User dmitri - MathOverflow

Answer by Dmitri for stability of Riemann surface with boundary

2010-08-01T08:57:25Z

Yes, it is equivalent. To prove, take the double of the surface along the boundary. The Euler charactericsitc will double. This proves that that if the Euler characteristics of the surfaces with boundary is negative then the surface is stable.

Suppose now that the Euler characteristic of the surface double is non-negative. Then we have 4 posibilities. This is a sphere with maximum 2 marked points or torus without marked points. This gives also the complete list of unstable surfaces with boundary:

Disk, a disk with one marked point (either in the exterior or in the interior), a disk with two marked points on the boudnary, and finally an annulus without marked points.

Answer by Dmitri for 4-manifolds in the 4-sphere such that it, and its complement have unsolvable word problem

2010-07-27T21:13:52Z

The answer to the question in the beginning should be YES and it follows from the answer to your previous question. We just need to use the fact that if $G$ has unsolvable word problem then $G*F$ too, where $F$ is a group and $G*F$ is the free product.

To construct the example take the solution to the previous question, namely a $4$-manifold $M^4 $embeddable in $S^4$ with unsolvable word problem. Now take an open ball $B^4$ in $M^4$ and cut from it a small copy of $M^4$, that we call $N^4$. Finally dig a wormhole that connects $N^4$ with $S^4\setminus M^4$. This divides the sphere into two connected 4-manifolds each of which is homotopic to a connected sum of two manifolds, one of which has same fundamental group as $M^4$.

Is there a reference containing standard mathematical notations?

2010-07-23T23:35:59Z

Suppose you are writing a mathematical text (say an article) and you want to call an object (for example, a set) by a letter. It would be cool then to have some reference (optimally available on the internet) where you could find some standard letters and notations of mathematical objects and pick one that you like. Does such a "notation dictionary" exist?

ADDED. Thanks everybody for interesting answers! Maybe it is worth to add that I had in mind rather basic things. The question was trigged by my attempt to find a good letter to denote a subset of the segment $[0,1]$. Finally I decided to call it $T$ (in the course of the proof it turns out that $T$ is equal $[0,1]$ :) ).

Answer by Dmitri for Illuminating piecewise-flat manifolds with geodesics

2010-07-21T21:43:50Z

I would like to propose a simple example of a flat surface of genus $3$ with dark points. This also gives an example in dimension $3$. It is based on the following simple observation.

Observation. Consider the torus $T^2=\mathbb R^2/\mathbb Z^2$. Suppose that there is a geodesic segment that joins points $(0,0)$ with $(\frac{1}{2},\frac{1}{2})$. Then it passes through one of four points $(\pm \frac{1}{4},\pm\frac{1}{4})$.

Now consider the double ramified cover $S$ of $T^2$ branching at points $(\pm \frac{1}{4},\pm\frac{1}{4})$. Then on $S$ there is no geodesic segment that goes from any of two preimages of the point $(0,0)$ to any of preimages of the point $(\frac{1}{2},\frac{1}{2})$. Indeed if there were such a segment it would project on $T^2$ to a segment that joins $(0,0)$ with $(\frac{1}{2},\frac{1}{2})$. Hence on $S$ it should pass through a branch point, which is forbidden.

Added. One can desribe give an alternative description of this example. Namely, we can take $8$ copies of squares of size $\frac{1}{2}\times \frac{1}{2}$ and glue a surface of genus $3$ from them in such a way, that at each vertex $8$ squars meet.

If you want a 3-dimensional example just multiply this example by $S^1$. But it should be of course possible to construct examples that are not products, using similar idea.

Contracting a geodesic on a space of curvature less than 1

2010-07-15T23:02:23Z

I would like to ask for a reference to the following statement (hopefully correct):

Let $M$ be a manifold of sectional curvature at most $1$ and let $\gamma$ be a closed geodesic. Suppose that $\gamma$ is contractible. Then for any contraction of this geodesic at some point its length will be equal to $2\pi$.

It would be even better if there is a reference for the case when $M$ a locally $CAT(1)$ space (not necessarily manifold)

Answer by Dmitri for Structure of Kähler cone

2010-07-17T11:20:57Z

Generalising the case of Hirzebrouch surface, you can say that toric varieties admit explicit description of Kahler cone. Also for each Fano variety its Kahler cone is polyhedral, i.e., it is spanned by a finite number of rays (but this does not mean, that the description is easy). If you leave the class of Fano varieties unpleasant things may start to happen. For example for a generic blow up of $\mathbb CP^2$ in $n\ge 10$ points the structure of Kahler cone it is still unknown (for $n<9$ we get Fano), this is related to Nagata conjecutre http://en.wikipedia.org/wiki/Nagata's_conjecture_on_curves

Morrison's conjecture states that for a Calabi-Yau manifold the quotient of the Kahler cone by the group of isometries of the manifold is polyhedral. The conjecture was proved only for surfaces, there is a recent very nice paper of Burt Totaro on this topic "The cone conjecture for Calabi-Yau pairs in dimension two", http://arxiv.org/abs/0901.3361

Curves of constant curvature on S^2

2010-07-14T23:06:10Z

Most probably this is a well known question Consider $S^2$ with a Riemannian metric. I would like to ask what is known about the structure of the set of simple (without self-intersections) closed curves on it of constant geodesic curvature.

Here is a series of questions.

1) Is this true that through each point of $S^2$ passes a simple closed curve of constant curvature? If not, can one estimate from below the proportion of the area of $S^2$ covered by such curves?

2) Is it true that for each value of curvature there are at least $2$ simple closed curves on $S^2$ of this curvature? Or maybe even more than $2$?

3) What can be said about the global structure of these curves on a generic $S^2$? Taking the union of all such closed curves we could try to cook up from them a surface (that maps naturally to $S^2$). Is something known about the topology of this surface?

Comments

1) The theorem of Birkhoff states that each Riemannian $S^2$ contains at least three simple closed geodesics, as Joseph remarks below.

2) For a generic metric on $S^2$ the set of such curves this set should be one dimensional. Indeed for each fixed value of curvature you can consider an analogue of the geodesic flow on the space of unite tangent vectors to $S^2$ and you expect that closed orbits will be isolated.

Answer by Dmitri for Closed 3-manifolds with free abelian fundamental groups

2010-07-14T21:20:45Z

Only $\mathbb Z$ and $\mathbb Z^3$ (for $T^3$) are free abelian groups that appear as fundamental groups of $3$-manifolds. Hopefully the following is an approximative proof.

The manifold must be prime (otherwise the $\pi_1$ is not ableian), hence it is $K(\pi,1)$. Hence its cohomology are just cohomology of the group $\mathbb Z^n$. So the can not get $\mathbb Z^2$ since $H^3(\mathbb Z^2)=0$, and we can not get $\mathbb Z^n$ with $n>3$ since $H^n(\mathbb Z^n)=\mathbb Z$.

Answer by Dmitri for Does negative Kodaira dimension imply uniruled?

2010-07-12T21:20:14Z

As far as I understand not all experts in birational geometry would agree that Siu settles in his preprint Abundance conjecture, and this conjecture is considered for the moment as open. When Kawamata mentions in his recent paper the work of Siu

On the abundance theorem in the case $\nu=0$

http://arxiv.org/PS_cache/arxiv/pdf/1002/1002.2682v3.pdf

his says that in his work Siu using analytic methods reproved the abundance conjecture for a non-minimal algebraic variety whose numerical Kodaira dimension is equal to 0. I would guess Kawamata would not write his paper if it were accepted that Siu gives complete proof of abundance conjecture.

So for the moment it should be conisdered that for $X$ of dimesnion $4$ and higher it is unknown if $H^0(nK_X)=0$ for all $n$ impies that $X$ is unirulled. As for analysis been stronger than algebra, there are examples when this is not quite the case. A famous one is bend and break, that uses characteristics $p$ and is a corner stone of minimal model program http://www.math.ens.fr/~debarre/Grenoble.pdf The existence of minimal models is still unknown even for 3-dimensional Kahler manifolds that are not algebraic...

Retraction of a Riemannian manifold with boundary to its cut locus

2010-07-14T00:42:20Z

This question is edited following the comment of Joseph. He pointed out that the main object of the first vesrion of this question is the cut locus.

Recall that the cut locus of a set $S$ in a geodesic space $X$ is the closure of the set of all points $p$ that have two or more distinct shortest paths in $X$ from $S$ to $p$. http://en.wikipedia.org/wiki/Cut_locus

A simple lemma shows that for a disk $D^2$ with a Riemannian metric the cut locus of $D^2$ with respect to its boundary is a tree. A picture of such tree can be found on page 542, figure 17 of the article of Thurston "Shapes of polyhedra". The tree is of white colour. http://arxiv.org/PS_cache/math/pdf/9801/9801088v2.pdf For an ellipse on the 2-plane the tree is the segment that joins its focal points.

More generically for a Riemannian manifold $M^n$ with a boundary the cut locus of $\partial M$ should be a deformation retract of $M$ (I guess it is a $CW$ complex of dimension less than $n$). To prove this lemma notice that $M^n\setminus Cat locus(\partial M^n)$ is canonically foliated by geodesic segments that join $X$ with $\partial M$.

I wonder if this lemma has a name or maybe it is contained in some textbook on Riemannian geometry?

Answer by Dmitri for A question about regularity of foliations

2010-07-11T22:19:35Z

As far as I understand, the answer to the question is no, you can check this in Handbook of Dynamical Systems, Volume 1, Part 1 By Boris Hasselblatt, Anatole Katok, page 173. This question is identical to the following -- suppose we have a diffeo of S^1 toplogically conjugate to an irrational rotation, can we make this conjugation a diffeo? Here is the sitation from the book:

For smooth and analytic circle diffeomrophisms with extremely well approximable rotation number, the conjugacy to a roation and hence the invariant measure tend to be singular. Arnold's theorem exposed sigularity of the conjugacies as a generic phenomenon in typical one-parameters families of real-analityc maps

Answer by Dmitri for Applications of non-reductive GIT

2010-07-08T20:00:04Z

On 80's Atiyah conference Kirwan spoke about one application. Namely she stated in her talk that there is an application to Green-Griffits conjecture. You can download the talk here http://www.maths.ed.ac.uk/~aar/atiyah80.htm and the slides are here http://www.icms.org.uk/downloads/GandP/Kirwan.pdf I am not sure if this was written down somewhere.

Answer by Dmitri for Osculating conics and cubics and beyond

2010-06-15T14:11:56Z

These highly osculating curves were studied, in particular by V.I. Arnol'd. One of the important refferences will be:

Topological invariants of plane curves and caustics. Dean Jacqueline B. Lewis Memorial Lectures presented at Rutgers University, New Brunswick, New Jersey. University Lecture Series, 5. American Mathematical Society, Providence, RI, 1994.

More precisely, what was studdied are the points of the curve, where the level of it tangency with (say) conics is higher than expected. I guess these are exactly the points that (using your terminology) separate elliptic part of the curve form hyperbolic.

The key words for these research are Extactic points (therminology proposed by D. Esenbud). Using google scholar you can find a complete text of Arnol'd, called

Remarks on the extatic points of plane curves V.I. Arnold - The Gelfand Mathematical Seminars, 1993-1995.

These article contains some genearlisations of four vertex theorem. http://en.wikipedia.org/wiki/Four-vertex_theorem

One more nice refference is a paper of Tabachnikov and Timorin http://arxiv.org/PS_cache/math/pdf/0602/0602317v2.pdf

De Rham decomposition theorem, generalisations and good references

2010-06-13T16:38:54Z

De Rham decomposition theorem states that every simply-connected Riemannian manifold $M$ that admits complementary sub-bundles $T'(M)$ and $T''(M)$ of its tangent bundle parallel with respect to the levi-Chevita connection is isometric to the direct product of two Riemanninan manifolds $M'\times M''$.

Question 1. In the first place I would like to have a good reference for a clear "modern" and complete proof of this theorem, if it exists (more recent than Kobayshi-Nomizu pp. 187-193) (Note, that Besse 10.44 claimed that no simple proof exists yet).

Eddited.

Question 2. Secondly it seems to me that there should be some statement much more general than de Rham theorem. Namely, suppose we have a metric space $X$ that is locally decomposable as an isometric product of two in such a way that this decomposition is "coherent" in a appropriate sense, i.e. forms something like a presheaf. When will we be able to say that $X=Y\times Z$? (I am interested only in the cases when this will work, not when this will fail). As corollary of such a general statement one should be able to deduce de Rham theorem for example, for Finsler of polyhedral manifold, ect.

How to prove that a projective variety is a finite CW complex?

2010-06-03T14:30:08Z

Let $X$ be a (singular) projective variety, in other words something given by a collection of polynomial equations in $\mathbb CP^n$ or $\mathbb RP^n$. How one proves that it is a finite $CW$ complex?

Similar question. Suppose that $X$ affine (i.e. given by polynomial equations in $\mathbb C^n$, or $\mathbb R^n$). How one proves that its one point compactification is a finite $CW$ complex?

These questions are sequel to the discussions here:

http://mathoverflow.net/questions/26374/for-which-classes-of-topological-spaces-euler-characteristics-is-defined

Answer by Dmitri for Can one bound the todd class of a 3-dimensional variety polynomially in c_3

2010-05-31T16:05:39Z

The case of complex manifolds of higher dimension is very different from the case of complex surfaces. So the answer to the question about complex $3$ folds is no, there exists a real 6-dimensional simply connected manfiold with integrable complex structures $J_m$ for all $m\in \mathbb Z^+$ such that $c_1c_2=48m$. This is a theorem A from the acticle of Leubrun. Though, the manifolds that he constructs are not algebraic

Topology versus Chern Numbers for Complex 3-Folds

http://arxiv.org/PS_cache/math/pdf/9801/9801133v1.pdf

The question for algebraic manifolds was studied by Kostchik, you may be interested this the following two articles:

TOPOLOGICALLY INVARIANT CHERN NUMBERS OF PROJECTIVE VARIETIES

http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.1587v1.pdf

CHERN NUMBERS AND DIFFEOMORPHISM TYPES OF PROJECTIVE VARIETIES

http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.2857v2.pdf

Finally, it should be added that a systematic attempt to construct various complex 3-fold is given in a very nice article of Okonek and Van de Ven "CUBIC FORMS AND COMPLEX 3-FOLDS" of Okonek, Ch. / Van de Ven, A, L'Enseignement Mathématique Volume / Année: 41 / 1995. The link is given in the comments

Answer by Dmitri for Are spaces of holomorphic maps manifolds?

2010-05-31T13:54:47Z

Here are some remarks about the last part of the question.

I wonder if the component of the space of maps $Hom(Y,Y)$ of degree $1$ and higher is not always smooth (it is not clear straight away how to consturct a contre-example). In the case of degree $1$ maps the space is clearly smooth since it is a Lie group of dimension $H^0(TY)$. In general, manifolds that admit slef-maps of degree higher than $1$ are not so common. For example if you take in $\mathbb CP^n$ ($n>3$) a hypersurface of degree $2$ and higher it does not admit self-maps of degree higher than $1$, this result is discussed in a nice article of Beauville http://math.unice.fr/~beauvill/pubs/endo.pdf.

On the other hand, it is not hard to construct manfiolds for which some irreducible components of the space of self-maps of zero degree will be non-smooth. It is sufficient to take $Y$ such that $Hom(\mathbb CP^1,Y)$ is non smooth, and conisder $X=\mathbb CP^1\times Y$. Then you just conisder maps $X\to \mathbb CP^1\to X$.

Answer by Dmitri for Shrinking Fano surfaces to a point in Calabi-Yau 3-folds

2010-05-29T17:53:14Z

Corrected. In the first version I stated that CY 3-fold containing Fano surfaces are unknown, this is not at all true as Mohammad pointed out. For example, one can take a product $E\times E\times E=X$ with $E$ an elliptic curve admitting a $\mathbb Z_3$ action, and then take crepant resolution of the quotient of diagonal action $X/\mathbb Z_3$. On the resolved manifold there will be $27=3^3$ copies of $\mathbb CP^2$. On the contrary, as Zhiyu pointed out, on smooth quintic in $\mathbb CP^4$ one can not find a plane $\mathbb CP^2$, since the second integral co-homology of the quintic is generated by a hyperplane section (in the previous "correction" of this answer I was claiming the opposite).

The rest of the answer was correct, and here it is.

It is also important that $D$ is Fano. In this case the normal budnle to it is negative (i.e. its inverse is ample), and so we can contract holomorphically the divisor by Grauert theorem. Notice that you can find a Kahler form that shrinks $D$ to zero only if it can be contracted holomorphically.

In order to see what singularity you get you just need to study case by case Fano surfaces. For example in the case of $\mathbb CP^2$ you get obrifold singularity $\mathbb C^3/\mathbb Z_3$ (with diagonal action of $\mathbb{Z}_3$ on $\mathbb C^3$). In the case of $\mathbb CP^1\times \mathbb CP^1$ you get the sinuglariy $x^2+y^2+z^2+t^2=0 / \pm 1$. In the case when $D$ is a cubic surface you get just $P_3(x,y,z,t)=0$, where $P_3$ is a homogenious polynomial of degree $3$. All other cases of cours were studied and some descripiton of singularities exists. In praticular you can find the model of the singularity if you construct an anti-canonical embedding of your Fano surface in some $\mathbb CP^n$.

No, why can you deform the Kahler form? This is because on the total space of the canonical bundle of a Fano surface the set of Kahler forms is connected, and in particular some of these forms are in the cohomology class $-D+\pi^*\omega$, where D is the zero section, and $w$ is the pullback of a Kahler form from the total space of the $K$ with zero section contracted. Replacing $-D$ with $-tD$ $(0< t\le 1)$ you get a one parameter family cohomology classes that can be represented by Kahler forms. Finally, for $t=0$ you get a cohomology class vanishing on $D$ (by definition).

For which classes of topological spaces Euler characteristics is defined?

2010-05-29T19:19:47Z

I would like to know something more than what is written on wikipedia http://en.wikipedia.org/wiki/Euler_characteristic

What would be some large (largest?) class of topological spaces for which $\chi$ is defined, so that all standard properties hold, for example that $\chi(X)=\chi(Y)+\chi(Z)$ if $X=Y \cup Z$, ($Y\cap Z=0$).

ADDED. The answer of Algori indicates that a reasonably large class of spaces for which Euler characteristics can be defined are locally compact spaces $X$, whose one point compactification $\bar X$ is a CW complex. Then we can define $\chi(X)=\chi(\bar X)-1$. For example, the Euler characteristics of an open interval according to this definition is $-1$. This definition rases a second (maybe obvious) question.

Question 2. Suppose $X$ is a locally compact space whose 1 point compactification is a $CW$ complex, and $Y$ is a subspace of $X$ such that both $Y$ and $X\setminus Y$ have this property. Is it ture that $\chi(X)=\chi(Y)+\chi(X\setminus Y)$?

Also, I was thinking, that Euler characteristics is more fundamental then homology.But can it be defined for spaces, where homology is not defined?

Finally, Quiaochu pointed out below that a very similar question was already discussed previously on mathoverflow.

Deformations of sheaves via automorphisms. How to express $Ext^1$?

2010-05-28T20:40:29Z

Let $X$ be a complex manifold (for example $\mathbb CP^n$), let $v$ be a holomorphic vector field on $X$, and let $F$ be a coherent sheaf (for example a vector bundle or a structure sheaf of a point). Then $v$ defines an element in $Ext^1(F,F)$. Indeed $v$ generates an action of $\mathbb C$ on $X$ and taking pull-backs of $F$ under this action we get a deformation of $F$, hence an element of $Ext^1(F,F)$.

Question. Is there any fancy (or not fancy) way to express the corresponding element of $Ext^1(F,F)$ in terms of $v$ and $F$? Maybe there is some construction with jets? (I understand, this is a bit vague)

Added. Two equally nice and far leading answers were given to this question. I would like summarise here what I understood from David's answer in down to earth terms. So, we want to associate an extension $F\to E\to F$ to a vector field $v$. Suppose (just for the sake of been very much down to earth), that $F$ is a vector bundle, and $v$ has no zeros. Then we can consider $1$-jets of sections of $F$ in the direction of $v$ (or if you want, along trajectories of $v$). It is not hard to see that this is a bundle of rank $2rank(F)$, and this is exactly $E$, that we are looking for.

Answer by Dmitri for Examples of the moduli space of X giving facts about a certain X

2010-05-27T22:01:21Z

Here is one example.

Let $M^3$ be a hyperbolic manifold. Consider the moduli space curvature $-1$ metrics on $M^3$ modulo $Diff_0(M^3)$. This is a point. Conclusion: every diffeomorphism of $M^3\to M^3$ is homotopic to an isometry.

Answer by Dmitri for Dolbeault cohomology of Hopf manifolds

2010-05-23T22:50:19Z

Maybe the following article by Soenke Rollenske (and refferences their), could be usefull for you:

Some very non-Kähler manifolds: the Frölicher spectral sequence can be arbitrarily non degenerate

http://arxiv.org/abs/0709.0481

Answer by Dmitri for limiting behaviour of converging loops on a torus

2010-05-22T16:14:03Z

EDITED. The following example should not be conisdered as a contre-example, it is just a well known example of a pathology that can happen.

EXAMPLE. $L$ is composed of a union of the vertical circle $(0, S^1)$ and a disjoint $R^1$ that is emdedded in $T^2$ in such a way, that it accumulates to $(0, S^1)$ from both sides. This $R^1$ projects one-to one to the horisontal circle without a point and can be represented a as graph (function of $x$), than if is given by the following formula:

$y= (sin(1/x))$

The point is that such $L$ can be Hausodrff approximated by a sequence of circles $s_n$ that are not null-homotopic ($s_n$ wiggles more an more near the vertical circle $(0, S^1)$ as $n\to \infty$).

The problem with this example, is that the topology on the union of $S^1$ and $R^1$ that we should take (I guess) is the toplogy induced from $T^2$. And for this induced topology, I guess the projection to the horisontal cicle is non null-homotopic...

Stein manifolds isomorphic at infinity

2010-05-06T22:14:15Z

Suppose $M$ and $N$ are two Stein manifolds of dimension at least $3$ with compact subsets $U$ and $V$ such that $M\setminus U$ is biholomorphic to $N \setminus V$. It it true that $M$ is biholomorphic to $N$?

It this is not true, what is the simplest example? And if this is true, what would be the reference for such a statement?

Answer by Dmitri for Looking for a 3-fold with some property?

2010-04-20T16:32:56Z

Let me give an attempt of a proof of the fact that such example does not exist.

Proof. Suppose $X$ is such a $3$-fold and let $L$ be the line bundle corresponding to the divisor $\mathbb CP^2$. First we will prove that $Pic(X)=\mathbb Z$ and then will get a contradiction.

Notice that $L$ has a lot of sections. In particular in a neighbourhood of $\mathbb CP^2$ for every two points $x,y$ there is section of $L$ that contains $x$ but does not contain $y$. And also notice that for every point of $X$ there is a section that does not contain it. Hence we have a morphism $X\to P(H^0(L)^*)$. This morphism can not contract anything since $L$ is ample. The map is an embedding on the neighbourhood of $D$ and so the image can have singularities at most in codimension $3$.

From this it should follow (I guess), that we can apply Lefshetz that says $Pic (X)=Pic(D)=\mathbb Z$. So $X$ is a Fano with $Pic=\mathbb Z$. Torsten Ekedahl explained how now one can deduce contrudiction, see his comment.

Answer by Dmitri for Holomorphic vector fields acting on Dolbeault cohomology

2010-04-19T14:16:55Z

Take a complex nilpotent or solvable group $G$ with the right action by a co-compact lattice $\Gamma$ and conisder the quotient $G/\Gamma$. On this quotient right-invariant $1$-forms give a subspace of $H^{1,0}$. The group $G$ is acting on $G/\Gamma$ on the left and if it would presrve all the $1$-forms, $G$ would be abelian.

Torsten Ekedahl expained that what is following IS NOT CORRECT (the article of Hasegawa tells something different)

In fact, the simplest example of this kind is given by primary Kodaira surfaces (http://en.wikipedia.org/wiki/Kodaira_surface), they have two holomorphic $1$-forms. These surfaces are described as quotinets of sovlable groups, for example, in an article of Keizo Hasegawa http://arxiv.org/PS_cache/math/pdf/0401/0401413v1.pdf

Answer by Dmitri for Extension of strictly plurisibharmonic functions on a Kahler manifold.

2010-04-19T08:39:41Z

It is instructive to conisder the case of Kahler metrics invariant under torus action. In this case your question becomes a certain (nontivial) question on convex functions.

Recall first, that Kahler metrics on $(\mathbb C^*)^n$ invariant under the action of $(S^1)^n$ have global potential that is given by a convex function $F$ on $\mathbb R^n$. Here $\mathbb R^n$ is identified with the quotient

$(\mathbb C^*)^n/(S^1)^n$ and we take coordinates $log|z_i|$ on $\mathbb R^n$. So we can translate your original question as follows

QESTION. Suppose you have a smooth convex function $F$, defined on $\mathbb R^n$ outside compact $\Omega$. Is it possible to extend $F$ to a smooth convex function on the whole $\mathbb R^n$?

It easy to construct an example of a non-convex $\Omega$ on $\mathbb R^2$, with convex $F$ defined on $\mathbb R^2\setminus \Omega$, so that $F$ can not be extended. For the moment I don't see how to make such an example when $\Omega$ is the unite disk, but it sounds plausible that such examples exist.

Answer by Dmitri for Hamiltonian circle actions and Lefschetz pencils

2010-04-16T23:40:50Z

I think, that in order to answer this question it is worth to conisder the complex algebraic analog of this question. Namely, suppose we have a $\mathbb C^*$ action on a projective manifold $V^n$. Can we find an invariant Lefshetz pencil?

It is sufficient to conisder the case of (complex) surfaces to spot some problems. Namely, the action of $\mathbb C^*$ in a neighborhood of an isolated fixed point should be of one of the following $3$ types: $$(z,w)\to (tz,tw), \;\; (z,w)\to (tz,t^{-1}w),\;\; (z,w)\to (tz,w)$$

where $t\in \mathbb C^*$ and $z,w$ are local coordinates at the fixed point.

There are not so many examples of $\mathbb C^*$ actions on surfaces that have only these three types of fixed points. But here are two examples: first is $\mathbb CP^2$ with the action that fixes one point and a separate line. Second is the action on $\mathbb CP^1\times \mathbb CP^1$ that fixes $4$ points. In both cases there is an (obvious) invariant Lefshetz pencil. You can also take the first example and blow up several distinct points on the fixed line in $\mathbb CP^2$. Maybe this is the complete list... Surelly, from all these examples one gets also the symplectic Lefshetz pencil of the kind that you wish to have.

I don't see why there will be much more examples if you will go into symplectic cathegory.

Lecture notes on representations of finite groups

2009-11-21T19:30:41Z

Next term I am supposed to teach a course on representation of finite groups. This is a third year course for undegrads. I was thinking to use the book of Grodon James and Martin Liebeck "Representations and characters of groups", but also looking for other refferences.

The question is: could you advise some other books (or lecture notes)? Maybe you had a nice experience of teaching or listening to a course with a similar title? It would be really nice if this book (notes) has also exercises.

ADDED. I would like to thank everybody who answered the question, very helpfull answers!!! The answer of John Mangual below contains a "universal" refference. For the moment my favourites are Serre (very clear and short introduction of main ideas), some bits from notes of Teleman and Martin, and Etingof for beautifull exposition. My last problem is to have enough of exercises, in particular to write down a good exam. So I would like to ask if there are some additional refferences for exercises (with or without solutions)?

Answer by Dmitri for Lecture notes on representations of finite groups

2009-12-27T17:44:00Z

Here is one more reference (kindly communicated to me by one of my colleagues), the course of Iain Gordon. You can find the photos of all blackboards (I still have to go through this course, but it seem a bit similar in spirit to the courses of Teleman and Martin).

http://www.maths.ed.ac.uk/~igordon/4rt/rt2008.htm

ADDED.

I just read the article "Representation Theory" of Ian Grojnowski in the book "Princeton companion of mathematics". I find it really vivid, inspiring and amazingly well written. The first 6 pages are on finite groups, then it proceeds to compact Lie groups and non-compact Lie groups, and smoothly finishes with Langlands correspondence :). The beginning will serve perfectly for the introductory lecture of my course :).

Comment by Dmitri

2010-08-01T09:20:21Z

Dear Makhalan, could you please explain what "inertia of some point" means?

Comment by Dmitri

2010-07-31T22:34:06Z

Lemega, after googling a bit I found, there was a discussion of this question on matheoverflow previously, and there is a "refference" there in the answer of Michael Thaddeus: mathoverflow.net/questions/27249/…

Comment by Dmitri

2010-07-27T20:41:11Z

Alex, on the contrary, I think that for large curvature there will be closed curves of constant geodesic curvature near the max and the min of the Gaussian cuvature of the sphere.

Comment by Dmitri

2010-07-21T22:49:15Z

@Joseph, this is terminology from the theory of Riemman surfaces, en.wikipedia.org/wiki/Ramified_covering_map . Double ramified cover is a ramified cover of degree two, i.e. generic point has two preimages. An example of such a cover is $z\to z^2$, $z\in \mathbb C^1$. The example that I proposed can be understood without this treminology. You can glue $S$ from 8 squares of the size $1/2 \times 1/2$, at each vertex on $S$ $8$ squares should meet.

Comment by Dmitri

2010-07-17T11:53:59Z

Artie, thanks a lot! You are right, I was thinking about the blow up at 10 points, I corrected the answer.

Comment by Dmitri

2010-07-17T09:22:25Z

+1 because, this question is about something that we thought to be pretty, but it turned out to be wrong. This is an interesting phenomena, that (I guess) every mathematician experiences in his life, (I guess) not only when he is a student. Maybe this question is quite personal, so it is hard to give really genuine answers, but the question is not empty. It is different from the question of Gowers (common faults believes) -- Gowers's question is about common things, and this question is about individual. "most interesting math mistakes" is also different.

Comment by Dmitri

2010-07-16T23:18:40Z

@macbeth this is a very nice comment, thanks a lot! Also, this is a nice reference, indeed this flow seems to be the magnetic flow as you describe.

Comment by Dmitri

2010-07-15T23:39:02Z

The soulution, thanks to rbell : warwick.ac.uk/~masgak/abstracts/lco.html

Comment by Dmitri

2010-07-15T23:37:14Z

Amazing, this is doing exactly what I want :). Here is the link to the abstract: warwick.ac.uk/~masgak/abstracts/lco.html

Comment by Dmitri

2010-07-15T23:32:30Z

@rbell, thanks a lot! I will have a look at this article.

Comment by Dmitri

2010-07-15T14:40:32Z

Wadim, that was just my guess... I don't understand the paper. It could be that the geuss if wrong.

Comment by Dmitri

2010-07-15T13:29:29Z

Tony, I am not sure if I understand what is your expectation about these manifolds. What about non-algebraic K3 surfaces that don't have line bundles apart from the trivial one? This can be deformed to an algebraic manifold, so the abelian category is not stable (If you impose that the whole deformation stay in the class of "simple" manifolds, you can multiply this family of K3 by a simple complex torus)

Comment by Dmitri

2010-07-15T13:20:06Z

@Jose, thanks, indeed the link does not work, but explained now how to retirve the paper

Comment by Dmitri

2010-07-15T10:03:38Z

Non-algebraic Kahler manifolds are not as intensively studied as algebraic one, so one could not exclude that this terminology "simple" is not common. In a certain sense it is a bit contradictory, these manifolds don't seem to be simple at all :)

Comment by Dmitri

2010-07-14T12:07:33Z

Gianni, I adjusted my answer according to what you say (I was not answering question 1 of the unknown :)

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