User janos erdmann - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:34:52Z http://mathoverflow.net/feeds/user/12653 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/126771/quantum-coordinate-algebras-at-roots-of-unity-and-non-standard-irrep-types Quantum Coordinate Algebras at Roots of Unity and Non-Standard Irrep Types Janos Erdmann 2013-04-07T12:57:15Z 2013-04-11T15:01:46Z <p>Let $\frak{g}$ be a complex semi-simple Lie algebra of rank $n$, and $U_q(\frak{g})$ the corresponding Drinfeld-Jimbo algebra. As is well-known, for $q$ not a root of unity, the irreducible finite-dimensional representations of $U_q(\frak{g})$ are divided into $2^n$ types, corresponding to elements of the $n^{\text{th}}$-Cartesian power of the set {$+,-$}. The classical representations are recovered as those corresponding to $(1,1, \dots, 1)$, and are called <strong>Type 1</strong>. One defines the coordinate algebra ${\cal O}_q[G]$ to be the direct sum of the coordinate functions of the Type 1 irreducible representation. When $q = 1$, the algebraic Peter-Weyl tells us that we recover the coordinate algebra of the connected compact semi-simple Lie group $G$ corresponding to $\frak{g}$. </p> <p>With the background recalled, let me now enquire:</p> <p>(i) What happens when one takes the direct sum of the coordinate algebras all the irreducible representations, ie irreps of all types? One should recover the Hopf dual of $U_q(\frak{g})$. Thus, I would guess that this means that ${\cal O}_q[G]$ is strictly smaller that the Hopf dual?</p> <p>(ii) What happens when $q$ is a root of unity? The irrep theory becomes a lot more complicated, so it is not clear that one can still define ${\cal O}_q[G]$ as one does for roots of unity. However, the construction of ${\cal O}_q[G]$ via $R$-matrices still makes sense (if I have understood correctly). So can one still define ${\cal O}_q[G]$ as some subalgebra of the Hopf dual? </p> http://mathoverflow.net/questions/126706/hopf-duals-and-matrix-coefficients Hopf Duals and Matrix Coefficients Janos Erdmann 2013-04-06T14:20:25Z 2013-04-06T17:16:48Z <p>One defines the <strong>finite dual</strong> of a Hopf algebra $A$ as $$H^o := {f \in H^* ~|~ f(I) = 0, \text{ for some ideal I of H with } \dim_C(H/I) &lt; \infty }.$$ As is well-known, $H^o$ has a well-defined Hopf algebra structure obtained by dualizing the Hopf structure of $H$. </p> <p>On the other hand, for any finite-dimensional $H$-module $V$, and element $v \in V$, and a functional $f \in V^*$, we can define a functional $c_{f,v} \in H^*$ according to $$c_{f,v}(h) := f(hv).$$ One usually calls any such functional a <strong>matrix coefficient</strong> of $H$. It is not difficult to see that the set of matrix coefficients of $H$ forms a Hopf subalgebra of $H^o$, which we will denote by Mat$(H)$.</p> <p>What I would like to know is when do we have the equality $$H^o = \text{Mat}(H)?$$</p> http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles From Topological to Smooth and Holomorphic Vector Bundles Janos Erdmann 2013-02-07T11:31:56Z 2013-02-10T23:11:51Z <p>In the last weeks I have been think of the transition from topological vector bundles to smooth and holomorphic vector bundles. This has resulted in a few questions (with a common thread) as follows: Always $\pi:E \to B$ is a topological (complex) vector bundle over a compact base,</p> <p>(A) For any given smooth manifold structure on $B$, can there exist more than one differential structure on $E$ giving $\pi:E \to B$ the structure of a smooth vector bundle. If so, what is an example?</p> <p>(B) Same question as above but replacing smooth by holomorphic.</p> <p>(C) For a choice of smooth vector bundle structure on $\pi:E \to B$, does the de Rham complex of $E$ have an easy relationship with the de Rham complex of $B$. A (very) naive guess would be that $$\Omega^{\bullet}(E) = \Gamma^{\infty}(E) \otimes_{C^\infty(B)}\Omega^{\bullet}(B),$$ but I can't see that there is a well-defined way to define the differential.</p> <p>(D) Same question as above but for holomorphic structures and the Dolbeault complex</p> http://mathoverflow.net/questions/95371/de-rham-vs-dolbeault-cohomology de Rham vs Dolbeault Cohomology Janos Erdmann 2012-04-27T16:14:03Z 2012-04-28T12:08:48Z <p>For a complex manifold $M$, one can consider (A) its de Rham cohomology, or (B) its Dolbeault cohomology. I'm looking for some motivation as to why one would bother introducing Dolbeault cohomology. To be more specific, here are some straight questions.</p> <ol> <li>What can Dolbeault tell us that de Rham can't?</li> <li>Does there exist some simple relationship between these two cohomologies? </li> <li>When are they equal? </li> <li>Do things become simpler for the Kahler case?</li> <li>What happens for the projective spaces?</li> <li>Why does nobody talk about the holomorphic cohomology?</li> </ol> http://mathoverflow.net/questions/57025/down-to-earth-uses-of-de-rham-cohomology-to-convince-a-wide-audience-of-its-usefu Down-To-Earth Uses of de Rham Cohomology to Convince a Wide Audience of its Usefulness Janos Erdmann 2011-03-01T19:30:52Z 2012-01-03T15:21:47Z <p>I'm soon giving an introductory talk on de Rham cohomology to a wide postgraduate audience. I'm hoping to get to arrive at the idea of de Rham cohomology for a smooth manifold, building up from vector fields and one-forms on Euclidean space. However, once I've got there I'm not too sure how to convince everyone that it was worth the journey. What down-to-earth uses could one cite to prove the worth of the construct?</p> http://mathoverflow.net/questions/71586/haar-functionals-and-coquasi-triangular-structures Haar Functionals and Coquasi-triangular Structures Janos Erdmann 2011-07-29T16:13:25Z 2011-07-29T16:13:25Z <p>In this <a href="http://mathoverflow.net/questions/67537/compact-quantum-groups-from-hopf-algebras" rel="nofollow">question</a> it is mentioned that the coordinate algebra $C_q[G]$ Drinfeld--Jimbo algebras, for $G$ a compact semi-simple Lie group, admit a unique positive definite Haar functional. I was wondering if this can be shown to have anything to do with their coquasi-triangular structure?</p> http://mathoverflow.net/questions/53932/generators-of-the-odd-dimensional-quantum-spheres Generators of the Odd Dimensional Quantum Spheres Janos Erdmann 2011-01-31T23:30:51Z 2011-06-27T14:51:15Z <p>As is well-known, the $(2N-1)$-quantum sphere $S^{2N-1}_q$ is defined to be the invariant subalgebra of $SU_q(N)$ under the coaction $\Delta_R = (id \otimes \pi) \circ \Delta$, where $\Delta$ is the comultiplication of $SU_q(N)$, and $\pi: SU_q(N) \to U_q(N-1)$ is the Hopf algebra surjection defined by setting, for $i,j \neq 1$, $\pi(u^i_1)=\pi(u^1_j)=0$, $\pi(u^1_1)$ = det$_q^{-1}$, and $\pi (u^i_j) = u^{i-1}_{j-1}$. (Recall that the invariant subalgebra of a $H$-coaction $\Delta_R$ on vector space $V$ is the subspace of all elements $v$ for which $\Delta_R(v) = v \otimes 1$.) An oft quoted result is that $S^{2N-1}_q$ is generated, as an algebra, by the elements $u^i, S(u^1_i)$, for $i=1, \ldots N$. Now it is clear that these elements are invariant, but it is far from clear (at least to me) that generate all the invariant subspace. Can anyone see why? The usual references given are in Russian and, even at that, are unavailable on the web.</p> http://mathoverflow.net/questions/59073/twisting-spinor-bundles-with-line-bundles Twisting Spinor Bundles with Line Bundles Janos Erdmann 2011-03-21T15:50:11Z 2011-03-23T08:02:30Z <p>In a paper I am reading, the following framework was given: Let $S$ be a spinor bundle, over a Riemannian manifold $M$, with Clifford action $$c:S \otimes \Omega^1(M) \to S.$$ Moreover, let $E$ be a line bundle over $M$ with connection $\nabla$.</p> <p>The author then speaks of the <em>canonical</em> Dirac operator $D$ on $S \otimes E$. What does he mean by this? My guess is as follows: Let $s \in S$ and $e \in E$, such that $\nabla(e) = \sum_i e_i \otimes \omega_i$, for $\omega_i \in \Omega^1(M)$. Moreover, let $D_S$ be the Dirac operator on $S$. I would define $D$ by $$D(s \otimes e) = D_S(s) \otimes e + \sum_i c(s \otimes \omega_i) \otimes e_i.$$ Is this correct? If so, how does one define the Clifford action for $S \otimes E$. Finally, does this work for a twisting by any vector bundle?</p> http://mathoverflow.net/questions/57273/ore-extensions-and-the-construction-of-the-quantum-general-linear-group Ore Extensions and the Construction of the Quantum General Linear Group Janos Erdmann 2011-03-03T18:21:30Z 2011-03-03T19:17:49Z <p>In the usual (fomal) construction of the quantum general linear group $GL_q(N)$, an Ore extension is used. See for example Kassel. Why is this necessary? Surely one can just augment the set of generators with an element det$^{-1}$ and the set of relations with the relations det.det$^{-1} = 1$ and det$^{-1}$.det$=1$ and get the same thing without going to all the trouble of considering Ore extensions?</p> http://mathoverflow.net/questions/56679/dual-of-a-basis-for-a-hopf-algebra-conatined-in-all-dually-paired-hopf-algebras Dual of a Basis for a Hopf Algebra Conatined in all Dually Paired Hopf Algebras Janos Erdmann 2011-02-25T20:45:02Z 2011-02-26T00:49:01Z <p>For an infinite dimensional Hopf algebra $H$, a non-degenerate dually pairing Hopf algebra $H'$, and a choice of basis $e_i$ of $H$, is the dual basis $e^i$ (defined of course by $e^i(e_j) = \delta_{ij}$) contained in $H'$?</p> <p>I am interested in the specific case of $SU_q(N)$ and the dually paired Hopf algebra $\mathfrak{sl}_N$.</p> http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties Almost Complex Integrability and Algebraic Varieties Janos Erdmann 2011-02-21T15:22:46Z 2011-02-21T19:12:35Z <p>Let $J$ be an almost complex structure on an algebraic variety $V$. As we all know, $J$ comes from a complex structure if the Nijenhuis tensor of $J$ vanishes. What I would like to know is if there exists a simpler characterisation of integrability than this for varieties (as opposed to general manifolds). </p> http://mathoverflow.net/questions/126706/hopf-duals-and-matrix-coefficients Comment by Janos Erdmann Janos Erdmann 2013-04-06T16:38:33Z 2013-04-06T16:38:33Z Danke! Put your remark as an answer and I can accept it. http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121063#121063 Comment by Janos Erdmann Janos Erdmann 2013-02-23T16:08:51Z 2013-02-23T16:08:51Z I mean the de Rham cohomology of $E$ as a Riemannian manifold. I've never seen anyone discuss this, and I was wondering why it's not interesting. http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121063#121063 Comment by Janos Erdmann Janos Erdmann 2013-02-07T13:03:52Z 2013-02-07T13:03:52Z ..... or at least that there is no well-known relationship. http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121063#121063 Comment by Janos Erdmann Janos Erdmann 2013-02-07T12:54:38Z 2013-02-07T12:54:38Z Thank you a lot for your answer. Then for (C), does that mean that in general there is no relationship between the two de Rham complexes? http://mathoverflow.net/questions/59073/twisting-spinor-bundles-with-line-bundles Comment by Janos Erdmann Janos Erdmann 2011-03-21T16:25:38Z 2011-03-21T16:25:38Z So if I've understood correctly, what you've said gives the Dirac operator I proposed. No? http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties Comment by Janos Erdmann Janos Erdmann 2011-03-21T15:31:21Z 2011-03-21T15:31:21Z Great, thanks alot! http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties Comment by Janos Erdmann Janos Erdmann 2011-02-25T18:43:34Z 2011-02-25T18:43:34Z ... and how exactly does $\nabla$ act on $J$? http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties Comment by Janos Erdmann Janos Erdmann 2011-02-25T18:42:21Z 2011-02-25T18:42:21Z @ Johannes: So you mean, given a complex structure $J$ and a metric $g$ compatible with $J$, we have a Kahler manifold if, for the Levi-Civita connection $\nabla$ of $g$, we have $\nabla(J) = 0$, then the manifold is Kahler? http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties/56199#56199 Comment by Janos Erdmann Janos Erdmann 2011-02-21T19:09:01Z 2011-02-21T19:09:01Z Great, thanks a lot! http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties Comment by Janos Erdmann Janos Erdmann 2011-02-21T18:56:44Z 2011-02-21T18:56:44Z .... and do you have a reference for a proof that each of these conditions implies a complex structure? http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties Comment by Janos Erdmann Janos Erdmann 2011-02-21T18:29:33Z 2011-02-21T18:29:33Z P.S. Put your comment as an answer do I can accept it. http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties Comment by Janos Erdmann Janos Erdmann 2011-02-21T18:28:41Z 2011-02-21T18:28:41Z @Gunnar: No I'm not so worried about whether it's projective or not. You're first answer seems to be what I'm looking for. Just one question: How do you define $\partial$ and $\overline{\partial}$ from $J$ – Janos Erdmann 0 secs ago http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties Comment by Janos Erdmann Janos Erdmann 2011-02-21T17:46:18Z 2011-02-21T17:46:18Z @ Zsolt: By smooth do you mean non-singular? @Scott: By algebraic variety I mean algebraic variety in the simplest sense - no assumptions. To paraphrase my question, I am asking when you can have an almost complex structure on a variety without it being a complex manifold.