parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/3003 | 26 | Since there is no "free field generated by a set", it would seem that
1) there is no monad on Set whose algebras are exactly the fields
and
2) there is no Lawvere theory whose models in Set are exactly the fields
(Are 1) and 2) correct?)
Fields don't form a variety of algebras in the sense of universal algebra since the field axioms can´t be written as identities (since the axiom for multiplicative inverses has the restriction that the element be non-zero).
I guess fields are an algebraic theory in a more general universal algebra sense of being defined by operations on a single set with a set of first order axioms.
Is there any better sense in which they are algebraic or are fields just not really algebraic in nature?
| https://mathoverflow.net/users/644 | In what sense are fields an algebraic theory? | 1 and 2 are correct, for a simple reason. If *C* is a category satisfying either 1 or 2 then *C* has a terminal object. But there is no terminal object in the category of fields (and ring homomorphisms), because there are no maps between fields of different characteristic.
For the same reason, the category of fields is not an essentially algebraic theory (mentioned in Andrew's answer). An **essentially algebraic theory** can be defined as, simply, a small category with finite limits. A **model** or **algebra** for an essentially algebraic theory *T* is a finite-limit-preserving functor *T* --> **Set**. (Of course, you can consider models in other finite limit categories too.) And the category of models always has a terminal object.
This embodies the idea that Andrew was describing, of a theory where some operations are only partially defined, but (and this is crucial!) the domain of definition is itself defined by equations. You can see some rough connection between finite limits and this intuitive idea if you consider pullbacks in **Set**. A pullback in **Set** is, after all, the set of pairs satisfying some equation.
I don't know in what sense the theory of fields *is* algebraic. It's partly because of its failure to be algebraic in any of the usual senses that one often chooses to work with commutative rings rather than fields, in algebraic geometry and in topos theory, for instance.
| 22 | https://mathoverflow.net/users/586 | 3070 | 2,028 |
https://mathoverflow.net/questions/2989 | 10 | When I studied physics, we learned how to write down planar waves and spherical waves. But, when I turn on my flashlight, I see a cone of light. How can I see that there is a solution to the wave equation which describes a wave in a conical region, dropping off sharply outside that cone? And am I right that the wave equation cannot describe a cylindrical beam?
| https://mathoverflow.net/users/297 | Cone shaped solutions to wave equation | At human scales, the wavelength of visible light is so tiny (or equivalently, the frequency is so high) that the wave equation can be modeled by [geometric optics](http://en.wikipedia.org/wiki/Geometrical_optics) (this is the [high frequency limit](http://en.wikipedia.org/wiki/High_frequency_approximation) of the wave equation). The cone you see from the flashlight is then nothing more than the shadow that the casing of the light casts from the light bulb.
At larger wavelengths, diffractive effects become more pronounced (when was the last time you saw a water wave propagating along a cone?). The asymptotic behaviour of any localised wave disturbance (in odd dimensions) is then an outgoing spherical wave, modulated by a scattering amplitude that depends only on the direction of propagation (and which is basically a kind of Radon transform of the initial data); this can be seen from asymptotics of the fundamental solution. (It's slightly more complicated in even dimensions due to the failure of the sharp Huygens principle.)
| 17 | https://mathoverflow.net/users/766 | 3086 | 2,038 |
https://mathoverflow.net/questions/3078 | 14 | Many texts which praise the generality of the bar construction associated to a monad, say that Hochschild homology is an example of this.
What exactly is in this case the underlying endofunctor of the monad, on which category is it an endofunctor, what are the monad structure maps and, most important (since I think my confusion lies here), why do then the face maps look as on the [wikipedia page](http://en.wikipedia.org/wiki/Hochschild_homology)?
| https://mathoverflow.net/users/733 | How exactly is Hochschild homology a monad homology? | [Tyler, are you sure about this?](https://mathoverflow.net/questions/3078/how-exactly-is-hochschild-homology-a-monad-homology/3085#3085) I thought the bar construction comes from the adjunction between R-modules and k-modules for R a given k-algebra (i.e. relative Tor). Besides, what you say only makes sense if we're taking coefficients in R itself and not a general bimodule M.
If memory serves correctly, starting with a k-algebra R and looking at a simplicial resolution for it via the adjunction k-modules -- k-algebras leads to *cyclic* homology as in the paper of Feigin-Tsygan.
That wiki page also looks off to me: the Loday construction is for the Hochschild homology and decomposition for *commutative* algebras, and this isn't made very clear in the wiki.
**quick edit:** There's a good if terse discussion of the bar construction via monads in Weibel Chapter 8 (p.283). I suspect you could also extract the desired information out of the much more general machinery in [Jon Beck's thesis](http://www.tac.mta.ca/tac/reprints/articles/2/tr2abs.html), modulo some possible struggle with notation.
**encore une fois:** Consider the adjunction between k-mod and R-mod. If M is an object of R-mod then the simplicial construction provided by the adjunction looks like this
M <--- R\otimes M <--- R \otimes R \otimes M <----- etc
where I've not been able to draw in all the face maps, but hopefully you get what I mean. Now by taking the alternating sum of face maps in each degree, we get a split exact sequence of R-module maps
M <--- R\otimes M <--- R \otimes R \otimes M <----- etc
which is a resolution in the classical sense of M in R-mod-R by R-mod projectives - I'm assuming k is a field for sake of convenience. (So you can use it to calculate Tor^R if you wish.)
Now take M=R and note that we have a resolution of R by R^e-projectives. Apply Hom{R^e}(\_\_, X) where X is your coefficient module, and you get *precisely* the Hochschild chain complex as in the original papers.
Of course, we didn't have to take sums of face maps before applying the Hom functor. So, if we start with R regarded as an object of R-mod, the canonical simplicial construction (for M=R) would give us a contractible simplicial object in R-mod with M=R at the bottom, this object would in fact live in R^e-mod, and so is eligible to be hit with HomR^e(\_\_,X). If we do this, we get a simplicial object in k-mod, and said object *should* be the one described in the wiki article, corresponding to the Hochschild chain complex.
| 5 | https://mathoverflow.net/users/763 | 3094 | 2,046 |
https://mathoverflow.net/questions/3059 | 24 | The Hopf fibration is a famous map $S^3\to S^2$ with fiber $S^1$, which is the generator in $\pi\_3(S^2)$. We can model this map in terms simplicial sets by taking the singular simplicial sets of these spaces and the induced map of simplicial sets. But this model is **huge** and isn't really useful for doing calculations. Does anyone know a nice small model for this map in terms of simplicial sets? Something suitable for computations? This map is also the attaching map used to build $\mathbb{C}P^2$ out of $S^2$, so I would equivalently be interested in a small combinatorial model for $\mathbb{C}P^2$.
| https://mathoverflow.net/users/184 | Simplicial model of Hopf map? | There is a paper [[MathSciNet](https://mathscinet.ams.org/mathscinet-getitem?mr=1789059 "Geom. Dedicata 82 (2000), no. 1-3, 105–114. zbMATH review at https://zbmath.org/0965.57021")] of Madahar and Arkaria called *A minimal triangulation of the Hopf map and its application*. They find a triangulation from a 12-vertex 3-sphere to a 4-vertex 2-sphere. The minimality is in Section 6.a. I hope this is useful.
Now, this gives the map the structure of a map of simplicial complexes. Choose an ordering of the vertices such that the map in the paper respects the order. This then gives you a model of the map on finite simplicial sets.
| 15 | https://mathoverflow.net/users/100 | 3096 | 2,048 |
https://mathoverflow.net/questions/871 | 6 | The generalisation of the vanishing cycle formalism in SGA 7 is apparently since the 1970's [an issue](http://www.math.u-psud.fr/~illusie/vanishing1b.pdf "Illusie's article"), Morava [mentioned](http://arxiv.org/abs/math/0509001 "Morava's article") a connection with Bousfield localization. I find the Morava's remarks un-understandable - would someone explain?
| https://mathoverflow.net/users/451 | Higher vanishing cycles | My apologies if this is too much or too little; leave a comment and I can try and correct it. He's talking about a specific issue in homotopy theory that we'd like a better understanding of.
The stable homotopy category (implicitly localized at a prime p) has a stratification into "chromatic" layers, which correspond to a connection to formal group laws. We geometrically think of the stable homotopy category as some kind of category of sheaves on a moduli stack X which has a sequence of open substacks X(n) - these are the "E(n)-local categories", and there are Bousfield localization functors taking a general element M to its E(n)-localization LE(n) M, which you can think of as restricting to the open substack. (A general Bousfield localization will take some notion of "equivalence" and construct a universal new category where those equivalences become isomorphisms, but in an appropriately derived way.)
The difference X(n) \ X(n-1) between two adjacent layers is a closed substack of X(n), which in our language is the "K(n)-local category". There is also a Bousfield localization functor that takes an element M to its K(n)-localization LK(n) M. Bousfield localization is pretty general machinery and in the previous "open" situation it acted as restriction; in this "closed" situation it acts as a completion along the closed substack.
We have some general understanding of the K(n)-local categories. They act a lot like some kind of quotient stack of some Lubin-Tate space classifying deformations of a height n formal group law by the group scheme of automorphisms of said formal group law, which is the n'th Morava stabilizer group Sn. Geometrically we think about it as a point with a fairly large automorphism group (even though this is, of course, [the wrong way to think about things](https://mathoverflow.net/questions/1467/are-curves-with-fractional-points-uniquely-determined-by-their-residual-gerbes)). These are places where you can get dirty and do specific computations and examine one chromatic layer at a time.
There are two remaining pieces of data we need, then, to understand M itself from its localizations LK(n) M: we need to understand how they are patched together into the E(n)-localizations, and we need to understand the limit of the LE(n)M. The latter is a "chromatic convergence" question and not immediately relevant to the point under discussion.
In general there is a "patching" diagram, which is roughly something like the data you'd usually associate with a recollement. (My favorite reference for data in this kind of situation is Mazur's "Notes on etale cohomology of number fields".) We have a (homotopy) pullback diagram
```
LE(n) M -> LE(n-1) M
| |
V V
LK(n) M -> LE(n-1) LK(n) M
```
that tells us that a general E(n)-local object is reconstructed from a K(n)-local object (something concentrated on the closed stack), an E(n-1)-local object (concentrated on the open stack), and patching data (a map from the object on the open stack to the restriction of the complete object to the open stack). This roughly follows because the K(n)-localization of any E(n-1)-local object is trivial.
The functor that takes an object concentrated near the closed substack and restricts it (in a derived way) to the open substack is what Morava considers. Here, in the language of Bousfield localization, it is E(n-1)-localization applied to K(n)-local objects. What he seems to be proposing is that this general Bousfield localization setup should be one way of thinking about the vanishing cycles functors (and I concur with his dislike for the "vanishing" terminology) in which we can, in a fully derived way, view sheaves on a large stack as coming from patching data on an open-closed pair.
Just to close the loop, what we don't really understand at all in this picture is what this "trans-chromatic-layer" stuff really does. We have, for example, two stabilizer groups connected to formal group laws of adjacent heights, and we don't really understand what the specialization functor is really doing in this case.
| 8 | https://mathoverflow.net/users/360 | 3101 | 2,053 |
https://mathoverflow.net/questions/3107 | 8 | The well-known Hopf fibration $S^1 \rightarrow S^3 \rightarrow S^2$ has explicit constructions involving the geometry of $C^2$ and intersections of complex lines with the $3$-sphere. They don't seem to generalize easily to "higher" Hopf maps from $S^3 \rightarrow S^2$ with Hopf invariant not equal to one. Are there any simple expressions for those maps?
| https://mathoverflow.net/users/353 | Construction of maps $f:S^3 \to S^2$ with arbitrary Hopf invariant? | You can get them by precomposing with a degree $n$ map from $S^3$ to itself. In particular, this gives an interpretation in terms of the group structure: if $h:S^3 \to S^2$ is the Hopf map (which is just modding out by the subgroup $S^1=U(1)$ of $S^3=Sp(1)$, then a map of Hopf invariant n is given by $x \mapsto h(x^n)$, where $x^n$ is using the group multiplication on $S^3$.
| 14 | https://mathoverflow.net/users/75 | 3126 | 2,070 |
https://mathoverflow.net/questions/3119 | 11 | It seems to be a well-known fact that there is a "one-to-one correspondence'' between prestacks and fibered categories. Here a prestack (called a pseudo-functor in SGA1) means a contravariant lax functor $F$ on a small category taking values in the $2$-category of small categories in which the structure natural transformation $F(f)\circ F(g)\Rightarrow F(gof)$ is invertible.
For example, Vistoli says in [this note](http://arxiv.org/abs/math/0412512) that "the theory of fibered categories is equivalent to the theory of pseudo-functors" at the end of section $3.1$.
Is this "equivalence" an equivalence of 2-categories? If so, where can I find a proof?
| https://mathoverflow.net/users/377 | Prestacks and fibered categories | I don't have a reference right now, but I hope this answer is useful. If nothing else, perhaps you could comment on why this *doesn't* answer your question.
A pseudofunctor is exactly the same thing as a fibered category *with a choice of cleavage* (a cleavage is a choice of cartesian arrow over every morphism in the base category with given target in the fiber). That is, there is an *isomorphism* between the (2-)category of pseudofunctors and the (2-)category of fibered categories with cleavage (where the morphisms don't have to respect the cleavage).
By the axiom of choice, every fibered category has a cleavage, and any two choices of cleavage are canonically isomorphic (via the identity functor; remember that the functor need not respect the cleavage). So the category of fibered categories with cleavage is equivalent to the category of fibered categories, and this is an equivalence in the usual 1-categorical sense. That is, you have two functors (the forget-cleavage and choose-cleavage functors) whose compositions are naturally isomorphic to the the identity. I don't think you need to use any kind of 3-morphism even though you're dealing with 2-categories.
| 9 | https://mathoverflow.net/users/1 | 3132 | 2,075 |
https://mathoverflow.net/questions/3134 | 35 | Certain formulas I really enjoy looking at like the [Euler-Maclaurin formula](https://en.wikipedia.org/wiki/Euler_Maclaurin) or the [Leibniz integral rule](https://en.wikipedia.org/wiki/Leibniz_integral_rule). What's your favorite equation, formula, identity or inequality?
| https://mathoverflow.net/users/812 | What's your favorite equation, formula, identity or inequality? | $e^{\pi i} + 1 = 0$
| 42 | https://mathoverflow.net/users/262 | 3137 | 2,078 |
https://mathoverflow.net/questions/1464 | 24 | The matrix norm for an $n$-by-$n$ matrix $A$ is defined as $$|A| := \max\_{|x|=1} |Ax|$$ where the vector norm is the usual Euclidean one. This is also called the induced (matrix) norm, the operator norm, or the spectral norm. The unit ball of matrices under this norm can be considered as a subset of $\Bbb R^{n^2}$. What is the Euclidean volume of this set?
I'd be interested in the answer even in just the $2$-by-$2$ case.
| https://mathoverflow.net/users/814 | Euclidean volume of the unit ball of matrices under the matrix norm | Building on the nice answer of Guillaume: The integral
$$ \int\_{[-1,1]^n} \prod\_{i < j} \left| x\_i^2 - x\_j^2 \right| \, dx\_1 \dots dx\_n $$
has the closed-form evaluation
$$ 4^n \prod\_{k \leq n} \binom{2k}{k}^{-1}.$$
This basically follows from the evaluation of the [Selberg beta integral](http://en.wikipedia.org/wiki/Selberg_integral) Sn(1/2,1,1/2).
Combined with modding out by a typo, we now arrive at the following product formula for the volume of the unit ball of nxn matrices in the matrix norm:
$$ n! \prod\_{k\leq n} \frac{ \pi^k }{ ((k/2)! \binom{2k}{k})} .$$
In particular, we have:
* 2/3 π2 for n=2
* 8/45 π4 for n=3
* 4/1575 π8 for n=4
| 15 | https://mathoverflow.net/users/359 | 3151 | 2,089 |
https://mathoverflow.net/questions/3150 | 16 | The Gelfand transform gives an equivalence of categories from the category of unital, commutative $C^\*$-algebras with unital $\*$-homomorphisms to the category of compact Hausdorff spaces with continuous maps. Hence the study of $C^\*$-algebras is sometimes referred to as non-commutative topology.
All diffuse commutative von Neumann algebras acting on separable Hilbert space are isomorphic to $L^\infty[0,1]$. Hence the study of von Neumann algebras is sometimes referred to as non-commutative measure theory.
Connes proposed that the definition of a non-commutative manifold is a spectral triple $(A,H,D)$. From a $C^\*$-algebra, we can recover the "differentiable elements" as those elements of the $C^\*$-algebra $A$ that have bounded commutator with the Dirac operator $D$.
What happens if we start with a von Neumann algebra? Does the same definition give a "differentiable" structure? Is there a way of recovering a $C^\*$-algebra from a von Neumann algebra that contains the "differentiable" structure on our non commutative measure space? This would be akin to our von Neumann algebra being $L^\infty(M)$ for $M$ a compact, orientable manifold (so we have a volume form). Or are von Neumann algebras just "too big" for this?
One of the reasons I am asking this question is Connes' spectral characterization of manifolds ([arXiv:0810.2088v1](http://arxiv.org/abs/0810.2088v1)) which shows we get a "Gelfand theory" for Riemannian manifolds if the spectral triples satisfy certain axioms. Connes starts with the von Neumann algebra $L^\infty(M)$ instead of the $C^\*$-algebra $C(M)$.
| https://mathoverflow.net/users/351 | Non-commutative geometry from von Neumann algebras? | You definitely need some extra structure on your von Neumann algebra, but I'm not quite sure what you're asking for.
Intuitively I would think that just as different topological spaces share the same measure space structure, trying to extract NC-topological information out of a von Neumann algebra is going to need extra structure. (For instance, no one does topological K-theory of von Neumann algebras as far as I know.)
I see that on page 7 of that Connes paper, he shows that the WOT-closure A'' does remember the original algebra A *if* extra data are given (the Dirac operator and its interaction with A).
Although it's probably not what you want: if you're looking at group von Neumann algebras and looking at the "geometry of the dual group", then the original group can be recovered from a suitable coproduct on the original (Hopf) von Neumann algebra. This is vaguely on the lines of Weil's theorem that "essentially" recovers a locally compact group and its Haar measure from a measurable group.
| 9 | https://mathoverflow.net/users/763 | 3152 | 2,090 |
https://mathoverflow.net/questions/2314 | 44 | Hey. I have a few off the wall questions about topos theory and algebraic geometry.
1. Do the following few sentences make sense?
Every scheme X is pinned down by its Hom functor Hom(-,X) by the yoneda lemma, but since schemes are locally affine varieties, it is actually just enough to look at the case where "-" is an affine scheme. So you could define schemes as particular functors from CommRing^op to Sets. In this setting schemes are thought of as sheaves on the "big zariski site".
If that doesn't make sense my next questions probably do not either.
2 The category of sheaves on the big zariski site forms a topos T, the category of schemes being a subcategory. It is convenient to reason about toposes in their own "internal logic". Has there been much thought done about the internal logic of T, or would the logic of T require too much commutative algebra to feel like logic? Along these lines, have there been attempts to write down an elementary list of axioms which capture the essense of this topos? I am thinking of how Anders Kock has some really nice ways to think of differential geometry with his SDG.
3 What is it about the category of commutative rings which makes it possible to put such a nice site structure on it, but not other algebraic categories? Gluing rings together lead to huge advancements in algebraic geometry. What about gluing groups? Is there a nice Grothendieck topology you could put on Groups^op, and then you could start studying sheaves on this site? If not, why not - what about rings makes them so special?
4 Why do people work with the category of schemes instead of the topos of sheaves on CommRing^op - toposes have every nice categorical property you could possibly ask for.
About me: I am a 1st year grad student who is taking a first course on schemes, and I just have a lot of crazy ideas floating around. I don't feel comfortable engaging in such wild speculation with my professors. Could you offer any insight into these ideas?
| https://mathoverflow.net/users/1106 | Several Topos theory questions | About 1: Yes!
About 2: (Internal logic of Zariski topos) I don't think it has been done systematically. A glimpse of it is in Anders Kock, Universal projective geometry via topos theory, if I remember well, and certainly in some other places. But one point is that it is not at all easy to find formulas in the internal language which express what you have in mind. See my answer at ["synthetic" reasoning applied to algebraic geometry](https://mathoverflow.net/questions/606/synthetic-reasoning-applied-to-algebraic-geometry)
About 3:You can indeed glue all sorts of things:
* Things fitting into the axiomatic framework of "geometric contexts":
Look at the "master course on Algebraic stacks" here: <http://perso.math.univ-toulouse.fr/btoen/videos-lecture-notes-etc/>
This one is great reading to understand the functorial point of view on schemes and manifolds!
* Commutative Monoid objects in good monoidal (model) categories: <http://arxiv.org/abs/math/0509684>
* Commutative monads (here you can glue monoids, semirings and other algebraic structures mixing them all): <http://arxiv.org/abs/0704.2030>
* In Shai Haran's "Non-Additive Geometry" you can even glue the monoids and semirings etc. with relations (although I wouldn't know why)
* You can also glue things "up to homotopy instead" of strictly - this is roughly what Lurie's infinity-topoi are about, and also the model catgeory part of the 2nd point, or any oter approaches to derived algebraic geometry
* (Edit in 2017) The [PhD thesis by Zhen Lin Low](https://www.repository.cam.ac.uk/handle/1810/256998) is relevant. "The main purpose of this thesis is to give a unified account of this procedure of constructing a category of spaces built from local models and to study the general properties of such categories of spaces."
One of several good points of view on what a Grothendieck topology does, is to say it determines which colimits existing in your site should be preserved under the Yoneda embedding, i.e. what glueing takes already place among the affine objects. So, if you insist on glueing groups it could be a good idea to look e.g. for a topology which takes amalgamated products (for me this means glueing groups, you may want only selected such products, e.g. along injective maps) to pushouts of sheaves...
Then feel free to develop a theory on this and send me a copy!
About 4: (Why don't people work with sheaves instead of schemes) They do. One situation where they do is when taking the quotient of a scheme by a group action. The coequalizer in the category of schemes is often too degenerate. One answer is taking the coequalizer in the category of sheaves, the "sheaf quotient" (but sometimes better answers are GIT quotients and stack quotients).
| 25 | https://mathoverflow.net/users/733 | 3155 | 2,092 |
https://mathoverflow.net/questions/3154 | 27 | I'm trying to learn a bit about equivariant homotopy theory. Let G be a compact Lie group. I guess there is a cofibrantly generated model category whose objects are (compactly generated weak Hausdorff or whatever) topological spaces with G-action and whose morphisms are G-maps, in which the generating cofibrations are maps of the form G/H x Sn-1 → G/H x Dn (n ≥ 0, H a closed subset of G) and the generating acyclic cofibrations are the obvious analogous thing. Apparently the weak equivalences in this category are those maps which induce weak equivalence on H-fixed points for every closed subgroup H of G. I assume the corresponding (∞,1)-category is presentable. (My preliminary question is, does anyone know a good source for this paragraph?)
My real question is: Can you give an (∞,1)-categorical description of this category, say via a universal property, or built somehow from the category of spaces? For instance, what is an explicit presentation as a localization of a category of presheaves of spaces? (An example of the kind of answer I am looking for is "functors from BG to Spaces", but that describes a model category of G-spaces whose weak equivalences are simply weak equivalences of the underlying spaces.)
(My next question would be asking for an analogous description of the equivariant stable homotopy category. I imagine this would be easy if I knew how to answer the first question, but if something special happens in the stable situation, I would like to know about it.)
| https://mathoverflow.net/users/126667 | (∞, 1)-categorical description of equivariant homotopy theory | I think a good reference for the first paragraph is "Equivariant Homotopy and Cohomology Theory" by Peter May and a bunch of other people. Chapter 5 includes "Elmendorf's theorem" that this homotopy theory of G-spaces is equivalent to the homotopy theory of diagrams of spaces on the orbit category O(G) of G. In the latter homotopy theory, the weak equivalences are "levelwise" as is usual in the homotopy theory of diagrams.
I'm less sure about the (∞,1)-categorical versions, but I would expect that the (∞,1)-category associated to a levelwise model structure on O(G)-diagrams will be essentially the (∞,1)-category of functors from O(G) to the (∞,1)-category of spaces. That ought to imply that it is locally presentable as well.
One might guess that the equivariant stable homotopy category would be the "stabilization" of this (∞,1)-category, but that's not entirely obvious to me. The point at issue is that there are two kinds of G-spectra: "naive" G-spectra, which are indexed on integers, and "true" G-spectra, which are indexed on G-representations. It seems possible to me that the standard "stabilization" process of an (∞,1)-category will only stabilize with respect to integers.
| 15 | https://mathoverflow.net/users/49 | 3162 | 2,094 |
https://mathoverflow.net/questions/3165 | 16 | If I remember correctly, I read that given a presheaf $P:\mathcal{C}^{op} \to Set$, it is possible to describe it as a limit of representable presheaves. Could someone give a description of the construction together with a proof?
| https://mathoverflow.net/users/1261 | Presheaves as limits of representable functors? | You mean "colimit of representable presheaves", not limit. Any limits that C has are preserved by the Yoneda embedding. So if C is, say, a complete poset like • → •, so that it is small and has all limits, you won't be able to produce any non-representable presheaves by taking limits of representable ones.
The way to write any presheaf as a colimit of representables is, like all things Yoneda-related, somewhat tautological, and should be worked out for oneself; but anyways it's explained to some extent at [this nlab page](http://ncatlab.org/nlab/show/presheaf#properties_of_presheaves_4). Rather than write out formulas, I usually think of the example of simplicial sets: every simplicial set X can be formed as a colimit of its simplices, i.e., a diagram of representables which is indexed on the "category of simplices of X", whose objects are pairs (n, x) where n is in the indexing category and x is an object of Xn. The same works in any presheaf category.
| 25 | https://mathoverflow.net/users/126667 | 3170 | 2,099 |
https://mathoverflow.net/questions/3124 | 14 | Suppose G is an algebraic group with an action G×X→X on a scheme. Then many of the usual constructions you make when you talk about group actions on sets can be made scheme-theoretically. For example, if x∈X is a point (thought of as a map x:∗→X, where ∗ is Spec of a field or the base scheme), then the stabilizer Stab(x) is naturally a scheme because it is the fiber product
```
Stab(x) ----> G×X (g,y)
| | _
| | |
v (x,x) v v
∗ --------> X×X (gy,y)
```
1. Does the orbit of a point have a natural scheme structure?
2. Does the fixed locus (the set of points x∈X fixed by all of G) have a scheme structure?
For (1), if everything is sufficiently nice, then the morphism G×∗→X, given by g→g⋅x has a scheme-theortic closed image, and the actual image is constructible and invariant under the G-action, so the actual image is an open subset of its closure. Thus, the orbit gets the structure of an open subscheme of a closed subscheme of X. But this construction doesn't feel very natural.
For (2), you can obviously define the functor Fix(T)={t∈X(T)|t is fixed by every element of G(T)}. Is this functor always representable?
**Edit:** Given that Scott has given such an excellent (negative) answer to question (1) but not said anything about question (2), I've asked (2) as [a separate question](https://mathoverflow.net/questions/3190/is-the-fixed-locus-of-a-group-action-always-a-scheme).
| https://mathoverflow.net/users/1 | Do orbits and stable loci of group actions have natural scheme structures? | Question number 1: Let your base S be Spec k[x] (say k is an algebraically closed field), let X be Spec k[x,y], and let G be **G**a,S, with action over the point s given by gs(xs) = (sgs) + xs. This action is transitive away from zero, so the orbit of the zero section is a plane with a slit. This is not an open subscheme of a closed subscheme of X, because it is not a scheme.
| 8 | https://mathoverflow.net/users/121 | 3181 | 2,110 |
https://mathoverflow.net/questions/3184 | 145 | The Yoneda Lemma is a simple result of category theory, and its proof is very straightforward.
Yet I feel like I do not truly understand what it is about; I have seen a few comments here mentioning how it has deeper implications into how to think about representable functors.
What are some examples of this? How should one think of the Yoneda Lemma?
| https://mathoverflow.net/users/362 | "Philosophical" meaning of the Yoneda Lemma | One way to look at it is this:
for $C$ a category, one wants to look at presheaves on $C$ as being "generalized objects modeled on $C$" in the sense that these are objects that have a sensible rule for how to map objects of $C$ into them. You can "probe" them by test objects in $C$.
For that interpretation to be consistent, it must be true that some $X$ in $C$ regarded as just an object of $C$ or regarded as a generalized object is the same thing. Otherwise it is inconsistent to say that presheaves on $C$ are generalized objects on $C$.
The Yoneda lemma ensures precisely that this is the case.
I wrote up a more detailed expository version of this story at [motivation for sheaves, cohomology and higher stacks](http://ncatlab.org/nlab/show/motivation+for+sheaves%2C+cohomology+and+higher+stacks).
| 101 | https://mathoverflow.net/users/381 | 3185 | 2,112 |
https://mathoverflow.net/questions/2506 | 9 | This is related to Noah's recent [question](https://mathoverflow.net/questions/2396/solving-polynomial-equations-when-you-know-in-which-number-field-the-solutions-li) about solving quadratics in a number field, but about an even earlier and easier step.
Suppose I have a huge system of linear equations, say ~10^6 equations in ~10^4 variables, and I have some external knowledge that suggests there's a small solution space, ~100 dimensional. Moreover, the equations are sparse; in fact, the way I produce the equations gives me an upper bound on the number of variables appearing in each equation, ~10. (These numbers all come form the latest instance of our problem, but we expect to want to try even bigger things later.) Finally, all the coefficients are in some number field.
Which computer algebra system should I be using to solve such a system? Everyone knows their favourite CAS, but it's often hard to get useful comparisons. One significant difficulty here is that even writing down all the equations occupies a big fraction of a typical computer's available RAM.
I'll admit that so far I've only tried Mathematica; it's great for most of our purposes, but I'm well aware of its shortcomings, hence this question. A previous slightly smaller instance of our problem was within Mathematica's range, but now I'm having trouble.
(For background, this problem is simply finding the "low weight spaces" in a graph planar algebra. See for example [Emily Peter's thesis](http://euclid.unh.edu/~eep/A_planar_algebra_construction_of_the_Haagerup_subfactor_3August2009.pdf) for an explanation, or our [follow-up paper](http://tqft.net/EH/), with Noah Snyder and Stephen Bigelow.)
| https://mathoverflow.net/users/3 | Which computer algebra system should I be using to solve large systems of sparse linear equations over a number field? | It probably goes without saying that solving linear systems over number fields is probably far from the being among the most important user-level functionality of the main commercial computer algebra systems. That said, I do know that this functionality in Maple was written by someone with a specific interest in this sort of thing. If you have access to a recent version of Maple, take a look at the help page: ?SolveTools,Linear. 10^6 is pretty big, but it might still be within reach of the solver.
While, I do not know much about how Mathematica does these things, I do know that in Maple sparse linear systems are more efficiently solved as polynomials (rather than sparse-matrices) since underlying polynomial data-structure turns out to be well suited to sparse system solving.
If Maple does not work for you (or you do not have access to it), this strikes me as exactly the sort of problem that [MAGMA](http://magma.maths.usyd.edu.au/magma/) might be targeting.
| 3 | https://mathoverflow.net/users/858 | 3186 | 2,113 |
https://mathoverflow.net/questions/570 | 29 | For an algebraic group G and a representation V, I think it's a standard result (but I don't have a reference) that
* the obstruction to deforming V as a representation of G is an element of H2(G,V⊗V\*)
* if the obstruction is zero, isomorphism classes of deformations are parameterized by H1(G,V⊗V\*)
* automorphisms of a given deformation (as a deformation of V; i.e. restricting to the identity modulo your square-zero ideal) are parameterized by H0(G,V⊗V\*)
where the Hi refer to standard group cohomology (derived functors of invariants). The analogous statement, where the algebraic group G is replaced by a Lie algebra *g* and group cohomology is replaced by Lie algebra cohomology, is true, but the only proof I know is a big calculation. I started running the calculation for the case of an algebraic group, and it looks like it works, but it's a mess. Surely there's a long exact sequence out there, or some homological algebra cleverness, that proves this result cleanly. **Does anybody know how to do this, or have a reference for these results?** This feels like an application of cotangent complex ninjitsu, but I guess that's true about all deformation problems.
While I'm at it, I'd also like to prove that the obstruction, isoclass, and automorphism spaces of deformations of G *as a group* are H3(G,Ad), H2(G,Ad), and H1(G,Ad), respectively. Again, I can prove the Lie algebra analogues of these results by an unenlightening calculation.
Background: What's a deformation? Why do I care?
------------------------------------------------
I may as well explain exactly what I mean by "a deformation" and why I care about them. Last things first, why do I care? The idea is to study the moduli space of representations, which essentially means understanding how representations of a group behave *in families*. That is, given a representation V of G, what possible representations could appear "nearby" in a family of representations parameterized by, say, a curve? The appropriate formalization of "nearby" is to consider families over a local ring. If you're thinking of a representation as a matrix for every element of the group, you should imagine that I want to replace every matrix entry (which is a number) by a power series whose constant term is the original entry, in such a way that the matrices still compose correctly. It's useful to look "even more locally" by considering families over *complete* local rings (think: now I just take formal power series, ignoring convergence issues). This is a limit of families over Artin rings (think: truncated power series, where I set xn=0 for large enough n).
So here's what I mean precisely. Suppose A and A' are Artin rings, where A' is a square-zero extension of A (i.e. we're given a surjection f:A'→A such that I:=ker(f) is a square-zero ideal in A'). A representation of G over A is a free module V over A together with an action of G. A deformation of V to A' is a free module V' over A' with an action of G so that when I reduce V' modulo I (tensor with A over A'), I get V (with the action I had before). An automorphism of a deformation V' of V *as a deformation* is an automorphism V'→V' whose reduction modulo I is the identity map on V. The "obstruction to deforming" V is something somewhere which is zero if and only if a deformation exists.
I should add that the obstruction, isoclass, and automorphism spaces will of course depend on the ideal I. They should really be cohomology groups with coefficients in V⊗V\*⊗I, but I think it's normal to omit the I in casual conversation.
| https://mathoverflow.net/users/1 | Deformation theory of representations of an algebraic group | A representation of G on a vector space V is a descent datum for V, viewed as a vector bundle over a point, to BG. That is, linear representations of G are "the same" as vector bundles on BG. So the question is equivalent to the analogous question about deformations of vector bundles on BG. We could just as easily ask about deformations of vector bundles on any space X.
Given a vector bundle V on X, consider the category of all first-order deformations of V. An object is a vector bundle over X', where X' is an infinitesimal thickening (in the example, one may take X = BG x E where E is a local Artin ring and X' = BG x E' where E' is a square-zero extension whose ideal is isomorphic as a module to the residue field). A morphism is a morphism of vector bundles on X' that induces the identity morphism on V over X.
If X is allowed to vary, this category varies contravariantly with X. Vector bundles satisfy fppf descent, so this forms a fppf stack over X.
This stack is very special: locally it has a section (fppf locally a deformation exists) and any two sections are locally isomorphic. It is therefore a *gerbe*. Moreover, the isomorphism group between any two deformations of V is *canonically* a torsor under the group End(V) (this is fun to check).
Gerbes banded by an abelian group H are classified by H^2(X,H) (this is also fun to check); the class is zero if and only if the gerbe has a section. If the gerbe has a section, the isomorphism classes of sections form a torsor under H^1(X,H). The isomorphisms between any two sections form a torsor under H^0(X,H). (This implies that the automorphism group of any section is H^0(X,H).)
In our case, H = End(V), so we obtain a class in H^2(X,End(V)) and if this class is zero, our gerbe has a section, i.e., a deformation exists. In this case, all deformations form a torsor under H^1(X,End(V)), and the automorphism group of a deformation is H^0(X,End(V)).
All of the cohomology groups above are sheaf cohomology in the fppf topology. If you are using a different definition of group cohomology, there is still something to check.
| 27 | https://mathoverflow.net/users/32 | 3187 | 2,114 |
https://mathoverflow.net/questions/3204 | 55 | It's sort of folklore (as exemplified by [this old post](http://cornellmath.wordpress.com/2007/08/02/sum-divergent-series-iii/) at The Everything Seminar) that none of the common techniques for summing divergent series work to give a meaningful value to the harmonic series, and it's also sort of folklore (although I can't remember where I heard this) that the harmonic series is more or less the only important series with this property.
What other methods besides analytic continuation and zeta regularization exist for summing divergent series? Do they work on the harmonic series? And are there other well-known series which also don't have obvious regularizations?
| https://mathoverflow.net/users/290 | Does any method of summing divergent series work on the harmonic series? | One common regularization method that wasn't mentioned in the Everything Seminar post is to take the constant term of a meromorphic continuation. While the [Riemann zeta function](https://en.wikipedia.org/wiki/Riemann_zeta_function) has a simple pole at 1, the constant term of the Laurent series expansion is the Euler-Mascheroni constant gamma = 0.5772156649...
It is reasonable to claim that most divergent series don't have interesting or natural regularizations, but you could also reasonably claim that most divergent series aren't interesting. Any function with extremely rapid growth (e.g., the Busy Beaver function) is unlikely to have a sum that is regularizable in a natural way.
| 23 | https://mathoverflow.net/users/121 | 3211 | 2,130 |
https://mathoverflow.net/questions/3104 | 16 | Caveat: I don't really know anything about PDEs, so this question might not make sense.
In complex analysis class we've been learning about the solution to Dirichlet's problem for the Laplace equation on bounded domains with nice (smooth) boundary. My sketchy understanding of the history of this problem (gleaned from Wikipedia) is that in the 19th century everybody "knew" that the problem had to have a unique solution, because of physics. Specifically, if I give you a distribution of charge along the boundary, it has to determine an electric potential in the domain, which turns out to be harmonic. But Dirichlet's proof was wrong, and it wasn't until around 1900 that Hilbert found a correct argument for the existence and uniqueness of the solution, given reasonable conditions (the boundary function must be continuous, and the boundary really has to be sufficiently smooth).
Is the physical heuristic really totally meaningless from a mathematical point of view? Or is there some way to translate it into an actual proof?
| https://mathoverflow.net/users/412 | Can the "physical argument" for the existence of a solution to Dirichlet's problem be made into an actual proof? | Well, I don't understand the electrostatics, but here is another physical heuristic:
Impose a temperature distribution at the exterior, and measure (after some time has passed) the temperature in the interior. This gives a harmonic function extending the exterior temperature. [What's the electrostatic analogue? Formerly I had written "charge density",
but now I am not sure if that's right.]
I think this strongly suggests a mathematically rigorous argument: We are naturally led to model the time-dependence of temperature in the interior.
This satisfies a diffusion (or heat) equation, but in words:
"After a time \delta, the new temperature is obtained by averaging the old temperature along a circle of radius \sqrt{\delta}."
This process converges under reasonable conditions, as time goes to infinity, to the solution of the Dirichlet problem. Anyway, we are led to the Brownian-motion proof of the existence, which I personally find rather satisfying. Another personal comment: I think one should always take "physical heuristics" rather seriously.
[In response to Q.Y.'s comments below, which were responses to previous confused remarks that I made: neither the electric field nor the Columb potential is a multiple of the charge density on the boundary: the former is a vector, and in either
case imagine the charge on the boundary to be concentrated in a sub-region; neither the electric field nor the potential will be constant outside that sub-region.]
| 7 | https://mathoverflow.net/users/513 | 3215 | 2,134 |
https://mathoverflow.net/questions/3222 | 6 | Say that a projective variety V over Q satisfies the local-global principle up to finite obstruction (#) if there are only finitely many isomorphism classes of projective varieties over Q that are not isomorphic to V over Q despite being isomorphic to V over every completion of Q..
In section 7 of Barry Mazur's 1993 article titled *On the Passage From Local to Global in Number Theory*, Mazur describes his attempt to prove that (#) for abelian varieties over Q implies (#) for *all* projective varieties over Q, and a partial result that he, Yevsey Nisnevich and Ofer Gabber achieved in this direction. Has there been further progress in this direction since 1993?
My understanding is that an effective version of (#) for genus 1 curves (an effective bound on certain Tate-Shafarevich groups) gives a finite algorithm (of *a priori* bounded running time) for determining whether a genus 1 curve has a rational point, and also that such an effective bound on Tate-Shafarevich groups is expected.
Is an effective version of (#) for general projective varieties over Q expected? If so, how does this relate to Hilbert's 10th problem over Q (which Bjorn Poonen has conjectured to be undecidable)?
| https://mathoverflow.net/users/683 | Finiteness of Obstruction to a Local-Global Principle | "Has there been further progress in this area since 1993?"
So far as I know, there has been no direct progress. I feel semi-confident that I would know if there had been a big breakthrough: Mazur was my adviser, this is one of my favorite papers of his, and I still work in this field. Also, I just checked MathReviews and none of the citations to this paper makes a big advance on the problem, although two are somewhat relevant:
MR1905389 Thăńg, Nguyêñ Quoc On isomorphism classes of Zariski dense subgroups of semisimple algebraic groups with isomorphic $p$-adic closures. Proc. Japan Acad. Ser. A Math. Sci. 78 (2002), no. 5, 60--62.
MR2376817 (2009f:14040) Borovoi, M.; Colliot-Thélène, J.-L.; Skorobogatov, A. N. The elementary obstruction and homogeneous spaces. Duke Math. J. 141 (2008), no. 2, 321--364.
I'm not sure what you mean by an effective bound on Shafarevich-Tate groups (henceforth "Sha"). It is certainly expected that the Sha of any abelian variety over a global field is finite. If this is true, then in any given case one can, "in principle", give an explicit upper bound on Sha by the method of n-descents for increasingly large n. (In practice, even for elliptic curves reasonable algorithms have been implemented only for small values of n.) I really can't imagine any algorithm having to do with Sha that has "a priori bounded running time". What do you have in mind here?
As to the final question, let me start by saying that it seems reasonable at least that the set of "companion varieties" (i.e., Q-isomorphism classes of varieties everywhere locally isomorphic to the given variety) of a projective variety V/Q is finite: as above, we believe this for abelian varieties, and Barry Mazur proved in this paper a lot of results in the direction that the conjecture for abelian varieties implies it for arbitrary varieties. (For instance, quoting from memory, I believe he proved the implication for all varieties of general type.)
Here is a key point: suppose you are given a variety V/Q and you are wondering whether it has rational points. If V is itself a torsor under an abelian variety (e.g. a genus one curve), then if you can compute Sha of the Albanese abelian variety of V, you can use this to determine whether or not V has a Q-rational point. In general, the connection between computation of sets of companion varieties of V and deciding whether V has a Q-rational point is less straightforward. If V is a curve, then there are theorem in the direction of the fact that finiteness of Sha(Jac(V)) implies that the Brauer-Manin obstruction is the only one to the existence of rational points on V. In particular, people who believe this (including Bjorn Poonen, I think), believe that there is an algorithm for deciding the existence of rational points on curves. But nowadays we know examples of varieties where the Brauer-Manin obstruction is not sufficient to explain failure of rational points.
So, in summary, it is a perfectly tenable position to believe that companion sets are always finite, even effectively computable, but still there is no algorithm to decide the existence of Q-points on an arbitrary variety.
| 4 | https://mathoverflow.net/users/1149 | 3225 | 2,141 |
https://mathoverflow.net/questions/3232 | 3 | I'm looking for a reference which deals with limits of families of algebraic varieties as the degree increases (or at least keywords from this subject).
For the kind of example I have in mind, consider the exponential function as a convergent power series e^x=1+x+x^2/2!+... on the interval [-1;1] say, and consider the varieties A0:={(x,y)\in\R^2 s.t. y=1}, A1:={(x,y)\in\R^2 s.t. y=1+x}, A2:={(x,y)\in\R^2 s.t. y=1+x+x^2/2!}... which converge (in the euclidean topology) to the graph of the exponential (as in the picture [here](http://en.wikipedia.org/wiki/Exponential_function#Formal_definition)).
| https://mathoverflow.net/users/469 | limits of algebraic varieties | See [On the limit of families of algebraic subvarieties with unbounded volume](http://www.tecgraf.puc-rio.br/~lhf/ftp/doc/limits.pdf), to appear in Astérisque.
| 5 | https://mathoverflow.net/users/532 | 3238 | 2,150 |
https://mathoverflow.net/questions/3239 | 11 | There are many proofs based on a "tertium non datur"-approach (e.g. prove that there exist two irrational numbers a and b such that a^b is rational).
But according to Gödel's First Incompleteness Theorem, where he provides a constructive example of a contingent proposition, which is neither deductively (syntactically) true nor false, we know that there can be a tertium.
**My question:** Are all proofs that are based on that principle useless since now we know that a tertium can exist?
| https://mathoverflow.net/users/1047 | Is no proof based on "tertium non datur" sufficient any more after Gödel? | You are confused.
The best way out of your confusion is to maintain a very careful distinction between strings of formal symbols and their mathematical meanings. Godel's theorem is, on its most primitive level, a theorem about which strings of formal symbols can be obtained from other strings by certain formal manipulations. These formal manipulations are called proofs, and the strings which are obtainable in this way are called theorems. For clarity, I'll call them formal proofs and formal theorems.
In particular, let G be a string such that G is not a formal theorem and neither is NOT(G). It is still true that G OR NOT(G) is a formal theorem. Moreover, if G IMPLIES H and NOT(G) IMPLIES H are both formal theorems, then H will be a formal theorem; because there are rules of formal manipulation that allow you to take the first two strings and produce the third. I believe that Douglass Hofstader discusses this in a fair bit of detail when he goes over Godel's theorem.
The above is mathematics. Next, some philosophy. I don't find it helpful to say that G is neither true nor false. It find it more helpful to say that our systems of formal symbols and formal manipulation rules can describe more than one system. For example, Euclid's first four axioms can describe both Euclidean and non-Euclidean geometries. This doesn't mean that Euclid's fifth postulate has some bizarre third state between truth and falsehood. It means that there are many different universes (the technical term is models) described by the first four axioms, and the fifth postulate is true in some and false in others.
However, in any particular one of those universes, either the fifth postulate is true or it is false. Thus, if we prove some theorem on the hypothesis that the fifth postulate holds, and also that the fifth postulate does not hold, then we have shown that this theorem holds in every one of those universes.
There are fields of mathematical logic, called constructivist, where the law of the excluded middle does not hold. As far as I understand, that issue is not related to Godel's theorem.
| 43 | https://mathoverflow.net/users/297 | 3244 | 2,153 |
https://mathoverflow.net/questions/3084 | 6 | Suppose I have n finite sets A1 through An contained in some fixed set S, and I am given non-negative integers N and N1 through Nn such that each Ai has cardinality N, and each k-tuple intersection has cardinality less than or equal to Nk.
Can I use this to construct a good lower bound on the cardinality of the union of the Ai?
| https://mathoverflow.net/users/910 | Bound on cardinality of a union | I don't know your reason for asking this question, so it's unlikely that what I'm about to write will be helpful. Nevertheless, there's an easy method I like a lot for deducing a lower bound just from the knowledge that N\_2 is small. It may be contained in what has been said above -- I haven't checked.
The idea is to think of each set A\_ i as a 01-valued function on a set of size M. We then look at the ell\_ 2 norm of sum\_i A\_i. The square of the ell\_ 2 norm is sum\_ {i,j}|A\_ i cap A\_ j|, which by assumption is at most nN + n(n-1)N\_ 2. But we also have a lower bound: the ell\_ 1 norm of the sum is nN, from which it follows that the square of the ell\_ 2 norm is at least (nN)^2/M.
Putting these two bits of information together gives a lower bound for M of nN/(1+(n-1)N\_ 2/N). So, for example, if N\_ 2 = cN for some smallish positive constant c, then the lower bound is roughly N/c.
| 9 | https://mathoverflow.net/users/1459 | 3246 | 2,155 |
https://mathoverflow.net/questions/3247 | 7 | This is probably an easy question. Let C be a category with (finite) products.
An internal hom in C category is an object uhom(X, Z) which represents the functor:
Y |-----> hom(Y x X, Z)
here "uhom" is for "underlined hom" as that is how it is commonly denoted. Many example of categories with internal homs satisfy an a priori stronger for of adjunction:
uhom(Y x X. Z) = uhom(Y, uhom(X,Z))
Is this automatic for categories with internal homs? Is there an easily understood counter example?
(\*) This might not be the most general/best definition of internal hom, but it is valid for many examples.
| https://mathoverflow.net/users/184 | internal homs and adjunctions? | I think both uhom(Y × X, Z) and uhom(Y, uhom(X, Z))
represent the same functor:
hom(W, uhom(Y × X, Z)) = hom(W × Y × X, Z)
hom(W, uhom(Y, uhom(X, Z))) = hom(W × Y, uhom(X, Z)) = hom(W × Y × X, Z),
hence by Yoneda lemma they are isomorphic.
As a side remark, the right notion of inner hom in the absence of monoidal structure is the notion of closed category:
<http://en.wikipedia.org/wiki/Closed_category>
| 8 | https://mathoverflow.net/users/402 | 3255 | 2,160 |
https://mathoverflow.net/questions/3077 | 5 | I'm interested in producing explicit bases for the sections of a line bundle on an embedded genus 1 curve. Let me restrict to the first case that I don't know how to do, so that I can be as concrete as possible. Take a smooth genus 1 curve E defined over QQ by an explicit cubic equation C0 in QQ[x,y,z]. Let D be a divisor of degree 6 on E, and I\_D be defined by four cubic equations (C0,C1,C2,C3). Note that D does not lie on any conic. Riemann-Roch says that h^0(E,OE(D))=6, and I'd like to find an explicit basis of rational functions for this vector space. How do I find such a basis?
Note that if D sat on a conic defined by Q, then finding a basis is relatively easy: we could simply choose the functions x^2/Q, xy/Q, ..., z^2/Q as our rational functions.
| https://mathoverflow.net/users/4 | Sections of a divisor on elliptic curve | I think I know how to answer this now. The main point is that OE(D) is the dual of ID. Namely: OE(D)=sheafHom(ID, OE). Thus, H^0(E,OE(D))=Hom(ID, OE).
This can be computed explicitly in any computer algebra package. Or you can see how to compute it as follows. Take a free presentation of ID as an OE-module. In the case I asked about, this yields:
OE3(-4)-->OE3(-3)-->ID.
Label the first map F. Then Hom(ID, OE) is just the kernel of the map of free modules:
Hom(OE3(-3), OE)--> Hom(OE3(-4), OE)
induced by composition with F. Thus, computing a free presentation of the ideal sheaf ID yields a presentation of H^0(E,OE(D)) as the kernel of a map of free modules.
| 1 | https://mathoverflow.net/users/4 | 3268 | 2,172 |
https://mathoverflow.net/questions/3271 | 2 | I've recently come across a variant of the binomial polynomials, and I'm curious if anyone has seen these before. If so, I'd love a reference, a name, etc.
First recall the following. If z is a formal variable, then we can consider \binom{z}{k} as a polynomial in z by the standard formula: \binom{z}{k}= [z(z-1)...(z-k+1))]\*[k!]-1.
Here's the variant I came across. Let a and k integers, where a divides k, and we write k=ab. Consider the polynomial
F(a,k)(z)=[z(z-a)(z-2a)...(z-k+a))][k(k-a)(k-2a)...(a)]-1.
Has anyone this before? Or anything similar?
**Update:** Jonah helpfully identifies a typo in the numerator of F(a,k)(z) which I've now corrected.
| https://mathoverflow.net/users/4 | Variant of binomial coefficients | (For simplicity, you probably want the last term in the numerator to be z-k+a, right? That way F(1,k)(z)=\binom{z}{k}. I'll pretend that's what you meant.)
I haven't come across such polynomials, but they're easily expressed in terms of [multifactorials](http://en.wikipedia.org/wiki/Factorial#Multifactorials). Namely,
F(a,k)(z)=z!(a)/[(z-k)!(a)k!(a)]
Note that this isn't an integer when a doesn't divide z. EDIT: (And, as Qiaochu points out, when a does divide z it's just regular ol' z-choose-k.)
| 1 | https://mathoverflow.net/users/1060 | 3272 | 2,173 |
https://mathoverflow.net/questions/3274 | 24 | Are there any efficient algorithms for computing the [Euler totient function](http://en.wikipedia.org/wiki/Euler%27s_totient_function)? (It's easy if you can factor, but factoring is hard.)
Is it the case that computing this is as hard as factoring?
**EDIT**: Since the question was completely answered below, I'm going to add a related question. How hard is it to compute the number of prime factors of a given integer? This can't be as hard as factoring, since you already know this value for semi-primes, and this information doesn't seem to help at all. Also, determining whether the number of prime factors is 1 or greater than 1 can be done efficiently using Primality Testing.
| https://mathoverflow.net/users/1042 | How hard is it to compute the Euler totient function? | For semiprimes, computing the Euler totient function is equivalent to factoring. Indeed, if n = pq for distinct primes p and q, then φ(n) = (p-1)(q-1) = pq - (p+q) + 1 = (n+1) - (p+q). Therefore, if you can compute φ(n), then you can compute p+q. However, it's then easy to solve for p and q because you know their sum and product (it's just a quadratic equation).
If you believe factoring is hard for semi-primes, then so is computing the Euler totient function.
*Update*! Factoring and computing the Euler totient function are known to be equivalent for arbitrary numbers, not just semiprimes. One reference is "Riemann's hypothesis and tests for primality" by Gary L. Miller. There, the equivalence is deterministic, but assumes a version of the Riemann hypothesis. See also section 10.4 of "[A computational introduction to number theory and algebra](http://shoup.net/ntb/)" by Victor Shoup for a proof of probabilistic equivalence.
| 38 | https://mathoverflow.net/users/1079 | 3275 | 2,174 |
https://mathoverflow.net/questions/3278 | 36 | Recall that a category C is *small* if the class of its morphisms is a set; otherwise, it is *large*. One of many examples of a large category is **Set**, for Russell's paradox reasons. A category C is *locally small* if the class of morphisms between any two of its objects is a set. Of course, a small category is necessarily locally small. The converse is not true, as **Set** is a counterexample.
Now, I can construct categories that are not locally small. However, what's the most common or most reasonable such category?
| https://mathoverflow.net/users/1079 | What's a reasonable category that is not locally small? | The category of multi-spans spans (thanks to everyone below for correcting my terminology). The objects are sets, and a map from $A$ to $B$ is a set $X$ equipped with a map $X → A × B$. The composition of $X → A × B$ and $Y → B × C$ is $X ×\_B Y → A × C$.
I am stealing notation from algebraic geometry here: $X ×\_B Y$ is the limit of the diagram $X → B ← Y$.
Admittedly, I've never wanted to allow $X$ to be an arbitrary set. I usually want it to be something like a finite set, a finite simplicial complex or a scheme of finite type. But it is certainly natural to define the category without any restrictions.
| 27 | https://mathoverflow.net/users/297 | 3282 | 2,179 |
https://mathoverflow.net/questions/3283 | 13 | I've recently started my personal wiki to organize my notes and thoughts. I use the wiki program instiki which I believe is the same as the n-lab uses. Instiki can upload [svg](http://en.wikipedia.org/wiki/Scalable_Vector_Graphics)'s. I want to be able to create nice looking pictures of some (not to complicated) geometric objects, e.g. knots, pair of pants, etc. My question is twofold.
1. Do people draw these things using svg format? For example people who do diagram algebra, do you use svg format?
2. If so, what are some good free/open source programs for creating these pictures?
| https://mathoverflow.net/users/135 | Are there good programs to create mathematical pictures in svg format? | I am a huge fan of the open source program [Inkscape](http://www.inkscape.org). I mostly use it to produce pictures for my papers in the eps format, but its native format is svg.
| 22 | https://mathoverflow.net/users/317 | 3287 | 2,183 |
https://mathoverflow.net/questions/2650 | 23 | In a forthcoming paper with Venkatesh and Westerland, we require the following funny definition. Let G be a finite group and c a conjugacy class in G. We say the pair (G,c) is *nonsplitting* if, for every subgroup H of G, the intersection of c with H is either a conjugacy class of H or is empty.
For example, G can be the dihedral group of order 2p and c the class of an involution.
The case where c is an involution is o special interest to us. One way to construct nonsplitting pairs is by taking G to be a semidirect product of N by (Z/2^k Z), where N has odd order, and c is the conjugacy class containing the involutions of G. Are these the only examples? In other words:
**Question 1:** Is there a nonsplitting pair (G,c) with c an involution but where the 2-Sylow subgroup of G is not cyclic?
Slightly less well-posed questions:
**Question 2:** Are there "interesting" examples of nonsplitting pairs with c not an involution? (The only example we have in mind is G = A\_4, with c one of the classes of 3-cycles.)
**Question 3:** Does this notion have any connection with anything of pre-existing interest to people who study finite groups?
**Update**: Very good answers below already -- I should add that, for maximal "interestingness," the conjugacy class c should generate G. (This eliminates the examples where c is central in G, except in the case G = Z/2Z).
| https://mathoverflow.net/users/431 | Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups | **Answer:**
I cheated and asked Richard Lyons this question (or at least, the reformulation of the problem, conjecturing that (G,c) is nonsplitting for an involution c with `<`c`>` generating G if and only if there exists an odd A such that G/A = Z/2). His response:
---
Good question! This is a famous (in my circles) theorem - the Glauberman Z^`*-`Theorem. (Z^\*(G) is the preimage of the exponent 2 subgroup of the center of G/O(G), and O(G)=largest normal subgroup of G of odd order.)
Z^`*-`Theorem: If c is an involution of G then c\in Z^\*(G) iff [c,g] has odd order for all g\in G iff for any Sylow 2-subgroup S of G containing c, c is the unique G-conjugate of itself in S.
The last property is absolutely fundamental for CFSG. The proof uses modular character theory for p=2. Attempts to do it with simpler tools have failed.
George Glauberman, Central Elements in Core-free Groups, Journal of Algebra 4, 1966, 403-420.
---
**Older Remarks:**
**Comment 1**: Suppose that P = Z/2+Z/2 is a 2-Sylow. If x lies in P, then P clearly centralizes x, and thus the order of `<`x`>` divides #G/P, and is thus odd. By a theorem of Frobenius, G has an odd number of elements of order 2, and thus we see it has an odd number of conjugacy classes of elements of order 2. Yet, by the Sylow theorems, every element of order 2 is conjugate to an element of P. If c lies in P, then by nonsplitting, it is unique in its G-conjugacy class in P. Thus there must be exactly three conjugacy classes of elements of order 2, and thus no element of P is G-conjugate. By a correct application of Frobenius' normal complement theorem, we deduce that G admits a normal subgroup A such that G/A \sim P. Yet `<`c`>` generates G, and thus the image of `<`c`>` generates G/A. Yet G/A is abelian and non-cyclic, a contradiction.
**Comment 2**: Suppose that A is a group of order coprime to p such that p | #Aut(A).
Let G be the semidirect product which sits inside the sequence:
1 ---> A ---> G --(phi)--> Z/pZ --> 0;
Let c be (any) element of order p which maps to 1 in Z/pZ. If c is conjugate to
c^j, then phi(c) = phi(c^j). Hence c is not conjugate to any power of itself.
Let H be a subgroup of G containing c (or a conjugate of c, the same argument applies). The element c generates a p-sylow P
of H (and of G). It suffices to show that if gcg^-1 lies in H, then it is conjugate to c inside H. Note that gPg^-1 is a p-Sylow of H.
Since all p-Sylows of H are conjugate, there exists an h such that gPg^-1 = hPh^-1, and thus h c^j h^-1 = gcg^-1. Yet we have seen that c^j is not conjugate to c inside G unless j = 1. Thus gcg^-1 = hch^-1 is conjugate to c inside H.
I just noticed that you wanted `<`c`>` to generate G. It's not immediately clear (to me) what condition on A one needs to impose to ensure this. Something like the automorphism has to be "sufficiently mixing". At the very worst, I guess, the group G' generated by `<`c`>` still has the property, by the same argument.
This works more generally if p || G and no element of order p is conjugate to a power of itself. (I think you know this already if p = 2.)
The case where the p-Sylow is not cyclic is probably trickier.
Examples: A = (Z/2Z)+(Z/2Z), p = 3. (This is A`_`4).
A = Quaternion Group, p = 3. (This is GL`_`2(F`_`3) = ~A`_`4, ~ = central extension).
A = M^37, M = monster group, p = 37.
| 19 | https://mathoverflow.net/users/nan | 3297 | 2,191 |
https://mathoverflow.net/questions/3296 | 2 | I have a very specific question: does anyone know of a (non-trivial) example of a projective curve which is also a homogenous space (or just a principal bundle)? The trivial example being CP^1 = SU(2)/U(1).
| https://mathoverflow.net/users/1095 | Projective Curves which are Principal Bundles | CP^2 is not a curve. So you may have misstated your question. Nonetheless, here is my answer:
Every curve of genus 1 is a principal homogenous space for its Jacobian. Over an algebraically closed field, a principal homogenous space is just the group itself, and that is what happens in this case.
For genus g >= 2, no algebraic curve has more than 84(g-1) algebraic automorphisms. In particular, no curve can be a homogenous space.
EDIT The comment about CP^2 refers to an earlier version of the question.
| 4 | https://mathoverflow.net/users/297 | 3301 | 2,192 |
https://mathoverflow.net/questions/2842 | 6 | Clearly, it is possible to colour the edges of an infinite complete graph so that it does not contain any infinite monochromatic complete subgraph. Now what about the following?
>
> Let $G$ be the complete graph with vertex set the
> positive integers. Each edge of $G$ is then coloured *c* with probability $\frac{1}{2^c}$, for
> $c = 1, 2, \dots$ What is the probability
> that *G* contains an infinite
> monochromatic complete subgraph?
>
>
>
It is unclear for me if the answer should be $0, 1$, or something in between.
| https://mathoverflow.net/users/416 | Infinite Ramsey theorem with infinitely many colours | Every countably infinite random graph is almost surely the Rado graph which contains all finite and countably infinite graphs as induced subsets. So each color class almost surely contains the Rado graph and hence a infinite monochromatic subgraph. See the following for more details here:
<http://en.wikipedia.org/wiki/Random_graph>
There are also links in the article to other articles including one on the Rado graph.
| 10 | https://mathoverflow.net/users/1098 | 3302 | 2,193 |
https://mathoverflow.net/questions/3300 | 3 | What is the largest dimensional linear space of singular planar cubics? Is this known?
Think of the space of planar cubics as a PP^9 (parametrized by the coefficients). The discriminant \Delta is then a degree 12 polynomial on PP^9 whose vanishing parametrizes singular cubic curves. What is the dimension of the largest linear space inside the vanishing of \Delta? I attempted doing this on a computer, but the computations were too large to terminate.
| https://mathoverflow.net/users/4 | Vector spaces of singular planar cubics | According to Bertini's Theorem a linear system is smooth away from its base points.
Thus there is a point in PP^2 contained in the singular set of every cubic in your linear system. You need three conditions to impose a singularity at a point p (f(p) = f\_x(p) = f\_y(p)=0). Thus maximal projective spaces inside the discriminant have dimension 6.
**Edit** As David pointed out in another answer, this argument uses char 0.
| 5 | https://mathoverflow.net/users/605 | 3314 | 2,200 |
https://mathoverflow.net/questions/2853 | 7 | Let Mg-bar be the Deligne-Mumford compactification of genus g curves, and let δ1 be the divisor of degenerate curves of the form `genus 1 meeting a genus g-1 transversely".
>
> **Question 1:** What is the n-fold self intersection of δ1?
>
>
>
I've seen mentioned in the literature that the answer for n = g is `rational curves with g many genus 1 tails. This certainly seems reasonable -- this cycle has the right dimension and is`visually' obtained by g-1 many degenerations.
>
> **Question 2:** Let C be a curve in Mg-bar contained in δ1. How do I calculate the intersection of C and δ1?
>
>
>
I'm interested in the case where C is a family of rational curves with g many fixed elliptic tails (so that each elliptic tail is the same elliptic curve), each with at most two rational components. The answer is in the literature, but without proof. Does anyone know a reference or a quick way to do this calculation?
| https://mathoverflow.net/users/2 | How does one intersect non-transverse divisors on Mg-bar. | Both questions reduce to showing the 2-fold intersection of D\_1 is a the closure of the locus of
genus g-2 curve with two elliptic tails.
The first question follows from this claim by induction and using the map D\_1 -> M(g-1)^bar x M(1,1)^bar.
The second is a direct application of the claim.
Which brings us to the claim: Instead of working on Mg^bar - which is very hard, you can rigidify the question, and work on some compactified Hurwitz scheme. The Hurwitz scheme H(g,d) is the scheme of degree d covers of P^1 by genus g curves, with simple ramficiation only. The crucial facts here are:
* Given branch points on P^1, there are only finitely many curves with these branch points.
* You can compactify this space: when branch points colide you add a "buble": a copy of P^1, connected to the original P^1 where these points colided, and marked ramification points on the buble. You now have a node connecting the two P^1s; the rule is that the over this nodes, the ramification pattern is identical for the curves lying over the buble and the original P^1. Needless to say, you can add as many bubles as you want.
The pullback of D1 to H(g,d) is the closure of the locius of a pair of P^1's, such that: there are 2(g-2+d)-1 marked branch points on one of them, and 2d-1 marked branch points on the other. This is a purely combinatorial condition. Intersecting the condition with itself you get the closure of the loci of a chain of 3 P^1s, with 2d-1, 2(g-3+d) -2, 2d-1 points. Which is a pullback to H(g,d) of the expected class.
Note that we worked with pullbacks, so we have an equality of classes, not of the underlying reduced schenes.
| 3 | https://mathoverflow.net/users/404 | 3324 | 2,206 |
https://mathoverflow.net/questions/3308 | 1 | I have always been fascinated by the so called taxicab geometry first considered by Hermann Minkowski. The metric that has to be used here is a L1 distance which e.g. means that the lenght of the diagonal in a unit square is 2 and not √2 - and this holds true no matter how fine the mesh is! Basically the solution doesn't converge when approximated dicretely.
If you see the mesh as an analogy of solving a differential equation numerically the difference between their true solution and the numerical outcome is obviously huge and won't be acceptable in most applications.
**My questions:** Do you know of certain classes of differential equations where no closed form solutions exist AND that misbehave in an corresponding way? How are they called and where do I find more information on those? How do you identify them (what characteristics do they have) and how do you tackle such misbehaved equations?
***Addition:*** When I think about it: What could be even stranger is the case when a closed form solution exists but the differential equation can't be solved numerically in the abovementioned sense.
| https://mathoverflow.net/users/1047 | "Misbehaved" differential equations | You'd probably be interested in reading about discrete differential geometry, as put forth by Bobenko et al. [Here's](http://page.mi.fu-berlin.de/polthier/articles/convergence/convergence.pdf) one paper about when certain geometric quantities defined in the discrete sense converge to the analogous quantities in the continuous sense. Again, I don't know what connection you're trying to make to differential equations, (perhaps the results on convergence of [finite element methods](http://en.wikipedia.org/wiki/Finite_element_method#)?) so perhaps you should spend some time trying to make that precise and then edit your question again.
| 2 | https://mathoverflow.net/users/353 | 3326 | 2,207 |
https://mathoverflow.net/questions/2483 | 20 | Suppose that the convex hull of the Minkowski sum of several compact connected sets in $\mathbb R^d$ contains the unit ball centered at the origin and the diameter of each set is less than $\delta$. If $\delta$ is very small (this smallness may depend on $d$ but on nothing else), does it follow that the sum itself contains the origin?
| https://mathoverflow.net/users/1131 | Minkowski sum of small connected sets | I finally figured it out. My solution is [here](http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308020). I would repost it on mathoverflow but until LaTeX is enabled, it is quite hard for me to communicate such things here...
| 13 | https://mathoverflow.net/users/1131 | 3328 | 2,209 |
https://mathoverflow.net/questions/3289 | 9 | This painful question is inspired by the question
"[non-Lie subgroups](https://mathoverflow.net/questions/3157/non-lie-subgroups)" . Let $f$ be a discontinuous additive map from $\mathbb{R} \to \mathbb{R}$. Is it possible that the graph of $f$, inside $\mathbb{R}^2$ with the usual topology, is connected? Remember that the graph of a discontinuous function can be connected, as in the
[Toplogist's sine curve](http://en.wikipedia.org/wiki/Topologist%27s_sine_curve).
I was hoping to show that any connected subgroup of a Lie group is a Lie group, thus answering the previous question, and I couldn't get rid of this horrible example.
| https://mathoverflow.net/users/297 | Potential connected non-Lie subgroup | I think the answer is, yes, the graph can be connected.
By definition, if the graph G is not connected, then we can find disjoint nonempty open sets A and B, such that G is contained in A union B. In particular, this implies no point in G can be contained in the boundary of A. So if we can construct an additive function f whose graph intersects the boundary of any potential separating open set A, we'll have shown the graph is connected.
Before constructing this function, note a technical point. Not all open sets are candidates for separating G. If G = A union B for nonempty open sets A,B, then the projections proj(A) and proj(B) onto the x-axis are both open, and must intersect. In turn this implies the projection of the boundary of A contains an interval. Call open sets with this property "candidate sets".
To make a function f whose graph intersects the boundary of all candidate sets, consider a basis H for R as a vector space over Q. This set has cardinality of the reals. Now note that the set of all open sets in R^2 also has cardinality of the reals. (<http://en.wikipedia.org/wiki/Cardinality_of_the_continuum>)
Put these two sets (basis H, all open sets) in 1-1 correspondence, so for each h in H, we have an open set O(h). If O(h) is not a "candidate set," let f(h)=0. Otherwise, using the fact that O(h) is a candidate set, we can always find a nonzero rational q, and a real y such that (qh,y) is in the boundary of O(h). Define f(qh)=y. Doing this for all elements of H will determine a unique additive function f on the reals.
The graph of f, by construction, is connected since it intersects the boundary of every candidate separating open set in R^2. (And it's not continuous--if it were, it would miss the boundaries of a lot of open sets!)
| 7 | https://mathoverflow.net/users/1227 | 3335 | 2,212 |
https://mathoverflow.net/questions/3339 | 4 | Define the lens space L(m,n) as the quotient of S2m+1 by the action of the cyclic group ℤn⊂S1⊂ℂ\*. We can create the infinite lens space L(∞,n) by a telescoping construction on the lens spaces L(m,n) for fixed n, which has as an n-sheeted covering space S∞. The homotopy exact sequence is then
... --> π1(S∞) --> π1(L(∞,n)) --> π0(n points) --> π0(S∞) --> ...
Here, π1(S∞) is the trivial group, and π0(S∞) is a set with one point. The sequence is still exact at the π0 portion, once we specify a basepoint for each set and call its preimage the kernel of the map, meaning that whatever π1(L(∞,n)) is, it definitely has n elements. It's a group, too, so if n is prime then we have no choice but to conclude that π1(L(∞,n))=ℤn. Of course, even when n isn't prime, the fact of the matter is that this statement is still true.
But this is a little unsettling to me. It seems like we're only concluding that because the deck group of the universal cover happens to be ℤn (or, admitting the full extent of our complicity, because we're very nearly taking as a *definition* that L(∞,n)=S∞/ℤn). If we aren't working with the universal cover, then π1 of the cover isn't trivial, so even if the cover is connected it seems like we could run into the extension problem in trying to compute π1 of the base. Of course, this is all algebraic; perhaps there's something geometric that will save the day and tell us how to interpret this. Is that the case, or is there some other way to unambiguously determine π1 of the base here (perhaps from the definition of the connecting homomorphism via the covering homotopy property)?
| https://mathoverflow.net/users/303 | properly interpreting Pi_0 in the homotopy exact sequence | There is a bit more structure in this long exact sequence that you can use. If you have $F \to E \to B$, then on choosing a basepoint in $B$, then for any basepoint $e$ in $F$ you have the sequence
$\dots \to \pi\_1 E \to \pi\_1 B \to \pi\_0 F \to \pi\_0 E \to \pi\_0 B$.
The extra structure is the fact that $\pi\_1 B$ acts on $\pi\_0 F$, through deck transformations, if $E\to B$ is actually a covering map, and that:
\* the stabilizer group of $[e]$ in $\pi\_0F$ under this action is the image of $\pi\_1E \to \pi\_1B$, and
\* the set of orbits $\pi\_0F/\pi\_1B$ under this action is exactly the "kernel" of $\pi\_0E\to \pi\_0B$.
Thus, if you have some non-trivial extension and are trying to compute $\pi\_1B$, then you get to use the information about deck transformations. I don't know what more to say without an explicit example.
| 5 | https://mathoverflow.net/users/437 | 3349 | 2,219 |
https://mathoverflow.net/questions/3323 | 13 | What kind of additional properties and/or structures one needs to impose on the category
of (commutative or noncommutative) monoids of some monoidal category
so that one can recover the original monoidal category from this data?
What kind of additional properties and/or structures one needs to impose on a category
to ensure that it is the category of monoids of some monoidal category?
The example I have in mind is the category of (commutative or noncommutative) C\*-algebras (or von Neumann algebras).
Can we obtain one of these categories as the category of monoids of some monoidal category?
| https://mathoverflow.net/users/402 | Recovering a monoidal category from its category of monoids | Here is a characterization of categories of commutative monoids. I don't know the answer in the non-commutative case.
Let *C* be a category. Then *C* is the category of commutative monoids in some symmetric monoidal category if and only if *C* has finite coproducts.
For suppose that *C = CMon(M)* for some symmetrical monoidal category *M = (M, @, I)*. Then one can show that the tensor product @ of M also defines a tensor product on *C* --- and that this is, in fact, binary coproduct in *C*. (Example: if *M* is the category of abelian groups then *C* is the category of commutative rings, and the tensor product of commutative rings is the coproduct.) Similarly, the unit object *I* of *M* is a commutative monoid in a unique way, and is in fact the initial object of *C*. So *C* has finite coproducts.
Conversely, suppose that *C* has finite coproducts. Then (+, 0) defines a symmetric monoidal structure on *C*, and with respect to this structure, every object of *C* is a commutative monoid in a unique way. Thus, *C = CMon(C)*.
| 18 | https://mathoverflow.net/users/586 | 3356 | 2,224 |
https://mathoverflow.net/questions/2529 | 25 | What do Gromov-Witten invariants (of say a Calabi-Yau 3-fold) represent, or what are they supposed to represent, in terms of string theory? When I compute GW invariants, am I actually computing some interactions between some particles, or what ...?
Please be gentle, and use only undergraduate-level physics words, if possible. (Perhaps this is too much to ask! Ok well, I'd rather get a response that involves fancier physics words than no response at all.)
I suppose this question is more physics than math -- I hope that's ok.
| https://mathoverflow.net/users/83 | What are Gromov-Witten invariants in terms of physics? | Here is a very rough answer.
The Gromov-Witten invariants show up in a few a priori different
contexts within string theory. Let me focus on one particular place they show up that is
directly related to conventional physics, as opposed to topological
quantum field theory.
Type IIA string theory is formulated on a spacetime "background"
which is, in the simplest setup, just a Lorentzian 10-manifold. The
equations of motion of the theory require (at least in their leading
approximation) that the metric on this 10-manifold should be Ricci-flat.
A popular thing to do is to take this 10-manifold of the form
X x R^{3,1}, where X is a compact Calabi-Yau threefold.
We can simplify matters by taking X to be very small ---
smaller than the Compton wavelength of any of the particles we are able
to create. (Remember that in quantum mechanics particles have a
wavelike character, with wavelength inversely related to their energy;
since we only have limited energy available to us, we can't make
particles with arbitrarily short wavelength.) A little more precisely,
let's take X such that the first nonzero eigenvalue of the Laplacian is
larger than the energy scale we can access.
In this case we low-energy
observers will not be able to detect X directly in any experiments. To
us, spacetime will appear to be R^{3,1}.
What will be the physics we see on this R^{3,1}? We will see
various different species of particle. Each species of particle that
we see corresponds to some zero-mode of the Laplacian of X.
In particular, there are particles corresponding to classes in H^{1,1}(X).
The genus 0 Gromov-Witten invariants are giving
information about the interactions between these particles. (So if you want to calculate what will come out when you
shoot two of these particles at each other, one of the inputs to that calculation
would be the genus 0 Gromov-Witten invariants.) The higher genus Gromov-Witten
invariants are giving information about interactions which involve these particles
together with other particles related to the gravitational interaction.
| 33 | https://mathoverflow.net/users/580 | 3365 | 2,230 |
https://mathoverflow.net/questions/3352 | 15 | In *Invent. math.* 116, 645-649 (1994) Dinakar Ramakrishnan proves a theorem which I understand to imply that the following statement (in light of the fact that elliptic curves over $\mathbb{Q}$ are modular):
Theorem: Let $E\_1 / \mathbb{Q}$ and $E\_2 / \mathbb{Q}$ be elliptic curves and suppose that the (mod $p$) reductions of $E\_1$ and $E\_2$ have the same number of points for a set of primes of Dirichlet density $> 7/8$. Then $E\_1$ and $E\_2$ are isogeneous.
Is $7/8$ expected to be sharp here? Could it be that $7/8$ can be replaced by any $a > 0$ with the conclusion in the above Theorem unaltered?
I'm interested in answers to these questions both in the level of generality in which Ramakrishnan works and in the special case that I give above.
| https://mathoverflow.net/users/683 | Very strong multiplicity one for Hecke eigenforms | Let $E$ and $F$ be two elliptic curves. For convenience, suppose that they do not have CM. For each $p$, there is a Galois representation
\[\rho: = \rho\_E + \rho\_F : G\_{\mathbb{Q}} \to \mathrm{GL}\_2(\mathbb{F}\_p) \times \mathrm{GL}\_2(\mathbb{F}\_p)\]
given by the action of Galois on the $p$-torsion of $E$ and $F$. For sufficiently large $p$, the Galois groups $\mathbb{Q}(E[p])$ and $\mathbb{Q}(F[p])$ are $\mathrm{SL}\_2(\mathbb{F}\_p)$-extensions of $\mathbb{Q}(\zeta\_p)$. If $p$ is big enough, $\mathrm{PSL}\_2(\mathbb{F}\_p)$ is simple. Thus, either:
1). These extensions are disjoint, and $\mathrm{im}(\rho) = \ker(\det(x)/\det(y))$.
2). These extensions intersect in a have a common $\mathrm{PSL}\_2(\mathbb{F}\_p)$ extension of $\mathbb{Q}(\zeta\_p)$, and hence a $\mathrm{PGL}\_2(\mathbb{F}\_p)$ extension of $\mathbb{Q}$. This implies that their projective representations are the same. This implies that they are equal up to twisting by a character. Since their determinants are also the same, this character is either trivial or quadratic.
If $E[p] = F[p]$ infinitely often then $E$ is isogenous to $F$, using Falting's proof of the Tate conjecture.
If Case 1) occurs, one can count that the density of pairs of elements $(\sigma,\tau)$ in $\mathrm{im}(\rho)$ such that $\mathrm{Tr}(\sigma) = \mathrm{Tr}(\rho)$ is $\sim 1/p$. By Cebotarev, this implies that the number of primes $\ell$ such that $a(E,\ell) = a(F,\ell) \pmod{p}$ has density at most $\sim 1/p$. Hence, if 1) occurs infinitely many times, then $a(E,\ell) = a(F,\ell)$ for set of density zero.
If $E[p]$ is a non-trivial quadratic twist of $F[p]$ for infinitely many primes, looking at the ramification of $E$ and $F$ it is easy to see that $E[p] = F[p] \otimes \chi$ for some fixed quadratic character $\chi$ for infinitely many $p$. Yet then using Faltings again, $E$ is isogenous to the twist of $F$ by $\chi$.
PS: This is a pretty standard argument.
PPS: The argument requires the existence of Galois representations (EDIT: with big image!) attached to $E$ (and $F$), which do not exist in general (EDIT: for example, weight one forms, see Toby's answer). The argument basically works for any pair of classical modular forms $e,f$ of weight at least $2$.
| 9 | https://mathoverflow.net/users/nan | 3371 | 2,234 |
https://mathoverflow.net/questions/3366 | 6 | This question is a spin-off from [Sammy Black's question on super Temperley-Lieb](https://mathoverflow.net/questions/3299/). Please see there for the background. The short version is that Sammy defines the Temperley-Lieb at index d as the algebra of sl(2) intertwiners of d copies of the defining rep of sl(2) (and then asks about a supersymmetric version).
My question: in what ways does Sammy's story extend to other simple Lie groups, e.g. sl(3)? For example, I think that the Schur-Weyl duality still holds, so that HomUsl(3)(V⊗d) is a quotient of the group algebra of the symmetric group Sd (here V is the defining representation of sl(3)). Is it something like the Young Tableaux with three rows? And is there a diagrammatic description of this algebra?
| https://mathoverflow.net/users/78 | Is there a version of Temperley-Lieb using sl(3) rather than sl(2)? | Cvitanovic is great for the groups themselves. But if you care about the quantum group (and this is really where TL shines) some good references are [Kuperberg's work on Spiders](http://arxiv.org/abs/q-alg/9712003) and [Scott Morrison's thesis](http://arxiv.org/abs/0704.1503).
| 6 | https://mathoverflow.net/users/22 | 3372 | 2,235 |
https://mathoverflow.net/questions/3370 | 5 | After reading a recent post on Church's Thesis, I ran into Turing-Church's Strong Thesis, that may be potentially disproven by advances in Quantum Computing. Does anyone know of a good resource that gets into the potential of quantum computing on complexity theory? And how complexity calculations would be done in that environment?
I am also not well read in complexity theory in general, so I may have made an ill-posed question here.
| https://mathoverflow.net/users/429 | Quantum Computing Complexity? | This is a very active field of current research. The book "Quantum Computation and Quantum Information", By Michael Nielsen and Ike Chuang, is one of the standard references.
Peter Shor found the original application of quantum computation to factorization, and I believe (without having full knowledge) that this still stands as one of the few instances of such an application - unfortunately, factorization is one of the few natural NP problems that are *not* known to be NP complete, so it may be the case that quantum computation "only" helps this particular problem (and problems trivially equivalent to it). The Wikipedia article on [BQP](http://en.wikipedia.org/wiki/BQP) is another useful reference.
| 2 | https://mathoverflow.net/users/25 | 3375 | 2,237 |
https://mathoverflow.net/questions/3270 | 20 | In this question, all rings are commutative with a $1$, unless we explicitly say
so, and all morphisms of rings send $1$ to $1$.
Let $A$ be a Noetherian local integral domain. Let $T$ be a non-zero $A$-algebra
which, as an A-module, is finitely-generated and torsion-free.
Can one realise $T$ as a subring of the (not necessarily commutative)
ring $End\_A(A^n)$ for some $n \ge 1$?
| https://mathoverflow.net/users/1384 | Which rings are subrings of matrix rings? | A starting proviso: you didn't require that the map $T \rightarrow End\_A(A^n)$ send elements of A to their obvious diagonal representatives. I am going to assume you intended this.
A few partial results:
1) If $A=k[x,y]/(x^3-y^2)$, and $T$ is the integral closure of A, then this can not be done. Let $t$ be the element $y/x$ of $T$ and $M$ the matrix that is supposed to represent it. Then we must have $xM=y Id\_n$, which has no solutions. More generally, whenever A is a non-normal ring and $T$ its integral closure, there are no solutions.
2) If $A$ is a Dedekind domain the answer is yes. Let $V$ be the vector space $T \otimes Frac(A)$, and $V^{\ast}$ the dual vector space. Let $T^{\ast} \subset V^{\ast}$ be the vectors whose pairing with $T$ lands in $A$. Using the obvious action of $T$ on itself, we get an action of $T^{op}$ on $T^{\ast}$. Since $T$ is commutative, this is an action of $T$ on $T^{\ast}$. Now, $T \oplus T^{\ast}$ is free as an A-module, so this gives us the desired representation.
2') A conjectural variant of the above: I have a vague recollection that, if $A$ is a polynomial ring, $T^{\ast}$ is always free. Can anyone confirm or refute this?
3) A case which I think is impossible, but can't quite prove at this hour: Let $T = k[x,y]$ and let $A$ be the subring $k[x^2, xy, y^2]$. I am convinced that we cannot realize $T$ inside the ring of matrices with entries in $A$, but the proof fell apart when I tried to write it down.
| 10 | https://mathoverflow.net/users/297 | 3384 | 2,243 |
https://mathoverflow.net/questions/3376 | 11 | In short, I'm wondering whether the universal coefficient theorem can be understood/reinterpreted by using maps of Eilenberg-MacLane spaces. This is a wishy-washy idea and I don't have evidence to back it up, but it would be very nice if the "freebie" cohomology classes in $H^n(X; G)$ we get when we're changing our coefficients from $\mathbb{Z}$ to $G$ (i.e., those that come simply from tensoring with the new group) corresponded to elements of $[K(\mathbb{Z}, n), K(G, n)]$. Then, the other classes that arise from $\operatorname{Ext} /\operatorname{Tor}$ would correspond to elements of $[X, K(G,n)]$ which *don't* factor through $K(\mathbb{Z},n)$. Is anything like this even remotely true?
This question is in part motivated by the responses to [an earlier question of mine](https://mathoverflow.net/questions/461/understanding-steenrod-squares), which mentioned that viewing $H^n(X; G)$ as $[X, K(G,n)]$ helps us understand cohomology operations (in that case, Steenrod squaring). It seems as if the representability of cohomology is probably only useful for studying honest cohomology operations, but I don't think I understand exactly what that means well enough to deduce whether changing coefficients qualifies...
| https://mathoverflow.net/users/303 | Relationship between universal coefficient theorem and $[K(\mathbb{Z},n), K(G,n)]$? | You can indeed! By Yoneda, *any* natural way of getting cohomology classes can be realized on the Eilenberg-MacLane spaces. You can look at those "freebie" classes in $H^n(K(\mathbb{Z}, n); G)$ coming from the universal class in $H^n(K(\mathbb{Z}, n); \mathbb{Z})$, and this will give your desired map $K(\mathbb{Z}, n) \to K(G, n)$. However, there's a nice bigger story behind this specific case besides just Yoneda.
First, the Dold-Kan theorem says that (nonnegatively graded) chain complexes of abelian groups are equivalent to simplicial abelian groups. More precisely, given a simplicial abelian group, the alternating sum of the face maps turns it into a chain complex, and modding out by the image of the degeneracies gives the "normalized" chain complex. This turns out the be an equivalence of categories (the inverse is taking a chain complex and formally adding degenerate things to it). What's more, this equivalence preserves the usual notion of homotopy in the two categories: the derived category of abelian groups is the same as simplicial abelian groups with weak equivalences formally inverted.
Now $K(G, n)$ can be realized as the simplicial abelian group which corresponds to the chain complex with $G$ in degree $n$ and $0$ everywhere else. Maps of simplicial abelian groups (mod homotopy) from $K(G, n)$ to $K(H, m)$ are then the same as maps of chain complexes in the derived category, i.e. $\operatorname{Ext}^{m-n}(G, H)$. This is $0$ except for $m=n$ and $m=n+1$, and in those cases you get exactly the correspondence that the universal coefficient theorem gives you. The cohomology operations with $m=n+1$ (coming from the $\operatorname{Ext}$ part of the universal coefficient theorem) are called Bocksteins and can also be obtained as the connecting homomorphisms of a long exact sequence on cohomology coming from a short exact sequence of coefficient groups (namely, the group extension corresponding to your element of $\operatorname{Ext}$).
Note that there are lots of other operations on cohomology (eg, Steenrod operations) besides these. What makes these special is that they can be implemented by maps from $K(G, n)$ to $K(H, m)$ which are *group homomorphisms* with respect to the abelian group structures on the spaces. Note, however, that any cohomology operation which commutes with addition (which includes all Steenrod operations) is a group homomorphism *up to homotopy*, since the group structure on cohomology is just the group structure on $K(G, n)$ taken modulo homotopy. Nevertheless, it is impossible to straighten these out to be homomorphisms on the nose, except in the case of Bocksteins and in the case $n=m$.
| 14 | https://mathoverflow.net/users/75 | 3389 | 2,248 |
https://mathoverflow.net/questions/3390 | 6 | I know that there definitely are two topological spaces with the same homology groups, but which are not homeomorphic. For example, one could take $T^{2}$ and $S^{1}\vee S^{1}\vee S^{2}$ (or maybe $S^{1}\wedge S^{1}\wedge S^{2}$), which have the same homology groups but different fundamental groups. But are there any examples in the smooth category?
| https://mathoverflow.net/users/855 | Are there two non-diffeomorphic smooth manifolds with the same homology groups? | Sure -- there are an abundance of homology spheres in dimension 3 (the [wikipedia](http://en.wikipedia.org/wiki/Homology_sphere) article is pretty nice).
For other examples, in dimension 4 you can find smooth simply-connected closed manifolds whose second homology groups (the only interesting ones) are the same but which have different intersection pairings.
This last subject is very rich. For bathroom reading on it, I cannot recommend Scorpan's book "The Wild World of 4-Manifolds" highly enough.
| 14 | https://mathoverflow.net/users/317 | 3393 | 2,251 |
https://mathoverflow.net/questions/3401 | 9 | Consider an idealized classical particle confined to a two-dimensional surface that is frictionless. The particle's initial position on the surface is randomly selected, a nonzero velocity vector is randomly assigned to it, and the direction of the particle's movement changes only at the surface boundaries where perfectly elastic collisions occur (i.e. there is no information loss over time).
My question is - Does there exist such a bounded surface where the probability of the particle visiting any given position at some time 't', P(x,y,t), becomes equal to unity at infinite time? In other words, no matter where we initialize the particle, and no matter the velocity vector assigned to it, are there surfaces that will always be 'everywhere accessible'?
(Once again, I welcome any help asking this question in a more appropriate manner...)
| https://mathoverflow.net/users/774 | Surfaces that are 'everywhere accessible' to a randomly positioned Newtonian particle with an arbitrary velocity vector | I'll interpret you question to be asking about whether the particle paths are "equidistributed" in the sense of dynamical systems. There is a large literature on this sort of thing, though usually instead of "particles" the authors talk about "billiards". While I don't know the answer to your question as stated, I do know that there are many examples where the paths become equidistributed for "generic" choices of positions and initial directions (in other words, the "bad" choices form a set of measure zero).
Many examples and results of this form can be found in the wonderful survey "Rational billiards and flat structures" by Masur and Tabachnikov, which is available on Masur's [web page](http://math.uchicago.edu/~masur/).
EDIT : I forgot a nice reference! Serge Tabachnikov has written a very accessible book entitled "Geometry and Billiards" which is available on his webpage [here](http://www.math.psu.edu/tabachni/Books/books.html).
| 7 | https://mathoverflow.net/users/317 | 3404 | 2,257 |
https://mathoverflow.net/questions/3405 | 11 | It's well-known that that [a subgroup of a free group is free](http://ncatlab.org/nlab/show/Nielsen-Schreier+theorem). Is a subgroup of a free *abelian* group (may not be finitely generated) always a free abelian group?
| https://mathoverflow.net/users/1 | Is a subgroup of a free abelian group free abelian? | [Yes](https://en.wikipedia.org/w/index.php?title=Free_abelian_group&oldid=317358888#Subgroup_closure).
(EDIT: If you don't like following links, this is the Wikipedia article on Free abelian groups which, uncharacteristically, contains a complete (and correct) proof of precisely that statement).
| 12 | https://mathoverflow.net/users/25 | 3407 | 2,259 |
https://mathoverflow.net/questions/3242 | 36 | Please list some examples of common examples of algebraic structures. I was thinking answers of the following form.
"When I read about a [insert structure here], I immediately think of [example]."
Or maybe you think about a small number of examples. For example, when someone says "group", maybe you immediately think of one example of an abelian group and one example of a non-abelian group. Maybe you have a list of examples that you test new theorems on.
I am an analyst by training. When I read algebra, I can follow the logic line by line, but I don't have the repertoire of examples in my head that I have for analysis and so it's hard for me to picture anything.
| https://mathoverflow.net/users/136 | Canonical examples of algebraic structures | Neat question...
Abelian Group: Z or Z/nZ
Nonabelian Group: Dihedral groups or GL\_n
Commutative Ring: Z or C[x]
Noncommutative Ring: Matrix rings
Division Ring: Hamilton's Quaternions
Field: R or C
Lie Algebra: sl\_2
| 15 | https://mathoverflow.net/users/765 | 3415 | 2,265 |
https://mathoverflow.net/questions/3398 | 5 | This must be an elementary question: could somebody tell me a reference for the Mayer-Vietoris *homotopy* groups sequence of a pull-back of a fibration?
I'm working in the category of pointed simplicial sets. So I've a pull-back of a (Kan) fibration of pointed simplicial sets, and I've read that in this situation you have an associated Mayer-Vietoris sequence relating the homotopy groups of the simplicial sets of the pull-back that looks like the classical Mayer-Vietoris sequence for the singular *homology* of a pair of open sets covering a topological space.
I've been searching in May's "Simplicial objects in Algebraic Topology" and Goerss-Jardine's "Simplicial Homotopy Theory", but I couldn't find it.
| https://mathoverflow.net/users/1246 | Mayer-Vietoris homotopy groups sequence of a pull-back of a fibration | I don't know of a reference, but here is a quick argument. Suppose we want to compute the homotopy pullback P = X ×hZ Y of two maps f : X → Z and g : Y → Z of pointed simplicial sets. Assume for convenience that everything is fibrant. There is a fibration ZΔ[1] → Z∂Δ[1] = Z × Z with fiber ΩZ. Now P is the pullback of the diagram X × Y → Z × Z ← ZΔ[1]. In particular, P → X × Y is also a fibration with fiber ΩZ, and the Mayer-Vietoris sequence follows from the long exact sequence of homotopy groups of this fibration.
| 15 | https://mathoverflow.net/users/126667 | 3416 | 2,266 |
https://mathoverflow.net/questions/3420 | 15 | What is known about countable subgroups of compact groups? More precisely, what countable groups can be embedded into compact groups (I mean just an injective homomorphism, I don't consider any topology on the countable group)? In particular, can one embed S\_\infty^{fin} (the group of permutations with finite support) into a compact group? Any simple examples of a countable group that can't be embedded into a compact group?
| https://mathoverflow.net/users/896 | Countable subgroups of compact groups | Your questions are related to [Bohr compactification](http://en.wikipedia.org/wiki/Bohr_compactification), a left adjoint to the inclusion of compact (= compact Hausdorff) groups into all topological groups. A discrete group G can be embedded into a compact group iff the natural map from G to its Bohr compactification is an injection. Such groups are called "maximally almost periodic". Take a look at [this paper](http://www.math.wisc.edu/~kunen/gpbohr.ps) for a more in-depth treatment. An example from that paper of a countable group which cannot be embedded into a compact group is SL(n, K) for n ≥ 2 and K an infinite countable field.
| 18 | https://mathoverflow.net/users/126667 | 3421 | 2,268 |
https://mathoverflow.net/questions/3237 | 43 | I'd like to learn to read math articles in Japanese or Chinese, but I am not interested in learning these languages from usual textbooks. Exist suitable texts, specialized for the needs for reading mathematics? What do you suggest?
I look for something similar to ["Russian for the mathematician"](https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-28/issue-2/Miscellaneous-back-pages-Comm-Math-Phys-Volume-28-Number-2/cmp/1103858368.full "review"), which was very usefull when I was interested in some russian articles. In the language books I know, most of the vocabulary is irrelevant for reading mathematics, but needed terminology is missing. A collection of mathematical vocabulary and training texts with translation would be usefull. I know good books, e.g. Bowring "An introduction to modern Japanese" or Lewin "Textlehrbuch der japanischen Sprache" and could read articles about history or humanities after having read them, but not mathematics (resulting in forgetting the language by lack of training).
Edit: F. Orgogozo's [dictionary](https://web.archive.org/web/20081208025041/http://www.math.polytechnique.fr:80/~orgogozo/english.html "link"). (BTW, giving the direct link did not work, app. jap./chin. characters not accepted within a url by the MO-software)
Edit: [Zagier's dictionary](https://people.mpim-bonn.mpg.de/zagier/files/scanned/EnglishJapaneseDictMathTerm/fulltext.pdf "link").
| https://mathoverflow.net/users/451 | Japanese/Chinese for mathematicians? | Here are a few for chinese:
Commercial Press Staff. English-Chinese Dictionary of Mathematical Terms. New York: French & European Publications, Incorporated, 1980.
De Francis, John F. Chinese-English Glossary of the Mathematical Sciences. Reprint. Ann Arbor, MI: Books on Demand.
Dictionary of Mathematics. New York: French & European Publications, Incorporated, 1974.
He Xiuhuang. A Glossary of Logical Terms. Hong Kong: Chinese University Press, 1982.
Science Press Staff. English - Chinese Mathematical Dictionary. Second Edition. New York: French & European Publications, Incorporated, 1989.
Science Press Staff. Chinese-English Mathematical Dictionary. New York: French & European Publications, Incorporated, 1990.
Science Press Staff. New Russian - Chinese Dictionary of Mathematical Terms. New York: French & European Publications, Incorporated, 1988.
Silverman, Alan S. Handbook of Chinese for Mathematicians. Berkeley, CA: University of California, Institute of East Asian Studies, 1970.
Source: [here](http://hua.umf.maine.edu/China/dictspec.html#Link18)
I have never read any of these books, and I honestly doubt it that they have all the mathematical terms (especially in higher more sophisticated fields). Don't expect to be able to write "diffeomorphism between manifolds" in chinese or japanese immediately. I suggest you take a look at these references in your public library and get one that helps you the most. To be honest, I am also interested I have several chinese papers I really want to read. I would first try anything with the latest jedict/edict/cedict, and then try something else like the above references.
| 22 | https://mathoverflow.net/users/1245 | 3424 | 2,271 |
https://mathoverflow.net/questions/3419 | 2 | Are there any specialized Computer Algebra Systems (or packages for these) for finding closed form solutions to
a) partial differential equations,
b) stochastic differential equations?
If yes, what experiences do you have with these?
| https://mathoverflow.net/users/1047 | CAS for finding closed form solutions to PDEs and SDEs? | I have used Wolfram Mathematica extensively in my undergraduate course so far. Although the PDEs and systems of PDEs I have encountered have not been overly complicated, Mathematica is able to solve them in closed form most of the time. While not a "specialised" CAS for PDEs/SDEs, it gave me the closed form solutions I was looking for.
[This](http://reference.wolfram.com/mathematica/tutorial/DSolveIntroductionToPDEs.html) link may be useful in terms of gauging what you can do.
| 3 | https://mathoverflow.net/users/1076 | 3432 | 2,276 |
https://mathoverflow.net/questions/3299 | 19 | **Background** Let V denote the standard (2-dimensional) module for the Lie algebra sl2(C), or equivalently for the universal envelope U = U(sl2(C)). The Temperley-Lieb algebra TLd is the algebra of intertwiners of the d-fold tensor power of V.
>
> TLd = EndU(V⊗…⊗V)
>
>
>
Now, let the symmetric group, and hence its group algebra CSd, act on the right of V⊗…⊗V by permuting tensor factors. According to Schur-Weyl duality, V⊗…⊗V is a (U,CSd)-bimodule, with the image of each algebra inside EndC(V⊗…⊗V) being the centralizer of the other.
In other words, TLd is a quotient of CSd. The kernel is easy to describe. First decompose the group algebra into its Wedderburn components, one matrix algebra for each irrep of Sd. These are in bijection with partitions of d, which we should picture as Young diagrams. The representation is faithful on any component indexed by a diagram with at most 2 rows and it annihilates all other components.
So far, I have deliberately avoided the description of the Temperley-Lieb algebra as a diagram algebra in the sense that Kauffman describes it. Here's the rub: by changing variables in Sd to ui = si + 1, where si = (i i+1), the structure coefficients in TLd are all integers so that one can define a ℤ-form TLd(ℤ) by these formulas.
>
> TLd = C ⊗ TLd(ℤ)
>
>
>
As product of matrix algebras (as in the Wedderburn decomposition), TLd has a ℤ-form, as well: namely, matrices of the same dimensions over ℤ. These two rings are very different, the latter being rather trivial from the point of view of knot theory. They only become isomorphic after a base change to C.
---
There is a super-analog of this whole story. Let U = U(gl1|1(C)), let V be the standard (1|1)-dimensional module, and let the symmetric group act by signed permutations (when two odd vectors cross, a sign pops up). An analogous Schur-Weyl duality statement holds, and so, by analogy, I call the algebra of intertwiners the super-Temperley-Lieb algebra, or STLd.
Over the complex numbers, STLd is a product of matrix algebras corresponding to the irreps of Sd indexed by hook partitions. Young diagrams are confined to one row and one column (super-row!). In that sense, STLd is understood. However, idempotents involved in projecting onto these Wedderburn components are nasty things that cannot be defined over ℤ
---
**Question 1:** Does STLd have a ℤ-form that is compatible with the standard basis for CSd?
**Question 2:** I am pessimistic about Q1; hence, the follow up: why not? I suspect that this has something to do with cellularity.
**Question 3:** I care about q-deformations of everything mentioned: Uq and the Hecke algebra, respectively. What about here? I am looking for a presentation of STLd,q defined over ℤ[q,q-1].
| https://mathoverflow.net/users/813 | Does the super Temperley-Lieb algebra have a Z-form? | It depends what you mean by "compatible." For any Z-form of a finite-dimensional C-algebra, there's a canonical Z-form for any quotient just given by the image (the image is a finitely generated abelian subgroup, and thus a lattice). I'll note that the integral form Bruce suggests below is precisely the one induced this way by the Kazhdan-Lusztig basis, since his presentation is the presentation of the Hecke algebra via the K-L basis vectors for reflections, with the additional relations.
What you could lose when you take quotients is positivity (which I presume is one of things you are after). The Hecke algebra of `S_n` has a basis so nice I would call it "canonical" but usually called Kazhdan-Lusztig. This basis has a a very strong positivity property (its structure coefficients are Laurent polynomials with positive integer coefficients). I would argue that this is the structure you are interested in preserving in the quotient.
If you want a basis of an algebra to descend a quotient, you'd better hope that the intersection of the basis with the kernel is a basis of the kernel (so that the image of the basis is a basis and a bunch of 0's). An ideal in the Hecke algebra which has a basis given by a subset of the KL basis is called "cellular."
The kernel of the map to TLd, and more generally to End`U_q(sl_n)`(V⊗d) for any n and d, is cellular. Basically, this is because the partitions corresponding to killed representations form an upper order ideal in the dominance poset of partitions.
However, the kernel of the map to STLd is **not** cellular. In particular, every cellular ideal contains the alternating representation, so any quotient where the alternating representation survives is not cellular. So, while STLd inherits a perfectly good Z-form, it doesn't inherit any particular basis from the Hecke algebra.
I'm genuinely unsure if this is really a problem from your standpoint. I mean, the representation V⊗d still has a basis on which the image of any positive integral linear combination of KL basis vectors acts with positive integral coefficients. However, I don't think this guarantees any kind of positivity of structure coefficients. Also, Stroppel and Mazorchuk have a categorification of the Artin-Wedderburn basis of S\_n, so maybe it's not as bad as you thought.
Anyways, if people want to have a real discussion about this, I suggest we retire to the nLab. I've started a [relevant page](https://ncatlab.org/nlab/show/super+q-Schur+algebra) there.
| 6 | https://mathoverflow.net/users/66 | 3442 | 2,284 |
https://mathoverflow.net/questions/3329 | 15 | Calculating the homology of the loop space and the free loop space is reasonably doable. There exists the Serre spectral sequence linking the homology of the loop space and the homology of the free loop space. Furthermore, for finite CW complexes the James product construct a homotopy approximation to the loop space of a suspension.
One wonders how to extend these calculations to generalized (co)homology theories, like K-theory and K-homology. For which spaces do we know the K-theory and/or K-homology groups? How can one calculate these?
| https://mathoverflow.net/users/798 | What is known about K-theory and K-homology groups of (free) loop spaces? | There are a lot of computational methodologies from algebraic topology that you can apply here, moving from less to more complicated. Suppose E`*` and E`*` is a pair of a generalized homology theory and its cohomology theory, which has a commutative and associative product, and you have a space X where you are interested in the loop space ΩX and the free loop space LX, which live in fibration sequences ΩX -> PX -> X and ΩX -> LX -> X. (Here PX is contractible.)
There are Atiyah-Hirzebruch spectral sequences
```
Hp(Y; Eq(*)) => Ep+q(Y)
Hp(Y; Eq(*)) => Ep+q(Y)
```
which, because they are generic, do not have such stellar behavior except in "easy" cases.
If you have a good grip on the (co)homology of X, then there are the Serre spectral sequences associated to the path-loop fibration
```
Hp(X; Eq(ΩX)) => Ep+q(*)
Hp(X; Eq(ΩX)) => Ep+q(*)
```
and those associated to the free-loop fibration
```
Hp(X; Eq(ΩX)) => Ep+q(LX)
Hp(X; Eq(ΩX)) => Ep+q(LX)
```
The Serre spectral sequence is sometimes less-than-spectacular for loop spaces and free loop spaces, again because it's pretty generic, and because to use then to compute for the loop space you have to play the fiber off the base. This leads to nasty inductive arguments.
Then there are the Eilenberg-Moore spectral sequences for ΩX. If E`*`X is a flat E`*`-module and X is simply connected, then you get a spectral sequence
```
Tor\*\*E\*X(E\*,E\*) => E\*ΩX
```
where this is Tor of graded modules over a graded algebra and inherits a bigrading. This is usually much more straightforward than the standard technique of playing the Serre spectral sequence game to find the homology of the fiber. There's also a homology version but it involves CoTor for comodules over E`*`.
There's also an Eilenberg-Moore spectral sequence starting with Tor over the cohomology of X of the cohomology of LX with the ground ring, and converging to the cohomology of ΩX. This is often less useful because usually you want to go the opposite direction, but it exists.
Finally, there is the Hochschild homology spectral sequence for LX. If E`*`X is a flat E`*`-module and X is simply connected, then there is a spectral sequence
```
HHE\*(E\*X,E\* X) => E\*LX
```
where this is Hochschild homology of E`*`X (over the ground ring E`*`) with coefficients in itself. This is a graded algebra over a graded ring and the Hochschild homology recovers a bigrading. If you instead took coefficients in the ground ring E`*` you're recover the Eilenberg-Moore spectral sequence for the based loop space ΩX.
For example, if E`*` X is a polynomial algebra over E`*` on classes in even degree, the cohomology of the loop space is exterior and the cohomology of the free loop space is the de Rham complex. More complicated cohomology yields more complicated behavior.
If you have specific spaces in mind then there are more specialized results. For example, one major theorem is the Atiyah-Segal theorem relating the K-theory of the classifying space of a compact Lie group to a completion of its complex representation ring. This is very hard to extract from the above general methods.
(Somebody who is an expert in string topology should step in and talk about K-theory of free loop spaces!)
| 17 | https://mathoverflow.net/users/360 | 3451 | 2,290 |
https://mathoverflow.net/questions/3448 | 27 | If so, do people expect certain invariants (regulator, # of complex embeddings, etc) to fully 'discriminate' between number fields?
| https://mathoverflow.net/users/949 | Are there two non-isomorphic number fields with the same degree, class number and discriminant? | John Jones has computed [tables of number fields of low degree with prescribed ramification](http://math.asu.edu/~jj/numberfields/). Though the tables just list the defining polynomials and the set of ramified primes, and not any other invariants, it's not hard to search them to find, e.g., that the three quartic fields obtained by adjoining a root of $x^4 - 6$, $x^4 - 24$, and $x^4 - 12x^2 - 16x + 12$ respectively all have degree $4$, class number $1$, and discriminant $-2^{11} \cdot 3^3$. On the other hand these three fields are non-isomorphic (e.g. the regulators distinguish them, the splitting fields distinguish them...).
| 31 | https://mathoverflow.net/users/379 | 3452 | 2,291 |
https://mathoverflow.net/questions/3409 | 14 | Let U be a Grothendieck universe, and U+ its successor universe (assume Grothendieck's universe axiom).
Everybody agrees that a U-small category is a category whose sets of objects and morphisms are both elements of U. For the "next larger size" of categories which are not necessarily even locally small, call them just U-categories, there are two possible definitions:
* a category whose set of objects and Hom-sets are all subsets of U;
* a category whose set of objects and Hom-sets are all elements of U+ (U+-small categories).
I quite prefer the second notion, so that the category of U-categories is cartesian closed and we can form localizations. This is the usage of Dwyer-Hirschhorn-Kan-Smith, "Homotopy Limit Functors on Model Categories and Homotopical Categories". I think the first more closely corresponds to non-Grothendieck universe-based treatments of category theory using sets and classes, but I might be wrong about that.
For U-locally small categories there are again two possible definitions:
* a category whose set of objects is a subset of U and whose Hom-sets are elements of U,
* a category whose set of objects is an element of U+ and whose Hom-sets are elements of U.
I don't see a strong reason to prefer one over the other, except that the second is more parallel with my preference for U-categories. DHKS uses the first. As an example of the difference between them, if I have a U-locally small category C, I can form the category (poset) of full subcategories of C; this is U-locally small under the second definition, but not the first. Is this a good thing or a bad thing? Or are there no theorems I would care about that are affected by this difference? Does anyone have an opinion about these two definitions?
| https://mathoverflow.net/users/126667 | How should we define "locally small"? | You are correct that the first notion of U-category corresponds more
closely to non-Grothendieck-universe-based treatments, e.g. using NBG
or MK set-class theory. To be precise, if U is a universe, defining
"set" to mean "element of U" and "class" to mean "subset of U" gives a
model of MK set-class theory (and hence also NBG, which is weaker than
MK). A comparison of the relationships between different set-theoretic
treatments of large categories can be found in my expository paper
"[Set theory for category theory](http://www.arxiv.org/abs/0810.1279)."
Here is an example of one theorem that can (maybe) tell the difference
between the two notions of U-locally-small categories. Let C be a
U-category whose hom-sets are in U (i.e. a "U-locally-small category"
by your second definition). Then C has a Yoneda embedding C →
[Cº,Set] where Set is the U-category of U-small sets. Note that
[Cº,Set] is only a U-category by your second definition (i.e. a
U⁺-small category). We say that C is **lex-total** if this Yoneda
embedding has a left adjoint which preserves finite limits. It is a
theorem of Freyd, which can be found in Ross Street's paper "Notions
of topos," that if C is lex-total *and* also U-locally-small according
to your *first* definition (its set of objects is a subset of U), then C
is a Grothendieck topos (i.e. the category of U-small sheaves on some
U-small site). The converse is not hard to prove, so this gives a
characterization of Grothendieck toposes. As far as I know, it is
unknown whether there can be lex-total U-categories with very large
object sets that are not Grothendieck toposes.
I would personally be inclined to use your second definition of
"U-locally small," because as you say it matches your preferred
definition of large category relative to U (which I would prefer to
just call a "U⁺-small category", since its definition makes no reference
to U), and also because the term "U-locally small" sounds as if it
only imposes a smallness condition locally. Street uses "moderate"
for a category with at most a U-small set of isomorphism classes of
objects, so if one wants to state a theorem (such as the above) about
U-locally-small categories according to your first definition, one can
instead say "U-locally-small and U-moderate."
| 8 | https://mathoverflow.net/users/49 | 3464 | 2,297 |
https://mathoverflow.net/questions/3400 | 36 | Take a smooth closed curve in the plane. At each self-intersection, randomly choose one of the two pieces and lift it up just out of the plane. (Perturb the curve so there are no triple intersections.) I don't really know anything about knot theory, so I don't even know if I'm asking the right questions here, but I'm wondering: What is the probability that this is the trivial knot? What can we say about how knotted this knot might be, and with what probabilities? (Measure "knottedness" in whatever way you like.) More generally, can we say anything about the probability of the various possible values in the usual invariants that people use to study knots?
I only have an idea of how to approach the first question, and even then it's only by brute force. I was just playing around with the easiest cases, and I think that with 0, 1, or 2 intersections, all knots are trivial, and with 3 intersections the knot is trivial with probability 75%.
A general analysis should presumably involve calculating the probability that we can simplify using various Reidemeister moves, but I don't know how to incorporate this. I'd imagine a computer could brute-force the first few cases pretty easily (I'm not so bold as to venture an order-of-magnitude guess on whether it's the first few hundred or the first few million)...
| https://mathoverflow.net/users/303 | Probabilistic knot theory | One possible route to a model of random knots would be through the braid group. Every knot can be expressed (non-uniquely) as the closure of a braid. So, for example, you could apply the braid generators uniformly $n$ times across $k$ strands, close the braid using your favorite closure, and then ask this question sensibly. I don't think you can directly ask about the $n \to \infty$ limit for the braid group, though, because I don't think there is a notion of uniform measure for that group. Actually, perhaps I will post this as a separate question, but is the braid group amenable? I would wager that in this model, the probability of having the unknot decreases very quickly with $n$ and $k$.
To test if you have the unknot, it is conjectured that you just have to check the Jones polynomial. But even this is *still* hard in general, unless *even if* you happen to have a quantum computer. :)
(Edit: Thanks Greg Kuperberg, below, for the correction.)
| 12 | https://mathoverflow.net/users/1171 | 3465 | 2,298 |
https://mathoverflow.net/questions/3325 | 0 | Basically you'll find two versions of ito's lemma in the literature: an integral and a differential form. The integral form is based on an Riemann-Stieltjes-integral approach, the differential form is said to be the chain rule for stochastic processes.
**My questions:** Some purists tell you that only the integral form is valid and the differential form is a shortcut to that at most. Why do they think so? Others tell you that both forms are ok and are just two sides of the same coin. What is true now and why?
| https://mathoverflow.net/users/1047 | Ito's lemma in differential form | I think there's an issue with definitions here. Ito's lemma in differential form only makes sense in the context of the integral form. It makes absolutely no sense to speak of $dW(s)/ds$, where $W$ is Brownian motion since it's nowhere differentiable. We DEFINE $dX\_t = sdB\_t+mdt$
by saying that this is shorthand for $X(t)-X(0)= \int\_0^t s dB\_s + \int\_0^t m ds$ . To reiterate, the differential form is a shorthand way of writing the integral form. It's like the notion of defining weak derivatives. You can't speak with a straight face about the derivative (in the usual sense) of say, a dirac delta function, but you can define it through an integral.
| 7 | https://mathoverflow.net/users/934 | 3475 | 2,304 |
https://mathoverflow.net/questions/3468 | 14 | One of the reasons I study von Neumann algebras is that they always have plenty of projections. There are many projectionless $C^\ast$-algebras ($0$ and possibly $1$ are the only projections), but the von Neumann algebras they generate must have nontrivial projections (unless it's just the complex numbers, of course). A good example of this is the reduced group $C^\ast$-algebra of any free group $F\_n$. If $n=1$, then $C\_r^\ast(Z)\cong C(S^1)$ via the Gelfand transform, which is clearly projectionless. If $n\geq 2$, the proof is fairly complicated. See [Davidson's book](http://rads.stackoverflow.com/amzn/click/0821805991) for a proof when $n=2$.
If $G$ is a torsion-free group, is the reduced group $C^\ast$-algebra of $G$ projectionless? This $C^\ast$-algebra always contains the group algebra $C[G]$, so a simpler question is whether $C[G]$ is projectionless if $G$ is torsion-free.
Note that torsion-free is a necessary condition as one gets a projection from summing up the elements in the cyclic group generated by a torsion element and dividing by the order of the element.
EDIT: changed typestting. still some bugs... help please?
| https://mathoverflow.net/users/351 | Do torsion-free groups give projectionless group ($C^\ast$) algebras? | Heh, you've picked an open problem: this is the *Kadison-Kaplansky conjecture*... I would answer it, but first I have to find a sufficiently big margin in which to write the proof.
To be less flippant, it is known to follow (but I don't understand *exactly* how) from the [Baum-Connes conjecture](http://www.math.psu.edu/higson/GFA/bc/special_cases.html): thus, if a torsion-free discrete group satisfies BC, then its reduced group C\*-algebra contains no non-trivial projections.
Trying to answer this question was, I think, one of the original motivations of Connes and others in some of the older work on cyclic cohomology and souped-up versions thereof. See e.g.
M. Puschnigg, The Kadison-Kaplansky conjecture for word-hyperbolic groups. Invent. Math. 149 (2002), no. 1, 153--194.
for some relatively recent work on those lines. Since I'm not an expert, I'd suggest Googling some combination of Kadison-Kaplansky and Baum-Connes and going from there.
| 19 | https://mathoverflow.net/users/763 | 3478 | 2,305 |
https://mathoverflow.net/questions/3476 | 11 | One way to categorify the non-negative integers is to consider the category FinSet whose objects are finite sets and whose morphisms are functions. The isomorphism classes of objects in FinSet can be labeled "sets of cardinality 0, sets of cardinality 1," and so forth, so are a natural way of talking about the non-negative integers.
What I want is a good definition of a category FinSSet (where SSet stands for "super set") whose isomorphism classes can be thought of as "sets of cardinality k" for all integers k in a natural way. A natural candidate for the set of objects is the set of "Z2-graded sets," i.e. pairs of sets (S0, S1). What we want is to define the cardinality of such a set as card(S0) - card(S1), so we want the isomorphism classes of objects to only depend on this number.
Unfortunately, I can't find a definition of the morphisms that actually accomplishes this. What should it be? For what it's worth, I've read "From Finite Sets to Feynman Diagrams" and I think John Baez gives up a little too early on the integers.
| https://mathoverflow.net/users/290 | Is there a categorification of the integers in terms of "graded sets"? | Not sure if you've seen this already, but it looks like Baez talks about this in [one of his "This Week's Finds" columns](http://math.ucr.edu/home/baez/twf_ascii/week102), where he shows that the morphisms are given by tangles.
| 6 | https://mathoverflow.net/users/673 | 3484 | 2,308 |
https://mathoverflow.net/questions/3219 | 8 | Say I start with some a transitive model of a large fragment of ZFC (say enough to run Łoś' Theorem externally) and a specific set x∈M. Now let's say I'm going to pick some M-ultrafilter U on x. By M-ultrafilter, I mean that U measures the subsets of x which are in M.
My question is, by varying U, how much can I affect the ultrapower of M by U? Let's say I limit myself to a U which are countably complete, so that the ultrapower will be wellfounded. If this question is too vague or broad, I'd welcome any interesting examples of things that are possible or impossible.
| https://mathoverflow.net/users/27 | Controlling Ultrapowers | As much as you wish. Lowenheim Skolem give you such situations and then you can affect it too much. For instance, you can construct situations where the critical point is singular in the ultrapower. You cannot do much if U is amenable to M. Then it is like a real ultrafilter.
I don't know what you are asking actually.
An interesting question is whether you can have an ultrafilter on kappa such that the powerset of kappa^+ is in the ultrapower and James Cummings solved this by showing that you can. I don't know if it is interesting to look for ultrafilters that can code powerset(kappa^++) into the ultrapower. probably his proof already gives that.
| 4 | https://mathoverflow.net/users/20584 | 3486 | 2,310 |
https://mathoverflow.net/questions/3483 | 6 | Can anyone explain what a Jacobson radical is using an intuitive example? I can't quite understand [Wikipedia's](http://en.wikipedia.org/wiki/Jacobson_radical) explanation.
| https://mathoverflow.net/users/494 | Intuitive Example of a Jacobson Radical | I think my favourite characterization for rings with identity is that y is in the Jacobson radical of R if and only if 1-yx is right invertible for any x in R - so y is sufficiently "zero-like" that moving the unit by its multiples doesn't stop it being invertible.
In fact one can strengthen this to if and only if 1-zyx is actually a unit for any z,x (and deduce from this that the left and right radicals agree).
| 12 | https://mathoverflow.net/users/310 | 3487 | 2,311 |
https://mathoverflow.net/questions/3482 | 4 | Let $A$ be a commutative, unital Banach algebra and $I \subset A$ an ideal such that $I$ with the relative norm is a uniform Banach algebra and $A / I$ with the quotient norm is uniform as well.
Does it follow that $A$ is uniform?
I expect there to be a counter example involving the Banach algebra $C(X)$ with $X$ a compact Hausdorff space but couldn't quite manage to construct one yet.
| https://mathoverflow.net/users/2258 | uniformity for Banach algebras - a three space property? | I think A has to be *isomorphic* to a uniform algebra, by the following argument.
Let q be the quotient HM from A onto A/I. Let rA be the spectral radius in A, note that if x \in I then || x ||= rI(x)=rA(x).
Let a\in A have norm 1. I claim that rA(a) \geq 1/3.
For, let r > rA(a).
Since the spectral radius can't be increased by a homomorphism, we have
|| q(a) || = rA/I(q(a)) < r;
then, since q is a quotient homomorphism, there exists b \in A such that q(b)=q(a) and || b || < r.
Since a-b \in I we have
rA(a-b) = || a- b || > 1 -r.
But since A is commutative the spectral radius rA is subaddititive, hence
rA(a-b) \leq rA(a) + rA(b) < r + r = 2r.
Therefore 1-r < 2r, i.e. r > 1/3. It follows that rA(a)\geq 1/3. By rescaling, we deduce that ||a|| \geq rA(a) \geq || a||/3 , and thus the Gelfand transform of A is injective with closed range, as claimed.
I hope that does the trick. It's a nice problem, I haven't seen it before, but I'd be very surprised if the argument above - if correct - is either new or best possible.
**EDIT:** I've been asked to expand on some of the steps in the argument above.
Firstly: if q is a quotient map from a B space X to Y, then by defn, for every y\in Y and every \epsilon>0 there exists x\in X with q(x)=y and || x || \leq (1+\epsilon)|| y ||.
In this case X=A, Y=A/I and y=q(a). We know that || q(a) || = r(q(a)) < r, so choosing \epsilon appropriately, we can find b\in A such that || q(b) || < r.
Secondly: we end up showing that r > 1/3. But by definition, r was anything strictly greater than r\_A(a). It follows that r\_A(a) must be at least 1/3; for if it weren't, there would be room in between 1/3 and r\_A(a) to insert some r which satisfies 1/3 > r > r\_A(a), and we've just seen that's not possible.
It might help to look at the argument in a vaguer but more intuitive way (the 1/3 is a slight distraction). Suppose you could find an element a in A which had large norm but very small spectral radius. Then its image in A/I would also have very small spectral radius, hence by your assumption it would have small norm in A/I. By definition of the quotient norm, that means a is very close to I (in the sense of the distance from a point to a closed subspace) and so there exists a' \in I which is very close to a. In particular, a' should have large norm (since a does) and hence have large spectral radius by the assumption on I. But now a and a' are elements of A which are very close together, yet one has very small spectral radius and the other has large spectral radius. That shouldn't be possible, since the spectral radius is dominated by the norm.
Making everything precise above, one gets essentially the original argument I gave. It just so happens that large=1 and very small = 1/3 does the job.
| 4 | https://mathoverflow.net/users/763 | 3495 | 2,317 |
https://mathoverflow.net/questions/3498 | 43 | What is Chern-Simons theory? I have read the wikipedia [entry](http://en.wikipedia.org/wiki/Chern%E2%80%93Simons_theory), but it's pretty physics-y and I wasn't really able to get any sense for what Chern-Simons theory really is in terms of mathematics.
Chern-Simons theory is supposed to be some kind of TQFT. But what kind of TQFT exactly? When mathematicians say that it is a TQFT, does this mean that it's a certain kind of functor from a certain bordism category to a certain target category? If so, what kind of functor is it? What kind of bordism category is it? What kind of target category is it? How exactly is the functor defined?
Also, from attending talks of Michael Freeman, I know that Chern-Simons theory is supposed to describe some aspects of the fractional quantum Hall effect. How does this work? How do I take some sort of Chern-Simons computation on a 3-(or 4-?)manifold and extract from that some kind of physical prediction about some 2d electron gas? I've also heard that Witten has interpretted various knot invariants like the Jones polynomial in terms of Chern-Simons theory. So does this mean that the Jones polynomial of a knot has a physical interpretation? If so, what is it?
| https://mathoverflow.net/users/83 | What is Chern-Simons theory? | Have you read the [nLab entry](http://ncatlab.org/nlab/show/Chern-Simons+theory)? That might answer some of your questions.
| 9 | https://mathoverflow.net/users/66 | 3499 | 2,320 |
https://mathoverflow.net/questions/3502 | 1 | This question is probably very basic, but I've been away from school for a while and the answer eludes me.
I was tempted to prove that d/dx(e^x) = (e^x) for old times sake and that was easy enough. I just expressed e^x as a power series where n goes from 0 to infinity for ((x^n)/n!).
During the derivation I started to wonder, how did they know that a power series where n goes from 0 to infinity for A(subscript n)\*X^n would converge to the form of (someconstant)^x.
Is there some theorem that proves this?
This question is pretty probably trivial for the hardcore math types but it's been bothering me, so I thought I'd ask :-) ....
| https://mathoverflow.net/users/1319 | Why does the power series expressing e^x have the form of a constant raised to x ? | Some people would say that your question is trivial because we *define* the power function by a^x = exp(x log a).
However, that's not a very satisfying answer.
Clearly one wants the power series for exp(x) to satisfy exp(z+w) = exp(z) exp(w), and exp(0) = 1, from what we know the power function should be if z and w are integers. (I'm writing exp(x), not e^x, because I'm assuming exp(x) hasn't yet been shown to have this property.)
So say exp(x) = a0 + a1 x + a2 x^2 + a3 x^3 ... is a formal power series satisfying exp(z+w) = exp(z) exp(w).
Then since exp(0) = 1, we must have a0 = 0.
So exp(x) = 1 + a1 x + a2 x^2 + a3 x^3 + ...; therefore
exp(2x) = exp(x) exp(x) = (1 + a1 x + a2 x^2 + a3 x^3 + ...) (1 + a1 x + a2 x^2 + a3 x^3 + ...)
and expanding the rightmost member of this equation as a formal power series,
exp(2x) = 1 + 2a1 x + (2a2 + a1^2) + (2a3 + 2 a2 a1) x^3 + ...
However, exp(2x) = 1 + 2a1 x + 4a2 x^3 + 8a3 x^3 + ... by substituting 2x into the formal power series for exp(x).
By equating the coefficients of x, x^2, and x^3, you get
2 a1 = 2 a1
2 a2 + a1^2 = 4 a2
2 a3 + 2 a2 a1 = 8 a3
and so on. The first equation tells you nothing. The second tells you a1^2 = 2a2, so a2 = a1^2/2. The third becomes
2 a3 + 2 (a1^2 / 2) a1 = 8 a3
from which you get a1^3 = 6 a3, and a3 = a1^3/6. The pattern here continues, with an = a1^n/n!, as can be proven by induction.
This gives the series
exp(x) = 1 + a1 x + a1^2/2! x^2 + a1^3/3! x^3 + ...
and now we just have to choose a1. We pick 1 just because it's simple to do so.
| 5 | https://mathoverflow.net/users/143 | 3504 | 2,323 |
https://mathoverflow.net/questions/3494 | 3 | Let S be a set of integers and denote the characteristic function of S as $\chi\_{S}(n)$. Define an operator on the space of trig functions by the relation $\hat{Tf}(n) = \chi\_{S}(n) \hat{f}(n)$. Here $\hat{f}(n)$ is the n-th Fourier coefficient of f.
For $p\geq 2$ we'll call S a $L^p$ multiplier set (or just $L^p$ multiplier) if there is an inequality of the form $\Vert Tf\Vert\\_{p} \leq c \Vert f\Vert\\_p$. If this inequality holds for some p but fails for $p+\epsilon$ for every $\epsilon>0$, we'll say that S is a strict $L^p$ multiplier.
Note that every set is a $L^2$ multiplier and if S is a $L^p$ multiplier for some p then it is a $L^q$ multiplier for $2 \leq q \leq p$. Moreover, it follows from a result of Zygmund that almost every (in the obvious sense) set is a strict $L^2$ multiplier. (I also think you can prove this via Khintchine's inequality, but I haven't checked this argument.)
Do strict $L^p$ multiplier sets exist for every $p>2$? Note that this is similar to the $\Lambda(p)$ problem, however, I don't see how to transform a strict $\Lambda(p)$ set into a strict $L^p$ multiplier set.
| https://mathoverflow.net/users/630 | L^{p} multiplier sets | Let M\_p be the class of L^p muliplier sets, as considered in this equation. It is known:
1. This is an algebra of sets, but not a sigma-algebra. [Not sure of the reference.]
2. The inclusion M\_p \subsetneq M\_q is strict for 2\le p
MR1728363 (2001g:42013)
Mockenhaupt, Gerd(5-NSW-SM); Ricker, Werner J.(5-NSW-SM)
Idempotent multipliers for $L^p(\bf R)$.
| 7 | https://mathoverflow.net/users/1158 | 3505 | 2,324 |
https://mathoverflow.net/questions/2976 | 13 | Is there a version of the Löwenheim-Skolem theorem in intuitionistic logic? I'm particularly interested in the "downward" form. The standard proof I know uses the Tarski-Vaught test for elementary substructures, which in turn relies on the fact that "forall" is equivalent to "not exists not", and that fails intuitionistically.
| https://mathoverflow.net/users/49 | Intuitionistic Lowenheim-Skolem? | I have a reference that says the downward Löwenheim-Skolem theorem does not occur
in intuitionistic logic. In the words of the abstract "even a very powerful
version of intuitionistic set theory does not yield any of the usual forms of a countable
downward Löwenheim-Skolem theorem."
Charles McCarty & Neil Tennant, *Skolem's paradox and constructivism*.
Journal of Philosophical Logic.
Springer Netherlands.
Issue Volume 16, Number 2 / May, 1987.
<https://doi.org/10.1007/BF00257838>
Also
page 341 of
A Companion to Metaphysics
By Jaegwon Kim
"...there is no intuitionistically acceptable analogue
of the classical downward Löwenheim-Skolem theorem"
| 14 | https://mathoverflow.net/users/1098 | 3515 | 2,332 |
https://mathoverflow.net/questions/3519 | 17 | Let $M$ be a 4-manifold with a complex structure.
Does there exist a finite list of simply connected complex 4-manifolds $M\_1, ... , M\_n$ such that M is the quotient of some $M\_i$ by the action of a group acting discretely on $M$?
This would be an analog of the Poincare-Koebe uniformization theorem in (real) dimension 2. People who I've asked this question to think that it's unlikely that there is such a list, but haven't been able to offer an argument or a reference.
| https://mathoverflow.net/users/683 | Uniformization theorem in higher dimensions | There exist infinitely many holomorphically non-isomorphic complex structures on the unit ball of R^4 (or more generally R^2n): this is a beautiful theorem of Burns, Shnider, Wells ( Inventiones Math. 46, p. 237-253 ,1978) .
Since these complex manifolds are simply connected (even contractible) they are their own covering spaces and this proves that the set (hum, hum) of universal covers of complex manifolds of dimension 2 is infinite.
As a historical aside, Poincaré was first in noticing that the unit ball and the bidisk (DXD) in C^2 were not holomorphically isomorphic.
Here is the reference to Kang-Tae's nice survey on the subject.
<http://mathnet.kaist.ac.kr/mathnet/kms_tex/100248.pdf>
| 17 | https://mathoverflow.net/users/450 | 3523 | 2,337 |
https://mathoverflow.net/questions/3524 | 20 | Start with A an abelian category and form the derived category D(A).
Take a triangle (not necessarily distinguished) and take it's cohomology. We obtain a long sequence (not necessarily exact). If the triangle is distinguished it is exact. How about the converse: if the long sequence in cohomology is exact does it follow that the triangle is distinguished? (My guess is no, but I can't find a counter-example).
| https://mathoverflow.net/users/1328 | distinguished triangles and cohomology | An important property of the derived category is that distinguished triangles don't just produce long exact sequences in cohomology. If A -> B -> C -> A[1] is an exact triangle and E is another object in the derived category, then you get a long exact sequence
```
... Hom(A[1],E) -> Hom(C,E) -> Hom(B,E) -> Hom(A,E) -> Hom(C[-1],E) -> ...
```
where these Hom-sets are sets of maps in the derived category.
A particular counterexample is as follows. We can view any abelian group as a chain complex concentrated in degree zero. There is a distinguished triangle as follows:
```
ℤ -> ℤ -> ℤ/2 -> ℤ[1]
```
However, we can take the last map ℤ/2 -> ℤ[1] in the sequence (which is not zero in the derived category) and replace it with the zero map. This still gives us a long exact sequence on (co)homology groups. However, if we let E = ℤ/2, then applying maps in the derived category from our new non-distinguished triangle gives us the sequence
```
... 0 -> Hom(ℤ/2,ℤ) -> Hom(ℤ,ℤ) -> Hom(ℤ,ℤ) -> Ext(ℤ/2,ℤ) -> 0 ...
```
where the last map is induced by the *zero* map from ℤ/2[-1] to ℤ, and so it must be zero. This sequence can't possibly be exact, and so the new triangle is not distinguished.
**EDIT**: A more subtle question this suggests is: "Suppose I have a triangle and, for any E, applying Map(-,E) or Map(E,-) gives a long exact sequence. Is this a distinguished triangle?"
The answer to this is actually still no. Still considering chain complexes of abelian groups, take the distinguished triangle
```
ℤ -> ℤ -> ℤ/3 -> ℤ[1]
```
where I'll call the last map β (for Bockstein). You can take this distinguished triangle and replace β with its negative -β. The new triangle still induces long exact sequences on maps in or maps out (because the maps in the triangle have the same kernel and image). However, as an exercise show that this can't be a disntinguished triangle because it's not isomorphic to the original distinguished triangle.
| 29 | https://mathoverflow.net/users/360 | 3535 | 2,343 |
https://mathoverflow.net/questions/3539 | 4 | The Eckmann-Hilton argument is used to prove that a doubly monoidal 0-category is a commutative monoid. If (x) is horizontal composition and . is vertical composition, and assuming that 1(x)a=a=a(x)1, then
a(x)b=(1.a)(x)(b.1)=(1(x)b).(a(x)1)=b.a=(b(x)1).(1(x)a)=(b.1)(x)(1.a)=b(x)a
which shows first that the two kinds of multiplications are the same and second that they are commutative. In a weak 2-category, horizontal composition of 2-cells is only unital up to conjugation by some invertible 2-cells (the unitors). My question is how does one prove the assumption, that horizontal composition is strictly unital, in this situation?
| https://mathoverflow.net/users/1068 | Eckmann-Hilton argument | There are in fact *three* binary operations in play here. There are vertical composition . and horizontal composition \* (or "(x)"), as you say. But then there's also the operation @, as follows.
Your "*a*" and "*b*" here are 2-cells of a doubly degenerate weak 2-category. Let's call its unique 0-cell *x*; then *a* and *b* are 2-cells from 1*x* to 1*x*. Part of the structure of the weak 2-category is a coherence isomorphism lambda: 1*x* o 1*x* --> 1*x*. So, given *a* and *b*, we can form the vertical composite
*a* @ *b* = (lambda . (*a* \* *b*) . lambda^{-1}).
Now, 11*x* is a unit for both . and @, and they satisfy the interchange law, so the Eckmann-Hilton argument applies to tell you that . and @ are equal and commutative. It follows that \* = @. It *then* follows that horizontal composition \* is strictly unital, as required.
| 3 | https://mathoverflow.net/users/586 | 3542 | 2,348 |
https://mathoverflow.net/questions/3540 | 28 | Could someone show an example of two spaces $X$ and $Y$ which are not of the same homotopy type, but nevertheless $\pi\_q(X)=\pi\_q(Y)$ for every $q$? Is there an example in the CW complex or smooth category?
| https://mathoverflow.net/users/1049 | Are there two non-homotopy equivalent spaces with equal homotopy groups? | There are such spaces, for example $X = S^2 \times \mathbb{R}P^3$, $Y = S^3 \times \mathbb{R}P^2.$
(These are both smooth and CW-complexes.)
Whitehead's Theorem says that for CW-complexes, if a map $f : X \to Y$ induces an isomorphism on all homotopy groups then it is a homotopy equivalence. But, as the example above shows,
you need the map. Such a map is called a weak homotopy equivalence.
(Whitehead's Theorem is not true for spaces wilder than CW-complexes. The Warsaw circle
has all of its homotopy groups trivial but the unique map to a point is not a homotopy equivalence.)
| 52 | https://mathoverflow.net/users/1109 | 3543 | 2,349 |
https://mathoverflow.net/questions/2791 | 103 | I have heard people say that a major goal of number theory is to understand the absolute Galois group of the rational numbers $G = \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. What do people mean when they say this? The Kronecker-Weber theorem gives a good idea of what the abelianization of $G$ looks like. But in one of Richard Taylor's MSRI talks, Taylor said that he's never heard of anyone proposing a similar direct description of $G$ and that to understand $G$ one studies the representations of $G$.
I know that there is a strong interest in showing the Langlands reciprocity conjecture [Edit: What I had in mind in writing this is evidently Clozel's conjecture, not the Langlands reciprocity conjecture - see Kevin Buzzard's post below] - that $L$-functions attached to $\ell$-adic Galois representations coincide with $L$-functions attached to certain automorphic representations. And I've heard people refer to the Tannakian philosophy which I understand as (roughly speaking) asserting that $G$ is determined by all of its finite dimensional representations.
Here is a representation of $G$ understood not to be a representation of $G$ as an abstract group but as a group together with a labeling of some of the conjugacy classes of G by rational primes (the Frobenius elements)?
When people talk about "understanding $G$" do they mean proving [Edit: Clozel's conjecture] (in view of the Tannakian philosophy)? If not, what do they mean? If so, this conceptualization seems quite abstract to me. Is this what people mean when they say "understand $G$"? Can [Edit: Clozel's conjecture] be used to give more tangible statements about $G$?
Something that I have in mind as I write this is the inverse Galois problem (does every finite group occur as a Galois group of a normal extension of $\mathbb{Q}$?) and Gross' conjecture (mostly proven by now) that for each prime $p$ there exists a nonsolvable extension of $\mathbb{Q}$ ramified only at $p$. But I am open to and interested in other senses and respects in which one might "understand" $G$.
| https://mathoverflow.net/users/683 | "Understanding" $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ | What would it mean to understand this Galois group? You could mean several things.
You could mean trying to give the group in terms of some smallish generators and relations. This would be nice, and help to answer questions like the inverse Galois problem that Greg Muller mentioned, and having a certain family of "generating" Galois automorphisms would allow you to study questions about e.g. the representation theory in quite explicit terms. However, the Galois group is an uncountable profinite group, and so to give any short description in terms of generators and relations leads you into subtle issues about which topology you want to impose.
You could also ask for a coherent system of *names* for all Galois automorphisms, so that you can distinguish them and talk about them on an individual basis. One system of names comes from the dessins d'enfant that Ilya mentioned: associated to a Galois automorphism we have some associated data.
* We have its image under the cyclotomic character, which tells us how it acts on roots of unity. By the Kronecker-Weber theorem this tells us about the abelianization of the Galois group.
* We also have an element in the free profinite group on two generators, which (roughly speaking) tells us something about how abysmally acting on the coefficients of a power series fails to commute with analytic continuation.
These two names satisfy some relations, called the $2$-, $3$-, and $5$-cycle relation, which are conjectured to generate all relations (at least the last time I checked), but it is difficult to know whether they actually do so. If they do, then the Galois group is the so-called Grothendieck-Teichmüller group.
The problem with this perspective is that the names aren't very explicit (and we don't expect them to be: we may need the axiom of choice to show they exist, and there are only two Galois automorphisms of $\mathbb{C}$ that are measurable functions!) and it seems to be a difficult problem to determine whether the Grothendieck-Teichmuller group really is the whole thing. (Or it was the last time I checked.)
However, the cyclotomic character is a nice, and fairly canonical, name associated for Galois automorphisms. We could try to generalize this: there are Kummer characters telling us what a Galois automorphism does to the system of real positive roots of a positive rational number number (these determine a compatible system of roots of unity, or equivalent an element of the Tate module of the roots of unity). This points out one of the main difficulties, though: we had to make *choices* of roots of unity to act on, and if Galois theory taught us nothing else it is that different choices of roots of an irreducible polynomial should be viewed as indistinguishable. Different choices differ by conjugation in the Galois group.
This brings us to the point JSE was making: if we take the "symmetry" point of view seriously, we should only be interested in conjugacy-invariant information about the Galois group. Assigning names to elements or giving a presentation doesn't really mesh with the core philosophy.
So this brings us to how many people here have mentioned understanding the Galois group: you understand it by how it manifests, in terms of its representations (as permutations, or on dessins, or by representations, or by its cohomology), because this is how it's most useful. Then you can study arithmetic problems by applying knowledge about this. If I have two genus $0$ curves over $\mathbb{Q}$, what information distinguishes them? If I have two lifts of the same complex elliptic curve to $\mathbb{Q}$, are they the same? How can I get information about a reduction of an abelian variety mod $p$ in terms of the Galois action on its torsion points? Et cetera.
| 29 | https://mathoverflow.net/users/360 | 3545 | 2,351 |
https://mathoverflow.net/questions/3525 | 67 | If two different probability distributions have identical moments, are they equal? I suspect not, but I would guess they are "mostly" equal, for example, on everything but a set of measure zero. Does anyone know an example of two different probability distributions with identical moments? The less pathological the better.
**Edit:** Is it unconditionally true if I specialize to discrete distributions?
And a related question: Suppose I ask the same question about Renyi entropies. Recall that the Renyi entropy is defined for all $a \geq 0$ by
$$ H\_a(p) = \frac{\log \left( \sum\_j p\_j^a \right)}{1-a} $$
You can define $a = 0, 1, \infty$ by taking suitable limits of this formula. Are two distributions with identical Renyi entropies (for all values of the parameter $a$) actually equal? How "rigid" is this result? If I allow two Renyi entropies of distributions $p$ and $q$ to differ by at most some small $\epsilon$ independent of $a$, then can I put an upper bound on, say, $||p-q||\_1$ in terms of $\epsilon$? What can be said in the case of discrete distributions?
| https://mathoverflow.net/users/1171 | When are probability distributions completely determined by their moments? | Roughly speaking, if the sequence of moments doesn't grow too quickly, then the distribution is determined by its moments. One sufficient condition is that if the moment generating function of a random variable has positive radius of convergence, then that random variable is determined by its moments. See Billingsley, *Probability and Measure*, chapter 30.
A standard example of two distinct distributions with the same moment is based on the lognormal distribution:
$$f\_0(x) = \sqrt{2 \pi} x^{-1} e^{-(\log x)^2 /2}$$
which is the density of the lognormal, and the perturbed version
$$f\_a(x) = f\_0(x)(1 + a \sin(2 \pi \log x))$$
These have the same moments; namely the nth moment of each of these is exp(n2/2).
A condition for a distribution over the reals to be determined by its moments is that lim supk → ∞ (μ2k)1/2k/2k is finite, where μ2k is the (2k)th moment of the distribution. For a distribution supported on the positive reals, lim supk → ∞ (μk)1/2k/2k being finite suffices.
This example is from Rick Durrett, *Probability: Theory and Examples*, 3rd edition, pp. 106-107; as the original source for the lognormal Durrett cites C. C. Heyde (1963) On a property of the lognormal distribution, J. Royal. Stat. Soc. B. 29, 392-393.
| 49 | https://mathoverflow.net/users/143 | 3553 | 2,357 |
https://mathoverflow.net/questions/3309 | 27 | Can you find two finite groups G and H such that their representation categories are Morita equivalent (which is to say that there's an invertible bimodule category over these two monoidal categories) but where G is simple and H is not. The standard reference for module categories and related notions is [this paper](http://arxiv.org/PS_cache/math/pdf/0111/0111139v1.pdf) of Ostrik's
This is a much stronger condition than saying that C[G] and C[H] are Morita equivalent as rings (where C[A\_7] and C[Z/9Z] gives an example, since they both have 9 matrix factors). It is weaker than asking whether a simple group can be isocategorical (i.e. have representation categories which are equivalent as tensor categories) with a non-simple group, which was shown to be impossible by [Etingof and Gelaki](http://arxiv.org/abs/math/0007196).
Matt Emerton asked me this question when I was trying to explain to him why I was unhappy with any notion of "simple" for fusion categories. It's of interest to the study of fusion categories where the dual even Haagerup fusion category appears to be "simple" while the principal even Haagerup fusion category appears to be "not simple" yet the two are categorically Morita equivalent.
| https://mathoverflow.net/users/22 | Are there two groups which are categorically Morita equivalent but only one of which is simple | I think an answer to your question is given in [Naidu, Nikshych, and Witherspoon - Fusion subcategories of representation categories of twisted quantum doubles of finite groups](http://arxiv.org/abs/0810.0032), theorem 1.1.
Subcategories of the double $D(G)$ are given by pairs of normal subgroups $K$, $N$ in $G$ which centralize each other, together with the datum of a bicharacter $K\times N \to \mathbb C^\times$.
So in particular if $G$ has no normal subgroups and $H$ does, then you're going to find that $D(G)$ has no nontrivial subcategories, while $D(H)$ will (one can take $K$ the normal subgroup in $H$, $N=\{id\}$, and the bicharacter $K\to \mathbb C^\times$ to be trivial, I guess).
| 12 | https://mathoverflow.net/users/1040 | 3554 | 2,358 |
https://mathoverflow.net/questions/3557 | 11 | The two that come to mind are splitting epics in Set and taking the Skel of a category. Surely there are lots of other interesting (and maybe upsetting) places where this comes up.
| https://mathoverflow.net/users/800 | Where are some interesting places where the axiom of choice crops up in category theory? | Using the usual definition of "functor," almost any functor
constructed using only universal properties requires the axiom of
choice. For instance, if a category C has products, then one wants a
"product assigning" functor C×C → C, but in order to define this you
have to *choose* a product for each pair of objects. If C is a large
category, then you need an axiom of choice for proper classes.
However, this sort of thing is arguably not a "real" use of the axiom
of choice. It's more accurate to say that in the absence of the axiom
of choice the usual definition of "functor" is not sufficient, and one
must use [anafunctors](http://ncatlab.org/nlab/show/anafunctor)
instead. Proving that fully faithful + essentially surjective =
equivalence is the same. Most often in category theory when we want
to "choose" something, that thing is in fact determined up to unique
isomorphism (though not uniquely on the nose) and in that case using
anafunctors is sufficient to avoid choice.
The axiom of choice does, however, come up in the study of particular
properties of the category Set. One interesting consequence of the
fact that epics split in Set is that *all functors defined on Set
preserve epics*. I think this is an important part of Blass' proof
that the existence of nontrivial (left and right) exact endofunctors
of set is equivalent to the existence of measurable cardinals.
Another interesting consequence is that Set is its own "ex/lex
completion."
| 27 | https://mathoverflow.net/users/49 | 3563 | 2,365 |
https://mathoverflow.net/questions/3551 | 48 | I'm taking introductory algebraic geometry this term, so a lot of the theorems we see in class start with "Let k be an algebraically closed field." One of the things that's annoyed me is that as far as intuition goes, I might as well add "...of characteristic 0" at the end of that.
I know the complex numbers from kindergarten algebra, so I have a fairly good idea of how at least one algebraically closed field of characteristic 0 looks and feels. And while I don't have nearly the same handle on the field of algebraic numbers, I can pretty much do arithmetic in it, so that's two examples.
But I've never (*really*) seen an algebraically closed field of characteristic p > 0! I can build one just fine, and if you put a gun to my head I could probably even do some arithmetic in it, but there's no intuition of the sort that you get with the complex numbers. So: does anyone know of an intuitive description of such a field, that it's possible to get a real sense of in the same way as C?
| https://mathoverflow.net/users/382 | Algebraically closed fields of positive characteristic | It's certainly not too hard to understand everything there is to understand about the algebraic closure of Fp. Perhaps the reason this is unsatisfying as an example for founding intuition is because it doesn't really have a nice topological structure; it lacks anything like a natural metric. So here's an attempt to explain why what is in some sense the next simplest example puts you in a better situation, intuition-wise.
If you have some intuition about the p-adic numbers look and feel (for example, topologically), then you secretly have intuition for the t-adic topology on the complete local field K=Fp((1/t)). Now, as far as characteristic p fields go, this sort of puts you in the position of (in your parlance) a "preschooler" who knows about R but hasn't yet gotten to kindergarten to learn about C. Why is K like R? First, it is locally compact. Second, it is at least analogous to completing Fp(t), which is very much like Q with Fp[t] as the analogue of Z, at an "infinite" valuation, namely the degree or (1/t)-adic valuation, rather than a "finite" place like a prime polynomial in Fp[t]. (The (1/t)-adic valuation corresponds to the point at infinity on the projective line over F\_p. Likewise, number theorists love to say, perhaps partly to annoy John Conway, that the real and complex absolute values on Q correspond to "archimedean primes" or "primes dividing infinity". This is actually a pretty lame analogy, though, since K=Fp((1/t)) looks a lot more like Fp((t)), say, than R or C looks like Q2.)
Unfortunately there are two extra difficulties in the characteristic p case. First, upon passing to the algebraic closure L of K we lose completeness. Second, we make an infinite field extension, unlike the degree 2 extension C/R. Thus, while L is an algebraically closed field of characteristic p, it bears little resemblance to R. In fact, it's a lot more like an algebraically closed field of characteristic 0 that is a bit scarier (at least to me) than C, namely Cp, or what you get when you complete the algebraic closure of Qp with respect to the topology coming from the unique extension of the p-adic valuation.
While this may seem bad, I think it's actually good, because one can really get a handle on some of the properties of Cp. [Note that as another answerer pointed out, Cp = C as a field, but not as a topological or valued field, which is really a more interesting structure to consider from the viewpoint of intuition anyway.]
For example of some similarities, miraculously Cp turns out to still be algebraically closed, and I believe the same proof goes through for L above. Another property L and Cp share is that in addition to "geometric" field extensions K'/K obtained by considering function fields of plane curves over Fp, there are also "stupider" extensions coming from extending the coefficient field. This is like passing to unramified extensions of p-adic fields, where one ramps up the residue field. (In fact, it's exactly the same thing.) Both L and Cp are complete valued fields with residue field the algebraic closure of Fp. (But the valuation is NOT discrete; it takes values in Q.) There are some dangerous bends to watch out for topologically, however. Some cursory googling tells me that Cp is not locally compact, although it is topologically separable.
In addition, positive characteristic inevitably brings along the problem of inseparable field extensions sitting side L. This is, of course, an aspect where L/K is unlike Cp/Qp. Notwithstanding such annoyances, I would argue that the picture sketched above actually does give an example of an algebraically closed field of characteristic p for which it is possible to have some real intuition.
| 24 | https://mathoverflow.net/users/412 | 3568 | 2,369 |
https://mathoverflow.net/questions/3528 | 4 | Let Σ be an axiom system. Can there be a formula φ, s.t.
* Con(Σ) does not imply Con(Σ + φ) AND
* Con(Σ) does not imply Con(Σ + not φ)
If yes, can you give me an example for ZFC?
| https://mathoverflow.net/users/1330 | Is there a formula phi s.t. phi and not-phi have a stronger consistency? | No, it's impossible for any axiom system. If Σ is consistent, then by the Completeness theorem, it has some model M. In M, φ is either true or false. So M is a model of either (Σ+φ) or (Σ+not φ). So at least one of them is consistent. It might be that your metatheory doesn't know which one is consistent, but it knows that at least one of them is.
| 9 | https://mathoverflow.net/users/27 | 3576 | 2,374 |
https://mathoverflow.net/questions/3575 | 5 | If I understand FC's remark under the post "[Very strong multiplicity one for Hecke eigenforms](https://mathoverflow.net/questions/3352/very-strong-multiplicity-one-for-hecke-eigenforms)," in the course of Faltings's proof of the Tate conjecture, Faltings proves the following statement: let E/Q and F/Q be elliptic curves and write Q(E[p]) (resp. Q(F[p])) for the number field obtained by adjoining the x and y coordinates of the p-torsion of E to Q. Then if Q(E[p]) = Q(F[p]) for infinitely many primes p, E/Q and F/Q are isogenous.
Learning of this result prompted me to wonder: suppose P is a *finite* set of primes. Then do there exist E/Q and F/Q such that Q(E[p]) = Q(F[p]) for each p in P with E/Q and F/Q *not isogenous*?
If not in general, what is known about the particular P for which the above question does (or doesn't) have an affirmative answer?
| https://mathoverflow.net/users/683 | An inverse problem: Number fields attached to elliptic curves over Q | Most people would say no. Indeed, there's a conjecture, most commonly ascribed to Frey, that for p a SINGLE large enough prime, E[p] isomorphic to F[p] implies that E and F are isogenous. I believe p=37 is thought to be large enough, but don't hold me to it.
| 8 | https://mathoverflow.net/users/431 | 3577 | 2,375 |
https://mathoverflow.net/questions/3474 | 12 | There's a well-known decomposition of $L^2(G)$, a regular representation of compact complex group Lie $G$, called **[Peter-Weyl theorem](http://en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem)**.
Turns out for some reason I [automatically think](https://mathoverflow.net/questions/3446/tannakian-formalism/3466#3466) that there is a similar theorem that decomposes regular representation $k[G]$ of *algebraic* group $G$:
$$k[G] = \bigoplus\_R \ R^\* \otimes R$$
where sum goes over representations to $GL(n, k)$. For this to work I think we need $G$ to be a linear reductive group over, say, algebraically closed field $k$ of characteristic 0. Also, perhaps we need $\pi\_1(G) = 1$.
But perhaps this is not true — the search hasn't given me a reference yet, but I wasn't able to provide a counterexample either.
Consider, for example, the multiplicative group $\mathbb G\_m$. Then $k[\mathbb G\_m] = k[x, x^{-1}]$ where each summand $k\cdot x^n$ is a separate representation of $\mathbb G\_m$ into $\mathbb G\_m = GL(1, k)$, specifically the one given by $a \mapsto a^n$. So the identity works.
So, is there such a theorem? What's a reference or a counterexample?
| https://mathoverflow.net/users/65 | Decomposition of k[G] | This is true for reductive groups, more or less by definition. An algebraic representation of an algebraic group is a comodule V over the algebra of functions O(G) of the group. Therefore, every representation V induces a map
V -> V ⊗ O(G), or equivalently V^\* ⊗ V --> O(G) (call the source of this map C(V) for coefficient space of V). It is not hard to see that the latter is a map of G x G modules. If G is reductive, then its representation category is semi-simple, and thus so is the representation category of G x G. In this case the simples of G x G are external tensor product of simples of V, and Hom(A ⊗' B, C ⊗' D) = d(A,C) ⊗ d(B,D) where d(V,W)=0 if v \cong W, C else. Here ⊗' means external tensor product. There doesn't appear to be a ⊠
For non-reductive groups, you can still form O(G) in an analogous way:
Let A = ⊕V V^\* ⊗' V, where here the sum is over ALL finite dimensional modules V (not just isoclass representatives, and not just simples), and again the tensor product is external, so this lives in a completion of Rep(G) ⊗' Rep(G), and ⊗' means Deligne tensor product of categories.
Well this A is way too big, but now let's quotient A by the images of f^\* ⊗' id - id ⊗' f, for all f:V-->W. This cuts A back down, for instance it identifies C(V) and C(V') whenever V and V' are isomorphic. If the category Rep(G) is semi-simple, you can similarly use the projectors and inclusions of simple objects to reduce to a Peter-Weyl type decomposition.
One nice thing about this construction (even in the semi-simple case) is that it is basis free because you don't choose representatives of simple objects, and also it makes the multiplication structure completely trivial:
V^\* ⊗' V ⊗2 W^\* ⊗' W = V^\* ⊗ W^\* ⊗' V ⊗ W --> W^\* ⊗ V^\* ⊗' V ⊗ W,
using the braiding (tensor swap). It also works in braided tensor categories and explains the multiplication structure on the "covariantized" quantum group.
| 16 | https://mathoverflow.net/users/1040 | 3580 | 2,376 |
https://mathoverflow.net/questions/3605 | 2 | The following question is posed in the book "The USSR Olympiad Problem Book: Selected Problems and Theorems of Elementary Mathematics"
"Prove that if integers a\_1, ..., a\_n are all distinct, then the polynomial
(x-a\_1)^2(x-a\_2)^2...(x-a\_n)^2 + 1
cannot be written as the product of two other polynomials with integral coefficients"
I still haven't solved this in the elementary case, but I want to pose it in a more general setting. (I'll post the elementary answer in a bit, I want to try a bit more to figure it out)
EDIT: Now I have solved it, it's MUCH more trivial than I thought, major brain fart on my part
Suppose we have a ring R[x], and the polynomial written above is factorable in this ring. Suppose further that the coefficients are members of a subset of the field that the ring sits in, and that none of the a\_i are equal (which means that some finite fields are of course out). Under what conditions IS the polynomial factorable into a product of two polynomials such that their coefficients sit inside the subset? Does the subset have to be algebraically closed?
(Is factorability even remotely easy in a noncommutative ring? I don't see a priori why a factorization would be unique either in a noncommutative ring).
Motivation: I'm doing research in mathematics education, and am interested in the metacognitive faculties of early college and high school pupils, which, for those not versed in metacognition, is the ability to separate oneself from "nitty-gritty" of the problem and think in more general terms. Alan Schoenfeld is the standard reference on this. I'm looking for problems that are good to pose to students to try and understand their thinking skills, and so am looking through particularly hard problems that do not require a strong background in mathematics. In this particular case I'd like to understand the problem in greater depth myself, and hopefully use my more general knowledge of the situation to aid in my study of how students think about such problems.
Hope this is interesting to someone, and that it isn't too specific.
| https://mathoverflow.net/users/429 | An "Elementary" Math Question Generalized (Ring Theory Perhaps) | I don't think this is a good problem for metacognition. Solving it is too contingent on what people have taught you about irreducibility.
Anyway, as for your general question, I am sure you can find a trivial example over Z/mZ for some composite m. Also, I can't tell whether you want a solution to the specific question, so here it is: suppose that the given polynomial is equal to f(x) g(x), where f, g have integer coefficients. Then f(ak) g(ak) = 1 for all k. Since f(x) g(x) has no real roots, neither does f or g, so they cannot take both the values +1 and -1. Therefore suppose WLOG that f(ak) = g(ak) = 1 for all k. On the other hand, at least one of f, g has degree at most n. The problem is straightforward from here.
| 3 | https://mathoverflow.net/users/290 | 3608 | 2,396 |
https://mathoverflow.net/questions/2981 | 3 | Consider two topological spaces X,Y and a function f from X to Y.
Are the following concepts already in use? How are they called?
1) f sends open subsets of X to either open or closed subsets of Y.
2) f sends closed subsets of X to either open or closed subsets of Y.
3) Both 1) and 2) simultaneously.
1') The preimage of every open subset of Y is either open or closed in X.
2') The preimage of every closed subset of Y is either open or closed in X.
3') Both 1') and 2') simultaneously.
(Obviously, those can be seen as weak generalizations for the definitions of open, closed and continuous maps).
Are there some useful results about them? Who has studied them and where?
| https://mathoverflow.net/users/1234 | Has anyone studied the applications which map open sets to either open or closed sets? | 1' and 2' (and thus 3') are equivalent.
| 2 | https://mathoverflow.net/users/1119 | 3620 | 2,408 |
https://mathoverflow.net/questions/3624 | 11 | Let k be an algebraically closed field. It's well known that every complete curve, period, is projective. Also, that every smooth surface is, and that there are smooth 3-folds which are not, and people go to reasonable lengths to include these examples all over the place, so they're easy to find. However, Hartshorne does say that singular complete surfaces are not all projective. Is there a simple example? A complete normal surface that is not projective? Is there some "least singular" possible such surface? I suspect that normality is too much to hope for, but I can't quite phrase why I think this, so is every normal complete surface projective?
| https://mathoverflow.net/users/622 | Nonprojective Surface | There is a construction of a proper normal non-projective surface [here](http://reh.math.uni-duesseldorf.de/~schroeer/publications_pdf/on_non_proj.pdf) .
There is an example given by Nagata in his paper "Existence theorems for nonprojective complete algebraic varieties" in the Illinois Journal, but I don't know where this is available on the web.
Over a finite field complete + normal implies projective for surfaces.
| 18 | https://mathoverflow.net/users/310 | 3625 | 2,412 |
https://mathoverflow.net/questions/3477 | 74 | I find [Wikipedia's discussion](http://en.wikipedia.org/wiki/Symbol_of_a_differential_operator) of symbols of differential operators a bit impenetrable, and Google doesn't seem to turn up useful links, so I'm hoping someone can point me to a more pedantic discussion.
### Background
I think I understand the basic idea on $\mathbb{R}^n$, so for readers who know as little as I do, I will provide some ideas. Any differential operator on $\mathbb{R}^n$ is (uniquely) of the form $\sum p\_{i\_1,\dotsc,i\_k}(x)\frac{\partial^k}{\partial x\_{i\_1}\dots\partial x\_{i\_k}}$, where $x\_1,\dotsc,x\_n$ are the canonical coordinate functions on $\mathbb{R}^n$, the $p\_{i\_1,\dotsc,i\_k}(x)$ are smooth functions, and the sum ranges over (finitely many) possible indexes (of varying length). Then the **symbol** of such an operator is $\sum p\_{i\_1,\dotsc,i\_k}(x)\xi^{i\_1}\dotso\xi^{i\_k}$, where $\xi^1,\dotsc,\xi^n$ are new variables; the symbol is a polynomial in the variables $\{\xi^1,\dotsc,\xi^n\}$ with coefficients in the algebra of smooth functions on $\mathbb{R}^n$.
Ok, great. So symbols are well-defined for $\mathbb{R}^n$. But most spaces are not $\mathbb{R}^n$ — most spaces are formed by gluing together copies of (open sets in) $\mathbb{R}^n$ along smooth maps. So what happens to symbols under changes of coordinates? An **affine change of coordinates** is a map $y\_j(x)=a\_j+\sum\_jY\_j^ix\_i$ for some vector $(a\_1,\dotsc,a\_n)$ and some invertible matrix $Y$. It's straightforward to describe how the differential operators change under such a transformation, and thus how their symbols transform. In fact, you can forget about the fact that indices range $1,\dotsc,n$, and think of them as keeping track of tensor contraction; then everything transforms as tensors under affine coordinate changes, e.g. the variables $\xi^i$ transform as coordinates on the cotangent bundle.
On the other hand, consider the operator $D = \frac{\partial^2}{\partial x^2}$ on $\mathbb{R}$, with symbol $\xi^2$; and consider the change of coordinates $y = f(x)$. By the chain rule, the operator $D$ transforms to $(f'(y))^2\frac{\partial^2}{\partial y^2} + f''(y) \frac{\partial}{\partial y}$, with symbol $(f'(y))^2\psi^2 + f''(y)\psi$. In particular, the symbol did not transform as a function on the cotangent space. Which is to say that I don't actually understand where the symbol of a differential operator lives in a coordinate-free way.
### Why I care
One reason I care is because I'm interested in quantum mechanics. If the symbol of a differential operator on a space $X$ were canonically a function on the cotangent space $T^\ast X$, then the inverse of this Symbol map would determine a "quantization" of the functions on $T^\ast X$, corresponding to the QP quantization of $\mathbb{R}^n$.
But the main reason I was thinking about this is from Lie algebras. I'd like to understand the following proof of the PBW theorem:
>
> Let $L$ be a Lie algebra over $\mathbb{R}$ or $\mathbb{C}$, $G$ a group integrating the Lie algebra, $\mathrm{U}L$ the universal enveloping algebra of $L$ and $\mathrm{S}L$ the symmetric algebra of the vector space $L$. Then $\mathrm{U}L$ is naturally the space of left-invariant differential operators on $G$, and $\mathrm{S}L$ is naturally the space of symbols of left-invariant differential operators on $G$. Thus the map Symbol defines a canonical vector-space (and in fact coalgebra) isomorphism $\mathrm{U}L\to\mathrm{S}L$.
>
>
>
| https://mathoverflow.net/users/78 | What is the symbol of a differential operator? | The **principal symbol** of a differential operator
$\sum\_{|\alpha| \leq m} a\_\alpha(x) \partial\_x^\alpha$
is by definition the function $\sum\_{|\alpha| = m} a\_\alpha(x) (i\xi)^\alpha$
Here $\alpha$ is a multi-index (so $\partial\_x^\alpha$ denotes $\alpha\_1$ derivatives
with respect to $x\_1$, etc.)
At this point, the vector $\xi = (\xi\_1, \ldots, \xi\_n)$ is merely a formal variable.
The power of this definition is that if one interprets $(x,\xi)$ as variables in the
cotangent bundle in the usual way -- i.e. $x$ is any local coordinate chart, then
$\xi$ is the linear coordinate in each tangent space using the basis $dx^1, \ldots, dx^n$,
then the principal symbol is an invariantly defined function on $T^\*X$, where $X$ is
the manifold on which the operator is initially defined, which is homogeneous of degree
$m$ in the cotangent variables.
Here is a more invariant way of defining it: fix $(x\_0,\xi\_0)$ to be any point in $T^\*X$
and choose a function $\phi(x)$ so that $d\phi(x\_0) = \xi\_0$. If $L$ is the differential operator,
then $L( e^{i\lambda \phi})$ is some complicated sum of derivatives of $\phi$, multiplied together, but always with a common factor of $e^{i\lambda \phi}$. The `top order part' is the one which has a $\lambda^m$, and if we take only this, then its coefficient has only first derivatives of $\phi$ (lower order powers of $\lambda$ can be multiplied by higher derivatives of $\phi$).
Hence if we take the limit as $\lambda \to \infty$ of $\lambda^{-m} L( e^{i\lambda \phi})$
and evaluate at $x = x\_0$, we get something which turns out to be exactly the principal symbol of $L$ at the point $(x\_0, \xi\_0)$.
There are many reasons the principal symbol is useful. There is indeed a `quantization map'
which takes a principal symbol to any operator of the correct order which has this as its principal symbol. This is not well defined, but is if we mod out by operators of one order lower. Hence the comment in a previous reply about this being an isomorphism between filtered algebras.
In special situations, e.g. on a Riemannian manifold where one has preferred coordinate
charts (Riemann normal coordinates), one can define a total symbol in an invariant fashion
(albeit depending on the metric). There are also other ways to take the symbol, e.g. corresponding to the Weyl quantization, but that's another story.
In microlocal analysis, the symbol captures some very strong properties of the operator $L$.
For example, $L$ is called **elliptic** if and only if the symbol is invertible (whenever $\xi \neq 0$). We can even talk about the operator being elliptic in certain directions if the principal symbol is nonvanishing in an open cone (in the $\xi$ variables) about those directions. Another interesting story is wave propagation: the characteristic set of the operator is the set of $(x,\xi)$ where the principal symbol $p(L)$ vanishes. If its differential (as a function on the cotangent bundle) is nonvanishing there, then the integral curves of the Hamiltonian flow associated to $p(L)$, i.e. for the Hamiltonian vector field determined by $p(L)$ using the standard symplectic structure on $T^\*X$, ``carries'' the singularities of solutions of $Lu = 0$. This is the generalization of the classical fact that singularities of solutions of the wave equation propagate along light rays.
| 54 | https://mathoverflow.net/users/888 | 3630 | 2,417 |
https://mathoverflow.net/questions/2272 | 8 | Has there been any progress about constructing **strong pseudorandom generators**?
I'm not an expert on this topic, basically everything I know is a definition of a pseudorandom generator, the idea that they are related to one-way functions, as well as other standard parts of complexity theory.
I'll appreciate any related information, e.g. what it is equivalent to.
| https://mathoverflow.net/users/65 | Pseudorandom generators | Ilya,
It's possible that I'm misinterpreting what you're asking (since complexity theorists, applied cryptographers, and combinatorialists all tend to use *slightly* different definitions of "pseudorandom"), but from a theory standpoint, I think the question is essentially solved and has been for a while.
As you mentioned, PRNGs are related to one-way functions. In particular it's easy to see that a PRNG immediately provides you with a one-way function; just plug in some initial parameters and run the generator! (Whence the implication to P \neq NP; a one-way function is clearly hard-on-average, and a hard-on-average function is clearly hard in the worst case.) It was a longstanding open problem whether the converse was true, whether you could build a PRNG from a one-way function. About a decade ago, Hastad, Impagliazzo, Levin and Luby answered this question in the affirmative in a [massive and technically challenging paper](http://www.icsi.berkeley.edu/~luby/PAPERS/hill.ps). The HILL construction is hugely inefficient, but from a theory standpoint, it shows that the question of the existence of PRNGs is equivalent to the existence of one-way functions.
If you want a PRNG strong enough to derandomize BPP, the question actually becomes a bit easier, and certainly less technical. On the assumption that some problem in E requires exponential-size circuits, Impagliazzo and Wigderson [construct](http://www.math.ias.edu/~avi/PUBLICATIONS/MYPAPERS/IW97/proc.pdf) such a generator. (The I-W paper is fantastic -- essential reading for anyone interested in derandomization.) It's also known that this is essentially the best we can do, in that derandomizing BPP requires either proving circuit lower bounds for E, or showing that one of several things no one really believes is true. (E.g., P = NP.) This is a result of I think Kabanets and someone in {Impagliazzo, Wigderson, Goldreich}, although I can't remember or find the paper.
| 11 | https://mathoverflow.net/users/382 | 3651 | 2,431 |
https://mathoverflow.net/questions/3653 | 0 | Suppose I have $n-1$ distinguishable labels for internal nodes $A=\{a\_1, a\_2,\dots, a\_{n-1}\}$ and $n$ distinguishable labels for leaves $B=\{b\_1,b\_2,\dots, b\_n\}$ with $A$ and $B$ disjoint. What is the best way to iterate over all possible binary trees if I label without replacement?
| https://mathoverflow.net/users/812 | How do I iterate over binary trees? | Disclaimer: I have no computer science background, this is probably not the fastest method of solving your problem.
It is easy to iterate over all unlabeled binary trees of a given size. (I hope you agree.)
If what you're doing is computing some sum over binary trees, then the easiest way to reduce to this situation might be to first iterate over all unlabeled trees, and then for each unlabeled tree add
1/|Aut(T)|\*(sum over all (n-1)!\*n! possible labelings of the tree T)
where Aut(T) is the group of automorphisms of the tree. The cardinality of the automorphism group can be computed recursively: one defines a function (in pseudocode)
```
function Aut(T):
if T == {leaf}:
return 1
T1, T2 <- T.subtree(left), T.subtree(right)
if T1==T2:
return 2*Aut(T1)^2
else:
return Aut(T1)*Aut(T2)
```
When you compare whether T1 and T2 are equal, you can again use recursion.
```
function equals(T1,T2):
if T1 == {leaf}:
return T2 == {leaf}
if T2 == {leaf}:
return false
T11,T12,T21,T22 <- T1.subtree(left), T1.subtree(right), T2.subtree(left), T2.subtree(right)
if equals(T11,T21):
return equals(T12,T22)
if equals(T11,T22):
return equals(T11,T22)
return false
```
If you're not computing a sum but you really want an iterator over all labeled trees, one way could be to implement something similar to this. First iterate over all unlabeled trees, then for each internal vertex of your tree check whether there exists an automorphism switching the left and right subtree. If so, rigidify the tree by imposing the condition that the root at the left subtree should be labeled by a smaller element than the root at the right subtree; sum only over these labelings.
| 4 | https://mathoverflow.net/users/1310 | 3659 | 2,436 |
https://mathoverflow.net/questions/3668 | 4 | I wanted to know if the following family of graphs has a name in graph theory: A claw with paths of any length attached to the three free vertices of the claw. More formally, a connected acyclic graph, with 1 vertex of degree 3 and the rest of degree 2 or less.
They're interesting because they arise in the study of graph minors. (In particular, if a graph of this type is a minor of another graph G, then it is also a subgraph of G.)
| https://mathoverflow.net/users/1042 | A name for a claw-graph with paths attached to it | I don't see any reason not to call them "subdivisions of claws," since that's exactly what they are; people working in subfactors [apparently](https://mathoverflow.net/questions/86/a-name-for-star-graph-with-long-laces) call them "star-shaped," or I guess in this case "claw-shaped." I don't know of any other name for them, though.
Now that I think about it, aren't these exactly the trees with exactly three leaves? Do trees with a specified number of leaves have a name?
| 4 | https://mathoverflow.net/users/382 | 3673 | 2,447 |
https://mathoverflow.net/questions/3687 | 10 | This is a follow up to my question [about D-modules supported on the nilpotent cone](https://mathoverflow.net/questions/2971/d-modules-supported-on-the-nilpotent-cone). I got some good answers there but now I have a more basic question.
Consider an affine algebraic variety X, a closed subvariety i:Y-->X, and the intermediate extension of the structure sheaf on Y to all of X (do I denote this i!\*OY ? For that matter, explaining either the \* or ! extension instead would be a helpful start if its easier).
My question is this: Since X is affine, D(X) is just an associative algebra, generated by O(X) and Vect(X) by the usual construction. My question is how can I understand i!\*OY as a module the associative algebra D(X), supposing I understand D(X)?
In other words, what is the vector space underlying i!\*OY, how do functions in O(X) and vector fields act?
I probably need to invest some serious time with a textbook to answer this question myself, but any help getting started would be most appreciated!
| https://mathoverflow.net/users/1040 | Making D-modules on affine varieties more explicit | Have you looked at Bernstein's [lectures](http://www.math.uchicago.edu/~mitya/langlands/Bernstein/Bernstein-dmod.ps) on D-modules? He proves a result relevant to your question in Lec. 3, Sec. 14: for an affine embedding Y --> X with Y irreducible, if E is an OY-coherent DY module (i.e. a vector bundle with a flat connection) then the !\* direct image from E can be characterized as the unique irreducible subquotient of either the \* or ! direct image which has nonzero restriction to Y. Since it can be easy in such situations to compute the \* direct image, this may be a good way to get a handle on the other functors too. For example, I think one can see from this that if we take the embedding of the origin into the affine line, then both the \* and !\* direct images of OY=k (the ground field) are A(1)/A(1)t, where t is the coordinate on the affine line and A(1)=DX is the 1-dimensional Weyl algebra k[t,d/dt]. This quotient is rightly considered the "delta-function" A(1)-module, since its generator δ satisfies tδ=0. I'm not sure how similar to this case your general situation will be. But certainly by Kashiwara's theorem one knows that the \* direct image (and hence the !\* direct image) will be supported on the subvariety Y.
| 5 | https://mathoverflow.net/users/412 | 3692 | 2,462 |
https://mathoverflow.net/questions/3455 | 35 | For (suitable) real- or complex-valued functions $f$ and $g$ on a (suitable) abelian group $G$, we have two bilinear operations: multiplication -
$$(f\cdot g)(x) = f(x)g(x),$$
and convolution -
$$(f\*g)(x) = \int\_{y+z=x}f(y)g(z)$$
Both operations define commutative ring structures (possibly without identity) with the usual addition. (For that to make sense, we have to find a subset of functions that is closed under addition, multiplication, and convolution. If $G$ is finite, this is not an issue, and if G is compact, we can consider infinitely differentiable functions, and if $G$ is $\mathbb R^d$, we can consider the Schwarz class of infinitely differentiable functions that decay at infinity faster than all polynomials, etc. As long as our class of functions doesn't satisfy any additional nontrivial algebraic identities, it doesn't matter what it is precisely.)
My question is simply: **do these two commutative ring structures satisfy any additional nontrivial identities?**
A "trivial" identity is just one that's a consequence of properties mentioned above: e. g., we have the identity
$$f\*(g\cdot h) = (h\cdot g)\*f,$$
but that follows from the fact that multiplication and convolution are separately commutative semigroup operations.
**Edit:** to clarify, an "algebraic identity" here must be of the form $A(f\_1, ..., f\_n) = B(f\_1, ..., f\_n)$," where $A$ and $B$ are composed of the following operations:
* addition
* negation
* additive identity (0)
* multiplication
* convolution
(Technically, a more correct phrasing would be "for all $f\_1, ..., f\_n$: $A(f\_1, ..., f\_n) = B(f\_1, ..., f\_n)$," but the universal quantifier is always implied.) While it's true that the Fourier transform exchanges convolution and multiplication, that doesn't give valid identities unless you could somehow write the Fourier transform as a composition of the above operations, since I'm not giving you the Fourier transform as a primitive operation.
**Edit 2**: Apparently the above is still pretty confusing. This question is about identities in the sense of universal algebra. I *think* what I'm really asking for is the variety generated by the set of abelian groups endowed with the above five operations. Is it different from the variety of algebras with 5 operations (binary operations +, \*, .; unary operation -; nullary operation 0) determined by identities saying that $(+, -, 0, \*)$ and $(+, -, 0, \cdot)$ are commutative ring structures?
| https://mathoverflow.net/users/302 | Do convolution and multiplication satisfy any nontrivial algebraic identities? | I think the answer to the original question (i.e. are there any universal algebraic identities relating convolution and multiplication over arbitrary groups, beyond the "obvious" ones?) is negative, though establishing it rigorously is going to be tremendously tedious.
There are a couple steps involved. To avoid technicalities let's restrict attention to discrete finite fields G (so we can use linear algebra), and assume the characteristic of G is very large.
Firstly, given any purported convolution/multiplication identity relating a bunch of functions, one can use homogeneity and decompose that identity into homogeneous identities, in which each function appears the same number of times in each term. (For instance, if one has an identity involving both cubic expressions of a function f and quadratic expressions of f, one can separate into a cubic identity and a quadratic identity.) So without loss of generality one can restrict attention to homogeneous identities.
Next, by [depolarisation](http://en.wikipedia.org/wiki/Polarization_identity), one should be able to reduce further to the case of multilinear identities: identities involving a bunch of functions $f\_1, f\_2, ..., f\_n$, with each term being linear in each of the functions. (I haven't checked this carefully but it should be true, especially since we can permit the functions to be complex valued.)
It is convenient just to consider evaluation of these identities at the single point 0 (i.e. scalar identities rather than functional identities). One can actually view functional identities as scalar identities after convolving (or taking inner products of) the functional identity with one additional test function.
Now (after using the distributive law as many times as necessary), each term in the multilinear identity consists of some sequence of applications of the pointwise product and convolution operations (no addition or subtraction), evaluated at zero, and then multiplied by a scalar constant. When one expands all of that, what one gets is a sum (in the discrete case) of the tensor product $f\_1 \otimes ... \otimes f\_n$ of all the functions over some subspace of $G^n$. The exact subspace is given by the precise manner in which the pointwise product and convolution operators are applied.
The only way a universal identity can hold, then, is if the weighted sum of the indicator functions of these subspaces (counting multiplicity) vanishes. (Note that finite linear combinations of tensor products span the space of all functions on $G^n$ when $G$ is finite.)
But when the characteristic of $G$ is large enough, the only way that can happen is if each subspace appears in the identity with a net weight of zero. (Indeed, look at a subspace of maximal dimension in the identity; for $G$ large enough characteristic, it contains points that will not be covered by any other subspace in the identity, and so the only way the identity can hold is if the net weight of that subspace is zero. Now remove all terms involving this subspace and iterate.)
So the final thing to do is to show that a given subspace can arise in two different ways from multiplication and convolution only in the "obvious" manner, i.e. by exploiting associativity of pointwise multiplication and of convolution. This looks doable by an induction argument but I haven't tried to push it through.
| 28 | https://mathoverflow.net/users/766 | 3709 | 2,472 |
https://mathoverflow.net/questions/3695 | 1 | This question comes from my notes, heavily edited, thus slightly unusual structure.
---
For Lie groups one can reformulate character theory as saying that
>
> **C** ⊗ K(`G`\ `pt`) = **C**[`T`/`W`] = **C**[ `X`\* ]W
>
>
>
where `G` is the complex Lie group, `W` its Weyl group, `T` its torus. (subquestion: is this correct?)
For example, one can write the character theory of a torus and `SL``2` as
>
> K(`G``m`\ `pt`) = **Z**[`q`, `q`-1] `and` K(`SL`2 \ `pt`) = **Z**[`q` + `q`-1].
>
>
>
**Question:** it's interesting that we're allowed to write **Z** in the examples above; I wonder if this works for any `G` in the formula above, e.g. if the isomorphism is valid over **Z** or **Q** rather then **C**?
| https://mathoverflow.net/users/65 | Character theory over integers | The isomorphism works over Z. The proof is that the basis change between W-symmetrized monomials and characters is upper-triangular with 1's on the diagonal.
| 3 | https://mathoverflow.net/users/66 | 3711 | 2,474 |
https://mathoverflow.net/questions/1194 | 7 | **Problem.** How to partition R^3 into pairwise non-parallel lines?
A possible solution is to stack infinitely many ``concentric'' hyperboloids, by increasing radius and decreasing slope. And don't forget the line on the $z$ axis at the center. The prototype hyperboloid [looks like this](http://www-lm.ma.tum.de/archiv/sos04/la2lb04/Regel_Hyperboloid.gif).
I heard a talk to which I didn't understand a lot ; a solution was given using Hopf fibration. I'm not familiar to these notions, and at the end it went like ``Tadaa! And here is our partition!''. The speaker could not describe what the partition looks like.
I would be very glad to: (1) understand the math he did (article, book?), (2) see what his solution looks like, and (3) know what kind of solutions exist.
Thanks in advance!
| https://mathoverflow.net/users/728 | How to partition R^3 into pairwise non-parallel lines? | Take the complex lines in ℂ2, and intersect with a copy of ℝ3 not containing the origin. This gives a foliation of ℝ3 by lines, which is the projection (from the origin) of the Hopf fibration of the unit sphere in ℂ2 (which is the foliation of S3 by intersections with complex lines).
One may easily write this down in coordinates, thinking of ℝ3 =ℝ x ℂ =(1+ iℝ) x ℂ ⊂ ℂ2.
Then for a fixed z ∈ ℂ, the line is given by (t, (1+it) z), t ∈ ℝ. When z=0, you get a vertical axis. For |z|=r, you get a hyperboloid which is obtained by rotating the line through (0,r) about the axis. I think this is probably the foliation by lines described in the talk you attended.
Another remark is that since you're interested in partitions rather than foliations, on each hyperboloid there is two foliations by lines (which are mirror images). So you can "flip" the foliation on each hyperboloid independently to obtain uncountably (actually 2|ℝ|) many partitions of ℝ3 into non-parallel lines.
| 18 | https://mathoverflow.net/users/1345 | 3714 | 2,475 |
https://mathoverflow.net/questions/1527 | 4 | I recently finished reading [this](http://math.unice.fr/~beauvill/pubs/bnr.pdf) paper, and was wondering about a couple of things relating to theorem 1, which says that for any curve X there is a curve Y and f:Y->X such that pushforward is a dominant rational map from the Jacobian of Y to the moduli of semistable vector bundles on X (with numerical invariants fixed to make things more definite.) So I had two questions:
1) Given a morphism of curves f:Y->X, is there a good characterization of the line bundles L on Y with f\_\*(L) semistable (or not semistable, equivalently)?
2) Given a morphism of curves f:Y->X, is there a good characterization of which semistable bundles are in/not in the image of f\_\*?
| https://mathoverflow.net/users/622 | Pushforwards of Line Bundles and Stability | Re: first question.
For semistability, we need a homomorphism from a line bundle U -> X of a certain degree to the pushdown of a line bundle L, which is the same thing as having a section on Y of f ^ \* U ^ \* \otimes L.
This can be expressed in terms of special spaces of divisors, and you can find details worked out explicitly for an example in rank 2 (i.e. a holomorphic double covering f: Y -> X) in pages 103-105 of [[NJH,87]](http://plms.oxfordjournals.org/cgi/pdf_extract/s3-55/1/59).
| 2 | https://mathoverflow.net/users/1177 | 3717 | 2,477 |
https://mathoverflow.net/questions/3691 | 0 | First, some background. I was trying to read the article [Loop Spaces and Langlands Parameters](http://arxiv.org/abs/0706.0322) but I get immediately stuck at Theorem 2.1 in the introduction.
This was actually forward-referring to Chapter 5, and I am able to read until Chapter 4 inclusively, but Proposition 5.1 knocks me off, despite providing some examples. The short statement of it is
>
> The category `L_qcoh (Z/S^1)` is equivalent to the category of comodule objects
> in `L_qcoh(Z)` for the coalgebra `p_∗act^∗` in `End(L_qcoh (Z))`.
>
>
>
Now since there are some references included, I tried looking there, but I think I just don't "get" something about localization. In fact, I remember some physical examples (Chern-Simons theory, I believe) about localizing the path integral onto the fixed points of some torus -- I'm sure this is relevant here, but I don't know how.
So what I have in mind for this question is that perhaps somebody could provide some simpler examples of localization, or connect it to other places where it naturally arises, e.g. physics.
| https://mathoverflow.net/users/65 | Understanding a lemma in "Loop Spaces and Langlands Parameters" article | S^1 localization is a fascinating subject IMHO, but the statement you quote is actually about something else, namely descent (or in this case, equivariance). The question is how do you describe sheaves on a quotient X/G (aka G-equivariant sheaves on X) in terms of sheaves on X. The answer is sheaves on X for which the two pushforwards to G x X (under projection and action maps) are identified, plus associativity and higher coherences. Using adjunction,
one can rewrite this isomorphism as a coaction of functions on G (with coproduct given by the group structure). That's all that's being asserted (of course in an oo-categorical context, but all the juice is the theory of adjunctions, namely Lurie's oo-categorical version of the Barr-Beck theorem).
For classical S^1 localization you should look at Atiyah and Bott's localization theorem.. for the physics version, Witten's papers on the Atiyah-Singer index theorem and on elliptic genus. My favorite reference in the subject (which gives many other references) is the paper by Goresky-Kottwitz-MacPherson on Koszul duality, localization and equivariant cohomology (which inspired the paper you quote).
| 5 | https://mathoverflow.net/users/582 | 3718 | 2,478 |
https://mathoverflow.net/questions/3721 | 39 | Since some computer scientists use category theory, I was wondering if there are any programming languages that use it extensively.
| https://mathoverflow.net/users/1369 | Programming Languages Based on Category Theory | Yes. I think that [Haskell](http://en.wikipedia.org/wiki/Haskell_%28programming_language%29) is the canonical example. Go [here](http://www.haskell.org/haskellwiki/Category_theory) for more.
| 15 | https://mathoverflow.net/users/83 | 3722 | 2,481 |
https://mathoverflow.net/questions/3740 | 18 | Pretty much anyone who does algebra is familiar with group objects in categories, but what about cogroup objects? Most of what I've been able to find about them is that they "arise naturally in algebraic topology" (from [wikipedia](https://en.wikipedia.org/wiki/Group_object#Examples)) and that, somehow, the n-sphere is one ([nLab](https://ncatlab.org/nlab/show/cogroup)'s meager entry). Is there a reference for more on the stuff? Specifically wondering if the spaces of pointed maps from a topological space X to a pointed sphere are cogroups, and if anything is known about these "cohomotopy groups."
| https://mathoverflow.net/users/622 | Cogroup objects | Spheres are (homotopy) cogroups for the same reason that homotopy groups are groups. The comultiplication $S^n \to S^n \vee S^n$ is the map that collapses the equator, the same map that is used to define composition in homotopy groups. Note that this only satisfies the cogroup axioms up to homotopy, just as composition in the homotopy group is only a group operation because you are taking homotopy classes.
However, Maps$(X,S^n)$ does not inherit a natural cogroup structure from S^n, because Maps$(X,S^n \vee S^n)$ does not naturally map to Maps$(X,S^n)$ $\vee$ Maps$(X,S^n)$. However, there is something called (stable) cohomotopy groups: these are stable homotopy classes of maps from $X$ to $S^n$. This is a cohomology theory, the cohomology theory associated to the sphere spectrum. This cohomology theory is very hard to compute though; its value on a point is the stable homotopy groups of spheres!
The way you *can* get a cogroup from a sphere and an arbitrary space is by taking smash products, rather than mapping spaces: for any $X, X \wedge S^n$ (which is the n-fold suspension of $X$) is a homotopy cogroup. You can see this by using the same "collapse the equator" map on the suspension, or you can see it more categorically from the fact that we have a map $X \wedge S^n \mapsto X \wedge (S^n \vee S^n) = (X \wedge S^n) \vee (X \wedge S^n)$. More generally, if $C$ is a cogroup, so is $X \wedge C$ for any X (and Maps$(C,X)$ is a group), and if G is a group, so is Maps$(X,G)$ for any X.
| 14 | https://mathoverflow.net/users/75 | 3743 | 2,492 |
https://mathoverflow.net/questions/3742 | 13 | Keel-Mori's theorem says an algebraic stack with a finite diagonal over a scheme S has a coarse moduli space. What is an example of an algebraic stack without coarse moduli space?
| https://mathoverflow.net/users/1363 | Examples of algebraic stacks without coarse moduli space? | [A^1/Gm] is one example. You can check that any Gm invariant map from A^1 to a scheme is constant. Thus the map from [A^1/Gm] to the point is universal for maps to schemes, but is not a bijection on geometric points (since [A^1/Gm] has two geometric points).
Check out Jarod Alper's [thesis](http://arxiv.org/pdf/0804.2242v3) to learn more.
| 13 | https://mathoverflow.net/users/2 | 3750 | 2,498 |
https://mathoverflow.net/questions/3757 | 7 | In the category of groups, there are lots of "exact sequences", e.g. 4 → H → 2, that neither split nor cosplit, where H is the eight-element group of quaternions, and lots of sequences like 4 → D → 2 that split but do not cosplit, where D is the eight-element dihedral group. By "2" and "4" I mean the cyclic groups of those orders. By "exact sequence" A → B → C, I mean that A is the kernel of the quotient B → C (equivalently C is the cokernel of the subobject A → B). A sequence A → B → C "splits" if there is a map C → B so that the compotision C → B → C is the identity; cosplitting is on the other side.
So in groups, a split exact sequence does not necessarily cosplit. (In fact, I have a hard time thinking of any cosplit sequences.) On the other hand, my friends who do ring theory state definitions like "A ring is semisimple if any short exact sequence of modules splits". Why don't they ask for the sequence to cosplit? Does that come for free? (Or am I misremembering the definition?)
More generally, what conditions does one have to place on a category so that "splits" implies "cosplits"?
| https://mathoverflow.net/users/78 | When does "splits" imply "cosplits"? | Yes, it comes for free. A short exact sequence of *abelian* groups, or R-modules (R not necessarily commutative), splits iff it cosplits iff the middle term is the sum of the other two terms. The key here is that maps of modules and be added/subtracted:
Let the middle maps in 0→A→B→C→0 be f:A→B and g:B→C. Then if there's a splitting q:C→B, then (1B-qg):B→B is a projection onto A as a submodule of B, i.e. a cosplitting. Together the splitting and cosplitting exhibit B as the direct sum of A and C.
A dual trick shows that a cosplit sequences are split: if p:B→A is a cosplitting, then (1B-fp):B→B is a projection onto a submodule of itself which is isomorphic to C via g, i.e. a cosplitting, so again B is the sum of A and C via these maps.
More generally, this same trick works in any abelian category. One way to recognize this instantly is via [Freyd's Exact Embedding theorem](http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem), which roughly implies that you can pretend a diagram in an abelian category is a diagram of R-modules for some R.
... whoah... are you guys related?
| 7 | https://mathoverflow.net/users/84526 | 3760 | 2,504 |
https://mathoverflow.net/questions/3734 | 3 | What's a good way to find how fast the integral of a function is growing near a pole of the function? Here is what I mean on an example.
Look at 1/z.
If I want to find out how fast ∫0a 1/(z-ε)dz is growing when ε->0, ε∈C, I can do this:
∫0a 1/(z-ε)dz = ln((a-ε)/ε)=-ln(-ε)+ln(a)+ε/a+O(ε).
What if I have ∫0a f(z)/(z-ε) dz , where f(z) is finite?
| https://mathoverflow.net/users/1338 | How to do asymptotics for integrals? | If you just want to see how fast it blows up, it shouldn't be too hard. First integrate by parts:
>
> ∫01 f(z)/(z-ε) dz = f(1)log(1-ε) - f(0)log(-ε) - ∫01 f'(z)log(z-ε) dz.
>
>
>
For the integral on the right-hand side, note that when you set ε to 0, you get ∫01 f'(z)log(z) dz, which should converge (to a finite value) as long as f'(z) is bounded, so let's rewrite the integral as
>
> ∫01 f'(z)log(z-ε) dz = ∫01 f'(z)log(z) dz +
> ∫01 f'(z)(log(z-ε) - log(z)) dz.
>
>
>
The second integral looks like it should converge to 0 as ε goes to 0. To confirm this, it seems advantageous to deal with the singularity at z=0 first (there may be a cleaner way). Make the change of variables z = u2:
>
> ∫01 f'(z)(log(z-ε) - log(z)) dz = 2∫01f'(u2) u(log(u2- ε) - log(u2)) du,
>
>
>
and now it shouldn't be too hard to show that the integrand converges uniformly to 0 as ε goes to 0 if f'(z) is bounded. This gives the estimate
>
> ∫01 f(z)/(z-ε) dz = -f(0)log(-ε) - ∫01 f'(z)log(z) dz + o(1).
>
>
>
| 5 | https://mathoverflow.net/users/302 | 3770 | 2,507 |
https://mathoverflow.net/questions/3643 | 6 | Rosser's algorithm is typically invoked during discussions of equal temperament scales, and is a way to obtain good approximations for multiple irrational numbers simultaneously. Is there a nice, accessible modern treatment? Rosser's paper was published in 1950.
Some background:
In an equal temperament scale, the ratio of frequencies of adjacent notes is a constant. This is a desirable feature, in that it allows you to transpose a piece from one key into any other. Now, for various reasons (psychological, historical) we'd like have have certain ratios of frequencies, or close approximations. If we require the unison ratio (2:1) to be exact, then we require that our frequency ratio R satisfies R^n = 2 for some integer n. Thus, R is an integral root of 2. Next, if we want a good approximation to the perfect 5th (3:2), we conclude that we need R^r to be a good approximation to log2(3), for some integer r. This leads us into continued fractions. Now, if we'd also like a good approximation to the perfect 3rd (5:4), we would like R^s to be a good approximation to log2(5) for some integer s. We've now left the realm of continued fractions, as we're seeking to approximate two irrational numbers with powers of the same root of 2.
| https://mathoverflow.net/users/795 | Rosser's Algorithm - Musical Scales and Generalized Ternary Continued Fractions | This is a relatively recent list of references, originally posted by Chris Hillman on sci.math. I've also attached another sci.math posting from Chris Hillman on the subject of multidimensional continued fractions. It appears that this subject would generally be classified under the topic "geometry of numbers".
A recent book not included in the below on this topic is "Multidimensional Continued Fractions" by Fritz Schweiger (2000). Doug Hensley's "Continued Fractions" (2006) also covers this topic in chapter 6.
==
From: Chris Hillman
Newsgroups: sci.math
Subject: Re: Hyperfractions ? : approximants to sqrt(-1)
Date: Thu, 14 May 1998 18:32:09 -0700
On Thu, 14 May 1998, Peter Jack wrote:
>
> Hyperfractions ? : approximants of sqrt(-1)
>
>
> I have a problem I'm working on. Maybe someone can help.
>
>
> The basic idea is to find a representation of the
> hypercomplex numbers in terms of a sequence of
> rational numbers.
>
>
>
[... snip ...]
>
> Has anyone attempted such a construction before?
>
>
>
I don't know about the particular construction you outline, but you are
looking for a variety of "multidimensional continued fraction algorithm"
and there is an enormous literature on such things, mostly on continued
fractions in R^n but some concentrating on various hypercomplex numbers,
so it is quite possible someone has tried the approach you outline before.
Here are a few recent references which should give some idea of the
variety of approaches recently taken to this problem:
@article{djg:fn,
author ={David J. Grabiner},
title = {Farey Nets and Multidimensional Continued Fractions},
journal = {Monatshefte fur Mathematik},
volume = 114,
year = 1992,
pages = {35--60}}
@article{n:pcfa,
author = {A. Nogueira},
title = {The Three-Dimensional Poincare Continued Fraction Algorithm},
journal = {Israel Journal of Mathematics},
volume = 90,
year = 1995,
pages = {373--401}}
@book{s:fs,
author = {Fritz Schweiger},
title = {Ergodic Theory of Fibered Systems and Metric Number Theory},
publisher = {Clarendon Press},
address = {Oxford},
year = 1995}
@article{iko:jpa,
author = {S. Ito and M. Keane and M. Ohtsuki},
title = {Almost everywhere exponential convergence of the modified Jacobi-Perron algorithm},
journal = {Ergodic Theory and Dynamical Systems},
year = 1993,
volume = 13,
pages = {319--334}}
@unpublished{l:skp,
author = {Giles Lachaud},
title = {Sails and {K}lein Polyhedra},
journal = {Contemporary Mathematics},
note = {to appear}}
@unpublished{l:kpgf,
author = {Giles Lachaud},
title = {{K}lein Polygons and Geometric Diagrams},
journal = {Contemporary Mathematics},
note = {to appear}}
@article{l:gmcf,
author = {J. C. Lagarias},
title = {Geodesic multidimensional continued fractions},
journal = {Proc. London Math. Soc.},
volume = 69,
year = 1994,
pages = {464--488}}
Hope this helps!
Chris Hillman
==
From: [email protected] (Christopher Hillman)
Newsgroups: sci.math
Subject: Re: Multidimensional Continued Fractions
Date: 1 Jul 1997 11:49:40 GMT
Organization: "University of Washington, Mathematics, Seattle"
[email protected] (No Chance) writes:
>
> A few months ago, someone posted a reply to an article about the
> continued fraction expansion of pi. At the end of the article, the poster wrote
> that research was being done on mulitdimensional continued fractions. I was
> wondering if anyone could tell me anything about this subject and give me some
> refrences.
>
>
>
You are probably thinking of an article I posted (I didn't save a copy).
The ordinary continued fraction algorithm provides a way to expand a real
number in a way quite different from a "decimal" expansion wrt to some base,
one which reveals some algebraic/number theoretic structure much better.
By truncating the expansion after n, n+1, n+2, ... terms we obtain a
sequence of rational approximations.
A multidimensional CFA is just some algorithm which gives a sequence of
rational approximations to a d-tuple of real numbers. The best known
such algorithm is the Jacobi-Perron algorithm.
One would naturally hope to be able to find an algorithm with a theory
which works out just as nicely as the one dimensional algorithm, and
which yields not only a definite sequence of approximations which is
in some sense optimal but which also yields an "expansion" which
reveals something about the number theoretic properties of the d-tuple,
in particular whether the various components are rationally independent.
Alas, it turns out that in higher dimensions there are MANY competing
algorithms, all equally disappointing ;) Well, if not all equally
disappointing, certainly disappointing for one reason or another.
Algorithms which are good from one standpoint are often quite bad
according to another way of thinking. Yet in one dimension there
is essentially only one algorithm which is at all reasonable, and
this one turns out to be good for many purposes.
References: two books
A. J. Brentjes, Multidimensional Continued Fraction Algorithms,
Amsterdam: Mathematisch Centrum, 1981.
Fritz Schweiger, Ergodic Theory of Fibred Systems and Metric Number Theory,
Oxford U Press, 1995.
(Has a chapter on multidimensional CFA's and many references.)
Some papers:
Giles Lachaud, Sails and Klein Polyhedra, Contemporary Mathematics, to appear.
(According to Vershik, the notion of a sail is the best to come along in this
field for years.)
David J. Grabiner, Farey Nets and Multidimensional CFA's,
Mh. Math. 114 (1992) 35-60
J. C. Lagarias, Geodesic Multdimensional Continued Fractions,
Proc. Lon. Math. Soc 69 (1994) 464-488.
A. Nogueira, The Three-Dimensional Continued Fraction Algorithm,
Is. J. Math. 90 (1995) 373-401.
Shunji Ito and Makoto Ohtsuki, Parallelogram Tilings and Jacobi-Perron Algorithm,
Tokyo J. Math. 17 (1994): 33-58.
(The reason for my interest in this is that these CFA's turn out to be relevant
for studying the combinatorial properties of certain types of tilings which
are idealized models of quasicrystals.)
This should give a quick impression of the scope of current research--- there are
ALOT of ideas out there!
Chris Hillman
| 1 | https://mathoverflow.net/users/795 | 3771 | 2,508 |
https://mathoverflow.net/questions/3787 | 6 | Has anyone seen the new MSC 2010? I was browsing around and to my suprise there is another revision of MSC. Has anyone noticed any major changes in there? Do major journals already accept papers with MSC 2010 classification?
| https://mathoverflow.net/users/1245 | What is new in MSC 2010? | I think I got partial answer to my questions in the following sites:
<http://www.ams.org/mathscinet/msc/conv.html?from=2000>
and
<http://www.ams.org/mathscinet/msc/conv.html?from=2010>
| 3 | https://mathoverflow.net/users/1245 | 3788 | 2,522 |
Subsets and Splits