User diego matessi - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:34:59Z http://mathoverflow.net/feeds/user/6658 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103243/what-are-some-interesting-problems-in-the-intersection-of-algebraic-number-theory/103278#103278 Answer by Diego Matessi for What are some interesting problems in the intersection of Algebraic Number Theory and Algebraic Topology? Diego Matessi 2012-07-27T07:20:20Z 2012-07-27T07:26:24Z <p>This is something I just heard about yesterday. In this article</p> <p>Nodal quintics in P^4.<br> B. van Geemen and J. Werner In: Arithmetic of Complex Manifolds; W.-P. Barth, H. Lange Eds. Springer LNM 1399, (1989) 48-59.</p> <p>(find a copy here: <a href="http://users.mat.unimi.it/users/geemen/publ.html" rel="nofollow">http://users.mat.unimi.it/users/geemen/publ.html</a>).</p> <p>the authors compute the Betti numbers of certain smooth complex Calabi-Yau manifolds using a combination of techniques from algebraic topology and number theory. If I understand they first reduce the equations to some finite field, then combining the use of etale cohomology, the Lefschetz fixed point theorem and other things they reduce the problem to counting a finite set of points. </p> http://mathoverflow.net/questions/28612/do-names-given-to-math-concepts-have-a-role-in-common-mistakes-by-students Do names given to math concepts have a role in common mistakes by students? Diego Matessi 2010-06-18T11:01:45Z 2012-02-29T00:21:40Z <p>Perhaps this question overlaps with similar ones, ... but I want to focus on a particular possible cause of confusion. I notice that students are often confused by the concepts of "infinite" and "unbounded". So, when asked if the set of invertible matrices is compact, they reply "no, because there are an infinite number of matrices with non zero determinant, therefore the set is unbounded". Actually this happens in Italian, where the corresponding words ("infinito" and "illimitato") are almost synonyms in everyday language. Does this happen in English too, or other languages?. I wonder: what if we chose another name for the two concepts? Would they make this mistake anyway? One way to check this would be to compare with what happens in other languages, where perhaps the words chosen do not create the confusion. Do you have other examples of this situation? Can you suggest different math concepts which in one language are named with synonyms, but not in another? Do you know if this problem has been studied anywhere? </p> http://mathoverflow.net/questions/66657/quotient-singularities-with-no-crepant-resolution Quotient singularities with no crepant resolution? Diego Matessi 2011-06-01T12:36:16Z 2011-06-03T09:49:48Z <p>I read that in dimension $\geq 4$ there are Gorenstein abelian quotient singularities that have no crepant resolutions. What is the simplest example? I immagine that there should be toric examples. Is it the case that the cone does not admit a suitable subdivision in simplicial cones, corresponding to a crepant resolution? In your answers, please consider that I am only an amateur algebraic geometer, but I know some toric geometry. </p> http://mathoverflow.net/questions/61515/life-after-hartshorne-the-book-not-the-person/61525#61525 Answer by Diego Matessi for Life after Hartshorne (the book, not the person)... Diego Matessi 2011-04-13T09:09:49Z 2011-04-14T01:34:35Z <p>I'm far from having read all of Hartshorne, but if I did I would study <em>Compact Complex Surfaces</em>, by Barth, Peters, Van de Ven. Also Geometric Invariant Theory would be a nice topic (I know about the book by Mumford, are there other good books on this topic?).</p> <p>Ah, I forgot! How about derived categories? Someone suggested that for this topic a good reference is the book by Hartshorne <em>Residues and duality</em>. I had a look at some notes by Caldararu on the arxiv, "Derived categories of sheaves: a skimming", they seem well written.</p> http://mathoverflow.net/questions/13682/which-mathematical-ideas-have-done-most-to-change-history/58519#58519 Answer by Diego Matessi for Which mathematical ideas have done most to change history? Diego Matessi 2011-03-15T10:55:34Z 2011-03-15T10:55:34Z <p>Together with the decimal system, already proposed by Neel Krishnaswami, I would also put binary notation. </p> http://mathoverflow.net/questions/57520/examples-in-mirror-symmetry-that-can-be-understood/58101#58101 Answer by Diego Matessi for Examples in mirror symmetry that can be understood. Diego Matessi 2011-03-10T19:06:57Z 2011-03-12T09:55:12Z <p>A toy model for mirror symmetry is the following. Consider a real manifold (not necessarily compact) $B$ with an atlas of affine coordinates, i.e. such that the change of coordinate maps are of the type $x \mapsto Mx + b$ where $M$ and $b$ are constant. Then the tangent bundle $TB$ has natural complex coordinates given by $z = x +i y$, where $y$ are coordinates on the fibre. If further one assumes that $\det M = 1$ then $TM$ also has a nowhere vanishing holomorphic $n$-form. On the other hand $T^*B$ has it's usual simplectic structure. So $TB$ and $T^*B$ can be thought of being mirror. One can also twist the complex structure on $TB$ with a B-field. One can go a bit further too, in fact suppose $\gamma$ in $B$ is a affine curve (i.e. a straight line in affine coordinates). Then one can lift $\gamma$ to a Lagrangian submanifold $L_{\gamma}$ of $T^*B$ by adding the anhilator of $\gamma'(t)$ in the fibre at $\gamma(t)$. On the other hand the same curve lifts to a complex object in $TB$ by adding the line generated by $\gamma'$ inside the fibre $T_{\gamma} B$. If the affine structure is also integral (i.e. $M$ and $b$ have integral coefficients), then one can also partially compactify by taking latices $\Lambda \subset TB$ and its dual $\Lambda^'$ and then form torus bundles $X = TB / \Lambda$ and $X^* = T^*B / \Lambda^'$. This picture is too simple to work in the compact case, but it is expected that actual mirror symmetry is a perturbation of this. What I just described is the SYZ approach to mirror symmetry. </p> <hr> <p>I like this example. Consider a surface $V$ in $(\mathbb{C}^{\ast})²$ given by some Laurent polynomial $p(z)$, then the hyperfurface $X$ defined by $xy = p(z)$ is Calabi-Yau. Its mirror $\check X$ can be constructed by taking the Newton polygon $\Delta$ of $p$ and then consider the toric variety defined by the cone over ${ 1 } \times \Delta$ in $\mathbb{R}³$. $\check X$ is a resolution of this toric variety obtained from some subdivision of $\Delta$. Now, the surface $V$ (the one we started from) has a "tropical amoeba". This can be thought of a graph in $\mathbb{R}²$ which is the limit (in some sense) of the image of $V$ under the standard torus fibration $(\mathbb{C}^{\ast})² \rightarrow \mathbb{R}²$. The interesting thing is that this graph gives a subdivision of $\mathbb{R}²$ which is dual to the subdivision of $\Delta$ (this is related Gil Kalai's answer). More over this graph is also the locus of singular fibres of a Lagrangian torus fibration defined on $X$. Such a Lagrangian fibration induces on the base an affine structure as I said previously. A construction such as the one I described above can be used to construct many Lagrangian $S^3$'s in $X$ over the bounded regions defined by the graph. The mirror of these objects are the divisors $\check X$ corresponding to interior integral points of $\Delta$, or better, line bundles supported on such divisors. There are some interesting correspondences between intersection points of these spheres and cohomology of the line bundles, even without getting into $\mathcal{A}^{\infty}$ constructions. </p> http://mathoverflow.net/questions/33405/still-open-why-are-two-notions-of-gaussian-curvature-are-the-same-what-is-the/57248#57248 Answer by Diego Matessi for [STILL OPEN] Why are two notions of Gaussian curvature are the same - what is the simplest & most didactic proof? Diego Matessi 2011-03-03T12:38:44Z 2011-03-03T16:02:38Z <p>A discrete version of curvature may help highschool students. Take a polyhedron in $R^3$ and define the curvature at a vertex $v$ by $2 \pi K(v) = 2 \pi - \theta_1 - \ldots - \theta_k$, where $\theta_j$ is the angle at $v$ of the j-th 2-face containing $v$. This gives a measure of how sharp that vertex is. For simplicity, assume there are only three faces meeting at $v$. Let the normal vectors to the faces be $n_1$, $n_2$, $n_3$. They are the corners of a geodesic triangle on the sphere, which is an analog of the region spanned by the Gauss map on a surface. By spherical geometry we have that the area of this triangle is $A = \beta_1 + \beta_2 + \beta_3 - \pi$, where $\beta_j$'s are the angles. One then shows that $\beta_j = \pi - \theta_j$, so we have the nice formula $A = 2 \pi K(v)$. I think that similarily one can argue with parallel transport (see comment by pasquale below). I have not tried the computation, but it should work. Also the global Gauss-Bonnet theorem holds: if you sum the curvature of the vertices on a closed polyhedron the result is the Euler charcteristic (I think this was orginally proved by Descartes, but I'm not sure). </p> http://mathoverflow.net/questions/43131/why-are-the-following-varieties-symplectomorphic/43273#43273 Answer by Diego Matessi for Why are the following varieties symplectomorphic? Diego Matessi 2010-10-23T08:54:39Z 2010-10-23T08:54:39Z <p>I think that $F(n)$ with a Kahler form such that the base $\mathbb{P}^1$ has area 2 and the fibre has area 1, is symplectomorphic to $F(n+2)$ with the Kahler form such that base and fibre have area 1. </p> <p>There is a "visual proof" of this which uses an idea of Leung and Symington (in an article called "Almost toric manifolds", on the arxiv). The operation is called "nodal trade", and is pictured below. </p> <p>Figure A) is the image of the moment map of the toric variety $F(n)$. You can perturb the moment map to another Lagrangian fibration, so to replace a "corner" with a singular Lagrangian fibre (a pinched torus) over the red cross in figure B). Then you can slide this fibre along the dashed straight line until it hits the opposite edge and becomes another "corner". This gives figure C) which is the polytope of $F(n+2)$. </p> <p>Leung and Symington explain these operations carefully in the above paper.</p> <p><img src="http://www.freeimagehosting.net/uploads/a5a7b4ad8e.jpg" alt="alt text"></p> http://mathoverflow.net/questions/40062/roadmap-for-mirror-symmetry/40188#40188 Answer by Diego Matessi for Roadmap for Mirror Symmetry Diego Matessi 2010-09-27T17:42:51Z 2010-09-27T17:42:51Z <p>Auroux's notes for a course on mirror symmetry at Berkeley: <a href="http://math.berkeley.edu/~auroux/277F09/index.html" rel="nofollow">http://math.berkeley.edu/~auroux/277F09/index.html</a>. They look interesting and they cover a lot of material.</p> http://mathoverflow.net/questions/32812/is-the-limit-of-symplectic-diffeomophisms-a-diffeomorphism/32915#32915 Answer by Diego Matessi for Is the limit of symplectic diffeomophisms a diffeomorphism? Diego Matessi 2010-07-22T08:11:37Z 2010-07-22T08:11:37Z <p>To find explicit examples of what Greg Kuperberg suggests, maybe one can try the following. Consider $\mathbb{R}^{2n}$ as the cotangent bundle of $\mathbb{R}^n$. Then diffeomorphisms between opens subsets of $\mathbb{R}^n$ lift to symplectomorphisms of the "cylinders" over these open subsets. So you can find a sequence of diffeomorphisms of $\mathbb{R}^n$ converging to a diffeomorphism of $\mathbb{R}^n$ onto a proper open subset of $\mathbb{R}^n$. When $n =1$ you can take a sequence converging to $f(x) = e^x$ (for instance $f_n(x) = e^x + \frac{x}{n}$). The symplectomorphisms obtained as the lift of these maps should give you an example of what Greg Kuperberg suggested. </p> http://mathoverflow.net/questions/30000/do-you-understand-syz-conjecture/30029#30029 Answer by Diego Matessi for Do you understand SYZ conjecture Diego Matessi 2010-06-30T09:42:22Z 2010-06-30T09:42:22Z <p>Here are some papers on SYZ worth reading:</p> <ul> <li><p>Hitchin's "The moduli space of special Lagrangian submanifolds" <a href="http://arxiv.org/abs/dg-ga/9711002" rel="nofollow"> arXiv:dg-ga/9711002</a></p></li> <li><p>M. Gross's <a href="http://arxiv.org/abs/0802.3407" rel="nofollow"> survey</a></p></li> </ul> <p>Hitchin's paper was written shorly after Mirror Symmetry is T-duality and it is a matematical explanation of the paper. Essentially Maclean proved that the moduli space of sL submanifolds is unobstructed and its tangent space is the space of harmonic 1-forms on the sL submanifold. A natural metric which you can put on the moduli space is the $L^2$ metric on harmonic forms. When the sL submanifold is a torus, the moduli space also has an "affine structure". This was already known from integrable systems, they are called action coordinates. They are affine because they are defined up to affine transformations (with linear part having integral coefficients). Hitchin shows that with respect to these coordinates the metric can be expressed as the Hessian of a function. Hitchin also shows that the moduli space has two affine strutures (this is because of the "special" condition). The two affine structures are related by Legendre transform using the Hessian (i.e.the metric). So one could say that mirror symmetry is "Legendre transform". </p> <p>Things have developed a lot since Hitchin's paper, and M. Gross surveys these developements. How to do "quantum corrections" to the metric is a major open problem, there are many approaches. They seem all quite difficult.... Auroux in the paper mentioned above deals with it. I heard a talk of Fukaya where he wants do do it with Floer homology for families, but I do not know much about it. Then there is the program of Kontsevich and Soibelmann, using rigid analytic geometry and the Gross-Siebert program. It seems that quantum corrections could be understood in terms of "tropical geometry" in the moduli space of SL tori (an "affine manifold with sigularities"). In a recent paper of M. Gross ("Mirror symmetry for $\mathbb{P}^{2}$ and tropical geometry"), he explains how "period calculations" can be understood in terms of tropical geometry (at least for $\mathbb{P}^{2}$). <a href="http://www.math.ucsd.edu/~mgross/" rel="nofollow"> Here </a> you can find a link to a book of M. Gross where he explains the connection between tropical geometry and mirror symmetry. </p> http://mathoverflow.net/questions/29840/homology-of-lens-spaces-using-morse-theory/29929#29929 Answer by Diego Matessi for Homology of lens spaces using Morse theory? Diego Matessi 2010-06-29T15:26:31Z 2010-06-29T15:26:31Z <p>I tried a computation, I hope it's correct. I considered the map h = Re $z^p$ + Re $w^p$ on the sphere, and I found four sets of critical points: (1) ($\mu_{1}^{+} / \sqrt{2}$, $\mu_{2}^{+} / \sqrt{2}$), (2) ($\mu_{1}^{-}/ \sqrt{2}$, $\mu_{2}^{-} / \sqrt{2}$), (3) ($\mu_{1}^{+}$, 0) and (0, $\mu_{2}^{+}$), (4) ($\mu_{1}^{-}$, 0) and (0, $\mu_{2}^{-}$), where $\mu_{j}^+$ denotes a p-root of 1 and $\mu_{j}^-$ denotes a p-root of -1. </p> <p>The first set gives maxima (where h = 2), the second set gives minima (h=-2) and the other two give index 1 and 2 points (I think). The first and second set contribute to p points in the lens space, the other two contribute 2 points each. Perhaps doing Morse homology as Pietro suggested counting connecting flow lines gives you the correct result.... but I have not tried. </p> http://mathoverflow.net/questions/29913/map-of-manifolds-inducing-iso-on-top-cohomology-but-not-surjective-on-one-other/29915#29915 Answer by Diego Matessi for map of manifolds inducing iso on top cohomology, but not surjective on one other cohomology group Diego Matessi 2010-06-29T14:09:13Z 2010-06-29T14:09:13Z <p>I think I have one: consider a torus in $\mathbb{R}^{3}$ embedded as a surface of rotation, e.g. rotate a circle of radius $1$ in the zy plane with center $(0,2,0)$ around the $z$ axis. Now put a small sphere with center $(0,2,0)$ and radius $\epsilon$. Then the map $T \rightarrow S^{2}$ given by projection towards the center of the sphere should give an isomorphism in $H^{2}$ (since the degree is 1) but it is obviously not surjective in $H^{1}$</p> http://mathoverflow.net/questions/29507/coordinates-on-flag-manifolds/29514#29514 Answer by Diego Matessi for Coordinates on Flag Manifolds Diego Matessi 2010-06-25T13:26:19Z 2010-06-25T13:26:19Z <p>I think you can use wedge products. Choose $v \in V_1$, then $u \in V_2$, which is linearly independent. Map the flag to $([v], [v \wedge u]) \in (CP^{2})^2$. This should be well defined and holomorphic.</p> http://mathoverflow.net/questions/29241/pdes-on-the-klein-bottle-and-real-projective-plane/29243#29243 Answer by Diego Matessi for PDEs on the Klein bottle and real projective plane Diego Matessi 2010-06-23T14:54:41Z 2010-06-23T14:54:41Z <p>I do not know of a reference, but maybe the problem can be reduced to studying the problem on the sphere $S^2$ and on the torus $T$ and then looking for solutions with certain symmetries. For instance, if $f$ is a function on $RP^2$ then it comes from a function $g$ on $S^2$ such that $g(p) = g(-p)$. Solve $\nabla^{2} u = f$ on $S^2$, then I think that $u' = (u(p) + u(-p))/2$ is a solution of your problem on $RP^2$. </p> http://mathoverflow.net/questions/28967/a-characterization-of-convexity/28968#28968 Answer by Diego Matessi for A characterization of convexity Diego Matessi 2010-06-21T16:47:40Z 2010-06-21T16:47:40Z <p>In coordinates $(x,y,z)$, consider { $z \geq 0$} and remove from it the half line { $(t,0,0) | t > 0$}. You get something non convex, but it seems to me that it also satisfies your property. Maybe you need some boundedness?</p> http://mathoverflow.net/questions/28541/complexified-kahler-form/28608#28608 Answer by Diego Matessi for complexified kahler form Diego Matessi 2010-06-18T09:32:19Z 2010-06-18T09:32:19Z <p>Since B-fields are being discussed in the other question, i'll try to answer your question on the complexified Kahler moduli. The set of Kahler classes of a Kahler manifold is an open cone in $H^{1,1}(X, C) \cap H^2(X, R)$. There are results on the structure of this cone which roughly say that it is rational polyhedral away from some set (I think!). In the construction of the complexified Kahler cone you have to take a "framing" of this cone, in the end you get something isomorphic to $(\Delta^{\ast})^{h^{1,1}}$, where $\Delta^*$ is the punctured disc in $C$. This matches the moduli of complex structures of the mirror, near a large complex structure limit point. These things are discussed in the book "Calabi-Yau manifolds and related geometries", by Gross, Huybrechts and Joyce. </p> http://mathoverflow.net/questions/1726/how-should-i-think-about-b-fields/28607#28607 Answer by Diego Matessi for How should I think about B-fields? Diego Matessi 2010-06-18T09:05:06Z 2010-06-18T09:05:06Z <p>Let me try to add a different point of view on B-fields and mirror symmetry. Ideally in mirror symmetry, given a Calabi-Yau manifold X, you would like to "construct" its mirror X', where the symplectic form on X should give you the complex structure on X'. As already mentioned, classes of symplectic forms have moduli of real dimension $h^{1,1}(X)$ and complex structures on X' have moduli of complex dimension $h^{2,1}(X') = h^{1,1}(X)$. So the kahler class is not enough to determine all complex structures on X'. In the context of the Strominger-Yau-Zaslow conjecture there is a nice interpretation of the B-field. Suppose X = $T^*B / \Lambda$, where B is a smooth manifold and $\Lambda$ is locally the span over the integers of 1-forms $dy_1$, ..., $dy_n$ (here $y_1$, ..., $y_n$ are coordinates which change with affine transformations from one chart to the other). Then $X$ has a standard symplectic form. We can consider $X'= TB / \Lambda'$, where $\Lambda'$ is the dual lattice. Then X' has a natural complex structure defined as follows. In standard coordinates on TB, given by $(y,x)$ --> $x \partial_y$, the complex coordinates on X' are $z_k = e^{2\pi i(x_k + i y_k)}$, which are well defined due to the nature of the coordinates x and y. But the above complex coordinates can be twisted locally (on a coordinate patch) by $z_k (b) = e^{2\pi i(x_k + b_k + i y_k)}$, where $b = (b_1, \ldots, b_n)$ is some local data. But since on overlaps $U_i \cap U_j$ the coordinates have to match, we must have $b(i) - b(j) \in \Lambda$. It turns out that by putting $b_{ij} = b(i) - b(j)$ on overlaps, we get a cohomology class in $H^{1}(B, \Lambda)$, this is the B-field. The cohomology group $H^{1}(B, \Lambda)$ shoud coincide (in some cases at least) with $H^2(X, R/Z)$, which is what Kevin Lin mentioned. The elliptic curve case (mentioned by Kevin) can be seen from this point of view. </p> <p>This point of view is also called "mirror symmetry without corrections" and it only approximates what happens in compact Calabi-Yaus. I have learned this in papers by Mark Gross (such as "Special lagrangian fibrations II: geometry") or the book "Calabi-Yau manifolds and related geometries" by Gross, Huybrechts and Joyce. </p> <p>I would be interested to know how this interpretation connects to the other ones which have been described. </p> http://mathoverflow.net/questions/28391/definition-of-the-curvature-tensor/28404#28404 Answer by Diego Matessi for Definition of the curvature tensor Diego Matessi 2010-06-16T15:45:19Z 2010-06-16T17:52:28Z <p>If you consider a frame of your bundle E, say $E_1$...$E_k$ then there will be forms $\omega_{jk}$ such that</p> <p>$$\nabla E_j = \omega_{jk} E_k$$</p> <p>(summation over repeated indices). The notation $\nabla_X E_j$ becomes $\omega_{jk}(X) E_k$. So for instance $$\nabla_X \nabla_Y (E_j) = X(\omega_{jk}(Y))E_k + \omega_{jk}(Y)\omega_{kl}(X) E_l$$ where I have applyied the rules you mentioned. On the other hand $$\nabla \circ \nabla (E_j) = \nabla(\omega_{jk} E_k) =d\omega_{jk}E_k - \omega_{jk}\wedge (\nabla E_k) = (d\omega_{jl} - \omega_{jk} \wedge \omega_{kl}) E_l,$$ (again applying the rules). Now you can apply the above two forms to vectors X and Y. You will need the formula $$d\omega_{jl}(X,Y) = X(\omega_{jl}(Y)) - Y(\omega_{jl}(X)) - \omega_{jl}([X,Y]),$$ this is how the $\nabla_{[X,Y]}$ part comes out in the curvature. By applying all these ideas in the end you get the "usual" formula for the curvature.</p> <p>Reference: a good book on this is "From calculus to cohomology" by Madsen and Tornehave. </p> http://mathoverflow.net/questions/28250/checking-whether-the-image-of-a-smooth-map-is-a-manifold/28259#28259 Answer by Diego Matessi for Checking whether the image of a smooth map is a manifold Diego Matessi 2010-06-15T15:06:42Z 2010-06-15T15:06:42Z <p>If you consider the real map $(x_1, x_2)$-->$(x_1+x_2, x_1x_2)$ then the image of the circle is not a submanifold. Infact your map is not injective: it is symmetric with respect to swapping the two coordinates, so the circle is folded once onto itself and its image is homeomorphic to a closed segment. In the complex case, my guess is that the image is homeomorphic to the quotient of the sphere by the involution that swaps the two complex variables... The fixed point set of this involution is the circle given by intersection with the line $(z,z)$. Near this circle the quotient should be like the quotient of R x C by the involution $(t, z) \mapsto (t, -z)$, which is still a manifold. So I would think F(M) is a manifold...</p> http://mathoverflow.net/questions/27749/what-are-some-correct-results-discovered-with-incorrect-or-no-proofs/27813#27813 Answer by Diego Matessi for What are some correct results discovered with incorrect (or no) proofs? Diego Matessi 2010-06-11T13:18:54Z 2010-06-11T13:18:54Z <p>Just to complement Gerhard Paseman's answer. The story of how Girolamo Saccheri in early 1700's "almost" discovered hyperbolic geometry is quite amusing. Actually he died thinking he had proved the fifth postulate, but his argument was weak: "the hypothesis of the acute angle is absolutely false; because it is repugnant to the nature of straight lines". The sentence referes to his construction of a quadrilateral with two sides of equal length perpendicular to a given one. The acute angles are the ones opposite to the right ones. But Wikipedia explains this too...</p> <p>In this example an ideological bias prevented the discovery of beautiful mathematics... I wonder if this still happens now a days, probably yes. </p> http://mathoverflow.net/questions/27660/applications-of-compactness/27661#27661 Answer by Diego Matessi for Applications of compactness Diego Matessi 2010-06-10T08:04:22Z 2010-06-10T08:04:22Z <p>A map from a compact space to a Hausdorff space is a homeomorphism if and only if it is continuous and bijective. This is useful to prove, for example, that all simple and closed curves are homeomorphic to the circle. </p> http://mathoverflow.net/questions/27586/number-of-a-subset-of-monomials/27602#27602 Answer by Diego Matessi for Number of A Subset of Monomials Diego Matessi 2010-06-09T16:58:20Z 2010-06-09T17:39:50Z <p>Isn't it k times the number of monomials of degree n-1 in k-1 variables? Since in such a monomial you have x_j followed by a degree n-1 monomial in the other variables. </p> <p>For the number of monomials of degree n-1 in k-1 variables you can check Wikipedia, search for "monomials". </p> <p>.... i just realised this is wrong! for example $x_1 x_2^4 x_3$ would be counted twice in the way i said.</p> http://mathoverflow.net/questions/26585/applications-of-connectedness/27484#27484 Answer by Diego Matessi for Applications of connectedness Diego Matessi 2010-06-08T16:02:08Z 2010-06-08T16:02:08Z <p>As an application of connectedness you can prove the "Borsuk-Ulam" theorem in dimension 1, i.e. that for any continuous function $f$ from $S^1$ to the reals there are two radially symmetric points which are mapped to the same point. This is because the function g(v) = f(v) - f(-v) is either constant or has points where it is positive and points where it is negative, therefore it must have a point where it is zero.</p> <p>As an application of this fact you can show that for any pair of compact regions A and B inside the plane there is one line splitting each region in pieces of equal area (see the book by Kosniowski, A first course in algebraic topology, where this is referred to as a "pancake problem"). </p> http://mathoverflow.net/questions/66657/quotient-singularities-with-no-crepant-resolution/66694#66694 Comment by Diego Matessi Diego Matessi 2011-06-03T08:21:35Z 2011-06-03T08:21:35Z thank you very much for the example, it should be easy enough for me to understand it! http://mathoverflow.net/questions/66657/quotient-singularities-with-no-crepant-resolution/66660#66660 Comment by Diego Matessi Diego Matessi 2011-06-03T08:20:38Z 2011-06-03T08:20:38Z thanks for the reference, this will be useful. http://mathoverflow.net/questions/66657/quotient-singularities-with-no-crepant-resolution/66702#66702 Comment by Diego Matessi Diego Matessi 2011-06-03T08:19:50Z 2011-06-03T08:19:50Z thank you very much sandor! my knowledge of algebraic geometry is (sigh!) still too weak to understand all the words you mention (e.g. $\mathbb{Q}$-factorial) but i'll keep this as something to learn! http://mathoverflow.net/questions/46156/is-every-closed-set-of-q-the-intersection-of-some-connected-closed-set-of-r-wit/65132#65132 Comment by Diego Matessi Diego Matessi 2011-05-16T12:42:53Z 2011-05-16T12:42:53Z I thought about the same thing, but then I realized that the set you obtain may not be closed in $\mathbb{R}^2$. http://mathoverflow.net/questions/57520/examples-in-mirror-symmetry-that-can-be-understood Comment by Diego Matessi Diego Matessi 2011-03-11T13:31:36Z 2011-03-11T13:31:36Z I think that research in mirror symmetry goes in the opposite direction to what happened with Poincare'-Birkhoff discovery. In that case a simple statement lead to a beautiful rich theory. In mirror symmetry a very complicated statement (such as the counting formula for curves on the quintic), which no-one understood at first, lead to a theory which is slowly becoming clearer and enriched with simpler examples. http://mathoverflow.net/questions/33405/still-open-why-are-two-notions-of-gaussian-curvature-are-the-same-what-is-the/57248#57248 Comment by Diego Matessi Diego Matessi 2011-03-03T16:00:26Z 2011-03-03T16:00:26Z @pasquale zito. I think you are right on parallel transport! Also because projection may give zero. http://mathoverflow.net/questions/33405/still-open-why-are-two-notions-of-gaussian-curvature-are-the-same-what-is-the/57248#57248 Comment by Diego Matessi Diego Matessi 2011-03-03T15:54:04Z 2011-03-03T15:54:04Z @pasquale zito. $\theta_1$ is the angle between $n_3 \wedge n_1$ and $n_1 \wedge n_2$. I think what I said is right. Assume $n_1$, $n_2$ and $n_3$ are positively oriented. Draw spherical geodesic arcs from $n_3$ to $n_1$ and from $n_2$ to $n_1$ and denote the tangent vectors to these arcs at $n_1$ by $v_3$ and $v_2$ respectively. $n_3 \wedge n_1$ is $v_3$ rotated 90 degrees in the direction of $v_2$ and $n_1 \wedge n_2$ is $v_2$ rotated 90 degrees in the direction of $v_3$. This gives the formula. http://mathoverflow.net/questions/29840/homology-of-lens-spaces-using-morse-theory/29929#29929 Comment by Diego Matessi Diego Matessi 2010-07-01T07:03:49Z 2010-07-01T07:03:49Z There are $2p + 4$ of them... I have not checked that they are all non-degenerate. I suspect they are not... the set (3) and (4) seem too few to give the correct homology. http://mathoverflow.net/questions/30000/do-you-understand-syz-conjecture/30029#30029 Comment by Diego Matessi Diego Matessi 2010-07-01T06:53:42Z 2010-07-01T06:53:42Z the most difficult part is to understand how to modify the complex structure on the moduli space. As I said, Kontsevich-Soibelman and Gross-Siebert make a lot of progress on this using tropical geometry. I think the role of holomorphic disks is still unclear, although Auroux, Fukaya (and maybe others) are working on it. http://mathoverflow.net/questions/29840/homology-of-lens-spaces-using-morse-theory Comment by Diego Matessi Diego Matessi 2010-06-29T13:49:07Z 2010-06-29T13:49:07Z I tried this function h = Re $z^{p}$ + Re $w^p$. If I did the calculations right, on the sphere it should have only these isolated critical points: $( \mu_1 /\sqrt{2}, \mu_2 / \sqrt{2})$ where $\mu_1$ and $\mu_2$ are p-roots of $1$. http://mathoverflow.net/questions/29840/homology-of-lens-spaces-using-morse-theory Comment by Diego Matessi Diego Matessi 2010-06-29T09:59:21Z 2010-06-29T09:59:21Z Isn't $r$ always positive? I thought it was the radius of $z$. http://mathoverflow.net/questions/29840/homology-of-lens-spaces-using-morse-theory Comment by Diego Matessi Diego Matessi 2010-06-29T08:08:52Z 2010-06-29T08:08:52Z .... but I'm afraid it will have a whole critical circle when z=0, so it would not be Morse as Ryan Budney says. http://mathoverflow.net/questions/29840/homology-of-lens-spaces-using-morse-theory Comment by Diego Matessi Diego Matessi 2010-06-29T07:53:14Z 2010-06-29T07:53:14Z Are you sure you didn't mean $h = r^{p}cos p \theta$. Then your function would be smooth and well defined on L(p,q). http://mathoverflow.net/questions/29840/homology-of-lens-spaces-using-morse-theory Comment by Diego Matessi Diego Matessi 2010-06-29T07:44:01Z 2010-06-29T07:44:01Z I agree with Petya, your function is not smooth. If you consider the map p: C ---&gt; $S^3$ suggested above and take z=x real, then $h \circ p (x) = |x|$...Since here $\theta =0$ and $r = |x|$. http://mathoverflow.net/questions/28612/do-names-given-to-math-concepts-have-a-role-in-common-mistakes-by-students Comment by Diego Matessi Diego Matessi 2010-06-18T14:09:46Z 2010-06-18T14:09:46Z @Pete L. Clark. I don't know how to change the question. The reference to languages is just a suggestion on how one could find out whether it is the names which cause the confusion. Simply because other languages may have chosen better names. The answers that are coming in are in theme. Thank you all...