User eugene lerman - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:10:03Z http://mathoverflow.net/feeds/user/25355 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120738/is-a-measurable-homomorphism-on-a-lie-group-smooth/120783#120783 Answer by Eugene Lerman for Is a measurable homomorphism on a Lie group smooth? Eugene Lerman 2013-02-04T16:50:40Z 2013-02-04T16:50:40Z <p>Not every measurable function on the real line is continuous, let along smooth. Real line is a perfectly nice Lie group.</p> http://mathoverflow.net/questions/119393/unitary-representation-compact-lie-groups/119396#119396 Answer by Eugene Lerman for Unitary representation compact Lie Groups Eugene Lerman 2013-01-20T12:57:02Z 2013-01-20T12:57:02Z <p>You need to prove that you group $G$ has a faithful representation. This "is" Peter-Weyl theorem (or a corollary of, depending on how one states things).</p> http://mathoverflow.net/questions/118548/teaching-stacks-to-differential-geometry-students Teaching stacks to differential geometry students Eugene Lerman 2013-01-10T16:15:03Z 2013-01-10T20:12:51Z <p>Does anyone have any experience teaching stacks over the category of manifolds to students whose background is, say, a semester-long course on manifolds? Does anyone know of any publicly available notes on the subject, preferably in English? [My French is limited to the knowledge of the alphabet :). I can read Russian.]</p> <p>I am aware of a paper by Behrend and Xu, Metzler's paper in the arxiv, and notes by Heinloth. Hepworth has a nice exposition of vector fields on stacks, but his papers are rather terse. Vistoli's notes on descent are quite nice, but are clearly aimed at algebraic geometers. And there differences between the categories of manifolds and schemes --- fiber products of manifolds are badly behaved, for one thing.</p> <p>The challenges in teachign such a course seem many. For one thing I don't know how to talk about stacks without getting into 2-category theory. And most differential geometers don't know much of 1-category theory. But I don't want to start with a crash course on category theory.</p> http://mathoverflow.net/questions/113941/what-classification-theorems-have-been-improved-by-re-categorizing/114073#114073 Answer by Eugene Lerman for What classification theorems have been improved by re-categorizing? Eugene Lerman 2012-11-21T16:18:42Z 2012-11-22T02:51:08Z <p>My understanding of algebraic geometry is woefully inadequate, but here goes.</p> <p>My example is Borisov-Chen-Smith's definition of toric Deligne-Mumford stacks in terms of stacky fans. In other words, essentially a definition by construction. Fantechi-Mann-Nironi then gave a definition of toric DM stacks as closures of orbits of "stacky tori." I believe they also showed that different stacky fans may give rise to isomorphic toric DM stacks.</p> <p>I don't think the Fantechi-Mann-Nironi definition is the ultimate answer to the question of what a toric DM stack is. This is because on the symplectic side there are toric DM stacks whose generic stabilizer groups are non-abelian (see <a href="http://arxiv.org/abs/0908.0903" rel="nofollow">arXiv:0908.0903v2 [math.SG]</a>)</p> http://mathoverflow.net/questions/112689/analytic-lagrangian-submanifolds/112741#112741 Answer by Eugene Lerman for Analytic Lagrangian Submanifolds Eugene Lerman 2012-11-18T02:05:55Z 2012-11-18T02:05:55Z <p>Try "Semi-classical analysis" by Guillemin and Sternberg available at <a href="http://math.mit.edu/~vwg/semiclassGuilleminSternberg.pdf" rel="nofollow">http://math.mit.edu/~vwg/semiclassGuilleminSternberg.pdf</a></p> http://mathoverflow.net/questions/38575/colimits-of-manifolds/111018#111018 Answer by Eugene Lerman for Colimits of manifolds Eugene Lerman 2012-10-29T18:31:15Z 2012-10-30T00:19:16Z <p>This is a few years late, but here goes.</p> <p>Associated to an open cover ${U_i}$ of a (Hausdorff paracompact) manifold $M$ there is a cover groupoid, whose space of objects is the disjoint union ${\mathcal U}:= \sqcup U_i$ and whose space of arrows is the fiber product ${\mathcal U }\times _M {\mathcal U}$. The manifold $M$ is the quotient space of the cover groupoid, i.e., the colimit of ${\mathcal U }\times _M {\mathcal U}\rightrightarrows {\mathcal U}$. Note that the cover groupoid is <em>proper</em> so its quotient is Hausdorff. The orbit spaces of two Morita equivalent groupoids are isomorphic. Hence you could write a manifold as a colimit of a Lie groupoid which is Morita equivalent to a cover groupoid. I believe the conditions for being equivalent to a cover groupoid are being proper and having all isotropy groups trivial ( i.e., $Hom (x,x)$ is a trivial group for any object $x$ of your Lie groupoid). </p> http://mathoverflow.net/questions/110007/courant-algebroids-which-are-not-exact/110013#110013 Answer by Eugene Lerman for Courant algebroids which are not exact Eugene Lerman 2012-10-18T13:57:59Z 2012-10-18T13:57:59Z <p>The answer depends on what you mean by interesting."</p> <p>For example the paper "On the Geometric Structure of Hamiltonian Systems with Ports" by Jochen Merker, J Nonlinear Sci (2009) 19: 717–738 DOI 10.1007/s00332-009-9052-3, may be considered as dealing with interesting examples of Courant algebroids. The algebroids there are not of the form $TM\oplus T^*M \to M$.</p> <p>Grützmann's thesis (arXiv:1004.1487 [math.DG]) maybe another good place to look.</p> http://mathoverflow.net/questions/109645/can-stabilizer-groups-in-an-orbifold-have-global-twisting/109666#109666 Answer by Eugene Lerman for Can stabilizer groups in an orbifold have global twisting? Eugene Lerman 2012-10-15T01:13:35Z 2012-10-15T13:38:18Z <p>The short answer is "yes" if you think of orbifolds as groupoids or stacks. </p> <p>(begin edit) In particular, if you think of orbifolds as groupoids, the "purely ineffective" orbifolds, the issue that seems to be at stake here, can be seen as bundles of groups. That is, they are fiber bundles whose fibers are groups. They are not to be confused with principal bundles, whose fibers are homogeneous spaces (end edit).</p> <p>Both of your examples are $\mathbb{Z}/3$ bundles <del>gerbes</del> over $S^1$, one trivial, one not. So they are not the same as stacks.</p> <p>Another example I like comes from an ineffective action of $U(1)$ on $S^3$, say given by $\lambda \cdot z = \lambda^3 z$. The corresponding etale Lie groupoid can be thought of as a nontrivial $\mathbb{Z}/3$ <del>gerbe</del> bundle over $S^2$. Of course there is also a trivial one: $\mathbb{Z}/3$ acting trivially on $S^2$.</p> http://mathoverflow.net/questions/108228/is-there-a-natural-distance-between-skew-hermitian-matrices/108230#108230 Answer by Eugene Lerman for Is there a natural distance between skew hermitian matrices? Eugene Lerman 2012-09-27T11:27:22Z 2012-09-27T11:27:22Z <p>Since skew-Hermitian matrices form the Lie algebra of the unitary group $U(n)$, the space has a natural positive definite inner product given by $$(A, B) = - tr (AB).$$ This inner product, of course, defines a distance. The inner product is invariant under conjugation by elements of $U(n)$.</p> <p>Real skew-symmetric matrices form the Lie algebra of the orthogonal group. It also has a conjugation invariant positive definite inner product, in fact only one up to a scalar multiple. It is also $-tr(AB)$.</p> http://mathoverflow.net/questions/107824/how-to-find-the-induced-metric-on-an-orbit/107833#107833 Answer by Eugene Lerman for how to find the induced metric on an orbit? Eugene Lerman 2012-09-22T12:27:34Z 2012-09-22T12:27:34Z <p>The orbit is homogeneous, so it's enough to compute the metric at one point, call it $x$. The map from the Lie algebra to the tangent space to the orbit at $x$ is surjective. Concretely for a vector $X$ in the Lie algebra the corresponding tangent vector is $X_M (x) := d/dt \exp (tX) \cdot x$ ( $\cdot$ denotes the action). Now take another vector $Y$ in the Lie algebra and compute the inner product between $X_M (x)$ and $Y_M (x)$. This computes the induced "metric" on $G/G_x$ ($G_x$ is the stabilizer of $x$). The word "metric" is in scary quotes because it could be zero. </p> http://mathoverflow.net/questions/107302/contactization-and-symplectization/107388#107388 Answer by Eugene Lerman for 'Contactization' and Symplectization Eugene Lerman 2012-09-17T14:43:09Z 2012-09-17T14:43:09Z <p>The "pre-quantization" construction of a contact manifold out of symplectic manifold predates prequantization by a couple of decades: see Boothby, W. M.; Wang, H. C. On contact manifolds. Ann. of Math. (2) 68 1958 721–734. The analogue of the theorem for symplectic orbifolds is due to Thomas: Thomas, C. B. Almost regular contact manifolds. J. Differential Geometry 11 (1976), no. 4, 521–533.</p> <p>You may think of the Boothby-Wang construction as constructing a contact fiber bundle over a symplectic manifold with fiber $S^1$. If we look at the construction this way, it can be generalized. See my paper Contact fiber bundles. J. Geom. Phys. 49 (2004), no. 1, 52–66. </p> http://mathoverflow.net/questions/106864/what-are-the-possible-symplectic-structures-on-a-given-lie-groupoid/106960#106960 Answer by Eugene Lerman for What are the possible symplectic structures on a given Lie groupoid? Eugene Lerman 2012-09-11T21:35:25Z 2012-09-12T13:47:30Z <p>This is more of an extended comment, rather than an answer. Consider the Lie trivial groupoid of the form $M\Rightarrow M$, where $M$ is a closed oriented surface. Then any two symplectic forms on $M$ with the same total integral over $M$ are symplectomorphic (Moser deformation argument). Hence the volume of your area form is a symplectic invariant. So the fiber of your map contains the category is equivalent to the set $\mathbb{R} \smallsetminus {0}$. But there is more: take a covering space $\tilde{M}$ of $M$. The fundamental group $\pi_1 (M)$ acts on $\tilde{M}$ and the action groupoid $\pi_1 (M)\times \tilde{M}\Rightarrow \tilde{M}$ is Morita equivalent to $M$. It is also a symplectic groupoid in your fiber for any choice of a $\pi_1(M)$-invariant symplectic form on $\tilde{M}$... </p> <p>(edit) Unfortunately the form on $\pi_1 (M)\times \tilde{M}\Rightarrow \tilde{M}$ does not look multiplicative, as Daniele Sepe points out. Oops.</p> http://mathoverflow.net/questions/106360/examples-of-manifolds-with-effective-circle-actions/106440#106440 Answer by Eugene Lerman for Examples of manifolds with effective circle actions? Eugene Lerman 2012-09-05T14:44:37Z 2012-09-05T19:29:36Z <p>This is more of a comment than an answer. Any simplectic manifold with an effective circle action preserving the form is an example. So any smooth projective toric variety is an example of a manifold with an effective circle action preserving a volume form.</p> <p>Edit: I forgot to say what the volume is: it's the top power of the symplectic form.</p> http://mathoverflow.net/questions/97391/how-to-deal-with-the-singular-reduction-of-the-hamiltonian-n-body-problem/103797#103797 Answer by Eugene Lerman for How to deal with the singular reduction of the Hamiltonian n body problem? Eugene Lerman 2012-08-02T16:26:16Z 2012-08-29T19:54:51Z <p>My favorite paper on singular reduction is <a href="http://www.math.cornell.edu/~sjamaar/papers/stratified.pdf" rel="nofollow">"Stratified symplectic spaces and reduction"</a>. Admittedly it does not have much by way of examples, but <a href="http://www.math.cornell.edu/~sjamaar/papers/lms.pdf" rel="nofollow">"Examples of singular reduction"</a> does. Section 5 may be of particular interest. You may also want to look at <a href="http://xxx.lanl.gov/abs/dg-ga/9608010" rel="nofollow">this</a> old preprint. </p> http://mathoverflow.net/questions/103784/geometric-invariant-theory-for-geometers/103800#103800 Answer by Eugene Lerman for Geometric invariant theory for geometers Eugene Lerman 2012-08-02T17:19:02Z 2012-08-07T18:34:07Z <p>I would recommend a look at chapter 8 of the <strong>third</strong> edition of <em>Geometric invariant theory</em> by Mumford, Forgarty and Kirwan. It describes a connection between GIT and Hamiltonian group actions in symplectic geometry. </p> <p>(edit) You may also like <a href="http://arxiv.org/abs/0912.1132" rel="nofollow">Moment maps and geometric invariant theory</a> by Chris Woodward.</p> http://mathoverflow.net/questions/103837/is-there-a-categorical-treatment-of-dynamical-systems/103870#103870 Answer by Eugene Lerman for Is there a categorical treatment of dynamical systems? Eugene Lerman 2012-08-03T14:59:55Z 2012-08-03T14:59:55Z <p>An action of a group $T$ on a set $X$ defines the <a href="http://ncatlab.org/nlab/show/action+groupoid%20%22action%20groupoid%22" rel="nofollow">action groupoid</a> $T\times X \rightrightarrows X$ (If $T$ is a semigroup then $T\times X \rightrightarrows X$ is a category). Thinking of dynamical systems this way suggests that morphisms are functors between action groupoids. If there is a topology on $T$ and $X$ and the action is continuous, then the action groupoid is a topological groupoid. You may then take your morphisms to be continuous functors. However, this is not the best one can do. A better notion of morphism between topological groupoid is that of a <a href="http://ncatlab.org/nlab/show/bibundle%20%22bibundle%22" rel="nofollow">bibundle</a> also known as a ‘Hilsum–Skandalis map.’ If you go down this road you end up with bicategories since the composition of bibundles is associative only up to isomorphism (see <a href="http://arxiv.org/abs/0911.2483/" rel="nofollow">this paper</a> by Christopher Schommer-Pries for a nice discussion of the issues).</p> <p>If your dynamical systems are vector fields on manifolds then the morphisms are smooth maps intertwining the vector fields. That is, if $X$ is a vector field on $M$, $Y$ a vector field on $N$ then a morphism from $(M,X)$ to $(N, Y)$ is a smooth map $f:M\to N$ with $Y \circ f = Df \circ X$.</p> <p>There is also a large body of literature on the category of labelled transition systems and on the category of Petrie nets... </p> http://mathoverflow.net/questions/103659/finite-generation-of-g-equivariant-holomorphic-maps-by-polynomials/103805#103805 Answer by Eugene Lerman for finite generation of $G$-equivariant holomorphic maps by polynomials? Eugene Lerman 2012-08-02T18:09:10Z 2012-08-03T14:30:26Z <p>I believe the answer is yes for $C^\infty$ maps and actions of compact (not necessarily finite) Lie groups. I think it is due to Poénaru and can be found in his book <em>Singularités $C^\infty$ en présence de symétrie</em> Lecture Notes in Mathematics, Vol. 510. </p> <p>See also Lemma 6.6.1 in <em>Dynamics and symmetry</em> by Michael Field. (ICP Advanced Texts in Mathematics, 3. Imperial College Press, London, 2007. xiv+478 pp. ISBN: 978-1-86094-828-2)</p> <p>(edit to reply to Brett's comment): Poénaru's theorem is not holomorphic. However, I believe it should not be hard to mimic its proof to extract the holomorphic version. I should note that I am not much of an expert on this area of mathematics and I know it more or less as a collection of black boxes. My impression, however, is that in going from polynomial versions the results (which is classical invariant theory) to $C^\infty$ version the main difficulty is in dealing with smooth invariant functions that vanish to infinite order. Going from polynomials to power series is not hard. And holomorphic maps from $V$ to $W$ are power series, aren't they?</p> <p>Note also that in your example there is a big difference between complex $\mathbb Z/3$ invariant polynomials on $\mathbb C$ and real invariant polynomials on $\mathbb C$: $\mathbb C [\mathbb C]^{\mathbb Z/3}$ is generated by $z^3$ while $\mathbb R[\mathbb C]^{\mathbb Z/3}$ is generated by $Re(z^3), Im (z^3)$ and $|z|^2$.</p> http://mathoverflow.net/questions/96340/symplectic-boundary/103798#103798 Answer by Eugene Lerman for Symplectic boundary Eugene Lerman 2012-08-02T16:33:14Z 2012-08-02T16:33:14Z <p>There is a <a href="http://books.google.com/books?id=4d2faNW3GREC&amp;lpg=PA1&amp;ots=jtyKeWl0ks&amp;lr&amp;pg=PA1#v=onepage&amp;q&amp;f=false" rel="nofollow">book</a> by Guillemin, Gizburg and Karshon: <em>Moment Maps, Cobordisms, and Hamiltonian Group Actions.</em></p> http://mathoverflow.net/questions/103200/is-there-a-difference-between-a-picard-stack-and-a-stack-of-symmetric-monoidal-ca Is there a difference between a Picard stack and a stack of symmetric monoidal categories? Eugene Lerman 2012-07-26T14:29:24Z 2012-07-26T14:29:24Z <p>It seems to me that Picard stacks as defined by Deligne and used in algebraic geometry are stacks of symmetric monoidal categories. Am I missing something? Is there a difference between the two notions? The Picard stack I am interested in shows up in symplectic toric geometry (<a href="http://arxiv.org/abs/0908.2783" rel="nofollow">arXiv:0908.2783v2[math.SG]</a> )</p> http://mathoverflow.net/questions/90455/why-do-i-need-densities-in-order-to-integrate-on-a-non-orientable-manifold/90714#90714 Comment by Eugene Lerman Eugene Lerman 2013-02-12T18:00:50Z 2013-02-12T18:00:50Z Your terminology of &quot;k-densities&quot; nicely clashes with 1/2-densities used in geometric quantization. There are really two indices attached to densities: one is the number of vectors they eat. The other deals with how they transform, that is, a character of GL(n), which can be identified with a nonzero complex number. http://mathoverflow.net/questions/121076/pseudo-differentialforms Comment by Eugene Lerman Eugene Lerman 2013-02-07T15:13:03Z 2013-02-07T15:13:03Z I suspect you are looking for a definition of 1-densities. http://mathoverflow.net/questions/118548/teaching-stacks-to-differential-geometry-students Comment by Eugene Lerman Eugene Lerman 2013-01-10T19:25:32Z 2013-01-10T19:25:32Z @Liviu I had, when it first came out. I don't understand it. http://mathoverflow.net/questions/118548/teaching-stacks-to-differential-geometry-students Comment by Eugene Lerman Eugene Lerman 2013-01-10T19:23:46Z 2013-01-10T19:23:46Z @Misha Thank you for the suggestion. But presenting orbifolds as topological spaces with extra structure kind of defeats the purpose of explaining how to think of them as stacks, doesn't it? http://mathoverflow.net/questions/112689/analytic-lagrangian-submanifolds/112741#112741 Comment by Eugene Lerman Eugene Lerman 2012-11-18T14:05:20Z 2012-11-18T14:05:20Z Sorry, not really. I don't know the book this well. http://mathoverflow.net/questions/38575/colimits-of-manifolds/111018#111018 Comment by Eugene Lerman Eugene Lerman 2012-11-15T01:47:30Z 2012-11-15T01:47:30Z Yes. It's a proper etale Lie groupoid whose objects have no nontrivial automorphisms. In other words for for each object $x$ of this groupoid the group $Hom (x, x)$ of arrows from $x$ to itself is trivial. http://mathoverflow.net/questions/110229/is-there-a-notion-of-a-chain-complex-with-corners Comment by Eugene Lerman Eugene Lerman 2012-10-22T14:09:22Z 2012-10-22T14:09:22Z Kreck's stratifiolds <a href="http://www.him.uni-bonn.de/homepages/prof-dr-matthias-kreck/the-stratifold-page/" rel="nofollow">him.uni-bonn.de/homepages/prof-dr-matthias-kreck/&hellip;</a> may be close in spirit to what you are asking for. http://mathoverflow.net/questions/108983/unique-symplectic-form-in-an-adapted-complex-structure Comment by Eugene Lerman Eugene Lerman 2012-10-08T18:34:40Z 2012-10-08T18:34:40Z For lagrangian tori in Kaehler toric manfolds yes and yes. See Abreu's paper I pointed to you yesterday: <a href="http://arxiv.org/abs/math/0004122" rel="nofollow">arxiv.org/abs/math/0004122</a> Another good place is Burns and Guillemin, Potential functions and actions of tori on Kaehler manifolds (<a href="http://arxiv.org/abs/math/0302350" rel="nofollow">arxiv.org/abs/math/0302350</a>). http://mathoverflow.net/questions/108983/unique-symplectic-form-in-an-adapted-complex-structure Comment by Eugene Lerman Eugene Lerman 2012-10-08T11:33:14Z 2012-10-08T11:33:14Z Well, have you looked at the reference I gave you? There the real torus is Lagrangian. More concretely, take $M=S^1$. Then $T^*M$ is $\mathbb{C} -\{0\}$. There are many different Kaehler structures on $T^*M$. You can embed it in $\mathbb{C}$ with a flat metric. Or you can embed it in $\mathbb{C}P^1$ with the Fubini-Study metric. Basically toric manifolds provide plenty of examples. http://mathoverflow.net/questions/108983/unique-symplectic-form-in-an-adapted-complex-structure Comment by Eugene Lerman Eugene Lerman 2012-10-07T11:47:42Z 2012-10-07T11:47:42Z If you start with a <i>Riemannian</i> manifold $(M, g_0)$ then $T^*M = TM$ acquires a natural Riemannian metric $g$ from $g_0$; this is what I thought you meant. Since this is not the case, are you asking: &quot;I have a complex manifold $(N, J)$. What are all symplectic forms compatible with $J$.&quot; ? If this is the case, I believe there are lots of these forms. See <a href="http://arxiv.org/abs/math/0004122" rel="nofollow">arxiv.org/abs/math/0004122</a> which applies to your question when $M$ is a torus (so $T^*M = (C^\times)^n$). http://mathoverflow.net/questions/108983/unique-symplectic-form-in-an-adapted-complex-structure Comment by Eugene Lerman Eugene Lerman 2012-10-06T21:41:51Z 2012-10-06T21:41:51Z On a Kaehler manifold, two out of three structures --- $J$, $g$, $\omega$ --- determine the third. Since you are fixing $g$ and $J$, there doesn't seem to be any choice for $\omega$. http://mathoverflow.net/questions/108440/abelian-principal-bundles Comment by Eugene Lerman Eugene Lerman 2012-09-30T12:16:42Z 2012-09-30T12:16:42Z Konrad's last comment is the answer I would give to the question: principal $G$ bundles for $G$ abelian have a &quot;multiplication&quot;, and so form a Picard stack http://mathoverflow.net/questions/108228/is-there-a-natural-distance-between-skew-hermitian-matrices/108230#108230 Comment by Eugene Lerman Eugene Lerman 2012-09-27T13:23:43Z 2012-09-27T13:23:43Z Up to a scaler the inner product is the Killing form. I don't know that the corresponding norm is called. http://mathoverflow.net/questions/107897/reference-request-for-essential-equivalence-of-top-groupoids/107899#107899 Comment by Eugene Lerman Eugene Lerman 2012-09-23T15:07:14Z 2012-09-23T15:07:14Z (continued) A good place to look for info on topological stacks are papers of Behrang Noohi (<a href="http://www.maths.qmul.ac.uk/~noohi/" rel="nofollow">maths.qmul.ac.uk/~noohi</a>) http://mathoverflow.net/questions/107897/reference-request-for-essential-equivalence-of-top-groupoids/107899#107899 Comment by Eugene Lerman Eugene Lerman 2012-09-23T15:05:24Z 2012-09-23T15:05:24Z You should take Konrad's answer with a small grain of salt. Lemma 3.34 can be said to prove that an essential equivalence between Lie groupoids H and G defines an isomorphism of corresponding stacks. I presume the same proof works for topological groupoids. Then the result you want follows from a fact, the reference to which I don't know, that isomorphic stacks have homotopy equivalent classifying spaces (assuming that by $BG$ you mean the classifying space of a stack rather than the stack itself; the notation is ambiguous).