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https://mathoverflow.net/questions/8282	8	For $\mathbb{CP}^1$ the bundles of holomorphic and antiholomorphic forms are equal to the $\mathcal{O}(-2)$ and $\mathcal{O}(2)$ respectively. Do the holomorphic and antiholomorphic bundles of $\mathbb{CP}^2$ (or indeed) $\mathbb{CP}^n$) have a description in terms of line bundles. What happens in the Grassmannian setting?	https://mathoverflow.net/users/1648	Holomorphic and antiholomorphic forms of projective space	Greg's otherwise excellent answer gives the impression that computing Chern classes on projective space requires a computer algebra system. I'm writing to repell this impression. The cohomology ring of $\mathbb{P}^{n-1}$ is $\mathbb{Z}[h]/h^n$ where $h$ is Poincare dual to the class of a hyperplane. We have the short exact sequence $$0 \to S \to \mathbb{C}^n \to Q \to 0$$, where $S$ is the tautological line bundle, whose fiber over a point of $\mathbb{P}^{n-1}$ is the corresponding line in $\mathbb{C}^n$. The line bundle $S$ is also called $\mathcal{O}(-1)$, and has Chern class $1-h$. So $$c(Q) = 1/(1-h) = 1+h+h^2 + \cdots h^{n-1}.$$ As Greg explained, the tangent bundle is $\mathrm{Hom}(S, Q) = S^{\star} \otimes Q$. The formula for the Chern class of a general tensor product is painful to use in practice, but we can circumvent that here by tensoring the above exact sequence by $S^{\star}$. $$0 \to \mathbb{C} \to (S^{\star})^{\oplus n} \to T\_{\mathbb{P}^{n-1}} \to 0$$, So the Chern class of the tangent bundle to $\mathbb{P}^{n-1}$ is $$(1+h)^n = 1 + n h + \binom{n}{2} h + \cdots + n h^{n-1}.$$ Yes, I deliberately ended that sum one term early. Remember that $h^n$ is $0$. As Greg says, if this is to be expressible as a direct sum1 of line bundles, then this should be a product of $n-1$ linear forms, $c(T) = \prod\_{i=1}^{n-1} (1+ a\_i h)$. Since this is an equality of polynomials of degree $n-1$, we can forget that we are working modulo $h^n$ and just check whether the honest polynomial $f(x) := 1 + n x + \binom{n}{2} x + \cdots + n x^{n-1}$ factors in this way. The answer is it does not, except when $n=2$ (the case of $\mathbb{P}^1$.) The unique factorization of $f(x)$ is $$\prod\_{\omega^n=1,\ \omega \neq 1} (1+x (1-\omega)).$$ Thus does raise an interesting question. When $n$ is even, $(1+2x)$ divides $f(x)$. This suggests that $\mathcal{O}(2)$ might be a subquotient of $T\_{\mathbb{P}^{n-1}}$. I can't figure out whether or not this happens. --- The case of Grassmannians is going to be worse for three reasons. The two minor ones are that (1) we may honestly have to use the formula for the chern class of a tensor product. and (2) the polynomials in question will be multivariate polynomials. The big problem will be that $H^{\star}(G(k,n))$ has relations in degree lower than $\dim G(k,n)+1$, so we can't pull the trick of forgetting that $h^n=0$ and working with honest polynomials. I suspect that your question was more "give a nice description of the tangent bundle to the Grassmannian" than "can that tangent bundle be expressed as a direct sum of line bundles?". If you seriously care about the latter, I'll give it more thought. 1 Direct sum is the natural thing to ask for in the categories of smooth, or of topological, complex vector bundles. If you like the algebraic or holomorphic categories, as I do, it is more natural to ask for the weaker property that there is a filtration of the vector bundle, all of whose quotients are line bundles.	7	https://mathoverflow.net/users/297	8472	5,804
https://mathoverflow.net/questions/8468	10	What discrete processes/models have been proven to have scaling limits to $\text{SLE}(\kappa)$, for various $\kappa$? I know about loop-erased random walk and uniform spanning trees. What about conjectures in this direction? (Such as the double-dimer-cover cycles, which I read are conjectured to be $\text{SLE}(4)$) I'm very new to this, so if you please could, together with the answer refer to a paper/article/survey accompanying the result, that would be greatly appreciated!	https://mathoverflow.net/users/2384	Models with SLE scaling limit	There are several other models proved to converge to SLE: critical percolation on the triangular lattice, Gaussian Free Field, Harmonic Explorer, and recently also the critical Ising model. You can check the paper Kevin linked or [Schramm's slides](http://dbwilson.com/schramm/memorial/ICM.pdf) from ICM2006 for some highlights. Just keep in mind that this is a fast changing field at the moment so there has been some progress since 2006.	5	https://mathoverflow.net/users/1061	8476	5,807
https://mathoverflow.net/questions/8260	46	This question is related to (maybe even the same in intent as) [Intro to automatic theorem proving / logical foundations?](https://mathoverflow.net/questions/1017/intro-to-automatic-theorem-proving-logical-foundations), but none of the answers seem to address what I'm looking for. There are a lot of resources available for people who want to use proof assistants like Coq, Isabelle, …, to prove properties about programs—and that's no surprise, since a lot of the development of these programs is done by computer scientists. However, I am interested in resources, and especially in course materials (because I'm trying to put together an independent study for a CS student), involving the use of proof assistants to prove mathematical statements—see the work of [Hales](https://sites.google.com/site/thalespitt/) and [Weedijk](http://www.cs.ru.nl/%7Efreek) for examples. Does anyone know of any such?	https://mathoverflow.net/users/2383	Proof assistants for mathematics	I am interested in the same kind of stuff. [This article](https://doi.org/10.1007/978-3-642-02614-0_34) tells about work done to formalize group representation theory in Coq. In particular, they formalize the proof of Maschke's theorem (that $F[G]$ is semisimple when $G$ is a finite group). Some links to math courses using Coq are listed in [Cocorico](https://github.com/coq/coq/wiki/CoqInTheClassroom#Coq_as_a_teaching_tool_in_mathematics).	10	https://mathoverflow.net/users/400	8480	5,810
https://mathoverflow.net/questions/8483	8	Does a rigid-analytic surface defined over a nonarchimedean complete field have Weierstrass points (if its genus is big enough let's say)? Is there a good reference that (ideally) lists theorems for rigid-analytic spaces that are the analog of commonly known theorems about complex analytic spaces?	https://mathoverflow.net/users/2493	Weierstrass points on rigid-analytic surfaces	Quick note: I am going to assume you want to talk about complete curves. One can, of course, have a curve with punctures in algebraic geometry, and I'm not sure how you'd want to define a Weierstrass point on it. In rigid geometry, you have even more freedom: you can have the analogue of a Riemann surface with holes of positive area, and I think (not sure) you can also build the analogue of a Riemann surface of infinite genus. I'm going to assume you are not thinking about these issues. What you want is the rigid GAGA theorem. I'm not sure what the best reference is; I refreshed my memory from [Coleman's lectures](http://math.berkeley.edu/~coleman/Courses/Sp08/), numbers 23-25. Rigid GAGA says: Let $\mathcal{X}$ be a projective rigid analytic variety. Then (1) $\mathcal{X}$ is the analytification of an algebraic variety $X$. (2) The analytificiation functor from coherent sheaves on $X$ to coherent sheaves on $\mathcal{X}$ is an equivalence of categories. (3) The cohomology of a coherent sheaf is naturally isomorphic to that of its analytification. Thus, if we define Weierstrass points by the condition that the dimension of $H^0(\mathcal{O}(kp), X)$ is higher than expected, we will get the same points whether we work algebraically or analytically. It shouldn't be too hard to show that your favorite definition is equivalent to this. Of course, all I've done is tell you how to translate between analysis and algebra. The algebra itself may be very difficult, as Felipe Voloch points out.	9	https://mathoverflow.net/users/297	8491	5,817
https://mathoverflow.net/questions/8477	6	I recently launched a rocket with an altimeter that is accurate to roughly 10 ft. The recorded data is in time increments of 0.05 sec per sample and a graph of altitude vs. time looks pretty much like it should when zoomed out. The problem is when I try to calculate other values such as velocity or acceleration from the data, the accuracy of the measurements makes the calculated values pretty much worthless. What techniques can I use to smooth out the data so that I can calculate (or approximate) reasonable values for the velocity and acceleration? It is important that major events remain in place in time, most notably the 0 for for the first entry and the highest point during flight (2707). The altitude data follows and is measured in ft above ground level. The first time would be 0.00 and each sample is 0.05 seconds after the previous sample. The spike at the beginning of the flight is due to a technical problem that occurred during liftoff and removing the spike would be best. All help is greatly appreciated. ``` 00000 00000 00000 00076 00229 00095 00057 00038 00048 00057 00057 00076 00086 00095 00105 00114 00124 00133 00152 00152 00171 00190 00200 00219 00229 00248 00267 00277 00286 00305 00334 00343 00363 00363 00382 00382 00401 00420 00440 00459 00469 00488 00517 00527 00546 00565 00585 00613 00633 00652 00671 00691 00710 00729 00759 00778 00798 00817 00837 00856 00885 00904 00924 00944 00963 00983 01002 01022 01041 01061 01080 01100 01120 01139 01149 01169 01179 01198 01218 01238 01257 01277 01297 01317 01327 01346 01356 01376 01396 01415 01425 01445 01465 01475 01495 01515 01525 01545 01554 01574 01594 01614 01614 01634 01654 01664 01674 01694 01714 01724 01734 01754 01764 01774 01794 01804 01814 01834 01844 01854 01874 01884 01894 01914 01924 01934 01954 01954 01975 01995 01995 02015 02015 02035 02045 02055 02075 02075 02096 02096 02116 02126 02136 02146 02156 02167 02177 02187 02197 02207 02217 02227 02237 02237 02258 02268 02278 02278 02298 02298 02319 02319 02319 02339 02349 02359 02359 02370 02380 02380 02400 02400 01914 02319 02420 02482 02523 02461 02502 02543 02564 02595 02625 02666 02707 02646 02605 02605 02584 02574 02543 02543 02543 02543 02543 02543 02554 02543 02554 02554 02554 02554 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02543 02533 02543 02543 02543 02543 02543 02543 02543 02543 02533 02523 02523 02523 02523 02523 02523 02523 02523 02543 02523 02523 02523 02523 02523 02523 02523 02523 02513 02513 02502 02502 02492 02482 02482 02482 02482 02482 02482 02482 02482 02482 02482 02482 02482 02482 02482 02482 02472 02472 02472 02461 02461 02461 02461 02451 02451 02451 02461 02461 02451 02451 02451 02451 02451 02451 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02441 02431 02441 02431 02441 02431 02420 02431 02420 02420 02420 02420 02420 02420 02420 02420 02420 02420 02420 02420 02410 02420 02410 02410 02410 02410 02400 02400 02410 02400 02400 02400 02400 02400 02400 02400 02400 02400 02400 02400 02400 02390 02390 02390 02380 02380 02380 02380 02380 02380 02380 02380 02380 02380 02380 02380 02380 02370 02370 02380 02370 02359 02359 02359 02359 02359 02359 02359 02359 02359 02359 02359 02359 02359 02359 02349 02349 02349 02349 02349 02339 02339 02339 02339 02339 02339 02339 02339 02339 02339 02339 02339 02339 ```	https://mathoverflow.net/users/2489	Smoothing out Noisy Data	If you have data for the whole flight available to you then a good approach is Kalman smoothing. If you want estimates during the flight you want Kalman filtering. Seems like you're interested in the former. The difference is that Kalman smoothing uses data from the entire flight to estimate values at each point in time while Kalman filtering only uses the "past" to make its estimates. Some googling found a [readable looking paper](http://www-npl.stanford.edu/~byronyu/papers/derive_ks.pdf) on both Kalman filtering and smoothing. But you need to tune it with an estimate of the properties of the noise from your sensors. I think that in the real world engineers often guess these values. (Or maybe just the ones I know...)	6	https://mathoverflow.net/users/1233	8500	5,821
https://mathoverflow.net/questions/8473	7	this is related to [this question](https://mathoverflow.net/questions/6627/convex-hull-in-cat0) but is simpler, and hopefully is well-known. There are a number of references that say that the convex hull of a collection of points in a CAT(0) space need not be closed. I was wondering if anyone was aware of an explicit example ?	https://mathoverflow.net/users/972	Example of non-closed convex hull in a CAT(0) space	There are such examples already in Riemannian world! In fact in any generic Riemannian manifold of dimension $\ge3$ convex hull of 3 points in general position is not closed. BUT it is hard to make explicit and generic at the same time :) If it is closed then there are a lot of geodesics lying in its boundary --- that is rare! To see it do the following exercise first: Show that in generic 3-dimensional manifold, arbitrary smooth convex surface contains no geodesic. (Here geodesic = geodesic in ambient space.) To make word "generic" more clear: show that any metric admits $C^\infty$-perturbation such that above property holds. Semisolution: Assume that a geodesic $\gamma$ lies in the boundary of a convex set $K$ with smooth boundary. Let $N(t)$ be the outer normal vector to $K$ at $\gamma(t)$. Note that $N(t)$ is parallel. Further note that from convexity of $K$ we get that for any Jacoby field $J(t)$ such that $$\langle N(t\_0),J(t\_0)\rangle\le 0\ \text{and}\ \langle N(t\_1),J(t\_1)\rangle\le 0,$$ we have $$\langle N(t),J(t)\rangle\le 0\ \text{if}\ t\_0<t<t\_1.$$ Note that this condition does not hold if the curvature tensor on $\gamma$ is generic. P.S. Roughly it means that convex hulls in Riemannian world are too complicated. But I know one example where it is used, see Kleiner's An isoperimetric comparison theorem. But he is only using that Gauss curvature of non-extremal points on the boundary of convex hulls is zero... Appendix. (A construction of convex hull.) To construct convex hull you can do the following: start with some set $K\_0$ and construct a sequence of sets $K\_n$ so that $K\_{n+1}$ is a union of all geodesics with ends in $K\_n$. The union $W$ of all $K\_n$ is convex hull. Now assume it coincides with its closure $\bar w$. In particular if $x\in\partial\bar W$ then $x\in K\_n$ for some $n$. I.e. there is a geodesic in $\bar W$ passing through $x$ (if $x\not\in K\_0$). From convexity, it is clear that such geodesic lies in $\partial \bar W$... P.P.S. A more general statement is proved in our paper [About every convex set...](https://arxiv.org/abs/2103.15189)	10	https://mathoverflow.net/users/1441	8505	5,825
https://mathoverflow.net/questions/8509	4	Does anyone know of a good introduction to the theory of martingales and betting strategies from the point of view of statistics and/or probability theory? I'm looking for something basic, with lots of examples. Thanks.	https://mathoverflow.net/users/2231	Martingales and Betting Strategies	I don't know about statistics per se but the best introduction to martingales period is Williams' Probability with Martingales: [http://www.amazon.com/Probability-Martingales-Cambridge-Mathematical-Textbooks/dp/0521406056](http://rads.stackoverflow.com/amzn/click/0521406056)	14	https://mathoverflow.net/users/1847	8510	5,827
https://mathoverflow.net/questions/8495	4	What is the best way to find an actual divisor of an affine curve? I.E. if I am interested in finding a canonical divisor of a curve in two variables, is there a general way to go about it? Do I need to consider a projection on the x-axis? I should clarify. I'm assuming the field is characteristic 0, and the curve is affine of the form f(x,y)=0. I computed the closure in P2, it was smooth, and am now trying to compute a canonical divisor on this curve. Thanks for the comments am reading up on it now.	https://mathoverflow.net/users/2473	Finding divisors on a curve	As others have pointed out, there are several different ways to explicitly write down a divisor. So it would be helpful to know what kind of answer you're looking for. Anyhow, here's one answer. For a curve, the canonical divisor is the same as the sheaf of differential 1-forms. Let's assume that your curve $C$ in the affine plane is cut out by the equation $f(x,y)=0$. Theorem II 8.17 in Hartshorne's Algebraic Geometry book yields an exact sequence for computing $\Omega^1\_{C/k}$. Namely, we have an exact sequence: $ I/I^2 \to \Omega^1\_{\mathbb A^2\_k}\otimes \mathcal O\_C\to \Omega^1\_{C/k}\to 0 $ where $I$ is the defining ideal of the curve $C$. Since $I$ is generated by $f$, and since $\Omega^1\_{\mathbb A^2\_k}\otimes \mathcal O\_C$ is the free $\mathcal O\_C$ module with generators $dx, dy$, it follows that $\Omega^1\_{C/k}$ is generated by $dx$ and $dy$ modulo the relation $df$. This provides an explicit presentation for the canonical divisor as a module. When $C$ is smooth, this will be a locally free rank 1 $\mathcal O\_C$-module.	5	https://mathoverflow.net/users/4	8511	5,828
https://mathoverflow.net/questions/8521	70	As a student, I was taught that the Jordan curve theorem is a great example of an intuitively clear statement which has no simple proof. What is the simplest known proof today? Is there an intuitive reason why a very simple proof is not possible?	https://mathoverflow.net/users/2498	Nice proof of the Jordan curve theorem?	There's a short proof (less than three pages) that uses Brouwer's fixed point theorem, available here: [The Jordan Curve Theorem via the Brouwer Fixed Point Theorem](http://www.maths.ed.ac.uk/~aar/jordan/maehara.pdf) The goal of the proof is to take Moise's "intuitive" proof and make it simpler/shorter. Not sure whether you'd consider it "nice," though.	35	https://mathoverflow.net/users/1557	8522	5,832
https://mathoverflow.net/questions/8508	3	Suppose we're given a strict and small 2-category $C$, and an object of $C$ called $A$. Can we produce an internal category structure on $A$ in some canonical way (maybe by some sort of argument similar to the construction of a group object [see my comment])? If it's not possible in general, but it is possible in special (and nontrivial) situtations, what are the situations where it is possible? If it is possible, would this be functorial (pseudofunctorial maybe?) into 2-Cat, the 2-category of small categories, and if it is, is it full, faithful, both, or neither? If the answer to all of the above is true, does it generalize past the special case where n=2? I ask this because of the coherence result for bicategories, which fails to generalize past n=2. If this still works, can we drop the assumption of strictness?	https://mathoverflow.net/users/1353	Generation of object-internal structure in a strict 2-category	It's not entirely clear to me what you're asking, but I'll have a go at answering it anyway. An internal category in a category with pullbacks consists of objects $A\_0$ and $A\_1$, maps $s,t\colon A\_1\to A\_0$, $i\colon A\_0\to A\_1$, and $c\colon A\_1\times\_{A\_0}A\_1\to A\_1$ satisfying the usual axioms. An equivalent way to describe this is that for each object $X$ you have a category whose objects are $Hom(X,A\_0)$ and whose morphisms are $Hom(X,A\_1)$, which varies with $X$ in a natural way. Of course the latter makes sense even if the category has no pullbacks. This is all just like the case of groups, and it requires only an ambient category, not even a 2-category. Now if I have a strict 2-category $C$ which has strict finite 2-limits, in particular its underlying 1-category $C^1$ has pullbacks. Moreover, any object $A$ gives rise to a canonical internal category in $C^1$ with $A\_0 = A$ and $A\_1 = A^{\mathbf{2}}$, the cotensor with the "walking arrow" $\mathbf{2} = (\cdot \to \cdot)$. This construction defines a strict 2-functor from $C$ to the 2-category $Cat(C^1)$ of internal categories, functors, and natural transformations in the 1-category $C^1$. Moreover, I believe that this 2-functor is strictly 2-fully-faithful, i.e. an isomorphism on hom-categories. For this reason, Street has called a strict 2-category with strict 2-pullbacks and strict cotensors with $\mathbf{2}$ (which is all you really need for this argument) a "representable 2-category": all the 2-dimensional structure of its objects can be "represented" as internal categories in some 1-category. Cf. for instance "Fibrations and Yoneda's lemma in a 2-category." If $C$ is a non-strict 2-category, then it doesn't have an underlying 1-category, so it's not as clear how to write this down. But I think that morally, it should still be true, if one takes the care to phrase it correctly. The question of $n$-categories for $n>2$ is quite a different matter, though; I don't know if anyone's thought about it.	3	https://mathoverflow.net/users/49	8526	5,835
https://mathoverflow.net/questions/8525	1	We have our general notions of manifolds, schemes, et cetera, and other geometric "spaces", and we realize that a lot of these look like topological spaces with structure sheaves i.e. structured spaces. In part 5 of Lurie's DAG, he describes this notion in terms of (infinity,1)-topoi. However, I don't see myself being able to read DAG any time soon (I'm still going through HTT), so I have a somewhat easier question: If we consider the same topic in terms of the semiclassical constructions, that is, algebraic stacks and orbifolds, or even schemes and manifolds, can we describe precisely what our "structured spaces" should be to be useful? Why is it standard to talk about the sheaves on a scheme where we talk about bundles on a manifold? Aren't these concepts the same thing? How much of what we talk about in differential geometry and algebraic geometry can be jointly generalized to structured spaces? If this question is too vague, then just tell me where I can find out. Edit: To correct my previous vagueness, as Pete pointed out, we want local rings, or their appropriate counterparts for stacks. Edit 2: To clarify, I'm looking for either a book that's a dumbed-down version of DAG book 5, or someone to dumb down the idea from (infinity,1)-categories to plain categories or sets.	https://mathoverflow.net/users/1353	Semiclassical explanation of "Structured" spaces	I don't quite understand what you want, but here's a shot in the dark: SGA 4 has a notion of ringed topos, and this is generalized in Lurie's work. Schemes and stacks have their "underlying topoi" with respect to whatever topology you put on them. I think you can define notions like smoothness and tangents in this setting, but I haven't looked at it in a while.	1	https://mathoverflow.net/users/121	8527	5,836
https://mathoverflow.net/questions/8513	11	This is probably an insanely hard question, but given an abstract metric space, is there some way to determine whether it's a manifold with a Riemannian, or more generally a Finslerian, metric? If that's too hard, one could start off by assuming that the underlying space is a manifold. The example that got me thinking about this was the induced metric on the 2-sphere embedded in $R^3$...the underlying space is obviously a smooth manifold, and the metric should be smooth(the geodesics would even be great circles, as they are for the standard metric on $S^2$,) but I don't see how you could prove the triangle inequality pointwise, as you'd have to to show that it's Finslerian, let alone Riemannian. This example is already incredibly simple, since it's a homogeneous space with a the metric induced by being a subspace of another homogeneous space.	https://mathoverflow.net/users/2497	Characterization of Riemannian metrics	If $X$ is a metric space and $x$, $y\in X$, a segment from $x$ to $y$ is a subset $S\subseteq X$ such that $x$, $y\in S$ and $S$ is isometric to $[0,d(x,y)]$. Let now $n\in\{1,2,3\}$ and let $X$ be a metric space which is locally compact, $n$-dimensional and such that (i) every two points are the endpoints of a unique segment, (ii) if two segments have an endpoint and one other point in common, then one is contained in the other, and (iii) every segment can be extended, at either end, to a larger segment. Then $X$ is homeomorphic to $\mathbb R^n$. This is a metric characterization of $\mathbb R^n$ for small $n$. Using it locally, you get a characterization of topological manifolds of small dimension. I imagine higher dimensions or smoothness are harder to come by... The theorem above, along with quite a few other nice results, is proved in [Berg, Gordon O. Metric characterizations of Euclidean spaces. Pacific J. Math. 48 (1973), 11--28.] --- LATER... Of course, the paper [Nikolaev, I. G. A metric characterization of Riemannian spaces. Siberian Adv. Math. 9 (1999), no. 4, 1--58.] is relevant here! The paper gives purely metric characterizations of Riemannian manifolds of all smoothnesses up to $C^\infty$. The [MR review](http://www.ams.org/mathscinet-getitem?mr=MR1749850) gives a small history of the problem and references to earlier work.	19	https://mathoverflow.net/users/1409	8535	5,841
https://mathoverflow.net/questions/8484	5	What are the representations of $U(n)$ that induce (see [link text](https://mathoverflow.net/questions/5772/principal-bundles-representations-and-vector-bundles)) the bundles of holomorphic $\Omega ^{(1,0)}$ and antiholomorphic $\Omega ^{(0,1)}$ forms of $\mathbb{CP}^n$ (recalling the well-known fact that $\mathbb{CP}^n = SU(n+1)/U(n)$). Also, what are the representations that induce the line bundles?	https://mathoverflow.net/users/1648	Reps of $U(n)$ for the bundles of holomorphic and antiholomorphic forms of projective space	The group U(n-1) has an abelian factor U(1) and a semisimple factor SU(n-1). I'll answer first the second part of the question: Since line bundles are induced from one dimensional character representations, and the semismple component has no non-trivial character representations, thus the line bundles are induced from character representations of the U(1) component. These representations are indexed by a single integer. The answer to the first part of the question is that the bundle of holomorphic one forms is induced from the representation of U(n-1) which has a unit central charge in the U(1) part and equal to the contragredient fundamental n-1 dimensional representation of the SU(n-1) component. One of the ways to see that is say from the explicit expression of the U(n) action on CP(n-1) in the affine coordinates realized as a rational transformation. When restricted to U(n-1), one can see that the action of the U(1) is through its unit central charge representation and the action of SU(n-1) a multiplication of the affine coordinate vector by the reciprocal of the group element in the fundamental representation. For higher holomorphic form bundles, the bundle is induced from the (irreducible) antisymmetric powers of the contragredient representations of SU(n-1) accomanied by the representation with the corresponding multiple central charge of U(1).	3	https://mathoverflow.net/users/1059	8538	5,843
https://mathoverflow.net/questions/8494	9	Maybe this is a silly question (or not even a question), but I was wondering whether the cotangent bundle of a submanifold is somehow canonically related to the cotangent bundle of the ambient space. To be more precise: Let $N$ be a manifold and $\iota:M \hookrightarrow N$ be an embedded (immersed) submanifold. Is the cotangent bundle $T^\ast M$ somehow canonical related to the cotangent bundle $T^\ast N$. Canonical means, without choosing a metric on $N$. The choice of a metric gives an isomorphism of $TN$ and $T^\ast N$ and therefore a "relation", since the tangent bundle of the submanifold $M$ can be viewed in a natural way as a subspace of the tangent bundle of the ambient space $N$ ($\iota$ induces an injective linear map at each point $\iota\_\ast : T\_pM \rightarrow T\_pN$). I think this is not true for the cotangent space (without a metric) Moreover, the cotangent bundle $T^\ast N$ of a manifold $N$ is a kind of "prototype" of a symplectic manifold. The symplectic structure on $T^\ast N$ is given by $\omega\_{T^\ast N} = -d\lambda$, where $\lambda$ is the Liouville form on the cotangent bundle. (tautological one-form, canonical one-form, symplectic potential or however you want). The cotangent bundle of the submanifold $T^\ast M$ inherits in the same way a canonical symplectic structure. So, is there a relation between $T^\ast N$ and $T^\ast M$ respecting the canonical symplectic structures. (I think the isomorphism given by a metric is respecting (relating) these structures, or am I wrong?) As I said, this question is perhaps strange, but the canonical existence of the symplectic structure on the cotangent bundle is "quite strong". For example: A given diffeomorphism $f:X \rightarrow Y$ induces a canonical symplectomorphism $T^\ast f : T^\ast Y \rightarrow T^\ast X$ (this can be proved by the special "pullback cancellation" property of the Liouville form). So in the case of a diffeomorphism the symplectic structures are "the same". Ok, a diffeomorphism has more structure than an embedding, but perhaps there is a similar relation between $T^\ast M$ and $T^\ast N$? --- EDIT: Sorry fot the confusion, but Kevins post is exactly a reformulation of the problem, I'm interested in. To clarify things: with the notation of Kevin's post: > > When (or whether) are the pulled back symplectic structures the same ? Under what circumstances holds $a^\ast \omega\_{T^\ast N} = b^\ast \omega\_{T^\ast N}$ > > > I think this isn't true for any submanifold $M \subset N$, but what is a nice counterexample? Is it true for more restricted submanifolds as for example embedded submanifolds which are not just homoeomorphisms onto its image, but diffeomorphisms (perhaps here the answer is yes, using the diffeomorphism remark above?)?	https://mathoverflow.net/users/675	Cotangent bundle of a submanifold	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. 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 $U^\* \oplus U$. Now, let $U^0$ denote the annahilator of $U$ in $V^\$. 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 $V^\\oplus U$. 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 $$ (V^\\oplus U )/ (U^0\times\{0\}) $$ inherits a natural symplectic structure. Of course, the quotient is precisely $U^\*\oplus U$ and the symplectic form is nothing but $\omega\_U$.	10	https://mathoverflow.net/users/380	8555	5,853
https://mathoverflow.net/questions/8559	8	Is it possible to find a coupling of two Wiener processes $W^0, W^x$ (i.e. two Wiener processes defined on a common probability space). One starting from $0$ and the other from $x$ such that $W\_t^0 - W\_t^x \rightarrow\_t 0$ almost surely and in $L^1$. Using some random-walks considerations I suspect that it is not possible to have convergence in $L^1$ but I do not know how to prove it. --- The answer is: $W\_t^0 - W\_t^x$ cannot converge to $0$ in $L^1$ The proof is as follow (it is a slightly extended version of the proof by smalldeviations). Obviously $\|W\_t^0 - W\_t^x\|\_1\geq \inf \_{\gamma\in \Gamma } \{\int \_{\mathbf{R}^d\times \mathbf{R}^d} \|x-y\| \text{d}\gamma(x,y)\},$ where $\Gamma$ is the set of all couplings of $\mathcal{L} (W\_t^0)$ and $\mathcal{L} (W\_t^x)$ ($\mathcal{L}$ is the law of given variable). By the duality formula (see <http://en.wikipedia.org/wiki/Transportation_theory>) the right hand side is equal to $\sup \{ \int\_{\mathbf{R}^d} \phi (x) \mathrm{d} \mu (x) + \int\_{\mathbf{R}^d} \psi (y) \mathrm{d} \nu (y) \},$ where the supremum runs over all pairs of bounded and continuous functions such that $\phi(x)+\psi(y)\leq \|x-y\|$ and $\mu =\mathcal{L}(W\_t^x)$ and $\nu = \mathcal{L} (W\_t^0)$. In our case it is sufficent to take $\phi(x) = x, \psi(y)=y$. Then the expression under the sup is equal to: $\mathbf{E}(W^x\_t) - \mathbf{E}(W^0\_t) = x-0 =x$. Therefore $\|W\_t^0 - W\_t^x\|\_1\geq x$.	https://mathoverflow.net/users/1302	Coupling of Wiener processes	If there is no coupling s.t. the distance goes to 0 in $L^{1}$ (which I agree seems likely), you might want to look up an introduction to the Wasserstein-1 distance (which is exactly expected $L^{1}$ distance after an optimal coupling). This is the language that I've seen this type of problem most often discussed in. The field of finding optimal couplings for given metrics is 'optimal transport'. I vaguely recall that there are theorems about optimal couplings (in only certain $L^{p}$ only, of course!) never giving rise to crossing lines. In a 1-dimensional problem, such as the one you have, this would tell you what the optimal coupling is explicitly (in this case, if you construct your Brownian motion via Donsker's theorem, it says: whenever $W^{0}$ takes a move in the $\alpha$ percentile, make $W^{x}$ also move in the $\alpha$ percentile... in other words, my probably-misremembered theorem would imply coupling doesn't help your $L^{1}$ distance at all in this case). Cedric Villani has two excellent books on the subject, at least one of which was available for free download the last time I checked, and you should be able to find 'this sort' of theorem. Please don't take my word for the statements. Edit: Here is (I believe) a proof... though it might fit in the category of so-simple-its-wrong. First of all, we have starting points x,0 and add two normal (0,a) random variables X and Y to them. Plugging in the obvious cost functions, the "Kantorovitch Duality" formula tells us that the $L^{1}$ distance between x+X and 0+Y is at least x (while plugging in the independent coupling to the standard way of writing this metric tells us it is at most x). So, at time a, the $L^{1}$ distance between the brownian motions must be at least x (since at time a they have the same distribution as x+X and 0+Y, and we have found this lower bound for ALL couplings, and in particular all couplings that come from them both being brownian motions). In particular, the $L^{1}$ distance can't go to 0.	3	https://mathoverflow.net/users/2282	8575	5,866
https://mathoverflow.net/questions/8579	38	I remember learning some years ago of a theorem to the effect that if a polynomial $p(x\_1, ... x\_n)$ with real coefficients is non-negative on $\mathbb{R}^n$, then it is a sum of squares of polynomials in the variables $x\_i$. Unfortunately, I'm not sure if I'm remembering correctly. (The context in which I saw this theorem was someone asking whether there was a sum-of-squares proof of the AM-GM inequality in $n$ variables, so I'm not 100% certain if the quoted theorem was specific to that case.) So: does anyone know a reference for the correct statement of this theorem, if in fact something like it is true? (Feel free to retag if you don't think it's appropriate, by the way.)	https://mathoverflow.net/users/290	Are all polynomial inequalities deducible from the trivial inequality?	One interpretation of the question is [Hilbert's seventeenth problem](https://en.wikipedia.org/wiki/Hilbert%27s_seventeenth_problem), to characterize the polynomials on $\mathbb{R}^n$ that take non-negative values. The problem is motivated by the nice result, which is not very hard, that a non-negative polynomial in $\mathbb{R}[x]$ (one variable) is a sum of two squares. What is fun about this result is that it establishes an analogy between $\mathbb{C}[x]$, viewed as a quadratic extension by $i$ of the Euclidean domain $\mathbb{R}[x]$; and $\mathbb{Z}[i]$ (the Gaussian integers), viewed as a quadratic extension by $i$ of the Euclidean domain $\mathbb{Z}$. In this analogy, a real linear polynomial is like a prime that is 3 mod 4 that remains a Gaussian prime, while a quadratic irreducible polynomial is like a prime that is not 3 mod 4, which is then not a Gaussian prime. A non-zero integer $n \in \mathbb{Z}$ is a sum of two squares if and only if it is positive and each prime that is 3 mod 4 occurs evenly. Analogously, a polynomial $p \in \mathbb{R}[x]$ is a sum of two squares if and only if some value is positive and each real linear factor occurs evenly. And that is a way of saying that $p$ takes non-negative values. In dimension 2 and higher, the result does not hold for sums of squares of polynomials. But as the Wikipedia page says, Artin showed that a non-negative polynomial (or rational function) in any number of variables is at least a sum of squares of rational functions. In general, if $R[i]$ and $R$ are both unique factorization domains, then some of the primes in $R$ have two conjugate (or conjugate and associate) factors in $R[i]$, while other primes in $R$ are still primes in $R[i]$. This always leads to a characterization of elements of $R$ that are sums of two squares. This part actually does apply to the multivariate polynomial ring $R = \mathbb{R}[\vec{x}]$. What no longer holds is the inference that if $p \in R$ has non-negative values, then the non-splitting factors occur evenly. For instance, $x^2+y^2+1$ is a positive polynomial that remains irreducible over $\mathbb{C}$. It is a sum of 3 squares rather than 2 squares; of course you have to work harder to find a polynomial that is not a sum of squares at all.	42	https://mathoverflow.net/users/1450	8582	5,871
https://mathoverflow.net/questions/5063	14	Let $V$ be the irreducible representation of $GL\_n$ with highest $\lambda$, and $\|\lambda\|=n$. It is well known that the representation of $S\_n$ on the $(1,1,\ldots,1)$ weight space is the Specht module $S\_{\lambda}$. What is known about the representation of $S\_n$ on the rest of $V$? I am also interested in the case where $\|\lambda\| < n$.	https://mathoverflow.net/users/297	Restriction from $GL_n$ to $S_n$	I keep meaning to post this and forgetting: Richard Stanley sent me the following two references Enumerative Combinatorics, Volume II Exercise 7.74: If $V\_{\lambda}$ is a representation of $GL\_n$, and $S\_{\mu}$ a representation of $S\_n$, then the multiplicity of $S\_{\mu}$ in $\mathrm{Restriction}^{GL\_n}\_{S\_n} V\_{\lambda}$ is the coefficient of the Schur function $s\_{\lambda}$ in the symmetric formal power series $$s\_{\mu}(1, x\_1, x\_2, \ldots, x\_1^2, x\_1 x\_2, x\_1 x\_3, \ldots, x\_1^3, x\_1^2 x\_2, \ldots).$$ That is to say, we take the Schur function $s\_{\mu}$ and plug in every monomial. (Yes, for those who know the term, this is an example of plethysm.) [Gay, David A.](http://www.ams.org/mathscinet-getitem?mr=414794) "Characters of the Weyl group of $SU(n)$ on zero weight spaces and centralizers of permutation representations." Rocky Mountain J. Math. 6 (1976), no. 3, 449—455. Determines the representation of $S\_n$ on the zero weight space of $V\_{\lambda}$ when $\|\lambda\|$ is a multiple of $n$. I just found another reference, which I haven't digested yet: [Scharf, Thibon and Wybourne (1997)](http://iopscience.iop.org/0305-4470/30/19/030/).	7	https://mathoverflow.net/users/297	8589	5,876
https://mathoverflow.net/questions/8560	26	Background: a field is formally real if -1 is not a sum of squares of elements in that field. An ordering on a field is a linear ordering which is (in exactly the sense that you would guess if you haven't seen this before) compatible with the field operations. It is immediate to see that a field which can be ordered is formally real. The converse is a famous result of Artin-Schreier. (For a graceful exposition, see Jacobson's Basic Algebra. For a not particularly graceful exposition which is freely available online, see [http://alpha.math.uga.edu/~pete/realspectrum.pdf.](http://alpha.math.uga.edu/%7Epete/realspectrum.pdf.)) The proof is neither long nor difficult, but it appeals to Zorn's Lemma. One suspects that the reliance on the Axiom of Choice is crucial, because a field which is formally real can have many different orderings (loc. cit. gives a brief introduction to the real spectrum of a field, the set of all orderings endowed with a certain topology making it a compact, totally disconnected topological space). Can someone give a reference or an argument that AC is required in the technical sense (i.e., there are models of ZF in which it is false)? Does assuming that formally real fields can be ordered recover some weak version of AC, e.g. that any Boolean algebra has a prime ideal? (Or, what seems less likely to me, is it equivalent to AC?)	https://mathoverflow.net/users/1149	How much choice is needed to show that formally real fields can be ordered?	This is equivalent (in ZF) to the Boolean Prime Ideal Theorem (which is equivalent to the Ultrafilter Lemma). Reference: R. Berr, F. Delon, J. Schmid, Ordered ﬁelds and the ultraﬁlter theorem, Fund Math 159 (1999), 231-241. [online](http://matwbn.icm.edu.pl/ksiazki/fm/fm159/fm15933.pdf)	27	https://mathoverflow.net/users/932	8591	5,878
https://mathoverflow.net/questions/8550	13	Consider the category of Banach spaces with contractions as morphisms (weak, so $\\|T\\| \le 1$). Is this an algebraic theory? I suspect that this is true. The "operations" will be weighted sums, where the sum of the weights is at most $1$. The "free Banach space" on a set $X$ should be $\ell^1(X)$. (Note that the "underlying set" functor sends a Banach space to its unit ball.) So, part one: * Is this correct? If so, can anyone supply a reference (unfortunately, searching for "algebraic theory" and "Banach" doesn't turn up anything obvious). * Has anything useful/unusual come out of this point of view? * If this is correct, then the algebraic theory seems to be [commutative](http://ncatlab.org/nlab/show/commutative+algebraic+theory), in which case it's a symmetric closed category. Has this angle been used? * The norm can't be encoded as an operation, can it be categorically recovered? Part two says: can we do this for Hilbert spaces? Edit: Anyone even vaguely intrigued by this question should read the paper linked in Yemon's answer. In particular, it also answers a question that I was going to ask as a follow-up: what's the nearest algebraic theory to Banach spaces (the answer being totally convex spaces).	https://mathoverflow.net/users/45	Is the category of Banach spaces with contractions an algebraic theory?	I think this is some kind of infinitary algebraic theory, but that it is not a monadic adjunction. That is, if you take the "closed unit ball functor" $B$ from ${\bf Ban}\\_1$ to ${\bf Set}$ and the "free Banach space functor" $L: {\bf Set} \to {\bf Ban}\_1$, then $L$ is left adjoint to $B$ but this adjunction is not monadic (IIRC, and I often don't). See, for instance, the first few pages of Pelletier and Rosicky, [On the equational theory of $C^\$-algebras](https://doi.org/10.1007/BF01196099), Algebra Universalis 30* (1993) pp 275-28. You say something about a closed structure on ${\bf Ban}\_1$, if I understand this right then this is symmetric monoidal with the tensor being the projective tensor product of Banach spaces. That seems to be well known but little-used, although IMHO having this kind of perspective takes some of the tedium/clutter out of certain computations/constructions in my corner of functional analysis. I think the ball functor from (Hilbert spaces & contractions) to Set doesn't have a left adjoint, but that's more of a guess than an intuition. Certainly the `natural' attempt to build a left adjoint falls over. As for putting a closed structure on Hilb .... well, the fact that the natural norm on B(H) is not Hilbertian suggests to me that this won't work. (Put another way, the natural Hilbertian tensor product would dualise to only considering Hilbert-Schmidt class maps between your Hilbert spaces, which in infinite dimensions rules out the identity morphism.)	9	https://mathoverflow.net/users/763	8593	5,880
https://mathoverflow.net/questions/8583	-6	I'm sorry if this isn't an appropriate question for MO. I've been reading here for a while, but I still haven't got a good grasp of what's a good question. Given a field A and the polynomial ring A[x], we order the elements of A in any sequence and we define the isomorphism $f\colon A\to A[x]$ such that every element an$\mapsto$an xn, an $\in$ A, xn $\in$ A[x]. Can this be considered an alternate definition for A[x], is it just wrong, or is it the same as the canonical one? Andy	https://mathoverflow.net/users/1559	Can I define the polynomial ring A[x] with an isomorphism f: A ---> A[x]?	The question seems to involve a construction of a set-theoretic map, and the indexing (natural numbers?) suggests that A is assumed to have a countable underlying set. That map doesn't even yield a surjection of sets. I would like to reinterpret the question in the following way: How much structure do we need to forget in order for there to exist an isomorphism $A \to A[x]$? YBL pointed out that there is never an A-algebra isomorphism (if A is nonzero) and that there can be a ring-theoretic isomorphism if A is big enough. If A has an infinite underlying set, then there exist isomorphisms on the underlying sets. It is potentially interesting to ask when we get isomorphisms on the underlying additive groups: it is sufficient for A to have a polynomial ring structure, but that is far from necessary: e.g., A could be any field of infinite dimension over its prime field. Regarding your last question, you can define a polynomial ring using a sequence of embeddings $f\_n: a \mapsto ax^n$ together with a specified multiplication law. This is a special case of the monoid ring construction. I'm not sure if this was the construction you initially had in mind, but it doesn't yield an isomorphism, since it isn't a single map.	5	https://mathoverflow.net/users/121	8620	5,900
https://mathoverflow.net/questions/8608	19	I'm struggling through Lurie's Higher Topos Theory, since it appears that someone reading through the book is expected to be somewhat comfortable with simplicial homotopy theory. The main trouble I've had is computing things like the join, product, coproduct, pullback, pushout, and so forth. I understand them as far as their universal properties, and maybe have a little intuition because the category of simplicial sets is a presheaf category over the simplex category, but Lurie uses geometrical language, so I can't even compute like when working with presheaves. So, could you lot recommend some books or lecture notes, preferably with suggested sections, that won't go too deep into simplicial homotopy theory, but deep enough for me to learn how to compute? Note: Most of the time I waste is sitting around with that book is trying to make sense of the computations, while I understand the arguments just fine. So please, don't suggest an entire book detailing all of simplicial homotopy theory from start to finish. I have a goal in mind here, and I'm only trying to learn as much as is necessary for me to continue reading HTT.	https://mathoverflow.net/users/1353	Simplicial homotopy book suggestion for HTT computations	I'm not really sure what you are asking for. Colimits and limits are easy to compute in simplicial sets, because it's a presheaf category (as you say). But if you want "geometrical" intuition about simplicial sets (including "pictures" of joins, etc.), you want to know about the geometric realization functor from simplicial sets to spaces. In particular, the fact that geometric realization preserves colimits (it's a left adjoint), and that it preserves finite limits (essentially a theorem of Milnor). The limit result is proved in the book by Gabriel-Zisman. (Goerss-Jardine mention the result in chapter 1, but don't seem to prove it.)	14	https://mathoverflow.net/users/437	8621	5,901
https://mathoverflow.net/questions/8632	12	What is the cut rule? I don't mean the rule itself but an explanation of what it means and why are proof theorists always trying to eliminate it? Why is a cut-free system more special than one with cut?	https://mathoverflow.net/users/nan	cut elimination	Suppose I have a proof of B starting from assumption A. And a proof of C starting from assumption B. Then the cut rule says I can deduce C from assumption A. But I didn't need the cut rule. If I was able to deduce B from A I could simply "inline" the proof of B from A directly into the proof of C from B to get a proof of C from A. So the cut rule is redundant. That's a good reason to eliminate it. But eliminating it comes at a price. The proofs become more complex. Here's a [paper](http://www.ams.org/bull/1997-34-02/S0273-0979-97-00715-5/S0273-0979-97-00715-5.pdf) that quantifies just how much. So it's a hard rule to give up.	19	https://mathoverflow.net/users/1233	8634	5,912
https://mathoverflow.net/questions/8599	35	I am curious about what is a good approach to the machinery of cohomology, especially in number-theoretic settings, but also in algebraic-geometric settings. Do people just remember all the rules and go through the formal manipulations of the cohomology groups of class field theory mechanically, or are people actually "feeling" what is going on here. If it is the latter, could you give a list of mnemonics, or cheat-sheet, or a little fairy tale involving all the characters, so that it is easy to associate with the abstraction. Is there some strong intuition coming from algebraic-topology that would help here? Do people think of Cech covers when they do diagram chasing on crazy grids of exact sequences? I personally was extremely happy with the central simple algebra approach to class field theory, because it gave me a whole new kind of creature, a CSA, which I could learn to adapt to and love. On the other hand, I don't see myself ever making friends with a 2-cocycle.	https://mathoverflow.net/users/2515	Tips on cohomology for number theory	Many number theorists, including me, learned Galois cohomology first via the proof of the Mordell-Weil theorem. The last chapter of Joe Silverman's book The Arithmetic of Elliptic Curves is a good source for this. It's very concrete and when you understand the proof you'll understand a lot about why number theorists like the cohomological formalism. Edit: But note that you'll learn just about H^1, not anything higher. It's a start!	17	https://mathoverflow.net/users/431	8646	5,923
https://mathoverflow.net/questions/5303	20	Is it possible to exhibit a (Hamel) basis for the vector space l^infinity, given by the bounded sequences of real numbers?	https://mathoverflow.net/users/1172	Basis of l^infinity	The question is about the complexity of the simplest possible Hamel basis of $\ell^\infty$, and this is a perfectly sensible thing to ask about even in a context where one wants to retain the Axiom of Choice. That is, we know by AC that there is a basis---how complex must it be? Such a question finds a natural home in descriptive set theory, a subject concerned with numerous similar complexity issues. In particular, descriptive set theory provides tools to make the question precise. My answer is that one can never prove a negative answer to the question, because it is consistent with the ZFC axioms of set theory that Yes, one can concretely exhibit a Hamel basis of $\ell^\infty$. To explain, one natural way to measure the complexity of sets of reals (or subsets of $\mathbb R^\infty$) is with the projective hierarchy. This is the hierarchy that begins with the closed sets (in $\mathbb R$, say, or in $\mathbb R^\omega$), and then iteratively closes under complements and projections (or equivalently, continuous images). The Borel sets appear near the bottom of this hierarchy, at the level called $\Delta^1\_1$, and then the analytic sets $\Sigma^1\_1$ and co-analytic sets $\Pi^1\_1$, and so on up the hierarchy. Sets in the projective hierarchy are exactly those sets that can be given by explicit definition in the structure of the reals, with quantification only over reals and over natural numbers. If we were to find a projective Hamel basis, then it will have been exhibited in a way that is concrete, free of arbitrary choices. Thus, a very natural way of making the question precise is to ask: Question. Does $\ell^\infty$ have a Hamel basis that is projective? If the axioms of set theory are consistent, then they are consistent with a positive answer to this question. This is not quite the same as proving a positive answer, to be sure, but it does mean that no-one will ever prove a negative answer to the question. Theorem. If the axioms of ZFC are consistent, then they are consistent with the existence of a projective Hamel basis for $\ell^\infty$. Indeed, there can be such a basis with complexity $\Pi^1\_3$. Proof. I will prove that under the set-theoretic assumption known as the Axiom of Constructibility V=L, there is a projective Hamel basis. In my answer to question about [Well-orderings of the reals](https://mathoverflow.net/questions/6593), I explained that in Goedel's constructible universe $L$, there is a definable well-ordering of the reals. This well-ordering has complexity $\Delta^1\_2$ in the projective hierarchy. From this well-ordering, one can easily construct a well-ordering of $\ell^\infty$, since infinite sequences of reals are coded naturally by reals. Now, given the well-ordering of l^infty, one defines the Hamel basis as usual by taking all elements not in the span of elements preceding it in the well-order. The point for this question is that if the well-order has complexity $\Delta^1\_2$, then this definition of the basis has complexity $\Pi^1\_3$, as desired. QED OK, so we can write down a definition, and in some set-theoretic universes, this definition concretely exhibits a Hamel basis of $\ell^\infty$. There is no guarantee, however, that this definition will work in other models of set theory. I suspect that one will be able to find other models of ZFC, in which there is no projective Hamel basis of $\ell^\infty$. It is already known that there might be no projective well ordering of the reals (a situation that follows from large cardinals and other set theoretic hypotheses), and perhaps this also implies that there is no projective Hamel basis. In this case, it would mean that the possibility of exhibiting a concrete Hamel basis is itself independent of ZFC. This would be an interesting and subtle situation. To be clear, I am not referring here merely to the existence of a basis requiring AC, but rather, fully assuming the Axiom of Choice, I am proposing that the possibility of finding a projective basis is independent of ZFC. Conjecture. The assertion that there is a projective Hamel basis of $\ell^\infty$ is independent of ZFC. I only intend to consider the question in models of ZFC, so that $\ell^\infty$ has a Hamel basis of some kind, and the only question is whether there is a projective one or not. In this situation, the fact that AD seems to imply that there is no Hamel basis is not relevant, since that axiom contradicts AC. Apart from this, I also conjecture that there can never be a Hamel basis of $\ell^\infty$ that is Borel. This would be a lower bound on the complexity of how concretely one could exhibit the basis.	25	https://mathoverflow.net/users/1946	8647	5,924
https://mathoverflow.net/questions/8648	7	For an affine variety, I know how to compute the set of singular points by simply looking at the points where the Jacobian matrix for the set of defining equations has too small a rank. But what is the corresponding method for a variety that is a projective variety,and also a variety is a subset of a product of some projective space and affine space? The way I can think of is covering it by sets that are affine, and doing it for each affine set in this open cover - but that seems tedious for practical purposes (but fine for theoretical definitions & theoretical properties). Also for resolution of singularities, what is a simple method that is guaranteed to work? The way suggested in the definitions in Hartshorne and other books, is to blow up along the singular locus, then look at the singular locus of the blow-up, and blow up again, and so on - is that guaranteed to terminate? What are some more efficient methods? I have looked at the reference "Resolution of Singularities", a book by someone - that's what he also seems to suggest (though his proof is very general, and I didn't read all of it).	https://mathoverflow.net/users/2623	Easiest way to determine the singular locus of projective variety & resolution of singularities	The Jacobian condition for smoothness is valid also for projective varieties as well as affine varieties: you just take a homogeneous defining ideal and compute the rank of the Jacobian matrix at the point p, see e.g. p. 4 of <http://www.ma.utexas.edu/users/gfarkas/teaching/alggeom/march4.pdf> For a general variety: yes, I think the most computationally effective way to do it is to cover it by affine opens and apply the Jacobian condition on each one separately. About your question on resolution of singularities: this is tricky in high dimension! As I understand it, you can indeed resolve singularities just by a combination of blowups and normalizations (at least in characteristic 0), but starting in dimension 3 you have to be somewhat clever about where in and what order you perform your blowups. It is not the case that if you just keep picking a closed subvariety and blowing it up (and then normalizing) that you will necessarily terminate with a smooth variety. For more details presented in a user-friendly way, I recommend Herwig Hauser's article <http://homepage.univie.ac.at/herwig.hauser/Publications/The%20Hironaka%20Theorem%20on%20resolution%20of%20singularities/The%20Hironaka%20Theorem%20on%20resolution%20of%20singularities.pdf>	6	https://mathoverflow.net/users/1149	8650	5,926
https://mathoverflow.net/questions/8597	13	Answers can come in mathematical, physical, and philosophical flavors. Edit: There seems to be a consensus that this question is not formulated well. I must respectfully disagree. My interest in the question is immaterial to the question itself. It is manifestly not a "what is" question. I see no reason to write more than is necessary for the formulation of the question, and I invite nothing more than a sentence giving me something to think about or an idea of where to look.	https://mathoverflow.net/users/2206	What is the meaning of symplectic structure?	Here's how I understand it. Classical mechanics is done on a phase space M. If we are trying to describe a mechanical system with n particles, the phase space will be 6\n\-dimensional: 3\n\ dimensions to describe the coordinates of particles, and 3\n\ dimensions to describe the momenta. The most important property of all this is that given a Hamiltonian function $H:M\to \mathbb R$, and a point of the phase space (the initial condition of the system), we get a differential equation that will predict the future behavior of the system. In other words, the function $H$ gives you a flow $\gamma:M\times \mathbb R \to M$ that maps a point p and a time t to a point $\gamma\_t (p)$ which is the state of the system if it started at p after time t passes. Now, if we do classical mechanics, we are very interested in changes of coordinates of our phase space. In other words, we want to describe M in a coordinate-free fashion (so that, for a particular problem, we can pick whatever coordinates are most convenient at the moment). Now, if $M$ is an abstract manifold, and $H:M\to\mathbb R$ is a function, you cannot write down the differential equation you want. In a sense, the problem is that you don't know which directions are "coordinates" and which are "momenta", and the distinction matters. However, if you have a symplectic form $\omega$ on M, then every Hamiltonian will indeed give you the differential equation and a flow $M\times\mathbb R \to M$, and moreover the symplectic form is precisely the necessary and sufficient additional structure. I must say, that although this may be related to why this subject was invented a hundred years ago, this seems to have little to do to why people are studying it now. It seems that the main reason for current work is, first of all, that new tools appeared which can solve problems in this subject that couldn't be solved before, and secondly that there is a very non-intuitive, but very powerful, connection that allows people to understand 3- and 4-manifolds using symplectic tools. --- EDIT: I just realized that I'm not quite happy with the above. The issue is that the phase spaces that come up in classical mechanics are always a very specific kind of symplectic manifold: namely, the contangent bundlde of some base space. In fact, in physics it is usually very clear which directions are "coordinates of particles" and which are "momenta": the base space is precisely the space of "coordinates", and momenta naturally correspond to covectors. Moreover, the changes of coordinates we'd be interested in are always just changes of coordinates of the base space (which of course induce a change of coordinates of the cotangent bundle). So, a symplectic manifold is an attempts to generalize the above to spaces more general than the cotangent bundle, or to changes of coordinates that mess up the cotangent-bundle structure. I have no idea how to motivate this. UPDATE: I just sat in a talk by Sam Lisi where he gave one good reason to study symplectic manifolds other than the cotangent bundle. Namely, suppose you are studying the physical system of just two particles on a plane. Then, their positions can be described as a point in $P = \mathbb R^2 \times \mathbb R^2$, and the phase space is the cotangent bundle $M=T^\* P$. Notice, however, that this problem has a lot of symmetry. We can translate both points, rotate them, look at them from a moving frame, etc., all without changing the problem. So, it is natural to want to study not the space M itself, but to quotient it out by the action of some Lie group G. Apparently, $M/G$ (or something closely related; see Ben's comment below) will still be a symplectic manifold, but it is not usually the cotangent bundle of anything. The difference is significant: in particular, the canonical one-form on $M=T^\*P$ will not necessarily descend to $M/G$.	16	https://mathoverflow.net/users/2467	8652	5,928
https://mathoverflow.net/questions/8667	7	I want to ask a stupid question. Let $I$ be an infinite set and suppose $i$ belongs to $I$. I wonder whether following morphisms exist in general: Hom($A$,colim $B\_i) \to$ lim Hom($A,B\_i$) and Colim Hom($A,B\_i) \to$ Hom($A$,colim $B\_i$) What I know is: if we replace lim by infinite product and colim by infinite coproduct, it exists. But I am not sure in this general case above.	https://mathoverflow.net/users/1851	On limits and Colimits	For any diagram $B\_i$ and an object $A$ in a category, there are natural maps of sets: 1. colim Hom($A,B\_i) \to$ Hom($A$, colim $B\_i$) 2. colim Hom($B\_i,A) \to$ Hom(lim $B\_i, A$) These maps need not be isomorphisms, in general (neither even when the diagram is filtered, nor when it is finite). Nor are they isomorphisms for infinite products and coproducts, in general (for finite products and coproducts in an additive category they are isomorphisms, though). Besides, for any diagram $B\_i$ and an object $A$ there are natural isomorphisms of sets: 1. Hom($A$, lim $B\_i$) = lim Hom($A,B\_i$) 2. Hom(colim $B\_i, A$) = lim Hom($B\_i,A$) These isomorphisms hold for any diagram (it does not have to be filtered, nor does it have to be finite). Actually, they hold by the definition of lim and colim.	21	https://mathoverflow.net/users/2106	8669	5,937
https://mathoverflow.net/questions/8671	15	I am trying to get used to Hochschild cohomology of algebras by proving its properties. I am currently trying to show that the cup product is graded-commutative (because I heard this somewhere); however, my trouble is that I have no idea what the exact conditions are for this to hold. As always in Hochschild cohomology, we start with an algebra A and an A-bimodule M. Is A supposed to be unital? In the original article by Hochschild, it is not, and the proofs would even become harder if we suppose it to be. On the other hand, I have some troubles working with non-unital algebras - they seem just so uncommon to me. Is it still standard to use non-unital algebras 60 years after Hochschild's articles? Anyway, this is not the main problem. The main problem is that I have no idea what we require from A and M. Of course, M should be an A-algebra, not just a module, for the cup product to make sense. Now: * Must A be commutative? * Must M be commutative? * Must M be a symmetric A-bimodule? (That is, am = ma for all a in A and m in M.) I have some doubts that if we require this all, then we still get something useful (in fact, the most common particular case of Hochschild cohomology is group cohomology, and it is as far from the "symmetric A-bimodule" case as it can be), but I may be completely mistaken. As there seem to be different definitions of Hochschild cohomology in literature, let me record mine: For a k-algebra A and an A-bimodule M, we define the n-th chain group $C^n \left(A, M\right)$ as $\mathrm{Hom}\left(A^{\otimes n}, M\right)$, with differential map $\delta : C^n \left(A, M\right) \to C^{n+1} \left(A, M\right)$, $\left(\delta f\right) \left(a\_1 \otimes ... \otimes a\_{n+1}\right) = a\_1 f\left(a\_2 \otimes ... \otimes a\_{n+1}\right)$ $ + \sum\limits\_{i=1}^{n} \left(-1\right)^if\left(a\_1 \otimes ... \otimes a\_{i-1} \otimes a\_{i}a\_{i+1} \otimes a\_{i+2} \otimes ... \otimes a\_{n+1}\right) + \left(-1\right)^{n+1} f\left(a\_1 \otimes ... \otimes a\_n\right) a\_{n+1}$. The cohomology is then the homology of the resulting complex, and if M is an A-algebra, the cup product is given by $\left(f\cup g\right)\left(a\_1 \otimes ... \otimes a\_{n+m}\right) = f\left(a\_1 \otimes ... \otimes a\_n\right) g\left(a\_{n+1} \otimes ... \otimes a\_{n+m}\right)$. I am aware of [this article by Arne B. Sletsjøe](http://folk.uio.no/arnebs/Artikler/Produkter.pdf), but it defines Hochschild cohomology differently (by using $\left(-1\right)^{n+1} a\_{n+1} f\left(a\_1 \otimes ... \otimes a\_n\right)$ in lieu of $\left(-1\right)^{n+1} f\left(a\_1 \otimes ... \otimes a\_n\right) a\_{n+1}$); this definition is only equivalent to mine if M is a symmetric A-module, so it won't help me find out whether this is necessary to assume. Thanks for any help, and sorry if the counterexamples are so obvious that I am an idiot not to find them on my own...	https://mathoverflow.net/users/2530	Graded commutativity of cup in Hochschild cohomology	The cup product in Hochschild cohomology$H^\bullet(A,A)$ is graded commutative for all unitary algebras. If $M$ is an $A$-bimodule, then the cohomology $H^\bullet(A,M)$ with values in $M$ is a symmetric graded bimodule over $H^\bullet(A,A)$. (If $M$ itself is also an algebra such that its multiplication map $M\otimes M\to M$ is a map of $A$-bimodules, then in general $H^\bullet(A,M)$ is not commutative (for example, take $A=k$ to be the ground field, and $M$ to be an arbitrary non-commutative algebra! I do not know of a criterion for commutatitivity in this case) These results originally appeared in [M. Gerstenhaber, The cohomology structure of an associative ring, Ann. of Math. (2) 78 (1963), 267–288.], and they are discussed at length in [Gerstenhaber, Murray; Schack, Samuel D. Algebraic cohomology and deformation theory. Deformation theory of algebras and structures and applications (Il Ciocco, 1986), 11--264, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 247, Kluwer Acad. Publ., Dordrecht, 1988.] Both references give proofs of a rather computational nature. You can find an element-free proof of the graded commutativity in this [paper](http://arxiv.org/abs/math/0209029), which moreover applies to the cup-products of many other cohomologies. As for what happens with non-unital algebras, I do not know. But they are very much used today as they were before. One particular context in which they show up constantly is in the intersection of K-theory and functional analysis, where people study `algebras' which are really just ideals in rings of operators of functional spaces---one egregius example is the algebra of compact operators in a Hilbert space. By the way, you say that group cohomology is a special case of Hochschild cohomology: it is only in a sense... There is a close relationship between the group cohomology $H^\bullet(G,\mathord-)$ of a group $G$, and the Hochschild chomology $H^\bullet(kG,\mathord-)$ of the group algebra, but they are not the same. You can use the second to compute the first (because more generally if $M$ and $N$ are $G$-modules, then $\mathord{Ext}\_{kG}^\bullet(M,N)=H^\bullet(kG, hom(M,N))$, where on the right we have Hochschild cohomology of the group algebra $kG$ with values in the $kG$-bimodule $\hom(M,N)$), but the «principal» group cohomology $H^\bullet(G,k)$ is only a little part of the «principal» Hochschild cohomology $H^\bullet(kG,kG)$. Finally, in the paper by Sletsjøe the definition for the boundary is given as you say because he only considers commutative algebras and only principal coefficients, that is $H^\bullet(A,A)$.	11	https://mathoverflow.net/users/1409	8674	5,940
https://mathoverflow.net/questions/8663	10	Edit & Note: I'm declaring a convention here because I don't feel like trying to fix this in a bunch of spots: If I said model category and it doesn't make sense, I meant a model-category "model" of an (infinity,1)-category. Also, "model" in quotes means the English word model, whereas without quotes it has do do with model categories. At the very beginning of Lurie's higher topos theory, he mentions that there is a theory of $(\infty,1)$-categories that can be directly constructed by using model categories. What I'd like to know is: Where can I find related papers (Lurie mentions two books that are not available for download)? How dependent on quasicategories is the theory developed in HTT? Can the important results be proven for these $(\infty,1)$-model-categories by proving some sort of equivalence (not equivalence of categories, but some weaker kind of equivalence) to the theory of quasicategories? When would we want to use quasicategories rather than these more abstract model categories? And also, conversely, when would we want to look at model categories rather than quasicategories? Does one subsume the other? Are there disadvantages to the model category construction just because it requires you to have all of the machinery of model categories? Are quasicategories better in every way? The only "models" of infinity categories that I'm familiar with are the ones presented in HTT.	https://mathoverflow.net/users/1353	(infinity,1)-categories directly from model categories	You can't produce every ($\infty$,1)-category from a model category. The slogan is that every presentable ($\infty$,1)-category comes from a model category, and every adjoint pair between such comes from a Quillen pair of functors between model categories. The paper by Dugger on [Universal model categories](http://arxiv.org/abs/math/0007070) works out this formalism from the point of view of model categories. (A [companion paper](http://arxiv.org/abs/math/0007068) shows that a large class of model categories (the combinatorial ones) are "presentable" in this sense.) (I say it's a slogan, but I'm sure it's a theorem; I just don't have a reference for you.) Presentable ($\infty$,1)-categories are special among all ($\infty$,1)-categories; in particular, they are complete and cocomplete. For instance, you can define the notion of ($\infty$,1)-topos in terms of model categories, since ($\infty$,1)-topoi are presentable, and morphisms between such are certain kinds of adjoint functor pairs.	25	https://mathoverflow.net/users/437	8675	5,941
https://mathoverflow.net/questions/8665	29	Suppose we are given an embedding of $S^2$ in $\mathbb{CP}^2$ with self-intersection 1. Is there a diffeomorphism of $\mathbb{CP}^2$ which takes the given sphere to a complex line? 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. 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 prove 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.	https://mathoverflow.net/users/380	Embeddings of $S^2$ in $\mathbb{CP}^2$	The conjecture that every $S^2 \subseteq \mathbb{C}P^2$ is standard if it is homologous to flat is implied by the smooth Poincaré conjecture in 4 dimensions. It also implies a special of smooth Poincaré that is accepted as an open problem, the case of [Gluck surgery](http://atlas-conferences.com/c/a/b/o/02.htm) in $S^4$. I can't prove or disprove the question of course, but since the question is sandwiched between two open problems, I can "prove" that it is an open problem. It is easier to consider $\mathbb{C}P^2$ minus a tubular neighborhood of the $S^2$, rather than to "blow it down". The condition on the homology class is equivalent to the condition that the boundary of this tube is $S^3$; the projection to the core is a Hopf fibration. The blowdown consists of attaching a 4-ball to this 3-sphere; let's skip this step. As Joel had in mind, the complement of the $S^2$ is simply connected. In fact, it is a homotopy 4-ball with boundary $S^3$. Thus, Freedman's theorem implies that it is homeomorphic to a 4-ball and smooth Poincaré would imply that it is diffeomorphic to a 4-ball. When it is, this 4-ball is still standard with its Hopf-fibered boundary (the Hopf fibration is unique up to orientation), so the $S^2$ is unknotted. In the other direction, the $S^2$ could be the direct sum of a standard complex line in $\mathbb{C}P^2$ with a 2-knot $K$ in $S^4$. I argue that in this case, the blowdown is equivalent to the Gluck surgery along $K$. What is a Gluck surgery? It looks like Dehn surgery in 3 dimensions, except with peculiar behavior. The official definition is that you remove a neighborhood of $K$ (which here is $D^2 \times S^2$, not the twisted bundle in Joel's construction), then glue it back after applying the non-trivial diffeomorphism of $S^1 \times S^2$. That diffeomorphism comes from the non-trivial element in $\pi\_1(\text{SO}(3)) = \mathbb{Z}/2$. One thing that is peculiar is that the Gluck surgery does not change the homotopy type of its 4-manifold, which is why it produces many candidate counterexamples to smooth Poincaré. Again, it is easier to think about the closed complement to Joel's $S^2$ than the blowdown. The corresponding version of Gluck surgery is to remove all of $D^2 \times K$, but only glue back a thickened $D^2$ (a 2-handle) along an attaching circle, and not glue back in the remaining 4-ball along the rest of $K$. What is peculiar here is that the attaching circle does not change; it is still a vertical circle in $S^1 \times S^2$. What changes instead is that the framing of the attachment is twisted by 1. (Or it can be twisted by some other odd number, since $\pi\_1(\text{SO}(3)) = \mathbb{Z}/2$ and not $\mathbb{Z}$. More prosaically, the "belt trick" lets you change the twisting by an even number.) Anyway, if Joel's sphere is $K$ connect summed with a complex line $L$, then you can represent this crucial 2-handle with another complex line $J$ in $\mathbb{C}P^2$. The question is whether the framing of its attachment to $L$ is odd or even. The fact that a perturbation $J'$ of $J$ intersects $J$ once tells me that the framing is odd. So the result is Gluck surgery. The old version of this answer was less developed (and at first I made the $\pi\_1$ mistake that is corrected in the comments and the edit to the question). But it is still worth noting that there are many open special cases of smooth Poincaré that consist of just one homotopy 4-sphere. Some topologists interpret this as strong evidence that smooth Poincaré is false. Others suppose that we just might not be very good at finding diffeomorphisms with $S^4$. A few examples, including some Gluck surgeries, were shown to be standard only after many years, for instance in [this paper](http://www.math.msu.edu/~akbulut/papers/cs.pdf) by Selman Akbulut.	17	https://mathoverflow.net/users/1450	8683	5,946
https://mathoverflow.net/questions/8684	18	I have a good grasp of ordinary pullbacks and pushouts; in particular, there are many categorical constructions that can be seen as special cases: e.g., equalizers/coequalizers, kernerls/cokernels, binary products/coproducts, preimages,... I know the (a?) definition of homotopy pullbacks/pushouts, but I am lacking two things: examples and intuition. So here are my questions: 1. What are the canonical examples of homotopy pullbacks/pushouts? E.g., in the category of pointed topological spaces the loop space $\Omega X$ is a homotopy pullback of the map $\ast \to X$ along itself. 2. How should I think about homotopy pullbacks/pushouts? What is the intuition behind the concept?	https://mathoverflow.net/users/1797	Homotopy pullbacks and homotopy pushouts	You can think of the pushout of two maps f : A → B, g : A → C in Set as computing the disjoint union of B and C with an identification f(a) = g(a) for each element a of A. We could imagine forming this as either the quotient by an equivalence relation, or by gluing in a segment joining f(a) to g(a) for each a, and taking π0 of the resulting space. If two elements a, a' of A satisfy f(a) = f(a') and g(a) = g(a'), the pushout is unaffected by removing a' from A. The homotopy pushout is formed by gluing in a segment joining f(a) to g(a) for each a and not forgetting the number of ways in which two elements of B ∐ C are identified; instead we take the entire space as the result. It is the "derived" version of the pushout. In general you can think of the homotopy pushout of A → B, A → C as the "free" thing generated by B and C with "relations" coming from A. But it's important that the "relations" are imposed exactly once, since in the homotopical/derived setting we keep track of such things (and have "relations between relations" etc.) Another, possibly more familiar example: In a derived category, the mapping cone of a morphism f : A → B is the homotopy pushout of f and the zero map A → 0. This certainly depends on A, even when B is the zero object: it is the suspension of A.	22	https://mathoverflow.net/users/126667	8690	5,952
https://mathoverflow.net/questions/8656	2	What is a "nice" way of choosing coset representatives for the symplectic group $Sp\_{2k}(\mathbb{C})$ in the general linear group $GL\_{2k}(\mathbb{C})$?	https://mathoverflow.net/users/2623	Nicest coset representatives of the symplectic group in the general linear group	I suppose that "nice" is very much application-dependent, but let me give it a try. The first thing to notice is that, of course, you will not be able to find a global coset representative, since the principal bundle $$\mathrm{Sp}(2k,\mathbb{C}) \to \mathrm{GL}(2k,\mathbb{C}) \to M = \mathrm{GL}(2k,\mathbb{C})/\mathrm{Sp}(2k,\mathbb{C})$$ is not trivial and hence it has no global (continous) section. So the best you can do is find a section over some $U \subset M$. The subgroup $\mathrm{Sp}(2k,\mathbb{C})$ is the stabilizer of a symplectic structure $\Omega$ on $\mathbb{C}^{2k}$. A convenient choice is $$\Omega = \pmatrix{ 0 & -\mathbf{1} \cr \mathbf{1} & 0}$$ where $\mathbf{1}$ is the $k\times k$ identity matrix. The Lie algebra $\mathfrak{sp}(2k,\mathbb{C})$ consists of those $2k \times 2k$ complex matrices $X$ such that $\Omega X$ is symmetric. A complementary vector subspace of $\mathfrak{sp}(2k,\mathbb{C})$ in $\mathfrak{gl}(2k,\mathbb{C})$ is given by $$\mathfrak{sp}(2k,\mathbb{C})^\perp := \big\lbrace \Omega X \mid X \in \mathfrak{so}(2k,\mathbb{C})\big\rbrace$$ Explicitly and for our choice of $\Omega$ above, this subspace consists of the matrices of the form $$\pmatrix{B^t & C \cr A & B}$$ for $k\times k$ matrices $A,B,C$ with $A$ and $C$ skewsymmetric. You can now exponentiate these matrices to find a coset representative. Depending on the calculation, though, you might it easier to write the coset representative as a product of exponentials,...	3	https://mathoverflow.net/users/394	8698	5,956
https://mathoverflow.net/questions/8692	15	The closed string A-model is mathematically described by Gromov-Witten invariants of a compact symplectic manifold $X$. The genus 0 GW invariants give the structure of quantum cohomology of $X$, which is then an example of a so-called Frobenius manifold. The mirror genus 0 B-model theory on the mirror manifold (or Landau-Ginzburg model) is usually described, mathematically, in terms of variation of Hodge structure data, or some generalization thereof. Since the higher genus A-model has a nice mathematical description as higher genus GW invariants, I am wondering whether the higher genus B-model has a nice mathematical description as well. Costello's paper on TCFTs and Calabi-Yau categories gives a partial answer to this. One can say that GW theory is the study of algebras over the (homology) operad of compactified (Deligne-Mumford) moduli space; I say that Costello gives a partial answer because he only gives an algebra over the operad of uncompactified moduli space. Though, according to Kontsevich (see Kontsevich-Soibelman "Notes on A-infinity..." and Katzarkov-Kontsevich-Pantev), we can extend this to the operad of compactified moduli space given some assumptions (a version of Hodge-de Rham degeneration). There are also various results (e.g. Katzarkov-Kontsevich-Pantev, Teleman/Givental) which say that the higher genus theory is uniquely determined by the genus 0 theory. But --- despite these sorts of results, I still have not seen any nice mathematical description of the higher genus B-model which "stands on its own", as the higher genus GW invariants do. I have only seen the higher genus B-model described as some structure which is obtained formally from genus 0 data, or, as in the situation of Costello's paper, a Calabi-Yau category, e.g. derived category of coherent sheaves of a Calabi-Yau manifold, matrix factorizations category of a Landau-Ginzburg model, etc. So, my questions are: * Are there any mathematical descriptions of the higher genus closed string B-model which "stand on their own"? What I mean is something that can be defined without any reference to other structure, just as genus 47 GW invariants can be defined without any reference to genus 0 GW invariants or the Fukaya category. * Genus 0 theory is essentially equivalent to the theory of Frobenius manifolds. What about higher genus theory? Is there any nice geometric structure behind, say, the genus 1 theory, analogous to the Frobenius manifold structure that we get out of genus 0 theory?	https://mathoverflow.net/users/83	Higher genus closed string B-model	This is a great question I wish I understood the answer to better. I know two vague answers, one based on derived algebraic geometry and one based on string theory. The first answer, that Costello explained to me and I most likely misrepeat, is the following. The B-model on a CY X as an extended TFT can be defined in terms of DAG: we consider the worldsheet $\Sigma$ as merely a topological space or simplicial set (this is a reflection of the lack of instanton corrections in the B-model), and consider the mapping space $X^\Sigma$ in the DAG sense. For example for $\Sigma=S^1$ this is the derived loop space (odd tangent bundle) of $X$.. In this language it's very easy to say what the theory assigns to 0- and 1-manifolds: to a point we assign coherent sheaves on $X$, to a 1-manifold cobordism we assign the functor given by push-pull of sheaves between obvious maps of mapping spaces (see e.g. the last section [here](http://arxiv.org/abs/0805.0157)). For example for $S^1$ we recover Hochschild homology of $X$. Now for 2-manifold bordisms we want to define natural operations by push-pull of functions, but for that we need a measure -- and the claim is the Calabi-Yau structure (together with the appropriate DAG version of Grothendieck-Serre duality, which Kevin said Lurie provides) gives exactly this integration... Anyway that gives a tentative answer to your question: the B-model assigns to a surface $\Sigma$ the "volume" of the mapping space $X^\Sigma$, defined in terms of the CY form. More concretely, you chop up $\Sigma$ into pieces, and use the natural operations on Hochschild homology, such as trace pairing and identification with Hochschild cohomology (and hence pair-of-pants multiplication).. of course this last sentence is just saying "use the Frobenius algebra structure on what you assigned to the circle" so doesn't really address your question - the key is to interpret the volume of $X^\Sigma$ correctly. The second answer from string theory says that while genus 0 defines a Frobenius manifold you shouldn't consider other genera individually, but as a generating series -- i.e. the genus is paired with the (topological) string coupling constant, and together defines a single object, the topological string partition function, which you should try to interpret rather than term by term. (This is also the topic of Costello's paper on the partition function). BTW for genus one there is a concrete answer in terms of Ray-Singer torsion, but I don't think that extends obviously to higher genus. As to how to interpret it, that's the topic of the famous [BCOV paper](http://arxiv.org/abs/hep-th/9309140) - i.e. the Kodaira-Spencer theory of gravity. For one thing, the partition function is determined recursively by the holomorphic anomaly equation, though I don't understand that as "explaining" the higher genus contributions. But in any case there's a Chern-Simons type theory quantizing the deformation theory of the Calabi-Yau, built out of the Kodaira-Spencer dgla in a simple looking way, and that's what the B-model is calculating. A very inspiring POV on this is due to [Witten](http://arxiv.org/abs/hep-th/9306122), who interprets the entire partition function as the wave function in a standard geometric quantization picture for the middle cohomology of the CY (or more suggestively, of the moduli of CYs). This is also behind the Givental quantization formalism for the higher genus A-model, where the issue is not defining the invariants but finding a way to calculate them. Anyway I don't know a totally satisfactory mathematical formalism for the meaning of this partition function (and have tried to get it from many people), so would love to hear any thoughts. But the strong message from physics is that we should try to interpret this entire partition function - in particular it is this function which appears in a million different guises under various dualities (eg in gauge theory, as solution to quantum integrable systems, etc etc...)	15	https://mathoverflow.net/users/582	8708	5,961
https://mathoverflow.net/questions/8697	13	Is it true that any closed oriented $4$-dimensional manifold can be obtained as a result of the following construction: Take $S^4$ with a finite collection of immersed closed 2-manifolds (with transversal intersections and self-intersections) and construct ramified cover of $S^4$ with a ramification of order at most 2 only at these submanifolds. Comments: * Two related questions: [Ramified covers of 3-torus](https://mathoverflow.net/questions/5546/ramified-covers-of-3-torus), [Ramified covers of $S^n$](https://mathoverflow.net/questions/5618/ramified-covers-of-sn) * According to Feighn's [Branched covers according to J.W. Alexander](http://www.raco.cat/index.php/CollectaneaMathematica/article/view/56965) any closed oriented 4-manifold is a branched cover of $S^4$ with a ramification along 2-skeleton of 4-tetrahedron embedded in $S^4$ (which is not at all a 2-manifold).	https://mathoverflow.net/users/1441	Ramified cover of 4-sphere	The answer is yes, at least if we interpret your phrase "ramification of order 2" to mean "simple branched covering". See Piergallini, R., Four-manifolds as $4$-fold branched covers of $S^4$. Topology 34 (1995), no. 3, 497--508. Any closed, orientable PL 4-manifold can be expressed as a 4-fold simple branched covering of S4 branched along an immersed surface with only transverse double points. It is apparently still an open question whether the branch set can be chosen to be nonsingular. A simple branched covering of degree d is a branched covering in which each branch point is covered by d-1 points, only one of which is singular, of local degree 2.	11	https://mathoverflow.net/users/1822	8715	5,965
https://mathoverflow.net/questions/8716	11	My background on number theory is very weak, so please bear with me... Given two matrices $A$ and $B$ in $\mathbb{Z}^{n\times n}$. Assume that for every prime $p$, the images of $A$ and $B$ in $\mathbb{F}\_p^{n\times n}$ are similar to each other. Does this yield the existence of a matrix $X\in\mathrm{SL}\_n\left(\mathbb{Z}\right)$ satisfying $AX=XB$ ? What if we additionally assume $A$ and $B$ to be similar in $\mathbb{Q}^{n\times n}$ ?	https://mathoverflow.net/users/2530	Local-globalism for similar matrices?	Your question reminds me of a classical theorem of Latimer and MacDuffee (Annals of Math. 1933). To be sure, the theorem does not answer your question, but it seems relevant. A nice contemporary treatment of this theorem (with references) can be found at <http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/matrixconj.pdf> [At the moment that I write this, math.uconn.edu seems to be down. I presume this is only temporary!]	11	https://mathoverflow.net/users/1149	8718	5,966
https://mathoverflow.net/questions/8723	6	Is there a source in the literature for ribbon diagrams for the knot-table knots known to be ribbon knots? For example, I'm interested in doing a computation which needs as input a ribbon diagram for the knot $8\_{20}$ (Rolfsen knot table notation). This knot is known to be ribbon, but I don't know a ribbon diagram for the knot. Usually when I encounter a claim of the sort "knot X is ribbon" either the author supplies the ribbon diagram, or nothing. Citations to information of this sort seem kind of sparse. Or am I just unaware of a standard source for this type of information?	https://mathoverflow.net/users/1465	Generating ribbon diagrams for knots known to be ribbon knots	I think Kawauchi's book has tables that include ribbon diagrams, but I don't have a copy with me. Look at [Livingston and Cha](http://www.indiana.edu/~knotinfo/diagrams/8_20.png) . It is not hard to get a ribbon disk from this diagram: add a handle between the ears on the top and bottom right. Generally, I check [Livingston/Cha](http://www.indiana.edu/~knotinfo/) , [Bar-Natan,](http://www.math.toronto.edu/~drorbn/KAtlas/Knots/8.20.html) and [Saito](http://shell.cas.usf.edu/quandle/Invariants/database/database.php) for various information. @ears: there are a pair of symmetric clasps on the top and bottom of the diagram. Pull the top-most and bottom-most arc to the right, and then attach a band. The vertical arc that forms a triangle, and the right vertical arc from the band forms an obvious embedded circle.	7	https://mathoverflow.net/users/36108	8728	5,972
https://mathoverflow.net/questions/8741	15	Here is a topic in the vein of [Describe a topic in one sentence](https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence "Describe a topic in one sentence") and [Fundamental examples](https://mathoverflow.net/questions/4994/fundamental-examples) : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. A classical example is Galois theory for solving polynomial equations. Any examples for homological algebra ? For Fourier analysis ? For category theory ?	https://mathoverflow.net/users/2389	Justifying a theory by a seemingly unrelated example	[In front of a blackboard, in an office at Real College] Skeptic: And why should I care about holomorphic functions? Holomorphic enthusiast:$\;$ Can you compute $\quad$ $\sum\_{n={-\infty}}^{\infty} \frac{1}{(a+n)^2}$ ? Here $a$ is one of your cherished real numbers, but not an integer. Skeptic: Well, hm... Holomorphic enthusiast, nonchalantly: Oh, you just get $$\sum\_{n={-\infty}}^{\infty} \frac{1}{(a+n)^2}=\pi^2 cosec^2 \pi a $$ It's easy using residues. Skeptic: Well, maybe I should have a look at these "residues". Holomorphic enthusiast (generously): Let me lend you this introduction to Complex Analysis by Remmert, this one by Lang and this oldie by Titchmarsh. As Hadamard said: "Le plus court chemin entre deux vérités dans le domaine réel passe par le domaine complexe".You can look for a translation at Mathoverflow. They have a nice list of mathematical quotations, following a question there. Skeptic: Mathoverflow ?? Holomorphic enthusiast (looking a bit depressed) : I think we should have a nice long walk together now. [Exeunt]	38	https://mathoverflow.net/users/450	8752	5,988
https://mathoverflow.net/questions/8756	29	So there are easy examples for algebraic closures that have index two and infinite index: $\mathbb{C}$ over $\mathbb{R}$ and the algebraic numbers over $\mathbb{Q}$. What about the other indices? EDIT: Of course $\overline{\mathbb{Q}} \neq \mathbb{C}$. I don't know what I was thinking.	https://mathoverflow.net/users/96	Examples of algebraic closures of finite index	Theorem (Artin-Schreier, 1927): Let K be an algebraically closed field and F a proper subfield of K with $[K:F] < \infty$. Then F is real-closed and $K = F(\sqrt{-1})$. See e.g. Jacobson, Basic Algebra II, Theorem 11.14.	51	https://mathoverflow.net/users/1149	8759	5,993
https://mathoverflow.net/questions/8707	15	Take a closed $n$-manifold $M$ and fix a nice $n$-ball $B$ in $M$. How much information about $M$ does the set of knotted $(n-2)$-spheres of $B$ which are unknotted in $M$ give?	https://mathoverflow.net/users/1409	Knots that unknot in a manifold	Kevin Walker and a member of the forum (by e-mail) let me know about a flaw in the argument in my previous post. I'll erase it after putting up this post. One of my claims was the if $K$ unknots in $M$ then it unknots in $B$, where $K$ is a knot in a ball $B \subset M$. If $K$ is the unknot in $M$ it bounds a disk $D \subset M$, $K = \partial D$. In dimension $3$ we perturb the disc so that $D \cap \partial B$ is a collection of circles, this allows us to cap off $D \cap B$ with disc(s) parallel to $\partial B$ (an embedded surgery construction), constructing a disc in $B$ that bounds $K$. In higher dimensions this argument runs into trouble as it's not clear if you can perform the required embedded surgeries. I also claimed that the only new isotopy relation in general is that knots in $B$ can be isotopic (in $M$) to their mirror inverse if $M$ is non-orientable. I still suspect this is the case and I think I have an idea for an argument but I should think about it some more before posting. Here is an argument that works in dimension 3: Say $M$ is a compact boundaryless $3$-manifold, $K$ and $K'$ contained in a ball $B$ are isotopic in $M$. Perform the connect-sum decomposition on $M \setminus K$ and $M \setminus K'$. Since this is a unique decomposition (modulo issues in the non-orientable case), in both cases you get a list of prime factors that are identical except in the decomposition of $M \setminus K$ you get a summand that's $B \setminus K$ capped off with a disc, and in the other case you get a summand $B \setminus K'$ capped off with a disc. By the Gordon-Luecke theorem $K$ and $K'$ are either isotopic in $B$ or mirror images of each other. I'm still hopeful I have an argument that works in general. I'll edit this post again tomorrow once I'm certain. edit: final exams and deadlines are eating up my time at the moment but I'll get around to another revision... eventually.	6	https://mathoverflow.net/users/1465	8786	6,014
https://mathoverflow.net/questions/8784	29	Recall the following classical theorem of Cartan (!): Theorem (Lie III): Any finite-dimensional Lie algebra over $\mathbb R$ is the Lie algebra of some analytic Lie group. Similarly, one can propose "Lie III" statements for Lie algebras over other fields, for super Lie algebras, for Lie algebroids, etc. The proof I know of the classical Lie III is very difficult: it requires most of the structure theory of Lie algebras. But why should it be difficult? For example, for finite-dimensional Lie algebra $\mathfrak g$ over $\mathbb R$, the Baker-Campbell-Hausdorff formula (the power series given by $B(x,y) = \log(\exp x \exp y)$ in noncommuting variables $x,y$; it can be written with only the Lie bracket, no multiplication) converges in an open neighborhood of the origin, and so defines a unital associative partial group operation on (an open neighborhood in) $\mathfrak g$. What happens if one were to try to simply glue together copies of this open neighborhood? Alternately, are there natural variations of Lie III that are so badly false that any easy proof of Lie III is bound to fail?	https://mathoverflow.net/users/78	Why is Lie's Third Theorem difficult?	That gluing together of group chunks, constructed from the BCH formula is precisely more or less what Serre does to prove the theorem (in the first proof he gives in) his book on Lie groups and Lie algebras. [Serre, Jean-Pierre. Lie algebras and Lie groups. 1964 lectures given at Harvard University. Second edition. Lecture Notes in Mathematics, 1500. Springer-Verlag, Berlin, 1992. viii+168 pp.]	20	https://mathoverflow.net/users/1409	8788	6,016
https://mathoverflow.net/questions/8789	89	Let $M$ be a (real) manifold. Recall that an analytic structure on $M$ is an atlas such that all transition maps are real-analytic (and maximal with respect to this property). (There's also a sheafy definition.) So in particular being analytic is a structure, not a property. Q1: Is it true that any topological manifold can be equipped with an analytic structure? Q2: Can any $C^\infty$ manifold (replace "analytic" by "$C^\infty$" in the first paragraph) be equipped with an analytic structure (consistent with the smooth structure)?	https://mathoverflow.net/users/78	Can every manifold be given an analytic structure?	(similar to Mariano's post) Q1: no. There are topological manifolds that don't admit triangulations, let alone smooth structures. All smooth manifolds admit triangulations, this is a theorem of Whitehead's. The lowest-dimensional examples of topological manifolds that don't admit triangulations are in dimension 4, the obstruction is called the Kirby-Siebenmann smoothing obstruction. Q2: $C^1$ manifolds all admit compatible $C^\infty$ and analytic ($C^\omega$) structures. This is a theorem of Hassler Whitney's, in his early papers on manifold theory, where he proves they embed in euclidean space. The basic idea is that your manifold is locally cut out of euclidean space by $C^1$-functions so you apply a smoothing operator to the function and then argue that the level-set does not change (up to $C^1$-diffeomorphism), provided your smoothing approximation is small enough in the $C^1$-sense. I'm not sure who gets the original credit but you can go much further -- compact boundaryless smooth manifolds are all realizable as components of real affine algebraic varieties, planar linkages in particular. There's a Millson and Kapovich paper, [Universality theorems for configuration spaces of planar linkages](https://arxiv.org/abs/math/9803150) on the topic available. It seems people give a lot of credit to Bill Thurston. edit: Some time ago Riccardo Benedetti sent me some comments to append to my answer. They appear below, with some minor MO-formatting on my part. In the famous paper "Real algebraic manifolds" (Annals of Math 56, 3, 1952), John Nash just proved that: "Every compact closed smooth manifold M embedded in some $R^N$, with $N$ big enough (as usually $N=2Dim(M)+1$ suffices), can be smoothly approximated by a union of components, say $M\_a$, of the non-singular locus of a real algebraic subset $X$ of $R^N$." In the same paper he stated also some conjectures/questions, in particular whether one can get $M\_a = X$ (so that $X$ is a non-singular real algebraic model of $M$), or whether one can even get such an algebraic model which is rational. A.H. Wallace (1957) solved positively the first question under the assumption that $M$ is a boundary. Finally a complete positive answer was given by A. Tognoli (1973) by using, among other things, a so called "Wallace trick" and the fact (due to Milnor) that the smooth/un-oriented bordism group is generated by real algebraic projective non singular varieties. Starting from this Nash-Tognoli theorem, mostly in the 80's-90's of the last century, a huge activity has been developed about the existence of real algebraic models for several instances of smooth or polyhedral structures, with major contributions by S. Akbulut and H. King (in particular they proved that if M is embedded in $R^n$, an algebraic model can be realized in $R^{n+1}$; to my knowledge it is open if we can stay in the given $R^n$). If I am not wrong, the realization of real algebraic varieties via planar linkages (with related credit to Thurston) does not provide an alternative proof of Nash-Tognoli theorem. The "Nash rationality conjecture" is more intriguing and has been basically "solved" in dimension less or equal to 3. This is mentioned for instance in some answers to the questions: [What's the difference between a real manifold and a smooth variety?](https://mathoverflow.net/questions/23911/whats-the-difference-between-a-real-manifold-and-a-smooth-variety/23944#23944) "You might also be interested in some of the articles by Kolla'r on the Nash conjecture contrasting real varieties and real manifolds. such as "What are the simplest varieties?", Bulletin, vol 38. I like the pair of theorems 54, 51, subtitled respectively: "the Nash conjecture is true in dim 3", and "The Nash conjecture is false in dim 3". [What is known about the MMP over non-algebraically closed fields](https://mathoverflow.net/questions/23107/what-is-known-about-the-mmp-over-non-algebraically-closed-fields/41737#41737) "Another issue is the rational connectivity and its relation to Mori fiber spaces ...... To illustrate the difficulties, here is a conjecture of Nash (yes, that Nash): Let $Z$ be a smooth real algebraic variety. Then $Z$ can be realized as the real points of a rational complex algebraic variety. This, actually, turns out to be false. Kollár calls it the shortest lived conjecture as it was stated in 1954 and disproved by Comessatti around 1914 (I don't remember the exact year). However, even if the statement is false, just the fact that it was made and no one realized for 50 years that it was false should show that these questions are by no means easy. (Comessatti's paper was in Italian and I have no idea how Kollár found it.) Kollár showed more systematically the possible topology types of manifolds that can satisfy this statement. In particular, Kollár shows that any closed connected 3-manifold occurs as a possibly non-projective real variety birationally equivalent to $P^3$. In other words the way Nash's conjecture fails is on the verge of the difference between projective and proper again showing that these questions are not easy." (Sandor Kovacs) As I have been even more concerned with, let me add a few comments. Comessatti's result was certainly "well known" at least to some Italian people and also to the real algebraic Russian guys of Arnol'd's and mostly Rokhlin's school. Moreover (before Kollar's work) it had been rediscovered (via modern tools) for instance by R. Silhol. This allows the following 2D solution of the Nash conjecture: " (1) Every non orientable compact closed surface admits a rational projective non singular model which can be explicitly given by (algebraically) blowing-up $\mathbb RP^2$ at some points; (2) $S^2$ and $T^2$ are the only orientable surfaces that admit non singular projective rational models (Comessatti); (3) Every surface $S$ admits a projective rational model possibly having one singular point (to get it, first smoothly blow-up $S$ at one point $p$ getting a non orientable $S'$ containing a smooth exceptional curve $C$ over $p$. Take a projective non singular rational model $S'\_r$ of $S$. Finally one can prove that $C$ can be approximated in $S'\_r$ by a non singular algebraic curve $C\_a$ and that we can perform a "singular" blow-down of $C\_a$ producing the required singular rational model of $S$.)" Inspired by this 2D discussion, in the paper (with A. Marin) Dechirures de varietes de dimension trois et la conjecture de Nash de rationalite' en dimension trois Comment. Math. Helv. 67 (4) (1992), 514-545 (a PDF file is available in [http://www.dm.unipi.it/~benedett/research.html](http://www.dm.unipi.it/%7Ebenedett/research.html)) we got a formally similar 3D solution. More precisely we provide at first a complete classification of compact closed smooth 3-manifolds $M$ up to "flip/flop" along smooth centres (see below - these are the "Dechirures" - perhaps we were wrong to write the paper in French ... ). Summarizing (and roughly) the results are: * There is a complete invariant $I(M)$ for this equivalence relation. Depending only on $I(M)$, we explicitly produce a real projective non singular rational 3-fold $Z$ such that $I(M)=I(Z)$. * There is a smooth link $L$ in $M$ and a non singular real algebraic link $L\_a$ in $Z$ such that by smoothly (algebraically) blowing up $M (Z)$ along $L (L\_a)$ we get the same manifold $Z'$ and furthermore the disjoint real algebraic exceptional tori over $L\_a$ coincide with the exceptional tori over $L$ (thinking all within the smooth category, basically by definition this means that $Z$ and $M$ are equivalent up to flip/flop along smooth centres). * Clearly $Z'$ is projective rational non singular and algebraically dominates $Z$. It also smoothly dominates M. Finally we can convert the smooth blow-down onto $M$ to a singular algebraic blow-down producing a projective rational model $M\_r$ of $M$, possibly singular at a non singular real algebraic copy of the link $L$ in $M\_r$. * The invariant $I(M)$ is easy when $M$ is orientable; this is just the dimension of $H\_2(M;\mathbb Z/2)$. In this case $Z$ is obtained by algebraically blowing up $S^3$ at some points. $I(M)$ is much more complicated if M is non orientable and involves, among other things, certain quadratic forms on $Ker(i\_\:H\_1(S;\mathbb Z/2) \to H\_1(M;\mathbb Z/2))$, $S$ being any characteristic surface of $M$. A combination of our work with Kollar's one roughly gives: (3D "a la" Comessatti) In the projective framework, in general our singular rational models cannot be improved (singularity cannot be avoided, and in a sense we provided models with very mild singularities); in other words those blowing-down $Z'\to M\_r$ were intrinsically singular. * (Non projective non singular rational models) Starting from our real projective rational non singular 3-fold $Z'$, as above, Kollar proves that one can realize a non singular blow-down $Z'\to M'\_r$, provided that we leave the projective framework. Finally it is intriguing to note another important occurrence of the opposition projective singular vs non singular (related to the existence of intrinsically singular blow-down). Going back to the original Nash-Tognoli kind of problems, for a while it was conjectured and very "desidered" (for several reasons also related to the general question of characterizing say the compact polyhedron which admit possibly singular real algebraic models) that every M as above admits a "totally algebraic model" M\_a, i.e. such that $H\_\(M\_a;\mathbb Z/2)$ is generated by the $\mathbb Z/2$-fundamental class of algebraic sub-varieties of $M\_a$. On the contrary we constructed counterexamples in: (with M. Dedo') Counterexamples to representing homology classes by real algebraic subvarieties up to homeomorphism, Compositio Math. 53 (2) (1984), 143-151 (idem) This contrasts with a result by Akbulut-King that M admits singular* totally algebraic models.	87	https://mathoverflow.net/users/1465	8794	6,019
https://mathoverflow.net/questions/8790	1	I'm having trouble finding the closure in the definition of a blow-up. For example, take the following example, with a node at $(0,0)$ (and at some other points) (it's not a homework question, just a concept that I'm stuck on!). $xy=x^6+y^6$. Then the blow-up should be the closure of this set, taken over all $(x,y) \neq (0,0)$: $ \{ ((u,v), (x,y)) \in \mathbb{A}^{2} \times \mathbb{P}^{1} \| uy=vx, xy = x^6 + y^6 \}$. How do I explicitly find the closure of this set? I understand the fibre of the projection map at the singular point should consist of two points (both of which are non-singular) - why is this so, and what are those two non-singular points (in the smooth variety that is the resolution)?.	https://mathoverflow.net/users/2623	finding the closure when blowing a variety at a singularity	Look at the affine pieces: over the open subset $u \neq 0$, you have a local coordinate $z = v/u$ and your equations can be written as $y = zx$ and $xy = x^6 + y^6$. Substituting $y$ in the second equation gives you $x^2z = x^6 + x^6 z^6$. Now this equation factors as $x^2 = 0$ and $z = x^4(1 + z^6)$; the locus where the first one vanishes is the exceptional divisor, while the second one gives you the closure you are looking for. Non-singularity of the point over $(x, y) = (0,0)$ (which is $(x,z) = (0,0)$) follows from $$ \frac{d}{dz} \big[z - x^4(1 + z^6)\big] \big\|\_{(x,z) = (0,0)} = 1$$ The other point shows up when considering the other affine piece, $v \neq 0$. The reason why there are two points over $(x,y) = (0,0)$ is because at that point your curve has two branches. To see that, look at the lowest degree piece of the polynomial defining it (essentially, you are looking here at a neighborhood of the origin in the classical/analytic topology): this is $xy$, which is the union of the two axes. Blowing up pulls these two branches apart.	7	https://mathoverflow.net/users/1797	8795	6,020
https://mathoverflow.net/questions/8793	4	Recall that a Lie group is a group object in the category of C∞ manifolds. If I have a group object in the category of topological manifolds, can I necessarily equip it with a smooth structure so that all the group operations are smooth? If so, how unique is this structure? Is a continuous group-homomorphism between two Lie groups necessarily smooth?	https://mathoverflow.net/users/78	Is every group object in TopMan a Lie group?	As Greg Kuperberg indicated in the comments, this is Hilbert's 5th Problem. The answer is yes, a theorem of Gleason, Montgomery and Zippin from the 1950's.	5	https://mathoverflow.net/users/1149	8796	6,021
https://mathoverflow.net/questions/8811	1	Suppose you have a curve $C$ such that deg$K\_C =0$ and $\Gamma(C,\Omega\_C^1) \neq 0$. Does this automatically imply that $\vartheta\_C \equiv \Omega\_C^1$? My thought is yes, I've seen a proposition (Stanford AG course notes) that $\vartheta\_C \equiv \Omega\_C^1$ for a nonsingular plane cubic, but the proof is done in a particular case, and I must think there is easier way to show this. In particular, what is the morphism? Thank you, just trying to make sense of this concept.	https://mathoverflow.net/users/2557	Sheaf isomorphism.	On a complete nonsingular curve over an algebraically closed field, a line bundle of degree zero with a global section is necessarily the trivial bundle. This is lemma IV.1.2 in Hartshorne's Algebraic Geometry: if $\mathrm{deg} D = 0$ and $\mathcal{L} = O\_C(D)$ has a global section, then $D$ is linearly equivalent to an effective divisor $E$ of the same degree; but the only effective divisor of degree zero is the zero divisor. Hence $\mathcal{L} \cong \mathcal{O}\_C(E) = \mathcal{O}\_C$.	2	https://mathoverflow.net/users/1797	8814	6,033
https://mathoverflow.net/questions/8753	7	Is anyone aware of alternative axioms to induction? To be precise, consider peano axioms without induction PA-. Is there any axiom/axiom schema that is equiconsistent to induction, assuming PA-? If so, why does it appear that nobody investigating it? To contextualize this question, I refer to discovery of the equiconsistency of the non-Euclidean geometries. The geometry has flourished (or at least, developed) to its modern state with manifold theories and later developments was only possible with the recognition of non-Euclidean geometries. For a long time, Euclidean geometry and the Parallel Postulate was seen as the only geometry, very much as today's standard model for number theory ${\mathbb N}$ and the induction axiom is seen as the only number theory. (at least to the extent that i'm aware of, so that's why I'm asking this question.) I wonder why, therefore, that there is only one number theory, and no alternatives like non-Induction number theory. Note: I am aware that recursion theorists study fragments of induction. However, in the sense of which I'm asking, this does not count at alternative axioms to induction since ${\mathbb N}$ is still a model of these these fragments. Using my geometry analogy, these fragments amount to saying that the Parallel Postulate does not hold everywhere throughout space, something like space is not homogeneous but still has some symmetry.	https://mathoverflow.net/users/nan	Alternative axiom to induction	A crucial difference between non-Euclidean geometries and "non-inductive" models of PA- is that any model of PA- contains a canonical copy of the true natural numbers, and in this copy of $N$, the induction schema is true. In other words, PA is part of the complete theory of a very canonical model of PA-, and as such it seems much more natural (so to speak) to study fragments of PA rather than extensions of PA- which contradict induction. To make this a little more precise (and sketch a proof), the axioms of PA- say that any model $M$ has a unique member $0\_M$ which is not the successor of anything, and that the successor function $S\_M: M \to M$ is injective; so by letting $k\_M$ (for any $k \in N$) be the $k$-th successor of $0\_M$, the set $\{k\_M : k \in N\}$ forms a submodel of $M$ which is isomorphic to the usual natural numbers, $N$. Any extra elements of $M$ not lying in this submodel lie in various "Z-chains," that is, infinite orbits of the model $M$'s successor function $S\_M$. (In the language of categories: the "usual natural numbers" are an initial object in the category of all models of PA-, where morphisms are injective homomorphisms in the sense of model theory.) So, while PA seems natural, I'm not sure why there would be any more motivation to study PA- plus "non-induction" than there is to study any of the other countless consistent theories you could cook up, unless you find that one of these non-inductive extensions of PA- has a particularly nice class of models.	15	https://mathoverflow.net/users/93	8819	6,036
https://mathoverflow.net/questions/8812	17	I am not sure that all automorphism groups of algebraic varieties have natrual algebraic group structure. But if the automorphism group of a variety has algebraic group structure, how do I know the automorphism group is an algebraic group. For example, the automorphism group of an elliptic curve $A$ is an extension of the group $G$ of automorphisms which preserve the structure of the elliptic curve, by the group $A(k)$ of translations in the points of $A$, i.e. the sequence of groups $0\to A(k)\to \text{Aut}(A)\to G \to 0$ is exact, see [Springer online ref - automorphism group of algebraic variaties](http://eom.springer.de/a/a011740.htm#a011740_00c2). In this example, how do I know $\text{Aut}(A)$ is an algebraic group.	https://mathoverflow.net/users/2348	Why do automorphism groups of algebraic varieties have natural algebraic group structure?	It is not always true that the automorphism group of an algebraic variety has a natural algebraic group structure. For example, the automorphism group of $\mathbb{A}^2$ includes all the maps of the form $(x,y) \mapsto (x, y+f(x))$ where $f$ is any polynomial. I haven't thought through how to say this precisely in terms of functors, but this subgroup morally should be a connected infinite dimensional object, and is thus not a subobject of an algebraic group. On the other hand, I believe that the automorphism group of a projective algebraic variety, $X$, can be given the structure of algebraic group in a fairly natural way. This is something I've thought about myself, but not written down a careful proof nor found a reference for: For any automorphism $f$ of $X$, consider the graph of $f$ as a subscheme of $X \times X$, and thus a point of the Hilbert scheme of $X\times X$. In this way, we get an embedding of point sets from $\mathrm{Aut}(X)$ into $\mathrm{Hilb}(X\times X)$. I believe that it should be easy to show that (1) $\mathrm{Aut}(X)$ is open in $\mathrm{Hilb}(X\times X)$, and thus acquires a natural scheme structure and (2) composition of automorphisms is a map of schemes.	17	https://mathoverflow.net/users/297	8820	6,037
https://mathoverflow.net/questions/8816	2	What is the result of multiplying several (or perhaps an infinite number) of binomial distributions together? To clarify, an example. Suppose that a bunch of people are playing a game with k (to start) weighted coins, such that heads appears with probability p < 1. When the players play a round, they flip all their coins. For each heads, they get a coin to flip in the next round. This is repeated every round until they have a round with no heads. How would I calculate the probability distribution of the number or coins a player will have after n rounds? Especially if n is extremely high and p extremely close to 1?	https://mathoverflow.net/users/942	Result of repeated applications of the binomial distribution?	Here's how I interpret your example: there are a bunch of coins (k initially), each being flipped every round until it comes up tails, at which point the coin is "out," And you want to know, after n rounds, the probability that exactly j coins are still active, for j = 0, ..., k. (The existence of multiple players seems irrelevant.) In that case, this is pretty elementary: after n rounds, the probability of each individual coin being active is p^n, so you have a binomial distribution with parameter p^n, k trials. Since you want to send p to 1 and n to infinity, note that replacing p by its square root and doubling n gives you the same distribution. Your problem has a surprisingly fascinating generalization, which I believe is called the Galton-Watson process. Its solution has a very elegant representation in terms of generating functions, but I think there are very few examples in which the probabilities are simple to compute in general. Your instance is one of those. (The generalization: at each round, you have a certain number of individuals, each of which turns (probabilistically, independently) into a finite number of identical individuals. If the individuals are coins and each coin turns into one coin with probability p and zero coins with probability 1-p, and you begin with k coins, then we recover your example.)	5	https://mathoverflow.net/users/302	8826	6,041
https://mathoverflow.net/questions/8829	38	Real projective spaces $\mathbb{R}P^n$ have $\mathbb{Z}/2$ cohomology rings $\mathbb{Z}/2[x]/(x^{n+1})$ and total Stiefel-Whitney class $(1+x)^{n+1}$ which is $1$ when $n$ is odd, so it follows that odd dimensional ones are boundaries of compact $(n+1)$-manifolds. My question is: are there any especially nice constructions of these $(n+1)$-manifolds? I'm especially interested in the case $n=3$. I believe we can get an explicit example of a 4-manifold bound by $\mathbb{R}P^3$ using Rokhlin/Lickorish-Wallace but it doesn't look like that would generalize to higher dimensions at all easily. Are there a lot of different 4-manifolds with this property?	https://mathoverflow.net/users/1772	What manifolds are bounded by RP^odd?	$\mathbb RP^3$ double-covers the lens space $L\_{4,1}$, so it's the boundary of the mapping cylinder of that covering map. In general $\mathbb RP^n$ for $n$ odd double-covers such a lens space. So in general $\mathbb RP^n$ is the boundary of a pretty standard $I$-bundle over the appropriate lens space. To be specific, define the general $L\_{4,1}$ as $S^{2n-1} / \mathbb Z\_4$ where $Z\_4 \subset S^1$ are the 4-th roots of unity, and we're using the standard action of the unit complex numbers on an odd dimensional sphere $S^{2n-1} \subset \mathbb C^n$. Edit: generalizing Tim's construction, you have the fiber bundle $S^1 \to S^{2n-1} \to \mathbb CP^{n-1}$. This allows you to think of $S^{2n-1}$ as the boundary of the tautological $D^2$-bundle over $\mathbb CP^{n-1}$. You can mod out the whole bundle by the antipodal map and you get $\mathbb RP^{2n-1}$ as the boundary of the disc bundle over $\mathbb CP^{n-1}$ with Euler class $2$. So this gives you an orientable manifold bounding $\mathbb RP^{2n-1}$ while my previous example was non-orientable.	41	https://mathoverflow.net/users/1465	8830	6,044
https://mathoverflow.net/questions/8809	8	Suppose G is an algebraic group (over a field, say; maybe even over ℂ) and H⊆G is a closed subgroup. Does there necessarily exist an action of G on a scheme X and a point x∈X such that H=Stab(x)? Before you jump out of your seat and say, "take X=G/H," let me point out that the question is basically equivalent† to "Is G/H a scheme?" If G/H is a scheme, you can take X=G/H. On the other hand, if you have X and x∈X, then the orbit of x (which is G/H) is open in its closure, so it inherits a scheme structure (it's an open subscheme of a closed subscheme of X). †I say "basically equivalent" because in my argument, I assumed that the action of G on X is quasi-compact and quasi-separated so that the closure of the orbit (i.e. the scheme-theoretic closed image of G×{x}→X) makes sense. I'm also using Chevalley's theorem to say the the image is open in its closure, which requires that the action is locally finitely presented. I suppose it's possible that there's a bizarre example where this fails.	https://mathoverflow.net/users/1	Is every subgroup of an algebraic group a stabilizer for some action?	In his book "Linear algebraic groups", 6.8, p98, Borel shows that the quotient of an affine algebraic group over a field by an algebraic subgroup exists as an algebraic variety, and he notes p.105 that Weil proved a similar result for arbitrary algebraic groups.	11	https://mathoverflow.net/users/930	8831	6,045
https://mathoverflow.net/questions/8840	4	You have $N$ boxes and $M$ balls. The $M$ balls are randomly distributed into the $N$ boxes. What is the expected number of empty boxes? I came up with this formula: $\sum\_{i=0}^{N}i\binom{N}{i}\left(\frac{N-i}{N}\right)^{M}$ This seems to yield the right answer. However, it requires calculating large numbers, such as $\binom{N}{\frac{N}{2}}$. Is there a more direct way, perhaps using a probability distribution? It seems that neither the binomial nor the hypergeometric distributions fit the problem.	https://mathoverflow.net/users/1646	Probability Question	Let $X\_i$ be a random variable with value 1 when box $i$ is empty and 0 otherwise. Now $P(X\_i=1)=(1-\frac{1}{N})^{M}$. And the expected number of empty boxes is just $\mathbb{E}(\sum X\_i)=N\mathbb{E}(X\_1)\approx \frac{N}{e^M}$ EDIT: gave the answer in terms of M,N instead of the numerical values given originally...	5	https://mathoverflow.net/users/2384	8843	6,053
https://mathoverflow.net/questions/4636	2	There is a short section in the book [Locally Compact Groups](http://books.google.com/books?id=3_BPupMDRr8C&printsec=frontcover&source=gbs_v2_summary_r&cad=0#v=onepage&q=&f=false) by Markus Stroppel (Chapter B7) on the notion of a "Hausdorff Solvable Group", which he defines as a topological group with a descending chain of closed normal subgroups, but it ends rather abruptly. I've tried searching around for the Hausdorff Dervied Series but I can not find a reference other than in this book. Perhaps this goes by a different name in other texts, but assuming not: If we equip the automorphism group of a field extension with the compact-open topology (by assumption of material present in the same book, chapter C9), then this shares separation properties of the field (9.2). What is the significance of an automorphism group being Hausdorff Solvable, or being solvable but not Hausdorff Solvable? Can we construct some examples?	https://mathoverflow.net/users/1521	Hausdorff Derived Series	The point of this section of Stroppel's book is to show, ultimately, that nothing new happens. Stroppel shows that each term in the Hausdorff derived series is nothing other than the closure of the same term in the usual derived series. A topological group is Hausdorff-solvable if and only if it is solvable, and the solvable height equals the Hausdorff-solvable height. In a sense, you can't construct interesting examples. :-) One thing that you can do is make an example of a topological group whose commutator subgroup isn't closed. I cheated with Google to find this, but here goes anyway. There exists a finite group $G\_n$ which is 2-step nilpotent and such that the commutator subgroup requires a product of $n$ commutators. Namely, take a central extension of a $2n$-dimensional vector space $V$ over an odd finite field by its exterior square $\Lambda^2 V$, such that the commutator of $a,b \in V$ is $a \wedge b \in \Lambda^2 V$. The point is that you need $k$ commutators to reach a tensor in $\Lambda^2 V$ of rank $k$. Now let $G$ be the product of all $G\_n$ in the product topology. The algebraic commutator subgroup of $G$ isn't closed, because it does not include elements in the closed commutator subgroup whose commutator length in $G\_n$ is unbounded as $n \to \infty$. Amazingly, this group $G$ is even compact. The implication for a Galois algebraic field extension, say, is as follows. The Galois group $G$ of such a field extension is a topological group, in fact a profinite group. You might have wondered if the algebraic field extension is "solvable" in the group-theoretic sense, but without leading to solvability by radicals. Happily, it doesn't happen, because what you should do is replace the solvable series of $G$ by the closed solvable series.	3	https://mathoverflow.net/users/1450	8848	6,057
https://mathoverflow.net/questions/8817	4	If L and M are two local systems on a space X, what can we say about the cohomology groups $H^i(X,L\otimes M)$ in terms of the cohomology of L and M? For example, can we determine their dimensions. You can assume X to be an affine connected smooth curve, if that helps. For the same question for coherent locally free sheaves, it seems that Riemann-Roch is helpful sometimes. But I don't know if there is a Riemann-Roch type theorem in l-adic cohomology?	https://mathoverflow.net/users/370	cohomology groups of tensor product of sheaves	If $X$ can just be a topological space, there are examples in which $H^i(X;L)$ and $H^i(X;M)$ vanish entirely, but $H^i(X,;L \otimes M)$ does not. For example, in cohomology with real coefficients, maybe $X = \mathbb{R}P^3$ and $L = M$ is the local system corresponding to the tautological line bundle, or if you like the non-trivial representation of $\pi\_1(\mathbb{R}P^3) = \mathbb{Z}/2$. Then $H^i(X,L)$ vanishes, but $H^i(X;L \otimes L)$ is the usual cohomology of $\mathbb{R}P^3$ and does not vanish in degrees $0$ and $3$. One thing that exists (again for cohomology in the setting of topological spaces) is a well-defined cup product map: $$\cup:H^i(X;L) \otimes H^j(X;M) \to H^{i+j}(X;L \otimes M).$$ This map is not a direct statement about dimensions of anything, but it is an important map in the theory of cohomology with local coefficients. For instance, if $L$ and $M$ are inverse line bundles and $X$ is a closed manifold, then this cup product is the right way to set up Poincare duality. (To review, they have to be line bundles with flat connections to give you local coefficients.)	14	https://mathoverflow.net/users/1450	8855	6,061
https://mathoverflow.net/questions/8853	7	In logic modules, theorems like Soundness and completeness of first order logic are proved. Later, Godel's incompleteness theorem is proved. May I ask what are assumed at the metalevel to prove such statements? It seems to me that whatever is assumed at the metalevel should not be more than whatever is being formulated at the symbolic level. I'm also asking about methodology. at the meta level, it seems like classical logic is used. So if proving statements about other kinds of logic like paraconsistent logic, then isn't there a discrepancy between what methodology is being formulated and what methodology is being used to prove the statement?	https://mathoverflow.net/users/nan	What assumptions and methodology do metaproofs of logic theorems use and employ?	It depends on what you're trying to prove, and for what purpose you are proving these metatheorems. So, the notion of "more" you're appealing to in asking about the metalevel is not completely well-defined. One common definition of the strength of a logic is "proof-theoretic strength", which basically amounts to the strongest induction principle the logic can justify (in set-theoretic terms, this is the same as identifying the largest set the logic's consistency proves well-ordered). The theory of ordinals classifies logic by this idea. This is natural for two reasons, both arising from Godel's incompleteness theorem. The incompleteness theorem tells us that we cannot prove the consistency of a stronger theory using a weaker one, so to prove consistency it's always the case that you need to assume a stronger logic in the metatheory than of the logic you're proving consistency of. More abstractly, this fact gives rise to a natural partial order on formal logical systems. However, consistency proofs are not the only thing people are interested in! Another purpose for giving semantics of logic is to help understand what people who use a different language than you do mean, in your own terms. For example, giving a classical semantics to intuitionistic logic gives traditional mathematicians a way of understanding what an intuitionist means, in their own terms. Likewise, semantics of classical logic in intuitionistic terms explains to intuitionists what classical logics mean. For this purpose, it doesn't matter how much mathematical machinery you bring to bear, as long as it brings insight. This is the end that something that ends up having big mathematical payoffs. It can illuminate all sorts of surprising facts. For example, Brouwer didn't just have strong opinions about logic, he also made assertions about geometry -- for instance, that the continuum was indivisible -- that are flat-out false, in naive classical terms. A priori, it's not clear what this has to do with the excluded middle. But it turns out that he wasn't just a crazy Dutchman; the logic of smooth analysis is intuitionistic, and using intuitionistic logic exactly manages piles of side-conditions that you'd have to manage by hand if you worked explicitly in its model. Conversely, studying classical logic in intuitionistic terms gives you a way of exploring the computational content of classical logic. Often, non-classical arguments (such as the use of double-negation elimination) amount to an appeal to the existence of a kind of backtracking search procedure, and sometimes you can show that an apparently-classical proof is constructive because this search is actually decidable. Martin Escardo's work on "exhaustible sets" is a delightful example of this, like a magician's trick. He shows that exhaustive search over some kinds of infinite sets is decidable (it's related to the fact that e.g. the Cantor space is compact).	8	https://mathoverflow.net/users/1610	8862	6,067
https://mathoverflow.net/questions/8856	4	For a new learner of several complex variables, the many domains (eg holomorphically convex, pseduconvex, Stein) and the many extension theorems (eg Riemann) and the many functions (plurisubharmonic) can be confusing. Which domains, extension theorems and functions do you think are the most important for a learner to get up to reach in reading papers, and the most important relationships between these domains, extension theorems and functions? As I am a new learner myself, if I question is too restrictive (eg other kinds of theorems are important too to understanding these domains) or too broad, please edit my question. Thank you in advance.	https://mathoverflow.net/users/nan	Most important domains, extension theorems, and functions in several complex variables	Here are a few points to guide you into the beautiful subject you had the good taste to choose. 1) Hartogs extension phenomenon :given two concentric balls in $ \mathbb C^n$, any holomorphic function $ B(0;M) \setminus B(0;m) \to \mathbb C$ extends to a holomorphic function $ B(0;M) \to \mathbb C$.This really launched the subject and showed that function theory in several variables is not just an extension of the theory in one variable.You should study a few such classes of examples. Key words: Hartogs figures, Reinhardt domains. 2) Domains for which such extensions do not exist are called holomorphy domains: balls are holomorphy domains but as we just saw "shells" $ B(0;M) \setminus B(0;m)$ are not. If a region is not a domain of holomorphy, it has a holomorphic hull, but this is no longer included in $\mathbb C^n$ : you get étalé spaces ( Yes, you algebraic geometers out there, this is where they were introduced ! ). This is an important subject and you can test whether a domain is a holomorphy domain at its boundary. Key-words: Levi problem, plurisubharmonic functions. 3)Holomorphic manifolds and Stein manifolds: these are abstractions of domains of $\mathbb C^n$ and holomorphic domains respectively . In retrospect they were inevitable because of the nature of holomorphy hulls (cf. 2). Stein manifolds are highly analogous to the affine varieties of algebraic geometers. 4) Sheaf theory, cohomology: these are all powerful techniques that you MUST master if you want to read anything at all in the subject. In particular you must understand coherent sheaves, which have a flavour of Noetherianness in them, but are a more subtle notion. The most important result here is Cartan's theorem B : coherent sheaves have no cohomology in positive dimension. To help you learn all this , I would recommend: B.Kaup, L. Kaup: Holomorphic Functions of several Variables (de Gruyter). [Quite friendly] H.Grauert R.Remmert: Theory of Stein Spaces (Springer). [The ultimate source by the Masters] All my wishes for success in your study of complex geometry.	12	https://mathoverflow.net/users/450	8864	6,068
https://mathoverflow.net/questions/8863	2	I am looking for suitable algorithm to compute the eigenvalues and eigenvectors of a matrix. My matrix is sparse ( think of Finite Element Matrix), and it is very, very big ( think of hundreds of thousands or even million degrees of freedom). The leading candidate for this task seems to be Lanczos algorithm. The issue now is, how well Lanczos algorithm fare if the matrix is sparse? The reason I ask this is because I want to know if there are a lot of zero terms in a matrix, will Lanczos take advantage of this by storing only nonzero terms and operate on them? Since my matrix is big, I want to conserve as much memory space as possible.	https://mathoverflow.net/users/807	The application of Lanczos Algorithm on sparse matrix	Lanczos is independent of the representation of your matrix. It does not store or operate on the entries of your matrix. The input to the algorithm is not the matrix $A$ itself, but a black-box subroutine for matrix-vector multiplication: you provide a method to compute $Av$ given vectors $v$. That's the only way it needs to use your matrix. In other words, you can represent $A$ however you want.	10	https://mathoverflow.net/users/302	8867	6,069
https://mathoverflow.net/questions/8870	7	More specifically, suppose I have a rational curve on a complete intersection, and I know that the relative Hilbert Scheme is not smooth at the point corresponding to my rational curve. Is there any algorithm that will eventually tell me whether the Hilbert Scheme is reduced or not there? Just to make it harder, this curve happens to pass through the singular locus of the complete intersection.	https://mathoverflow.net/users/2363	Is there a way to check if a relative Hilbert Scheme is reduced?	Showing nonreducedness of a Hilbert scheme is a hard question in general. The most direct algorithm would involve producing a Grobner basis for the defining ideal of the component in question, and then computing its radical and seeing if the two are equal. But with the exception of rather simple cases, this computation is not going to be feasible. Here are two shortcut answers that I've seen in the past. 1. You could show that the entire component of the Hilbert scheme in question is nonreduced. Mumford produces a famous such example, in his "Further Pathologies" paper from 1962. I have also seen works by J.O. Kleppe and some by Scott Nollet which provide many further examples and techniques. 2. You could produce a global description of component of the Hilbert scheme in question. For instance, perhaps the component is parametrized by something that you understand (a Grassmanian, say). The best example of this that I've ever read is Vainsencher-Avritzer "Compactifying the space of ellpitic quartic curves".	7	https://mathoverflow.net/users/4	8886	6,079
https://mathoverflow.net/questions/8887	10	I scoured Silverman's two books on arithmetic of elliptic curves to find an answer to the following question, and did not find an answer: Given an elliptic curve E defined over H, a number field, with complex multiplication by R, and P is a prime ideal in the maximal order of H and E has good reduction at P. Is it legitimate to reduce an endomorphism of E mod P? In the chapter "Complex Multiplication" of the advanced arithmetic topics book by Silverman, a few propositions and theorems mention reducing an endomorphism mod P. A priori, this doesn't seem trivial to me. Sure, the endomorphism is comprised of two polynomials with coefficients in H. But I still don't see why if a point Q is in the kernel of reduction mod P, why is phi(Q) also there. When I put Q inside the two polynomials, how can I be sure that P is still in the "denominator" of phi(Q)? (\*) I looked at the curves with CM by sqrt(-1), sqrt(-2) and sqrt(-3), and it seems convincing that one can reduce the CM action mod every prime, except maybe in the case of sqrt(-2) at the ramified prime.	https://mathoverflow.net/users/2024	Legitimacy of reducing mod p a complex multiplication action of an elliptic curve?	I'm not sure if there's a trivial way to see this. One answer is to use the fact that every rational map from a variety X / $\mathbb{Z}\_p$ to an abelian scheme is actually defined on all of X (see for instance Milne's abelian varieties notes). Here, since the generic fiber is open in X you can apply this by viewing the map you started with as a rational map.	8	https://mathoverflow.net/users/2	8891	6,081
https://mathoverflow.net/questions/8885	11	Suppose $E/ \mathbb{Q}$ is an elliptic curve whose Mordell-Weil group $E(\mathbb{Q})$ has rank r. When can we realize E as a fiber of an elliptic surface $S\to C$ fibered over some curve, with everything defined over $\mathbb{Q}$, such that the group of $\mathbb{Q}$-rational sections of $S$ has rank at least r? Edit: Let me also demand that the resulting family is not isotrivial, i.e. the j-invariants of the fibers are not all equal.	https://mathoverflow.net/users/1464	Building elliptic curves into a family	If you require $C = P^1$ then it's probably not possible except for very small values of $r$. If you don't care about $C$, then here is something that might work. Suppose $E$ is given by $y^2=x^3+ax+b$ and $P\_i=(x\_i,y\_i),i=1,\ldots,r$ is a basis for the Mordell-Weil group. Let $C$ be the curve given by the system of equations $u\_i^2 = (t^i+x\_i)^3 + a(t^i+x\_i) + b + t, i=1,\ldots,r$ in $t,u\_1,\ldots,u\_r$ and $S$ be the family $y^2 = x^3 + ax + b+t$ pulled back to $C$. So above $t=0$, $C$ has a point with $u\_i=y\_i$ and and the fiber of $S$ above this point is $E$. Also $C$ is defined so that there are sections of $S$ with $x$-coordinate $x=t^i+x\_i$ and I bet they are independent. Finally the family is non isotrivial if $a \ne 0$. If $a=0$ adjust the construction is an obvious way.	10	https://mathoverflow.net/users/2290	8894	6,083
https://mathoverflow.net/questions/8890	13	Can someone give me a non-trivial example of a flat SU(2)-connection over a compact orientable hyperbolic 3-manifold? The literature on such bundles over 3-manifolds is huge and my naive searches don't seem to turn up specific examples. 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. Finally, I would also be happy with just being told precisely where I should go and look in the literature!	https://mathoverflow.net/users/380	Flat SU(2) bundles over hyperbolic 3-manifolds	Many (compact orientable) hyperbolic 3-manifolds have non-trivial $SU(2)$ representations. By Mostow rigidity, the representation of the fundamental group $\Gamma$ of a closed hyperbolic 3-manifold into $SL(2,\mathbb{C})$ (lifted from $PSL(2,\mathbb{C})$) may be conjugated so that it lies in $SL(2,K)$, for $K$ a number field (because transcendental extensions have infinitesimal deformations in $\mathbb{C}$). In particular, the traces of elements will always lie in a number field. One may take different Galois embeddings of $K$ into $\mathbb{C}$, and get new representations of $\Gamma$ into $SL(2,\mathbb{C})$. Sometimes, this representation is just conjugate to the original (e.g. if $K$ was chosen too large), but in other cases the new representation of $\Gamma$ lies in $SL(2,\mathbb{R})$ or in $SU(2)$. A nice class of examples of this type are arithmetic hyperbolic 3-manifolds. In fact, they are characterized by the fact that all traces of elements are algebraic integers, and non-trivial Galois embeddings lie in $SU(2)$ (you have to be a bit careful about what this means). Some arithmetic manifolds will only have the complex conjugate representation this way (basically, if squares of the traces lie in a quadratic imaginary number field), but otherwise you get a non-trivial $SU(2)$ representation. The simplest example is the Weeks manifold, with trace field a cubic field. I suggest the book by [MacLachlan and Reid](http://www.ams.org/mathscinet-getitem?mr=1937957) as an introduction to arithmetic 3-manifolds. The description I've given though is encoded in terms of quaternion algebras and other algebraic machinery. Another characterization of arithmeticity is in [this paper](http://www.ams.org/mathscinet-getitem?mr=1433117). The nice thing about these representations is that they are faithful. There is a very explicit way to see these representations for hyperbolic reflection groups (studied by Vinberg in the arithmetic case). Basically, given a hyperbolic polyhedron with acute angles of the form $\pi/q$, sometimes one can form a spherical polyhedron with corresponding angles which are $p\pi/q$, and get a representation into $O(4)$. Passing to finite index manifold subgroups, one can obtain $SU(2)$ reps. (since $SO(4)$ is essentially $SU(2)\times SU(2)$). There are other ways one has $SU(2)$ representations, but they are less explicit. [Kronheimer and Mrowka](http://www.ams.org/mathscinet-getitem?mr=2106239) have shown that any non-trivial integral surgery on a knot has a non-abelian $SU(2)$ representation. Also, any hyperbolic 3-manifold with first betti number positive or a smooth taut orientable foliation has non-abelian $SU(2)$ representations. Addendum: Another observation relating $SU(2)$ representations to hyperbolic geometry is via the observation that the binary icosahedral group (a $\mathbb{Z}/2$ extension of $A\_5$) is a subgroup of $SU(2)$. By an [observation of Long and Reid](http://www.ams.org/mathscinet-getitem?mr=1459136), every hyperbolic 3-manifold group has infinitely many quotients of the form $PSL(2,p)$, $p$ prime. These groups always contain subgroups isomomorphic to $A\_5 < SO(3)$, so one may find a finite-sheeted cover which has a non-abelian $SO(3)$ and therefore $SU(2)$ representation. I have no idea though whether these representations are detected by the Casson invariant or Floer homology.	18	https://mathoverflow.net/users/1345	8897	6,086
https://mathoverflow.net/questions/8746	21	Gale [famously showed](http://www.cs.cmu.edu/afs/cs/academic/class/15859-f01/www/notes/brouwer-hex.pdf) that the determinacy of n-player, n-dimensional Hex is equivalent to the Brouwer fixed point theorem in n dimensions. We can (and Gale does) view this as saying that if you d-color the vertices of a certain graph specifically, the graph with vertex set $[n]^d$ and two vertices $v, w$ adjacent iff the max norm of $v - w$ is 1 and all the nonzero components of $v - w$ have the same sign -- then there's a certain monochromatic path. Alternatively, you can think of d-coloring a d-dimensional $n \times \ldots \times n$ cube, and the determinacy of Hex/Brouwer fixed-point says that a certain "twisted path" must exist. Here's what I want to know: > > Is there a topological proof of the density version of the determinacy of Hex? > > > The density version ends up following from density [Hales-Jewett](http://en.wikipedia.org/wiki/Hales%E2%80%93Jewett_theorem), since combinatorial lines are paths in the underlying graph. But density Hales-Jewett is hard, and this seems like it should admit a proof along the lines of Gale's. What I mean by the "density version" is: for any $\delta > 0$, and fixed n, for sufficiently large dimension d any choice of $\delta n^d$ moves must connect two opposite sides of the hypercube/d-dimensional Hex board. (I'm fairly sure this is the correct statement, but it's possible I'm wrong. Let me know if this is for some reason utterly trivial or false.)	https://mathoverflow.net/users/382	The density hex	For a closely related question when you do not insist that all non zero components of v-w has the same sign, then the answer is known: See the following paper: B. Bollobas, G. Kindler, I. Leader, and R. O'Donnell, Eliminating cycles in the discrete torus, LATIN 2006: Theoretical informatics, 202{210, Lecture Notes in Comput. Sci., 3887, Springer, Berlin, 2006. Also: Algorithmica 50 (2008), no. 4, 446-454. This Graph is referred to as G\_\inf and there is a beautiful new proof via the Brunn Minkowski's theorem by Alon and Feldheim. For this graph a rather strong form of a density result follows, and the results are completely sharp. The paper by Alon and Klartag <http://www.math.tau.ac.il/~nogaa/PDFS/torus3.pdf> is a good source and it also studies the case where we allow only a single non zero coordinate in v-u. An even sharper result is given [in another paper](http://www.math.tau.ac.il/~nogaa/PDFS/torusone1.pdf) by Noga Alon. There, there is a log n gap which can be problematic if we are interested in the case that n is fixed and d large. See also this [post](http://gilkalai.wordpress.com/2009/05/27/answer-to-test-your-intuition-3/). As Harrison points out, the graph he proposes (that we can call the Gale-Brown graph) is in-between the two graphs. So the unswer is not known but we can hope that some discrete isoperimetric methods can be helpful. The statement is an isoperimetric-type result so this can be regarded as a quantitative version of the topological notion of connectivity. Two more remarks: 1) The Gale result seems to give an example of a graph where there might be a large gap between coloring number and fractional coloring. This is rare and an important other example is the Kneser graph where analyzing its chromatic number is a famous use (of Lovasz) of a topological method. 2) Hex is closely related to planar percolation and the topological property based on planar duality is very important in the study of planar percolation and 1/2 being the critical probability. (See eg [this paper](http://uk.arxiv.org/PS_cache/math/pdf/0508/0508580v2.pdf)) It seems that we might have here an interesting high dimensional extension with some special significance to chosing each vertex with probability 1/d.	6	https://mathoverflow.net/users/1532	8900	6,089
https://mathoverflow.net/questions/8882	5	Is it possible for SOME positive $c$, $c<1$ to find a pair of COMPACT hyperbolic manfiolds $M^3$ and $N^3$ with a positive degree map $$f: M^3 \to N^3,$$ such that $f$ is contacting with constant $c$? Are there may examples like this? One can ask the same question of Riemann surfaces, and it seems to me that this should be possible. For example we can take a double cover of Riemann surface with many points or ramification. Though I don't know a proof even in this case. Of course for non-ramified cover the best possible constant $c$ is $1$. ADDED. Following the answer of Sam Need, let me give an approximative "proof" of the fact that this works in dimesnion 2. Let us triangulate a hyperbolic surface $N^2$ in triangles of very small size, that have acute angles (this is always possible). We want to show that a double cover of $N^2$ with ramifications at vertices of the triangulation will do the job. For this we need a lemma (without a proof). Lemma. Suppose we have two hyperbolic trianlges, one very small and acute with angles $a$, $b$, $c$, and the over with angles a/2, b/2, c/2. Then there is a contacting map from the second triangle to the first one. The lemma is true, since the second trianlge will be large. Now on the double cover we can take a trangulation that comes from $N^2$ and glue it from these triangles with half angles. Half angles come from doble cover. Then we just need to "adjust" the map. Of course this is not a real proof, but I am 100% it can be made real.	https://mathoverflow.net/users/943	Contracting maps of hyperbolic manifolds	In general, for any non-zero degree map from one closed negatively curved manifold to another, there is a canonical map (due to Besson-Courtois-Gallot) called the "natural map". However, it's only known to be pointwise volume decreasing, not necessarily contracting. They call this the "real Schwarz-Lemma". Applying the Schwarz lemma for Riemann surfaces I think gives the contracting map in this case for branched covers. Think of the induced map on the universal cover, which is the unit disk, or $\mathbb{H}^2$. The Schwarz lemma says that any conformal map from the disk to the disk is contracting, unless it's an isometry. I thought of one (not very explicit) example in 3-D. Take two simplices in hyperbolic space. There is a canonical affine map (say in the Lorentzian model) taking one simplex to the other. This will be a contracting map for the hyperbolic metric if one simplex sits inside the other [Edit: actually I'm not sure about this now, but in the example below there exists a contracting map]. There are finitely many tetrahedra in $\mathbb{H}^3$ which give rise to fundamental domains for discrete reflection groups (see [Ratcliffe](http://books.google.com/books?id=hwQmbllvFMUC&lpg=PP1&dq=ratcliffe%2520hyperbolic&pg=PA296#v=onepage&q=&f=false)). Two of these have one dihedral angle $\pi/5$, with opposite edge angle $\pi/2$ and $\pi/4$, respectively, and all other angles the same. There is a 1-parameter family of polyhedra interpolating between these (basically, just "push" the two faces closer together along the dihedral angle $\pi/5$ edge) which decreases distances. Also, the orbifold fundamental group (i.e. reflection group) from the $\pi/4$ one maps to that of the $\pi/2$ one. So there's a distance decreasing map from one orbifold to the other. Using Selberg's lemma, one may find finite-sheeted manifold covers with the same property.	5	https://mathoverflow.net/users/1345	8904	6,091
https://mathoverflow.net/questions/8912	21	I've been driven up a wall by the following question: let p be a complex polynomial of degree d. Suppose that \|p(z)\|≤1 for all z such that \|z\|=1 and \|z-1\|≥δ (for some small δ>0). Then what's the best upper bound one can prove on \|p(1)\|? (I only care about the asymptotic dependence on d and δ, not the constants.) For the analogous question where p is a degree-d real polynomial such that \|p(x)\|≤1 for all 0≤x≤1-δ, I know that the right upper bound on \|p(1)\| is \|p(1)\|≤exp(d√δ). The extremal example here is p(x)=Td((1+δ)x), where Td is the dth Chebyshev polynomial. Indeed, by using the Chebyshev polynomial, it's not hard to construct a polynomial p in z as well as its complex conjugate z\*, such that (i) \|p(z)\|≤1 for all z such that \|z\|=1 and \|z-1\|≥δ, and (ii) p(1) ~ exp(dδ). One can also show that this is optimal, for polynomials in both z and its complex conjugate. The question is whether one can get a better upper bound on \|p(1)\| by exploiting the fact that p is really a polynomial in z. The fastest-growing example I could find has the form p(z)=Cd,δ(1+z)d. Here, if we choose the constant Cd,δ so that \|p(z)\|≤1 whenever \|z\|=1 and \|z-1\|≥δ, we find that p(1) ~ exp(dδ2) For my application, the difference between exp(dδ) and exp(dδ2) is all the difference in the world! I searched about 6 approximation theory books---and as often the case, they answer every conceivable question except the one I want. If anyone versed in approximation theory can give me a pointer, I'd be incredibly grateful. Thanks so much! --Scott Aaronson PS. The question is answered below by David Speyer. For anyone who wants to see explicitly the polynomial implied by David's argument, here it is: pd,δ(z) = zd Td((z+z-1)(1+δ)/2+δ), where Td is the dth Chebyshev polynomial.	https://mathoverflow.net/users/2575	Analogue of the Chebyshev polynomials over C?	I may be missing something obvious here. Let $f(z, z^{\})$ be the polynomial in $z$ and $z^{\}$ of degree $d$ which achieves $\exp(d \delta)$. Let $g(z)$ be the Laurent polynomial obtained from $f$ by replacing $z^{\*}$ by $z^{-1}$. On the unit circle, we have $f=g$. Now, let $h$ be the polynomial $z^d g$. This is an honest polynomial, because we multiplied by a high enough power of $z$ to clear out all the denominators and, for $z$ on the unit circle, we have $\|h\|=\|f\|$. Doesn't this mean that $h$ is a polynomial of degree $2d$, obeying your conditions, with $\|h(1)\| \sim \exp(d \delta)$?	31	https://mathoverflow.net/users/297	8915	6,097
https://mathoverflow.net/questions/8871	5	Hi, I see that the tetrad postulate: $\nabla\_{\mu}e\_{\nu}^{I}=\partial\_{\mu}e\_{\nu}^{I}-\Gamma\_{\mu\nu}^{\rho}e\_{\rho}^{I}+\omega\_{\mu J}^{I}e\_{\nu}^{J}=0$ Can be merely derived from writing a tensor in two different basis (pure natural-coordinates $\{\partial\_\mu\}$ and mixed $\{\partial\_\mu\} + \{e\_a\}$), my questions are: 1. Does this imply the metricity of the connexion or the inverse? 2. If it is the inverse, how? 3. In writing $g\_{\mu\nu}=\eta\_{IJ}e\_{\mu}^{I}e\_{\nu}^{J}$ do we need to impose a metricity on $\eta$? ps: metricity = metric compatible	https://mathoverflow.net/users/2566	Tetrad postulate: Implies or results from the metricity of the connection?	Having botched the first attempt at answering this question and not wanting to delete the evidence, let me try again here. The "tetrad postulate" is independent from metricity and from the condition that the connection be torsion-free. It is simply the equivalence (via the vielbein) of two connections on two different bundles. Here are the details. $M$ is a smooth $n$-dimensional manifold. First of all we have an affine connection $\nabla$ on $TM$ with connection coefficients $\Gamma^\rho\_{\mu\nu}$ relative to a coordinate basis -- that is, $$\nabla\_{\partial\_\mu} \partial\_\nu = \Gamma\_{\mu\nu}^\rho \partial\_\rho,$$ with $\partial\_\mu$ an abbreviation for $\partial/\partial x^\mu$ where $x^\mu$ is a local chart on $M$. Then we have a connection on an associated vector bundle to the frame bundle $P\_{\mathrm{GL}}(M)$. The frame bundle is a principal $\mathrm{GL}(n)$-bundle and given any representation $\rho: \mathrm{GL}(n) \to \mathrm{GL}(V)$ of $\mathrm{GL}(n)$ we can define a vector bundle $$P\_{\mathrm{GL}}(M) \times\_\rho V.$$ Take $V$ to be the defining $n$-dimensional representation and call the resulting bundle $E$. Relative to a local frame $e\_a$ for $E$, a connection $\hat\nabla$ defines connection one-form $\omega$ by $$\hat\nabla\_{\partial\_\mu} e\_a = \omega\_{\mu~a}^b e\_b.$$ Now the vielbein defines a bundle isomorphism $TM \buildrel\cong\over\longrightarrow E$ and all the "tetrad postulate" says is that the two connections $\nabla$ and $\hat\nabla$ correspond. In fact, the "tetrad postulate" is just the statement that the vielbein is a parallel section of the bundle $T^\M \otimes E$ relative to the tensor product connection. This works for any* affine connection $\nabla$ on any smooth manifold $M$. No metric is involved. A special case of this construction is when $(M,g)$ is a riemannian manifold and $\nabla$ is the Levi-Civita connection (i.e., the unique torsion-free, metric connection on $TM$). You can without loss of generality restrict to orthonormal frames, which defines a principal $\mathrm{O}(n)$ (or $\mathrm{O}(p,q)$ depending on signature) bundle. The representation $V$ restricts to an irreducible rep of the orthogonal group, possessing an invariant bilinear form $\eta$. This relates $g$ and $\eta$ as in your question.	0	https://mathoverflow.net/users/394	8927	6,105
https://mathoverflow.net/questions/8918	37	Suppose G is a group object in the category of algebraic spaces (over a field, if you like, or even over ℂ if you really want). Is G necessarily a scheme? My feeling is that the answer is "yes" because an algebraic space group which is not a scheme would be too awesome. Any group homomorphism from such a G to an algebraic group (a scheme group) would have to have infinite kernel since [an algebraic space which is quasi-finite over a scheme is itself a scheme](https://mathoverflow.net/questions/4573/when-is-an-algebraic-space-a-scheme/4574#4574). In particular, G would have no faithful representations or [faithful actions on projective varieties (probably)](https://mathoverflow.net/questions/8812/why-do-automorphism-groups-of-algebraic-varieties-have-natural-algebraic-group-st/8820#8820). There can't be a surjective group homomorphism from an algebraic group H to G, since that would identify G with H/K, [which is a scheme](https://mathoverflow.net/questions/8809/is-every-subgroup-of-an-algebraic-group-a-stabilizer-for-some-action). In particular, you cannot put a group structure on any étale cover of G.	https://mathoverflow.net/users/1	Is an algebraic space group always a scheme?	I think the answer is yes. If S is a noetherian scheme and G is a relative algebraic group space over S, then there is a stratification of S such that over each stratum, G is a group scheme (see K. Behrend, Derived \ell-adic categories for Algebraic Stacks, 5.1.1). When S is Spec k there is no non-trivial stratification.	15	https://mathoverflow.net/users/370	8929	6,107
https://mathoverflow.net/questions/8926	7	With regard to [my original question](https://mathoverflow.net/questions/7617/probability-question-closed): > > A subset of k vertices is chosen from the vertices of a regular N-gon. What is the probability that two vertices are adjacent? > > > I suppose that the responses that were elicited to my question were to be expected. You see, I have been in your position, and thought what you did, on many occasions. But mostly not in the area of mathematics. By way of providing background, I am not a student at all. In fact, I am a biochemist and part-time university instructor. I have often provided answers to students who post chemistry/biochemistry questions (samples will be provided upon request). And, like you, I hope that I have not (or had not!) become a vehicle for students to avoid thinking through THEIR chemistry homework assignments. The above question, believe it or not, comes from the local, Boston-based television program "extrahelp", hosted by a somewhat sarcastic character who went by the name "Mr. Math". It was intended for the K-12 demographic, but, according to the story behind the video, an M.I.T. student was listening, and asked the question, ostensibly to give this man his comeuppance. The video is no longer available online, but I can send you a copy, if you don't mind that it is 25.1 MB, and that I don't have access to any FTP server at the university where I teach. It is hilarious. When I saw this video (and after I stopped laughing), I was interested by the question itself. And since at least one of you asked for any work that I have done on my own, here is as far as I got before I made my original post: 1) N must, of course, be a positive integer >= 3. k must be a positive integer <= N. 2) When k = 1, the solution is trivial (p = 0). In fact, the non-trivial values of k are: 2 <= k <= (N \ 2), where "\" is integer division. For other values of k, p = 1. 3) For non-trivial values of k, the denominator is C(N,k). 4) For k = 2, the numerator is N. 5) For k = 3, the numerator is N(N-k). 6) For N even, and k = N / 2, the numerator is N – 2. For N odd, and k = N \ 2, the numerator is C(N,k) – N. 7) For k = (N \ 2) -1, the numerator may be C(N,k) – N(N-k). Where I am having trouble is, obviously, getting from here to a general solution. It has been suggested to me that I take the approach of finding the general expression for the probability of NOT selecting adjacent vertices, but the general answer of p = 1 - [C(N-k,k) / C(N,k)] is not confirmed by the examples that I am sure about (that is, where N is so small that the answers can be confirmed by enumeration) I am sure that you would agree that, by now, enough time has passed that, if this really were a homework question, that the due date for such homework would have passed. Further, I hope that I have convinced you beyond a reasonable suspicion that a) this is definitely not a homework question, and b) that I have worked on the problem myself to such a degree that I haven't just passed this off on the respondents without making some effort.	https://mathoverflow.net/users/2582	Probability vertices are adjacent in a polygon	It's not so hard to calculate the probability that no two points are adjacent: We may as well assume that the first vertex is chosen for us. So let's ignore it and unroll the rest of the polygon into a line. Now imagine that I write down a sequence of $0$s and $1$s along this line to indicate whether the corresponding vertices are chosen or not. The condition on this sequence is that: the first and last numbers in this sequence must not be $1$s, and any $1$ must be followed by a $0$. So, let's replace any subsequence of the form $01$ with an $x$ and ignore the last $0$ in the sequence, so for instance we'd have the transformation $01001010 \rightarrow x0xx$. Now there is no condition on the sequence of $0$s and $x$s, other than the length and number of $x$s. The length of the new sequence will be $(n-1)-(k-1)-1 = n-k-1$, and will contain $k-1$ $x$s, so the number of such sequences is just $\binom{n-k-1}{k-1}$, while the number of all sequences of $0$s and $1$s (including ones violating our condition) is $\binom{n-1}{k-1}$. So the probability you want is just $1-\frac{\binom{n-k-1}{k-1}}{\binom{n-1}{k-1}}=1-\frac{(n-k-1)!(n-k)!}{(n-1)!(n-2k)!}$. Just to be sure, let's check that with $n = 6, k = 3$. Our formula gives us a probability of $1-\frac{2!3!}{5!0!} = 1-\frac{12}{120} = \frac{9}{10}$. We can check by hand that there are two ways for the vertices to not all be adjacent - the two triangles in the hexagon, and there are a total of $\binom{6}{3} = 20$ ways to choose three vertices, and this gives us a probability of $1-\frac{2}{20} = \frac{9}{10}$. Hooray! No off by one errors!	5	https://mathoverflow.net/users/2363	8934	6,109
https://mathoverflow.net/questions/8940	2	f:X-->Y is flat and projective map between integral varieties over k, an algebraically closed field. Suppose every fiber at closed points of Y is still an integral variety. L is a line bundle on X, if f\_\*(L) is trivial line bundle on Y, and for every closed point y of Y, L\_y is also a trivial line bundle on fiber X\_y, can we deduce that L is also a trivial line bundle on X?	https://mathoverflow.net/users/2008	Is line bundle determined by the parameter space and fiber?	Yes, you can. This follows from the see-saw principle for instance. You can also argue directly as follows. The spaces of global sections $H^{0}(Y,f\_\L)$ and $H^{0}(X,L)$ are naturally isomorphic. Since $f\_\L$ is a trivial line bundle we can choose a global nowhere vanishing section $e$ of $f\_\*L$. Let $s \in H^{0}(X,L)$ be the section of $L$ corresponding to $e$ under the above isomorphism. To show that $L$ is trivial it suffices to check that $s$ does not vanish anywhere. But if $x \in X$ is a closed point where $s$ vanishes, then if we restrict $s$ to the fiber $X\_{f(x)}$, we will get a section of the structure sheaf of integral projective variety which vanishes at a point. Thus the restriction of $s$ to $X\_{f(x)}$ must be identically zero. This shows that $e$ vanishes at $f(x)$ which is a contradiction.	6	https://mathoverflow.net/users/439	8947	6,116
https://mathoverflow.net/questions/8907	4	I am doing some research on the Spearman Rank Correlation Coefficient; all the references I can find refer essentially to a sample statistic. That is, given a sample of the jointly distributed $(x\_i,y\_i)$, one can compute the Spearman Coefficient between $x$ and $y$; I am wondering if there is a population equivalent. My guess would be that it is defined as $$E[sign((x\_i - x\_j)(y\_i - y\_j))],$$ where $(x\_i,y\_i)$ and $(x\_j,y\_j)$ are i.i.d draws from the joint distribution. My question: 1. is there a widely accepted definition of the population Spearman? (references?) 2. does it match my intuition? 3. is the sample Spearman an unbiased estimator of the population Spearman? thanks,	https://mathoverflow.net/users/2570	Population Spearman Rank Correlation Coefficient	Let p(x,y) be the joint probability density function of the random variables X and Y. Let P\_x(x) and P\_y(y) the marginial cumulative distribution functions respectively. The key observation is that the normalized rank of a sample of x (i.e., its rank divided by the number of observations R(x\_i)/n) is just a sample of the random variable P\_x(X). Thus, it is not hard to convince oneself that the statistic: Rho = 1-6(P\_x(X)-P\_y(Y))^2 is an estimator of the Spearman rank correlation, and its population mean is the population's Spearman rank coefficient is given by: rho = 1 - 6 int ((P\_x(x)-P\_y(y))^2 p(x,y) dxdy) The following article performs the same calculation for a weighted version of the Spearman's correlation coefficient: <http://www.ine.pt/revstat/pdf/rs060301.pdf> I think that the sample Spearman is unbiased because of the averaging by n\(n-1)\(n+1), but I still don't know how to prove that. Please, notice that the population mean of the statistic (the population Spearman correlation coefficient) becomes zero when the random variables are independent, i.e., p(x,y) = p(x)\*p(y).	1	https://mathoverflow.net/users/1059	8955	6,122
https://mathoverflow.net/questions/8714	5	It's well-known that, for lots of concrete categories (but [by no means all](https://mathoverflow.net/questions/1166/can-the-objects-of-every-concrete-category-themselves-be-realized-as-small-catego)), we can think of the objects as themselves being small categories, and morphisms are the functors between these categories. Examples include Grp, Ab, Top... When we apply such a construction, we turn a 1-category into a (strict, I think?) 2-category. But 2-categories carry some extra structure, namely the notion of a natural transformation. When we "decategorify" back down, where does this extra structure go? I can work it out in some specific cases; for instance, if we categorify Ab in the obvious way, there are no nontrivial natural transformations. I don't have a characterization for when two morphisms of groups are naturally isomorphic as functors between the underlying categories, although I have a feel for how the question behaves. Are there any sort of general results on what natural transformations between morphisms look like if we categorify thusly? Is it at least independent of how we realize the objects as small categories? (I suspect the answer to the second question is "no," but don't have the skills to construct a counterexample. I hope I'm wrong, though.) More generally, if we can categorify an n-category into an (n+k)-category forgetting the higher morphisms, do the higher morphisms go downstairs to the n-category in any nice way?	https://mathoverflow.net/users/382	What are natural transformations in 1-categories?	Here is a counterexample for your next-to-last question. Let S be a set with more than one element and consider the two full subcategories of Cat on, respectively, the single category which is the discrete category on S, and the single category which is the codiscrete category on S. In each case, when viewing Cat as a 1-category, the resulting full subcategory has a single object with endomorphisms Hom(S, S). However, if we view Cat as a 2-category, the former subcategory has no nontrivial natural transformations and thus really is BHom(S, S), while the latter has a unique natural transformation between any two functors and thus is actually • up to 2-equivalence. Cat-the-1-category and Cat-the-2-category are very different constructs which unfortunately usually go by the same name. Even though they have "the same" objects, I suggest thinking of their objects as being different kinds of things. An object of Cat-the-1-category has more information than an object of Cat-the-2-category; we may talk about the cardinality of its set of objects, not just the cardinality of its set of isomorphism classes of objects. (This shouldn't seem too strange, since an object of Cat-the-0-category is a "specific" category, of which we may talk about the actual set of objects.) Put differently, an object of Cat-the-1-category is a "monoid with many objects", while an object of Cat-the-2-category is what we more often think of when thinking about categories (especially large ones). In your example, you expressed Ab as a full subcategory of Cat-the-1-category. The full subcategory of Cat-the-2-category on the same objects is not Ab, since it has nontrivial natural automorphisms, as others have pointed out. It only becomes Ab after truncation—replacing each Hom-category by its set of isomorphism classes of objects. For Grp, the situation is worse, since distinct group homomorphisms may be naturally isomorphic as functors. The usual way to repair this is to work with "pointed categories", as described at [this nlab page](http://ncatlab.org/nlab/show/k-tuply+monoidal+n-category#__9). But of course this is a kind of extra structure on a category, and if I'm allowed to introduce arbitrary extra structure then the question is too easy. Anyways, I'm not sure that one should expect various concrete categories to naturally be full subcategories of either Cat-the-1-category or Cat-the-2-category.	5	https://mathoverflow.net/users/126667	8959	6,124
https://mathoverflow.net/questions/8957	144	Several times I've heard the claim that any Lie group $G$ has trivial second fundamental group $\pi\_2(G)$, but I have never actually come across a proof of this fact. Is there a nice argument, perhaps like a more clever version of the proof that $\pi\_1(G)$ must be abelian?	https://mathoverflow.net/users/2510	Homotopy groups of Lie groups	I don't know of anything as bare hands as the proof that $\pi\_1(G)$ must be abelian, but here's a sketch proof I know (which can be found in Milnor's Morse Theory book. Plus, as an added bonus, one learns that $\pi\_3(G)$ has no torsion!): First, (big theorem): Every (connected) Lie group deformation retracts onto its maximal compact subgroup (which is, I believe, unique up to conjugacy). Hence, we may as well focus on compact Lie groups. Let $PG = \{ f:[0,1]\rightarrow G \| f(0) = e\}$ (I'm assuming everything is continuous.). Note that $PG$ is contractible (the picture is that of sucking spaghetti into one's mouth). The projection map $\pi:PG\rightarrow G$ given by $\pi(f) = f(1)$ has homotopy inverse $\Omega G = $Loop space of G = $\{f\in PG \| f(1) = e \}$. Thus, one gets a fibration $\Omega G\rightarrow PG\rightarrow G$ with $PG$ contractible. From the long exact sequence of homotopy groups associated to a fibration, it follows that $\pi\_k(G) = \pi\_{k-1}\Omega G$ Hence, we need only show that $\pi\_{1}(\Omega G)$ is trivial. This is where the Morse theory comes in. Equip $G$ with a biinvariant metric (which exists since $G$ is compact). Then, following Milnor, we can approximate the space $\Omega G$ by a nice (open) subset $S$ of $G\times\cdots \times G$ by approximating paths by broken geodesics. Short enough geodesics are uniquely defined by their end points, so the ends points of the broken geodesics correspond to the points in $S$. It is a fact that computing low (all?...I forget)\* $\pi\_k(\Omega G)$ is the same as computing those of $S$. Now, consider the energy functional $E$ on $S$ defined by integrating $\|\gamma\|^2$ along the entire curve $\gamma$. This is a Morse function and the critical points are precisely the geodesics\\. The index of E at a geodesic $\gamma$ is, by the Morse Index Lemma, the same as the index of $\gamma$ as a geodesic in $G$. Now, the kicker is that geodesics on a Lie group are very easy to work with - it's pretty straight forward to show that the conjugate points of any geodesic have even index. But this implies that the index at all critical points is even. And now THIS implies that $S$ has the homotopy type of a CW complex with only even cells involved. It follows immediately that $\pi\_1(S) = 0$ and that $H\_2(S)$ is free ($H\_2(S) = \mathbb{Z}^t$ for some $t$). Quoting the Hurewicz theorem, this implies $\pi\_2(S)$ is $\mathbb{Z}^t$. By the above comments, this gives us both $\pi\_1(\Omega G) = 0$ and $\pi\_2(\Omega G) = \mathbb{Z}^t$, from which it follows that $\pi\_2(G) = 0$ and $\pi\_3(G) = \mathbb{Z}^t$. Incidentally, the number $t$ can be computed as follows. The universal cover $\tilde{G}$ of $G$ is a Lie group in a natural way. It is isomorphic (as a manifold) to a product $H\times \mathbb{R}^n$ where $H$ is a compact simply connected group. H splits isomorphically as a product into pieces (all of which have been classified). The number of such pieces is $t$. (edits) \- it's only the low ones, not "all", but one can take better and better approximations to get as many "low" k as one wishes. \\*- I mean CLOSED geodesics here	144	https://mathoverflow.net/users/1708	8961	6,126
https://mathoverflow.net/questions/8935	1	When we have a variety and a resolution of singularities, but it is not semi-small (i.e. the dimensions of the fibres do not satisfy the right conditions), then what can we say about the intersection cohomology? Someone was telling me something about this with shifting the IC or something, but I cannot remember the precise statement.	https://mathoverflow.net/users/2623	intersection cohomology when the resolution is not semi-small	I'm not sure exactly what question you're asking. I think you may be looking for the following answer. By the decomposition theorem, the intersection cohomology of the variety is a direct summand of the cohomology of the resolution. I'm not sure there's anything more specific you can say than that in general.	1	https://mathoverflow.net/users/916	8968	6,130
https://mathoverflow.net/questions/8972	38	I have been thinking about which kind of wild non-measurable functions you can define. This led me to the question: Is it possible to prove in ZFC, that if a (Edit: measurabel) set $A\subset \mathbb{R}$ has positive Lebesgue-measure, it has the same cardinality as $\mathbb{R}$? It is obvious if you assume CH, but can you prove it without CH?	https://mathoverflow.net/users/2097	Do sets with positive Lebesgue measure have same cardinality as R?	I found the answer in the paper ["Measure and cardinality" by Briggs and Schaffter](http://www.jstor.org/pss/2320153). In short: not if I interpret positive measure to mean positive outer measure. A proof is given that every measurable subset with cardinality less than that of $\mathbb{R}$ has Lebesgue measure zero. However, they then survey results of Solovay that show that there are models of ZFC in which CH fails and every subset of cardinality less than that of $\mathbb{R}$ is measurable, and that there are models of ZFC in which CH fails and there are subsets of cardinality $\aleph\_1$ that are nonmeasurable. So it is undecidable in ZFC. If it was intended that our sets are assumed to be measurable, then the answer would be yes by the first part above. Edit: In light of the comment by Konrad I added a couple of lines to clarify.	42	https://mathoverflow.net/users/1119	8975	6,134
https://mathoverflow.net/questions/8970	41	I've heard that the problem of counting topologies is hard, but I couldn't really find anything about it on the rest of the internet. Has this problem been solved? If not, is there some feature that makes it pretty much intractable?	https://mathoverflow.net/users/1353	Number of valid topologies on a finite set of n elements	It's wiiiiiiide open to compute it exactly. As far as I know the "feature that makes it intractable" is that there's no real feature that makes it tractable. Very broadly speaking, if you want to count the ways that a generic type of structure can be put on an n-element set, there's no efficient way to do this -- you basically have to enumerate the structures one by one. This is essentially because "given a description of a structure type and $n$, count the number of structures on an n-element set" is a ridiculously broad problem which ends up reducing to lots and lots of different counting and decision problems. Alternatively I think you can argue via Kolmogorov complexity and all that, but that's not my style? So in any case, the burden of proof is on the person who claims that an efficient counting algorithm should exist. (If you believe some crazy things about complexity theory, like P = PSPACE, this starts to become less true since the structure types hard to enumerate will usually be hard to describe. But if you believe that you're a lost cause in any case :P) It's still reasonable to ask for further justification, though. I'd attempt to give some, but I've been awake for like 30 hours and it would be even more handwavey than the above. The short version: If you do enough enumerative combinatorics, you start to see that nice formulas for enumerating structures arise from one of a few situations. Some really big ones are: 1. The ability to derive a sufficiently nice recurrence relation. This is a rather nebulous property, and some surprising structure types have cute recurrence relations. I can't really tell you a good solid reason why the number of finite topologies doesn't admit a nice recurrence relation; if you work with it a while, it just doesn't feel like it does. 2. A classification theorem such that the structures in each class have nice formulas. Sometimes these are deep, sometimes they aren't. One non-deep example is in the usual (non-linear-algebra) proof of Cayley's formula on the number of trees. Point-set topology is weird, and it's pretty weird even in the finite case. This is out of the question. 3. If the set of structures on a set of n elements is very rigid, there may be an "algebraic" way of counting them. Again, point-set topology is too weird for this to kick in. So my answer boils down to: It's intractable 'cause it is. Not a particularly satisfying reason, but sometimes that's the way combinatorics works. Sorry if you read that whole post -- I meant for it to be shorter and have more content, but it ended up like most tales told by idiots. But hopefully you learned something, or at least had fun with it?	38	https://mathoverflow.net/users/382	8980	6,138
https://mathoverflow.net/questions/8983	12	Given a two-dimensional cubic Bézier spline defined by 4 control-points as described in [the Wikipedia entry](https://en.wikipedia.org/wiki/B%C3%A9zier_curve#Cubic_B.C3.A9zier_curves), is there a way to solve analytically for the parameter along the curve (ranging from 0 to 1) which yields the point closest to an arbitrary point in space? $$ \mathbf{B}(t) = (1-t)^3 \,\mathbf{P}\_0 + 3(1-t)^2 t\,\mathbf{P}\_1 + 3(1-t) t^2\,\mathbf{P}\_2 + t^3\,\mathbf{P}\_3, ~~~~~ t \in [0,1] $$ where $\mathbf P\_0$, $\mathbf P\_1$, $\mathbf P\_2$, and $\mathbf P\_3$ are the four control-points of the curve. I can solve it pretty reliably and quickly with a divide-and-conquer algorithm, but it makes me feel dirty…	https://mathoverflow.net/users/2201	Closest point on Bézier spline	If you have a Bézier curve $(x(t),y(t))$, the closest point to the origin (say) is given by the minimum of $f(t) = x(t)^2 + y(t)^2$. By calculus, this minimum is either at the endpoints or when the derivative vanishes, $f'(t) = 0$. This latter condition is evidently a quintic polynomial. Now, there is no exact formula in radicals for solving the quintic. However, there is a really nifty new iterative algorithm based on the symmetry group of the icosahedron due to Doyle and McMullen ([Solving the quintic by integration](https://math.dartmouth.edu/%7Edoyle/docs/icos/icos.pdf)). They make the point that you use a dynamical iteration anyway to find radicals via Newton's method; if you think of a quintic equation as a generalized radical, then it has an iteration that it just as robust numerically as finding radicals with Newton's method. Contrary to what lhf [said](https://mathoverflow.net/questions/8983/closest-point-on-bezier-spline#comment12433_8983), Cardano's formula for the cubic polynomial is perfectly stable numerically. You just need arithmetic with complex numbers even if, indeed exactly when, all three roots are real. There is also a more ordinary approach to finding real roots of a quintic polynomial. (Like Cardano's formula, the Doyle–McMullen solution requires complex numbers and finds the complex roots equally easily.) Namely, you can use a cutoff procedure to switch from divide-and-conquer to Newton's method. For example, if your quintic $q(x)$ on a unit interval $[0,1]$ is $40-100x+x^5$, then it is clearly close enough to linear that Newton's method will work; you don't need divide-and-conquer. So if you have cut down the solution space to any interval, you can change the interval to $[0,1]$ (or maybe better $[-1,1]$), and then in the new variable decide whether the norms of the coefficients guarantee that Newton's method will converge. This method should only make you feel "a little dirty", because for general high-degree polynomials it's a competitive numerical algorithm. (Higher than quintic, maybe; Doyle–McMullen is really pretty good.) See also the related MO question [Can Gröbner bases be used to compute solutions to large, real-world problems?](https://mathoverflow.net/questions/5760/can-grobner-bases-be-used-to-compute-solutions-to-large-real-world-problems) on the multivariate situation, which you would encounter for bicubic patches in 3D. The multivariate situation is pretty much the same: You have a choice between polynomial algebra and divide-and-conquer plus Newton's method. The higher the dimension, the more justification there is for the latter over the former.	14	https://mathoverflow.net/users/1450	8997	6,151
https://mathoverflow.net/questions/8999	15	Given a graph $G$ we will call a function $f:V(G)\to \mathbb{R}$ discrete harmonic if for all $v\in V(G)$ , the value of $f(v)$ is equal to the average of the values of $f$ at all the neighbors of $v$. This is equivalent to saying the discrete Laplacian vanishes. Discrete harmonic functions are sometimes used to approximate harmonic functions and most of the time they have similar properties. For the plane we have Liouville's theorem which says that a bounded harmonic function has to be constant. If we take a discrete harmonic function on $\mathbb{Z}^2$ it satisfies the same property (either constant or unbounded). Now my question is: If we take a planar graph $G$ so that every point in the plane is contained in an edge of $G$ or is inside a face of $G$ that has less than $n\in \mathbb{N}$ edges, does a discrete harmonic function necessarily have to be either constant or unbounded? I know the answer is positive if $G$ is $\mathbb{Z}^2$, the hexagonal lattice and triangular lattice, I suspect the answer to my question is positive, but I have no idea how to prove it. Edited the condition of the graph to "contain enough cycles". (So trees are ruled out for example)	https://mathoverflow.net/users/2384	Discrete harmonic function on a planar graph	The answer is no. I first describe the graph $G$. Let $N\_i$ be a sequence of positive integers; we will choose $N\_i$ later. Let $T$ be an infinite tree which has one root vertex, the root has $N\_1$ children; the children of that root have $N\_2$ children, those children have $N\_3$ children and so forth. Let $V\_0$ be the set containing the root, $V\_1$ be the set of children of the root, $V\_2$ the children of the elements of $V\_1$, and so forth. To form our graph, take $T$ and add a sequence of cycles, one going through the vertices of $V\_1$, one through $V\_2$ and so forth. (In the way which is compatible with the obvious planar embedding of $T$.) Every face of $G$ is either a triangle or a quadrilateral. We will build a harmonic function $f$ on $G$ as follows: On the root, $f$ will be $0$. On $V\_1$, we choose $f$ to be nonzero, but average to $0$. On $V\_i$, for $i \geq 2$, we compute $f$ inductively by the condition that, for every $u \in V\_{i-1}$, the function $f$ is constant on the children of $u$. Of course, we may or may not get a bounded function depending on how we choose the $N\_i$. I will now show that we can choose the $N\_i$ so that $f$ is bounded. Or, rather, I will claim it and leave the details as an exercise for you. Let $a\_i$ be a decreasing sequence of positive reals, approaching zero. Take $N\_i = 6/(a\_{i+1} - a\_i)$. Exercise: If $f$ on $V\_1$ is taken between $-1+a\_1$ and $1-a\_1$, then $f$ on $V\_i$ will lie between $-1+a\_i$ and $1-a\_i$. In particular, $f$ will be bounded between $-1$ and $1$ everywhere.	15	https://mathoverflow.net/users/297	9003	6,154
https://mathoverflow.net/questions/8976	13	The class of all ordinal numbers $\mathbf{Ord}$, aside being a proper class, can be thought of an ordinal number (of course it contains all ordinal numbers that are sets, not itself). Then one could consider $\mathbf{Ord}+1$, $\mathbf{Ord}+\mathbf{Ord}$, $\mathbf{Ord} \cdot \mathbf{Ord}$ and so on. Does this extension of ordinals make sense/is interesting? Maybe it was described by someone? Could it go deeper - to create a "superclass" of all ordinals that are classes?	https://mathoverflow.net/users/158	Ordinals that are not sets	Yes. This is both studied by set theorists and interesting. I personally find some of the related questions below extremely interesting, connected with some very deep questions about the nature of mathematical existence. While the term "ordinal" is usually used only to refer to objects that are sets, one can nevertheless consider class relations that are well-ordered. For example, it is easy to define a relation on $\mathrm{ORD}$ that has order type $\mathrm{ORD}+1$; one can make order type $\mathrm{ORD}+\mathrm{ORD}$ by defining a relation that puts all even ordinals before all odd ordinals, and so on. One can get on a long way with this method, past $\mathrm{ORD}^{\mathrm{ORD}}$, $\mathrm{ORD}^{\mathrm{ORD}^{\mathrm{ORD}}}$, and so on, simply by defining more complicated well-ordered relations on $\mathrm{ORD}$. The flavor here is something like the manner that one considers computable well-ordered relations on the natural numbers, and defines the class of computable ordinals. The analogy is indeed very strong, for the supremum of the computable ordinals is known as $\omega\_1^{CK}$, pronounced "omega 1 Church Kleene", and this is precisely the smallest admissible ordinal, which means that the corresponding level of the constructible universe satisfies the Kripke-Platek axioms of set theory, a weak fragment of ZFC. Similarly, for the situation with the definable class well-ordered relations on $\mathrm{ORD}$, if one looks at the corresponding (constructible) universe of sets that arises from these super-ordinals, it also can be admissible. Another way to describe the situation is that models of ZFC can extend to models of KP, by adding precisely the sets on top coded by a class well-founded relation on $\mathrm{ORD}$. The question looming in the background here, is the extent to which every model of set theory is an initial segment of a much taller model of set theory. Can we extend every model of ZFC to an initial segment of a taller model of ZFC? Of ZFC-? Of KP? Part of the answer is that every model of Kelly-Morse set theory arises similar to this way. Namely, every model of KM is the $V\_\kappa$ of a model of ZFC- in which $\kappa$ is an inaccessible cardinal. Indeed, the theory KM is mutually interpretable with the theory ZFC- + "there is a largest cardinal, which is inaccessible." To have a model of KM is simply to have a model of ZFC that is the $V\_\kappa$ for some inaccessible cardinal $\kappa$ in a model of ZFC- in which $\kappa$ is the largest cardinal. Another part of the answer is that every countable computably-saturated model of ZFC is precisely the $V\_\alpha$ of another model of ZFC, to which it is externally isomorphic. In particular, every such model of ZFC is (externally) isomorphic to a rank initial segment of itself. Another part of the answer is that every model of ZFC elementarily embeds into another model, which is the $V\_\alpha$ of a much taller model. This can be proved by an easy compactness argument. Given any model $M$ of ZFC, write down the theory consisting of ZFC+the elementary diagram of $M$, relativized to $V\_\delta$, in the language with an additional constant $\delta$. This theory is finitely realized by $M$ itself, on account of the Reflection Theorem, and so it has a model $N$. The original model $M$ embeds into $V\_\delta^N$ because of the elementary diagram, so $N$ is as desired. Finally, note that if $M$ has the property that it is point-wise definable (every object is definable in $M$ without parameters), and there are definitely such models of ZFC if there are any at all, then $M$ cannot be the $V\_\alpha$ of any model $N$ of ZFC, since $N$ would recognize that $M$ is pointwise definable, and thus $N$ would have only countably many reals, a contradiction. Let me point out that Harry's observation (now apparently removed) about Grothendieck universes amounts to a special case of the question I pose, where one considers only transitive models of set theory, under the assumption that there is a proper class of inaccessible cardinals (which is equivalent to the Universe axiom). In this case, obviously every transitive model of set theory extends to one of the $V\_\kappa$, for $\kappa$ inaccessible above, and this is essentially what he observed. But actually, one doesn't need this strength to make the conclusion, since the large cardinal consistency strength of the assertion that every transitive set is an element of a larger transitive model of set theory is strictly weaker than even one inaccessible cardinal, let alone a proper class of them.	20	https://mathoverflow.net/users/1946	9010	6,157
https://mathoverflow.net/questions/9007	16	Let X be a topological space, let $\mathcal{U} = \{U\_i\}$ be a cover of X, and let $\mathcal{F}$ be a sheaf of abelian groups on X. If X is separated, each $U\_i$ is affine, and $\mathcal{F}$ is quasi-coherent, then Cech cohomology computes derived functor cohomology; in general one only gets a spectral sequence $$ H^p(\mathcal{U},\underline{H}^q(\mathcal{F})) \Rightarrow H^{p+q}(X,\mathcal{F}) $$ where $\underline{H}^q(\mathcal{F})$ is the presheaf $U \mapsto H^q(U,\mathcal{F}\|\_U)$. > > Question: For q > 0, $\underline{H}^q(\mathcal{F})$ sheafifies to 0. > > > For a quasi-coherent sheaf $\mathcal{F}$ this is clear because cohomology vanishes on affines. Is this really true in general? Brian Conrad states this in the introduction to his [notes](http://math.stanford.edu/~conrad/papers/hypercover.pdf) on cohomological descent.	https://mathoverflow.net/users/2	Cech to derived spectral sequence and sheafification	Yes, this is true in general. It suffices to show the stalks vanish. Pick $x \in X$ and take an injective resolution $0 \to {\cal F} \to I^0 \to \cdots$. For any open $U$ containing $x$, we get a chain complex $$0 \to I^0(U) \to I^1(U) \to \cdots$$ whose cohomology groups are $H^p(U,{\cal F}\|\_U)$. Taking direct limits of these sections gives the chain complex $$0 \to I^0\_x \to I^1\_x \to \cdots$$ of stalks, which has zero cohomology in positive degrees because the original complex was a resolution. However, direct limits are exact and so we find $$0 = {\rm colim}\_{x \in U} H^p(U,{\cal F}\|\_U) = {\underline H}^p({\cal F})\_x$$ as desired. Generally, cohomology tells you the obstructions to patching local solutions into global solutions, and this says that locally those obstructions vanish.	23	https://mathoverflow.net/users/360	9012	6,158
https://mathoverflow.net/questions/8924	19	Surgery theory aims to measure the difference between simple homotopy types and diffeomorphism types. In 3 dimensions, geometrization achieves something much more nuanced than that. Still, I wonder whether the surgeons' key problem has been solved. Is every simple homotopy equivalence between smooth, closed 3-manifolds homotopic to a diffeomorphism? In related vein, it follows from J.H.C. Whitehead's theorem that a map of closed, connected smooth 3-manifolds is a homotopy equivalence if it has degree $\pm 1$ and induces an isomorphism on $\pi\_1$. Is there a reasonable criterion for such a homotopy equivalence to be simple? One could, for instance, ask about maps that preserve abelian torsion invariants (e.g. Turaev's).	https://mathoverflow.net/users/2356	Diffeomorphism of 3-manifolds	Turaev [defined a simple-homotopy invariant](https://mathscinet.ams.org/mathscinet-getitem?mr=970081) which is a complete invariant of homeomorphism type (originally assuming geometrization). Here is the Springer link if you have a subscription: [Towards the topological classification of geometric 3-manifolds](https://doi.org/10.1007/BFb0082780) He claims in the paper that a map between closed 3-manifolds is a homotopy equivalence if and only if it is a simple homotopy equivalence, but he says that the proof of this result will appear in a later paper. I'm not sure if this has appeared though (I haven't searched through his later papers on torsion, and there's no MathScinet link).	20	https://mathoverflow.net/users/1345	9015	6,159
https://mathoverflow.net/questions/9022	9	Sorry if this question is below the level of this site: I've read that the quotient of a Hausdorff topological group by a closed subgroup is again Hausdorff. I've thought about it but can't seem to figure out why. Is it obvious? A simple yes or no (with reference is possible) is all I need.	https://mathoverflow.net/users/2612	Quotient of a Hausdorff topological group by a closed subgroup	Edit: Below I expand my crude original answer "[Yes](http://en.wikipedia.org/wiki/Hausdorff_space)" as requested by the community. --- Yes. Let $G$ be the group and $H$ be the closed subgroup. The [kernel](http://en.wikipedia.org/wiki/Kernel_of_a_function) of the quotient map $G \to G/H$ is equal to $\Delta^{-1}(H)$ where $\Delta : G \times G \to G$ is the continuous function $\Delta(x,y)= x- y$. Hence the kernel is closed. According to [this](http://en.wikipedia.org/wiki/Hausdorff_space#Properties) $G/H$ is Hausdorff.	7	https://mathoverflow.net/users/605	9023	6,163
https://mathoverflow.net/questions/9006	17	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! Background ---------- Let $E \to X$ be a holomorphic vector bundle over a complex manifold. A connection $A$ in $E$ is called holomorphic if in local holomorphic trivialisations of $E$, $A$ is given by a holomorphic 1-form with values in End(E). Notice that the curvature of $A$ is necessarily a (2,0)-form. In particluar, holomorphic connections over Riemann surfaces are flat. This will be important for my question. The Question ------------ 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 not fix) and that we choose a nowhere vanishing section $v$ of $\Lambda^2 E$. (So I do 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). 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? 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! Motivation ---------- A naive dimension count shows that in fact the two spaces have the same dimension: 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 $6g-6$. 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 $6g-6$. A curious remark ---------------- 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.	https://mathoverflow.net/users/380	Representations of surface groups via holomorphic connections	This question is addressed in a very recent paper of Bogomolov-Soloviev-Yotov (I don't think it is on the web yet). Among many interesting things they prove that the map from the moduli space of pairs $(C,\nabla)$ where $\nabla$ is a holomorphic connection on the trivial rank two bundle on some smooth curve $C$ is submersive whenever $\nabla$ is irreducible and $C$ is generic. With regard to Jack Evans' comment: this is a very different question than the question of determining respresentations in a real form (which has been extensively studied by Hitchin, Goldman, Garcia-Prada, etc.). It is about a holomorphic subvariety in the moduli of representations. A better analogy will be to look at the moduli space of opers which is the moduli space of holomorphic flat connections on a fixed (non-trivial) rank two vector bundle, namely, the 1-st jet bundle of a theta characteristic on the curve.	13	https://mathoverflow.net/users/439	9034	6,173
https://mathoverflow.net/questions/9031	8	Is there a simple proof that 3-dimensional [graph manifolds](http://en.wikipedia.org/wiki/Graph_manifold) have residually finite fundamental groups? By "simple" I mean the proof that does not use any hard 3d topology. I care because I wish to generalize this to higher-dimensional analogs of graph manifolds.	https://mathoverflow.net/users/1573	Residual finiteness for graph manifold groups	As far as I'm aware, every proof of this fact is essentially the same as [Hempel's original proof](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=AUCN&pg7=ALLF&pg8=ET&review_format=html&s4=hempel&s5=residual&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=2&mx-pid=895623). I don't know whether it's "simple" enough for you! The key point is that the fundamental group G of a Seifert-fibred piece has the following property. Property. There exists an integer K such that for any positive integer n there is a finite-index normal subgroup Gn of G such that any peripheral subgroup P intersects Gn in KnP. It's not too hard to prove. There's a nice account in [a paper by Emily Hamilton](http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&s1=629944&vfpref=html&r=5&mx-pid=1851085) (which generalizes Hempel's result). The other important fact is that peripheral subgroups in Seifert-fibred manifold groups are separable (ie closed in the profinite topology, for any non-experts out there). Using these two pieces of information, you can piece together finite quotients of Seifert-fibred pieces into a virtually free quotient of π1 of the graph manifold in which your favourite element doesn't die. Note on separability of peripheral subgroups. Of course, Scott proved that Seifert-fibred manifold groups are LERF. But, by a pretty argument of [Long and Niblo](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=AUCN&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=long&s5=niblo&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq), a subgroup is separable if and only if the double along it is residually finite. In particular, you can deduce peripheral separability from the easier fact that Seifert-fibred manifold groups are residually finite.	9	https://mathoverflow.net/users/1463	9036	6,175
https://mathoverflow.net/questions/8889	28	I am not interested in the philosophical part of this question :-) When I look at mathematics, I see that lots of different logics are used : classical, intuitionistic, linear, modal ones and weirder ones ... For someone new to the field, it is not easy to really see what they have in common for justifying the use of the word "logic". Is it just because of a filiation with classical logic ? I have attempted to find an answer in the literature. Some papers are telling me that a logic is a pre-order. It is not a satisfactory answer to me. I imagined that it may be related to the use of some specific connectors : but linear logic is telling me it is not so simple. I imagined that it may be related to some symmetry properties of the rules of the system : but it is dependent on how the logic is formalized. Then, I had the crazy idea (after discovering the Curry-Howard isomorphism) that it may be related to the computational content of the system. But, it is obviously wrong. So, I have not progressed and I am still wondering if there may be a point of view allowing to see what all these systems have in common. I have avoided the use of the word "truth" in this question. I am expecting a mathematical answer if there is one. There are too many philosophical problems related to the notion of truth. But, perhaps my question is a naive one ...	https://mathoverflow.net/users/1173	What is a logic?	I know nothing about this but I happened to come across it while reading about closure operators on wikipedia: [Universal Logic](http://en.wikipedia.org/wiki/Universal_logic).	12	https://mathoverflow.net/users/1233	9042	6,180
https://mathoverflow.net/questions/9043	27	This question is related to the question [Is an algebraic space group always a scheme?](https://mathoverflow.net/questions/8918/is-an-algebraic-space-group-always-a-scheme) which I've just seen which was posted by Anton. His question is whether an algebraic space which is a group object is necessarily a group scheme, and the answer appears to be YES. Now my naive idea of what an algebraic space is is that it is the quotient of a scheme by an étale equivalence relation, but I seem to be confusing myself. I'm hoping someone will help lead me out of my confusion. Let me first consider an analogous topological situation, where the answer is no. We can consider the category of smooth spaces, by which I mean the category of sheaves on the site of smooth manifolds which are quotients of manifolds by smooth equivalence relations (with discrete fibers). Here is an example: If we have a discrete group $G$ acting (smoothly) on a space $X$, we can form the equivalence relation $R \subset X \times X$, where $R$ consists of all pairs of points $(x,y) \in X \times X$ where $y= gx$ for some $g \in G$. If R is a manifold, then the sheaf $[X/R]$ (which is defined as a coequalizer of sheaves) is a smooth space. Here is an example of a group object in smooth spaces: We start with the commutative Lie group $S^1 = U(1)$. Now pick an irrational number $r \in \mathbb{R}$ which we think of as the point $w = e^{2 \pi i r}$. We let $\mathbb{Z}$ act on $S^1$ by "rotation by $r$" i.e. \begin{align\} \mathbb{Z} \times S^1 & {}\to S^1\\ (n, z) & {}\mapsto w^n z. \end{align\} This gives us an equivalence relation $R = \mathbb{Z} \times S^1 \rightrightarrows S^1$, where one map is the action and the other projection. The fibers are discrete and the quotient sheaf is thus a smooth space, which is not a manifold. However the groupoid $R \rightrightarrows S^1$ has extra structure. It is a group object in groupoids, and this gives the quotient sheaf a group structure. The group structure on the objects $S^1$ and morphisms $R$ are just given by the obvious group structures. Incidentally, this group object in groupoids serves as a sort of model for the "quantum torus": Blohmann–Tang–Weinstein, Hopfish structure and modules over irrational rotation algebras, [arXiv:math/0604405](https://arxiv.org/abs/math/0604405). Now what happens when we try to copy this example in the setting of algebraic spaces and schemes? Let's make it easy and work over the complex numbers. An analog of $S^1$ is the group scheme $$ \mathbb{G}\_m / \mathbb{C} = \operatorname{Spec} \mathbb{C}[t,t^{-1}]. $$ Any discrete group gives rise to a group scheme over $\operatorname{Spec} \mathbb{C}$ by viewing the set $G$ as the scheme $$ \bigsqcup\_G \operatorname{Spec} \mathbb{C}. $$ So for example we can view the integers $\mathbb{Z}$ as a group scheme. This (commutative) group scheme should have the property that a homomorphism from it to any other group scheme is the same as specifying a single $\operatorname{Spec} \mathbb{C}$-point of the target (commutative) group scheme. A $\operatorname{Spec} \mathbb{C}$-point of $\operatorname{Spec} \mathbb{C}[t,t^{-1}]$ is specified by an invertible element of $\mathbb{C}$. Let's fix one, namely the one given by the element $w \in S^1 \subset \mathbb{C}^\times$. So this gives rise to a homomorphism $\mathbb{Z} \to \mathbb{G}\_m$ and hence to an action of $\mathbb{Z}$ on $\mathbb{G}\_m$. Naïvely, the same construction seems to work to produce a group object in algebraic spaces which is not a scheme. So my question is: where does this break down? There are a few possibilities I thought of, but haven't been able to check: 1. Does $R = \mathbb{Z} \times \mathbb{G}\_m$ fail to be an equivalence relation for some technical reason? 2. Do the maps $R \rightrightarrows \mathbb{G}\_m$ fail to be étale? 3. Is there something else that I am missing?	https://mathoverflow.net/users/184	Why is this not an algebraic space?	If $X$ denotes the quotient sheaf $\mathbf{G}\_m / \mathbf{Z}$ then the inclusion $\mathbf{G}\_m \times\_X \mathbf{G}\_m \rightarrow \mathbf{G}\_m \times \mathbf{G}\_m$ can be identified with the action map $\mathbf{G}\_m \times \mathbf{Z} \rightarrow \mathbf{G}\_m \times \mathbf{G}\_m$. Since this map is not quasi-compact, $X$ is not quasi-separated, so it is not an algebraic space by Knutson's definition.	21	https://mathoverflow.net/users/32	9044	6,181
https://mathoverflow.net/questions/8948	2	I've just finished my first course in differential geometry, so forgive me if this is maybe a silly or well-known question, but given any, say, diffeomorphism of $n$-manifolds $\phi:M\rightarrow N$, I was wondering whether the map $f:M\rightarrow GL\_n(\mathbb{R})$ defined by $f(p)=d\phi\_p$ has any important or nice properties? One guess I had was that further sending $p$ to det$(d\phi\_p)$ could say something about orientation? I had been thinking about the Gauss map and how it's a map from any surface $S$ to the sphere $\mathbb{S}^2$, and I was trying to come up with other canonical ways of creating maps of surfaces and/or manifolds to "special" ones.	https://mathoverflow.net/users/1916	Special map from a manifold to GL_n(R)?	The question can be interpreted to ask, given a diffeomorphism $\phi:M \to N$, what topological information in its derivative, interpreted as a bundle map? If the bundle map is used carefully, the answer is that there is a lot of information. There is a principle that, in dimension $n \ne 4$, there is enough information to completely tell when a homeomorphism can be improved to a diffeomorphism. That is a hard result if you are comparing PL homeomorphisms to diffeomorphisms, and a very hard result if you are comparing topological homeomorphisms to diffeomorphisms. It is easier to sketch that in principle you get important obstructions to smoothing a homeomorphism. The obstructions exist in all dimensions, the only difference being that in dimension 4 they aren't everything. Suppose that $M$ and $N$ are smooth $n$-manifolds, or topological $n$-manifolds, or PL $n$-manifolds, and $\phi$ is an equivalence. Then in all of these cases, $\phi$ has a formal "derivative" that is locally a germ of a (continuous, PL, or smooth) homeomorphism. The "derivative" is a map between "tangent" bundles (which are actually most natural as microbundles) to match the germ picture, rather than ordinary fiber bundles. If you're afraid of germs (who isn't), a germ is no more than an equivalence class of local maps: Two local maps are the same if they agree on a neighborhood. The germ derivative of a map is basically a fake, tautological derivative. On the other hand, the group $\text{Diff}(n)$ of germs of diffeomorphisms that fix the origin is homotopy equivalent to the true derivative group $GL(n,\mathbb{R})$, which is homotopy equivalent to the orthogonal group $O(n)$. The other two groups of germs are called $PL(n)$ and $TOP(n)$. Now suppose, to take the best-behaved case, that $\phi$ is a PL homeomorphism between two smooth manifolds $M$ and $N$ with smooth triangulations. It has a formal derivative $D\phi$ that is a map between tangent $PL(n)$ microbundles. You could ask, roughly speaking, whether you can homotop $D\phi$ to make it a true smooth derivative in the available $\text{GL}(n,\mathbb{R})$ bundle, or equivalently an $O(n)$ bundle. This is a necessary condition to be able to improve $\phi$ itself to a diffeomorphism. The more subtle result is that it is also a sufficient condition. In this sense, the relative homotopy theory $PL(n)/O(n)$ exactly measures the non-uniqueness of smooth structures on manifolds. (I learned about this stuff from Rob Kirby, but it looks like it is covered in some form by [Jacob Lurie](http://www.math.harvard.edu/~lurie/937.html).) There is an older situation in which the derivative of a map gives you an obstruction to something, and it turns out to be the only obstruction. Namely, if $M$ is a smooth $n$-manifold, you can ask whether it immerses in $\mathbb{R}^k$; or you could ask whether two immersions are equivalent through an isotopy of immersions. Clearly if $\phi:M \to \mathbb{R}^k$ is an immersion, the homotopy type of the derivative $d\phi$ (among maximum rank choices) is an invariant of $\phi$, and clearly there is no immersion if there is no available homotopy class. The theorem of Smale and Hirsch is that these are the only obstructions to existence and equivalence of immersions, unless $M$ is closed and the dimensions $n$ and $k$ are equal. This theorem is illustrated very nicely in the math movie [Outside In](http://video.google.com/videoplay?docid=-6626464599825291409#). Another way to say it is that you can formaly weaken both questions by decoupling maps from their derivatives, but still get the correct answer.	9	https://mathoverflow.net/users/1450	9057	6,191
https://mathoverflow.net/questions/9051	3	I´m trying to solve a problem of cancellation of reflexive finitely generated modules over normal noetherian domains. When $R$ is regular domain with $\dim R \le 2$, for finitely generated modules, reflexive is equivalent to projective. Now I´m studying the case $\dim R=2$ and $R$ normal. In this hypothesis, reflexive modules are maximal Cohen-Macaulay modules. I´m looking for references about this topic, with especial emphasis in lifting of homomorphism between factors of maximal CM modules: something like "... an homomorphism $M/IM\to N/IN$ can be lift to an homomorphism $M\to N...$"; indescomponibles maximal CM modules are welcome too.	https://mathoverflow.net/users/2040	About maximal Cohen-Macaulay modules	You probably want to look at this paper: <http://www.springerlink.com/content/8r44x50448644568/> on deformations of MCM modules and the references there.	4	https://mathoverflow.net/users/2083	9062	6,195
https://mathoverflow.net/questions/9066	14	Is that true that there is no rational curves contained in an Abelian variety? If it's true, is that because abelian varieties are not uniruled? How do I know whether an abelian variety is not uniruled?	https://mathoverflow.net/users/2348	Is there any rational curve on an Abelian variety?	There are no rational curves in an abelian variety, this is much stronger than not being uniruled. If there is a map $P^1 \to A$, $A$ abelian, the map would factor through the Albanese variety of $P^1$, by definition. However, for curves, the Albanese is the Jacobian (from general theory of the Jacobian) and the Jacobian of $P^1$ is a point.	35	https://mathoverflow.net/users/2290	9069	6,199
https://mathoverflow.net/questions/9065	10	It's easy to see that in bipartite maximal triangle free graphs (n vertices), the maximum degree is at least $\lceil n/2 \rceil$. What about mtf graphs in general? Must there always be some vertex of high degree? Before I jump in with both feet, is there an obvious reason why there cannot be an absolute constant $c$ such that for every mtf $G$, $\Delta(G) > cn$? A handy counterexample to c = 1/4?	https://mathoverflow.net/users/2588	Maximum degree in maximal triangle free graphs	There is a sequence of [Kneser graphs](http://en.wikipedia.org/wiki/Kneser_graph), generalizing the [Petersen graph](http://en.wikipedia.org/wiki/Petersen_graph), that comprises a counterexample. Let $k \ge 1$ be an integer and let $G$ be a graph whose vertices are subsets of size $k$ of $\{1,2,\ldots,3k-1\}$. Connect two vertices $A$ and $B$ by an edge if they are disjoint as subsets. Then $G$ has no triangles, because there isn't room for three disjoint subsets. On the other hand, if $A$ and $B$ are not connected by an edge, which is to say they are not disjoint, there is room for a third set $C$ which is disjoint from both of them. Thus if you add any edge $(A,B)$ to this graph $G$, it forms a triangle with $(A,C)$ and $(B,C)$. Now let's count vertices and edges. The graph $G$ has $\binom{3k-1}{k}$ vertices, and each vertex has $\binom{2k-1}{k}$ edges. QED (It is the Petersen graph when $k=2$.) --- This is partly plagiarizing from David's insightful answer below, but I can't resist an addendum to his remarks. In the paper, [The triangle-free process](http://dx.doi.org/10.1016/j.aim.2009.02.018), Tom Bohman simplifies the construction of Kim that David cites. He makes a maximal triangle-free graph on $n$ vertices in using simplest plausible method: The random greedy algorithm. The result is a graph that is statistically very predictable. Its independence number is almost always $\Theta(\sqrt{n \log n})$, and therefore so is its maximum valence. Its average valence is also in the same class. You can easily make graphs like this yourself with a simple computer program and see their properties. It's ironic, but a common theme, that a very simple random algorithm is more efficient than a highly symmetric construction such as the Kneser graphs. As David explains, you get an immediate lower bound of maximum valence $\Omega(n^{1/2})$ for any graph of diameter 2 or even radius 2. The Kneser graphs above have valence $O(n^\alpha)$ with $\alpha = \log\_{27/4}(4) \approx 0.726$. So the Kim-Bohman result is much stronger, and that's why he pointed out Kim's paper. In fact, this result closes a circle for me. A triangle-free graph on $n$ vertices is also a type of "packing design" in which each triple of vertices only has room for two edges. The original paper that introduced the random greedy algorithm in the topic of packing designs is by [Gordon, Kuperberg, Patashnik, and Spencer](http://arxiv.org/abs/math/9511224). In that paper, we were looking at packing designs at the opposite end, for instance choosing triples of points with a random greedy algorithm such that no edge is contained in more than one triple. (The paper says covering designs, but at our end of the asymptotics, they are almost the same as packing designs.) It's the same highly predictable phenomenon at both ends. One of the ideas in this paper was to simplify a fancier construction using "Rödl nibbles" to the random greedy algorithm. Bohman is doing the same thing (but with much stronger asymptotics than in our paper), because Kim also uses Rödl nibbles.	9	https://mathoverflow.net/users/1450	9070	6,200
https://mathoverflow.net/questions/9073	29	Does anyone have examples of when an object is positive, then it has (or does not have) a square root? Or more generally, can be written as a sum of squares? Example. A positive integer does not have a square root, but is the sum of at most 4 squares. (Lagrange Theorem). However, a real positive number has a square root. Another Example. A real quadratic form that is postive definite (or semi-definite) is, after a change of coordinates, a sum of squares. How about rational or integral quadratic forms? Last Example. A positive definite (or semidefinite) real or complex matrix has a square root. How about rational or integral matrices? Do you have other examples?	https://mathoverflow.net/users/nan	When does 'positive' imply 'sum of squares'?	For many examples of this kind, see Olga Taussky, "Sums of squares", Amer. Math. Monthly 77 (1970) 805-830.	29	https://mathoverflow.net/users/143	9075	6,204
https://mathoverflow.net/questions/4953	24	Do we know any problem in NP which has a super-linear time complexity lower bound? Ideally, we would like to show that 3SAT has super-polynomial lower bounds, but I guess we're far away from that. I'd just like to know any examples of super-linear lower bounds. I know that the time hierarchy theorem gives us problems which can be solved in O(n^3) but not in O(n^2), etc. Thus I put the word "natural" in the question. I ask for problems in NP, because otherwise someone would give examples of EXP-complete problems. I know there are time-space tradeoffs for some problems in NP. I don't know if any of them imply a super-linear time complexity lower bound though. (To address a question below about machine models, consider either multitape Turing machines or the RAM model.)	https://mathoverflow.net/users/1042	Super-linear time complexity lower bounds for any natural problem in NP?	Sorry I am so late to the discussion, but I just registered... There are non-linear time lower bounds on multitape Turing machines for NP-complete problems. These lower bounds follow from the fact that the class of problems solvable in nondeterministic linear time is not equal to the class of those solvable in (deterministic) linear time, in the multitape Turing machine setting. This is proved in: > > [Wolfgang J. Paul, Nicholas Pippenger, Endre Szemerédi, William T. Trotter: On Determinism versus Non-Determinism and Related Problems (Preliminary Version) FOCS 1983: 429-438](http://www-wjp.cs.uni-saarland.de/publikationen/PPST83.pdf) > > > In fact, unraveling the proof shows that there must be some problem solvable in nondeterministic linear time that is not solvable in $o(n \cdot (\log^\* n)^{1/4})$ time (again, on a multitape Turing machine). Note the \* in the logarithm; this is just "barely" above linear. One known application of this result is that a natural NP-complete problem in automata theory cannot be solved in $o(n \cdot (\log^\* n)^{1/4})$ time: > > Etienne Grandjean: A Nontrivial Lower Bound for an NP Problem on Automata. SIAM J. Comput. 19(3): 438-451 (1990) > > > Unfortunately the lower bound of Paul et al. relies crucially on the geometry that arises from accessing one-dimensional tapes. We don't know how to prove a non-linear lower bound even if you allow the Turing machine to have a constant number of two-dimensional tapes. We can prove time lower bounds for NP problems on general computational models if you severely restrict the extra workspace used by the machine. (This is getting into my own work so I won't say more unless you're truly interested.) As for the comment above me: the sorting lower bound holds only in a comparison-based model, which is extremely restricted. The claim that sorting requires Omega(n log n) time on general computational models is false. There are faster algorithms for sorting integers. See for example: > > Yijie Han: Deterministic sorting in O(n log logn) time and linear space. J. Algorithms 50(1): 96-105 (2004) > > >	29	https://mathoverflow.net/users/2618	9081	6,210
https://mathoverflow.net/questions/9013	4	I don't know if this is an easy question for specialists in the field. Consider the following interpolation problem : let $\varepsilon >0$, $X$ be a finite set of real numbers and $g$ be a real-valued function on $X$. The goal is to find a function $f$, defined on an interval containing $X$, that coincides with $g$ on $X$, and admits a continuous second derivative $f''$ bounded by $\varepsilon$ : $\|f''\|< \varepsilon$. A necessary condition is that, for any $x< y < z$ in $X$, $(\) \|\frac{(y-x)g(z)-(z-x)g(y)+(z-y)g(x)}{(z-x)(z-y)(y-x)}\| < \frac{\varepsilon}{2}$. Is that condition also sufficient ? If $X$ has just three elements then (\) says precisely that the Lagrange interpolating polynomial $f$ satisfies $\|f''\|< \varepsilon$, so that equivalence holds in this case.	https://mathoverflow.net/users/2389	Interpolation by a function whose second derivative is bounded	It isn't sufficient. Suppose that the elements of $X$ are $$x\_1 < x\_2 < \cdots < x\_n,$$ and suppose for simplicity that $\epsilon = 1$. Then the values $f(x\_k)$ carry the same information as the integrals $$\langle p, f'' \rangle = \int\_{x\_1}^{x\_n} p(x) f''(x) dx,$$ where $p(x)$ is a continuous, piecewise-linear function with $p(x\_1) = p(x\_n) = 0$. You can convert between such an integral and linear combinations of values of $f(x)$ by integrating by parts twice. For any such test function $p$, the inequality $$\|\langle p, f'' \rangle\| \le \int\_{x\_1}^{x\_n} \|p(x)\| dx \qquad \qquad$$ holds. Moreover, many of these inequalities are logically independent. More precisely, the extremal choices of $p(x)$ are those for which $p(x\_k)$ alternate in sign for $i \le k \le j$, and $p(x\_k) = 0$ when $k < i$ or $k > j$. The idea to show that such as $p$ is extremal is to make a dual choice of $f''$ that switches between $1$ and $-1$ when $p$ crosses the $x$ axis. (I apologize for skipping some of the logic in the extremality calculation, but I think that this is correct.) The resulting set of values for $f(\vec{x})$ is not a polytope, but a certain convex region with both flat and curved boundary. For example, consider your statistic, $$f\_k = \frac{(x\_{k+1}-x\_k)f(x\_{k+1})-(x\_{k+1}-x\_{k-1})f(x\_k)+(x\_k-x\_{k+1})f(x\_{k-1})}{(x\_{k+1}-x\_k)(x\_{k+1}-x\_{k-1})(x\_k-x\_{k-1})},$$ for consecutive triples of points, and suppose that $n=4$. Then the pair $(f\_2,f\_3)$ must lie inside the square $[-\frac12,\frac12]^2$; you identified two other pairs of inequalities, but they do not clip the square further. What actually happens is that two of the corners of the square are rounded by parabolas. The supporting lines of the parabolas come from the test function $p$ that is a piecewise-linear interpolation of $(0,-1,a,0)$ with $a > 0$.	3	https://mathoverflow.net/users/1450	9085	6,212
https://mathoverflow.net/questions/9055	3	So a really simple way of describing a digital computer is to say that it is a device for performing boolean operations. You feed it a bunch of bit strings, which is a description of the problem and its parameters in binary, the computer performs a bunch of boolean operations like $\wedge$, $\vee$, $\neg$ and gives you back another bit string which hopefully is the encoded version of the answer you were looking for. From this description it is not hard to see the connection of classical computation to discrete dynamical systems and classical logic, the operations provide the dynamics by flipping bits around in a controlled fashion and since we restrict the operations to a certain subset we get classical logic. My question is about analog and quantum computation. Is there a simple description of a quantum computer or an analog computer that makes the connection of that notion of computation to the other branches of mathematics a little more obvious? What is the most basic kind of enconding of a problem that can be fed into a quantum or analog computer and what are the most basic operations performed on this encoding? Edit: Downvotes should come with comments so I know what to change in order to make my question clearer. Being a novice I'm just trying to piece together some themes about computation, logic and algebra. Edit: alpheccar provided a link to a paper by John Baez and Mike Stay that pretty much answers my question and even puts it in a much wider context. Here's the [link](http://math.ucr.edu/home/baez/rosetta.pdf) alpheccar provided.	https://mathoverflow.net/users/nan	computation, algebra, logic	> > From this description it is not hard to see the connection of classical computation to discrete dynamical systems and classical logic, the operations provide the dynamics by flipping bits around in a controlled fashion and since we restrict the operations to a certain subset we get classical logic. > > > Careful -- this observation only works for purely finite-state systems! If your idealized model of a digital computer can handle potentially unbounded quantities of input, then the connection to classical logic is lost. Happily, it fails in a way that reveals connections to topology, and explains why topological models of intuitionistic logic exist. The basic idea is that we view a computer as realizing a function $f$. Then for any input value (from the input set $A$), if it returns an answer in a finite amount of time, it can have observed at most a finite amount of information about the input. This means that we can equip our input set with a topology in the following way: suppose our basic observations are a collection of predicates on $A$. Then we get the topological structure from the following fact: we can only take finite intersections because we can only make finitely many observations, and so can only conclude the conjunction of finitely many predicates. We can take infinite unions (i.e., existentials) because we can "get lucky" and guess the correct branch of the union (i.e., witness to the existential). Then, amazingly, we can use continuity as a stand-in for computability! This is pretty much the idea Dana Scott had when he invented domain theory (which gets a different name because the topologies that we get this way are "weird" -- e.g., they're typically not Hausdorff -- and so the theorems you want develop a bit differently).	7	https://mathoverflow.net/users/1610	9096	6,222
https://mathoverflow.net/questions/9046	12	Every commutative $C^\*$-algebra is isomorphic to the set of continuous functions, that vanish at infinity, of a locally compact Hausdorff space. Every commutative finite dimensional Hopf algebra is the group algebra of some finite group. Does there exist a characterisation of the finitely generated commutative Hopf algebras that arise as group algebras?	https://mathoverflow.net/users/1095	Hopf algebras arising as Group Algebras	Since the questioner starts off asking about $C^\$ algebras, I am going to assume that he only cares about Hopf algebras over $\mathbb{C}$. Every finitely generated, commutative $\mathbb{C}$-Hopf-algebra is the polynomial functions on an algebraic group $G$.* As Ben says, we can just take $G$ to be Spec of the Hopf algebra, and then the comultiplication gives a group structure on this Spec. There are several ways that working over $\mathbb{C}$ makes things nicer than working over arbitrary fields. Other answers have pointed them out but, IMO, have done so in a way that makes things sound more confusing. Let me instead point out how nice algebraic groups over $\mathbb{C}$ are: 1. You might worry that $G$ would not be reduced. This happens over fields of characteristic $p$, but not characteristic zero. See [my earlier question](https://mathoverflow.net/questions/1967/hopf-algebra-reference). Also, you might worry that $G$ has singularities, but it doesn't. In short, $G$ is a complex Lie group. 2. Over non-algebraically closed fields $k$, the behavior of the $k$-points of $G$ may not determine the behavior of $G$. For example, let $X = \operatorname{Spec} \mathbb{R}[x]/(x^n-1)$. Then $X$ can be equipped with the structure of an algebraic group of order $n$; the coproduct is $x \mapsto x \otimes x$. Intuitively, you should think $x\_1 \times x\_2 \mapsto x\_1 x\_2$, so this group is the $n$-th roots of $1$ under multiplication. By the Nullstellensatz, this can't happen over $\mathbb{C}$. The $\mathbb{C}$-points will be dense (in both the ordinary and Zariski topologies) and any map will be determined by what it does to the $\mathbb{C}$-points.	14	https://mathoverflow.net/users/297	9104	6,226
https://mathoverflow.net/questions/9111	6	Suppose I have a morphism of schemes for which I know the relative cotangent complex is trivial, and the map on reduced subschemes is an isomorphism. Is the map an isomorphism? More generally, given a morphism of schemes with zero relative cotangent complex, which is of finite presentation on the reduced points. Is the map of finite presentation, and thus etale? (Maybe a better way to phrase this is - what's the reference for these statements? are they in SGA or in Illusie somewhere?)	https://mathoverflow.net/users/582	Detecting etale maps on reduced points	Illusie, Complexe cotangent et deformations I, Prop. 3.1.1 (p. 203) is essentially the second thing you asked. Just a technical point: I don't think people would use the term "etale" unless the morphism is locally finitely presented or something like that (you seem to be wanting to assume that only at the level of reduced schemes or something?). Without thinking too hard, though, Prop. 3.1.2 (same page) says that L\_{X/Y} of perfect amplitude in [0,0] implies f is formally smooth...surely also your condition implies it's formally etale, which is what you're asking in general (without finiteness assumption), no?	4	https://mathoverflow.net/users/2628	9112	6,232
https://mathoverflow.net/questions/9100	21	I'm currently working through Frenkel's beautiful paper: <http://arxiv.org/PS_cache/hep-th/pdf/0512/0512172v1.pdf>. I'm looking for a good example of a projective curve to get my hands dirty, and go through the general constructions that Frenkel shows there and try to do them manually for this example of a curve. Are there any good instructive examples for doing this? (Or does it always get out of hand very quickly?)	https://mathoverflow.net/users/2623	A good example of a curve for geometric Langlands	Unfortunately I don't think geometric Langlands is very easy on any curve. The only curve where the objects are readily accessible is $P^1$, but even there the general statement is kind of tricky (see Lafforgue's note [here](http://people.math.jussieu.fr/~vlafforg/geom.pdf)). I would look at Frenkel's writings on the Gaudin model, which is a concrete illustration of the Beilinson-Drinfeld-Feigin-Frenkel approach to geometric Langlands for $P^1$ with several punctures. Also Arinkin and Lysenko worked out explicitly a case of geometric Langlands (in a stronger sense) on $P^1$ minus 4 points -- see the first four papers on a mathscinet search for Arinkin. So the answer is try $P^1$ with some punctures, but don't be surprised if things are rather tricky already there. (I also think geometric Langlands on an elliptic curve should be accessible, but as far as I know it hasn't been worked out very explicitly.)	21	https://mathoverflow.net/users/582	9113	6,233
https://mathoverflow.net/questions/9000	22	Wikipedia says that the [intermediate value theorem](https://en.wikipedia.org/wiki/Intermediate_value_theorem) “depends on (and is actually equivalent to) the completeness of the real numbers.” It then offers a simple counterexample to the analogous proposition on ℚ and a proof of the theorem in terms of the completeness property. Does an analogous result hold for the computable reals (perhaps assuming that the function in question is computable)? If not, is there a nice counterexample?	https://mathoverflow.net/users/2599	Intermediate value theorem on computable reals	Let me assume that you are speaking about computable reals and functions in the sense of [computable analysis](https://en.wikipedia.org/wiki/Computable_analysis), which is one of the most successful approaches to the topic. (One must be careful, since there are several incompatible notions of computability on the reals.) In computable analysis, the [computable real numbers](https://en.wikipedia.org/wiki/Computable_real_number) are those that can be computed to within any desired precision by a finite, terminating algorithm (e.g. using Turing machines). One should imagine receiving rational approximations to the given real. In this subject, functions on the reals are said to be computable, if there is an algorithm that can compute, for any desired degree of accuracy, the value of the function, for any algorithm that produces approximations to the input with sufficient accuracy. That is, if we want to know $f(x)$ to within $\epsilon$, then the algorithm is allowed to ask for $x$ to within any $\delta$ it cares to. The Computable Intermediate Value Theorem would be the assertion that if $f$ is a computable continuous function and $f(a)\lt c\lt f(b)$ for computable reals $a$, $b$, $c$, then there is a computable real $d$ with $f(d)=c$. The book [Computable analysis: an introduction](https://books.google.com/books?id=OPolVWVFDJYC&dq=computable+analysis&printsec=frontcover&source=bl&ots=_QlUjkXhSy&sig=5m7E0aXQ-CVl8HcF77tzW2AW5Pk&hl=en&ei=Eh8pS93oN4Oj8AbDoIW1DQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0CEIQ6AEwCQ#v=onepage&q=&f=false) by Klaus Weihrauch discusses exactly this question in Example 6.3.6. The basic situation is as follows. The answer is Yes. If f happens to be increasing, then the usual bisection proof of existence turns out to be effective. For other $f$, however, one can use a trisection proof. Theorem 6.3.8 says that if $f$ is computable and $f(x)\cdot f(z)\lt 0$, then f has a computable zero. This implies the Computable Intermediate Value theorem above. In contrast, the same theorem also says that there is a non-negative computable continuous function f on $[0,1]$, such that the sets of zeros of $f$ has Lebesgue measure greater than $\frac{1}{2}$, but $f$ has no computable zero. In summary, if the function crosses the line, you can compute a crossing point, but if it stays on one side, then you might not be able to compute a kissing point, even if it is kissing on a large measure set.	31	https://mathoverflow.net/users/1946	9116	6,235
https://mathoverflow.net/questions/9115	8	Is there a notion of algebraic geometry for these objects? If we take the dual category of the category of cocommutative corings with counit, is there geometry in it in a sense dual to affine schemes? Can we look at the set of coideals of a coring, put a space structure on it and sheaves (maybe cosheaves) of sections?	https://mathoverflow.net/users/2300	Algebraic geometry for cocommutative corings with counit.	Let's consider coalgebras over a field rather than corings. There is a theorem that every (coassociative) coalgebra over a field is the union of its finite-dimensional subcoalgebras. So the category of coalgebras over a field k is the category of ind-objects in the category of finite-dimensional coalgebras, while the latter is the opposite category to the category of finite-dimensional algebras. Now if we restrict ourselves to the commutative case, then the category of finite-dimensional commutative algebras with unit over k is the opposite category to the category of 0-dimensional schemes of finite type over k, or just schemes finite over Spec k. Combining it all, the category of cocommutative coalgebras with counit over k is equivalent to the category of ind-0-dimensional ind-schemes of ind-finite type over k, or just ind-finite ind-schemes over Spec k. This explains, in particular, that the "underlying topological space" functor maps cocommutative coalgebras to discrete sets rather than anything else, and the coalgebra itself is simply the infinite direct sum of the coalgebras sitting at the points of this set.	15	https://mathoverflow.net/users/2106	9121	6,240
https://mathoverflow.net/questions/9122	9	A recursive presentation of a group is a one in which there is a finite number of generators and the set of relations is recursively enumerable. I found the following quote in Lyndon-Schupp, chapter II.1: "This usage may seem a bit strange, but we shall see that if G has a presentation with the set of relations recursively enumerable, then it has another presentation with the set of relations recursive." It is not clear to me however how one proves it. Does it go through Higman theorem? I.e. one first proves that group with a recursive presentation embeds in a finitely presented group and then one proves that every subgroup of a finitely presented group has a presentation with a recursive set of relations? And in any case, can one see it somehow directly that having a presentation with a recursively enumerable set of relations (i.e. being recursive) implies having a presentation with a recursive set of relations?	https://mathoverflow.net/users/2629	Recursive presentations	The answer is a simple trick. Essentially no group theory is involved. Suppose that we are given a group presentation with a set of generators, and relations R\_0, R\_1, etc. that have been given by a computably enumerable procedure. Let us view each relation as a word in the generators that is to become trivial in the group. Now, the trick. Introduce a new generator x. Also, add the relation x, which means that x will be the identity. Now, let S\_i be the relation (R\_i)x^(t\_i), where t\_i is the time it takes for the word R\_i to be enumerated in the enumeration algorithm. That is, we simply pad R\_i with an enormous number of x's, depending on how long it takes for R\_i to be enumerated into the set of relations. Clearly, S\_i and R\_i are going to give the same group, once we have said that x is trivial, since the enormous number of copies of x in S\_i will all cancel out. But the point is that the presentation of the group with this new presentation becomes computably decidable (rather than merely enumerable), because given a relation, we look at it to see if it has the form Sx^t for some t, then we run the enumeration algorithm for t steps, and see if S has been added. If not, then we reject; if so, then we accept. One can get rid of the extra generator x simply by using the relation R\_0 from the original presentation. This gives a computable set of relations in the same generating set that generates the same group. The essence of this trick is that every relation is equivalent to a ridiculously long relation, and you make the length long enough so that one can check that it really should be there.	23	https://mathoverflow.net/users/1946	9123	6,241
https://mathoverflow.net/questions/9089	10	Hilbert's 17th problem asked if a nonnegative real polynomial is the sum of squares of rational functions. It was answered affirmative by Artin in around 1920. However, in his speech, he also asked if the rational functions could have coefficients over Q rather than over R. Here is the relavant part of his speech "At the same time it is desirable, for certain questions as to the possibility of certain geometrical constructions, to know whether the coefficients of the forms to be used in the expression may always be taken from the realm of rationality given by the coefficients of the form represented." Does anyone know what these "certain geometrical constructions" are? It seems maybe to me that Hilbert was attempting to embed rational projective space into higher dimensional rational projective space via these polynomials. Briefly, given a nonnegative homogeneous function $f(x\_0,\ldots, x\_n)$ with rational coefficients induces a metric on $QP^n$. Suppose Hilbert's dream holds that $f=p\_0^2+\cdots + p\_N^2$ where $p\_i$'s are polynomials with rational coefficients. Then the map $p: QP^n\to QP^N$ where $p(x)=(p\_0(x),\ldots, p\_N(x))$ is an isometric embedding (almost!) where the metric induced by $f$ is the pullback back of the Euclidean metric on $QP^N$. The above is just my hazard. But I would be delighted if anyone is aware of what exactly Hilbert's intended "geometrical constructions" are.	https://mathoverflow.net/users/nan	what was Hilbert's geometric construction in his 17th problem?	Actually the answer is in the sections 36 to 39 of Hilbert's "Foundations of geometry", which can be found on the web. The constructions are construction with "straightedge" (ruler) and "transferrer of segments". I quote a result from Hilbert's book : Theorem 41. A problem in geometrical construction is, then, possible of solution by the drawing of straight lines and the laying off of segments, that is to say, by the use of the straight-edge and a transferrer of segments, when and only when, by the analytical solution of the problem, the co-ordinates of the desired points are such functions of the co-ordinates of the given points as may be determined by the rational operations and, in addition, the extraction of the square root of the sum of two squares. This result explains relatively clearly why this kind of geometrical constructions leads to the question of the determination of those functions of $x\_1,\ldots,x\_n$ which can be written as sums of squares of rational functions with rational coefficients.	10	https://mathoverflow.net/users/2630	9124	6,242
https://mathoverflow.net/questions/9125	19	The use of the term "spectrum" to denote the prime ideals of a ring originates from the case that the ring is, say, $\mathbb{C}[T]$ where $T$ is a linear operator on a finite-dimensional vector space; then the prime spectrum (which is equal to the maximal spectrum) is precisely the set of eigenvalues of $T$. The use of the term "spectrum" in the operator sense, in turn, seems to have originated with Hilbert, and was apparently not inspired by the connection to atomic spectra. (This appears to have been a coincidence.) A cursory Google search indicates that Hilbert may have been inspired by the significance of the eigenvalues of Laplacians, but I don't understand what this has to do with non-mathematical uses of the word "spectrum." Does anyone know the full story here?	https://mathoverflow.net/users/290	What is the origin of the term "spectrum" in mathematics?	Hilbert, in fact, got the term from Wilhelm Wirtinger (the first one to propose it according to, say <http://www.mathphysics.com/opthy/OpHistory.html>) the paper of Wirtinger is "Beiträge zu Riemann’s Integrationsmethode für hyperbolische Differentialgleichungen, und deren Anwendungen auf Schwingungsprobleme" (1897). In <http://jeff560.tripod.com/s.html> it says "...Wirtinger drew upon the similarity with the optical spectra of molecules when he used the term "Bandenspectrum" with reference to Hill’s (differential) equation." I haven't read Wirtinger's paper, nor do I know how reliable these sources are :)	29	https://mathoverflow.net/users/2384	9126	6,243
https://mathoverflow.net/questions/9143	15	In 'Cours d'arithmetique', Serre mentions in passing the following fact (communicated to him by Bombieri): Let P be the set of primes whose first (most significant) digit in decimal notation is 1. Then P possesses an analytic density, defined as $\lim\_{s \to 1^+} \frac{\sum\_{p \in P} p^{-s}}{\log(\frac{1}{s-1})}$. This is an interesting example since it's easy to see that this set does not have a 'natural' density, defined simply as the limit of the proportion of elements in P to the # of all primes up to $x$, as $x$ tends to infinity. Therefore the notion of analytic density is a genuine extension of the naive notion (they do coincide when both exist). How would one go about proving that P has an analytic density? EDIT: This question has been [asked again](https://mathoverflow.net/q/278209/25) a few years ago, and it is my fault. I did accept Ben Weiss' answer but I couldn't back then check the papers, and it turns out that they don't actually answer my question! So, please refer to the newer version of the question for additional information.	https://mathoverflow.net/users/25	Analytic density of the set of primes starting with 1	I think instead of posting my own explanation (which will only lose something in the translation) I'll instead refer you to two very interesting papers (thanks for posting this question, I haven't thought about this stuff in a couple years, and these papers were interesting reads to solve your problem.) The first (among other things) proves that the density of primes with leading coefficient $k$ is $\log\_{10}\left(\frac{k +1}{k}\right).$ [Prime numbers and the first digit phenomenon](https://doi.org/10.1016/0022-314X(84)90061-1) by Daniel I. A. Cohen\* and Talbot M. Katz in Journal of number theory 18, 261-268 (1984) The second is a more general statement about first digits. It is [The first digit problem](https://doi.org/10.1080/00029890.1976.11994162) by Ralph Raimi in American Math Monthly vol 83 No 7 Hope this all helps.	13	https://mathoverflow.net/users/2043	9145	6,255
https://mathoverflow.net/questions/8800	167	K-theory sits in an intersection of a whole bunch of different fields, which has resulted in a huge variety of proof techniques for its basic results. For instance, here's a scattering of proofs of the Bott periodicity theorem for topological complex K-theory that I've found in the literature: * Bott's original proof used Morse theory, which reappeared in Milnor's book Morse Theory in a much less condensed form. * Pressley and Segal managed to produce the homotopy inverse of the usual Bott map as a corollary in their book Loop Groups. * Behrens recently produced a novel proof based on Aguilar and Prieto, which shows that various relevant maps are quasifibrations, therefore inducing the right maps on homotopy and resulting in Bott periodicity. * Snaith showed that $BU$ is homotopy equivalent to $CP^\infty$ once you adjoin an invertible element. (He and Gepner also recently showed that this works in the motivic setting too, though this other proof relies on the reader having already seen Bott periodicity for motivic complex K-theory.) * Atiyah, Bott, and Shapiro in their seminal paper titled Clifford Modules produced an algebraic proof of the periodicity theorem. EDIT: Whoops x2! They proved the periodicity of the Grothendieck group of Clifford modules, as cdouglas points out, then used topological periodicity to connect back up with $BU$. Wood later gave a more general discussion of this in Banach algebras and Bott periodicity. * Atiyah and Bott produced a proof using elementary methods, which boils down to thinking hard about matrix arithmetic and clutching functions. Variations on this have been reproduced in lots of books, e.g., Switzer's Algebraic Topology: Homotopy and Homology. * A proof of the periodicity theorem also appears in Atiyah's book K-Theory, which makes use of some basic facts about Fredholm operators. A differently flavored proof that also rests on Fredholm operators appears in Atiyah's paper Algebraic topology and operations on Hilbert space. * Atiyah wrote a paper titled Bott Periodicity and the Index of Elliptic Operators that uses his index theorem; this one is particularly nice, since it additionally specifies a fairly minimal set of conditions for a map to be the inverse of the Bott map. * Seminaire Cartan in the winter of '59-'60 produced a proof of the periodicity theorem using "only standard techniques from homotopy theory," which I haven't looked into too deeply, but I know it's around. Now, for my question: the proofs of the periodicity theorem that make use of index theory are in some vague sense appealing to the existence of various Thom isomorphisms. It seems reasonable to expect that one could produce a proof of Bott periodicity that explicitly makes use of the facts that: 1. The Thom space of the tautological line bundle over $CP^n$ is homeomorphic to $CP^{n+1}$. 2. Taking a colimit, the Thom space of the tautological line bundle over $CP^\infty$ (call it $L$) is homeomorphic to $CP^\infty$. 3. The Thom space of the difference bundle $(L - 1)$ over $CP^\infty$ is, stably, $\Sigma^{-2} CP^\infty$. This seems to me like a route to producing a representative of the Bott map. Ideally, it would even have good enough properties to produce another proof of the periodicity theorem. But I can't find anything about this in the literature. Any ideas on how to squeeze a proof out of this -- or, better yet, any ideas about where I can find someone who's already done the squeezing? Hope ~~this isn't~~ less of this is nonsense! -- edit -- Given the positive response but lack of answers, I thought I ought to broaden the question a bit to start discussion. What I was originally looking for was a moral proof of the periodicity theorem -- something short that I could show to someone with a little knowledge of stable homotopy as why we should expect the whole thing to be true. The proofs labeled as elementary contained too much matrix algebra to fit into parlor talk, while the proofs with Fredholm operators didn't seem -- uh -- homotopy-y enough. While this business with Thom spaces over $CP^\infty$ seemed like a good place to look, I knew it probably wasn't the only place. In light of Lawson's response, now I'm sure it isn't the only place! So: does anyone have a good Bott periodicity punchline, aimed at a homotopy theorist? (Note: I'll probably reserve the accepted answer flag for something addressing the original question.)	https://mathoverflow.net/users/1094	Proofs of Bott periodicity	Here is my attempt to address Eric's actual question. Given a real $n$-dimensional vector bundle $E$ on a space $X$, there is an associated Thom space that can be understood as a twisted $n$-fold suspension $\Sigma^E X$. (If $E$ is trivial then it is a usual $n$-fold suspension $\Sigma^n X$.) In particular, if $E=L$ is a complex line bundle, it is a twisted double suspension. In particular, if $X = \mathbb{C}P^\infty$, the twisted double suspension of the tautological line bundle $L$ satisfies the equation $$\Sigma^L\mathbb{C}P^\infty = \mathbb{C}P^\infty.$$ As I understand it, Eric wants to know whether this periodicity can be interpreted as a Bott map, maybe after some modification, and then used to prove Bott periodicity. What I am saying matches Eric's steps 1 and 2. Step 3 is a modification to make the map look more like Bott periodicity. I think that the answer is a qualified no. On the face of it, Eric's map does not carry the same information as the Bott map. Bott periodicity is a theorem about unitary groups and their classifying spaces. What Eric has in mind, as I understand now, is [a result of Snaith](http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=2083072) that constructs a spectrum equivalent to the Bott spectrum for complex K-theory by modifying $\mathbb{C}P^\infty$. Snaith's model has been called "Snaith periodicity", but the existing arguments that it is the same are a use and not a proof of Bott periodicity. (In that sense, Snaith's model is [stone soup](http://en.wikipedia.org/wiki/Stone_soup), although that metaphor is not really fair to his good paper.) For context, here is a quick definition of Bott's beautiful map as Bott constructed it [in the Annals](http://www.jstor.org/stable/1970106) is beautiful. In my opinion, it doesn't particularly need simplification. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact [symmetric space](http://en.wikipedia.org/wiki/Symmetric_space) with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. For complex K-theory, Bott uses $M = U(2n)$, $p = q = I\_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I\_n \oplus \exp(i t) I\_n$, with $0 \le t \le 2\pi$. The map is then $$\Sigma (U(2n)/U(n)^2) \to U(2n).$$ The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group. (If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi\_2(G)$ is trivial; c.f. [this related MO question](https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups).) At first glance, Eric's twisted suspension is very different. It exists for $\mathbb{C}P^\infty = BU(1)$, and of course $\mathbb{C}P^\infty$ is a $K(\mathbb{Z},2)$ space with a totally different homotopy structure from $BU(\infty)$. Moreover, twisted suspensions aren't adjoint to ordinary delooping. Instead, the space of maps $\Sigma^L X \to Y$ is adjoint to sections of a bundle over $X$ with fiber $\mathcal{L}^2 Y$. The homotopy structure of the twisted suspension depends on the choice of $L$. For instance, if $X = S^2$ and $L$ is trivial, then $\Sigma^L S^2 = S^4$ is the usual suspension. But if $L$ has Chern number 1, then $\Sigma^L S^2 = \mathbb{C}P^2$, as Eric computed. However, in Snaith's paper all of that gets washed away by taking infinitely many suspensions to form $\Sigma\_+^{\infty}\mathbb{C}P^\infty$, and then as Eric says adjoining an inverse to a Bott element $\beta$. (I think that the "+" subscript just denotes adding a disjoint base point.) You can see what is coming just from the rational homotopy groups of $\Sigma^\infty \mathbb{C}P^\infty$. Serre proved that the stable homotopy of a CW complex $K$ are just the rational homology $H\_\*(K,\mathbb{Q})$. (This is related to the theorem that stable homotopy groups of spheres are finite.) Moreover, in stable, rational homotopy, twisted and untwisted suspension become the same. So Snaith's model is built from the fact that the homology of $\mathbb{C}P^\infty$ equals the homotopy of $BU(\infty)$. Moreover, there is an important determinant map $$\det:BU(\infty) \to BU(1) = \mathbb{C}P^\infty$$ that takes the direct sum operation for bundles to tensor multiplication of line bundles. Snaith makes a moral inverse to this map (and not just in rational homology). Still, searching for a purely homotopy-theoretic proof of Bott periodicity is like searching for a purely algebraic proof of the [fundamental theorem of algebra](http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra). The fundamental theorem of algebra is not a purely algebraic statement! It is an analytic theorem with an algebraic conclusion, since the complex numbers are defined analytically. The best you can do is a mostly algebraic proof, using some minimal analytic information such as that $\mathbb{R}$ is real-closed using the intermediate value theorem. Likewise, Bott periodicity is not a purely homotopy-theoretic theorem; it is a Lie-theoretic theorem with a homotopy-theoretic conclusion. Likewise, the best you can do is a mostly homotopy-theoretic proof that carefully uses as little Lie theory as possible. The [proof by Bruno Harris](http://dx.doi.org/10.1016/0021-8693(80)90194-5) fits this description. Maybe you could also prove it by reversing Snaith's theorem, but you would still need to explain what facts you use about the unitary groups. (The answer is significantly revised now that I know more about Snaith's result.)	44	https://mathoverflow.net/users/1450	9151	6,259

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