation like the one above, but using "staples" instead of square brackets, and introducing morphisms after objects by a vertical rectangular block (a bit like a closed staple). 2-morphisms could then be introduced in a similar way by a double block (a block with a vertical line through it). While the answers convince me that such notation is often unnecessary and maybe unhelpful sometimes, I'm not convinced such notation would be worthless.</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/47581#47581 Answer by David MJC for What makes four dimensions special? David MJC 2010-11-28T12:36:58Z 2010-11-28T12:36:58Z <p>(Riemannian geometry) Four is the only dimension $n$ in which the adjoint representation of SO($n$) is not irreducible. Since the adjoint representation is isomorphic to the representation on 2-forms, this means that the bundle of 2-forms on an oriented Riemannian manifold decomposes into self-dual and anti-self-dual forms. 2-forms are particularly significant, since the curvature of a connection is a 2-form. In particular the curvature of the Levi-Civita connection is a 2-form with values in the adjoint bundle, so it has a 4-way decomposition into self-dual and anti-self-dual pieces. Hence there are natural curvature conditions on Riemannian 4-manifolds which have no analogue in other dimensions (without imposing additional structure).</p> <p>The impact of self-duality includes: special properties of Einstein metrics, Yang-Mills connections, and twistor theory for (anti-)self-dual Riemannian manifolds.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46688#46688 Answer by David MJC for Sexy vacuity .... David MJC 2010-11-19T22:34:53Z 2010-11-19T22:34:53Z <p>I regard "negative thinking" in category theory as an example of cool vacuity: see e.g. <a href="http://ncatlab.org/nlab/show/negative+thinking" rel="nofollow">http://ncatlab.org/nlab/show/negative+thinking</a>. As category theory is not set theory, such vacuity does not necessarily involve the empty set directly, but the same principle of backwards generalization is used.</p> <p>The fact that a set is uniquely determined by its elements (i.e., has no additional structure beyond the equality relation between its elements) is summarized by saying that a (-1)-category is a truth value: a morphism between two elements in a set is either true (the elements are the same) or false (they are not). So the morphisms in a 0-category (a set) are (-1)-categories (either true or false), just as the morphisms in a (1-)category are 0-categories (sets). This admits generalizations to situations where "truth" is a more subtle concept (e.g. parameter dependent).</p> http://mathoverflow.net/questions/45185/pseudonyms-of-famous-mathematicians/45617#45617 Answer by David MJC for Pseudonyms of famous mathematicians David MJC 2010-11-10T23:07:46Z 2010-11-10T23:07:46Z <p>Arthur L. Besse - after the round tables held at Besse in France. (The "L." is for Lancelot.)</p> http://mathoverflow.net/questions/42410/what-is-the-dual-concept-to-annihilator-called-and-do-any-linear-algebra-textb What is the dual concept to "annihilator" called, and do any linear algebra textbooks discuss this concept first? David MJC 2010-10-16T20:07:13Z 2010-11-10T22:58:22Z <p>When introducing dual spaces for the first time, most linear algebra textbooks proceed in what seems to me a rather backwards fashion: the annihilator <code>$\{f\in V^*: f(u)=0\quad \forall u\in U\}$</code> of a subset/space <code>$U$</code> of a vector space <code>$V$</code> is introduced before the dual concept of the "joint kernel"(?) <code>$\{v\in V: f(v)=0\quad \forall f\in W\}$</code> of a subset/space <code>$W$</code> of the dual space <code>$V^*$</code>. The latter notion, despite corresponding to the intuitive idea of the solution space of a homogeneous linear system of equations, is then introduced indirectly by mapping <code>$V$</code> into <code>$V^{**}$</code> and considering the annihilator of the system of equations <code>$W$</code> of <code>$V^*$</code>, which is <code>$\{\phi\in V^{**}: \phi(f)=0\quad \forall f\in W\}$</code>. This seems a pity as many students find the isomorphism of a finite dimensional vector space with its double dual difficult to grasp!</p> <p>Worse, as the "(?)" above suggests, there appears to be no common terminology or notation for the concept dual to annihilator. Possibilities include "(joint) kernel, null/zero space, pre-annihilator, solution space". I'd be grateful for any pointers to textbooks which introduce the concept directly, or any suggestions for terminology and notation.</p> <p><strong>Update</strong>: Many thanks for the suggestions so far. I am still rather surprised and disappointed by the lack of references to elementary linear algebra text books which discuss solution spaces (aka null spaces, joint kernels) before or on the same footing as annihilators. I am unconvinced by the arguments that have so far been made for this omission.</p> <p>Instead it seems to me that the current situation is the result of inertia. The "annihilator" is a term that caught on, and has been carried forward in the absence of a cool name for the dual concept. In this respect I like the invention of the "vanquished".</p> <p>The question is not moot as I am currently lecturing this material. I am going to stick to "solution/null space" and "joint kernel". I will then discuss duality and row rank = column rank without ever mentioning the double dual (the double dual will come later in the course).</p> <p>In preparing this, I noticed the fact that solution spaces and annihilators provide a Galois connection between the posets of subsets/subspaces of a vector space and its dual (that is, solution spaces and annihilators are contravariant adjoint functors between these posets). I've seen discussion of this in the context of annihilators in ring theory, but it seems to me to be at the heart of the concept of duality for vector spaces. I hope this rings some bells among MO readers.</p> <p>Such categorical thinking leads to a fairly clean analysis of solution spaces and annihilators, which has in turn simplified some results subsequently.</p> http://mathoverflow.net/questions/42929/suggestions-for-good-notation/45377#45377 Answer by David MJC for Suggestions for good notation David MJC 2010-11-08T23:00:55Z 2010-11-08T23:00:55Z <p><code>$f_*$</code> and <code>$f^*$</code> for direct and inverse image. We really should use this right from the beginning, for functions $f\colon X\to Y$, where <code>$f_*\colon P(X)\to P(Y)$</code> ($P(X)$ being the power set) and <code>$f^*\colon P(Y) \to P(X)$</code> instead of the awful notations $f(A)$ and $f^{-1}(B)$ for subsets $A$ of $X$ and $B$ of $Y$.</p> http://mathoverflow.net/questions/43889/proof-synopsis-collection/43891#43891 Answer by David MJC for Proof synopsis collection David MJC 2010-10-27T22:22:30Z 2010-10-27T23:38:11Z <p>(1) Proof of a theorem in finite dimensional linear algebra.</p> <p>The hypotheses invite us to construct a linearly independent set: extend this set to a basis and conclude.</p> <p>(2) Proof of a theorem related to compact topological spaces.</p> <p>The hypotheses invite us to construct an open cover: observe this open cover has a finite subcover and conclude.</p> http://mathoverflow.net/questions/43579/examples-of-categorification/43597#43597 Answer by David MJC for Examples of categorification David MJC 2010-10-25T23:54:09Z 2010-10-27T21:27:20Z <p>I like one of the simplest and most well known examples: the category of finite sets and <s>bijections</s> functions (see below for comments) categorifies the natural numbers. Or rather it un-de-categorifies the de-categorification that led to much of mathematics in the first place. That makes it pretty special, even if it is rather basic compared with other examples.</p> http://mathoverflow.net/questions/43237/generalized-geometry/43239#43239 Answer by David MJC for Generalized geometry David MJC 2010-10-22T22:05:49Z 2010-10-26T00:23:27Z <p>Generalized geometry (in Hitchin's sense, following Courant and Dorfman) is adapted to the physical motion of string-like particles in the same way that traditional geometry is adapted to the physical motion of point-like particles. More general generalized geometries are useful in connection with higher dimensional objects such as membranes (and hence also M-theory). Pavol Severa's first <a href="http://sophia.dtp.fmph.uniba.sk/~severa/letters" rel="nofollow"> letter to Alan Weinstein</a> is a nice early reference point for the basic idea.</p> <p><strong>Update 1</strong>. I won't be able to add significantly to this post until Tuesday perhaps, but I want to indicate some of the relationships between my answer (as it is and to come) and Urs'. Ignoring higher dimensional objects than strings for now, generalized geometry initially concerns geometry on the generalized tangent bundle <code>$T\oplus T^*$</code> (where $T=TM$ is the tangent bundle of a manifold $M$). The bundle <code>$T\oplus T^*$</code> has a natural symmetric form with respect to which both $T$ and <code>$T^*$</code> are maximal isotropic.</p> <p>However, generalized geometry takes the point of view that <code>$T\oplus T^*$</code> is an extension of $T$ by <code>$T^*$</code>, and is thus an example of a Courant algebroid $CA$, in that there is a short exact sequence <code>$0\to T^*\to CA\to T\to 0$</code>, where $CA$ has a symmetric form and other structure (the Courant bracket) making it isomorphic to <code>$T\oplus T^*$</code> for suitable isotropic splittings of the exact sequence. A Dirac structure is such an isotropic splitting.</p> <p>From the naive <code>$T\oplus T^*$</code> viewpoint, a Dirac structure is given by an orthogonal involution of <code>$T\oplus T^*$</code> whose eigenspaces do not meet <code>$T^*$</code>. Generalized complex geometry is a subfield of generalized geometry, in which one studies orthogonal complex structures on <code>$T\oplus T^*$</code> instead of involutions. However, there are interesting structures on <code>$T\oplus T^*$</code> which involve neither Dirac structures nor generalized complex structures. This should not be surprising: there is more to ordinary geometry than involutions and complex structures.</p> http://mathoverflow.net/questions/42929/suggestions-for-good-notation/43236#43236 Answer by David MJC for Suggestions for good notation David MJC 2010-10-22T21:20:16Z 2010-10-22T21:20:16Z <p>I like to interpret $f(x)$ as meaning $f\circ x$, otherwise known as the pullback $x^*f$. For instance $x$ could be the standard real valued coordinate on a line. This makes rigorous sense of the concept of a "variable" and hence also dependent and independent variables ($y=f(x)$). In the example of functions on a line, $f'=dy/dx$ is simply a ratio of 1-forms.</p> <p>Such an interpretation also answers the common complaint that $f=f(x)$ confuses a function with its values. Instead it represents the very common shorthand of omitting pullbacks!</p> http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/43229#43229 Answer by David MJC for Awfully sophisticated proof for simple facts David MJC 2010-10-22T20:28:05Z 2010-10-22T20:28:05Z <p>A quiver whose unoriented graph is the affine D4 Dynkin diagram is tame. Therefore the moduli space of four points on a projective line is one dimensional.</p> http://mathoverflow.net/questions/50610/what-are-classical-groups Comment by David MJC David MJC 2010-12-30T22:24:54Z 2010-12-30T22:24:54Z I agree that there is unlikely to be one definitive answer to this question, but would suggest that the term &quot;classical group&quot; is misleading in the same way as the term &quot;holonomy group&quot;, in that one has in mind not only a group, but a particular linear (or possibly projective) representation of that group. http://mathoverflow.net/questions/50610/what-are-classical-groups/50689#50689 Comment by David MJC David MJC 2010-12-30T13:42:17Z 2010-12-30T13:42:17Z Correct: I was hoping someone would notice that :) I claim it is entirely appropriate to regard G_2 as a classical group for this reason, and it can be manipulated classically in its 7d fundamental representation. It just happens that Hermann Weyl did not consider it so. http://mathoverflow.net/questions/49181/what-is-a-convenient-shorthand-notation-for-a-category Comment by David MJC David MJC 2010-12-14T00:24:57Z 2010-12-14T00:24:57Z In 20-30 years time we might want to introduce categorical thinking more systematically at an undergraduate level (as set theory becomes an outdated foundation). For most mathematical purposes, I don't care what is the element of a 1-element set, nor do I care precisely what an ordered pair or a disjoint union is, as long as the universal properties are satisfied. I agree that describing sets by elements is uncategorical, but that applies to morphisms too, and we aren't going to introduce infinity categories in one breath are we? http://mathoverflow.net/questions/49181/what-is-a-convenient-shorthand-notation-for-a-category Comment by David MJC David MJC 2010-12-14T00:16:56Z 2010-12-14T00:16:56Z I like David's answer about just giving the morphisms, as one almost invariably has to specify the objects in the process, but I'm concerned here about introducing beginners to categorical thinking, not experts. http://mathoverflow.net/questions/49181/what-is-a-convenient-shorthand-notation-for-a-category Comment by David MJC David MJC 2010-12-12T23:27:46Z 2010-12-12T23:27:46Z My motivation is not computer readability, or indeed any desire to make categories 1-dimensional. Rather, in the 21st century, we ought to be able to describe simple examples of categories concisely, without saying &quot;the objects are this&quot; and &quot;the morphisms between objects X and Y are that&quot; every single time. The impetus for this post was an attempt to describe a groupoid, namely the value of a global quotient stack [X/G] on a scheme S. Of course, one can easily say what [X/G](S) is in words, but shouldn't we also have notation for categories like this? http://mathoverflow.net/questions/49181/what-is-a-convenient-shorthand-notation-for-a-category/49185#49185 Comment by David MJC David MJC 2010-12-12T23:20:50Z 2010-12-12T23:20:50Z Can you give an example? Thanks http://mathoverflow.net/questions/49181/what-is-a-convenient-shorthand-notation-for-a-category Comment by David MJC David MJC 2010-12-12T23:18:30Z 2010-12-12T23:18:30Z I'm definitely <i>not</i> proposing any notation! I also tried a double pipe while formulating this question. However, morphisms are closely related to weak quotients (as the example of a point/G illustrates). http://mathoverflow.net/questions/49181/what-is-a-convenient-shorthand-notation-for-a-category/49183#49183 Comment by David MJC David MJC 2010-12-12T23:10:54Z 2010-12-12T23:10:54Z Typesetting inline is not a requirement, but note that I could have typeset the commutative diagram inline if I had the energy! Two and three dimensional diagrams can often be represented concisely on paper, hence inside a &quot;Category of...&quot; notation. http://mathoverflow.net/questions/48675/various-definitions-of-connections-on-bundles/48807#48807 Comment by David MJC David MJC 2010-12-09T22:41:51Z 2010-12-09T22:41:51Z In other words, the omega's in the second calculation are minus the omega's in the first. Hence the sign difference in the transition formulae. http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/47581#47581 Comment by David MJC David MJC 2010-11-28T23:16:18Z 2010-11-28T23:16:18Z Note also Torsten Ekedahl's response to the question above (which I missed when posting this): in any even dimension, middle dimensional forms are not irreducible for the complexified special orthogonal group. This accounts not only for the special features of four dimensions in Riemannian geometry, but also dimensions 2 and 6, where 1-forms and 3-forms play a special role. Further, Lorentzian geometry in four dimensions is special because the bundle of 2-forms has a natural complex structure: this underpins the Petrov Classification of spacetimes, for example. http://mathoverflow.net/questions/45951/sexy-vacuity/46150#46150 Comment by David MJC David MJC 2010-11-16T08:59:38Z 2010-11-16T08:59:38Z Scott, I think you are talking about the empty (n-1)-manifold there, not the empty set :) http://mathoverflow.net/questions/46151/which-flag-manifolds-are-symmetric Comment by David MJC David MJC 2010-11-15T23:22:57Z 2010-11-15T23:22:57Z Another name for these spaces is &quot;symmetric R-spaces&quot;. http://mathoverflow.net/questions/45951/sexy-vacuity Comment by David MJC David MJC 2010-11-15T22:45:15Z 2010-11-15T22:45:15Z I like examples where a trivial/vacuous case helps determine the &quot;right&quot; definition in the general case, e.g., the norm of a morphism of normed spaces below. I like examples which begin an induction in a surprisingly vacuous case, yet the induction still works. And I like examples where the same principle gives a useful answer in multiple cases - e.g. in an order lattice, the sup/join of the empty set is the least element, and the inf/meet is the greatest: for the extended real numbers, $\inf\emptyset=\infty$ and for subsets of a set X, the empty intersection is X. http://mathoverflow.net/questions/45951/sexy-vacuity/46150#46150 Comment by David MJC David MJC 2010-11-15T22:03:36Z 2010-11-15T22:03:36Z This is an example I would have given. In particular, in the extended reals, inf A is only less than or equal sup A if A is nonempty. The same logic explains why the zero polynomial has degree $-\infty$ and why the empty set has dimension $-\infty$. http://mathoverflow.net/questions/45951/sexy-vacuity/46074#46074 Comment by David MJC David MJC 2010-11-15T21:56:19Z 2010-11-15T21:56:19Z $-\infty$ crops up here (and in other places such as the dimension of the empty set) because it is the supremum of the empty set: in this case it is the supremum of the set of powers with nonzero coefficients.