User gunnar magnusson - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T03:19:05Z http://mathoverflow.net/feeds/user/4054 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107073/translation-of-kahlers-ber-eine-bemerkenswerte-hermitesche-metrik Translation of Kähler's "Über eine bemerkenswerte Hermitesche Metrik" Gunnar Magnusson 2012-09-13T09:37:23Z 2013-05-09T02:13:10Z <p>Has anyone translated Erich Kähler's "<a href="http://www.springerlink.com/content/521l6v51h370264m/" rel="nofollow">Über eine bemerkenswerte Hermitesche Metrik</a>" into English or French? (Preferably, but I'll take anything.)</p> http://mathoverflow.net/questions/57141/is-the-deformation-limit-of-ricci-flat-kahler-manifolds-kahler Is the deformation limit of Ricci-flat Kahler manifolds Kahler? Gunnar Magnusson 2011-03-02T17:42:19Z 2013-01-13T04:20:12Z <p>Let $X$ be a compact complex Kahler manifold with first real Chern class $c_1 = 0$. Consider a family $\pi : \mathcal X \to \Delta$ over the unit disc in $\mathbb C$, where the fibers $X_s$ are compact Kahler with $c_1(X_s) = 0$ for $s \not= 0$. Do we know that the central fiber $X_0$ is Kahler with $c_1 = 0$?</p> <p>Some remarks:</p> <p>a) The condition on the Chern class is topological, so the question is really if the central fiber is Kahler.</p> <p>b) This is true for complex tori and K3 surfaces, though for K3 surfaces being Kahler is a consequence of the topological condition of having even first Betti number.</p> <p>c) Kuranishi constructed an example of non-Kahler deformations of projective manifolds, but the manifolds in question were not Kahler-Einstein so that example does not apply here.</p> <p>d) There exist non-Kahler compact complex manifolds with $c_1 = 0$, like the Iwasawa manifold. The Iwasawa manifold does not have the right first Betti number to provide a counterexample to the question. However, I've heard physicists have found many examples of non-Kahler manifolds of Calabi-Yau type, and maybe one of those does at least not have topological obstructions to being a counterexample?</p> <p> Two more remarks:</p> <p>e) For the special case of Calabi-Yau manifolds, Popovici gives that the central fiber is Moishezon. One could hope that a Moishezon manifold with the Hodge numbers of a Calabi-Yau manifold is Kahler, but this is false by an example of Oguiso. His example is however not homeomorphic to a projective manifold, leaving the question open.</p> <p>f) One could try looking at Calabi-Yau threefolds, of which many examples are apparently known (I only know of complete intersections of appropriate degree in projective space). The case of a Calabi-Yau hypersurface in $\mathbb P^4$ is uninteresting, as they are rigid, so the central fiber is isomorphic to the general fiber.</p> http://mathoverflow.net/questions/118251/is-the-cup-product-of-holomorphic-n-forms-with-a-fixed-class-injective Is the cup product of holomorphic $n$-forms with a fixed class injective? Gunnar Magnusson 2013-01-07T08:42:04Z 2013-01-07T23:45:07Z <p>Let $X$ be a compact Kahler manifold of complex dimension $n$. Fix a nonzero class $u \in H^1(X,T_X)$. This gives a linear morphism $$\phi_u : H^0(X,\Omega^n) \to H^{1}(X,\Omega^{n-1}), \quad \sigma \mapsto u \cup \sigma.$$ Is $\phi_u$ injective?</p> <p>It is so for manifolds with $\Omega^n_X = \mathcal O_X$; the proof I've got is not hard but uses Ricci-flat Kahler metrics and the hard Lefschetz theorem so it cannot generalize to other situations. In the examples I know (curves, hypersurfaces in $\mathbb P^n$) we have $h^{n,0} \leq h^{n-1,1}$, so I haven't stumbled upon an obvious counterexample yet.</p> http://mathoverflow.net/questions/69792/can-a-metric-conformal-to-a-kahler-metric-be-kahler Can a metric conformal to a Kahler metric be Kahler? Gunnar Magnusson 2011-07-08T13:01:27Z 2013-01-05T05:53:29Z <p>Let $X$ be a non-compact complex manifold of dimension at least 2 equipped with a Kahler metric $\omega$. Take a smooth positive function $f : X \to \mathbb R$, and define a new hermitian metric on $X$ by $\tilde \omega = f \omega$. If $f$ is non-constant, then can this new metric ever be Kahler?</p> <p>If $\dim_{\mathbb C} X = 1$ the new metric is automatically Kahler because of dimension. If $\dim_{\mathbb C} X \geq 2$ and if $X$ is compact the new metric is never Kahler. Indeed, we have that $d \tilde \omega = d f \wedge \omega$ is zero if and only if $df$ is zero by the hard Lefschetz theorem, so $f$ must be constant if $\tilde \omega$ is Kahler.</p> <p>If $X$ is not compact, then to the best of my knowledge we do not have the hard Lefschetz theorem, but does the conclusion on metrics conformal to a Kahler metric still hold?</p> http://mathoverflow.net/questions/112627/why-is-gauss-credited-with-his-connection Why is Gauss credited with his connection? Gunnar Magnusson 2012-11-17T00:41:21Z 2012-11-17T13:05:19Z <p>Let $\pi : X \to B$ be a family of compact Kähler manifolds over a smooth base $B$. We then have a local system $\mathcal R^k \pi_* \mathbb Z$ (for your favorite $k$) of abelian groups over $B$, whose fiber over a point $b$ is the cohomology group $H^k(X_b, \mathbb Z)$.</p> <p>We can tensor this system by $\mathcal O_B$ and obtain a holomorphic vector bundle $E^k \to B$. This bundle is equipped with a flat connection $\nabla$, that is induced by the exterior derivative $d$ in local coordinates. This connection is called the Gauss-Manin connection of the bundle $E^k$.</p> <p>Now: Why do we call this the <em>Gauss</em>-Manin connection, when it seems that nothing that Gauss could have worked on relates to it?</p> http://mathoverflow.net/questions/112504/complex-manifold-with-non-finitely-generated-canonical-ring Complex manifold with non-finitely generated canonical ring Gunnar Magnusson 2012-11-15T17:41:00Z 2012-11-15T17:41:00Z <p>P.M.H. Wilson has an example of a compact non-Kahler manifold whose canonical ring is not finitely generated; see <a href="http://www.numdam.org/item?id=CM_1981__43_3_365_0" rel="nofollow">his article</a> and <a href="http://mathoverflow.net/questions/87860/example-of-a-compact-kahler-manifold-with-non-finitely-generated-canonical-ring" rel="nofollow">this MO question</a>. I'm trying to understand his construction and have trouble understanding how he conludes that his ring is not finitely generated.</p> <p>I'll briefly describe his construction. First, Wilson fabricates a projective surface $\widetilde{\mathbb{P}}$ with a divisor $D$ such that the ring $$R(\widetilde{\mathbb{P}} ,D) := \bigoplus_{m \geq 0} H^0(X,mD)$$ is not finitely generated by following ideas of Zariski: Let $C \subset \mathbb P^2$ be an elliptic curve and $H$ a line. Blow up 12 points in general position on $C$ and a point outside of $C$ to get $\widetilde{\mathbb{P}}$. Let $C'$ be the proper transform of $C$, $E$ the exceptional divisor of the point outside of $C$ and $H' = f^*H - E$. If $D := C' + H'$, then $R(\widetilde{\mathbb{P}},D)$ is not finitely generated.</p> <p>Second, Wilson makes a double cover $S$ of $\widetilde{\mathbb{P}}$ ramified over a general element of $|6C' + 6H'|$ and desingularizes to get $\alpha : \tilde S \to \widetilde{\mathbb{P}}$. Then $K_{\tilde S} \sim \alpha^*(2C' + 3H' + E)$.</p> <p>Third, he makes a nontrivial torus bundle $\pi : W \to \widetilde{\mathbb{P}}$, that is therefore non-Kahler. The fibers are of dimension 2, so $W$ is a fourfold. One remarks that $K_{W / \widetilde{\mathbb{P}}} = \pi^*\mathcal O(-2H')$.</p> <p>Fourth, take the fibered product $V = W \times_{\widetilde{\mathbb{P}}} \tilde S$ and let $g : V \to \widetilde{\mathbb{P}}$ be the induced morphism. As $\alpha : \tilde S \to \widetilde{\mathbb{P}}$ is finite of degree 2, this is a compact non-Kahler fourfold. We see that $K_V = g^* \mathcal O(2C' + H' + E)$. </p> <p>Until now, I'm mostly fine with his construction and can either see or am willing to take on faith how each step is necessary. But here Wilson claims that $R(V,K_V)$ is not finitely generated, and I don't see how. I assume there's some link between $R(V,K_V)$ and $R(\widetilde{\mathbb{P}},D)$, but it has escaped me so far. How do we see this?</p> http://mathoverflow.net/questions/53185/noncompact-kahler-manifolds-with-nonzero-ricci-tensor-but-vanishing-scalar-curvat/108321#108321 Answer by Gunnar Magnusson for Noncompact Kahler manifolds with nonzero Ricci tensor but vanishing scalar curvature Gunnar Magnusson 2012-09-28T07:08:47Z 2012-09-28T07:08:47Z <p>Dear Peter, I don't think one can say anything about such manifolds because the scalar curvature is too weak an invariant to be of use. Here is an infinite family of non-diffeomorphic compact examples to support my claim; for non-compact ones, remove a subvariety.</p> <p>Let $Y$ be a projective manifold of dimension $n$ with ample canonical bundle. By the Calabi-Yau theorem, $Y$ admits a Kahler-Einstein metric $\omega_Y$ with $Ric \omega_Y = - \omega_Y$. Recall that the projective space $\mathbb P^n$ admits the Fubini-Study metric $\omega_{FS}$ that has $Ric \omega_{FS} = \omega_{FS}$. We set $X = \mathbb P^n \times Y$ and equip this space with the product metric $\omega = \omega_{FS} \oplus \omega_Y$. (Here and everywhere we should write $pr_1^\ast\omega_{FS} \oplus pr_2^*\omega_Y$ for the appropriate projection maps.) By varying $Y$ among projective manifold with ample bundle (which are legion) we get non-diffeomorphic $X$.</p> <p><strong>Claim.</strong> The space $X$ has non-zero Ricci curvature but zero scalar curvature.</p> <p><em>Proof.</em> The dimension of $X$ is $2n$. We have $\omega^{2n} = \binom{2n}{n} \omega_{FS}^n \wedge \omega_Y^n$. A calculation in local coordinates then gives that $$Ric \omega = Ric \omega_{FS} + Ric \omega_Y = \omega_{FS} - \omega_Y \not= 0.$$ The scalar curvature $s$ of $\omega$ satisfies $$2n s dV = Ric \omega \wedge \omega^{2n-1} / (2n-1)!,$$ where $dV = \omega^{2n}/(2n)!$ is the volume form of $\omega$. Since $$\omega^{2n-1} = \binom{2n-1}{n} \bigl( \omega_{FS}^{n-1}\wedge \omega_Y^n + \omega_{FS}^n \wedge \omega_Y^{n-1} \bigr)$$ we get $$2n s dV = \frac{1}{(2n-1)!}\binom{2n-1}{n}\bigl( (2n)! dV - (2n)! dV\bigr) = 0,$$ whence $s = 0$.</p> http://mathoverflow.net/questions/107836/automorphism-group-of-ruled-surface Automorphism group of ruled surface Gunnar Magnusson 2012-09-22T13:14:21Z 2012-09-22T13:59:20Z <p>Let $C$ be an elliptic curve over the complex numbers. Consider a nontrivial extension $$0 \to \mathcal O_C \to E \to \mathcal O_C \to 0$$ of rank 2 of the structure sheaf of $C$. This defines a ruled surface $X = \mathbb P(E)$ over $C$.</p> <p>Is the automorphism group of $X$ transitive?</p> <p>I ask because I'm looking for a compact Kahler manifold $X$ with nef $-K_X$ whose automorphism group is not transitive. The Albanese variety of $X$ should also be nontrivial, ruling out the obvious Fano candidates. The ruled surface above has nef $-K_X$ and its Albanese variety is $C$ itself, so it almost fits the bill.</p> http://mathoverflow.net/questions/107064/kahler-manifold-with-ricci-flat-kahler-form/107066#107066 Answer by Gunnar Magnusson for Kähler manifold with Ricci-flat Kähler form Gunnar Magnusson 2012-09-13T06:49:18Z 2012-09-13T06:49:18Z <p>You need some more hypotheses for the existence of the $(n,0)$-form, in general it will exist only up to finite torsion. For example, an Enriques surface does not admit a nowhere zero holomorphic $(2,0)$-form. You can also construct explicit examples with products of two or three elliptic curves quotiented by a group generated by well chosen involutions.</p> <p>When such an $n$-form exists, the usual way to prove that it is parallel is to use a Bochner technique. This is sketched in the first pages of Beauville's <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1214438181" rel="nofollow">"Variétés Kahleriennes dont la premiere classe de Chern est nulle"</a> (the proof might work for the general case, i.e., for $(n,0)$-forms that have some zeros by using that the norm in the Bochner technique will not be zero at all points). If I recall correctly, a more complete version of the same argument is given in <a href="http://www.google.fr/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=4&amp;ved=0CDoQFjAD&amp;url=http%3A%2F%2Fcarlossicoli.free.fr%2FG%2FGross_M._Huybrechts_D.%2C_Joyce_D.-Calabi-Yau_Manifolds_and_Related_Geometries(2003).pdf&amp;ei=MIFRUIKPOMi90QWs_YCgCA&amp;usg=AFQjCNFXTtawQZcYLHXK0v4iHprjtN64tg&amp;sig2=MKSGMfInjBlr221Yt1MUZw&amp;cad=rja" rel="nofollow">"Calabi-Yau manifolds and their friends"</a> and in the opening chapters of <a href="http://www.google.fr/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=1&amp;cad=rja&amp;ved=0CCAQFjAA&amp;url=http%3A%2F%2Fmath.stanford.edu%2F~vakil%2Ffiles%2Fmirrorfinal.pdf&amp;ei=rYFRUJK9M-qw0QX8oIGQDg&amp;usg=AFQjCNFNtSEM20sKW-oD0BH4KOQHHSm6fA&amp;sig2=UZ5hLCAvX08K6xDi7ODA-Q" rel="nofollow">"Mirror symmetry"</a>.</p> http://mathoverflow.net/questions/104145/tangent-sheaf-of-a-hom-scheme Tangent sheaf of a hom scheme Gunnar Magnusson 2012-08-06T20:52:53Z 2012-08-07T12:57:32Z <p>I apologize if this question is too basic, but I haven't been able to work this out for myself.</p> <p>Let $X$ and $Y$ be projective schemes, say over the complex numbers. There exists a scheme $Hom(X,Y)$ parameterizing morphisms $f : X \to Y$. In Kollar's "Rational curves on algebraic varieties" it is proved that the stalk of the tangent sheaf of $Hom(X,Y)$ at a point $[f]$ is $$T_{Hom(X,Y),[f]} = H^0(X, Hom(f^*\Omega^1_Y, \mathcal O_X)).$$ I suppose that if $X$ and $Y$ are sufficiently nice, this coincides with $H^0(X,f^*T_Y)$, but that's not important for now.</p> <p>Can we obtain the tangent sheaf of $Hom(X,Y)$ "globally"? What I mean is, suppose we consider the evaluation morphism $ev : X \times Hom(X,Y) \to Y$ given by $(x,f) \mapsto f(x)$ and the projection $p : X \times Hom(X,Y) \to Hom(X,Y)$. The morphism $p$ is proper since $X$ is compact. Then the sheaf $$\mathcal T := p_{\ast} Hom({ev^\ast} \Omega^1_Y,\mathcal O_X)$$ over $Hom(X,Y)$ is coherent, since it is the direct image of a coherent sheaf on the product space. It also has the same stalks as the tangent sheaf of $Hom(X,Y)$. Is $\mathcal T$ the tangent sheaf of $Hom(X,Y)$?</p> http://mathoverflow.net/questions/31982/simple-kahler-manifolds "Simple" Kahler manifolds Gunnar Magnusson 2010-07-15T09:01:42Z 2012-07-24T20:15:57Z <p>I have some lecture notes from Demailly on Kahler geometry where he talks about "variétés Kahleriennes simples", which are defined as Kahler manifolds $X$ such that for very generic points $x_0$ in $X$ there exists no irreducible submanifold $Y$ of $X$ of dimension $0 &lt; \dim Y &lt; \dim X$ which contains $x_0$.</p> <p>Examples of these kinds of manifolds are very general complex tori and quotients thereof, and they're interesting because they give counterexamples to the Hodge conjecture in the analytic category.</p> <p>I thought I'd take a look at these things, but I can't find any mention of "simple Kahler manifolds" either here or on google. Did I get the name wrong? Do any of you know what I'm talking about and know of some references?</p> http://mathoverflow.net/questions/102681/ample-vector-bundles-on-complex-tori/102717#102717 Answer by Gunnar Magnusson for Ample vector bundles on complex tori Gunnar Magnusson 2012-07-20T06:04:27Z 2012-07-20T06:04:27Z <p>Dear Simone,</p> <p>This is just a comment.</p> <p>The answer to your question, if one exists, is surely that there exist universal polynomials $p_j^r$ ($1 \leq j \leq r = {\rm rank} E$) in the Chern classes of the vector bundle $E \to X$ such that $$\int_X p_j{}^r(c_1(E), \dots, c_j(E)) \wedge \omega^{n-j} > 0$$ for all $j$ implies that the bundle $E$ is ample. Here "universal" means that the polynomials in question only depend on the dimension $n$, but otherwise not on the variety $X$. For line bundles, these polynomials are known and are, as you wrote, $p_j{}^1(x) = x^j$ for all $j$.</p> <p>The problem is that the condition you wrote for line bundles is a corollary of a more general theorem of father, one that characterizes Kahler classes amongst real $(1,1)$-classes. Our hope of approaching the problem should thus be to find suitable positivity criterion for higher degree classes. The ideal outcome would be a higher rank version of the Kodaira condition; so we'd know that if a "degree vector" $(u_1, \dots,u_r)$ of integral cohomology classes satisfies some conditions, then there exists an ample vector bundle $E$ of rank $r$ such that $c_j(E) = u_j$ (compare with $L$ ample iff $c_1(L)$ Kahler and integral). The trouble is that finding these conditions amounts to finding the universal polynomials $p_j{}^r$, and thus answering a much more general question.</p> <p>In short, we have no idea what a "positivity condition" for a collection of cohomology classes $(u_1, \ldots, u_r)$ looks like. I agree with your approach of simplifying the problem and starting the search for these on complex tori. However, I think that if we knew the answer on complex tori, we'd know it on general complex manifolds too.</p> http://mathoverflow.net/questions/101644/fiction-books-about-mathematicians/101655#101655 Answer by Gunnar Magnusson for Fiction books about mathematicians? Gunnar Magnusson 2012-07-08T12:33:23Z 2012-07-08T12:33:23Z <p>You could try "Solar" by Ian McEwan. It's about a senior researcher that hasn't done good work in years but gets a break one day.</p> <p>Somewhat less seriously, there's also the delightfully named "Advanced calculus of murder" by Eric Rosenthal.</p> http://mathoverflow.net/questions/99581/recognizing-kahler-manifolds-of-complex-dimension-n/99593#99593 Answer by Gunnar Magnusson for recognizing Kahler manifolds of complex dimension n Gunnar Magnusson 2012-06-14T10:49:43Z 2012-06-14T10:49:43Z <p>I thought this had already been answered here on MO but my searches didn't turn anything up. It might be good to have a general purpose answer here somewhere.</p> <p>I'll restrict myself to compact manifolds, since the question is already complicated enough there!</p> <p>In complex dimension one every manifold is Kahler, simply because the exterior derivative of any 2-form is zero, so every hermitian metric is Kahler.</p> <p>In complex dimension two a surface is Kahler if and only if its first Betti number is even. This is known since the 80's by classification of surfaces and hard results of Siu, but modern proofs by Buchdahl and Lamari that do not make use of classification of surfaces appeared in 1999.</p> <p>The condition that the odd Betti numbers of the manifold be even for it to be Kahler is necessary by the Hodge decomposition theorem. Interestingly this is sufficient in dimension two, implying that the Kahler condition is topological there. This fails in higher dimensions, for there is an example due to Hironaka of Kahler manifolds of dimension 3 that can be deformed to a non-Kahler manifold. Since the underlying smooth manifolds of these complex manifolds are all diffeomorphic, a sufficient condition for being a Kahler manifold cannot be read off the topological or smooth structure of a given complex manifold.</p> <p>The list of necessary conditions that a manifold must satisfy to be Kahler is long and growing. First among these are the multiple conditions that the Hodge decomposition and the hard Lefschetz theorem impose on its cohomology ring. The properties of this ring are not yet well understood, considering how recently Voisin constructed examples demonstrating that there are Kahler manifolds whose homotopy type differs from that of projective manifolds.</p> <p>Another active area of research is on the fundamental groups of Kahler manifolds. I can't say I know much about this, but just to give an idea then examples of Serre show that any finite group can be the fundamental group of a Kahler manifold (projective surface, even). Finer results are available under additional hypotheses, for example Paun has shown that the fundamental group of a Kahler manifold with nef anticanonical bundle has polynomial growth.</p> <p>One can say more about necessary conditions for a manifold to be Kahler (strictly speaking any theorem that begins "Let $X$ be a compact Kahler manifold..." can be considered one!), but by contrast almost nothing is known about sufficient conditions in higher dimensions. With the examples of Hironaka in mind such conditions are likely to be extremely subtle, if they exist at all.</p> http://mathoverflow.net/questions/99436/calculating-a-second-fundamental-form-in-the-space-of-hermitian-metrics Calculating a second fundamental form in the space of hermitian metrics Gunnar Magnusson 2012-06-13T12:24:50Z 2012-06-13T12:24:50Z <p>Let $X$ be a compact Kahler manifold and let $\mathcal M$ denote the space of hermitian metrics on $X$. We'll identify a hermitian metric with a smooth, real and positive $(1,1)$-form $\omega$. Let $\mathcal K$ be the subspace of $\mathcal M$ defined by Kahler metrics, that is, those hermitian metrics $\omega$ such that $d \omega = 0$.</p> <p>It is possible to equip $\mathcal M$ with a Riemannian metric, for example by using the Hodge $L^2$ metric defined by $$G(U,V)_\omega = \int_X \langle U,V \rangle dV_\omega,$$ where the inner product inside the integral is the one defined by $\omega$ on $(1,1)$-forms on $X$. Here $U$ and $V$ are vectors tangent to $\mathcal M$, or in other words, they are smooth real $(1,1)$-forms on $X$. Despite some technical issues (the fibers of the tangent space are not complete) this metric admits a Levi-Civita connection, and a curvature tensor, and it induces a metric on $\mathcal K$ by restriction.</p> <p><strong>Q:</strong> How can we calculate the second fundamental form of $\mathcal K$ in $\mathcal M$?</p> <p>It is tempting to try to do this by considering the exterior derivative as a linear map from $\mathcal M$ to the space of 3-forms on $X$, and saying that $\mathcal K$ is its fiber over $0$. If we were talking about a smooth function $f : \mathcal M \to \mathbb R$, then the second fundamental form of the fiber $f^{-1}(0) \subset \mathcal M$ (assuming smoothness) would be given by the Hessian $-\nabla^2 f$ (see Lang's "Fundamentals of differential geometry, Prop. 2.1, p. 376). Is there a similar formula when the submanifold in question is defined by a map $f : \mathcal M \to \mathcal A$ where the target space is an infinite-dimensional manifold? </p> <p>An alternative approach would be to use the orthogonal projection onto the normal bundle of $\mathcal K$ in $\mathcal M$, but this projection is expressed using the Laplacian and Green operator associated to the metric $\omega$, so this road promises to be quite bumpy if at all usable. Any references or remarks would be greatly appreciated.</p> http://mathoverflow.net/questions/99292/question-regarding-riemann-hurwitz-formula-proof/99311#99311 Answer by Gunnar Magnusson for Question Regarding Riemann-Hurwitz Formula Proof Gunnar Magnusson 2012-06-11T18:40:04Z 2012-06-11T18:40:04Z <p>I'll try to sketch a proof of Riemann-Hurwitz using the Leray spectral sequence. It has the feel of a fun exercise.</p> <p>To fix notation, let $X$ and $Y$ be compact Riemann surfaces and let $f : X \to Y$ be a finite surjective morphism of degree $d$. We want to compare the topological Euler numbers of these surfaces; these are the numbers defined as $$\chi(X) = h^0(X,\mathbb C) - h^1(X,\mathbb C) + h^2(X,\mathbb C)$$ and similarly for $Y$. We'll use arbitrarily fancy facts of sheaf cohomology to do this.</p> <p>If $\mathcal F$ is a sheaf on $X$, then the first terms of the Leray spectral sequence read $$E_2^{p,q} = H^q(Y, \mathcal R^p f_* \mathcal F) \Rightarrow H^{p+q}(X,\mathcal F).$$ As the fibers of $f$ are 0-dimensional, we have $\mathcal R^p f_* \mathcal F = 0$ for any $p \geq 1$. Combined with the annihilation of cohomology on $Y$ for dimension reasons, we find that $E^{p,q}_2 = 0$ for any $p \geq 1$ and $q \geq 3$. The second page of the Leray spectral sequence is thus just $$E_2^{0,0} \qquad E_2^{0,1} \qquad E_2^{0,2}$$ and all other entries are zero, so the sequence degenerates at the $E_2$-level. It follows that $H^k(Y,f_*\mathcal F) = H^k(X,\mathcal F)$ for any $k$.</p> <p>Consider now a point $y$ on $Y$ that is not in the image of the ramification locus of $f$, in other words the preimage $f^{-1}(y)$ consists of $d$ distinct points. Then we see that $f_{\ast} {\mathbb C} = {\mathbb C}^{\oplus d}$. This line of though yields a short exact sequence $$0 \longrightarrow f_* \mathbb C \longrightarrow \mathbb C^{\oplus d} \longrightarrow \mathcal G \longrightarrow 0$$ where $\mathcal G$ is a skyskraper sheaf supported on the image of the ramification divisor of the morphism $f$. Taking Euler characteristics we get $$d\,\chi(Y) = \chi(f_*\mathbb C) + \chi(\mathcal G) = \chi(X) + h^0(Y,\mathcal G).$$ Expressing $h^0(Y,\mathcal G)$ in terms of the degrees of $f$ at its ramification points, and thus showing that it has the expected form, should not be a source of great trouble.</p> <p>The thing that makes this proof relatively painless is that the Leray spectral sequence degenerates straight away (at least without recourse to heavy machinery) and that calculating the cohomology of a sheaf supported on a finite number of points is easy. The spectral sequence will again degenerate at the $E_2$-level in the case of a morphism between surfaces, but there a finer analysis is needed to calculate the cohomology of the corresponding sheaf $\mathcal G$. In any case the proof points the way to a similar statement for finite surjective morphisms between higher dimensional varieties, though it also seems to indicate that this is not a path one wants to take unless one really needs to.</p> http://mathoverflow.net/questions/97061/adjoint-of-a-connection-using-the-hodge-map/97095#97095 Answer by Gunnar Magnusson for Adjoint of a Connection Using the Hodge Map? Gunnar Magnusson 2012-05-16T08:10:37Z 2012-05-16T08:22:02Z <p>Unfortunately this does not work, because the Hodge $*$ operator commutes with the Levi-Civita connection. Indeed, we have $$\nabla (\langle u,v \rangle dV) = \langle \nabla u, v \rangle dV + (-1)^m \langle u, \nabla v \rangle dV = \nabla u \wedge * v + (-1)^m u \wedge *\nabla v$$ because $\nabla dV = 0$ since the metric $g$ is parallel with respect to $\nabla$. The left hand side of this formula is again $$\nabla (u \wedge * v) = \nabla u \wedge *v + (-1)^m u \wedge \nabla(*v),$$ from which we get $\nabla * = * \nabla$. It follows that $* \nabla * = ** \nabla = (-1)^l \nabla$ for some $l$.</p> <p>There is an expression for $\nabla^*$ in terms of a local orthonormal frame in Werner Ballmann's <a href="http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/kaehler0609.pdf" rel="nofollow">Lectures on Kahler manifolds</a> (Proposition 1.27, Chapter 1, p. 11) that says that if $(X_1, \ldots, X_n)$ is such a frame, and if $\hat\nabla$ is the dual connection, then $$\nabla^*u = - \sum_j X_j \llcorner \hat\nabla_{X_j} u.$$ I don't know if this is what you're looking for, if not you might have more luck with Bocher-Weitzenböck type identities.</p> http://mathoverflow.net/questions/96118/what-the-kahler-cone-of-one-point-blow-up-of-mathbbcpn-looks-like/96119#96119 Answer by Gunnar Magnusson for what the Kahler cone of one point blow-up of $\mathbb{C}P^n$ looks like? Gunnar Magnusson 2012-05-06T08:47:39Z 2012-05-06T08:47:39Z <p>The Kahler cone of any compact manifold is described by a theorem of <a href="http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/kahlercone.pdf" rel="nofollow">Demailly and Paun</a>. If $X$ is a compact Kahler manifold, then its Kahler cone is one of the connected components of the set $$\mathcal P = \lbrace \alpha \in H^{1,1}(X,\mathbb R) \mid \int_Z \alpha^p > 0 \rbrace$$ where $Z$ runs through all the $p$-dimensional closed complex subspaces of $X$. If $X$ is projective, then the Kahler cone is actually this set.</p> <p>Since the blowup of a projective variety in a point is projective, and the cohomology ring of $\mathbb P^n$ blown up in a point is pretty explicit, this lets us calculate the Kahler cone. Indeed, by some fun manipulations one gets $$\mathcal P \simeq \lbrace (a,b) \in \mathbb R^2 \mid a > 0, \quad b > 0, \quad a > b \rbrace$$ where $(a,b) \mapsto aH - bE$. Here $H$ is the divisor of a general hyperplane in $\mathbb P^n$, pulled back to the blowup, and $E$ is the exceptional divisor of the blowup. You seem to be missing the $a, b > 0$ conditions, since $aH - bE > 0$ is equivalent to $a^n - b^n = (a-b) \cdot (a^{n-1} + \ldots + b^{n-1}) > 0$, and one can fulfill this condition with zero or negative $a$ or $b$, which would place us outside of the nef, or even pseudoeffective, cones.</p> http://mathoverflow.net/questions/94180/are-there-hermitian-metrics-with-the-volume-form-of-a-kahler-metric Are there hermitian metrics with the volume form of a Kahler metric? Gunnar Magnusson 2012-04-16T05:24:29Z 2012-04-16T10:51:12Z <p>Let $X$ be a compact Kahler manifold of complex dimension $n$. The Aubin--Calabi--Yau theorem says that if we fix a smooth form $\rho$ in the Chern class $c_1(X)$, then every Kahler class on $X$ contains a unique Kahler metric $\omega$ whose Ricci-form is $\rho$. Alternatively, one may fix a volume form $dV$ on $X$, then the theorem gives the existence of a unique metric $\omega$ in each Kahler class whose volume form is a constant multiple of $dV$, or $dV_\omega = c dV$ where $c > 0$ is a constant:</p> <p>Indeed, if we have $\rho$, let $dV = dV_\omega$ for any Kahler metric $\omega$ whose Ricci-form is $\rho$. If we have $dV$, consider the smooth hermitian metric $h$ on the canonical bundle $K_X$ defined by the equality $i^{n^2} \alpha \wedge \overline \beta = h(\alpha,\overline \beta) dV$, and take $\rho$ to be its curvature form.</p> <p>Since there are at least three ways to define the Ricci tensor of a hermitian metric, but the volume form of any hermitian metric $\omega$ is $dV_\omega = \omega^n/n!$, we'll fix a volume form $dV$ such that $Vol(X,dV) = 1$.</p> <p><strong>Question:</strong> The ACY theorem gives Kahler metrics $\omega$ with $dV_\omega = dV$. Can there be non Kahler metrics on $X$ whose volume form is $dV$?</p> http://mathoverflow.net/questions/94180/are-there-hermitian-metrics-with-the-volume-form-of-a-kahler-metric/94189#94189 Answer by Gunnar Magnusson for Are there hermitian metrics with the volume form of a Kahler metric? Gunnar Magnusson 2012-04-16T07:45:20Z 2012-04-16T07:45:20Z <p>I really should think more about these things before asking. The answer is "yes".</p> <p>K. Yang considers the flag manifold $F := F_{1,2,3} := SU(3)/S(U(1)^3)$ in <a href="http://journals.cambridge.org/action/displayAbstract?fromPage=online&amp;aid=4787044" rel="nofollow">Invariant Kahler metrics and projective embeddings of the flag manifolds</a>. He shows that the space of hermitian metrics on $F$ is parametrized by $(\mathbb R_+)^3$, that is, any hermitian metric on $F$ is given by $$h_{a,b,c} = a^2 \theta_1\otimes\overline\theta_1 + b^2 \theta_2\otimes\overline\theta_2 + b^2 \theta_3\otimes\overline\theta_3$$ where $\theta_j$ are holomorphic 1-forms on $F$ and $a$, $b$ and $c$ are positive constants. He also shows that if $h$ is a Kahler metric on $F$, then $h$ is a constant multiple of $h_{1,\sqrt 2,1}$.</p> <p>Note that the volume form of $h_{a,b,c}$ is $(abc)^2 dV$, where $dV$ is the volume form obtained by wedging the $(1,1)$-forms $(i/2)\theta_j \wedge \overline\theta_j$ together. A Kahler metric $h = \lambda h_{1,\sqrt 2,1}$ thus has the volume form $dV_h = 2\lambda^2 dV$ and we can find lots of hermitian metric $h_{a,b,c}$ with that same volume form.</p> http://mathoverflow.net/questions/94023/example-of-a-variety-with-explicit-cohomology-ring-and-kahler-cone Example of a variety with explicit cohomology ring and Kahler cone Gunnar Magnusson 2012-04-14T11:08:40Z 2012-04-14T21:15:57Z <p>I'm looking for some fairly explicit varieties to use as (counter?-)examples for my thesis and I'd appreciate any suggestions. I need a smooth projective variety $X$ of general type that satisfies:</p> <ul> <li>The Hodge number $h^{1,1}$ is at least 2.</li> <li>The cohomology ring of $X$ (or at least the subring generated by $(1,1)$-classes) is explicit.</li> <li>The Kahler cone of $X$ is known(-ish).</li> </ul> <p>I want to calculate the sectional curvatures of the Riemannian metric on the Kahler cone of $X$ which is defined by the intersection product. These curvatures may be expressed by the intersection product on the subring $A$ of $H^*(X)$ genereated by degree $(1,1)$ classes, which explains the conditions I impose.</p> <p>The condition on the Hodge number is necessary, since when $h^{1,1} = 1$ one ends up with the metric $g(x,y)(t) = xy/t$ on the half-line $\mathbb R_+$ and not many interesting things remain unsaid about this case. This excludes most hypersurfaces in $\mathbb P^n$, except perhaps for those in $\mathbb P^3$. </p> <p>I must also exclude the example of a blowup of several points of a variety $X$ with $h^{1,1} = 1$, since one can calculate explicitly what happens in this case. Are there other relatively easy examples?</p> http://mathoverflow.net/questions/93556/finding-topological-obstructions-for-a-complex-manifold-to-be-kaehler/93563#93563 Answer by Gunnar Magnusson for Finding topological obstructions for a complex manifold to be Kaehler Gunnar Magnusson 2012-04-09T11:19:37Z 2012-04-09T11:19:37Z <p>The cohomology ring of $X$ is probably a good place to look for candidates for such invariants. For example, let $$B = \lbrace \alpha \in H^{1,1}(X,\mathbb R) \mid \text{Vol}(X,\alpha) := \int_X \alpha^n/n! > 0 \rbrace$$ be the big cone of $X$. If $X$ is Kahler, then it contains the Kahler cone $K$ of $X$, but it is in general larger.</p> <p>The big cone admits a (in general only pseudo-)Riemannian metric $g$, given by the Hessian of the smooth function $- \log \text{Vol}$. Conjecturally, the sectional curvature of this metric is seminegative if $X$ is Kahler (see Wilson's <a href="http://arxiv.org/abs/math/0307260" rel="nofollow">http://arxiv.org/abs/math/0307260</a> for the first version of this question).</p> <p>So, suppose we have a compact manifold $X$ and that we know its cohomology ring and its intersection product. Then we can calculate the big cone and the Riemannian metric $g$ and check if its sectional curvature is seminegative. If not, then $X$ should not be Kahler.</p> <p>Of course this may not be so easy to check in practice, but Wilson's article contains an interesting prototype of an example (Propositon 5.3) where one should obtain the non-existence of a certain type of Kahler structure on a differentiable manifold through these means, that one does not get by the more traditional invariants.</p> http://mathoverflow.net/questions/79383/calculating-a-curvature-tensor-by-polarization Calculating a curvature tensor by polarization Gunnar Magnusson 2011-10-28T12:50:00Z 2012-01-28T23:54:51Z <p>I'm reading some articles by Siu and Nannicini on the Weil-Petersson metric associated to families of compact Kahler-Einstein manifolds. In each article Siu and Nannicini construct a Weil-Petersson metric $h$, show that it is Kahler, and obtain results on the holomorphic sectional curvature by heroic calculations.</p> <p>From what I can piece together, the curvature estimates go as follows: 1) show that the holomorphic sectional curvature, given by $R_{jjjj}$, is negative, 2) use a <em>polarization trick</em> to calculate the general tensor $R_{jklm}$, 3) then a miracle occurs, 4) so the holomorphic sectional curvature $R_{jjkk}$ is negative.</p> <p>I'm trying to fill in the gaps in my understanding between the first and fourth steps, and I'm stuck on the second one. For Kahler manifolds the holomorphic sectional curvature determines the entire curvature tensor (see for example Lemma 7.19 of Zheng's "Complex differential geometry"), so I'm perfectly willing to believe that knowing $R_{jjjj}$ lets us calculate $R_{jklm}$. The problem is that I don't know how to do it.</p> <p>This is a purely algebraic calculation, so I imagine it's written down somewhere, but the only results I've found are of the same type as in Zheng's book, i.e. they show that these calculations are theoretically possible but don't say how to do them.</p> <p><strong>Question:</strong> Is there a reference where this calculation is done explicitly?</p> <p> As Deane pointed out in the comments, one needs to know $R(X,\bar X,X,\bar X)$ for all holomorphic tangent fields $X$ to know the curvature tensor, just knowing $R_{j\bar jj\bar j}$ doesn't cut it. This makes two phrases from Siu's and Nannicini's papers a bit mysterious:</p> <p>Siu: "We now polarize the expression for $R_{i\bar i i\bar i}$ to get the expression for $R_{i \bar j k \bar l}$." in "Curvature of the Weil-Petersson metric in the moduli space of compact Kahler-Einstein manifolds of negative first Chern class" (beginning of paragraph 5.4).</p> <p>Nannicini: "The complete expression for the Riemann tensor can now be obtained by polarization of $R_{i\bar ii\bar i}$" in "Weil-Petersson metric in the moduli space of compact polarized Kahler-Einstein manifolds of zero first Chern class" (the page before Theorem 1).</p> <p><strong>Revised question:</strong> What exactly are Siu and Nannicini doing, if not applying the lemma on the holomorphic sectional curvature?</p> http://mathoverflow.net/questions/52960/which-almost-complex-manifolds-admit-a-complex-structure Which almost complex manifolds admit a complex structure? Gunnar Magnusson 2011-01-23T14:37:31Z 2012-01-27T10:19:40Z <p>I was reading Yau's list of problems in geometry, and one of them is to prove that any almost complex manifold of complex dimension $n \geq 3$ admits a complex structure. It's been some time since Yau's list was published, so what is the status of this problem today?</p> <p>Obviously it isn't hasn't been shown to be true, because we're still looking for complex structures on the six-sphere, but I have a vague feeling of having read that this doesn't hold. So do we know any counterexamples to this question? If not, then is anyone working on this problem?</p> <p>Also, Yau only stated the problem for manifolds of dimension $n \geq 3$. We know this is true in dimension one, because there we have isothermal coordinates which give complex structures, but why didn't Yau mention almost complex surfaces? Do we know this holds there, or are there counterexamples in dimension 2?</p> http://mathoverflow.net/questions/86555/how-many-points-determine-a-line How many points determine a line? Gunnar Magnusson 2012-01-24T18:11:12Z 2012-01-24T22:13:22Z <p>Consider the affine space $\mathbb C^n$ and then, because of reasons, compactify it to obtain the projective space $\mathbb P^n$. One of the most basic axioms or propositions of geometry is that through two distinct points in $\mathbb P^n$ there passes a single projective line.</p> <p>Somehow this seems to be very much a fact of projective space. Thus it seems natural to ask how this fact changes when we consider a different ambient space.</p> <p>So, for simplicity, let's consider a compact complex surface $X$. Again, for simplicity, assume that it admits some rational curves<strike>, of degree 1 even</strike>. How many of those pass through two (or more?) points of $X$? (We do not assume that a line passes through any two points. We just want to know that given two points that lie on a line, could another line pass through those points?)</p> <p>From what I understand, basic intersection theory was developed to answer exactly these problems, so one should be able to obtain the answer by intersecting relevant Chern classes on the right vector bundles. Any reasonable attempt at this kind of answer leads to the badly understood problem of calculating the Chow groups and intersection products of surfaces or manifolds. Thus we arrive at a perhaps answerable</p> <p><strong>Question:</strong> Is there an example of a surface (or manifold) $X$ on which we know that more than one projective line passes through two given points? In that case, do we know how many points are necessary to determine a line?</p> http://mathoverflow.net/questions/81235/wanted-an-example-of-a-natural-non-k-ahler-metric-on-a-kahler-manifold Wanted: an example of a natural non-K\"ahler metric on a Kahler manifold Gunnar Magnusson 2011-11-18T08:51:15Z 2011-11-21T08:41:43Z <p>Let $X$ be a Kahler manifold. Associated to any hermitian metric $h$ on $X$ is a smooth real $(1,1)$-form $\omega = -\text{Im } h$, called the Kahler form of $h$. One of several equivalent conditions for the metric $h$ to be a Kahler metric is that $\omega$ is symplectic, or that $\text{d} \omega = 0$.</p> <p>Morally speaking, this implies that Kahler metrics are rare. In a sense, they are contained in a proper subspace of the cone of hermitian metrics on $X$.</p> <p><strong>Question:</strong> Does anyone know an example of a manifold $X$ and a natural metric $h$ on $X$ which is not Kahler?</p> <p><strong>For extra credit:</strong> Can we find such an example where $X$ is compact?</p> <p>I'll elaborate a bit on what I mean by "natural" and then provide some motivation for the questions. The following paragraphs are not meant to be mathematically rigorous, but rather heuristic, so please be gentle when you see any inaccuracies.</p> <p><em>"Naturality":</em> It is easy to give explicit examples of non-Kahler metrics on a Kahler manifold. Just take any Kahler metric $h$ and multiply it by a positive non-constant function $f$: the Kahler form of the new metric will then satisfy $\text{d} (f \omega) = \text{d} f \wedge \omega + f \text{d} \omega = \text{d} f \wedge \omega$. As $\omega$ is symplectic the wedge product $\text d f \wedge \omega$ can only be zero if $\text d f$ is zero, so the new metric is not Kahler.</p> <p>This feels like cheating to me. It's like starting a book on linear algebra by defining vector spaces axiomatically and then only giving the trivial space as an example. The example does not advance our understanding in any significant way.</p> <p>I would like to see an example where the metric $h$ arises in a geometric way or is somehow an obvious candidate for a metric on $X$. For example, consider a Hopf surface $X$, which arises as a quotient of $\mathbb C^2 \setminus {0}$ by a group $G$. The naive way to give an example of a metric on $X$ is to find a metric on $\mathbb C^2 \setminus {0}$ which is invariant under the action of $G$, and it is perfectly possible to give an explicit example of such a metric by some calculations (see [1] for an example). If only the Hopf surface were Kahler I would accept this as a "real" example.</p> <p><em>Motivation:</em> Given a hermitian metric $h$ there are several equivalent definitions of it being Kahler. One can say that its Kahler form is closed, that one can approximate the euclidean metric to the second degree in local coordinates, or that the Chern and Levi-Civita connections of $h$ are the same. This last condition is the one I like the most, because with good will one can interpret it as saying that the complex and Riemannian geometries defined by the metric are the same.</p> <p>This is all well and good, and I feel I understand the different definitions and the links between them. However, given an explicit metric, I have absolutely no intuition for if it is Kahler or not. I can't look at a metric and just go "Aha!", I have to "fly blind" and calculate.</p> <p>For example, take the Fubini-Study metric on $\mathbb P^n$. It can be obtained by considering a scalar product on $\mathbb C^{n+1}$ and saying that the scalar product of lines in that space is the "angle" between the lines (-ish). This is a very pretty and geometric way of obtaining a metric. Now, the only way I know to show that the metric obtained in this beautiful way is Kahler is by long and violent calculations. I can't give you an a priori plausibility argument for it being Kahler. The same is true for any explicit example of a Kahler metric on any manifold.</p> <p>I see this as failure on my part, and a sign that I have not really understood Kahler metrics. I think that an explicit example of a natural non-Kahler metric would help me understand complex geometry better.</p> <hr> <p>[1] <a href="http://mathoverflow.net/questions/57535/examples-of-non-kahler-surfaces-with-explicit-non-kahler-metric" rel="nofollow">http://mathoverflow.net/questions/57535/examples-of-non-kahler-surfaces-with-explicit-non-kahler-metric</a></p> http://mathoverflow.net/questions/81235/wanted-an-example-of-a-natural-non-k-ahler-metric-on-a-kahler-manifold/81485#81485 Answer by Gunnar Magnusson for Wanted: an example of a natural non-K\"ahler metric on a Kahler manifold Gunnar Magnusson 2011-11-21T08:41:43Z 2011-11-21T08:41:43Z <p>Thanks Claudio, Francesco and José for your interesting answers.</p> <p>This is an "answer in absence" of Demailly; he doesn't use this site, but I thought his remark was nice enough to share. What follows is only his sketch, there are details to fill in that I haven't yet had the time to take care of.</p> <p>Let $X$ be a smooth projective variety, embedded in $\mathbb P^N$. Take $k$ big enough so that the vector bundle $T_X \otimes \mathcal O(k)$ is generated by its global sections. The Fubini-Study metric on $\mathbb P^N$ now gives a hermitian metric on $X$ by restriction, and on $\mathcal O(k)$ by taking powers of the determinant metric. These induce an $L^2$ metric on the global sections of $T_X \otimes \mathcal O(k)$.</p> <p>If we tensor by $\mathcal O(-k)$, then we have a surjective bundle map</p> <p>$$H^0(X, T_X \otimes \mathcal O(k)) \otimes \mathcal O(-k) \to T_X \to 0.$$</p> <p>The $L^2$ metric and the metric induced by Fubini-Study on $\mathcal O(-k)$ now gives a metric on the tensor product on the left. This induces a quotient metric $h$ on $T_X$. Despite its algebraic origins, this metric $h$ should (almost) never be a Kahler metric.</p> <p>Moreover, by applying approximation theorems of Tian, Demailly and others, one should be able to prove that these non-Kahler metrics are dense in the cone of hermitian metrics on $X$ -- i.e. starting from any metric on $X$ and using that to define the $L^2$ metric, it should be possible to fabricate a series of non-Kahler metrics as above which converges to the given metric. The process should in fact generalize and yield similar metrics on any holomorphic vector bundle over a projective manifold.</p> <p>This should imply that any heuristic of the form "my metric is geometric/algebraic, thus Kahler" is doomed to failure.</p> http://mathoverflow.net/questions/81161/software-for-calculating-products-and-sums-of-kronecker-deltas Software for calculating products and sums of Kronecker deltas Gunnar Magnusson 2011-11-17T10:14:50Z 2011-11-17T10:49:52Z <p>I am looking at a Kahler metric $g$ on a certain manifold $M$, which has the good taste to be invariant under a transitive group of isometries, and I want to say something about its holomorphic sectional curvature.</p> <p>Now, I can calculate the curvature tensor $R$ of $g$ explicitly at the center of a holomorphic coordinate system. If $(X_1, \ldots, X_n)$ is an orthonormal frame of tangent vectors at a given point, then the entries of the tensor are of the form</p> <p>$$R_{j \overline k l \overline m} := R(X_j, \overline X_k, X_l, \overline X_m) = \text{term 1} \cdot \text{term 2} - \text{term 3}$$</p> <p>where the terms are sums of Kronecker deltas in $j$, $k$, $l$ and $m$. So if $X = \sum_j a_j X_j$ is a holomorphic tangent vector, which we may assume is of unit norm, the holomorphic sectional curvature in the direction of $X$ is</p> <p>$$H(X) = \sum_{j,k,l,m} a_j \overline a_k a_l \overline a_m R_{j \overline k l \overline m}.$$</p> <p>This is where the pain begins. So far I haven't been able to make any sense of the herd of Kronecker deltas which comes out (except in exceptional cases, like for $R_{j \overline j k \overline k}$), and I'd be quite happy if I could just let a computer work the damn thing out for me.</p> <p><strong>Question:</strong> Is there software that calculates this sort of thing? Or is there a language particularily well adapted to hacking out a script that will calculate this?</p> http://mathoverflow.net/questions/78255/finding-the-action-of-the-symplectic-group-on-the-siegel-half-plane Finding the action of the symplectic group on the Siegel-half plane Gunnar Magnusson 2011-10-16T08:01:28Z 2011-10-16T15:58:03Z <p>Let $S$ be the Siegel-half plane of dimension $n$, i.e. the set of complex $n \times n$ matrices $Z$ which are symmetric and whose imaginary part is positive-definite. In dimension 1 we can identify $S$ with the Poincaré half-plane.</p> <p>In dimension 1 there is an action of the symplectic group $Sp_2(\mathbb R)$ on $S$, given by fractional linear transformations. One can extend this action to higher dimensions; we write an element $M$ of the symplectic group $Sp_{2n}(\mathbb R)$ as a block matrix with blocks $A$, $B$, $C$, $D$, and then set</p> <p>$$M \cdot Z = (AZ + B)(CZ + D)^{-1}.$$</p> <p>Now, in dimension 1 one sees that this action is induced by the action of the general linear group on the projective line and the embedding $GL_2(\mathbb R) \hookrightarrow GL_2(\mathbb C)$. However, this does not seem to be true in higher dimensions, as $GL_{2n}(\mathbb C)$ acts on the projective space of dimension $2n$, and not on the space of dimension $n$.</p> <p>The definition of the action of $Sp_{2n}(\mathbb R)$ on $S$ in higher dimensions thus seems very ad-hoc, which motivates:</p> <p><strong>Question:</strong> How can one find this action naturally in higher dimensions?</p> <p>[Edit:] <strong>Question 2:</strong> David's answer is very nice, but I'd like to iterate the question before accepting it. Instead of the Siegel half-plane above, consider the space $U$ of complex $(1,1)$-forms whose imaginary part is positive-definite. In a basis we can identify this with the set of $n \times n$ matrices $Z$ whose antihermitian part is positive-definite. The action of the symplectic group now makes sense on this space as well.</p> <p>In a basis the Siegel half-plane is a closed subspace of $U$ if $n > 1$. It is easiest to define the action of the symplectic group in a basis, but we can define it without reference to a basis by choosing a hermitian inner product on our vector space. Now, can we find the action of the symplectic group on $U$ in the same fashion as on the Siegel half-plane?</p> http://mathoverflow.net/questions/77688/is-a-certain-composition-of-harmonic-forms-again-harmonic Is a certain composition of harmonic forms again harmonic? Gunnar Magnusson 2011-10-10T12:26:54Z 2011-10-10T16:56:15Z <p>Let $(X,\omega)$ be a compact Kahler manifold, and let $\alpha$ and $\beta$ be smooth $(1,1)$-forms on $X$ that are harmonic (with respect to $\omega$). I can consider each of my $(1,1)$-forms as an antilinear vector bundle morphism $T_X \to \overline T_X^*$. Then I can fabricate a new $(1,1)$-form on $X$ by setting $F = \alpha \circ \omega^{-1} \circ \overline \beta$.</p> <p><strong>Question:</strong> Is the form $F$ harmonic?</p> <p>I was hoping there was some general theory available to answer this quickly, but I haven't found anything. Calculations in local coordinates also quickly degenerated into filth.</p> <p>For motivation, if you take $Tr_\omega(F)$ then you get the value of the scalar product on $(1,1)$-forms induced by $\omega$ of $\alpha$ and $\beta$. Thus knowing that $F$ is harmonic lets one conclude that the map $x \mapsto \langle \alpha(x),\beta(x) \rangle_\omega$ is constant.</p> http://mathoverflow.net/questions/119940/tensor-contraction-and-covariant-derivative/119959#119959 Comment by Gunnar Magnusson Gunnar Magnusson 2013-01-26T20:43:48Z 2013-01-26T20:43:48Z It's such a pleasure to read your answers. http://mathoverflow.net/questions/118314/trace-formula-for-psdos Comment by Gunnar Magnusson Gunnar Magnusson 2013-01-16T12:44:21Z 2013-01-16T12:44:21Z +1 for the &quot;Pseudo-differential-opera&quot; tag. http://mathoverflow.net/questions/118251/is-the-cup-product-of-holomorphic-n-forms-with-a-fixed-class-injective/118256#118256 Comment by Gunnar Magnusson Gunnar Magnusson 2013-01-07T09:25:46Z 2013-01-07T09:25:46Z I'd have upvoted just for &quot;$10^{100}$ points (just to be safe)&quot; alone. :D http://mathoverflow.net/questions/19632/question-about-kahler-cone-of-a-compact-kahler-manifold/20256#20256 Comment by Gunnar Magnusson Gunnar Magnusson 2013-01-02T13:52:36Z 2013-01-02T13:52:36Z @Tim: It's not that counterintuative; consider an open cone in $\mathbb R^2$ defined by two rays that form an angle $\theta &lt; \pi /2$. Picking two vectors in the cone at random gives a basis of $\mathbb R^2$, but the basis will never be orthonormal. One such example is (pretty much) the Kahler cone of the product of two curves. http://mathoverflow.net/questions/19632/question-about-kahler-cone-of-a-compact-kahler-manifold/20256#20256 Comment by Gunnar Magnusson Gunnar Magnusson 2013-01-02T13:48:23Z 2013-01-02T13:48:23Z @Tim: No, two Kahler classes are never orthogonal to each other on any compact manifold. You can see this by representing both by smooth forms and calculating their scalar product in coordinates that diagonalize both forms at a point. What comes out will necessarily be positive, so its integral over the manifold (ie the scalar product of the Kahler classes) will be positive. http://mathoverflow.net/questions/115001/open-problems-in-birational-geometry-after-bchm Comment by Gunnar Magnusson Gunnar Magnusson 2012-11-30T18:06:22Z 2012-11-30T18:06:22Z Finite generation of the canonical ring of a compact Kahler manifold is still open (you know, if you believe in things like Kahler manifolds). http://mathoverflow.net/questions/113924/line-bundle-on-s2 Comment by Gunnar Magnusson Gunnar Magnusson 2012-11-20T14:27:09Z 2012-11-20T14:27:09Z Huh. And yet there are nontrivial vector bundles on $S^$. http://mathoverflow.net/questions/112627/why-is-gauss-credited-with-his-connection Comment by Gunnar Magnusson Gunnar Magnusson 2012-11-17T08:29:12Z 2012-11-17T08:29:12Z @Gerard: Sure, Gauss was a hipster of differential geometry, but it seems a long way between theorema egorium and flat connections on bundles arising from local systems. :) http://mathoverflow.net/questions/57072/a-remark-of-connes Comment by Gunnar Magnusson Gunnar Magnusson 2012-11-16T14:02:46Z 2012-11-16T14:02:46Z @Yemon: Well played, sir. http://mathoverflow.net/questions/57072/a-remark-of-connes Comment by Gunnar Magnusson Gunnar Magnusson 2012-11-15T16:04:09Z 2012-11-15T16:04:09Z Ah, c'est vraiment une remarque conne ! (I'm sorryyyyy! It had to be done.) http://mathoverflow.net/questions/111618/rotation-in-hyperkahler-manifolds Comment by Gunnar Magnusson Gunnar Magnusson 2012-11-06T09:29:55Z 2012-11-06T09:29:55Z I suppose it depends on how one interprets the question. The hyperkahler manifold comes equipped with a hyperkahler metric $g$. Do you want $J$ to be compatible with $g$ or not? http://mathoverflow.net/questions/109987/is-a-simply-connected-ricci-flat-kaehler-manifold-a-calabi-yau-manifold Comment by Gunnar Magnusson Gunnar Magnusson 2012-10-18T09:15:53Z 2012-10-18T09:15:53Z Dear Jos&#233;, don't we need simple connectedness to know that the reduced holonomy group is the entire holonomy group for this argument to work? I think a finite quotient of a simply connected C-Y manifold may not have any nonzero holomorphic $(n,0)$-form, while still admitting Ricci-flat metrics. http://mathoverflow.net/questions/107901/calabi-yau-and-ricci-flat-metrics Comment by Gunnar Magnusson Gunnar Magnusson 2012-09-24T14:03:46Z 2012-09-24T14:03:46Z For extra credit, figure out why this doesn't work in general: Locally we can write $\omega^n/n! = c_n \Omega \wedge \overline \Omega$ for <i>any</i> Kahler (or hermitian) metric $\omega$ with $c_n$ a constant and $\Omega$ an $(n,0)$-form. Why doesn't this imply that all metrics are Ricci flat? http://mathoverflow.net/questions/107901/calabi-yau-and-ricci-flat-metrics Comment by Gunnar Magnusson Gunnar Magnusson 2012-09-24T10:00:17Z 2012-09-24T10:00:17Z That's my fault, I forgot the logarithm. The real formula is $Ric \omega = -i \partial \bar \partial \log \det \omega_{jk}$. http://mathoverflow.net/questions/107901/calabi-yau-and-ricci-flat-metrics Comment by Gunnar Magnusson Gunnar Magnusson 2012-09-24T08:40:49Z 2012-09-24T08:40:49Z With the edit the question is trivial and you should work it out yourself. Hint: $Ric \omega = -i\partial \bar \partial \det \omega_{jk}$ in local coordinates. Now use the hypothesis to write $\det \omega_{jk}$ as the modulus of a holomorphic function.