User john mccarthy - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:59:50Z http://mathoverflow.net/feeds/user/1095 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120528/determing-hodges-maps-by-their-essential-algebraic-properties Determing Hodges Maps by their Essential Algebraic Properties John McCarthy 2013-02-01T16:02:36Z 2013-03-18T10:22:00Z <p>Let $M$ be a complex manifold of real dimension $2N$, equipped with a Hermitian metric. The corresponding Hodge map $\ast$ has the following properties:</p> <p>(i) It is a ${\bf C}$-linear map $\ast:\Omega^k(M) \to \Omega^{2N-k}(M)$;</p> <p>(ii) $\ast(\Omega^{(p,q)}(M)) = \Omega^{(N-p,N-q)}$(M);</p> <p>(iii) $\ast^2 = (-1)^{k}$ on $\Omega^k(M)$.</p> <p>Now I would guess that there exist other maps on $\Omega(M)$ with these properties which do not arise as Hodge maps from some Hermitian metric. So my question is, do there exist extra (algebraic) properties of $\ast$, which when put together with $(i),(ii)$, and $(iii)$, determine all the Hodge maps, but without ever explicitly mentioning metrics.</p> http://mathoverflow.net/questions/123427/coherent-sheaves-and-holomorphic-vector-bundles Coherent Sheaves and Holomorphic Vector Bundles John McCarthy 2013-03-02T17:42:33Z 2013-03-04T14:56:09Z <p>For a complex manifold $M$, I'm trying to understand (from a differential geometry point of view) what its <em>category of coherent sheaves</em> is. If I understand correctly, then the sheaf of holomorphic sections of a complex vector bundle is a coherent sheaf for $M$. Is there some case when the category of coherent sheaves is equivalent to the category of holomorphic vector bundles? In other words can I just forget about the algebraic geometry and understand the definition in terms of complex geometry? Or alternatively, can one somehow <em>"generate"</em> the category of coherent sheaves from the category of holomorphic vector bundles</p> http://mathoverflow.net/questions/120033/why-are-the-holomorphic-line-bundle-sections-finite-dimensional Why are the holomorphic line bundle sections finite dimensional? John McCarthy 2013-01-27T16:15:59Z 2013-02-03T15:50:03Z <p>I'm trying to understand the <a href="http://en.wikipedia.org/wiki/Borel%25E2%2580%2593Weil_theorem" rel="nofollow">Borel--Weil theorem</a> at the moment (not the whole Bott--Borel--Weil theorem as has been asked elsewhere). However, I am having a little difficulty finding a direct proof, so I started to try and reconstruct it for myself. The question I can't seem to find an answer for is why should the space of holomorphic sections should be a finite dimensional space in the first place? Is there any easy way to see why this should be the case? It seems to me that if one can answer this question, then the theorem follows directly from the classification of the finite dimensional reps of $G$.</p> <p>As I mention in a comment below I am more (but not exclusively!) interested in algebraic ways of proving finite dimensionality.</p> http://mathoverflow.net/questions/67537/compact-quantum-groups-from-hopf-algebras Compact Quantum Groups from Hopf Algebras John McCarthy 2011-06-11T20:14:31Z 2013-01-03T15:53:10Z <p>For a compact quantum group $C_q[G]$, it was shown by Woronowicz that $C_q[G]$ contains a dense Hopf algebra generalising the algebra of representations of $G$. I am interested in the other way around, ie given a Hopf algebra $H$ (say a Drinfeld--Jimbo algebra if it makes things easier) can it always be completed to give a compact quantum group? If so, is this completion unique in anyway, or are there many ways to get a cqg from a Hopf algebra?</p> http://mathoverflow.net/questions/70670/yetter-drinfeld-modules-and-braidings Yetter--Drinfeld Modules and Braidings John McCarthy 2011-07-18T20:47:10Z 2011-12-21T02:22:12Z <p>Let $H$ be a Hopf algebra, $M$ a right $H$-comodule for a coaction $\Delta_R$, and $\triangleleft$ a right $H$-action on $M$ such that $M$ is a Yetter--Drindfeld module. We know from general theory that we have a corresponding braiding $$\sigma: M \to M, ~~~~~~~~~~ m \otimes n \mapsto n_{(0)} \otimes (m \triangleleft n_{(1)}),$$ where $n_{(0)} \otimes n_{(1)}$ is the image of $n$ under $\Delta_R$. </p> <p>Questions: (1) Can there exist another right action giving $M$ (still endowed with $\Delta_R$) the structure of a Yetter--Drinfeld module and producing the same braiding?</p> <p>(2) Can there exist another right coaction giving $M$ (still endowed with $\triangleleft$) the structure of a Yetter--Drinfeld module and producing the same braiding?</p> <p>(3) Do the answers to these questions change if $M$ is assumed to be finite dimensional?</p> http://mathoverflow.net/questions/10581/kontsevich-and-geometric-quantization-and-the-podles-sphere Kontsevich, and Geometric, Quantization and the Podles sphere John McCarthy 2010-01-03T13:01:38Z 2011-10-21T10:59:39Z <p>There exist a large family of noncommutative spaces that arise from the quantum matrices. These algebraic objects $q$-deform the coordinate rings of certain varieties. For example, take quantum $SU(2)$, this is the algebra $&lt; a,b,c,d >$ quotiented by the ideal generated by $$ab−qba, ~~ ac−qca, ~~ bc−cb, ~~ bd−qdb, ~~ cd−qdc, ~~ ad−da−(q−q^{−1})bc,<br />$$ and the "q-det" relation $$ad−qbc−1$$ where $q$ is some complex number. Clearly, when $q=1$ we get back the coordinate ring of $SU(2)$. In the classical case $S^2 = SU(2)/U(1)$ (the famous Hopf fibration). This generalises to the q-case: the $U(1)$-action generalises to a $U(1)$-coaction with an invariant subalgebra that q-deforms the coordinate algebra of $S^2$ - the famous Podles sphere. There exist such q-matrix deformations of all flag manifolds. </p> <p>Since all such manifolds are Kahler, we can also apply Kontsevich deformation to them to obtain a q-defomation. My question is: What is the relationship between these two approaches?</p> <p>Alternatively, we can apply Kostant-Souriau geometric quantization to a flag manifold. How does alegbra relate to its q-matrix deformation? </p> http://mathoverflow.net/questions/78036/maximal-ideals-and-kahler-differentials Maximal Ideals and Kahler Differentials John McCarthy 2011-10-13T17:20:01Z 2011-10-14T22:46:22Z <p>For an algebraic variety $V$, denote its ring of regular functions by ${\cal O}(V)$. The <em>Kahler differentials</em> of $V$ are the quotient of the kernel $M$ of the multiplication map $$m: {\cal O}(V) \otimes {\cal O}(V)\to {\cal O}(V)$$ by the ideal $M^2$. </p> <p>What can be said about the maximal proper submodules of $M$?</p> <p>Is there any sense/specific-case in which the submodule $M^2$ is maximal?</p> <p>I am particularly interested in the homogeneous variety case, specifically the flag variety case. For example, is $M^2$ a maximal right $G$-invariant proper submodule when $V$ a a $G$-homogeneous variety.</p> http://mathoverflow.net/questions/73587/classification-of-the-k-ahler-structures-on-the-sphere Classification of the K\"ahler Structures on the Sphere John McCarthy 2011-08-24T16:59:08Z 2011-08-24T16:59:08Z <p>Is there a classification of the K\"ahler structure on the sphere? More generally, is there a classification of the K\"ahler structures on the complex projective spaces? Even more generally, what about the flag manifolds?</p> http://mathoverflow.net/questions/64204/harmonic-forms-for-complex-projective-space Harmonic forms for Complex Projective Space. John McCarthy 2011-05-07T15:13:53Z 2011-05-08T12:48:39Z <p>For complex projective space with the Fubini-Study metric and associated Laplace-de Rham operator $dd^\ast+d^\ast d$. How does one find a concrete description of the space of harmonic forms? That is, how does one find a basis of the space of forms $\omega$ for which $(dd^\ast+d^\ast d)(\omega)=0$?</p> http://mathoverflow.net/questions/63410/uses-of-the-chern-connes-pairing Uses of the Chern--Connes Pairing? John McCarthy 2011-04-29T13:11:38Z 2011-04-29T14:01:59Z <p>The backbone of Connes' approach to noncommutative geometry is the Chern--Connes pairing. By discovering the cyclic homology of an algebra and then pairing it the $K$-theory of that algebra, Connes showed how to find numerical invariants of $K$-theory classes. This is analaogous to the way in which the Chern character map gives numerical invariants of vector bundles. I would like to know what has the Chern--Connes pairing been used for? Besides its intrinsic interest as a noncommutative version of a classical phenomenon, why is it important?</p> http://mathoverflow.net/questions/59588/quantum-mathematics/59594#59594 Answer by John McCarthy for Quantum mathematics? John McCarthy 2011-03-25T18:47:52Z 2011-03-25T18:47:52Z <p>I would hold that the term non-commutative algebra is usually used to refer to the study of general noncommutative algebras. Quantum algebra involves the study of certain types of non-commutative algebras, not all non-commutative algebras. It's not black and white, but reasonably well-defined subfamily. The algebras quite often involve a parameter $q$ st when $q=1$ or $0$ the algebra is commutative - take for example Drinfeld--Jimbo algebras. The parallels with quantum theory here are obvious.</p> http://mathoverflow.net/questions/59211/diracs-original-operator-and-the-hodge-dirac-operator Dirac's Original Operator and the Hodge--Dirac Operator John McCarthy 2011-03-22T17:55:56Z 2011-03-23T08:23:40Z <p>For the usual $4$-dimensional Minkowski space $M$, the standard Dirac operator is given by $$D: C^{\infty}(M) \to C^{\infty}(M), ~~~~~ f \mapsto \sum_{i=1}^4 \gamma_i\frac{\partial f}{\partial x_i},$$ where the $\gamma_i$ are the usual <a href="http://en.wikipedia.org/wiki/Gamma_matrices" rel="nofollow">gamma matrises</a>. As we all know, $D$ was originally constructed as a square root of the Lapalcian $\sum_{i=1}^4 \frac{\partial^2 }{\partial x_i^2}$. Indeed, routine calculation will verify that $D$ does indeed square to give the Laplacian.</p> <p>Moreover, there exists another square root of the Laplacian, namely $$d + \ast d \ast: C^{\infty}(M) \to \Omega^1(M),$$ where $d$ is the usual exterior derivative, and $\ast$ is the Hodge $\ast$-operator. </p> <p>How do these two square roots of the Laplacian relate to each other? Are they two separate objects, or just two ways of looking at the same thing?</p> http://mathoverflow.net/questions/57714/history-of-the-odd-dimensional-quantum-spheres History of the Odd Dimensional Quantum Spheres John McCarthy 2011-03-07T18:42:54Z 2011-03-10T18:32:35Z <p>After reading this <a href="http://mathoverflow.net/questions/53932/generators-of-the-odd-dimensional-quantum-spheres" rel="nofollow">question</a>, I began to wonder about the history of quantum $(2N-1)$-spheres. Basically I have two questions:</p> <p>(1) Who first introduced the $(2N-1)$-spheres, and who first introduced them as invariant subalgebras of quantum $SU_N$? I know that Podles is often credited with introducing quantum spheres, but as far as I can see (his paper is difficult to find on the web) he only introduced a family of deformations of $2$-spheres, which are not of course of the form $S^{2N-1}_q$. I know also that something appeared in the early FRT-papers, but these are also difficult to find, and are in Russian.</p> <p>(2) The proof in Klimyk and Schmudgen of the generators and relations result is based on a 1993 paper by Noumi, Yamada, and Mimachi. (This is also difficult to find.) Was this the first generators and relations description for the for $S^{2N-1}_q$? Did it generalise an earlier result, ie for $S^5_q$?</p> <p>P.S. I would also be very interested in hearing about the history of quantum projective spaces. Who first introduced them? Who first introduced them as invariant subalgebras of quantum $SU_N$? Who first introduced them as invariant subalgebras of the quantum spheres? </p> http://mathoverflow.net/questions/56293/global-definition-of-the-almost-complex-structure-of-a-complex-manifold Global Definition of the Almost Complex Structure of a Complex Manifold John McCarthy 2011-02-22T15:58:01Z 2011-02-23T23:56:22Z <p>Motivated by this <a href="http://mathoverflow.net/questions/56182/almost-complex-integrability-and-algebraic-varieties" rel="nofollow">question</a>, I began to wonder if there is a global definition of the almost complex structure of a complex manifold. It is (almost) always presented as multiplication by complex $i$ on the tangent space, and then globalized. Using the formulae given earlier $$\overline{\partial}\omega = \frac{1}{2}(\text{d}\omega + i \text{d}(J\omega)),$$ and $$\partial \omega = \frac{1}{2}(\text{d}\omega - i \text{d}(J\omega)),$$ it is easy to see that $$-\frac{i}{2}d\omega = d(J\omega).$$ Thus, $J\omega = -\frac{i}{2}\omega + \omega'$, where $\omega'$ is some closed form. What this $\omega'$ is, however, I cannot see.</p> http://mathoverflow.net/questions/55037/restriction-of-the-levi-civita-connection-to-a-connection-on-the-anti-holomorph Restriction of the Levi-Civita Connection to a Connection on the (Anti-)Holomorphic Forms John McCarthy 2011-02-10T14:49:43Z 2011-02-10T15:07:07Z <p>For a Riemannian manifold $(M,g)$, that is also a complex manifold, when does the Levi-Civita $\nabla_g$ connection restrict to a connection on the holomorphic forms $\Omega^{(\cdot,0)}$, and when does it restrict to a connection on the anti-holomorphic forms $\Omega^{(0,\cdot)}$?</p> <p>I would assume there are some sufficient and neccessary conditions for the interaction of the metric $g$ and the complex structure.</p> http://mathoverflow.net/questions/54781/dual-riemannian-metric-and-the-dual-metric-form Dual Riemannian metric and the Dual Metric Form John McCarthy 2011-02-08T16:00:36Z 2011-02-09T15:47:33Z <p>Let $M$ be a Rieamnnian manifold with metric $g: X(M) \times X(M) \to C^{\infty}(X)$, where $X(M)$ are the vector fields of $X$.</p> <p>As is well known, we can induce a bilinear pairing $$\langle \cdot , \cdot \rangle_g: \Omega^1(M) \times \Omega^1(M) \to C^{\infty}(M)$$ by setting $$\langle \omega, \omega' \rangle_g = g(\omega^{\sharp}, (\omega')^{\sharp}),$$ where, as usual, $\sharp$ is defined by $g(\omega^\sharp, X) = \omega(X)$, for $X \in X(M)$.</p> <p>On the other hand, as is also well known, there exists a unique element $\omega_g \in \Omega(M) \otimes \Omega(M)$ such that, for $(X,Y) \in X(M) \times X(M)$, $$\omega_g (X,Y) = g(X,Y),$$ where $\omega_g$ is applied to $(X,Y)$ in the obvious way.</p> <p>Thus, we have a pairing on $T^\ast(M)$ and an element of $\Omega(M) \times \Omega(M)$ both coming from $g$. I would like to know if there exists a simple relationship between these two objects. (By simple, I suppose I mean something global and algebraic, free from messy local expressions.)</p> <p>Moreover, what does metric compatibility for a connection look like for either of these?</p> http://mathoverflow.net/questions/50867/embedding-quantum-sl2-into-the-quantum-matrices Embedding Quantum SL(2) into the Quantum Matrices John McCarthy 2011-01-01T17:12:19Z 2011-01-11T19:11:47Z <p>Let $M_q[2]$ be the algebra of quantum matrices over the complex numbers with the usual generators $a,b,c,d$ and the relations $ab = qba$, ... etc. Moreover, let $SL_q(2)$ be the quotient of $M_q(2)$ by the ideal generated by det$_q-1$, where det$_q = ad - qbc$. Given a basis of $SL_q(2)$ it is easy to construct an embedding of $SL_q(2)$ into $M_q(2)$. What I would like to know is: Can anyone see a canonical way of embedding $SL_q(2)$ into $M_q(2)$?</p> http://mathoverflow.net/questions/46503/lie-group-principal-embedding Lie Group Principal Embedding John McCarthy 2010-11-18T16:43:46Z 2010-11-19T00:35:26Z <p>I'm reading a paper on complex semi-simple algebraic group geometry at the moment, but finding the going a bit tough since I'm missing alot of the prerequisites. Firstly, the author refers to a <em>principal embedding</em> of one Lie group into another. I guess that this is an embedding of one group into another that is maximal in some sense. The paper states that in the dual Lie algebra setting this means that ${\frak g}$ is obtained from ${\frak g_0}$ by adding a node to the Dynkin diagram. The family of examples used is that of the embedding of $U_{N-1}$ into $SU_{N}$. I've searched the web but can't find a principal embedding defn, could someone point me in the right direction.</p> <p>Secondly, if $G_0$ is principal embedded into $G$, is there some result about the quotient $G/G_0$ having a complex structure? This works for example $CP^{N-1} = SU_N/U_{N-1}$.</p> http://mathoverflow.net/questions/45854/explicit-coquasi-triangular-quantised-coordinate-algebra-of-a-complex-semi-simple Explicit Coquasi-Triangular Quantised Coordinate Algebra of a Complex Semi-Simple Lie Group? John McCarthy 2010-11-12T18:28:02Z 2010-11-13T00:38:22Z <p>Let $SL_q(N)$ be usual quantised coordinate algebra of the special linear group. As is well-known, this is co-quasi-triangular algebra with coquasi-triangular structure given by $$R(u^i_j \otimes u^k_l) = q^{-\frac{1}{2}}.(q^{\delta_{ij}}\delta_{im}\delta_{jn} + (q-q^{-1})\theta (i-j)\delta_{in}\delta_{jm}),$$</p> <p>Now consider the much more general case of $G_q$ the quantised coordinate algebra of a complex semi-simple Lie group. This is defined dually in terms of a Drinfield-Jimbo quantised enveloping Lie algebra $\mathfrak{g}$. As is also well-known, these algebras are also co-quasi-triangular. My question is does there exist a general formula for the co-quasi-triangular structure $R(u^i_j \otimes u^k_l)$ in terms of the Cartan data of $\mathfrak{g}$?</p> http://mathoverflow.net/questions/43331/quantum-group-calculations-in-mathematica Quantum Group Calculations in Mathematica John McCarthy 2010-10-23T19:08:01Z 2010-10-25T20:46:27Z <p>I'm trying to learn how to do algebraic manipulations in Mathematica but not finding the help very helpful. I'm going to ask about a specific quantum group example related to a previous question of mine. For $SU_q(N)$, how would I use Mathematica to show that<br> $$S(u^1_2)u^3_1 = q^{-1}u^3_1S(u^1_2).$$ I am, of course, assuming that such a thing can be done in Mathematica. If I am wrong in this assumption, could someone please direct me a package that can do this calculation? Gap, Magma? </p> http://mathoverflow.net/questions/41628/the-relation-su1-iuj-1-q-1uj-1su1-i The relation $S(u^1_i)u^j_1 = q^{-1}u^j_1S(u^1_i)$ John McCarthy 2010-10-09T20:38:11Z 2010-10-21T18:50:55Z <p>In the Hopf algebra $SL_q(N)$, it can be shown, using direct calculations, that $S(u^1_i)u^j_1 = q^{-1}u^j_1S(u^1_i)$. Can anyone see a more elegant way of establishing this?</p> <p>Moreover, does anyone know of a similar relation in the more general case of $S(u^1_r)u^i_j$?</p> <p>Edit (referneces): By $SL_q(N)$ I mean the quantized coordinate algebra (not the quantized enveloping algebra). I am using the conventions of Klimyk and Schmudgen, Chpt 4 for the N=2 case, or Chpt 9 for the general case. </p> http://mathoverflow.net/questions/41804/varieties-frechet-completions-and-regular-functions Varieties, Frechet Completions, and Regular Functions John McCarthy 2010-10-11T16:18:26Z 2010-10-14T08:15:25Z <p>Take an algebraic variety $V$, and its set of smooth functions $C^{\infty}(V)$. One can endow $C^{\infty}(V)$ with a canonical locally convex topology (the seminorms are defined using the local coordinate patches of the variety). With respect to this topology the space is a Frechet space (this means that, amongst other properties, that it is metrisable and complete with respect to the metric). </p> <p>I have two questions:</p> <p>(1) If $O(V)$ is the set of regular functions of $V$, is $O(V)$ dense in $C^{\infty}(V)$ with respect to the Frechet topology?</p> <p>(2) Can one caracterize these locally convex topologies on $O(V)$ that are induced by a differential structure?</p> http://mathoverflow.net/questions/36219/the-killing-form-for-co-quasi-triangular-hopf-algebras The Killing Form for Co-Quasi-Triangular Hopf Algebras John McCarthy 2010-08-20T20:39:03Z 2010-08-27T18:45:47Z <p>For a co-quasi-triangular Hopf algebra $H$, with universal $r$-form $r$, there exists an important map $Q$ defined by <code>$$Q:H \otimes H \to k, ~~~~~~h \otimes g \mapsto r(g_{(1)}\otimes h_{(1)})r(h_{(2)}\otimes g_{(2)}).$$</code> The map is usually called the <em>quantum Killing form</em>.</p> <p>In some papers I have read, it seems that the authors have tacitly assumed that the kernel of $Q$ is a right ideal. Is this true? If so, why?</p> http://mathoverflow.net/questions/35997/action-of-co-quasi-triangular-universal-r-form-on-a-otimes-1 Action of Co-quasi-triangular Universal r-form on $a \otimes 1$ John McCarthy 2010-08-18T18:03:46Z 2010-08-19T17:28:59Z <p>A very easy question I can't seem to answer: For a universal r-form on a co-quasi-triangular Hopf algebra why is $r(a \otimes 1) = r(1 \otimes a) = \epsilon(a)$?</p> http://mathoverflow.net/questions/35057/map-constructed-from-the-coquasitriangular-structure-of-slq2-which-appears-not Map constructed from the coquasitriangular structure of SLq(2) which appears not to respect the standard commutation relations John McCarthy 2010-08-09T22:35:05Z 2010-08-18T13:01:25Z <p>Let $A$ be a Hopf algebra dually paired with a quasi-triangular Hopf algebra $B$. If $x$ is some fixed element of $A$, then we can define a linear map $$P_x: A \to \mathbb{C}$$ by setting $$P_x:a \mapsto \langle R,x \otimes a \rangle.$$</p> <p>Let us take the case $A = SL_q(2)$, $B = U_q({\mathfrak sl}_2)$, and let $R$ be the standard universal $R$-matrix for $U_q({\mathfrak sl}_2)$, for which <code>$$\langle R, u^i_m \otimes u^j_n \rangle = R^{ij}_{mn} = q^{-\frac{1}{2}}.(q^{\delta_{ij}}\delta_{im}\delta_{jn} + (q-q^{-1})\theta (i-j)\delta_{in}\delta_{jm}),$$</code> where $\theta$ is the Heaviside symbol. If we take $x=u^k_l$, then $$P_{u^k_l}(a) = \langle R, u^k_l \otimes a \rangle.$$ Now since $ab = qba$, we should have $$P_{u^k_l}(u^1_1u^1_2) = q P_{u^k_l}(u^1_2u^1_1), \qquad \qquad \text{ for all } \quad k,l = 1,2.$$ However, <code>$$P_{u^2_1}(u^1_1u^1_2) = \langle R, u^2_1 \otimes u^1_1u^1_2 \rangle = \sum_{z=1}^2 \langle R,u^2_z \otimes u^1_1 \rangle \langle R, u^z_1 \otimes u^1_2 \rangle = \sum_{z=1}^2 R^{21}_{z1}R^{z1}_{12}.$$</code> From the formula for $R^{ij}_{mn}$, we get that <code>$$P(u^1_1u^1_2) = \sum_{z=1}^2 R^{21}_{z1}R^{z1}_{12} = R^{21}_{11}R^{11}_{12} + R^{21}_{21}R^{21}_{12} = 0.0 + q^{-\frac{1}{2}}.1.q^{-\frac{1}{2}}.(q-q^{-1}) = q^{-1}(q-q^{-1}).$$</code></p> <p>On the other hand, we have <code>$$qP_{u^2_1}(u^1_2u^1_1) = \langle R,u^2_1 \otimes u^1_2u^1_1 \rangle = \sum_{z=1}^2 \langle R,u^2_z \otimes u^1_2 \rangle \langle R,u^z_1 \otimes u^1_1 \rangle =\sum_{z=1}^2R^{21}_{z2}R^{z1}_{11}.$$</code> From the formula for $R^{ij}_{mn}$, we now get that <code>$$qP_{u^2_1}(u^1_2u^1_1) = q\sum_{z=1}^2R^{21}_{z2}R^{z1}_{11} = qR^{21}_{12}R^{11}_{11} + qR^{21}_{22}R^{21}_{11} = q.q^{-\frac{1}{2}}.(q-q^{-1}).q^{-\frac{1}{2}}.q + q.0.0 = q(q-q^{-1}).$$</code></p> <p>Thus, the two results are not equal, but instead differ by a factor of $q^2$. A similar problem arises for the action of $P_{u^2_1}$ on $bd - qdb$. We get $$P_{u^2_1}(u^1_2u^2_2) = q^{-1}(q-q^{-1}),$$ whereas $$qP_{u^2_1}(u^2_2u^1_2) = q(q-q^{-1}).$$</p> <p>I've checked and rechecked everything very carefully but can't seem to spot my error. Can anyone see what is going wrong here?</p> http://mathoverflow.net/questions/35057/map-constructed-from-the-coquasitriangular-structure-of-slq2-which-appears-not/35806#35806 Answer by John McCarthy for Map constructed from the coquasitriangular structure of SLq(2) which appears not to respect the standard commutation relations John McCarthy 2010-08-16T22:54:21Z 2010-08-16T22:54:21Z <p>I'm going to put my comment to Damien's answer as an answer since there's not enough room to place it as a comment. Firstly, thank you for pointing the typos. I have corrected them and apologise for not checking what I had written thoroughly enough at the start.</p> <p>With regard to the normalisation factor $q^{\frac{-1}{2}}$, I tacitly dropped it because it cancels out for the calculation I'm interested in. However, you're right, it should be included in my definition and I've changed it.</p> <p>With regard to the commutation relations of $SL_q(2)$ there are two conventions: one is as I have written, with, for example, $ab=qba$, and another has $ab=q^{-1}ba$, as you have written. Both algebras are of course isomorphic. I have taken my conventions from Klimyk and Schmuedgen, both for the relations (Chapter 4) and for the definition of $R^{ij}_{nm}$ (Chapter 9).</p> <p>I don't have Kassel's book at hand, so I can't really comment at the moment on his conventions. I will try to have a look tomorrow though.</p> <p>With regard to the $R^{ij}_{mn}$ calculations, I just use the fact that the only non-zero entries are <code>$$R^{11}_{11} = R^{22}_{22} = q^{\frac{1}{2}}, \qquad R^{12}_{12}=R^{21}_{21}=q^{-\frac{1}{2}}, \qquad R^{21}_{12} = q^{-\frac{1}{2}}(q-q^{-1}).$$</code> (But I think it was my typos that were causing the confusion here.) </p> http://mathoverflow.net/questions/29810/weyl-character-formula-for-quantum-groups Weyl Character Formula for Quantum Groups John McCarthy 2010-06-28T17:03:25Z 2010-06-29T15:26:35Z <p>How much is known about the Weyl character formula for quantum groups? More specifically, has the formula been generalized to the general setting of deformed coordinate algebras $\mathbb{C}[G_q]$ of semi-simple Lie groups and their associated flag varieties? I am most interested in the non-root of unity case.</p> http://mathoverflow.net/questions/25602/formula-for-the-matrix-elements-of-the-inverse-of-special-linear-universal-r-matr Formula for the Matrix Elements of the Inverse of special linear Universal R-Matrix of Uq(sln) John McCarthy 2010-05-22T21:00:16Z 2010-05-22T23:35:51Z <p>Motivated by <a href="http://mathoverflow.net/questions/25423/the-inverse-of-a-universal-r-matrix-for-quantized-universal-enveloping-algebra-of" rel="nofollow">this question</a>, I'm going to ask a much more direct one: Let $R$ be a universal R-matrix for the quasitriangular Hopf algebra ${\cal U}_q ({\mathfrak sl}_N)$, $~$ $R ^{-1}$ its inverse, and $\langle \cdot,\cdot \rangle$ the dual pairing between ${\cal U}_q ({\mathfrak sl}_N)$ and $SU_q[N]$, do we have a formula for $$\langle R^{-1},u^i_j \otimes u^r_s\rangle$$ analogous to the formula<br> $$\langle R,u^i_j \otimes u^r_s\rangle = q^{\delta_{ir}}\delta_{ij}\delta_{rs} + (q-q^{-1})\theta (i-r)\delta_{is}\delta_{jr},$$ where $\theta$ denotes the Heaviside symbol. </p> http://mathoverflow.net/questions/15363/examples-for-which-it-is-not-known-if-grothendiecks-standard-conjectures-hold Examples for which it is not known if Grothendieck's Standard Conjectures hold. John McCarthy 2010-02-15T21:04:37Z 2010-02-16T14:22:22Z <p>What is the simplest example of a projective variety for which it is not known if Grothendieck's Standard Conjectures hold?</p> <p>What is the simplest example of a flag manifold for which it is not known if Grothendieck's Standard Conjectures hold? </p> http://mathoverflow.net/questions/14849/what-would-a-moral-proof-of-the-weil-conjectures-require/14933#14933 Answer by John McCarthy for What would a "moral" proof of the Weil Conjectures require? John McCarthy 2010-02-10T18:44:13Z 2010-02-10T18:44:13Z <p>This is both an answer and a question:</p> <p>As part of a response to a previous <a href="http://mathoverflow.net/questions/3430/castelnuovo-positivity-rewrite-of-weils-original-proof-for-fp2" rel="nofollow">question</a> of mine, David Speyer wrote that:</p> <blockquote> <p>... it is known how to adapt Weil's proof of the Riemann hypothesis to higher dimensional S, if one had an analogue of the Hodge index theorem for $S \times S$ in characteristic p. I've been told that a good reference for this is <a href="http://www.ams.org/mathscinet-getitem?mr=292838" rel="nofollow">Kleiman's Algebraic Cycles and the Weil Conjectures</a>...</p> </blockquote> <p>So perhaps a "moral" proof would require a Hodge index theorem in characteristic p.</p> <p>However, David later writes that Grothendieck's standard conjectures assert that the Hodge theorem holds. So is this possible proof the same as "Grothendieck's envisaged" one? </p> http://mathoverflow.net/questions/123427/coherent-sheaves-and-holomorphic-vector-bundles Comment by John McCarthy John McCarthy 2013-03-02T20:00:15Z 2013-03-02T20:00:15Z @Liviu: Thanks for the comment. So if you take the derived category of the locally free sheaves, what would the corresponding subcategory of the Fukaya category under mirror symmetry? http://mathoverflow.net/questions/120033/why-are-the-holomorphic-line-bundle-sections-finite-dimensional/120035#120035 Comment by John McCarthy John McCarthy 2013-01-27T20:43:00Z 2013-01-27T20:43:00Z @ unknown &amp; Roy: Thanks for your answers. However, I was hoping that there might exist a more algebraic reason for finite dimensionality. Is it possible to show this without appealing to analysis? http://mathoverflow.net/questions/78036/maximal-ideals-and-kahler-differentials/78066#78066 Comment by John McCarthy John McCarthy 2011-10-13T22:09:24Z 2011-10-13T22:09:24Z Are the flag manifolds simple homogeneous spaces? http://mathoverflow.net/questions/78036/maximal-ideals-and-kahler-differentials Comment by John McCarthy John McCarthy 2011-10-13T17:55:12Z 2011-10-13T17:55:12Z All fixed now. Thanks for pointing out the mistakes. http://mathoverflow.net/questions/67537/compact-quantum-groups-from-hopf-algebras/67599#67599 Comment by John McCarthy John McCarthy 2011-06-13T11:04:32Z 2011-06-13T11:04:32Z So the Drin-Jim QUE do not fit in the category, but their their dual quantised function algebras do? http://mathoverflow.net/questions/67537/compact-quantum-groups-from-hopf-algebras/67599#67599 Comment by John McCarthy John McCarthy 2011-06-12T20:03:49Z 2011-06-12T20:03:49Z Do the Drinfeld--Jimbo algebras fit into this category? http://mathoverflow.net/questions/59230/when-does-the-anti-holomorphic-chain-complex-exist-for-non-kahler-manifolds/59292#59292 Comment by John McCarthy John McCarthy 2011-03-23T18:51:37Z 2011-03-23T18:51:37Z .. sorry, forgot the $-$ again. http://mathoverflow.net/questions/59230/when-does-the-anti-holomorphic-chain-complex-exist-for-non-kahler-manifolds/59292#59292 Comment by John McCarthy John McCarthy 2011-03-23T18:51:14Z 2011-03-23T18:51:14Z Before I posted the comment, I was thinking that this might be the case. Then I looked in my edition of Griffiths and Harris, and page 82 says $\overline{\partial}^\ast = \ast \overline{\partial} \ast$ and got confused. What's going on here? http://mathoverflow.net/questions/59230/when-does-the-anti-holomorphic-chain-complex-exist-for-non-kahler-manifolds/59292#59292 Comment by John McCarthy John McCarthy 2011-03-23T14:54:38Z 2011-03-23T14:54:38Z ... sorry that should be $\overline{\partial}^\ast = -\ast \overline{\partial}\ast$. But the problem still holds. http://mathoverflow.net/questions/59230/when-does-the-anti-holomorphic-chain-complex-exist-for-non-kahler-manifolds/59292#59292 Comment by John McCarthy John McCarthy 2011-03-23T14:53:08Z 2011-03-23T14:53:08Z Ok I'm confused. Let's assume that, as you say, $\ast(\Omega^{(0,k)}) \subset \Omega^{(N-k,N)}$ and look at the action of $\overline{\partial}^\ast$ on some $\omega \in \Omega^{(0,k)}$. Since $\overline{\partial}$ is zero on $\omega^{(N-k,N}$, we must have that $\overline{\partial}^\ast(\omega) = \ast (\overline{\partial} (\ast(\omega))) = 0$. Since $k$ is arbitrary, our chain complex is trivial. http://mathoverflow.net/questions/59211/diracs-original-operator-and-the-hodge-dirac-operator/59250#59250 Comment by John McCarthy John McCarthy 2011-03-23T13:19:41Z 2011-03-23T13:19:41Z What's the explicit action of $i$ on a smooth function? http://mathoverflow.net/questions/59211/diracs-original-operator-and-the-hodge-dirac-operator/59250#59250 Comment by John McCarthy John McCarthy 2011-03-23T12:26:42Z 2011-03-23T12:26:42Z I see that identifying the Clifford and exterior bundles allows us to view the two operators as operating on the same space. However, I don't see that they are equal: $D$ sends zero forms to zero forms, while $d+d^{\dagger}$ sends zero forms to one forms; how can they be equal? http://mathoverflow.net/questions/53932/generators-of-the-odd-dimensional-quantum-spheres Comment by John McCarthy John McCarthy 2011-03-07T17:35:22Z 2011-03-07T17:35:22Z Does anyone know where this result was first proved? http://mathoverflow.net/questions/56293/global-definition-of-the-almost-complex-structure-of-a-complex-manifold Comment by John McCarthy John McCarthy 2011-02-22T20:51:18Z 2011-02-22T20:51:18Z Ok, I see now what's going on. Thanks a lot Johannes. Sorry for asking a question before I understood waht I was asking about. I think the best thing to do with this question would be to <b>close it</b>. http://mathoverflow.net/questions/56293/global-definition-of-the-almost-complex-structure-of-a-complex-manifold Comment by John McCarthy John McCarthy 2011-02-22T17:06:13Z 2011-02-22T17:06:13Z I'm looking at the $J$ as an operator on $\Omega^1(M)$. Then if $f$ is a smooth function, $\partial(f)$ is a (1,0)-form. Operating on $\partial(f)$ by $J$ will, if I understand it correctly, will send it to the $\Omega^{(0,1)}(M)$ forms, which are spanned by elements of the form $\overline{\partial}(g)$, for $g$ also a smooth function. Thus, if $J$ is just multiplication by $i$, we get my statement above.