User jean delinez - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:41:50Z http://mathoverflow.net/feeds/user/1648 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131497/equivariant-k-theory-singular-vectors-and-flag-manifolds Equivariant $K$-theory, singular vectors, and flag manifolds Jean Delinez 2013-05-22T19:02:13Z 2013-05-23T10:47:17Z <p>For a homogeneous space $M = G/B$, with $G$ a (complex) semi-simple Lie group, it is very well-known that equivariant vector bundles $E$ over $M$ correspond to representations $(V_{\lambda},\lambda)$ of $B$, and the differential operators on $E$ are closely linked to the representation theory of $G$.</p> <p>For the special case of a flag manifold, which is to say, when $B$ is a Borel subgroup of $G$, differential operators from $E$ to itself correspond to homomorphisms of the Verma module $U({\frak g})\otimes_{U({\frak b})} V_{\lambda}$. These homomorphisms are in turn classified by the so-called <strong>singular vectors</strong> of $V_{\lambda}$, which is to say the vectors killed by the action of the positive niradical. Moreover again, these singular vectors correspond to solutions of certain hyper-geometric functions.</p> <p>What I would like to know is how all this relates to equivariant K-theory. Is there some characterization of the singular vectors correspond to a Fredholm operator. Also, can the defining equivalence relation of the equivariant K-theory group $K^0$ be nicely reformulated in terms of representation theory and singular vectors?</p> http://mathoverflow.net/questions/122401/classifying-globally-generated-holomorphic-line-bundles-over-a-flag-manifold Classifying Globally Generated Holomorphic Line Bundles over a Flag Manifold Jean Delinez 2013-02-20T13:56:19Z 2013-02-23T14:49:49Z <p>I was recently looking back at an <a href="http://mathoverflow.net/questions/14296/indexing-the-line-bundles-of-a-flag-manifold" rel="nofollow">old question</a> of mine, where I asked about the classification of the line bundles over a general complex flag manifold. Pavel Etingof gave the following excellent answer:</p> <blockquote> <p>The partial flag manifold X which you mentioned is the set of flags $0 ⊂ V_1 ⊂ \ldots ⊂ V_m=ℂ^n$, such that $\dim(V_j/V_{j−1})=k_j$. So we have vector bundles $V_j$ on $X$ and line bundles $L_j$ on $X$ which are the top exterior powers of the bundles $V_j$, $j = 1,...,m−1$. Any line bundle on $X$ is a tensor product of powers of these line bundles, and the powers are uniquely determined. This is why line bundles on $M$ are labeled by a set of $m−1$ integers.</p> </blockquote> <p>I then began to wonder how this relates to a more recent <a href="http://mathoverflow.net/users/1648/jean-delinez" rel="nofollow">question</a> of mine regarding the definition of a globally generated holomorphic vector bundle. So my question is: With respect to the above line bundle classification, and the obvious choice of holomorphic structure for these bundles (I am assuming that there is just one obvious choice here), which of these bundles are globally generated?</p> <p>For the simplest case of complex projective space, we have of course that the line bundles are classified by the integers, and the positive ones are globally generated (or the negative, depending on convention). Now for the Grassmannians the line bundles are again classified by the integers, and I would again guess that the globally generated ones are exactly those of positive charge. So how does this extend?</p> http://mathoverflow.net/questions/121345/when-are-the-smooth-sections-of-a-bundle-generated-as-a-module-over-smooth-funct When are the Smooth Sections of a Bundle Generated as a Module (over Smooth Functions) by the Holomorohic Sections Jean Delinez 2013-02-09T18:58:41Z 2013-02-10T06:30:37Z <p>For a holomorphic vector bundle $E$ over a complex manifold $M$, we denote its space of smooth sections by $\Gamma^{\infty}(E)$, and its space of holomorphic sections by $\Gamma^{hol}(E)$. Now I've been looking at the line bundles $L_k$ over the complex projective spaces ${\bf C} P^N$, and I have managed to show that $\Gamma^{\infty}(L_k)$ is generated as a $C^{\infty}({\bf C} P^N)$-module by $\Gamma^{hol}(E)$, which is to say that every element $\Gamma^{\infty}(E)$ is a sum of elements of the form $ef$, where $e \in \Gamma^{hol}(E)$, and $f \in C^{\infty}({\bf C} P^N)$. </p> <p>I am guessing that this result is extremely well known, and an example of a well understood general phenomenon. So I would like to ask if there is a characterization of the manifolds for which this result holds, for both the case of line bundles alone, and holomorphic vector bundles of general dimension? </p> http://mathoverflow.net/questions/120984/rep-theory-consequences-of-bott-weil-borel Rep Theory Consequences of Bott--Weil--Borel Jean Delinez 2013-02-06T16:18:48Z 2013-02-07T15:51:49Z <p>I've been getting interested in the (Bott--)Borel--Weil theorem lately. As a (mainly) geometer it is very interesting to see representation appearing (from nowhere as far as I can see) in the theory of complex geometry. </p> <p>What I can't seem to find out, though, is if this fact has been used to prove any interesting results in representation theory, ie, does the geometric realization of the representation allow us to prove anything we didn't know already? </p> http://mathoverflow.net/questions/120901/group-representations-and-holomorphic-vectors-bundles-over-homogeneous-spaces-ex Group Representations and Holomorphic Vectors Bundles over Homogeneous Spaces (extending Borel--Weil) Jean Delinez 2013-02-05T19:11:18Z 2013-02-06T14:14:28Z <p>For a flag manifold $F$ of a group $G$, the Borel--Weil theorem deals with representations of $G$ on the holomorphic sections of the line bundles over $F$. </p> <p>Let us consider a general framework than the flag manifolds: Take a homogeneous space $M := G/K$, with $G$ compact, such that $M$ has been given a complex structure. In this setting, does $G$ still have a representation on the holomorphic sections of the holomorphic vector bundles over $M$? It seems to me that this will only happen if the action of $G$ on $M$ is holomorphic. However, I can't see for sure that nothing goes wrong, and I don't see that this condition is obviously true in the special case of the flag manifolds. Can anyone help?</p> http://mathoverflow.net/questions/111723/cohomology-ring-of-the-flag-manifolds-cartan-subalgebras-and-weyl-groups Cohomology Ring of the Flag Manifolds, Cartan Subalgebras, and Weyl Groups Jean Delinez 2012-11-07T13:18:32Z 2012-11-07T13:44:58Z <p>I've recently read the following line in an interesting paper:</p> <blockquote> <p>It is well-known that the cohomology ring of a flag variety $G/B$ is isomorphic to the quotient ring of the ring of polynomial functions on the Cartan sub algebra $\frak{h}$ by the ideal generated by the fundamental invariants $f_1 , . . . , f_r, r =$rank$({\frak h})$, of the Weyl group W, i.e. $$H^∗(G/B,{Q}) \simeq {\text Sym}_Q{\frak h}^∗/(f_1, . . . , f_r).$$</p> </blockquote> <p>I would like to ask:</p> <p>(1) Does this result extend to other fields, i.e. the complex and real case?</p> <p>(2) What is a good understandable reference for learning about this result? </p> http://mathoverflow.net/questions/109392/jets-of-equivariant-vector-bundles Jets of Equivariant Vector Bundles Jean Delinez 2012-10-11T16:14:33Z 2012-10-15T08:11:32Z <p>Let $M$ be a (compact) $G$-homogeneous space with fibre group $H$, and let ${\cal E}$ be a $G$-equivariant $k$-dimensional vector bundle over $M$ with corresponding representation $\pi:H \to$R$^k$. What I would like to know is whether all the jet bundles (see <a href="http://en.wikipedia.org/wiki/Jet_bundle" rel="nofollow">here</a> for a definition of jet bundle) of $\cal E$ are also $G$-equivariant, and if so, can one construct their corresponding representations from $\pi$?</p> http://mathoverflow.net/questions/102819/finite-field-varieties-and-the-de-rham-complex-of-kahler-differentials Finite Field Varieties and the de Rham Complex of Kähler Differentials Jean Delinez 2012-07-21T14:13:23Z 2012-08-08T10:34:26Z <p>In an answer to a previous <a href="http://mathoverflow.net/questions/76932/kahler-structure-for-projective-varieties-over-a-finite-field" rel="nofollow">question</a> I asked, where the Kahler differentials of a variety over a finite field were discussed, it was stated that:</p> <blockquote> <p>You can certainly define de Rham cohomology using Kähler differentials, but over a field of characteristic p>0 . . . it is somewhat pathological: the Poincaré lemma can fail etc</p> </blockquote> <p>(1) What does this "etc" mean here? That is to say, what else goes wrong?</p> <p>(2) Is the de Rham complex for Kähler differentials useful for anything? Or is it just a curiosity?</p> http://mathoverflow.net/questions/103705/global-definition-of-the-dolbeault-complex-of-a-vector-bundle Global Definition of the Dolbeault Complex of a Vector Bundle Jean Delinez 2012-08-01T16:15:47Z 2012-08-02T12:03:55Z <p>For an $2n$-dimensional complex manifold $M$, and a smooth vector bundle $E$ over $M$, it is well-known (see Voisin, Huybrechts) that there exists an operator $\overline{\partial}$, built locally from the usual anti-holomorphic derivative, that acts on $\Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,\bullet)}(M)$ so as to give a complex $$\Gamma^{\infty}(E) \overset{\overline{\partial}}{\to} \Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,1)}(M) \overset{\overline{\partial}}{\to} \cdots \overset{\overline{\partial}}{\to} \Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,n)}(M)\overset{\overline{\partial}}{\to} 0.$$</p> <p>As I prefer global constructions, I began to wonder how one would construct this globally. To apply id $\otimes \overline{\partial}$ to $\Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,\bullet)}(M)$ is of course not well-defined since we are tensoring over ${C^{\infty}(M)}$. So what can one do . . .?</p> <p>P.S. Does such a complex exist in the purely real case, and if not, then why not?</p> http://mathoverflow.net/questions/97074/is-the-susy-algebra-isomorphic-for-all-kahler-manifolds Is the SUSY Algebra isomorphic for all Kahler Manifolds? Jean Delinez 2012-05-16T01:23:56Z 2012-05-16T01:59:49Z <p>For a Kahler manifolds, the graded algebra generated by $\partial,\overline{\partial},\partial^*,\overline{\partial}^\ast$, the Lefschetz operator, and the dual Lefschetz operator, is called the <em>supersymmetric algebra</em>, or in physics parlance the <em>$N=(2,2)$ SUSY algebra</em>. I am correct in understanding that one gets the same algebra for all Kahler manifolds?</p> http://mathoverflow.net/questions/75864/invariant-metrics-on-the-sphere Invariant Metrics on the Sphere Jean Delinez 2011-09-19T16:07:22Z 2012-03-28T18:53:10Z <p>I've been thinking about $SU(n)$-invariant metrics on the odd-dimensional spheres $S^{2n-1} \simeq SU(n)/SU(n-1)$. For $S^1$, all such metrics are in correspondence with the positive reals. For $S^{3} \simeq SU(2)$, the tangent space is parallelizable, and so, such metrics are in correspondence with metrics on $R^2$. Does such a characterization exist for $S^5$, or higher. </p> http://mathoverflow.net/questions/92029/equivariant-tangent-bundle-decomposition Equivariant Tangent Bundle Decomposition Jean Delinez 2012-03-23T18:46:26Z 2012-03-23T19:31:07Z <p>Given a $G$-homogeneous space $M$, for $G$ a (Lie) group, we have a canonical $G$-action on the tangent bundle $T(M)$ of $M$. If $M$ is a complex manifold, then we have a decomposition of $T(M) \otimes_{R} C$ into $T^{(1,0)}(M)$ and $T^{(0,1))}(M)$. If this decomposition is equivariant with respect to the $G$-action, then we say that we have an equivariant complex structure. As far as I can see, the set of all equivariant decompositions of $T(M) \otimes_{R} C$ into two ismorphic parts, corresponds to set of all possible equivariant almost-complex structures for $M$.</p> <p>More generally though, what can we say about the relationship between equivariant decompositions of $T(M)$ (ot $T(M) \otimes_{R} C$) and the structure of $M$. When do they exist? Does the number of components in an irreducible equivariant decomposition of $T(M)$ have any geometric meaning?</p> http://mathoverflow.net/questions/46674/square-of-the-dirac-and-the-laplacian-on-a-k-ahler-manifold Square of the Dirac and the Laplacian on a K\"{a}hler Manifold Jean Delinez 2010-11-19T19:58:25Z 2012-03-14T11:27:48Z <p>In the Euclidean setting, the Dirac operator was constructed so as to give the square of the Laplacian. Now for a K\"{a}hler manifold with a spin$^c$ structure we have the a corresponding Dirac operator $D$. Moreover, we have a Laplacian $(d+d^{\ast})$, where $d^{\ast}$ is the coadjoint $\ast d \ast$, for $\ast$ the Hodge $\ast$-mapping. Now in the case where the manifold is also symmetric we get a relationship between the square of the Dirac and the Laplacian that involves an extra curvature term. Does this extend to all K\"{a}hler manifolds, and if it does, what is the exact relationship?</p> http://mathoverflow.net/questions/90126/global-lichnerowicz-formula-proof-in-the-kahler-case Global Lichnerowicz Formula Proof (in the Kahler case)? Jean Delinez 2012-03-03T16:24:50Z 2012-03-06T11:33:36Z <p>For a Kahler manifold $M$, let us denote its Dirac operator $\overline{\partial} + \overline{\partial}^\ast$, with respect to a metric $g$, by $D$. Moreover, let us dentoe the Levi-Civita connection wrt $g$ by $\nabla$, and the corresponding connection Laplacian by $\nabla^*\nabla$.</p> <p>If I am not mistaken, then the Lichnerowicz formula says that $$D^2 = \nabla^{\ast} \nabla + O,$$ where $O$ is a zero order operator.</p> <p>I have two questions:</p> <p>(i) Does there exist a global version of this proof? The two versions I've looked at are in Andre Moroianu's notes, and Thomas Friedrich's book. Both are given in local terms, but both seem realtively algebraic, causing me to suspect that there may exist a global version. I am right here?</p> <p>(ii) I know that the Lichnerowicz formula can be used to prove that $D$ has compact resolvent, wrt the standard Hilbert space completion of the exterior algebra. However, I'm finding it difficult to see the wood for the trees in proofs I have to hand. Could someone explain why a relation between the Dirac-Laplacian and the connection Laplacian tells me some about $(1+D)^{-1}$? Is it some clear that $\nabla^{\ast} \nabla$ has nice properties as Hilbert space operator?</p> http://mathoverflow.net/questions/83743/hilbert-nullstellsatz-and-non-complete-fields Hilbert Nullstellsatz and Non-Complete Fields Jean Delinez 2011-12-17T21:20:01Z 2011-12-18T19:13:29Z <p>The Hilbert nullstellsatz tells us that for a complete field $K$, there is a bijective correspondence between K-varieties and finitely generated algegras over $K$.</p> <p>When the field is not complete, eg $K=R$, what goes wrong here? We still have a map from the set of varieties over $K$ to finitely generated algebras over $K$. So it not to give us a bijection it must fail to be bijective or injective. Which of these happens? Does it fail to be surjective or injective or both? </p> http://mathoverflow.net/questions/83445/hyper-complex-and-quaternionic-kahler-geometry Hyper-Complex and quaternionic Kahler Geometry Jean Delinez 2011-12-14T18:03:12Z 2011-12-15T15:53:07Z <p>What is the exact relationship <strong>hyper-complex</strong> and <strong>quaternionic Kahler manifolds</strong>? From Wikipedia we get that hyper-Kahler manifolds are both hyper-complex and quaternionic Kahler. Thus, the two families have a non-empty intersection. But this is all I can conclude.</p> <p>Moreover, <strong>quaternionic projective space</strong> is quaternionic Kahler, but is it also hyper-Kahler? Is it even hyper-complex?</p> http://mathoverflow.net/questions/5772/principal-bundles-representations-and-vector-bundles Principal bundles, representations, and vector bundles Jean Delinez 2009-11-17T03:43:20Z 2011-12-10T20:38:20Z <p>What is the exact relationship between principal bundles, representations, and vector bundles?</p> http://mathoverflow.net/questions/76932/kahler-structure-for-projective-varieties-over-a-finite-field Kahler Structure for Projective Varieties over a Finite Field Jean Delinez 2011-10-01T16:35:10Z 2011-10-01T18:03:54Z <p>(i) In 1960 Serre proved a famous <a href="http://www.jstor.org/stable/1970088?origin=crossref" rel="nofollow">analogue</a> of the Weil conjectures for Kähler manifolds. This poses an obvious question: Does there exist an analogue of a Kahler structure for (non-singular) projective varieties over a finite field. That is, do there exist things like almost complex structures, Lefschetz operators, Kahler identities, etc?</p> <p>(ii) Moreover, the study of generalised Hodge structures is a well-studied field. Does there exist a subfield of generalised Kahler structures?</p> http://mathoverflow.net/questions/73293/when-is-a-form-a-kahler-form When is a Form a Kahler Form? Jean Delinez 2011-08-20T20:57:47Z 2011-08-21T13:56:48Z <p>Let $M$ be a complex manifold, and $\omega$ a closed $2$-form. When is $\omega$ a Kahler form? I mean, when does there exist a Kahler metric for which $\omega$ is the corresponding form.</p> <p>I would (wildly) guess that necessary and sufficient conditions might be got from the K\"ahler identities.</p> http://mathoverflow.net/questions/16207/why-can-the-dolbeault-operators-be-realised-as-lie-algebra-actions Why can the Dolbeault Operators be Realised as Lie Algebra Actions Jean Delinez 2010-02-23T22:50:45Z 2011-08-06T20:03:34Z <p>I've been looking at an example in the non-commutative geometry literature and I'm having trouble figuring out what the classical motivation is. I'll just describe the classical case here: Recall that $\mathbb{CP}^2 = SU(\mathbb{C},3)/U(\mathbb{C},2)$. Recall also that since $T_\mathbb{C}^*(\mathbb{CP}^2)$ can be viewed as a vector bundle associated to the $SU(2)$-bundle $SU(\mathbb{C},3) \to \mathbb{CP}^2$, we can view $\Omega_\mathbb{C}^1(\mathbb{CP}^2)$ as a subset of $\mathcal{O}(\mathbb{CP}^2) \oplus \mathcal{O}(\mathbb{CP}^2)$. Now in the example, the action of each of the Dolbeault operators $\partial,\overline{\partial}$ is given in terms of two mappings, $\partial_1,\partial_2$ and $\overline{\partial_1},\overline{\partial_2}$, such that $$\partial(f) = (\partial_1(f),\partial_2(f)) \in \mathcal{O}(\mathbb{CP}^2) \oplus \mathcal{O}(\mathbb{CP}^2),$$ and similarly for $\overline{\partial}$. The mappings ${\partial}_i, \overline{\partial}_i$ are constructed using an action of the Lie algebra $\mathfrak{su}(3)$ on $\mathcal{O}(SU(3))$ constructed using the canonical pairing between Lie algebras and coordinate algebras.</p> <p>I have a feeling this is a complicated incarnation of a simple classical object. Does any of this ring any bells with anyone?</p> <p>Please ask if you would like more details.</p> <p>Edit: This question has been superseded by <a href="http://mathoverflow.net/questions/16880/dolbeault-operators-for-cp1-as-mathfraksu2-actions" rel="nofollow">this question</a> and I am voting to close. I would ask others to do likewise.</p> http://mathoverflow.net/questions/71354/did-kahler-say-a-long-list-of-miracles-occur Did Kahler say "a long list of miracles occur"? Jean Delinez 2011-07-26T20:40:51Z 2011-07-27T12:34:06Z <p>I've been reading Moroianu's Kahler geometry notes and found a unattributed quote that says that if the Kahler properties hold, then "a long list of miracles occur"</p> <p>I am guessing that this quote belongs to Kahler himself, but I can't back this up. Does anyone know?</p> http://mathoverflow.net/questions/71006/semi-simple-kahler-groups Semi-Simple Kahler Groups? Jean Delinez 2011-07-22T16:57:45Z 2011-07-23T12:13:30Z <p>We say that a Kahler manifold is a <em>Kahler group</em> if it is also a Lie group. I would like to know which semi-simple Lie groups are also Kahler groups?</p> http://mathoverflow.net/questions/67532/atiyah-singer-for-the-complex-projective-line Atiyah--Singer for the Complex Projective Line Jean Delinez 2011-06-11T19:03:06Z 2011-06-11T19:03:06Z <p>I'm trying to understand Atiyah--Singer by looking at the usual starting point of $CP^1$ and the Dirac--Dolbeault operator. If I've reduced everything down correctly, then in this case the theorem gives $$\text{Index}(\overline{\partial} + \overline{\partial}^*) = \frac{1}{2}\int_{CP^1} \text{ch}_1(T(CP^1)),$$ where ch$_1(T(CP^1))$ is the first Chern class of the tangent bundle $CP^1$. I would like a direct explanation of why this is true. </p> http://mathoverflow.net/questions/64520/global-algebraic-proof-of-the-kahler-identities Global Algebraic Proof of the Kahler Identities? Jean Delinez 2011-05-10T17:48:18Z 2011-05-10T20:02:25Z <p>I'm looking at Kahler geometry at the moment and admiring how it manages to do so much with clean global algebraic arguments. One of the big exceptions to all this, however, is the proof of the Kahler identities $$[\Lambda,\overline{\partial}]=-i \partial^\ast, ~~~~~~ [\Lambda,\partial]=-i \overline{\partial}^\ast.$$ In the two standard references, Voisin, and Griff + Harris, the identities are proved using arguments that are local and somewhat analaytic. Does there exists anywhere a nice global algebraic proof?</p> http://mathoverflow.net/questions/63414/nash-style-embedding-theorem-for-connections "Nash Style" Embedding Theorem for Connections Jean Delinez 2011-04-29T14:07:53Z 2011-04-29T14:23:55Z <p>The strong Whitney embedding theorem states that any smooth (Hausdorff and second-countable) manifold can be smoothly embedded in Euclidean space. John Nash went on to show that any Riemannian manifold could be embedded into Euclidean space in such a way that the metric of the manifold would coincide with the standard dot product. This means that the Levi-Civita connection for the manifold maps to the standard connection. More generally, for a manifold $M$ with connection $\nabla$, when does there exist an embedding into Euclidean space such that the connection is mapped to the standard connection? </p> http://mathoverflow.net/questions/54434/when-can-a-connection-induce-a-riemannian-metric-for-which-it-is-the-levi-civita When can a Connection Induce a Riemannian Metric for which it is the Levi-Civita Connection? Jean Delinez 2011-02-05T18:38:34Z 2011-04-29T13:54:43Z <p>As we all know, for a Riemannian manifold $(M,g)$, there exists a unique torsion free connection $\nabla_g$, the Levi-Civita connection, that is compatible witht metric.</p> <p>I was wondering if one can reverse this situation: Given a manifold with $M$ with connection $\nabla$, when does there exist a Riemannian metric $g$ for which $\nabla$ is the Levi-Civita connection. </p> <p>If this were true for complex projective manifolds it would make me be very happy.</p> http://mathoverflow.net/questions/59230/when-does-the-anti-holomorphic-chain-complex-exist-for-non-kahler-manifolds When does the Anti-Holomorphic Chain Complex Exist for Non-Kahler Manifolds? Jean Delinez 2011-03-22T20:53:07Z 2011-03-28T15:03:04Z <p>Given an $N$-dimensional Riemannian manifold $M$, with associated Hodge $\ast$-mapping $\ast$, we have the chain complex $$\Omega^{0} {\buildrel {\text d}^\ast \over \longleftarrow} \Omega^{N} {\buildrel {\text d}^\ast \over \longleftarrow}\cdots \Omega^{N}<br>$$ For a Kahler manifold $M$ of complex dimension $N$, with associated Hodge $\ast$-mapping $\ast$, we have a second chain complex $$\Omega^{(0,0)} {\buildrel \overline{\partial}^\ast \over \longleftarrow} \Omega^{(0,1)} {\buildrel \overline{\partial}^\ast \over \longleftarrow} \cdots \Omega^{(0,N)}.$$ What I would like to know is for which other Hermitian manifolds does this complex exist? I suppose I'm asking what condition on the Hermitian metric will give a Hodge $\ast$-map for which $\overline{\partial}^\ast(\Omega^{(0,k)}) \subset \Omega^{(0,k-1)}$ and $\overline{\partial}^2=0$.</p> http://mathoverflow.net/questions/59452/relation-between-the-de-rham-and-hodge-laplacians-on-the-exterior-algebra Relation between the de Rham and Hodge Laplacians on the Exterior Algebra Jean Delinez 2011-03-24T18:03:34Z 2011-03-24T18:38:56Z <p>For a Kahler manifold $M$ we have two well-known Laplacians: the de Rham Laplacian $\Delta_{\text{d}} = ($d$+$d$^\ast)^2$, and the Dolbeault Laplacian $\Delta_{\overline{\partial}} = (\overline{\partial} + \overline{\partial}^\ast)^2$. Now on smooth functions, these two operators are related by the well-known formula $$\Delta_{\text{d}} = 2\Delta_{\overline{\partial}}$$ Now both these operators act on the exterior algebra. Does there exist a similar formula in this more general setting?</p> http://mathoverflow.net/questions/58194/holonomy-groups-and-the-hopf-fibration Holonomy Groups and the Hopf Fibration Jean Delinez 2011-03-11T19:40:42Z 2011-03-11T23:27:09Z <p>I am trying to understand holonomy groups at the moment and am focusing on the example of the Hopf fibration $SU_2 \to S^2$. Since $S^2$ is path connected we can talk about the holonomy group of a connection for the fibration - ie it's the same at all points. For the usual monopole connection $\omega$, what is the holonomy group Hol$_p(\nabla)$ of $\omega$ at a point $p$. The holonomy bundle will of course be a principal Hol$_p(\nabla)$-bundle over $S^2$, what is the total space of this bundle, how does Hol$_p(\nabla)$ act on it, what is the projection?</p> http://mathoverflow.net/questions/44474/complex-projective-space-spin-and-dirac-part-ii Complex Projective Space Spin and Dirac: Part II Jean Delinez 2010-11-01T19:14:52Z 2010-11-10T15:56:07Z <p>(1) Let $M$ be a complex manifold of real dimension $2n$, and denote the line bundle of complex $(N,0)$-forms by $\Omega^{(N,0)}(M)$. When $M = CP^N$, the line bundles are indexed by the integers, and so, $\Omega^{(N,0)}(CP^N)$ must correspond to a integer. What is this integer? In the $N=1$ case, the corresponding integer is $2$. This suggests, that in general, $\Omega^{(N,0)}(CP^N)$ corresponds to $2N$. Is this true?</p> <p>(2) For the anti-canonical spin$^c$ structure of $CP^N$, the spinor bundle is isomorphic to $$S := (\Omega^{(0,0)}(CP^N) + \cdots + \Omega^{(0,N)}(CP^N)) \otimes S_N,$$ where $S_k$ is the square root of $\Omega^{(N,0)}(CP^N)$ (square root wrt tensoring as multiplication). What does the square root mean when when line bundle integer is odd? In the $N=2$ case, this is seen to reduce to $\cal{E}_{-1} \otimes \cal{E}_1$, where $\cal{E}_p$ is the line bundle corresponding to the integer $p$. Is this the $Z2$ grading on the spinor bundle? If so, what does this look like in higher dimensions?</p> <p>(3) Finally, for a given spin connection $\nabla$, to define a Dirac operator we would need a Clifford representation, ie a map $$C:(\Omega^{(1,0)} \oplus \Omega^{(0,1)}) \otimes S \to S.$$ For $N=2$, this should be given by a map $$C: (\Omega^{(1,0)} \oplus \Omega^{(0,1)}) \otimes (\cal{E}_{-1} \oplus \cal{E}_1) \to \cal{E}_1 \oplus \cal{E}_1.$$ What is this rep? What does it look like for higher order $N$? Note: the first subindex in the second $\cal{E} \oplus \cal{E}$ above should be $-1$, I'm having tex problems when I try to write it as such though.</p> http://mathoverflow.net/questions/131497/equivariant-k-theory-singular-vectors-and-flag-manifolds Comment by Jean Delinez Jean Delinez 2013-05-23T11:40:28Z 2013-05-23T11:40:28Z Yes, the is true. Some good references can be found here <a href="http://mathoverflow.net/questions/109392/jets-of-equivariant-vector-bundles" rel="nofollow" title="jets of equivariant vector bundles">mathoverflow.net/questions/109392/&hellip;</a> http://mathoverflow.net/questions/131497/equivariant-k-theory-singular-vectors-and-flag-manifolds Comment by Jean Delinez Jean Delinez 2013-05-23T10:48:35Z 2013-05-23T10:48:35Z Sorry, that's been fixed now. And no there is no reason why I don't assume $G/B is a flag variety from the start. http://mathoverflow.net/questions/122401/classifying-globally-generated-holomorphic-line-bundles-over-a-flag-manifold/122405#122405 Comment by Jean Delinez Jean Delinez 2013-02-26T13:49:30Z 2013-02-26T13:49:30Z @Jim Great that seems to answer my question. It's the full flag manifolds I'm really interested in any way. Please put your comment as an answer so I can accept it. http://mathoverflow.net/questions/122401/classifying-globally-generated-holomorphic-line-bundles-over-a-flag-manifold Comment by Jean Delinez Jean Delinez 2013-02-20T15:54:20Z 2013-02-20T15:54:20Z Good point! I've changed it. http://mathoverflow.net/questions/3512/top-specialized-journals/9912#9912 Comment by Jean Delinez Jean Delinez 2013-02-13T17:02:48Z 2013-02-13T17:02:48Z What about the Journal of Geometry and Physics? http://mathoverflow.net/questions/121345/when-are-the-smooth-sections-of-a-bundle-generated-as-a-module-over-smooth-funct/121378#121378 Comment by Jean Delinez Jean Delinez 2013-02-11T17:13:55Z 2013-02-11T17:13:55Z Ora capisco grazie! http://mathoverflow.net/questions/121345/when-are-the-smooth-sections-of-a-bundle-generated-as-a-module-over-smooth-funct/121378#121378 Comment by Jean Delinez Jean Delinez 2013-02-11T16:13:44Z 2013-02-11T16:13:44Z .... but I can't see why every vector is not globally generated. Surely, for$v \in E_p$as above, we have a local section for which$s(p) = v$, why can one not just use a partition of unity argument to extend$s$to a global section, and hence conclude that your vector bundle is globally generated? I can't see the problem in my logic here. http://mathoverflow.net/questions/121345/when-are-the-smooth-sections-of-a-bundle-generated-as-a-module-over-smooth-funct/121378#121378 Comment by Jean Delinez Jean Delinez 2013-02-11T16:06:14Z 2013-02-11T16:06:14Z Sorry, but I would like to get this totally clear for my mind. What you are saying is that there exists for$v \in E_p$, for any$p \in M$, an element$s \in \Gamma^{\infty}(E)$of the smooth sections (or global sections as you call them), such that$v = s(p)$. Is this correct? http://mathoverflow.net/questions/121345/when-are-the-smooth-sections-of-a-bundle-generated-as-a-module-over-smooth-funct/121378#121378 Comment by Jean Delinez Jean Delinez 2013-02-10T15:56:35Z 2013-02-10T15:56:35Z Excuse my ignorance, but what is a globally generated vector bundle? http://mathoverflow.net/questions/120984/rep-theory-consequences-of-bott-weil-borel/120999#120999 Comment by Jean Delinez Jean Delinez 2013-02-06T18:40:39Z 2013-02-06T18:40:39Z Great answer! Thanks a lot. http://mathoverflow.net/questions/120901/group-representations-and-holomorphic-vectors-bundles-over-homogeneous-spaces-ex Comment by Jean Delinez Jean Delinez 2013-02-06T15:44:09Z 2013-02-06T15:44:09Z That seems to answer my question. Please enter it as an answer and so I can accept it. One more thing thing though: Is this an iff situation, ie if the complex structure is not$G$-invariant, then will it happen that the holomorphic sections no longer carry a representation of$G$? http://mathoverflow.net/questions/109392/jets-of-equivariant-vector-bundles/109630#109630 Comment by Jean Delinez Jean Delinez 2012-10-16T17:32:41Z 2012-10-16T17:32:41Z Yes, that fills the gap! http://mathoverflow.net/questions/109392/jets-of-equivariant-vector-bundles/109630#109630 Comment by Jean Delinez Jean Delinez 2012-10-16T13:00:14Z 2012-10-16T13:00:14Z So then what is the explicit representation of$H$on$\frak{U(g)}\otimes _{\frak{U(h)}}E^*$. I'm guessing that its the tensor product of the dual rep of$H$on$E$with some rep of$H$on$\frak{U(g)}$. http://mathoverflow.net/questions/103705/global-definition-of-the-dolbeault-complex-of-a-vector-bundle/103714#103714 Comment by Jean Delinez Jean Delinez 2012-08-01T21:00:40Z 2012-08-01T21:00:40Z So in the Kahler case all holomorphic structures on$E\$ induce the same set of holomorphic sections? http://mathoverflow.net/questions/97074/is-the-susy-algebra-isomorphic-for-all-kahler-manifolds Comment by Jean Delinez Jean Delinez 2012-05-16T03:16:36Z 2012-05-16T03:16:36Z ah ok - then that's good I suppose - thanks