User joel fine - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:59:45Z http://mathoverflow.net/feeds/user/380 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19210/holomorphic-vector-fields-acting-on-dolbeault-cohomology Holomorphic vector fields acting on Dolbeault cohomology Joel Fine 2010-03-24T16:34:55Z 2012-02-08T13:26:12Z <h2>The question.</h2> <p>Let $(X, J)$ be a complex manifold and $u$ a holomorphic vector field, i.e. $L_uJ = 0$. The holomorphicity of $u$ implies that the Lie derivative $L_u$ on forms preserves the (p,q) decomposition and also that it commutes with $\bar{\partial}$. From this it follows that $u$ acts infinitesimally on the Dolbeault cohomology groups $H^{p,q}(X)$ of $X$. My question is, does anyone know of an example in which this action is non-trivial?</p> <h2>Some context.</h2> <p>To give some context, first note that the analgous action for de Rham cohomology is always trivial: If $M$ is any smooth manifold and $v$ any vector field, then the formula $L_v = d \circ i_v + i_v \circ d$ shows that the infinitesimal action of $v$ on de Rham cohmology is trivial. (This is an instance of the more general fact that homotopic maps induce the same homomorphisms on singular cohomology. The field $v$ generates diffeomorphisms which are by construction isotopic to the identity map.)</p> <p>Returning to Dolbeault cohomology, suppose we know that each Dolbeault class is represented by a $d$-closed form. (For example, this is true if $X$ is a compact Kähler manifold, by Hodge theory.) Then the action is necessarily trivial. The proof is as follows. Let $\alpha$ be a $\bar{\partial}$-closed (p,q)-form which is also $d$-closed. Then we know that $L_u \alpha = d(i_u \alpha)$ is also of type (p,q). So, $$L_u\alpha = \bar{\partial}\left((i_u\alpha)^{p, q-1}\right) + \partial\left((i_u \alpha)^{p-1, q}\right)$$ and the other contributions $\bar{\partial}((i_u\alpha)^{p-1,q}$) and $\partial((i_u\alpha)^{p,q-1})$ vanish. Now the fact that $\bar\partial((i_u\alpha)^{p-1,q}) = 0$ and our hypothesis imply that there is a (p-1, q-1)-form $\beta$ such that $$(i_u\alpha)^{p-1,q}+ \bar\partial \beta$$ is closed. Hence $$\partial \left((i_u\alpha)^{p-1,q}\right) = \bar\partial \partial \beta$$ and so $$L_u\alpha = \bar \partial \left( (i_u \alpha)^{p,q-1} + \partial \beta\right)$$ which proves the action of $u$ on $H^{p,q}(X)$ is trivial.</p> http://mathoverflow.net/questions/68316/extremal-fano-with-non-constant-scalar-curvature-vs-kaehler-einstein-fano-manifol/68402#68402 Answer by Joel Fine for Extremal Fano with non constant scalar curvature vs Kaehler-Einstein Fano manifolds Joel Fine 2011-06-21T16:10:17Z 2011-06-21T16:15:30Z <p>I'm not 100% sure what you're looking for because of the wording of your question. Do you want a Kähler manifold for which <i>every</i> Kähler class has an extremal representative, but no constant scalar curvature representative? Or examples of manifolds with just one such Kähler class?</p> <p>If it's the latter, then note that if there is a metric of constant scalar curvature in the class there can be no genuinely extremal metrics (since the Futaki invariant must vanish). (If I misunderstood your question and you already knew this then sorry for teaching you to suck eggs!) For plenty of concrete examples of extremal metrics you could look in the article of Arezzo-Pacard-Singer for blow-ups:</p> <p><a href="http://arxiv.org/abs/math/0701028" rel="nofollow">http://arxiv.org/abs/math/0701028</a></p> <p>Alternatively, you can try the paper of Chen-Li-Sheng, which (building on work of Donaldson in the case of constant scalar curvature) settles the problem completely for toric surfaces (of which $F_1$ is an example):</p> <p><a href="http://arxiv.org/abs/1008.2607" rel="nofollow">http://arxiv.org/abs/1008.2607</a></p> <p>From here you might be able to find more toric surfaces which have the property that all Kähler classes have extremal but not constant scalar curvature representatives. It will come down to some calculations involving polygons, but they could well be very difficult. Perhaps trying them in the case of $F_1$ would show how to find another example. </p> http://mathoverflow.net/questions/66046/which-nonlinear-pdes-are-of-interest-to-algebraic-geometers-and-why/67047#67047 Answer by Joel Fine for Which nonlinear PDEs are of interest to algebraic geometers and why? Joel Fine 2011-06-06T15:42:41Z 2011-06-06T15:42:41Z <p>One important use of PDEs in algebraic geometry is in so-called "Hitchin-Kobayashi correspondences". The original example of this is the following theorem.</p> <p><strong>Theorem (Donaldson, Uhlenbeck-Yau)</strong> Let $L \to X$ be an ample line bundle over a compact complex manifold and $\omega$ a Kähler metric representing $c_1(L)$. A holomorphic vector bundle $E \to X$ is slope polystable with respect to $L$ if and only if it admits a Hermitian-Einstein metric with respect to $\omega$.</p> <p>To spell this out in detail: the slope of a bundle $\mu(E)$ is its degree divided by its rank, where degree is $\langle c_1(E) \cup c_1(L)^{n-1}, [X] \rangle$. A bundle is slope stable if all proper coherent subsheaves have strictly smaller slope. A bundle is called slope polystable if it is the sum of stable bundles of equal slope. Meanwhile, a Hermitian metric in $E$ is called Hermitian-Einstein if its curvature $F$ satisfies the equation $(F,\omega) = c \cdot \mathrm{Id}$, for a constant $c$. (Here we are taking the innerproduct on 2-forms, the result being an endomorphism of $E$.) This is a non-linear PDE on the Hermitian metric.</p> <p>Notice that the slope polystability is purely algebraic - it makes no mention whatsoever of the metric $\omega$. What is remarkable is that this is equivalent to the existence of a solution of a non-linear PDE. The story behind this theorem is quite a long one. It can be seen as an infinite dimensional example of the equivalence between quotients via GIT and symplectic quotients. The PDE plays the role of the moment map.</p> <p>Since this result was proved there have been many other versions involving bundles with additional data, say, e.g., a Higgs field, which appears both in the definition of stability and the corresponding PDE. The corresponding "Hitchin-Kobayashi correspondences" play important roles in the study of the moduli of the algebraic objects. For example the fact that one can always solve the relevant PDE leads directly to an interesting Kähler metric on the moduli space of stable objects. In the case of Higgs fields, this is a hyperkähler metric. This metric is one of the starting points of the approach to geometric Langlands proposed by Kapustin and Witten. There are also other applications of the PDE point of view here leading to, amongst other things, strong restrictions on the fundamental groups of Kähler manifolds. This subject sometimes goes by the name "Non-abelian Hodge theory". (I should stress that this is a long way from my expertise!)</p> <p>In a different vein, one version of the Hitchin-Kobayashi correspondence (which is still conjectural) concerns not metrics on bundles but rather metrics on the manifold itself. Many people in Kähler geometry are currently working on understanding both the conjecture and its ramifications. The idea (originally due to Yau, later refined by Donaldson and Tian) is that given an ample line bundle $L \to X$, one should be able to find a so-called "extremal" Kähler metric in $c_1(L)$ if and only if the polarised variety is "stable". Here a metric is "extremal" if it's a critical point of the $L^2$-norm of the curvature tensor, restricted to metrics in $c_1(L)$. This turns out to be equivalent to the gradient of the scalar curvature being holomorphic, a sixth-order fully non-linear PDE. "Simple" examples are Kähler-Einstein metrics (when $L$ is a multiple of the canonical bundle). The correct definition of stability, known as K-polystability, is a little too involved to give neatly here, but it is important to mention that, just as for slope polystabilty of a vector bundle, it is a purely algebraic concept. This whole subject is vast, and I could write about it for pages and pages, but I've probably already said too much for one answer!</p> http://mathoverflow.net/questions/63439/how-can-we-detect-the-existence-of-almost-complex-structures/63476#63476 Answer by Joel Fine for How can we detect the existence of almost-complex structures? Joel Fine 2011-04-29T21:46:16Z 2011-05-03T20:45:39Z <p><strong>Edit:</strong> Now updated to include reference and slightly more general result. <strong>Edit 2:</strong> Includes remark about integrability.</p> <p>Similar to Francesco Polizzi's answer, there is the following Theorem concerning 6-manifolds.</p> <p>A closed oriented 6-dimensional manifold $X$ without 2-torsion in $H^3(X,\mathbb{Z})$ admits an almost complex structure. There is a 1-1 correspondence between almost complex structures on $X$ and the integral lifts $W \in H^2(X, \mathbb{Z})$ of $w_2(X)$. The Chern classes of the almost complex structure corresponding to $W$ are given by $c_1 = W$ and $c_2 = (W^2 - p_1(X))/2$.</p> <p>In fact, a necessary and sufficient condition for the existence of an almost complex structure is that $w_2(X)$ maps to zero under the Bockstein map $H^2(X,\mathbb{Z}_2) \to H^3(X,\mathbb{Z})$. </p> <p>I think the reason for results such as this and the one mentioned by Francesco is the following. To find an almost complex structure amounts to finding a section of a bundle over $X$ with fibre $F_n=SO(2n)/U(n)$. The obstructions to such a section existing lie in the homology groups $H^{k+1}(X, \pi_k(F_n))$. When $n$ is small I would guess we can compute these homotopy groups and so have a good understanding of the obstructions. For example, in the case mentioned above, n=3, $F_n = \mathbb{CP}^3$ and so the only non-trivial homotopy group which concerns us is $\pi_2 \cong \mathbb{Z}$. This is what leads to the above necessary and sufficient condition concerning 2-torsion. On the other hand when $n$ is large I don't know what $F_n$ looks like, let alone its homotopy groups...</p> <p>For the proof of the above mentioned result see the article "Cubic forms and complex 3-folds" by Okonek and Van de Ven. (I highly recommend this article, it's full of interesting facts about almost complex and complex 3-folds.)</p> <p>It is worth pointing out that in real dimension 6 or higher there is <em>no known obstruction to the existence of an integrable complex structure</em>. In other words, there is no known example of a manifold of dimension 6 or higher which has an almost complex structure, but not a genuine complex structure. By the classification of compact complex surfaces, those 4-manifolds admitting integrable complex structures are well understood.</p> http://mathoverflow.net/questions/62307/if-x-fails-to-be-holmorphic-what-is-lie-x-bar-d-bar-d-lie-x/62670#62670 Answer by Joel Fine for If X fails to be holmorphic, what is Lie_X \bar\d - \bar\d Lie_X ? Joel Fine 2011-04-22T21:56:55Z 2011-04-22T21:56:55Z <p>I'll answer the simpler question. Since $J^2=-1$, it follows that for any vector field $X$, the endomorphism $L_X(J)$ anti-commutes with $J$. In other words, $L_X(J)$ is a section of $\bar{T}^* \otimes T$. This is the same space where $\bar{\partial}X$ lives and in fact they are equal, at least up to a constant factor which I forget right now.</p> <p>Now we can work out $[L_X, \bar{\partial}]$ in terms of $[L_X, J]$: $$[L_X, \bar{\partial}] = \frac{1}{2}[L_X, d + i J d] = \frac{i}{2}[L_X,J]\circ d$$ since $[L_X,d] = 0$. So up to a factor then, $[L_X, \bar{\partial}]$ is just $\bar\partial X \circ d$. From this it seems that the $C^k$-norm of <code>$[L_X,\bar\partial](\alpha)$</code> is controlled by the $C^k$ norm of $\bar\partial X$ and the $C^{k+1}$-norm of $\alpha$.</p> http://mathoverflow.net/questions/59738/the-complex-version-of-nashs-theorem-is-not-true/60520#60520 Answer by Joel Fine for "The complex version of Nash's theorem is not true" Joel Fine 2011-04-04T03:45:55Z 2011-04-04T03:45:55Z <p>This is not so much an answer to the original question, more an addition to the answer of diverietti and part of the answer Robert Bryant. Both mention the following analogue of Nash's embedding problem:</p> <p>Given a Kähler manifold $(X, \omega)$, is there a projective embedding $X \to \mathbb{CP}^n$ for which the Fubini-Study metric pulls back to give $\omega$?</p> <p>As Robert says, taken literally the answer is no. However, there is a beautiful theorem of Tian which says that the answer is yes, <i>provided one is willing to let $n$, the dimension of the projective space, tend to infinity.</i> </p> <p>More precisely, let $L \to X$ be a positive holomorphic line bundle on $X$. This means there is a Hermitian metric $h$ in $L$ whose curvature is a Kähler form $\omega$ in $c_1(L)$. (Moreover, all Kähler forms in $c_1(L)$ arise this way.) With this metric $h$ and the volume form $\omega^n$ you can define an $L^2$-inner-product on the space of holomorphic sections of $L^k$, where $k$ is a large integer. Let $s_0, \ldots , s_n$ be an orthonormal basis of holomorphic sections (where $n$ depends on $k$ roughly $n \sim k^m$ where $m$ is the dimension of $X$). Then (for large $k$) the map $$f_k(x) = [s_0(x) : \cdots : s_n(x) ]$$ defines an embedding to $\mathbb{CP}^n$ which has the following property: if we restrict the Fubini-Study metric from projective space to $X$ via the map $f_k$, rescale by $1/k$ (to keep the total volume fixed) and then take the limit as $k \to \infty$ we get the original metric $\omega$.</p> http://mathoverflow.net/questions/60201/what-is-a-lagrangian-submanifold-intuitively/60519#60519 Answer by Joel Fine for What is a Lagrangian submanifold intuitively? Joel Fine 2011-04-04T03:26:13Z 2011-04-04T03:26:13Z <p>I want to answer the question "why should I care about Lagrangians" by showing how they arise when you try to answer a very natural question which in some sense lies at the heart of classical mechanics. </p> <p>Suppose we consider some physical system in which the possible configurations correspond to points in a manifold $M$. Given two points $p, q \in M$ a natural question is to ask, is there some initial momentum one can select at $p$ under which after 1 second the configuration will have evolved into the configuration $q$?</p> <p>In the Hamiltonian formulation of classical mechanics, the position and momentum of a system is described by the points of the cotangent bundle $T^*M$. This is a symplectic manifold in a natural way. The physics of the system are described by a so-called Hamiltonian function $H \colon T^*M \to \mathbb{R}$ and the evolution of the system is given by the flow of the Hamiltonian vector field $X_H$ associated to $H$. We can now rephrase our question as follows: is there a point $x$ in $T^*_pM$ such that the time-1 flow of $X_H$ starting at $x$ lies on the fibre $T^*_qM$? In other words, if I look at the image of $T^*_pM$ under the time-1 flow of $X_H$, does it intersect $T^*_qM$?</p> <p>What does this have to do with Lagrangians? Well the fibres $T^*_pM$ and $T^*_qM$ are special cases of Lagrangian submanifolds. The image of $T^*_pM$ under the flow of $X_H$ is certainly no longer a fibre but, since the flow of $X_H$ preserves the symplectic form, the image of $T^*_pM$ remains a Lagrangian submanifold. </p> <p>From this point of view you can think of a Lagrangian submanifold as a generalisation of "the set of possible initial momenta of a given point in configuration space". Moreover, this gerenalisation is forced on you, even if you only care about the very natural question described above. Finally, I hope this answer explains not only why Lagrangians are interesting, but why the possible intersections of different Lagrangians is interesting. </p> <p>(Credit where credit is due: I guess this picture goes back at least to Arnold. I heard it in a seminar the other week given by Fukaya. The study of intersections of Lagrangians has become a huge industry recently, building essentially on the pioneering ideas of Floer.)</p> http://mathoverflow.net/questions/19308/is-there-an-analogue-of-curvature-in-algebraic-geometry/19319#19319 Answer by Joel Fine for Is there an analogue of curvature in algebraic geometry? Joel Fine 2010-03-25T15:44:51Z 2011-01-07T00:58:35Z <p>One well-studied example is ampleness of an invertible sheaf and positive curvature for a holomorphic line bundle. A theorem of Kodaira says that a holomorphic line bundle over a complex manifold is ample if and only if it admits a metric of positive curvature. </p> <p>Related to this, in the special case when the line bundle is the canonical bundle and also invoking Yau's proof of the Calabi conjecture, one has the following equivalences:</p> <ul> <li><p>A compact complex manifold admits a Kähler metric of positive Ricci curvature only if it is Fano i.e, has ample anticanonical bundle.</p></li> <li><p>A compact complex manifold admits a Kähler metric of negative Ricci curvature if and only if it has ample canonical bundle.</p></li> <li><p>A compact complex manfiold admits a Ricci flat Kähler metic if and only if it has torsion first Chern class.</p></li> </ul> http://mathoverflow.net/questions/35334/integration-and-stokes-theorem-for-vector-bundle-valued-differential-forms/35345#35345 Answer by Joel Fine for Integration and Stokes' theorem for vector bundle-valued differential forms? Joel Fine 2010-08-12T12:18:15Z 2010-08-12T18:36:51Z <p>For a general vector bundle, I there is no "$E$-valued integration" as you put it. You are trying to add up elements in the fibres of $E$, but since the fibres over different points are not the same vector space you can't add their elements. </p> <p>For the trivial bundle $M \times \mathbb{R}^k$ - and with a fixed choice of trivialisation! - you can carry out the integral component by component, But if you change the trivialisation you will get a different answer. Moreover, you can change the trivialisation in a way that varies over the manifold, so there is no hope that the integral will just change by a linear map of $\mathbb{R}^k$. You can see this behaviour even with ordinary functions. A function is a section of the trivial<i>ised</i> rank 1 bundle. If you change the trivialisation, but insist on regarding ordinray $p$-forms as bundle valued, you multiply all your forms by a fixed nowhere vanishing function. This can change the integral over a $p$-cycle in a more-or-less arbitrary way.</p> <p>What you <i>can</i> integrate $E$-valued forms against is <code>$E^*$</code>-valued ones. Given $a \in \Omega^p(E)$ and <code>$b \in \Omega^q(E^*)$</code> their wedge product is an ordinary $(p+q)$-form which you can then integrate over a $(p+q)$-cycle. Now you have a version of Stokes theorem. If you have a connection $A$ in $E$ then you can check that $$d(a \wedge b) = d_A(a) \wedge b \pm a \wedge d_A(b).$$ So Stokes theorem gives $$\int_{M}d_A(a) \wedge b \pm a \wedge d_A(b) = \int_{\partial M} a \wedge b.$$ In the case when $b$ is a covariant constant section of $E$ and $M$ has dimension one more than the degree of $a$, we get $$\int_M \langle d_A(a) , b \rangle=\int_{\partial M} \langle a, b \rangle$$ This is just the usual Stokes theorem, for the component of $a$ in the direction $b$. Since $b$ is covariant-constant, $\langle d_A(a), b \rangle = d \langle a, b \rangle$.</p> http://mathoverflow.net/questions/32810/are-the-eigenvalues-of-the-laplacian-of-a-generic-kahler-metric-simple Are the eigenvalues of the Laplacian of a generic Kähler metric simple? Joel Fine 2010-07-21T15:52:36Z 2010-08-01T23:42:22Z <p>It is a <a href="http://www.jstor.org/stable/2374041" rel="nofollow">theorem of Uhlenbeck</a> that for a generic Riemannian metric, the Laplacian acting on functions has simple eigenvalues, i.e., all the eigenspaces are 1-dimensional. (Here "generic" means the set of such metrics is the complement of a <a href="http://en.wikipedia.org/wiki/Meagre_set" rel="nofollow">meagre set</a> in the space of all metrics on a given manifold.)</p> <p>I would like to know whether this property is also true for a generic Kähler metric. More precisely, given a compact complex manifold with a fixed choice of Kähler class, is it true that for the generic representative of this class, the Laplacian acting on functions has simple eigenvalues?</p> <p>(To be honest, on a first reading it seems as if one might be able to apply similar arguments to those of Uhlenbeck to prove this result. I guess I would like to know if someone has already done this, or if there is a cunning counter-example that I'm missing, before I commit the time and energy to try!)</p> http://mathoverflow.net/questions/33896/how-are-these-two-ways-of-thinking-about-the-cross-product-related/33903#33903 Answer by Joel Fine for How are these two ways of thinking about the cross product related? Joel Fine 2010-07-30T09:03:50Z 2010-07-30T10:27:45Z <p>To expand on Victork Protsak's comment, if V is an n-dimensional real vector space with inner-product, the inner-product gives an isomorphism $V\to V^*$ and hence $V\otimes V \to \mathrm{End}(V)$. Under this isomorphism, $\Lambda^2(V)$ is identified with skew-adjoint endomorphisms of $V$, which is precisely the Lie algebra $\mathfrak{so}(V)$.</p> <p>In the case dim V =3, the Hodge star gives an isomorphism $\Lambda^2(V) \to V$ and so in total we see that $V$ is canonically isomorphic to $\mathfrak{so}(V)$. A more direct way to see this isomorphism is to send the vector $v \in V$ to the generator of the right-handed rotation about the axis in the direction of $v$ with speed $|v|$. </p> <p>The use of the phrase "right-handed" makes it clear that in order to identify $V$ and $\mathfrak{so}(V)$ we have used an orientation on $V$; indeed, you need that for the Hodge star. What is interesting is that if you reverse the orientation on $V$, the map to $\mathfrak{so}(V)$ changes sign. This means that what ever orientation you chose on $V$, the push-forward to $\mathfrak{so}(V)$ is the same. Conclusion: $\mathfrak{so}(3)$ is <i>naturally oriented</i>. This is analogous to the natural orientation on $\mathbb{C}$. A more prosaic way to describe the orientation is to pick two independent elements $x,y \in \mathfrak{so}(3)$ and then use $[x,y]$ to complete them to an oriented basis. (Of course, you then need to check that this doesn't depend on your choice of $x,y$.)</p> http://mathoverflow.net/questions/22039/are-any-two-k3-surfaces-over-c-diffeomorphic/22043#22043 Answer by Joel Fine for Are any two K3 surfaces over C diffeomorphic? Joel Fine 2010-04-21T11:46:09Z 2010-04-21T11:46:09Z <p>I think this was first proved by Kodaira. See <a href="http://www.jstor.org/pss/2373157" rel="nofollow">On the structure of complex analytic surfaces, 1</a>. There Kodaira proves that any K3 surface is a deformation of a non-singular quartic surface in $\mathbb{CP}^3$. In particular, they are all diffeomorphic.</p> http://mathoverflow.net/questions/21939/what-can-you-do-with-a-compact-moduli-space/21997#21997 Answer by Joel Fine for What can you do with a compact moduli space? Joel Fine 2010-04-20T21:39:32Z 2010-04-20T21:47:09Z <p>The answers here are all excellent examples of things that can only be proved once a moduli space is compactified. I would like to add a perhaps more basic reason for compactifying moduli spaces, involving something simpler than theoretical applications such as defining enumerative invariants. The moral is the following: </p> <blockquote> <p>If you study families of geometric objects then either you are almost certain to encounter the boundary of the moduli space, or you must have some very good reason to rule it out. </p> </blockquote> <p>For example, to find a non-trivial compact family of smooth complex curves is actually quite awkward and such families are very rare. (The first examples were due to Atiyah and Kodaira.) From this point of view the "ubiquity of the compactification" amounts to the fact that the boundary divisor of singular curves in the compactified moduli space is positive in a certain sense, so it intersects almost all curves in the moduli space. It is this positivity of the boundary which forces us to study it!</p> <p>Some more examples explain - I hope! - the way compactification enters when considering pseudoholomorphic curves as in Gromov-Witten theory, without ever coming close to trying to define an enumerative invariant. Just by looking at a conic in $\mathbb{CP}^2$, which degenerates into two lines, one sees that when moving a pseudoholomorphic curve around, one is almost certain to encounter bubbling, unless one has a very good reason to know otherwise. Understanding how to compactify the moduli space, we see that this bubbling phenomenon is the main thing which can go wrong. What is interesting here is that often one tries to prove this compactification is <i>not actually necessary</i>, by ruling out bubbling somehow. Two examples follow - taken from Gromov's original use of pseudoholomorphic curves in his Inventiones paper - which exploit this idea. </p> <p>Firstly, Gromov's proof of his non-squeezing theorem. Here the key point in the argument is that one can find a certain pseudoholomorphic disc for a standard almost complex structure on $\mathbb{C}^n$. One would like to know that as one deforms the almost complex structure the disc persists so that we have such a disc for a special non-standard almost complex structure. It is standard in this kind of "continuity method" that you can always deform the disc for a little while because the problem is elliptic. But to push the deformation indefinitely you need to show compactness - why doesn't the disc break up? Thanks to our knowledge of the compactification of the moduli space, we understand that the only thing that can go wrong is bubbling and in this case bubbles cannot form because the symplectic structure is exact.</p> <p>The second example is of the following type: suppose one knows the existence of <i>one</i> pseudoholomorphic curve in a symplectic manifold; then one can try and use it to investigate the ambient space, moving it around and trying to sweep out as much of the space as possible. In this way you can prove, for example, that any symplectic structure on $\mathbb{CP}^2$ which admits a symplectic sphere with self-intersection 1 must be the standard symplectic structure. The reason is you can find an almost complex structure which makes this sphere a pseudoholomorphic curve. Then you move the curve around until is sweeps out the whole space, doing it carefully enough to give a symplectomorphism with the standard $\mathbb{CP}^2$. Here you can push the curve wherever you want because it wont break. Bubbles can't form because the curve has symplectic area 1 and so there is no "spare area" to make bubbles with.</p> http://mathoverflow.net/questions/20373/the-central-issues-in-complex-geometry/20436#20436 Answer by Joel Fine for the central issues in complex geometry Joel Fine 2010-04-05T23:22:23Z 2010-04-05T23:22:23Z <p>One major area of research is that of canonical metrics on Kahler manifolds. The original definition of "canonical" is due to Calabi. One considers all metrics in a fixed Kahler class and attempts to minimise the L^2-norm of the curvature tensor. The Euler-Lagrange equations say that a metric is a critical point for this functional precisely when its scalar curvature has holomorphic gradient. Such metrics are called <i>extremal</i>. Special cases include Kahler-Einstein metrics and constant scalar curvature metrics. </p> <p>Enormous amounts of research have been done in this direction. In one direction, one tries to prove existence. A random selection of results: Aubin and Yau's work on Kahler-Einstein metrics of negative and zero scalar curvature, Tian's work on Kahler-Einstein surfaces of positive scalar curvature, more recently Donaldson's work on extremal metrics on toric surfaces.</p> <p>In the opposite direction, one tries to find obstructions to existence. Futaki found one such obstruction, involving holomorphic vector fields, which was then vastly generalised by Tian and Donaldson. This gives examples of many manifolds and Kahler classes which can never admit extremal metrics. </p> <p>There is a conjecture (due in various forms to Donaldson, Tian and Yau) which says that the known obstructions are the only obstructions. When they vanish, an extremal metric exists. This would be a beautiful result because the obstructions are purely algebro-geometric (at least when the Kahler class is integral) yet the conclusion is analytic - we can find a metric which solves a PDE. If you know of the Hitchin-Kobayashi correspondence for Hermitian Yang-Mills metrics on vector bundles, this conjecture can be seen as the analogue for Kahler metrics. </p> <p>At the moment, the conjecture is known to hold for toric surfaces (Donaldson) and is close to being settled for Kahler-Einstein metrics. This is due to Aubin-Yau for negative scalar curvature, and Yau for zero scalar curvature - in these cases the metrics always exist. The positive scalar curvature case is far more delicate. Donaldson recently announced significant progress in this direction.</p> <p>If you are looking for a link with Hormander's L^2 estimates, then look no further than projective embeddings via higher and higher powers of a positive line bundle L over your Kahler manifold X. For each power of L we consider the following problem: find a basis of holomorphic sections of L so that the image of X in CP<sup>n</sup> has zero centre of mass (to define centre of mass we think of CP<sup>n</sup> as a coadjoint orbit in the linear space su(n+1)<sup>*</sup> equiped with an inner-product via the Killing form). This can be thought of as a finite dimensional analogue of the problem of finding an extremal metric representing the first Chern class of L. As the power of L tends to infinity, these projective problems converge in some sense to the problem of finding an extremal metric. Understanding the precise nature of this convergence involves difficult questions concerning the Bergman kernel and, ultimately, Hormander's estimate. (Again, this part of the story is due to Donaldson.) </p> <p>To get started you could look at:</p> <ul> <li><i>Extremal Kahler metrics 1 and 2</i>, by Calabi.</li> <li><i>Canonical metrics in Kahler geomtery</i>, by Tian.</li> <li><i>Remarks on gauge theory, complex geometry and 4-manifold topology</i>, by Donaldson.</li> <li><i>Scalar curvature and stability of toric varieties</i>, by Donaldson.</li> <li><i>Scalar curvature and projective embeddings 1</i>, by Donaldson.</li> </ul> http://mathoverflow.net/questions/19635/gauge-theory-construction-of-moduli-of-vector-bundles/19662#19662 Answer by Joel Fine for Gauge theory construction of moduli of vector bundles Joel Fine 2010-03-28T20:20:13Z 2010-03-28T21:38:24Z <p>As Charlie Frohman says, for curves this is the Narasimhan-Seshadri correspondonce. For Kahler manifolds of higher dimensions it is called the Hitchin-Kobayashi correspondence, proved by Donaldson-Uhlenbeck-Yau. You search for a Hermitian metric on your vector bundle which is Hermitian-Yang-Mills, i.e. it's curvature form is orthogonal to the Kahler form. Such a metric exists if and only if the bundle is a sum of slope stable bundles. </p> <p>In general this can be seen as an infinite dimensional version of GIT vs symplectic reduction. The condition that a metric be HYM says that a certain moment map vanishes for this connection.</p> <p>To start, I would recommend the following articles:</p> <ul> <li><p>"The Yang-Mills equations over Riemann surfaces" by Atiyah-Bott.</p></li> <li><p>"A new proof of a theorem of Narasimhan-Seshadri" by Donaldson.</p></li> <li><p>"Anti-self-dual Yang-Mills connections over complex algebraic surfaces" by Donaldson. </p></li> </ul> http://mathoverflow.net/questions/19258/topological-results-from-geometry/19298#19298 Answer by Joel Fine for Topological results from geometry Joel Fine 2010-03-25T12:34:26Z 2010-03-25T12:55:25Z <p>A nice topic to read about is Chern-Weil theory. This is the generalisation of Gauss-Bonnet to higher dimensions and to vector bundles other than the tangent bundle. Put very briefly, topological invariants of a vector bundle over a manifold (its characteristic classes - certain classes in the cohomology of the base) can be computed using the curvature tensor of any choice of connection in the bundle. </p> <p>The prototype is Gauss-Bonnet in which, as you know, the Euler characteristic of a (compact orientable) surface is equal to a fixed constant times the integral of the scalar curvature of any Riemannian metric on the surface.</p> http://mathoverflow.net/questions/17736/way-to-memorize-relations-between-the-sobolev-spaces/17770#17770 Answer by Joel Fine for Way to memorize relations between the Sobolev spaces? Joel Fine 2010-03-10T22:24:28Z 2010-03-10T22:24:28Z <p>Exactly as Dan Lee says, the key is the "weight" $k-n/p$ of the Sobolev space $W^{k,p}$. To remember the weight, you can use scaling, just as the other answers and comments suggest, but there is a cheap trick which is much simpler than changing the function whose norms you are considering. </p> <p>Instead, multiply the <i>metric</i> by a constant R. This will scale the $W^{k,p}$-norm by $R^{n/p-k}$ and the $C^{m,\alpha}$-norm by $R^{-(m+\alpha)}$. From here you see immediately the relevance of the weights and the direction in which the embeddings must go. </p> http://mathoverflow.net/questions/13020/is-there-a-name-for-this-differential-operator-and-or-its-corresponding-spectrum/13036#13036 Answer by Joel Fine for Is there a name for this differential operator and/or its corresponding spectrum? Joel Fine 2010-01-26T11:12:06Z 2010-01-26T11:12:06Z <p>There is an simple explicit formula for your operator in terms of known operators. To see this, note that $\delta E_f(g)$ (the differential of $E$ at the function $f$ in the direction $g$) is equal to $$2 \int X(f) X(g) dV = 2\int \left[L_X(g X(f) dV )-g\left(X(X(f)) dV+X(f)L_X(dV)\right)\right]$$ where $L_X$ is the Lie deriviative with respect to $X$. Now, $\int L_X(\alpha) = 0$ for any top degree form $\alpha$, so we get $$\delta E_f(g) = - 2 \int_X g \left[X(X(f)) + X(f) div(X)\right]dV$$ where the divergence of $X$ is defined by $div(X)= L_X(dV)/dV$. So, using the $L^2$-inner product on functions, we can interpret the 1-form $\delta E$ as the differential operator $$D :f \mapsto - X(X(f)) - X(f) div(X).$$ (This is the same way that the differential of Dirichlet energy is seen as the Laplacian.) </p> <p>Note that the leading order part of D is just $X^2$ and so, in particular, $D$ is not elliptic. You'd expect this, of course, because $E$ only sees the change of $f$ in the $X$-direction.</p> http://mathoverflow.net/questions/12500/nontrivial-isotrivial-family-of-elliptic-curves/12529#12529 Answer by Joel Fine for (nontrivial) isotrivial family of elliptic curves Joel Fine 2010-01-21T11:42:04Z 2010-01-21T16:19:03Z <p>(<b>Edit:</b> added motivation for my "answer" even though it isn't exactly what was asked for. It shows the moduli space of genus 1 curves cannot be fine - see last paragraph.) </p> <p>The Hopf surface is a (non-algebraic) example of a non-trivial family of isomorphic elliptic curves. Here I use "elliptic curve" to mean "smooth complex curve of genus 1", which is probably not what you mean. (In particular my curves have no distinguished base point.)</p> <p>The Hopf surface $X$ is a quotient of $\mathbb{C}^2\setminus0$ by the action of $\mathbb Z$ generated by $z \mapsto 2z$. Since this action commutes with the action of $\mathbb{C}^*$ there is a map $X \to \mathbb{CP}^1$. Each fibre is a copy of $\mathbb{C}\setminus{0}$ divided by the action $z\sim 2z$. So all the fibres are isomorphic elliptic curves. (If you want base points on each curve, this example doesn't work because the fibration $X \to \mathbb{CP}^1$ doesn't have a section.) On the other hand, its easy to see that $X$ is diffeomorphic to $S^1\times S^3$ and so you can't trivialise the fibration even topologically. Replacing $z\mapsto 2z$ by $z\mapsto \lambda z$ for $\lambda\in \mathbb{C}$ non-zero, I guess you can get any genus-1 curve to appear as all fibres of a topologically non-trivial family. </p> <p>Even though this is not eaxctly what you're looking for, I thought it worth giving the example anyway, because it's a very simple way to see that the moduli space of genus 1 curves cannot be fine. (Fine moduli spaces carry a universal family and all other families are pulled back from the universal family via the map to moduli space - in particular a family of isomorphic objects in such a moduli space is pulled back by the constant map and so must be trivial. One often sees that objects cannot be parametrised by a fine moduli space by giving examples of special objects with "extra" automorphisms. The Hopf surface is an alternative to that approach in this situation.)</p> http://mathoverflow.net/questions/11846/has-mathoverflow-yet-led-to-mathematical-breakthroughs/11848#11848 Answer by Joel Fine for Has mathoverflow yet led to mathematical breakthroughs? Joel Fine 2010-01-15T11:41:58Z 2010-01-15T11:41:58Z <p>I have an example almost of the sort you are looking for, but not quite with the happiest ending (at least for me!). I was struggling to solve a problem. I identified a statement that would be helpful. <a href="http://mathoverflow.net/questions/9006/representations-of-surface-groups-via-holomorphic-connections" rel="nofollow">I asked on mathoverflow</a>. <a href="http://mathoverflow.net/questions/9006/representations-of-surface-groups-via-holomorphic-connections/9034#9034" rel="nofollow">I got an answer</a>. It turns out that both the statement and my intended application were about to appear in an article by some other authors. </p> <p>In this instance mathoverflow has saved me a great deal of effort simply by allowing me to not work on the problem and read instead someone else's solution. Without mathoverflow it is very unlikely that I would have heard of the article before it was published. (It doesn't seem to be on the arxiv.)</p> <p>In general, it seems to me that mathoverflow is an extremely efficient way of getting expert advice on research-level questions. </p> http://mathoverflow.net/questions/11289/geometry-meaning-of-higher-cohomology-of-sheaves/11314#11314 Answer by Joel Fine for Geometry meaning of higher cohomology of sheaves? Joel Fine 2010-01-10T11:21:52Z 2010-01-10T11:21:52Z <p>One way to think about higher cohomology groups of, say, holomorphic vector bundles is via the Dolbeault isomorphism $$H^q(X, \mathcal O(E)) \cong H^{0,q}(X,E)$$ (and also the more general $H^{q}(\mathcal O( \Lambda^pT^*X\otimes E)) \cong H^{p,q}(E)$.)</p> <p>If we also choose a Kähler metric on X and Hermitian metric on E, then Hodge theory says that cohomology is represented by harmonic forms. So we can think of the q<sup>th</sup> cohomology group of the sheaf of sections of E as the space of harmonic (0,q)-forms with values in E.</p> <p>One can interpret Kodaira vanishing from this point of view. Pick a holomorphic line bundle L with a Hermitian metric over a Kähler manifold X. Then one has a connection in L and so an induced connection on L-valued forms. This gives a "rough" Laplacian $\nabla^* \nabla$. The Weitzenbock formula tells us how this differs from the standard Laplacian. On a (p,q)-form with values in L, $$\Delta = \nabla^*\nabla + F$$ where F is an endomorphism of the bundle $\Lambda^{p,q}\otimes L$ where the L-valued forms take their values. F depends on (p,q) and on the metrics of both L and X.</p> <p>The hypothesis for Kodaira vanishing is that there is a Hermitian metric on L whose curvature is a Kahler form. If we use this metric on L and also this Kahler form on X then the operator F has a certain sign: when p+q>n, the dimension of X, F is a positive definite endomorphism of the bundle $\Lambda^{p,q}\otimes L$. From here we can prove Kodaira vanishing. A harmonic (p,q) form $\alpha$ with values in L has $\Delta \alpha = 0$ so, by integrating the Weitzenbock formula against $\alpha$, we see $$\int |\nabla \alpha|^2 + \int (F(\alpha), \alpha) = 0$$ Now positivity of $F$ means both terms here are non-negative and so must each vanish. This forces $\alpha$ to vanish and so $H^{p,q}(L)=0$ when $p+q >n$.</p> http://mathoverflow.net/questions/9466/why-these-particular-numerical-factors-in-the-definition-of-gaussian-curvature/9467#9467 Answer by Joel Fine for Why these particular numerical factors in the definition of Gaussian curvature? Joel Fine 2009-12-21T10:21:00Z 2009-12-22T12:48:25Z <p>First, I guess it should say "geodesic disc" rather than "circle". At least to me, a geodesic circle is a closed geodesic loop in your surface, whereas a geodesic disc of radius r is all the points distance r from a fixed point (at least for r smaller than the injectivity radius). Note the boundary of a geodesic disc is <i>not</i> a geodesic. </p> <p>As for the factors in those formulae, well, there's no absolute scale for Gaussian curvature. People have just agreed on the convention that the curvature of the unit sphere should be 1. (<b>EDIT:</b> As Greg Kuperberg points out in his answer, there are some good reasons for this convention. E.g., Gauss-Bonnet.) That then forces those factors to be what they are. It amounts to the statement that, for a small geodesic disc on the unit sphere of radius r , $$C(r) \sim 2\pi\left( r - \frac{1}{6} r^3\right),$$ and a similar formula for the area. There really is no deeper reason than that.</p> <p>So, to see if the factors are right (and you should never trust what you read on the internet!) I would suggest doing exactly those calculations for the unit sphere. I've checked the first formula involving the circumference and it looks good to me. If you have problems with the calculation, leave a comment and I'll write my version down for you, but I have a feeling it's best to do these things yourself.</p> http://mathoverflow.net/questions/9391/higher-dimensional-gromov-witten-theories/9417#9417 Answer by Joel Fine for Higher Dimensional Gromov-Witten Theories Joel Fine 2009-12-20T10:48:09Z 2009-12-20T10:48:09Z <p>I know that you are thinking firmly about the integrable world, but I thought it worth adding that for symplectic manifolds, there is no obvious generalisation of Gromov-Witten theory to higher dimensional subvarieties. This is because to define "holomorphic" you use a non-integrable almost complex structure and non-integrability means that there are no higher dimensional holomorphic objects. The fact that there are holomorphic curves can be thought of as an instance of the fact that all almost complex structures over 2-manifolds are automatically integrable. (E.g., since there are no (2,0)-forms, the space where the Nijenhuis tensor should live is zero.)</p> http://mathoverflow.net/questions/9297/curves-on-elliptic-ruled-surfaces/9331#9331 Answer by Joel Fine for Curves on elliptic ruled surfaces? Joel Fine 2009-12-19T00:35:31Z 2009-12-19T09:03:08Z <p>To find higher genus curves without using a specific embedding $S \subset \mathbb{P}^n$, it could help to think first about the case when your surface is actually a product $S=\mathbb{P}^1 \times E$. Let $C$ be a curve which admits two branched covers, $f\colon C \to E$ and $g \colon C \to \mathbb{P}^1$. Then the product $f \times g \colon C \to S$ maps into the surface $S$. If the branch points of $f$ and $g$ are different then $f \times g$ will even be an embedding. </p> <p>In general, let $V \to E$ be your rank-two vector bundle, so $S=\mathbb{P}(V)$. Given a banched cover $f \colon C \to E$, you pull back $V$ to a bundle $V' \to C$. Now every time you have a line sub-bundle $L$ of $V'\to C$ you get a section of $\mathbb{P}(V')$ which plays the role of $g$ in the first paragraph. It can be combined with $f \colon C \to E$ to give a map $C \to S$. Depending on how much you know about $E$ and $V$, hopefully this should help you find plenty of explicit curves in $S$.</p> http://mathoverflow.net/questions/9006/representations-of-surface-groups-via-holomorphic-connections Representations of surface groups via holomorphic connections Joel Fine 2009-12-15T17:30:43Z 2009-12-15T22:36:18Z <p>EDIT: Tony Pantev has pointed out that the answer to this question will appear in forthcoming work of Bogomolov-Soloviev-Yotov. I look forward to reading it! </p> <h2>Background</h2> <p>Let $E \to X$ be a holomorphic vector bundle over a complex manifold. A connection $A$ in $E$ is called <i>holomorphic</i> if in local holomorphic trivialisations of $E$, $A$ is given by a holomorphic 1-form with values in End(E).</p> <p>Notice that the curvature of $A$ is necessarily a (2,0)-form. In particluar, holomorphic connections over Riemann surfaces are <i>flat</i>. This will be important for my question.</p> <h2>The Question</h2> <p>I am interested in the following situation. Let $E \to S$ be a rank 2 holomorphic vector bundle over a Riemann surface of genus $g \geq 2$. I suppose that $E$ admits a global holomorphic trivialisation (which I do <i>not</i> fix) and that we choose a nowhere vanishing section $v$ of $\Lambda^2 E$. (So I <i>do</i> fix a trivialisation of the determinant bundle.) I want to consider holomorphic connections in $E$ which make $v$ parallel. The holonomy of such a connection takes values in $\mathrm{SL}(2,\mathbb{C})$ (modulo conjugation). </p> <p>My question: if I allow you to change the complex structure on $S$, which conjugacy classes of representations of $\pi_1(S)$ in $\mathrm{SL}(2,\mathbb C)$ arise as the holonomy of such holomorphic connections?</p> <p>EDIT: As jvp points out, some reducible representations never arise this way. I actually had in mind irreducible representations, moreover with discrete image in $\mathrm{SL}(2,\mathbb{C})$. Sorry for not mentioning that in the beginning!</p> <h2>Motivation</h2> <p>A naive dimension count shows that in fact the two spaces have the same dimension: </p> <p>For the holomorphic connections, if you choose a holomorphic trivialisation of $E\to S$, then the connection is given by a holomorphic 1-form with values in $sl(2, \mathbb C)$. This is a $3g$ dimensional space. Changing the trivialisation corresponds to an action of $\mathrm{SL(2,\mathbb C)}$ and so there are in fact $3g-3$ inequivalent holomorphic connections for a fixed complex structure. Combined with the $3g -3$ dimensional space of complex structures on $S$ we see a moduli space of dimension $6g-6$.</p> <p>For the representations, the group $\pi_1(S)$ has a standard presentation with $2g$-generators and 1 relation. Hence the space of representations in $\mathrm{SL}(2,\mathbb{C})$ has dimension $6g-3$. Considering representations up to conjugation we subtract another 3 to arrive at the same number $6g-6$.</p> <h2>A curious remark</h2> <p>Notice that if we play this game with another group besides $\mathrm{SL}(2,\mathbb{C})$ which doesn't have dimension 3, then the two moduli spaces do not have the same dimension. So it seems that $\mathrm{SL}(2,\mathbb{C})$ should be important in the answer somehow.</p> http://mathoverflow.net/questions/8890/flat-su2-bundles-over-hyperbolic-3-manifolds Flat SU(2) bundles over hyperbolic 3-manifolds Joel Fine 2009-12-14T17:16:26Z 2009-12-15T18:39:18Z <p>Can someone give me a non-trivial example of a flat SU(2)-connection over a compact orientable hyperbolic 3-manifold?</p> <p>The literature on such bundles over 3-manifolds is huge and my naive searches don't seem to turn up specific examples. </p> <p>Roughly speaking, the Casson invariant counts flat bundles over 3-manifolds, so in principal I suppose I would be happy with an example of a hyperbolic 3-manifold with non-zero SU(2) Casson invariant (and surely such things are known). In practice, I would really like to see the non-trivial bundle (or corresponding representation) more-or-less explicitly. </p> <p>Finally, I would also be happy with just being told precisely where I should go and look in the literature!</p> http://mathoverflow.net/questions/8665/embeddings-of-s2-in-mathbbcp2 Embeddings of $S^2$ in $\mathbb{CP}^2$ Joel Fine 2009-12-12T11:27:50Z 2009-12-13T20:14:05Z <p>Suppose we are given an embedding of $S^2$ in $\mathbb{CP}^2$ with self-intersection 1. Is there a diffeomorphism of <code>$\mathbb{CP}^2$</code> which takes the given sphere to a complex line?</p> <p>Note: I suspect that either it is known that there is such a diffeomorphism, or the problem is open. This is because if there was an embedding for which no such diffemorphism existed, you could use it to produce an exotic 4-sphere. To see this, reverse the orientation on $\mathbb{CP}^2$ then blow down the sphere.</p> <p>EDIT: for a counter-example, it is tempting to look for the connect-sum of a line and a knotted $S^2$. The problem is to <i>prove</i> that the result cannot be taken to a complex line. For example, the fundamental group of the complement $C$ is no help, since it must be simply connected. This is because the boundary of a small neighbourhood $N$ of the sphere is $S^3$ and so $\mathbb{CP}^2$ is the sum of $N$ and $C$ across $S^3$ and so in particuar $C$ must be simply-connected. </p> http://mathoverflow.net/questions/8494/cotangent-bundle-of-a-submanifold/8555#8555 Answer by Joel Fine for Cotangent bundle of a submanifold Joel Fine 2009-12-11T11:36:09Z 2009-12-11T11:36:09Z <p>It is possible to see the cotangent bundle of the submanifold as a kind of symplectic reduction of the cotangent bundle of the ambient manifold. I think it might be enough to explain the analogous fact from linear algebra.</p> <p>Let V be a vector space and U a subspace. There is a natural symplectic form $\omega_V$ on $V^*\oplus V$ given by $$\omega_V((\alpha,u),(\beta,v)) = \alpha(v) - \beta(u)$$ where greek letters are elements of $V^*$ and roman letters are elements of $V$. (This is just d of the Louville form in this situation.) There is an analogous form $\omega_U$ on $U^* \oplus U$. </p> <p>Now, let $U^0$ denote the annahilator of $U$ in <code>$V^*$</code>. Consider the subspace $$U^0 \times{0} \subset V^* \oplus V$$ This subspace is isotropic for $\omega_V$. Its symplectic complement is the coisotropic subspace <code>$V^*\oplus U$</code>. </p> <p>Now it is a standard fact in symplectic geometry that if you divide a coisotropic subspace by its symplectic complement the result is naturally a symplectic vector space. (This is the linear algebra behind symplectic reduction.) Applying this idea here we see that the quotient <code>$$(V^*\oplus U )/ (U^0\times\{0\})$$</code> inherits a natural symplectic structure. Of course, the quotient is precisely $U^*\oplus U$ and the symplectic form is nothing but $\omega_U$.</p> http://mathoverflow.net/questions/8247/one-step-problems-in-geometry/8378#8378 Answer by Joel Fine for One-step problems in geometry Joel Fine 2009-12-09T18:01:59Z 2009-12-09T18:01:59Z <p>Here's a cute question which Frederic Bourgeois asked me on a train journey recently. He was asked it by Givental, if my memory serves correctly, but I've no idea where it came from originally. Anyway, the question:</p> <p>There is a mountain of frictionless ice in the shape of a perfect cone with a circular base. A cowboy is at the bottom and he wants to climb the mountain. So, he throws up his lasso which slips neatly over the top of the cone, he pulls it tight and starts to climb. If the mountain is very steep, with a narrow angle at the top, there is no problem; the lasso grips tight and up he goes. On the other hand if the mountain is very flat, with a very shallow angle at the top, the lasso slips off as soon as the cowboy pulls on it. The question is: what is the critical angle at which the cowboy can no longer climb the ice-mountain?</p> <p>To solve it, you should think like a geometer and not an engineer. (And yes, it needs just one trick which is certainly applicable elsewhere.)</p> <p>P.S. When I was asked the question, I failed miserably!</p> http://mathoverflow.net/questions/7787/intuition-behind-moduli-space-of-curves/7795#7795 Answer by Joel Fine for Intuition behind moduli space of curves Joel Fine 2009-12-04T18:22:38Z 2009-12-05T18:03:47Z <p>(EDIT 1: Replaced hand-waving argument in third paragraph with a hopefully less incorrect version)</p> <p>(EDIT 2: Added final paragraph about obtaining all conformal deformations for surfaces other than sphere.)</p> <p>I think it is possible to see the infinitesimal rigidity of the sphere, even if it does involve a PDE as Dmitri says. I think you can also try and see if for other embedded surfaces, all infinitesimal deformations of conformal structure are accounted for by deformations of the embedding in a similar way.</p> <p>For the case of S<sup>2</sup>, what you want is to do is take a normal vector field V (i.e. infinitesimal change of embedding) and produce a tangent vector field X such that flowing along X gives the same infinitesimal change in conformal structure as flowing along V. This should amount to solving a linear PDE, so as Dmitri says a PDE is definitely involved, but probably not as hard as proving the existence of isothermal coordinates (which from memory is non-linear). For the standard embedding of S<sup>2</sup> there can't be too many choices for this linear differential operator given that it has to respect the SO(3)-symmetry. </p> <p>I guess we're looking for a first-order equivariant linear operator from normal vector fields to tangent vector fields. If we identify normal fields with functions then two possible candidates are to take X=grad V or X to be the Hamiltonian flow generated by V. I can't think of any others and probably it's possible to prove these are the only such ones. (Assuming it's elliptic, the symbol of the operator must be an SO(3)-equivariant isomorphism from T<sup>*</sup>S<sup>2</sup> to TS<sup>2</sup> and there can't be too many choices! Using the metric leads to grad and using the area form leads to the Hamiltonian flow.) Then you just have to decide which one to use.</p> <p>For the case of a general embedded surface $M$, you can ask "is it possible to obtain all deformations of conformal structure by deforming the embedding into R<sup>3</sup>?" To answer this we can again think of a normal vector field as a function V on the surface. There is a second-order linear differential operator $$D\colon C^\infty(M) \to \Omega^{0,1}(T)$$ which sends a normal vector field to the corresponding infinitesimal change of conformal structure. This operator will factor through the hessian with a homomorphism from <code>$T^* \otimes T^*$</code> to <code>$T^{*0,1}\otimes T^{1,0}$</code>. The operator $D$ will not be onto, but what we want to know is whether every <i>cohomology class</i> in $H^{0,1}(T)$ has a representative in the image of $D$. At least, this is how I would try and approach the question; I'm sure there are other methods.</p> http://mathoverflow.net/questions/119586/what-foliations-are-symplectic-foliations Comment by Joel Fine Joel Fine 2013-01-23T12:14:07Z 2013-01-23T12:14:07Z Thanks Robert, thats very clear. (And not so hard after all!) http://mathoverflow.net/questions/119586/what-foliations-are-symplectic-foliations Comment by Joel Fine Joel Fine 2013-01-22T21:53:34Z 2013-01-22T21:53:34Z I agree with Tim that, as currently phrased, this question is an extremely hard open problem, way beyond current techniques. Perhaps though with different hypotheses, it might become a hard, open, but not completely impossible question. For example, what if the leaves are 2-dimensional? (Lefschetz fibrations play nicely with symplectic topology, after all.) What if we're allowed foliations and Poisson structures with singularities or degeneracies? (Perhaps along the lines of broken pencils on 4-manifolds?) http://mathoverflow.net/questions/100630/symplectic-form-for-the-spinors Comment by Joel Fine Joel Fine 2012-06-25T21:05:31Z 2012-06-25T21:05:31Z Isn't tr(XY)=tr(YX), meaning w is zero? http://mathoverflow.net/questions/19210/holomorphic-vector-fields-acting-on-dolbeault-cohomology/87889#87889 Comment by Joel Fine Joel Fine 2012-02-08T16:13:45Z 2012-02-08T16:13:45Z Yes, sorry, I forgot to specify that I was really looking for compact examples. http://mathoverflow.net/questions/59738/the-complex-version-of-nashs-theorem-is-not-true/59771#59771 Comment by Joel Fine Joel Fine 2011-12-21T11:41:43Z 2011-12-21T11:41:43Z I think this version of your answer is much better! Just one small technical point: the convergence of the sequence you mention is in fact in the C-infinity topology. Tian originally proved C^2 convergence but this has subsequently been improved. (I guess this extension is due to Ruan, but I'm a little hazy on the exact history. There is also important work of Zelditch and Catlin on this.) http://mathoverflow.net/questions/69980/give-a-restriction-to-ensure-a-surgery-of-a-balanced-manifold-is-still-balanced Comment by Joel Fine Joel Fine 2011-07-12T08:32:09Z 2011-07-12T08:32:09Z There are at least two totally different definitions of &quot;balanced&quot; for a complex manifold. Also, what do you mean by &quot;surgery&quot;? There are many different things you could mean, most of which do not keep your manifold complex (at least not in a natural way). http://mathoverflow.net/questions/69703/examples-of-non-simply-connected-manifolds-with-trivial-h1 Comment by Joel Fine Joel Fine 2011-07-07T14:27:49Z 2011-07-07T14:27:49Z You've almost answered your own question. You can realise any finitely presented group as the fundamental group of a closed 4-manifold. You take your 2-dimensional CW complex, embed it in R^5 (dim 5 is necessary so you can separate all the 2-cells) and then take the boundary of a tubular neighbourhood (and perhaps smooth off any corners). The result is a closed 4-manifold with the same fundamental group as the CW complex, unless I'm mistaken. http://mathoverflow.net/questions/68316/extremal-fano-with-non-constant-scalar-curvature-vs-kaehler-einstein-fano-manifol/68402#68402 Comment by Joel Fine Joel Fine 2011-06-21T16:17:20Z 2011-06-21T16:17:20Z It's just dawning on me which Yann you probably are. Salut! In which case you can skip the first two paragraphs of my answer, and you'd probably already thought of the second two yourself... http://mathoverflow.net/questions/66046/which-nonlinear-pdes-are-of-interest-to-algebraic-geometers-and-why/67047#67047 Comment by Joel Fine Joel Fine 2011-06-10T13:46:53Z 2011-06-10T13:46:53Z @mathphysicist: Sorry to take so long to get back to you. Good places to read about this are the two articles of Donaldson: &quot;A new proof of a theorem of Narasimhan and Seshadri&quot; in the Journal of Differential Geometry and &quot;Anti-self-dual connections over complex algebraic surfaces and stable vector bundles&quot; in Proceedings of the London Mahtematical Society. Another great article is &quot;The Yang-Mills equations over Riemann surfaces&quot; by Atiyah and Bott, in, I think, Transactions of the Royal Society. http://mathoverflow.net/questions/64029/if-a-manifold-suspends-to-a-sphere/64036#64036 Comment by Joel Fine Joel Fine 2011-05-06T00:31:58Z 2011-05-06T00:31:58Z @Igor again, I'm sorry, having reread all of the comments more carefully I see that this line of reasoning was clearly already apparent to you! http://mathoverflow.net/questions/64029/if-a-manifold-suspends-to-a-sphere/64036#64036 Comment by Joel Fine Joel Fine 2011-05-06T00:30:16Z 2011-05-06T00:30:16Z @Igor, no me neither! I guess it's the case that $M$ must be a homotopy sphere and so it follows from Poincar&#233;. But since we are assuming a homeomorphism it feels like we're giving ourselves strictly more information than in the Poincar&#233; conjecture. Although I could easily be wrong on that. http://mathoverflow.net/questions/64029/if-a-manifold-suspends-to-a-sphere/64036#64036 Comment by Joel Fine Joel Fine 2011-05-05T22:43:30Z 2011-05-05T22:43:30Z I'm no expert but perhaps it's not necessary to invoke all of these deep theorems to prove this. I'm fairly sure the following is true. Given an $n$-manifold $M$, consider the cone $CM$ on $M$. Then the vertex of $CM$ has a neighbourhood homeomorphic to an open set in $\mathbb{R}^{n+1}$ if and only if $M$ is the $n$-sphere. If this really is true then the only manifold whose suspension is again a manifold is $S^n$. Am I right here? http://mathoverflow.net/questions/63439/how-can-we-detect-the-existence-of-almost-complex-structures/63476#63476 Comment by Joel Fine Joel Fine 2011-05-03T20:44:13Z 2011-05-03T20:44:13Z Yep that's right: there is no known example of a manifold of dimension 6 or higher which admits an almost complex structure but does not admit a complex structure. I agree my original wording was very poor, so I've changed it so it actually makes sense! http://mathoverflow.net/questions/62492/non-kahler-manifolds-and-the-ddc-lemma/62561#62561 Comment by Joel Fine Joel Fine 2011-04-22T03:26:53Z 2011-04-22T03:26:53Z No access to Gray's paper where I am, so no idea if he does the following, but here's a short proof that $h^{0,1}\geq 1$ for those who are curious. The argument holds whenever both integer cohomology in degree 2 and $b_n$ vanish on a complex $n$-fold. Since $H^2(Z)=0$, the exponential sequence shows $H^{0,1}$ surjects onto the group of line bundles. So it's enough to find a non-trivial holomorphic line bundle. Now the canonical bundle is nontrivial, for if not a holomorphic $n$-form $\Omega$ would have $\int \Omega \wedge \bar{\Omega} &gt;0$ so $[\Omega]$ would be non-zero contradicting $b_n=0$. http://mathoverflow.net/questions/62492/non-kahler-manifolds-and-the-ddc-lemma/62507#62507 Comment by Joel Fine Joel Fine 2011-04-22T02:26:10Z 2011-04-22T02:26:10Z @Yang-Mills, from what you say it follows that $h^{0,1} = 0$ DOES imply the $dd^c$-lemma for $(1,1)$-forms. For if $h^{0,1}=0$ then it follows that $b_1=0$ hence certainly $b_1 = 2h^{0,1}$. But there is a much more prosaic way to see it. Let $a = db$ be an exact real (1,1) form. Then $\bar{\partial}b^{0,1} = 0$ so $b^{0,1} = \bar{\partial} f$. From here it follows that $a = 2 i \partial\bar{\partial} g$ where $g$ is the imaginary part of $f$.