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https://mathoverflow.net/questions/7782 | 40 | Let $H$ be a group. Can we find an automorphism $\phi :H\rightarrow H$ which is not an inner automorphism, so that given any inclusion of groups $i:H\rightarrow G$ there is an automorphism $\Phi: G\rightarrow G$ that extends $\phi$, i.e. $\Phi\circ i=i\circ \phi$?
| https://mathoverflow.net/users/2224 | Are the inner automorphisms the only ones that extend to every overgroup? | The answer is that the inner automorphisms are indeed characterized by the property of the existence of extensions to larger groups containing the original group. I learned as much from
[this blog entry](http://mathreader.livejournal.com/22934.html) (in Russian). The reference is *Schupp, Paul E.*, [**A characterization of inner automorphisms**](http://dx.doi.org/10.2307/2045986), Proc. Am. Math. Soc. 101, 226-228 (1987). [ZBL0627.20018](https://zbmath.org/?q=an:0627.20018)..
| 39 | https://mathoverflow.net/users/2106 | 7792 | 5,338 |
https://mathoverflow.net/questions/7794 | 9 | I am puzzled with the following problem:
Given $n$ real numbers it is to obtain a Yes/No answer to: "whether it is possible to arrange different points in the Euclidean $\mathbb{R}^3$ so that every of the given numbers represents a shortest distance which belongs to a distinct pair of points?"
What is an efficient algorithm to solve such problem? If I understand properly, first I have to find $m$ from $n=\frac{m(m+1)}{2}$ which is the number of such points. But what is the next step? Should I deal with checking the triangle inequality (which seems to be very inefficient) or what?
Thank you in advance!
| https://mathoverflow.net/users/2266 | Feasibility of a list of prescribed distances in R^3 | Let us label the points $x\_0,x\_1,\dots,x\_m$.
For each pair $(x\_i,x\_j)$ prescribe the a value from your list;
denote it by say $d(x\_i,x\_j)$.
If there are no repititions then all together you have $N=[m{\cdot}(m+1)]!$ ways to do this.
Note that $d(x\_i,x\_j)$ completely determine "$m\times m$-matrix of scalar products"
$$a\_{ij}=\tfrac12\cdot[d^2(x\_0,x\_i)+d^2(x\_0,x\_j)-d^2(x\_i,x\_j)].$$
You need to answer two questions for each of obtained $N$ matrices:
* $(a\_{ij})$ defines a non-negative quadratic form
* $\mathop{\rm rank}(a\_{ij})\leqslant 3$.
If for one matrix the answers are YES to both then the answer to your question is also YES (and vise versa).
**P.S.** Apparently the statement I used is called *Schoenberg criterion*; thanks to Piotr Hajlasz for pointing this out.
| 18 | https://mathoverflow.net/users/1441 | 7799 | 5,341 |
https://mathoverflow.net/questions/7627 | 17 | Every foundational system for mathematics I have ever read about has been a set theory, from ETCS to ZFC to NF. Are there any proposals for a foundational system which is not, in any sense, a set theory? Is there any alternative foundation which is not a set-theory?
| https://mathoverflow.net/users/2266 | Set theory and alternative foundations | Bill Lawvere has suggested axiomatizing the category of categories as a foundation of mathematics, and there is no sense in which this could be thought of as a set theory. Colin McLarty is one person who has done some work on achieving such an axiomatization.
| 15 | https://mathoverflow.net/users/1106 | 7802 | 5,343 |
https://mathoverflow.net/questions/7800 | 7 | Let $R$ be a commutative ring, like the ring of integers $\mathbb Z$ or the ring of $p$-adic integers $\mathbb Z\_p$. Let $G$ be a finite group; let us consider permutational representations of $G$ over $R$, i.e., $R[G]$-modules of the form $R[G/H]$, where $H\subset G$ is a subgroup, and direct sums of such modules. These are free $R$-modules where $G$ acts so that there exists a basis of the module preserved (as a whole) by the action.
I am interested in finite exact sequences of representations of the above type, particularly in those of them that are not very long. There are some beautiful examples, e.g., for $G=\mathbb Z/2$ and $R=\mathbb Z/2$ there is an exact sequence
$$
0\rightarrow R\rightarrow R[G]\rightarrow R\rightarrow 0,
$$
while for $G=\mathbb Z/n$ and $R=\mathbb Z$ there is an exact sequence
$$
0\rightarrow R\rightarrow R[G]\rightarrow R[G]\rightarrow R\rightarrow 0.
$$
For the fourth symmetric group $G=\mathbb S\_4$ and $R=\mathbb Z$ there is an exact sequence
$$
0\rightarrow R\rightarrow R[\mathbb X\_4]\oplus R\rightarrow
R[\mathbb X\_6]\oplus R\rightarrow R[\mathbb X\_3]\rightarrow0,
$$
where $\mathbb X\_4$ is the four-element set that $\mathbb S\_4$ permutes, $\mathbb X\_6$ is the set of all two-element subsets of $\mathbb X\_4$, and $\mathbb X\_3$ is the quotient set of $\mathbb X\_6$ by the obvious involution. Dihedral groups also have some four-term exact sequences of permutational representations.
Where is one supposed to get such exact sequences? There are some obvious ways, like e.g. one can take cones of morphisms of exact sequences of this type, or one can do restriction or induction from one group to another one. Are there any other constructions?
For constructions to be interesting, they should of course be removed far enough from the trivial case when $|G|$ is invertible in $R$. E.g., to have $R=\mathbb Z\_p$ and $|G|$ a large $p$-group would be perhaps most highly nontrivial.
EDIT: One of the commenters asked where does the sequence for $\mathbb S\_4$ come from, so let me say a few words about this. Not that I really understand it, but there is a geometric construction using a CW complex, and not quite of the kind that Greg suggests in his second answer below.
Represent the group $\mathbb S\_4$ as the group of rotations of the $3$-dimensional cube. Consider the quotient CW complex of the cube's surface by the central symmetry involution. The group $\mathbb S\_4$ still acts on the quotient. The set of vertices of the quotient is the $\mathbb S\_4$-set $\mathbb X\_4$, the set of edges is the $\mathbb S\_4$-set $\mathbb X\_6$, and the set of faces is the $\mathbb S\_4$-set $\mathbb X\_3$.
Now consider the map $R[\mathbb X\_4]\rightarrow R[\mathbb X\_6]$ assigning to a vertex the sum of the three edges ending in it (without any signs!) and also the map $R[\mathbb X\_6]\rightarrow R[\mathbb X\_3]$ assigning to an edge the sum of the two faces bordering on it (also without any signs!). The composition of these two maps is not zero, of course; what it does is taking every vertex to twice the sum of all the three faces. One somehow transforms this pair of arrows into a four-term exact sequence by adding the trivial $\mathbb S\_4$-module direct summands $R$ in several degrees.
| https://mathoverflow.net/users/2106 | Exact sequences of permutational representations? | [Here is a preprint](http://math.ucsc.edu/~boltje/publications/p09y.pdf) by Boltje and Hartmann that constructs a conjectured resolution of Specht modules of $S\_n$ (over $\mathbb{Z}$) by Young modules. This is presumably a related tool.
(An earlier version of this answer had some out-of-step comments about resolutions that were either not helpful or already addressed in the original question.)
| 5 | https://mathoverflow.net/users/1450 | 7806 | 5,346 |
https://mathoverflow.net/questions/7808 | 5 | Suppose that I have a map of simplicial spaces,
$ f: X\_\* \to Y\_\*$,
and that I know that the map on zero spaces $f\_0: X\_0 \to Y\_0$ is n-connected. Can I conclude anything about the connectivity of the map of geometric realizations?
$ |f|: |X| \to |Y|$
Are there any reasonable conditions I can place on the simplicial spaces X and Y that would allow me to conclude something along these lines? I'm especially interested in knowing when the map is 0-connected (i.e. a surjection on $\pi\_0$).
| https://mathoverflow.net/users/184 | Connectivity after Geometric Realization? | Yes it is a surjection on $\pi\_0$, because each component of $|Y|$ has at least one component of $Y\_0$.
Beyond that there are no restrictions. For instance, you can get any homotopy type for $|X|$ and $|Y|$ and any homotopy type for the map between them with $X\_0$ and $Y\_0$ just one point, as long as you ask that $\pi\_0(|X|)$ and $\pi\_0(|Y|)$ are trivial.
(I'm taking "simplicial space" to mean a simplicial object in the category of topological spaces, say the compact Hausdorff ones.)
| 4 | https://mathoverflow.net/users/1450 | 7809 | 5,348 |
https://mathoverflow.net/questions/7793 | 17 | My question is motivated by [Are the inner automorphisms the only ones that extend to every overgroup?](https://mathoverflow.net/questions/7782/are-the-inner-automorphisms-the-only-ones-that-extend-to-every-overgroup)
What are the auto-equivalences of the category of groups? What kind of structure do they form?
There are certainly some not-so-trivial examples (which are still trivial if one just focus on a single group), e.g. the functor send an abstract group G to the group G^op with the same underlying set with multiplication g,h ---> hg, where hg is the product of h and g in G. [The notation G^op will make sense if you think G as the category BG]
| https://mathoverflow.net/users/1657 | What are the auto-equivalences of the category of groups? | Suppose $F:\mathrm{Grp}\to\mathrm{Grp}$ is an equivalence. The object $\mathbb{Z}\in\mathrm{Grp}$ is a minimal generator (it is a generator, and no proper quotient is also a generator), and this property must be preserved by equivalences. Since there is a unique minimal generator, we can fix an isomorphism $\phi:\mathbb Z\to F(\mathbb Z)$. Now $F$ must preserve arbitrary coproducts, so for all cardinals $\kappa$, the isomorphism $\phi$ induces an isomorphism $\phi\_\kappa:L\_\kappa\to F(L\_\kappa)$, where $L\_\kappa$ is the free product of $\kappa$ copies of $\mathbb Z$. In particular, if $1$ is the trivial group, $\phi\_0:1\to F(1)$ is an isomorphism.
Next pick a group $G\in\mathrm{Grp}$, and consider a free presentation $L\_1\to L\_0\to G\to1$, that is, an exact sequence with the $L\\_i$ free. (For simplicity, we can take $L\\_0=L(G)$ the free group on the underlying set of $G$, and $L\_1$ to be the free group on the underlying subset the kernel of the obvious map $L\_1\to G$; this eliminates choices) Since $F$ is an equivalence, we have another exact sequence $F(L\_1)\to F(L\_0)\to F(G)\to F(1)$. Fixing bases for $L\_1$ and $L\_0$ we can use $\phi$ to construct isomorphisms $L\_i\to F(L\_i)$ for both $i\in\{0,1\}$. *Assuming* we can prove the square commutes, one gets an isomorphism $\phi\_G:G\to F(G)$—this should not be hard, I guess.
The usual arguments prove then in that case the assignment $G \mapsto \phi\_G$ is a natural isomorphism between the identity functor of $\mathrm{Grp}$ and $F$.
| 8 | https://mathoverflow.net/users/1409 | 7818 | 5,356 |
https://mathoverflow.net/questions/7817 | 8 | I'm hoping that the following are true. In fact, they are probably easy, but I'm not seeing the answers immediately.
Let $M$ be a smooth $m$-dimensional manifold with chosen positive smooth density $\mu$, i.e. a chosen (adjectives) volume form. (A *density* on $M$ is a section of a certain trivial line bundle. In local coordinates, the line bundle is given by the transition maps $\tilde\mu = \left| \det \frac{\partial \tilde x}{\partial x} \right| \mu$. When $M$ is oriented, this bundle can be identified with the top exterior power of the cotangent bundle.) **Hope 1: Near each point in $M$ there exist local coordinates $x: U \to \mathbb R^m$ so that $\mu$ pushes forward to the canonical volume form $dx$ on $\mathbb R^m$.**
Hope 1 is certainly true for volume forms that arise as top powers of symplectic forms, for example, by always working in Darboux coordinates. If Hope 1 is true, then $M$ has an atlas in which all transition maps are volume-preserving. My second Hope tries to describe these coordinate-changes more carefully.
Let $U$ be a domain in $\mathbb R^m$. Recall that a change-of-coordinates $\tilde x(x): U \to \mathbb R^m$ is *oriented-volume-preserving* iff $\frac{\partial \tilde x}{\partial x}$ is a section of a trivial ${\rm SL}(n)$ bundle on $U$. An *infinitesimal change-of-coordinates* is a vector field $v$ on $U$, thought of as the map $x \mapsto x + \epsilon v(x)$. An infinitesimal change-of-coordinates is necessarily orientation-preserving; it is volume preserving iff $\frac{\partial v}{\partial x}(x)$ is a section of a trivial $\mathfrak{sl}(n)$ bundle on $U$. **Hope 2: The space of oriented-volume-preserving changes-of-coordinates is generated by the infinitesimal volume-preserving changes-of-coordinates, analogous to the way a finite-dimensional connected Lie group is generated by its Lie algebra.**
Hope 2 is not particularly well-written, so Hope 2.1 is that someone will clarify the statement. Presumably the most precise statement uses infinite-dimensional Lie groupoids. The point is to show that a certain *a priori* coordinate-dependent construction in fact depends only on the volume form by showing that the infinitesimal changes of coordinates preserve the construction.
***Edit:*** I have preciseified Hope 2 as [this question](https://mathoverflow.net/questions/7853/is-the-space-of-volume-preserving-maps-path-connected).
| https://mathoverflow.net/users/78 | Normal coordinates for a manifold with volume form | Yes for "hope" 1. This theorem was proven by Moser using volume-preserving flows. A manifold with a volume form is the same thing as a manifold with an atlas of charts modeled on the volume-preserving diffeomorphism pseudogroup. He found an argument that can be adapted to either the symplectic case or the volume case. I cited this result in my paper *[A volume-preserving counterexample to the Seifert conjecture](https://arxiv.org/abs/math/9504230)* (Comment. Math. Helv. 71 (1996), no. 1, 70-97), where I also established a similar result for the volume-preserving PL pseudogroup. In the PL case, the corresponding decoration on the manifold is a piecewise constant volume form.
In my opinion, the most exciting result on this theme is the Ulam-Oxtoby theorem. (But Moser's version is the most useful one and the most elegant.) The theorem is that if you have a topological manifold with any Borel measure that has no atoms and no bald spots, then it is modeled on the pseudogroup of volume-preseserving homeomorphisms. For example, you can start with Lebesgue measure in the plane and add uniform measure on a circle, and there is a homeomorphism that takes that measure to Lebesgue measure.
For a long time I have wondered about the pseudogroup of volume-preserving Lipschitz maps. The question is whether there is a corresponding cone of measures, and if so, how to characterize it.
| 9 | https://mathoverflow.net/users/1450 | 7819 | 5,357 |
https://mathoverflow.net/questions/22 | 47 | [Pablo Solis](http://math.berkeley.edu/~pablo/) asked this at a recent [20 questions seminar](http://scratchpad.wikia.com/wiki/20qs) at Berkeley. Is there a positive integer $N$, not of the form $10^k$, such that the digits of $N^2$ are all 0's and 1's?
It seems very unlikely, but I don't have a proof. It's easy to see that such a number must end in 1 or 9, and then easy to see that it must end in 01, 49, 51 or 99, and you can continue recursively for as long as you like, determining possible "suffixes". Using this, I had a computer check for me that there are no such N up to about $10^{24}$.
If you pretend that the digits of $N^2$ are randomly distributed, and $N$ has $n$-digits, there's a $(2/10)^{2n}$ chance of satisfying this condition. There are only $10^n$ $n$-digit numbers, so you might expect a $(4/10)^n$ chance of having a some $n$-digit number. This suggests we shouldn't expect to find anything.
(If you try the same problem in other bases, where the probabilities are better, you do find a few: in base 5, 222112144, 22222111221444 and 100024441003001 work.)
| https://mathoverflow.net/users/3 | Can $N^2$ have only digits 0 and 1, other than $N=10^k$? | In the interest of completeness, here is what I put on the 20-questions wiki — we might as well repeat it here in the $\infty$-questions site. I had basically the same idea as Ilya, do a branched search to look for the digits of $N^2$. However, the code that I wrote in Python works from the 10-adic end, while Ilya's works from the Archimedean end. Both programs support a heuristic model that implies that solutions in the integers are very unlikely. If you wanted an optimized search in the integers, you would work from both ends and try to match the partial solutions. And you would probably want to implement the algorithms in C++ rather than in Python.
```
maxmod = 10**24
def check(x):
if not str(x**2).replace('0','').replace('1',''):
print 'Eureka:',x,x**2
def search(x,mod):
x %= mod
if mod == maxmod:
check(x)
check(mod-x)
return
top = -(x**2/mod) % 10
x += (top + top%2)/2 * mod
search(x,mod*10)
search(x + 5*mod,mod*10)
search(1,10) # Solution is either 1 or 9 mod 10
```
Also, it's tempting to mark the problem as open rather than as a "puzzle". I did find several papers that analyzed the congruence structure of integers with restricted digits, and one that looked at prime factors. Two are by Erdos, Mauduit, and Sarkozy [1](http://dx.doi.org/10.1006/jnth.1998.2229) [2](http://dx.doi.org/10.1016/S0012-365X%2898%2900331-8), and two are by Banks and Shparlinski (one also with Conflitti) [3](http://www.math.missouri.edu/~bbanks/papers/char_sums_restr_digits.pdf) [4](http://www.math.missouri.edu/~bbanks/papers/arith_props_restr_digits.pdf). Presumably some of these authors can say whether the problem should be called open.
| 16 | https://mathoverflow.net/users/1450 | 7825 | 5,361 |
https://mathoverflow.net/questions/7828 | 9 | A Hilbert space is a complete vector space equipped with scalar product, i.e. a symmetric positive definite bilinear form.
What if we replace 'bilinear' by 'n-linear'? One might wonder, whether the $l^3$-norm might be induced by a trilinear form in a similar fashion like the $l^2$-norm by a bilinear form is.
Is there any interesting theory on this?
| https://mathoverflow.net/users/2082 | Hilbert spaces are induced by a bilinear form. How about n-linear forms? | As long as the form is positive definite and the unit ball is convex, you get a perfectly good Banach space using any symmetric $n$-linear form on a real vector space $V$. The degree $n$ is necessarily even. It is equivalent to defining the norm as the $n$th root of a homogeneous degree $n$ polynomial. $\ell^p$ is an example for any even integer $p$. There are many other examples. I found a paper, [Banach spaces with polynomial norms](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-82/issue-1/Banach-spaces-with-polynomial-norms/pjm/1102785074.full), by Bruce Reznick, that studies these norms. He obtains various results; the most appealing one to me at a glance is that these Banach spaces are all reflexive.
Off-hand I can't think of any simple way to recover positive definiteness starting with odd polynomials. The cube of the norm on $\ell^3$ is a polynomial in the absolute values of the coordinates rather than the coordinates themselves.
Addendum: To address Darsh's comment, what you would look at in the complex case is self-conjugate polynomials of degree $(n,n)$. Equivalently, as with all complex Banach norms, the realification is a real Banach norm which is invariant under complex scalar rotation.
| 8 | https://mathoverflow.net/users/1450 | 7832 | 5,366 |
https://mathoverflow.net/questions/7836 | 55 | I have this question coming from an earlier [Qiaochu's post](https://mathoverflow.net/questions/5243/why-is-it-a-good-idea-to-study-a-ring-by-studying-its-modules). Some answers there, especially David Lehavi's one, were drawing the analogy bundles and varieties versus modules and rings. So I am just wondering, is there any big reason why the study of bundles would give information about varieties? (I suppose that for this matter, I should actually replace varieties by manifolds?)
I have heard about some invariants, like the Picard group for complex manifolds. But given my inexperience in these concepts, I don't really know why they should be important. So for those who are thinking about "cooking up invariants", some more detailed explanations on why they are useful (and hopefully some elementary examples!) would be appreciated. Thanks!
| https://mathoverflow.net/users/nan | Why is it useful to study vector bundles? | Well, in algebraic geometry, here's a couple of reasons:
1) Subvarieties: Take a vector bundle, look at a section, where is it zero? Lots of subvarieties show up this way (not all, see [this question](https://mathoverflow.net/questions/1614/when-is-a-scheme-a-zero-set-of-a-section-of-a-vector-bundle)) but generally, we can get lots of information out of vector bundles regarding subvarieties.
2) Invariants of spaces: The Picard group of Line bundles and more generally the Grothendieck group/ring is a useful invariant for differentiating spaces and analyzing the geometry indirectly. On smooth spaces, in fact, complexes of vector bundles can be used to replace coherent sheaves entirely (I believe by the Syzygy Theorem).
3) Maps into Projective Space: This one is line bundle specific. Let $V\to\mathbb{P}^n$ be any imbedding, say, then the pullback of $\mathcal{O}(1)$ is a line bundle on $V$. The nice thing is, the global sections of this line bundle determine and are determined by the map (we can get degenerate mappings by taking subspaces, but lets ignore that, and base loci for the moment). It turns out that we can define a line bundle to be ample, a condition just on the bundle, and that suffices to say that a power of it gives a morphism to $\mathbb{P}^n$, so understanding maps into projective space is the same thing as studying ample line bundles on a variety.
Hope that helps, there's a lot more, but those are the first three things that came to mind.
| 33 | https://mathoverflow.net/users/622 | 7837 | 5,368 |
https://mathoverflow.net/questions/7824 | 6 | Why exactly is the unique factorization of elements into irreducibles a natural thing to look for? Of course, it's true in $\mathbb{Z}$ and we'd like to see where else it is true; also, regardless of whether something is natural or not, studying it extends our knowledge of mathematics, which is always good. But the unique factorization of elements - being specifically a question of **elements** - seems completely counter to the category theory philosophy of characterizing structure via the maps between objects rather than their elements. Indeed, I feel like unique factorization of ideals into prime ideals is less a generalization of unique factorization of elements into irreducibles than the latter is a messier, unnatural special case of the former, a "purer" question (ideals, being the kernels of maps between rings, I feel meet my criteria for being a category-theoretically acceptable thing to look at). Certainly, the common theme in algebra (and most of mathematics) is to look at the decomposition of **structures** into simpler **structures** - but quite rarely at actual elements.
Now, for nice cases like rings of integers in number fields, we can characterize being a UFD in terms of the class group and other nice structures and not have to mess around with ring elements, but looking at the Wikipedia page on UFDs and the alternative characterizations they list for general rings, they all appear to depend on ring elements in some way (the link to "divisor theory" is broken, and I don't know what that is, so if someone could explain it and/or point me to some resources for it, it'd be much appreciated).
Sorry about the rambling question, but I was wondering if anyone had any thoughts or comments? Is "being a UFD" equivalent to any property which can be stated entirely without reference to ring elements? Should we care whether it is or not?
---
EDIT: Here's a more straightforward way of saying what I was trying to get at: The structure theorem for f.g. modules over a PID, the Artin-Wedderburn theorem, the Jordan-Holder theorem - these are structural decompositions. Unique factorization of elements is not, **because elements are not a structure**. My feeling is that this makes it a fundamentally less natural question, and I ask whether being a UFD can be characterized in purely structural terms, which would redeem the concept somewhat, I think.
| https://mathoverflow.net/users/1916 | Factorization of elements vs. of ideals, and is being a UFD equivalent to any property which can be stated entirely without reference to ring elements? | This is sort of an anti-answer, but: my instinct is that ZC is taking the categorical perspective too far.
To start philosophically, I think it is quite appropriate to, when given a mathematical structure like a topological space or a ring -- i.e., a set with additional structure -- refrain from inquiring as to exactly what sort of object any element of the structure is. There is a famous essay "What numbers could not be" by Paul Benacerraf, in which he pokes fun at this idea by imagining two children who have been taught about the natural numbers by two different "militant logicists". Their education proceeds well until one day they get into an argument as to whether 3 is an element of 17. (The writing is very nice here and unusually witty for an essay on mathematical philosophy: the names of the children are Ernie and Johnny, an allusion to Zermelo and von Neumann, who had rival definitions of ordinal numbers.) The point of course is that it's a silly question, and a mathematically useless one: it won't help you to understand the structure of the natural numbers any better.
On the other hand, to deny that a set is an essential part of certain (indeed, many) mathematical structures seems to be carrying things too far. As far as I know, it is not one of the goals of category theory to eliminate sets (though one occasionally hears vague mutterings in this direction, I have never seen an explanation of this or, more critically, of the need for this).
Coming back to rings, it seems to me that very few properties of rings can be expressed without elements. You also seem to implicitly suggest that it is "more structural" to think about things in terms of ideals than elements. Can you explain this? It would seem that speaking of ideals involves *more* set theoretic machinery than speaking about elements: this is certainly true in model theory in the language of rings.
It seems wrong to say that unique factorization of ideals into primes is a "generalization" of unique factorization of elements, since neither property implies the other.
Finally a positive remark: it sounds like you might like the characterization of UFDs as Krull domains with trivial divisor class group.
| 3 | https://mathoverflow.net/users/1149 | 7839 | 5,370 |
https://mathoverflow.net/questions/7823 | 14 | I understand the heuristic reason why Gromov-Witten invariants can be rational; roughly it's because we're doing curve counts in some stacky sense, so each curve $C$ contributes $1/|\text{Aut}(C)|$ to the count rather than $1$.
However, I don't understand why or how Gromov-Witten invariants can be negative. What is the meaning of a negative GW invariant? What are some simple examples?
| https://mathoverflow.net/users/83 | Negative Gromov-Witten invariants | Gromov--Witten invariants are designed to count the "number" of curves in a space in a deformation invariant way. Since the number of curves can change under deformations, the Gromov--Witten invariants won't have a direct interpretation in terms of actual numbers of curves, even taking automorphisms into account.
Here is an example of how a negative number might come up, though strictly speaking it isn't a Gromov--Witten invariant. Let M be the moduli space of maps from P^1 to a the total space of O(-4) on P^1. Call this space X. Note that I said maps from P^1, not a genus zero curve, so the source curve is rigid. That's why this isn't Gromov--Witten theory. Any such map factors through the zero section (since O(-4) has no nonzero sections), so this space is the same as the space of maps from P^1 to itself. I just want to look at degree one maps, so the moduli space is 3 dimensional.
We could also compute the dimension using deformation theory: the deformations of a map f are classified by $H^0(f^\ast T)$ where T is the tangent bundle of the target. The target in this case is O(-4), not just P^1, and the tangent bundle restricts to O(2) + O(-4) on the zero section. Thus $H^0(f^\ast T)$ is indeed 3-dimensional, as we expected. However, the Euler characteristic of $f^\ast T$ is not 3 but 0, which means that the "expected dimension" is zero.
The meaning of expected dimension is rather vague. Roughly speaking, it is the dimension of the moduli space for a "generic" choice of deformation. The trouble is that such a deformation might not actually exist. Nevertheless, we can still pretend that a generic deformation does exist and, if the expected dimension is zero, compute the number of curves that it "should" have.
What makes this possible is the obstruction bundle E on M. Any deformation of X gives rise to a section of E and the vanishing locus of this section is the collection of curves that can be deformed to first order along with X. Even though a generic deformation might not exist, the obstruction bundle does still exist, and we can make sense of the vanishing locus of a generic section by taking the top Chern class.
In our situation, the (fiber of the) obstruction bundle is $H^1(f^\ast T)$. Since O(2) does not contribute to H^1, the obstruction bundle is $R^1 p\_\ast f^\ast O\_{P^1}(-4) = R^1 p\_\ast O\_{P^3 \times P^1}(-4, -4)$ where $p : P^3 \times P^1 \rightarrow P^3$ is the projection. By the projection formula, this is $O(-4)^{\oplus 3}$ and the top Chern class is -64. This is the "Gromov--Witten invariant" of maps from P^1 to $O\_{P^1}(-4)$.
Unfortunately, I don't have anything to say about what this -64 means...
| 14 | https://mathoverflow.net/users/32 | 7841 | 5,371 |
https://mathoverflow.net/questions/7840 | 9 | What is an example of a finite local rings, that has length 2 or 3?
I want something different from $F\_{q}[x] / x^{i}$ for $i=2, 3$; I'm looking for something more interesting. If you can give me examples of higher length, yet have "simple structure" (e.g. $F\_{q}[x]/x^{i}$), that would be nice too.
I know this is related to [Classification of finite commutative rings](https://mathoverflow.net/questions/7133/classification-of-finite-commutative-rings), but I didn't completely understand the answer there.
| https://mathoverflow.net/users/2623 | Examples of finite local rings of length 2 or 3 | There are a bunch of different notions of length/depth in ring theory: Projective length, Artinian length, local depth, etc. If we take length to mean Artinian length, then Charles is right: The Artinian length of a finite-dimensional commutative algebra is just its dimension. Every such algebra is a direct sum of local ones, and you can chip away at each local summand of the ring from the bottom end, one dimension at a time.
The local algebras that have a description that looks as nice as $\mathbb{F}[x]/(x^n)$ are the toric ones. These local algebras are $\mathbb{F}[\vec{x}]$ divided by an ideal generated by monomials and a basis of monomials. You can make a diagram of the exponents of the monomials that aren't killed by the ideal. If the ring has $n$ generators, then the diagram is a stable stack of blocks in the $n$-dimensional orthant. For example, the algebra $\mathbb{F}[x,y]/(x^3,x^2y,y^3)$ has a basis of seven monomials: 1, $x$, $x^2$, $y$, $xy$, $y^2$, $xy^2$. The diagram of these monomials looks like this:
```
#
###
1##
```
I have put a 1 at the corner in the diagram corresponding to the monomial 1.
What is easy to forget is that all finite-dimensional local algebras with one generator in $m/m^2$ are of this form, but with more variables these are just special examples.
---
I was looking at the second part of the question first, interesting higher-dimensional examples. Here are some non-isomorphic local rings $R$ (not necessarily algebras) that have length 2 or 3 and such that $R/m = \mathbb{Z}/p$:
Length 2:
1. $(\mathbb{Z}/p)[x]/(x^2)$
2. $\mathbb{Z}/p^2$
Length 3:
1. $(\mathbb{Z}/p)[x]/(x^3)$
2. $(\mathbb{Z}/p)[x,y]/(x^2,xy,y^2)$
3. $(\mathbb{Z}/p^2)[x]/(px,x^2)$
4. ~~$(\mathbb{Z}/p^2)[\sqrt{p}]$~~ $\mathbb{Z}[\sqrt{p}]/p^{3/2} = \mathbb{Z}[x]/(x^2-\lambda p,x^3)$
5. $\mathbb{Z}[x]/(x^2-\lambda p,x^3)$ where $\lambda \in \mathbb{Z}/p$ is not a square. (Noted by Jonathan Wise.)
6. $\mathbb{Z}[x]/(x^2+2x,4)$ (Similar idea to previous, in characteristic 2.)
7. $\mathbb{Z}/p^3$
(**Edit:** My notation for #4 was not strictly correct.)
~~I *think*, although I can't really speak with authority, that these are all of them.~~ I thought that I knew all of these rings, but that was naive. One point is that among algebras over $\mathbb{Z}/p$, the length is too small to see anything non-toric. But you can also have local rings that look like these toric local algebras (which I listed first), but have carries. The most creative one is the fourth one of length 3, namely $(\mathbb{Z}/p^2)[\sqrt{p}]$. You can express an element of this ring as three digits in base $p$, say $d\_2d\_1d\_0$. Then addition carries from $d\_0$ to $d\_2$.
I would also guess that all of these generalize to $\mathbb{F}\_q$, using [the Witt vector](http://en.wikipedia.org/wiki/Witt_vector) construction in the cases with carries. And maybe it is again all of them.
| 5 | https://mathoverflow.net/users/1450 | 7843 | 5,373 |
https://mathoverflow.net/questions/7859 | 0 | I have a friend with dyscalculia and was teaching her some some mathematics (namely, solving a linear equation, simplifying certain expressions, and what (affine linear) functions are).
She understood solving equations of the form $ax + b = 0$ by first adding $-b$ to both sides and then diving by $a$. Dealing with negative $a$ and with expression $b - b$ was something of a problem, but I hope she figured it out, also.
Adding slightly more complexity created more problems. For example: $2x + 3 = -7$. We subtract 3 from both sides, getting $2x = -7 - 3$. She has great trouble seeing that $-7-3 = -10$.
How to communicate and teach the concepts here? I tried using the thermometer analogy, explaining how $a - a = a + (-a) = 0$ and, somewhat poorly, that $-7 - 3 = - (7 + 3) = -10$. How to justify the last attempt in a useful way? What other models or intuitions are there for understanding the negative numbers and particularly summing them?
| https://mathoverflow.net/users/1445 | How to teach addition of negative numbers? | It would help to know a little more about the nature of your friend's dyscalclia. The sense of the integer line comes from the inferior parietal cortex (and causes difficulty with problems like "what number is halfway between 7 and 11?") while rote memory-type problems ("what is 4 times 7?") are associated with the basal ganglia. It is possible to have trouble with one but not the other.
If your friend has difficulty with rote memory, you need to invoke their visual intuition (that is, rely heavily on a number line). Because the inferior parietal cortex is associative (that is, it's located in the brain at a convergence of senses) I would recommend also including their kinesthetic sense by making a physical number line and having them practice by walking along it to do number problems (I use playing cards, black for positive and red for negative).
If your friend instead has difficulty with mathematical intuition, I would present a systemized set of rules that can be incorporated into rote memory. I wouldn't stop there though, if they can master the rules that way -- I would then go back to associating those rules with mathematical intutition. It is possible to teach someone mathematical intuition who is normally impaired this way; this is a technique using neuroplasticity which involves "rewiring" around problematic parts in the brain.
| 15 | https://mathoverflow.net/users/441 | 7866 | 5,387 |
https://mathoverflow.net/questions/7847 | 3 | The following is from this talk: <http://www.maths.usyd.edu.au/u/anthonyh/piecestalk.pdf>, Slide 14.
The Springer correspondence gives bijections
SO2n+1 \ N(so2n+1) ↔ {(μ; ν) | μi ≥ νi − 2, νi ≥ μi+1},
Sp2n \ N(sp2n) ↔ {(μ; ν) | μi ≥ νi − 1, νi ≥ μi+1 − 1},
obtained from the previous parametrizations by taking 2-quotients.
What I don't understand, is given a partition of say, $2n$, that is symplectic (odd parts occur with even multiplicity), how to construct a bijection to the set above; and same with orthogonal partitions (even parts occur with even multiplicity).
| https://mathoverflow.net/users/2623 | A bijection between "symplectic" partitions and bi-partitions via Springer correspondance | From looking at the slides, it sure looks like you wrote it in your answer: take 2-quotients.
I'm not sure if there's a standard reference for n-quotients of partitions, but they're described in [this paper](http://arxiv.org/abs/math.CO/0609175), for example.
| 2 | https://mathoverflow.net/users/66 | 7871 | 5,391 |
https://mathoverflow.net/questions/7858 | 2 | What is the standard resolution of singularities, for the nilpotent cone (of the adjoint representation) for the symplectic group? I know how to do this for the general linear group, but am having trouble finding a good reference for the symplectic group. I understand it uses Richardson orbits.
(I do know what the closure ordering should be though).
| https://mathoverflow.net/users/2623 | Resolution of singularities for nilpotent cone of the symplectic group | Ben gave the general answer above. If you care specifically about the symplectic group and are interested in a "flag-like" description of its flag variety, then one exists. It is given by all half-flags of isotropic subspaces (this is just like for $SL\_n$, the symplectic group acts transitively and the stabilizer of the standard half-flag will be the standard Borel). With this description, it's just as straightforward computing Springer fibers and the like as it is for the $SL\_n$ case, which you're presumably familiar with.
A reference for these flag-like descriptions can be found in the section of Fulton and Harris on "Homogeneous Spaces" (there's a similar description for the special orthogonal groups).
| 2 | https://mathoverflow.net/users/916 | 7874 | 5,394 |
https://mathoverflow.net/questions/7881 | 16 | Let $n,k$ be positive integers. What is the smallest value of $N$ such that for any $N$ vectors (may be repeated) in $(\mathbb Z/(n))^k$, one can pick $n$ vectors whose sum is $0$?
My guess is $N=2^k(n-1)+1$. It is certainly sharp: one can pick our set to be $n-1$ copies of the set $(a\_1,...,a\_k)$, with each $a\_i=0$ or $1$. The case $k=1$ is some math competition question (I think, but can't remember the exact reference). Does anyone know of some references? Thanks.
---
Thank you all! I wish I could accept all the answers, they are very helpful!
| https://mathoverflow.net/users/2083 | Sum of $n$ vectors in $(\mathbb Z/n)^k$ | Your guess is correct for k=1 and 2, but when k is bigger, things get more complicated. For instance, when k=n=3, N=19. For a summary of some known results, see:
[Elsholtz, C. Lower Bounds For Multidimensional Zero Sums. *Combinatorica* **24**, 351–358 (2004).](https://doi.org/10.1007/s00493-004-0022-y)
| 11 | https://mathoverflow.net/users/1013 | 7890 | 5,408 |
https://mathoverflow.net/questions/7892 | 14 | Is it known whether $O(4) \to PL(4)$, the map from the orthogonal group to the group of piecewise linear homeomorphisms of $\mathbb{R}^4$, is a homotopy equivalence? By smoothing theory for PL manifolds, this is equivalent to whether the space of smooth structures on a PL 4-manifold is contractible. (I think it's known that this map is at least 4-connected, which shows that the space of smooth structures on any PL 4-manifold is nonempty and connected.)
| https://mathoverflow.net/users/2327 | Smooth structures on PL 4-manifolds | Very little is known about that question, the same smoothing theory gives something that I'm trying to get people to call "The Cerf-Morlet Comparison Theorem"
$$ Diff(D^n) \simeq \Omega^{n+1}(PL(n)/O(n)) $$
$Diff(D^n)$ is the group of diffeomorphisms of the $n$-ball where the diffeomorphisms are pointwise fixed on the boundary. Nobody knows if $Diff(D^4)$ is path-connected or not. Very little is known about the homotopy-type of $Diff(D^4)$, no seriously informative statements other than that homotopy-equivalence. I wrote up a paper where I described in detail the iterated loop-space structure and how it arrises naturally. Moreover, I described how that iterated loop-space structure relates to various natural maps. That's my main relation to to topic. The paper is called "Little cubes and long knots" and is on the arXiv. I elaborate on some of these issues in the paper "A family of embedding spaces", also on the arXiv.
There are several natural connections here, one of the big ones being that $Diff(D^n)$ has the homotopy-type of the space of round metrics on $S^n$ -- ie the subspace of the affine-space of Riemann metrics on $S^n$, the subspace is specified by the condition that "$S^n$ with this metric is isometric to the standard $S^n$."
| 10 | https://mathoverflow.net/users/1465 | 7894 | 5,410 |
https://mathoverflow.net/questions/4454 | 29 | Is there a *natural* measure on the set of statements which are true in the usual model (i.e. $\mathbb{N}$) of Peano arithmetic which enables one to enquire if 'most' true sentences are provable or not? By the word 'natural' I am trying to exclude measures defined in terms of the characteristic function of the set of true sentences.
| https://mathoverflow.net/users/1508 | How many of the true sentences are provable? | It seems to me that the probability that a statement is provable and that it is undecidable should both be bounded away from 0, for any reasonable probability distribution.
Let $C\_n$ be the number of grammatical statements of length $n$. For any statement $S$, the statement
>
> $S$, or $1=1$
>
>
>
is a theorem. So the number of provable statements of length $n$ is bounded below by $C\_{n-k}$, where $k$ is the number of characters needed to tag on "or $1=1$".
On the other hand, let $G$ be an undecidable sentence, and $S$ any sentence. Then
>
> Either $S$ and $1 \neq 1$, or else $G$
>
>
>
is undecidable. So the number of undecidable sentences of length $n$ is bounded below by $C\_{n-\ell}$, for some constant $\ell$. For any reasonable grammar, the ratios $C\_{n-k}/C\_n$ and $C\_{n- \ell}/C\_n$ should both be bounded away from 0.
My computation is seemingly incompatible with the [paper of Calude and Jurgensen](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6W9D-4F9F850-2&_user=501045&_rdoc=1&_fmt=&_orig=search&_sort=d&_docanchor=&view=c&_acct=C000022659&_version=1&_urlVersion=0&_userid=501045&md5=dde512d656cc98a3a5a0c7819de84cf9) cited by [Konrad](https://mathoverflow.net/questions/4454/how-many-of-the-true-sentences-are-provable/7856#7856). I sent some time trying to figure out why and my view is that I am right and Calude and Jurgensen are wrong but, of course, I could be doing something dumb.
---
This answer is getting discussed again, so let me mention a way that one might be able to salvage this, though. There is extensive literature on the behavior of random boolean statements, of which the most common model is "random $3$-SAT". The way that this works is that you have $n$ boolean variables $x\_1$, $x\_2$, ..., $x\_n$. A "three-term clause" is a statement of the form $p\_1 \vee p\_2 \vee p\_3$ where each $p$ is either $x\_i$ or $\neg(x\_i)$ for some $i$. One samples $k$ of the three-term clauses independently at random and asks whether they can all be satisfied at once. Generally, one takes $k = cn$ for fixed $c$ and asks about behavior as $n \to \infty$.
When I first heard this problem, my intuition was that the probability would be bounded away from $1$, for the same sort of reason as this answer: Some positive density of $3$-SAT instances would be of the form $P \wedge Q$, where $P$ was some explicit finite list of incompatible $3$-term clauses. (For example, $P$ could be the AND of the eight $3$-term clauses involving $(x\_1, x\_2, x\_3)$.) But this is wrong! Because all of the $n$ variables are equally likely, for any finite number of $3$-term clauses, the probability that they will have any variables in common goes to $0$.
The flaw in this answer's argument, when it comes to random $3$-SAT, is the following: Because the size of our alphabet is allowed grow with the length of the statement, the number of $3$-SAT instances with $cn$ clauses and $n$ variables grows like $n^{3cn}$, not like an exponential! It is actually really important whether writing down "$x\_i$" counts as $O(1)$ characters, or as $O(\log i)$ characters!
So this makes me wonder whether some similar phenomenon might occur for decidability in first order logic. I dug around a little, but I couldn't find any research.
| 31 | https://mathoverflow.net/users/297 | 7902 | 5,415 |
https://mathoverflow.net/questions/7897 | 11 | I heard people mentioned this in one sentence, but don't see the reason.
Why a (smooth) variety of general type, i.e. an algebraic variety X with K\_X big, is hyperbolic, i.e. has no non-constant map from the complex number into it?
I don't know what are the necessary assumption on the variety, do we need properness or smoothness?
Edit: according to David Lehavi's reply, we should certainly put some more condition on it. What's the correct statement of the fact?
| https://mathoverflow.net/users/1657 | Why is a variety of general type hyperbolic? | You must be thinking of Lang's conjecture which predicts that a smooth projective variety is (Brody) hyperbolic if and only if all of its irreducible subvarieties are of general type.
This is still not known in general but there are many special cases that are known. A good example is McQuillan's theorem - a smooth surface of general type which satisfies $c\_{1}^{2} > c\_{2}$ and does not contain any rational or elliptic curves is Brody hyperbolic.
| 9 | https://mathoverflow.net/users/439 | 7905 | 5,416 |
https://mathoverflow.net/questions/7911 | 19 | I am looking for a counter-example of two functors F : C -> D and G : D->C such that
1) F is left adjoint to G
2) F is right adjoint to G
3) F is not an equivalence (ie F is not a quasi-inverse of G)
| https://mathoverflow.net/users/2330 | Is a functor which has a left adjoint which is also its right adjoint an equivalence ? | The answer of Ben Webster, can be made easier. Consider the functor F : (A-mod) -> (A-mod) which maps any A-module on (0). Then, F is a left adjoint to F ; and so, is a also a right adjoint to F. This is clear because for all A-modules N, M, one has Hom\_A(0,N)=Hom\_A(M,0). But, F is not an equivalence.
| 13 | https://mathoverflow.net/users/2330 | 7922 | 5,427 |
https://mathoverflow.net/questions/7926 | 2 | I´m looking for information about the intersection of two vector bundles (principally trivial bundles, but no necessarily). I´m trying to make a picture (literally) of reflexive finite generated modules.
Another related topic is sub-budles of a vector bundles.
All suggestions are wellcome!
---
**Edit**: I try to be more specific.
Suppose two sub-bundles of a bundle over a topological space X. We can do the intersection of their vector fibers in each point of base space and collect this intersection vector fibers along the base space. Then we can give it a topology, restriction of the bigger vector bundle.
What´s about of this "submodule of sections"?
Is it another vector bundle?
Has another interesting structure?
| https://mathoverflow.net/users/2040 | About the intersection of two vector bundles | You need to refine the question to get better answers, but here are some thoughts:
1) Over a normal variety, you can think of line bundles as divisors, and "intersect" them.
2) A vector bundle can be represented by a reflexive sheaf, but being reflexive is a lot weaker.
| 3 | https://mathoverflow.net/users/2083 | 7932 | 5,433 |
https://mathoverflow.net/questions/7942 | 2 | Let R a normal domain, that is an integrally closed noetherian domain, like Dedekind domains, UFD, etc
Let A=(a i j ) a matrix with elements in R and dimension n x m.
Suppose
* rank A=1 ↔ all 2 x 2 minors are =0.
* J:= ideal generated by a i j verify (R:(R:J))=R ↔ J is not included in any prime ideal with height 1.
If R is an UFD, with the preview conditions, we can write A like product of a n x 1 vector column C=(c i ) and a 1 x m vector file F=(f j ), that is a i j =c i ·f j .
I conjecture that is true in the general case, but I cannot make any progress.
Have you contraexemples with normal rings?
I´m grateful for your advices!
| https://mathoverflow.net/users/2040 | Splitting matrix of rank one | I am having trouble understanding your English. But, if I understand you correctly, the following is a counter-example:
Let $k$ be a field and let $R$ be the ring $k[a,b,c,d]/(ab-cd)$. Then $R$ is normal and $\left( \begin{smallmatrix} a & c \\\\ d & b \end{smallmatrix} \right)$ has rank 1. However, we can not write this matrix as $\left( \begin{smallmatrix} w \\\\ x \end{smallmatrix} \right) \left( \begin{smallmatrix} y & z \end{smallmatrix} \right)$ for any $w$, $x$, $y$, $z \in R$.
I think your condition should almost imply that the ring is a UFD. If I have any non-unique factorization $ab=cd$, I can use it to build a counter-example like this one.
**UPDATE** Here are two more examples: $R=k[a,b,c]/(ac-b^2)$ and $\left( \begin{smallmatrix} a & b \\\\ b & c \end{smallmatrix} \right)$.
$R=\mathbb{Z}[\sqrt{-5}]$ and $\left( \begin{smallmatrix} 2 & 1+\sqrt{-5} \\\\ 1-\sqrt{-5} & 3 \end{smallmatrix} \right)$.
These examples rule out most attempts I could think of to find a class of rings larger than UFDs for which the result holds.
| 3 | https://mathoverflow.net/users/297 | 7945 | 5,440 |
https://mathoverflow.net/questions/7951 | 4 | Is there a poset P with a unique least element, such that every element is covered by finitely many other elements of P (and P is locally finite -- actually, per David Speyer's example, let's say that it satisfies the descending chain condition), and P has countably infinite automorphism group?
The question is motivated by extensions of Sperner's theorem and the LYM inequality to infinite posets. In particular I'm interested in whether you can extend Bollobas' (I believe) probabilistic proof of LYM to the infinite setting in general -- you can for some specific posets. But a prerequisite for a direct extension is for the automorphism group of the poset to be compact, and at the very least we want it to have some nice topological properties. So a poset with countably infinite automorphism group would be a very interesting case.
| https://mathoverflow.net/users/382 | Is there a poset with 0 with countable automorphism group? | It seems unlikely (once you assume d.c.c.). Define the height of an element $x$ in $P$ to be the length of the shortest unrefinable chain from $x$ to $0$.
Let $P\_n$ denote the elements of $P$ whose height is at most $n$. Since each element has a finite number of covers, the number of elements in $P\_n$ is finite.
By d.c.c., every element of $P$ is in some $P\_n$.
Let $G$ denote the automorphisms of $P$ and let $G\_n$ denote the automorphisms of $P\_n$. $G$ is the inverse limit of the system $G\_n$. Let $H\_n$ denote the image of $G$ inside $G\_n$. (Note that this might not be all of $G\_n$, since there could be automorphisms of $P\_n$ that don't extend to $P$.) $G$ is also the inverse limit of the system $H\_n$.
If the system $H\_n$ stabilizes, then $G$ is finite. On the other hand, if $H\_n$ doesn't stabilize, then the cardinality of $G$ is an infinite product, i.e. uncountable.
| 5 | https://mathoverflow.net/users/468 | 7955 | 5,447 |
https://mathoverflow.net/questions/7907 | 10 | How can I determine the integer points of a given elliptic curve if I know its rank and its torsion group?
I read same basic books on elliptic curves but as a non-professional I didn't understand everything. Is it true that if rank is 0 and torsion group is isomorphic to a group of order $n$ then the number of integer points is $n-1$? And what is a good reference to learn to determine the integer points if the rank is positive?
I tried to read the book Rational Points on Elliptic Curves but I didn't found an explicit algorithm. I just heard something like take some point and use group law to find the rest. But how can I be sure that I have found every point?
The curve I had on my mind is $2x^3 + 385x^2 + 256x - 58195 = 3y^2$. I'm not even sure if this is an elliptic curve. I mean why it is projective and why it is isomorphic to a closed subvariety of $\mathbb{P}\_{\mathbb{Q}}^2$? And why it contain the priviledged rational point $(0,1,0)$?
| https://mathoverflow.net/users/2120 | How to find all integer points on an elliptic curve? | Finding all the integral points on an elliptic curve is a non-trivial computational problem. You say you are a "non-professional" so here is a non-professional answer: get hold of some mathematical software that does it for you (e.g. MAGMA), and then let it run until it either finds the answer or runs out of memory. Alternatively, do what perhaps you should have done at the start if you just have one curve and want to know the answer: post the equation of the curve, and hope that someone else does it for you. Here's another example of an algorithm currently used in these sorts of software (a Thue one was mentioned above but here's a different approach): find generators for the group (already computationally a bit expensive at times, depending on your luck and/or the size of sha), invoke Baker-like theorems saying "if the coordinates of the point are integral then it must be of the form sum\_i n\_i P\_i with the n\_i at most ten to the billion", and then use clever congruence techniques to massively cut down the search space by giving strong congruences for all the n\_i. Then just do a brute force search.
Whether or not this will work for you, I cannot say, because it all depends on how big the coordinates of your curve are. The only clue you give so far is that the conductor is "bigger than 130000" [Edit: that was written before the OP edited the question to tell us which curve he was interested in] which of course does not preclude it being bigger than 10^10^10. Also, you need an expert to decide which of the algorithms is best for you. I'd rather do a massive amount of arithmetic in a field of tiny discriminant than a small amount of arithmetic in a field whose discriminant is so large that I can't even factor it, for example.
So in short the answer is that you're probably not going to be able to do it with pencil and paper, but there are programs around that will do it, if all you want to know is the answer.
EDIT: you posted the equation of the curve. Magma V2.15-10 says the integral points are
[ <-23, -196>, <19, 182>, <61, 784>, <-191, 28>, <103, -1442>, <-19, -144>,
<-67, 592>, <23, 242>, <-49, -454>, <-157, -742>, <817, 21196>, <521, 11364>,
<3857, 200404>, <10687, -910154>, <276251, -118593646> ]
plus what you get if you change all the y's to -y's.
| 9 | https://mathoverflow.net/users/1384 | 7979 | 5,460 |
https://mathoverflow.net/questions/7870 | 0 | How does one linearize and analyze such a system?
Just noticed I could edit this, so from my comment below:
I am trying to get a feel for what analysis us used beyond the introduction I have had. The equations for the double pendulum were derived from the second derivative of the equations for position of each mass and the tension of the mass against the rods. So far the only understanding I have been able to get from it is using numerically generated phase planes and plots of motion made in maple.
I have seen an overview of the analysis mentioned on wolfram's site and am reading into that, but these books on mechanics are fairly broad, and I am looking to see if there is a specific field I need to look into.
Edit in response to comment: Not looking for this to be simplified to the level of a pendulum, I just need to know how to approach the problem. I do not know what analytical tools would be used to better understand the system(as I understand it, there is no solving for such a complicated system in the same form that a regular nonlinear pendulum can be).
| https://mathoverflow.net/users/2320 | Stability analysis of a system of 2 second order nonlinear differential equations | This is an answer to Charles' restatement of the question.
Recall that equation F(x,x',x'') = 0 (e.g. x'' + sin x = 0) can be written as a system
X' = f(X), where X = (x,x')^T (e.g. f(x',x) = (-sin x, x')^T) and that system can be linearized about an equilibrium E = (x\_*,x'\_*)^T to obtain a linear equation
X' = AX where A is the 2 x 2 matrix given by the derivative of f at E.
So too a larger system F(x,x',x'',y,y',y'') = 0 can be written as a system
X' = f(X) where X = (x,x',y,y')
Given an equilibrium E = (x\_*,x'\_*,y\_*,y'\_*), the linearization of X' = f(X) about E is
again X' = AX where X = (x',x,y',y) and A is the derivative of f evaluated at E. A is a 4 by 4 matrix. If all eigenvalues of A have negative real part, the system is stable. If one eigenvalue has positive real part, the system is unstable. If there are no eigenvalues with positive real part, and there are eigenvalues which lie on the imaginary axis, then the equilibrium is "spectrally stable," and further analysis is required to determine the nonlinear stability.
With regard to the energy of the system, you are looking for a function of the form
V(x,x',y,y') whose time-derivative along solutions is constant. A good place to start is with the guess 1/2((x')^2 + (y')^2) + F(x,y). You should be able to figure out what F needs to be in this particular example.
For future reference, this forum is (I believe) intended primarily for questions from students and practitioners who are a little further along in their study. I decided to answer the question because I can imagine being very frustrated at not having some information that it would take an expert five minutes to explain and the question didn't strike me as the kind which would encourage others to attempt to turn this forum into a homework help site. If the moderators disagree, I apologize.
| 3 | https://mathoverflow.net/users/1281 | 7993 | 5,469 |
https://mathoverflow.net/questions/7968 | 2 | Let $N\_3$ be the genus three non orientable surface. Do we have an analogous 3d manifold as the solid torus and the solid Klein bottle for $N\_3$? I don't see how to extend the ideas related to the 3d lens spaces. Any feedback would be super-welcome
| https://mathoverflow.net/users/2196 | Two solid N_3 glued by its boundary | It is a general fact that a closed manifold of odd Euler characteristic cannot bound a compact manifold. This can be deduced pretty easily from the fact that a closed manifold of odd dimension has Euler characteristic zero (a consequence of Poincaré duality) as follows. Suppose N is the boundary of a compact manifold P. Let M be the double of P, the union of two copies of P glued along N. Then the Euler characteristics of M, N, and P are related by:
$\chi(M)=\chi(P)+\chi(P)-\chi(N)$
Thus $\chi(M)$ and $\chi(N)$ are congruent mod 2. If the dimension of N is even, then M is a closed manifold of odd dimension so $\chi(M)=0$, hence $\chi(N)$ is even. And if the dimension of N is odd then $\chi(N)=0$ anyhow.
I should have put this in my book!
| 13 | https://mathoverflow.net/users/23571 | 8005 | 5,478 |
https://mathoverflow.net/questions/8014 | 5 | I'm a bit stuck, and I'm hoping someone can help me out. I have a vector bundle $E$ on an algebraic curve (the ones I am interested in are holomorphic, but I'm sure that doesn't matter so much for the purposes of this question...), and I also have an endomorphism $\Psi\in\Gamma(\mbox{End}(E)).$
How **in general** does one compute the determinant of $\Psi$? If $E$ splits as a sum of rank-1 bundles then so too does the endomorphism bundle associated with $E$, and then it's obvious to me how to compute the determinant in that case. But, when $E$ is more general, what is an invariant way to go about computing the determinant? (I'm looking for something similar to the invariant way of computing the trace, for which I can rewrite $\Psi$ as a map $E\otimes E^\*\rightarrow\mathcal{O}$ and then take the image of the identity morphism of $E$.)
I'm sure there is a quick answer to this, that I am just not seeing. Any help is greatly appreciated.
| https://mathoverflow.net/users/2347 | Endomorphisms of vector bundles | The determinant of $\Psi$ is the top exterior power $\bigwedge^{\mathrm{rk}(E)}\Psi$ (identifying endomorphisms of the line bundle $\bigwedge^{\mathrm{rk}(E)}E$ with functions).
| 10 | https://mathoverflow.net/users/2035 | 8018 | 5,487 |
https://mathoverflow.net/questions/8023 | 13 | I'm looking for an easily-checked, local condition on an $n$-dimensional Riemannian manifold to determine whether small neighborhoods are isometric to neighborhoods in $\mathbb R^n$. For example, for $n=1$, all Riemannian manifolds are modeled on $\mathbb R$. When $n=2$, I believe that it suffices for the scalar curvature to vanish everywhere (this is certainly necessary). But my intuition is poor for higher-dimensional structures.
Put another way: given a Riemannian structure $g$ on a smooth manifold, when can I find coordinates $x^1,\dots,x^n$ so that $g\_{ij}(x) = \delta\_{ij}$?
| https://mathoverflow.net/users/78 | When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$? | If the Riemannian metric is twice differentiable in some co-ordinate system, then this holds in any dimension if and only if the Riemann curvature tensor vanishes identically.
In dimension 2, it suffices for the scalar curvature to vanish.
In dimension 3, it suffices for the Ricci curvature to vanish.
In higher dimensions, you need to have the full Riemann curvature vanish.
| 19 | https://mathoverflow.net/users/613 | 8026 | 5,493 |
https://mathoverflow.net/questions/7998 | 16 | Is there a sort of structure theorem for pairwise independent random variables or a very general way to create them?
I'm wondering because I find it difficult to come up with a lot of examples of nontrivial pairwise independent random variables. (by 'nontrivial', i mean not mutually independent)
one example (three r.v.):
=========================
X = face of dice 1
Y = face of dice 2
Z = X + Y mod 6
another example (three events) from some book:
==============================================
Throw three coins. A = the number of heads is even, B = the first two flips are the same, C = the second two flips
are heads.
another example:
================
$A\_{ij}$ = dice i and dice j having the same face
($A\_{ij}$, $i\neq j$) form a set of pairwise independent events, but the triple ($A\_{ij}$, $A\_{jk}$, $A\_{ki}$) is not mutually independent.
| https://mathoverflow.net/users/1354 | most general way to generate pairwise independent random variables? | I'm sure that Gil's answer is wise and that it is a good idea to look at Alon and Spencer's book. Here also is a quick summary of what is going on.
Suppose that $X\_1,\ldots,X\_n$ are random variables, and suppose for simplicity that they take finitely many values. Suppose that you prescribe the distribution of each $X\_i$, and suppose also that you want the random variables to be pairwise independent or $k$-wise independent. Then the constraints on the joint distribution are a finite list of equalities and inequalities. The solution set is a polytope whose dimension is fairly predictable, and the fully independent distribution is always in the interior of this polytope. If you are interested in $k$-wise independent distributions that are far from $k+1$-wise independent, then it can be difficult to determine what is achievable because the polytope is complicated. (The vertices are a particularly interesting and non-trivial class: $k$-wise independent distributions with small support. These are called "weighted orthogonal arrays".) However, if you're just intersted in examples, it is much easier to write down a small deviation of the fully independent distribution. The deviation just satisfies linear equations.
For example, suppose that $X,Y,Z$ are three unbiased Bernoulli random variables (coin flips) that take values $0$ and $1$. Then there are 8 probabilities $p\_{ijk}$, one for each outcome $(X,Y,Z) = (i,j,k)$. Then you can set
$$p\_{ijk} = \frac18 + (-1)^{i+j+k}\epsilon. \qquad\qquad\qquad \text{(1)}$$
to get a pairwise independent but not independent distribution. In this simple example, there is a 1-dimensional space of deviations and it is easy to compute how far you can vary the independent solution. (Up to $|\epsilon| = \frac18$.) In larger cases, the variations can be multidimensional and the polytope of deviations can be more complicated.
**Addendum:** If I have not made a mistake, all deviations for any finite list of discrete random variables are linear combinations of those of the form (1). More precisely, given discrete random variables $X\_1,\ldots,X\_n$, let $f\_i$ be some function of the value of $X\_i$ which is 1 for one value, $-1$ for another value, and $0$ otherwise. Then you can make deviations proportional to $\prod f\_{i\_j}$ as long as there are at least $k+1$ factors. It looks like all deviations are a linear combination of those of this form.
| 7 | https://mathoverflow.net/users/1450 | 8027 | 5,494 |
https://mathoverflow.net/questions/8003 | 19 | The following is a theorem of [Elkies](http://www.ams.org/mathscinet-getitem?mr=903384):
Let $X$ be an elliptic curve over $\mathbb{Q}$. Then there are infinitely many primes $p$ such that the action of Frobenius on $H^1(\mathcal{O}, X)$ is zero.
Allen Knutson and I would like a similar theorem for higher dimensional Calabi-Yau varieties. Unfortunately, we've been told that this is probably open. (References for the fact that it is open are appreciated.) But we don't need the full strength of Elkies' result. It would be enough for us to know the following:
Let $X$ be an $n$-dimensional, smooth, complete Calabi-Yau variety over $\mathbb{Q}$, for $n>0$. Write $X/p$ for the fiber of $X$ over $p$. Let $T(p)$ be the action of Frobenius on $H^n(\mathcal{O}, X/p)$. Are there infinitely many $p$ for which $T(p) \neq 1$? Also, in the same generality, are there infinitely many primes for which $T(p) \neq 0$?
| https://mathoverflow.net/users/297 | Elkies' supersingularity theorem in higher dimension | If I understand your question correctly, the "easy" version of the question you ask is unknown in dimension $\ge 3$, and is basically easy for surfaces of interest.
These questions are certainly unknown in dimension $\ge 3$, or else the Sato-Tate conjecture for weight $>3$ would have been known $12$ months ago, rather than $4$ months ago.
I take it you are only asking that the action of Frobenius is neither almost always trivial ($T(p) = 0$) or acts as the identity ($T(p) = 1$), which is much weaker than asking that there exist any supersingular primes.
Assume otherwise. Fix a prime $\ell$, and Let $V$ be the $\ell$-adic etale cohomology
$H^2(X)$ with the usual action of $\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$. Let $F\_p(T)$ denote the characteristic polynomial of Frobenius. It has coefficients in $\mathbf{Z}[x]$. Say one wants to show that there exist infinitely many primes $p$ such that $T(p) \ne 0$. You are imposing the condition that $F\_p(T) \equiv T^n \mod p$ for all but finitely many $p$, where $n = \mathrm{dim}(V)$.
Choose a prime $L > 2n$. There will a positive density of primes such that $F\_p(T) \equiv (T-1)^n \mod L$.
If $p \equiv 1$ modulo $L$, These conditions together imply that the trace of $F\_p(T)$, which is *a* *priori* an integer, is $np \mod pL$. By the Weil conjectures, the roots of $F\_p(T)$ have absolute value $p$, and thus, since $L > 2n$, they must all equal $p$. By Cebatarev density, it follows that a finite index subgroup of the Galois group acts (on the semisimplification of $V$) via the cylotomic character on $V$. By looking at Hodge-Tate weights this implies that $h^{2,0} = h^{0,2} = 0$, and so (for example) one has a contradiction for a K3-surfaces.
If one wants to rule out that $T(p) = 1$ for (almost all) $p$, one is imposing the condition that $F\_p(T) = (T-1)^n \mod p$. In the same way one obtains an open subgroup of the Galois group such that the trace of Frobenius is always $n$, and then one deduces that $h^{0,2} > 0$ and $h^{2,0} = 0$, which can never happen.
EDIT: Forgot to mention that it was Ogus who proved in ~70's that Abelian surfaces had infinitely primes of ordinary reduction, presumably by a very similar argument. Well, except for the comparison theorem of Faltings I used above...
| 17 | https://mathoverflow.net/users/nan | 8035 | 5,500 |
https://mathoverflow.net/questions/8031 | 12 | The Koebe–Andreev–Thurston theorem gives a characterization of planar graphs in terms of disjoint circles being tangent. For every planar graph $G$ there is a circle packing whose graph is G. What happens when circles are replaced by spheres? By spheres of higher dimension? What is known about the graphs that are formed in this case? I know that they must contain all planar graphs but that is about it.
Let me update this. Based on the answer to this question I am beginning to think about the maximalchromatic number of this graph. If its edge density is less than 7 than for all graphs of this type there must be a point with 13 or less edges. Assume that there is such a graph with chromatic number greater than 14 then look at a graph with the minimum number of points with chromatic number greater than 14. One point must have 13 or less edges. Remove it and since the graph had minimum number of points to have chromatic number greater than 14 the new graph must have chromatic number 14 or less. Color it with 14 or less colors. Add the removed point and look at the colors of the thirteen points it is adjacent to give it the remaining color then we have a 14 coloring of the graph and hence a contradiction. So the chromatic number must be 14 or less. We have an lower bound of 4 as the graphs of tangent spheres contain the graphs of tangent circles which are the planar graphs which contain graphs with chromatic number four.
As we increase the dimensions of the tangent spheres the chromatic number goes to infinity just take $d+1$ tangent spheres. But I think we can do better than that for one thing we should be able to improve the lower bound from 4. Also I think there should be examples from lattices that give better lower bounds on the chromatic number possibly exponential.
Finally based on the existing chromatic numbers I am wondering if it is possible to answer this question. Is there a dimension where the chromatic number of the unit distance graph is different from the chromatic number of the graphs in that dimension of tangent spheres.
The unit distance graph is the set of all points in the $n$-dimensional space with two points connected if their distance is one. For dimension two the chromatic number is known to be in the range from 4 to 7. For dimension three the range is 6 to 15. For the graphs of tangent circles we have a chromatic number of 4 and for spheres a range from 4 to 15. So the possibility that the chromatic numbers of the two types of graphs are the same has not yet been eliminated. So the specific question is what is known and what can be proved about the chromatic number of the graphs of tangent spheres.
| https://mathoverflow.net/users/1098 | Graphs of Tangent Spheres | The number of edges in such a graph is linear in the number of vertices, and they can be split into two equal-sized subgraphs by the removal of $O(n^{\frac{d-1}{d}})$ vertices. See e.g.
A deterministic linear time algorithm for geometric separators and its applications.
D. Eppstein, G.L. Miller, and S.-H. Teng.
[Fundamenta Informaticae 22:309-330, 1995](http://www.ics.uci.edu/~eppstein/pubs/EppMilTen-FI-95.ps.gz).
Unlike in the 2d case (where we know that the maximum number of edges such a graph can have is 3n-6) the precise maximum edge density is not known even in 3d. There's a lower bound of roughly $\frac{3828n}{607}\approx 6.3064n$ in another of my papers,
Fat 4-polytopes and fatter 3-spheres.
D. Eppstein, G. Kuperberg, and G. Ziegler.
arXiv:math.CO/0204007.
Discrete Geometry: In honor of W. Kuperberg's 60th birthday, Pure and Appl. Math. 253, Marcel Dekker, pp. 239-265, 2003.
and an upper bound of $(4+2\sqrt3)n \approx 6.8284n$ in
Greg Kuperberg and Oded Schramm, Average kissing numbers
for non-congruent sphere packings, Math. Res. Lett. 1 (1994),
no. 3, 339–344, arXiv:math.MG/9405218.
| 13 | https://mathoverflow.net/users/440 | 8036 | 5,501 |
https://mathoverflow.net/questions/7938 | 4 | How do you compute in characteristic $0$, intersection cohomology of partial flag varieties (corresponding to a fixed partition $\lambda$)? I understand the answer involves Kazhdan-Lusztig polynomials; all I can find is a reference for characteristic $p$ (<http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.0207v2.pdf>), I'm looking for the paper by Kazhdan & Lusztig: Schubert varieties and Poincare duality, which I cannot find.
I'm specifically trying to compute the intersection cohomology of a subspace of the product of two flag varieties $(V\_{i}), (W\_{j})$ where the intersections $dim(V\_{i} \cap W\_{j})$ have fixed dimension. This problem isn't known or studied right? Is there anything to be said about intersection cohomology of homogeneous spaces?
| https://mathoverflow.net/users/2623 | Intersection cohomology of flag varieties/Schubert varieties | First, let me rephrase your question in a slightly pedantic manner.
To establish some notation, for a point $p$ on the flag variety $G/B$, let $V\_1(p)\subset\cdots V\_{n-1}(p)$ be the flag in $\mathbb{C}^n$ that it corresponds to. (Be careful. There are no flags actually in the flag variety, just points. Rather, the points in the flag variety correspond to flags. If this confuses you you need a live person to straighten you out.)
You are asking for the intersection cohomology of the subvariety $X\subset G/B \times G/B$ consisting of points $(p,q)$ such that $\dim(V\_i(p)\cap V\_j(q))=a\_{ij}$ (for some specified $a\_{ij}$).
Now an answer:
Your variety $X$ has a projection onto the second factor, and this map is a fiber bundle whose base space is smooth (since it is the entire flag variety). Therefore, the local intersection cohomology for the whole space is determined entirely by the local intersection cohomology of the fibers.
If the conditions $a\_{ij}$ are conditions that determine a Schubert variety, then the fibers are Schubert varieties, and hence local intersection cohomolgy Betti numbers are precisely given by Kazhdan--Lusztig polynomials.
If the conditions $a\_{ij}$ are not conditions determining a Schubert variety, then your fibers will be unions of Schubert varieties. I don't know if anyone has bothered to do this, but I would think that if you take any of the definitions of Kazhdan--Lusztig polynomials $P\_{u,v}(q)$ and modify it in the obvious way (if there is one) to allow $v$ to be an arbitrary lower ideal in Bruhat order rather than a principal lower ideal you should get the right thing.
| 3 | https://mathoverflow.net/users/3077 | 8047 | 5,508 |
https://mathoverflow.net/questions/8039 | 23 | This question comes out of the answers to [Ho Chung Siu's](https://mathoverflow.net/questions/7836/why-is-it-useful-to-study-vector-bundles) question about vector bundles. Based on my reading, it seems that the definition of the term "section" went through several phases of generality, starting with vector bundles and ending with **any** right inverse. So admittedly I'm a little confused about which level of generality is the most useful.
Some specific questions:
* Why can we think of sections of a bundle on a space as generalized functions on the space? (I'm being intentionally vague about the kind of bundle and the kind of space.)
* What's the relationship between sections of a bundle and sections of a sheaf?
* How should I think about right inverses in general? I essentially only have intuition for the set-theoretic right inverse.
Pointers to resources instead of answers would also be great.
| https://mathoverflow.net/users/290 | What is a section? | **To your first question,** "function on a space" $X$ usually means a morphism from $X$ to one of several "ground spaces" of choice, for example the reals if you work with smooth manifolds, Spec(A) if you work with schemes over a ring, etc. (This is a fairly selective use of the word "function" which used to confuse me.) A section $\gamma$ of a (some-kind-of) bundle $E\to X$ is thought of as a "generalized function" on $X$ by thinking of it as a funcion with "varying codomain", i.e. at each point $x\in X$, it takes value in the fibre
$E\_x\to x$. If one is talking about locally free / locally trivial bundles, meaning $E$ is locally (over open sets
$U\subset X$) isomorphic to some product $U\times T$, then we can locally identify the fibres with $T$. Thus locally a section just looks like a function with codomain $T$, which is often required to be nice.
**To your second question,** I generally take the "right-inverse" or "pre-inverse" definition from category theory, because it relates back to others in the following precise way:
Say $\pi: Y\to X$ is a space over $X$ (intentionaly vague). The word "over" is used to activate the tradition of suppressing reference to the map $\pi$ and refering instead to the domain $Y$. For $U\subseteq X$ open, the notation $\Gamma(U,Y)$ denotes *sections of the map $\pi$ over $U$*, i.e. maps $U\to Y$ such that the composition $U \to Y\to X$ is the identity (thus necessarily landing back in $U$). It's not hard to see that
$\Gamma(-,Y)$ actually forms a sheaf of sets on $X$.
Conversely, given any sheaf of sets $F$ on a space $X$, one can form its [espace étalé](http://en.wikipedia.org/wiki/Sheaf_%28mathematics%29#The_.C3.A9tale_space_of_a_sheaf), a topological space over $X$, say $\pi: \acute{E}t(F) \to X$. Then for an open $U\subseteq X$, the elements of $F(U)$ correspond precisely to sections of the map $\pi$, which by the above notation is written $\Gamma(U,\acute{E}t(F)$. That is to say,
$F(-)\simeq\Gamma(-,\acute{E}t(F))$ as sheaves on $X$. This explains why people often refer to sheaf elements as "sections" of the sheaf.
Moreover, what we now denote by $\acute{E}t(F)$ actually used to be the definition of a sheaf, so people tend to identify the two and write $\Gamma(-,F)$ a instead of $\Gamma(-,\acute{E}t(F))$. This explains the otherwise bizarre tradition of writing $\Gamma(U,F)$ instead of the the more compact notation $F(U)$.
$\Big($**Unfortunate linguistic warning:** Many people incorrectly use the term "étale space". However, the French word "étalé" means "spread out", whereas "étale" (without the second accent) means "calm", and they were not intended to be used interchangeably in mathematics. This is unfortunate, because the espace étalé has very little to with with étale cohomology. More unfortunate is the annoying coincidence that when dealing with schemes the projection map from the espace étalé happens to be an étale morphism, because it is locally on its domain an isomorphism of schemes, a much stronger condition.$\Big)$
**To your third question,** I think the observation that $\Gamma(-,Y)$ forms a sheaf on $X$ gives a nice context in which to think of sections $X$ to $Y$: they "live in" the sheaf $\Gamma(-,Y)$ as its globally defined elements.
| 25 | https://mathoverflow.net/users/84526 | 8049 | 5,510 |
https://mathoverflow.net/questions/8038 | 7 | Chebyshev polynomials have real roots and satisfy a recurrence relation. I was wondering if one can find a sequence of polynomials with integral or rational roots with similar properties. More precisely, one is looking for a sequence of polynomials $(f\_n),f\_n\in\mathbf{Q}[t]$ such that
1. $\deg f\_n\to\infty$ as $n\to\infty$;
2. $\sum\_{n=0}^\infty f\_n(t) x^n$ is (the Taylor series of) a rational function $F$ in $x$ and $t$.
3. All roots of any $f\_n$ are integer and have multiplicity 1. (A weaker version: the roots are allowed to be rational and are allowed to have multiplicity $>1$ but there should be an $a>0$ such that the number of distinct roots of $f\_n$ is at least $a\deg f\_n$.)
| https://mathoverflow.net/users/2349 | Chebyshev-like polynomials with integral roots | Here's one thought. For each integer k, f\_n(k) satisfies a recurrence relation. If the roots of f\_n are all integers, then f\_n(k) and f\_n(k+1) have to be "in sync" in the sense that they never have opposite sign. This is a strong condition! For instance, suppose the sequences f\_n(k) and f\_n(k+1) each have unique largest eigenvalue: then these eigenvalues would have to have the same argument.
**Update:** Qiaochu's answer suggests that in fact working mod p would be just better than the "archimedean" picture sketched above, since it is really F\_q[t], not Z[t] or Q[t], that is analogous to the integers. Let F\_n(t) be the reduction of f\_n(t) to F\_p[t]. If f\_n(t) has all roots rational for every n, then the reduction of f\_n(t) mod p has the same property. But now we are saying something quite strong; that f\_n(t) lies in a finitely generated subgroup of F\_q(t)^\*! This is presumably ruled out by Mason's theorem (ABC for function fields.) Indeed, you could probably prove in this way that not only are the roots of f\_n(t) not rational, but f\_n(t) has irreducible factors of arbitrarily large degree.
I don't think this approach would touch a harder question along the same lines like [this one](https://mathoverflow.net/questions/5994/number-theoretic-spectral-properties-of-random-graphs/6053#6053).
| 3 | https://mathoverflow.net/users/431 | 8051 | 5,512 |
https://mathoverflow.net/questions/7921 | 27 | Smoothing theory fails for topological 4-manifolds, in that a smooth structure on a topological 4-manifold $M$ is not equivalent to a vector bundle structure on the tangent microbundle of $M$. Is there an explicit compact counterexample, i.e., are there two compact smooth 4-manifolds which are homeomorphic, have isomorphic tangent bundles, but are not diffeomorphic? (The uncountably many smooth structures on $\mathbb{R}^4$ should give a noncompact counterexample, since $Top(4)/O(4)$ does not have uncountably many components.)
Addendum to question, added 12/11/09:
I'm also interested in the other type of counterexample, of a nonsmoothable topological 4-manifold whose tangent microbundle does admit a vector bundle structure. Does someone know such an example? Tim Perutz's answer to my first question, below, says that homeomorphic smooth 4-manifolds have isomorphic tangent bundles. If it's not true that all topological 4-manifolds have vector bundle refinements of their tangent microbundle, what is the obstruction in the homotopy of $Top(4)/O(4)$?
| https://mathoverflow.net/users/2327 | Failure of smoothing theory for topological 4-manifolds | For a pair of smooth, simply connected, compact, oriented 4-manifolds $X$ and $Y$,
* Any isomorphism of the intersection lattices $H^2(X)\to H^2(Y)$ comes from an oriented homotopy equivalence $Y\to X$ (Milnor, 1958).
* Any oriented homotopy equivalence is a tangential homotopy equivalence (Milnor, Hirzebruch-Hopf 1958).
* Any oriented homotopy equivalence comes from an h-cobordism (Wall 1964).
* Any oriented homotopy equivalence comes from a homeomorphism (Freedman).
* It need not be the case that $X$ and $Y$ are diffeomorphic (Donaldson). Many examples are now known: e.g., Fintushel-Stern knot surgery on a K3 surface gives a family of exotic K3's parametrized by the Alexander polynomials of knots.
Here's a sketch of why homotopy equivalences preserve tangent bundles: $X$ and $Y$ have three characteristic classes: $w\_2$, $p\_1$ and $e$. However, $e[X]$ is the Euler characteristic, and $p\_1[X]$ three times the signature. By the Wu formula, $w\_2$ is the mod 2 reduction of the coset of $2H^2(X)$ in $H^2(X)$ given by the characteristic vectors, hence is determined by the lattice. In trying to construct an isomorphism of tangent bundles over a given homotopy equivalence, the obstructions one encounters are in $H^2(X;\pi\_1 SO(4))=H^2(X;Z/2)$ and in $H^4(X;\pi\_3 SO(4))=Z\oplus Z$, and these can be matched up with the three characteristic classes.
| 24 | https://mathoverflow.net/users/2356 | 8054 | 5,514 |
https://mathoverflow.net/questions/8032 | 9 | It seems a well-known fact that subvarieties of a variety of general type containing a general point are also of general type. This fact is an essential property used to prove some extension theorems of pluricanonical forms on algebraic varieties of general type, see for example the very nice survey on extension of pluricanonical forms
<http://www.iecl.univ-lorraine.fr/~Gianluca.Pacienza/notes-grenoble.pdf> ([Internet Archive](http://web.archive.org/web/20190320105238/http://www.iecl.univ-lorraine.fr/~Gianluca.Pacienza/notes-grenoble.pdf))
I want to know why this is true? Is there any thing more we can say about the relation between the canonical line bundle of a variety and that of subvarieties of codimension no less than 2.
| https://mathoverflow.net/users/2348 | Why a subvariety of a variety of general type is of general type | Let me reproduce the relevant bit from the [reference](http://www.iecl.univ-lorraine.fr/~Gianluca.Pacienza/notes-grenoble.pdf) above ([Internet Archive](http://web.archive.org/web/20190320105238/http://www.iecl.univ-lorraine.fr/~Gianluca.Pacienza/notes-grenoble.pdf)):
>
> **Exercise 3.1.** Let $X$ be a variety of general type. Prove that a
> subvariety of $X$ passing through a
> general point is of general type.
>
>
>
Bogomolov proved in his paper "Families of curves on a surface of general type" that surfaces of general type satisfying $c\_1^2 > c\_2$ contain at most a finite number of rational and elliptic curves. It is unknown if the same holds true for an arbitrary surface of general type. Thus in the exercise above general point does not mean outside a closed subvariety. Instead it means outside a countable union of closed subvarieties. Of course, the author is considering all his varieties defined over $\mathbb C$.
Standard arguments reduces the exercise to the following
>
> **Exercise.** Let $X$ be a variety of
> general type. If $Y \rightarrow T$ is a family
> of irreducible general type subvarieties of $X$ parametrized by
> a irreducible complex variety $T$ then the
> natural map $Y\to X$ is not dominant.
>
>
>
| 4 | https://mathoverflow.net/users/605 | 8055 | 5,515 |
https://mathoverflow.net/questions/8052 | 56 | I sort of understand the definition of a spectral sequence and am aware that it is an indispensable tool in modern algebraic geometry and topology. But why is this the case, and what can one do with it? In other words, if one were to try to do everything without spectral sequences and only using more elementary arguments, why would it make things more difficult?
| https://mathoverflow.net/users/344 | Why are spectral sequences so ubiquitous? | 1. Let's say you have a resolution $0\to A\to J^0\to J^1\to\dots$ (of a module, a sheaf, etc.) If $J^n$ are acyclic (meaning, have trivial higher cohomology, resp. derived functors $R^nF$), you can use this resolution to compute the cohomologies of $A$ (resp. $R^nF(A)$). If $J^n$ are not acyclic, you get a spectral sequence instead, and that's the best you can do.
2. Let us say you have two functors $F:\mathcal A\to\mathcal B$ and $G:\mathcal B\to \mathcal C$. Let us say you know the derived functors for $F$ and $G$ and would like to compute them for the composition $GF$. Answer: Grothendieck's spectral sequence.
1 and 2 account for the vast majority of applications of spectral sequences, and provide plenty of motivation -- I am sure you will agree.
The reason for the spectral sequences in both cases is the same. Intuitively, $A$ in case 1 (resp. $F(A)$ in case 2) is made of parts which are not themselves elementary. Instead, they are made (via an appropriate filtration) from some other elementary, "acyclic" objects.
So there is a 2-step process here. You can do the first step and the second step separately but they are not exactly independent of each other. Instead, they are entangled somehow. The spectral sequence gives you a way to deal with this situation.
| 44 | https://mathoverflow.net/users/1784 | 8057 | 5,516 |
https://mathoverflow.net/questions/8042 | 32 | The title pretty much says it all: suppose $R$ is a commutative integral domain such that every countably generated ideal is principal. Must $R$ be a principal ideal domain?
More generally: for which pairs of cardinals $\alpha < \beta$ is it the case that: for any commutative domain, if every ideal with a generating set of cardinality at most $\alpha$ is principal, then any ideal with a generating set of cardinality at most $\beta$ is principal?
Examples: Yes if $2 \leq \alpha < \beta < \aleph\_0$; no if $\beta = \aleph\_0$ and $\alpha < \beta$:
take any non-Noetherian Bezout domain (e.g. a non-discrete valuation domain).
My guess is that valuation domains in general might be useful to answer the question, although I promise I have not yet worked out an answer on my own.
| https://mathoverflow.net/users/1149 | Do there exist non-PIDs in which every countably generated ideal is principal? | No such ring exists.
Suppose otherwise. Let $I$ be a non-principal ideal, generated by a collection of elements $f\_\alpha$ indexed by the set of ordinals $\alpha<\gamma$ for some $\gamma$. Consider the set $S$ of ordinals $\beta$ with the property that the ideal generated by $f\_\alpha$ with $\alpha<\beta$ is not equal to the ideal generated by $f\_\alpha$ with $\alpha\leq \beta$.
$I$ is generated by the $f\_\beta$ with $\beta \in S$, so if $S$ is finite, then $I$ is finitely generated and thus is principal.
On the other hand, if $S$ is infinite, then take a countable subset $T=
\{\beta\_1<\beta\_2<\dots\}$ of $S$. If the ideal generated by the corresponding set of $f\_\beta$'s were principal, its generator would have to be in some $\langle f\_{\beta\_k}
\mid k\leq i \rangle$ for some $i$ (since any element of $\langle f\_{\beta}\mid \beta \in T\rangle$ is a finite combination of $f\_\beta$'s and therefore lies in some such ideal).
Now no $\beta\_j$ with $j>i$ could be in $T$.
\*\*
The same argument shows that all rings for which any countably generated ideal is finitely generated, have all their ideals finitely generated.
\*\*
Corrected thanks to David's questions.
| 38 | https://mathoverflow.net/users/468 | 8067 | 5,524 |
https://mathoverflow.net/questions/8088 | 0 | For P a partially ordered set, let S be a subset of P such that if:
a,c\in S and b\in P and a<=b<=c then b\in S
Is there a name for a subset with this property? The term "dense" subset is already taken and means something else.
| https://mathoverflow.net/users/2361 | name for "solid" subset of a partially ordered set? | A set with this property is called *convex*.
See e.g. Quasi-uniform spaces, Volume 77 of Lecture notes in pure and applied mathematics, Peter Fletcher, William F. Lindgren, Marcel Dekker, 1982, p.84.
| 11 | https://mathoverflow.net/users/440 | 8089 | 5,536 |
https://mathoverflow.net/questions/8095 | 2 | I was reading a paper, and it said that the following were equivalent using the Axiom of Choice, but I tried working it out, and I wasn't sure how: an algebra $A$ is primitive; $A$ has a proper left ideal $B$ such that $A = B +C$ for any non-trivial two-sided ideal $C$ of $A$. I've tried reasoning it out, and I'm not sure how - can anyone help? I know it should be really easy, but I seem to be missing a key step. If someone could give me a hint or two to work it out, that would be great.
| https://mathoverflow.net/users/2623 | Some equivalent statements about primitive algebras | Lam (A first course in noncommutative rings, 2ed) does it for (unital) rings $R$ in Lemma 11.28 (page 186):
>
> If such a $B$ exists, we may assume (after an application of Zorn's Lemma) that it is a maximal left ideal. The annihilator of the simple left $R$-module $R/B$ is an ideal in $B$, and so it must be zero. This shows that $R$ is left primitive. Conversely, if $R$ is left primitive, there exists a faithful simple left $R$-module, which we may take to be $R/B$ for some (maximal) left ideal $B \subsetneq R$. A nonzero ideal $C$ cannot lie in $B$ (for otherwise $C$ annihilates $R/B$) and so must be comaximal with $B$.
>
>
>
Here, the statement equivalent to the Axiom of Choice is [Zorn's Lemma](http://en.wikipedia.org/wiki/Zorn_lemma/); it says that:
>
> Every partially ordered set in which every chain (i.e. totally ordered subset) has an upper bound contains at least one maximal element.
>
>
>
In this proof we get to see one of its most common uses: it assures that any **unital** ring has a maximal ideal (see the Wikipedia page for more information).
| 3 | https://mathoverflow.net/users/1234 | 8098 | 5,541 |
https://mathoverflow.net/questions/8097 | 16 | Most of the number theory textbooks I've dealt with take a very classical approach to the subject. I'm looking for a textbook that's something like a first course in number theory for people who have a decent command of modern algebra (at the level of something like Lang's Algebra). Does such a book exist, and if it does, what is it called?
Edit: As I posted in a comment below:
In the introduction to Ireland and Rosen, they note something that was bugging me for a while, "Nevertheless it is remarkable how a modicum of group and ring theory introduces unexpected order into the subject."
This is precisely the perspective I was looking for, so if anyone passes by this topic looking for a book that approaches number theory in this way, I feel like this quote should point him (her?) in the right direction.
| https://mathoverflow.net/users/1353 | Number theory textbook with an algebraic perspective | There are probably many such books, for instance "[Fundamentals of Number Theory](http://store.doverpublications.com/0486689069.html)" by LeVeque, "[Elementary Number Theory](http://store.doverpublications.com/0486458075.html)" by Bolker and "[A Classical Introduction to Modern Number Theory](http://www.springer.com/math/numbers/book/978-0-387-97329-6)" by Ireland and Rosen.
| 10 | https://mathoverflow.net/users/532 | 8099 | 5,542 |
https://mathoverflow.net/questions/4790 | 15 | So, many of us know the answer to "what kind of structure on an algebra would make its category of representations braided monoidal": your algebra should be a quasi-triangular Hopf algebra (maybe if you're willing to weaken to quasi-Hopf algebra, maybe this is all of them?).
I'm interested in the categorified version of this statement; let's say I replace "algebra" with a monoidal category (I'm willing to put some kind of triangulated/dg/$A\_\infty$/stable infinity structure on it, if you like), and "category of representations" with "2-category of module categories" (again, it's fine if you want to put one the structures above on these). What sort of structure should I look for in the algebra which would make the 2-category braided monoidal?
Let me give a little more background: Rouquier and Khovanov-Lauda has defined a monoidal category which categories the quantized universal enveloping algebra $U\_q(g)$. It's an open question whether the category of module categories over this monoidal category is itself braided monoidal, and I'm not sure I even really know what structure I should be looking for on it.
However, one thing I know is that the monoidal structure (probably) does not come from just lifting the diagrams that define a Hopf algebra. In particular, I think I know how to take tensor products of irreducible representations, and the result I get is not the naive tensor products of the categories in any way that I understand it (it seems to really use the categorified $U\_q(g)$-action in the definition of the underlying category).
To give people a flavor of what's going on, if I take an irreducible $U\_q(g)$ representation, it has a canonical basis. It's tensor product also has a canonical basis, but it is *not* the tensor product of the canonical bases. So somehow the category of "$U\_q(g)$ representations with a canonical basis" is a monoidal category of it's own which doesn't match the usual monoidal structure on "vector spaces with basis." I want to figure out the right setting for categorifying this properly.
Also, I started an [n-lab page](http://ncatlab.org/nlab/show/braided+monoidal+2-category) on this question, though there's not much to look at there right now.
| https://mathoverflow.net/users/66 | What structure on a monoidal category would make its 2-category of module categories monoidal and braided? | Here is one set of data that will be sufficient. To get the monoidal structure you don't actually need a (monoidal) *functor* $C \to C \boxtimes C$. It is sufficient to have a bimodule category M from C to $C \boxtimes C$. You will also need a counit $C \to Vect$, and these will need to give C the structure of a (weak) comonoid in the 3-category of tensor categories, bimodule categories, intertwining functors, and natural transformations.
To compute what the induced tensor product does to two given module categories you will have to "compose" the naive tensor product with this bimodule category. This can be computed by an appropriate (homotopy) colimit of categories. It is basically a larger version of a coequalizer diagram. This is right at the category number where you will start to see interesting phenomena from the "homotopy" aspect of this colimit, which I think explains the funny behavior you're noticing with regards to bases.
Finally, you may get a braiding by having an appropriate isomorphism of bimodule categories,
$ M \circ \tau \Rightarrow M$
which satisfies the obvious braiding axioms. Here $\tau$ is the usual "flip" bimodule.
| 8 | https://mathoverflow.net/users/184 | 8103 | 5,546 |
https://mathoverflow.net/questions/8091 | 95 | You and I decide to play a game:
To start off with, I provide you with a frictionless, perfectly spherical sphere, along with a frictionless, unstretchable, infinitely thin magical rope. This rope has the magical property that if you ever touch its ends to each other, they will stick together and never come apart for all eternity. You only get one such rope, but you are allowed to specify its length.
Next, I close my eyes and plug my ears as you do *something* to the rope and the sphere. When you are done with whatever you have decided to do, you give me back the sphere and rope. Then I try my best to remove the rope from the sphere (i.e., make the smallest distance from a point on the rope to a point on the sphere at least 1 meter). Of course, since the rope is not stretchable, the total length of the rope cannot increase while I am trying to remove it from the sphere.
If I succeed in removing the rope from the sphere, I win. Otherwise, you win. Who has the winning strategy?
EDIT: To clarify, Zeb is looking for an answer with a finite length, piecewise smooth rope, and the sphere should be rigid.
| https://mathoverflow.net/users/2363 | Is it possible to capture a sphere in a knot? | Without loss of generality we can assume that rope is everywhere tangent to the sphere.
Then some infinitesimal Möbius tranform of its surface will shorten the wrapping length while preserving the crossing pattern (and therefore will loosen the rope without causing it to pass through itself).
Once it is proved, moving in this direction will eventually allow the sphere to escape.
**Proof.** let $u$ be conformal factor.
Since Möbius tranform preservs total area $\oint u^2=1$ .
Thus, $\oint u<1$.
It follows that for a sutable rotation of $S^2$, we get length decreasing family of Möbius tranforms.
**Comments**
* The same proof works for link made out of 3 circles.
* It is easy capture sphere in a link from 4 circles. (4 strings go around 4 faces of regular tetrahedron on $S^2$ and link at the vertexes as in the answer of Anton Geraschenko, [the first picture](https://i.stack.imgur.com/zbBabm.jpg))
* BTW, [Can one capture a convex body in a knot?](https://mathoverflow.net/questions/360066/which-convex-bodies-can-be-captured-in-a-knot)
| 63 | https://mathoverflow.net/users/440 | 8111 | 5,553 |
https://mathoverflow.net/questions/8113 | 7 | What are interesting, illustrative examples of Borel sets, situated in Borel hierarchy higher than $\Sigma^{0}\_{2}$ /$\Pi^{0}\_{2}$?
| https://mathoverflow.net/users/158 | Higher-rank Borel sets | The Borel hierarchy is, of course, strictly increasing at
every step, so there will be sets of every countable
ordinal rank. Furthermore, all of these sets are relatively
concrete, obtained as countable unions of sets having lower
rank. But you asked for natural examples, so let me give a
few.
1) The collection of true statements of number theory, in
the ring of integers Z. This is well known to have
complexity Sigma^0\_omega, which is a large step up from the
complexity you mentioned. This is, of course, a light-face
notion. The corresponding bold-face notion would be truth in arbitrary countable structures. That is, the
set of pairs (A,phi), where A is a first-order structure
and phi is true in A. This has complexity Sigma^0\_omega for
the same reason as (1). (I regard this as a highly natural example, but Kechris' remark was about natural examples specifically from topology and analysis.)
2) The set of countable graphs with finitely many connected
components. This has complexity Sigma^0\_3, since you must
say: there is a finite set of vertices, for all other
vertices, there is a path to one of them.
There are numerous other statements about finiteness that
also arise this way.
3) The set of (reals coding) finitely generated groups.
This is Sigma^0\_3, since you say: there is a finite set in
the group, such that for all other elements of the group,
there is a generating word.
4) finitely-generated other structures, etc. etc.
5) The set of pairs (A,p), such that A is a real and p is a
Turing machine program with oracle A that accepts all but
finitely many input strings. Sigma^0\_3, since "There is a
finite set of strings, for all other strings, there is a
computation showing them to be accepted."
| 5 | https://mathoverflow.net/users/1946 | 8121 | 5,560 |
https://mathoverflow.net/questions/8110 | 4 | I am trying to understand some construction done by Lusztig in his book on quantum groups. Given some Cartan datum, let $U=U\_q(\mathfrak{g})$ the standard quantized enveloping algebra of the Kac-Moody algebra $\mathfrak{g}$. Its negative part $U^{-}$ (the subalgebra of $U$ generated by the $f\_i$'s) has a canonical basis $B$. In his book [Introduction to Quantum Groups](http://rads.stackoverflow.com/amzn/click/0817647163) (don't be fooled by the title!), Lustzig constructs the non-unital extension $\dot{U}$ and proves that it has a canonical basis $\dot{B}$.
As a set, $\dot{B}$ is in bijection with $B\times X \times B$ (where $X$ is the lattice of weights, normally called $P$ in any other books on quantum groups), and its elements are described with the rather cryptic notation $b\diamond\_\zeta b''$. The definition of those elements is however very obscure and non-explicit. I would appreciate finding an easier, more explicit, description, like the one given in section 25.3 in Lustzig's book for $U\_q(\mathfrak{sl}\_2)$, also described in great detail by Lauda in the first part of his paper [A categorification of quantum sl(2)](http://arxiv.org/abs/0803.3652). It is not clear to me how this description can be extended to more complicated quantum groups.
Does anyone know any similar simple description of the canonical basis $\dot{B}$, even in some other particular cases? I am also very interested of knowing if is there any relation with crystals, akin to the equivalence between $B$ and Kashiwara's crystal basis $B(\infty)$.
| https://mathoverflow.net/users/914 | Canonical basis for the extended quantum enveloping algebras | The basis $\dot{B}$ is usually hard to compute, in the same way the Kazhdan-Lusztig basis for the Hecke algebra is. On the other hand, at least for type A, there is another way to get at this basis via what you might call "quantum Schur-Weyl duality": there's a geometric construction of finite dimensional quotients of the quantum group of type sl\_n due to Beilinson-Lusztig-MacPherson, each of which has a canonical basis. More recent work of Lusztig, Ginzburg and Schiffmann-Vasserot showed that these finite dimensional quotients could be put into a compatible family, from which you can recover the whole modified quantum group, and it's canonical basis. How much more explicit this realization is probably depends on what sort of question you're interested in I imagine.
With regard to the crystal structure, there's a paper of Kashiwara "Crystal bases of modified quantized enveloping algebras" in Duke which studies this. He shows that the crystal structure reflects the $B\times X\times B$ parametrization you mention, and investigates a crystal basis of a "dual algebra" (a kind of quantum coordinate algebra) for which the crystal structure is a kind of Peter-Weyl theorem (at least for the finite type case). Lusztig investigates something similar in his book.
| 6 | https://mathoverflow.net/users/1878 | 8122 | 5,561 |
https://mathoverflow.net/questions/8115 | 11 | Let $T\_1$ and $T\_2$ be two Grothendieck topologies on the same small category $C$, and let $T\_3 = T\_1 \cup T\_2$ (by which I mean the smallest Grothendieck topology on $C$ containing $T\_1$ and $T\_2$). If $X$ is a $T\_1$-sheaf, is its $T\_2$-sheafification still a $T\_1$-sheaf (and therefore a $T\_3$-sheaf)?
If the answer is "not always," then are there conditions one can impose on $T\_1$ and $T\_2$ to make it true? Does it matter if $X$ is already $T\_2$-separated?
(I'm really interested in the analogous question for stacks, but I'm guessing the answers will be pretty much the same.)
| https://mathoverflow.net/users/49 | Does sheafification preserve sheaves for a different topology? | I think my answer to [this question](https://mathoverflow.net/questions/6592/localizing-model-structures) provides a counterexample: Let C be the category a → b, and consider the topologies T1 generated by the single covering family {a → b} and T2 generated by declaring the empty family to be a covering of a. (Caution: In my other answer I used covariant functors for some reason, so I hope I didn't err in translating the example.)
Note that if I switch T1 and T2, though, the condition you ask for is satisfied, even though neither class of sheaves contains the other. So there may be some interesting conditions under which it is true.
| 10 | https://mathoverflow.net/users/126667 | 8123 | 5,562 |
https://mathoverflow.net/questions/8134 | 6 | It is a standard result of elementary homological algebra that to every R-module $A$ there exists a projective resolution. It is often said that the category of R-modules has "enough projectives." In which other categories is this also true? In particular is it true for abelian categories?
| https://mathoverflow.net/users/2376 | Existence of projective resolutions in abelian categories | Among the standard examples of abelian categories without enough projectives, there are
1. the categories of sheaves of abelian groups on a topological space (as VA said), or sheaves of modules over a ringed space, or quasi-coherent sheaves on a non-affine scheme;
2. the categories of comodules over a coalgebra or coring.
No abelian category where the functors of infinite product are not exact can have enough projectives. In Grothendieck categories (i.e. abelian categories with exact functors of small filtered colimits and a set of generators) there are always enough injectives, but may be not enough projectives.
Among the standard examples of abelian categories with enough projectives, there are
1. the category of functors from a small category to an abelian category with enough projectives (as VA said), or the category of additive functors from any small additive category to the category of abelian groups (this class of examples includes the categories of modules over any rings);
2. the category of pseudo-compact modules over a pseudo-compact ring (see Gabriel's dissertation);
3. the category of contramodules over a coalgebra or coring (see Eilenberg-Moore, "Foundations of relative homological algebra").
| 18 | https://mathoverflow.net/users/2106 | 8139 | 5,575 |
https://mathoverflow.net/questions/8137 | 4 | We know that every surface of genus ($g\geq 2$) admits a pair of pants decomposition. And there is the Fenchel Nielsen Coordinates on the Teichmuller space associated to such a decomposition where we have the length functions of the geodesics boundaries and the twisting parameters for gluing these boundaries.
My question is the following:
We choose again a pair of pants decomposition. Then there arises the trivalent graph representing this decomposition.
We could associate a fixed a pair of pants to each of vertices, say of boundary geodesic length (1,1,1). And then glue flat cylinders of both boundary length 1 and height $h$ to form a Riemann surface(of course we will also need the twisting parameter for the sewing, but it seems that we should only need the relative twisting parameter as cylinder itself has a rotation symmetry.).
The height of these cylinders and the relative twisting parameter together form 6g-6 parameters. Will they also be a parametrization of Teichmuller space?
| https://mathoverflow.net/users/2380 | Coordinates on Teichmuller space | You are describing the "grafting construction". I am not an expert: however if you google "grafting a Riemann surface" there are many references available.
If you take $h$ to be non-negative you cannot reach all of Teichmuller space in the way you describe. This is because the cuffs of your pair of pants decomposition will always have annular neighborhoods of definite modulus (ie of definite width). There are Riemann surfaces where the cuffs only have very thin annular neighborhoods.
| 4 | https://mathoverflow.net/users/1650 | 8140 | 5,576 |
https://mathoverflow.net/questions/7200 | 2 | What results are known about representations of reductive groups over finite rings in general? Here by finite rings I usually mean an algebra over $F\_q$, I guess.
I know Lusztig has a paper generalizing Deligne-Lusztig theory to finite commutative rings of the form $F\_q[x]/x^{\epsilon}$ for some integer $\epsilon > 0$ (so the quotient of that polynomial ring by that ideal), though even that theory (as best I understand) is not complete yet.
There would be no way of reducing representation theory of all commutative finite rings to the particular case of those rings (along with perhaps other "simple" rings, e.g. $\mathbb{Z} / p^{k} \mathbb{Z}$)?
Are any results known about representation theory of finite commutative rings that satisfy the Frobenius endomorphism condition, i.e. $t^q = t$ for all $ t \in R$? In that case, some aspects of Deligne-Lusztig theory can perhaps be salvaged.
What about non-commutative finite rings? In general, this seems to require something much more powerful than Deligne-Lusztig theory however, the Frobenius endomorphism breaks down.
| https://mathoverflow.net/users/2623 | Representations of reductive groups over finite rings | Lusztig's results mentioned above have been generalised to groups over arbitrary finite local rings, see *Unramified representations of reductive groups over finite rings*, Represent. Theory 13 (2009), 636-656.
There is a recent paper extending the above construction to certain ramified maximal tori, cf. *Extended Deligne-Lusztig varieties for general and special linear groups*, arXiv:0911.4593v1. It is shown here that the above "unramified" construction omits certain interesting representations. Some of these missing representations are connected with ramified maximal tori, and it is therefore desirable to have generalised Deligne-Lusztig varieties attached also to such tori.
In addition to the cohomological approaches to constructing representations, there are some partial purely algebraic constructions. There is a paper by G. Hill (*Regular Elements and Regular Characters of* $\text{GL}\_n(\mathcal{O})$, which gives a construction of many so-called *regular* representations. These include most of the interesting representations, such as the strongly cuspidal ones.
There is also an approach, due to U. Onn, which defines a new type of induction functor (called infinitesimal induction) which complements the classical parabolic induction. This construction considers general automorphism groups of finite modules over DVRs with finite residue field. So far, this approach has led to a classification of all the representations in the rank-2 case (which includes $\text{GL}\_2(\mathcal{O}/\mathfrak{p}^r)$).
Since any finite commutative ring is the direct product of finite local rings, it is enough to study representations of groups over the latter. This is however a very hard problem, and one cannot expect an explicit list of all the representations in general. As in the representation theory of most infinite groups, the most fruitful approach seems to be to define a nice category of representations which is both possible to control, and at the same time is rich enough to include most (or all) the representations one is interested in. A candidate for such a category for $\text{GL}\_n(\mathcal{O})$ is the one consisting of regular representations, mentioned above. One can ask interesting questions about the regular representations, such as which of them are given by the cohomological constructions, or which of them are types for supercuspidal representations of $\text{GL}\_n(F)$, where $F$ is a local field with ring of integers $\mathcal{O}$.
| 2 | https://mathoverflow.net/users/2381 | 8141 | 5,577 |
https://mathoverflow.net/questions/8145 | 16 | A stick knot is a just a piecewise linear knot. We could define "stick isotopy" as isotopy that preserves the length of each linear piece.
Are there stick knots which are topologically trival, but not trivial via a stick isotopy?
| https://mathoverflow.net/users/3 | Are there piecewise-linear unknots that are not metrically unknottable? | Yes, there are. See "Locked and Unlocked Polygonal Chains in 3D", T. Biedl, E. Demaine, M. Demaine, S. Lazard, A. Lubiw, J. O'Rourke, M. Overmars, S. Robbins, I. Streinu, G. Toussaint, S. Whitesides, [arXiv:cs.CG/9910009](http://arxiv.org/abs/cs.CG/9910009), figure 6.
| 22 | https://mathoverflow.net/users/440 | 8146 | 5,581 |
https://mathoverflow.net/questions/8147 | 1 | I´m looking for homomorphisms between exterior powers of a **free** module M of rank m
ΛmR M → Λm-1R M
Exactly, I´m looking for an **explicit** isomorphism
M → Hom R (ΛmR M , Λm-1R M)
I compare the ranks and the things go, but I can not imagine a concrete expression.
Suggestions are welcome
| https://mathoverflow.net/users/2040 | Homomorphism between exterior powers of a free module of finite rank | If $R$ is commutative, let $e\_1, \ldots, e\_m$ be an $R$-basis of $M$. Then $\Lambda^m M$ is a free $R$-module of rank one generated by the vector $e\_1 \wedge e\_2 \wedge \ldots \wedge e\_m$ and $\Lambda^{m-1}M$ is free of rank $m$ with basis $$\lbrace e\_1 \wedge \ldots \wedge \hat{e\_i} \wedge \ldots \wedge e\_m \rbrace\_{i=1}^{m}$$ where the hat denotes an omitted basis vector. Here is an explicit isomorphism: $$e\_i \in M\mapsto \Big(e\_1 \wedge e\_2 \wedge \ldots \wedge e\_m \mapsto e\_1 \wedge \ldots \wedge \hat{e\_i} \wedge \ldots \wedge e\_m \Big) \in \mathrm{Hom}\_R(\Lambda^m M, \Lambda^{m-1} M)$$
| 6 | https://mathoverflow.net/users/1797 | 8149 | 5,583 |
https://mathoverflow.net/questions/7316 | 4 | I am now learning induction problems in representation theory. I know David Vogan's book cohomological induction and unitary representation theory might be good references,but it is too thick.
I wonder whether there is some good and detailed notes on using derived functor to construct irreducible representations. It seems that Zuckermann ever gave a series of talks at Yale to introduce this method, but unfortunately, it was not published.
It seems that this method was motivated by Schmid's phd thesis. But I did not have schmid's phd thesis either.
All the related comments are welcomed
| https://mathoverflow.net/users/1851 | On the materials about cohomological induction | You might look at Vogan's orange book "Unitary Representations of Reductive Lie Groups," which is a much thinner book, or (thinner yet) chapters 5-6 of "Dirac Operators in Representation Theory," by Jing-Song Huang and Pavle Pandzic. These both have descriptions of cohomological induction, though off the top of my head I forget what information they include specifically about irreducible representations.
| 4 | https://mathoverflow.net/users/1322 | 8154 | 5,588 |
https://mathoverflow.net/questions/8075 | 6 | I've been reading about the Langlands program (the paper by Torsten Wedhorn "Local langlands correspondence for GL(n) over p-adic fields, to be precise), and I want to get my hands dirty with examples.
What are some interesting (yet not too nasty) examples of admissible representations of $GL\_{n}(K)$, where $K$ is a $p$-adic field?
Taking Schur functors of an admissible representation, should still generally give you something admissible, right?
| https://mathoverflow.net/users/2623 | examples of admissible representations of $GL_{n}(K)$ over p-adic field | I second L Spice's recommendation of the book by Bushnell and Henniart, called "The local Langlands conjectures for GL(2)."
After you master the principal series representations, it's not too hard to tinker with some supercuspidals. Easiest among these are the tamely ramified supercuspidals. To construct these, let's start with the unramified quadratic extension $L/K$, with corresponding residue fields $\ell/k$. Choose a character $\theta$ of $L^\times$ which has these properties:
(a) The character $\theta$ is trivial on $1+\mathfrak{p}\_L$, so that $\theta\vert\_{\mathcal{O}\_L^\times}$ factors through a character $\chi$ of $\ell^\times$.
(b) $\chi$ is distinct from its $k$-conjugate. (In other words, $\chi$ does not factor through the norm map to $k^\times$.)
It's a standard fact that there's a corresponding representation $\tau\_\chi$ of $\text{GL}\_2(k)$, characterized by the identity $\text{tr}\tau\_\chi(g)=-(\chi(\alpha)+\chi(\beta))$ whenever $g\in\text{GL}\_2(k)$ has eigenvalues $\alpha,\beta\in\ell\backslash k$. (This is somewhere in Fulton and Harris, for instance.)
Inflate $\tau\_\chi$ to a representation of $\text{GL}\_2(\mathcal{O}\_K)$, and extend this to a representation $\tau\_\theta$ of $K^\times\text{GL}\_2(\mathcal{O}\_K)$ which agrees with $\theta$ on the center. Finally, let $\pi\_\theta$ be the induced representation of $\tau\_\theta$ up to $\text{GL}\_2(K)$; then $\pi\_\theta$ is an irreducible supercuspidal representation.
By local class field theory, our original character $\theta$ can be viewed as a character of the Weil group of $L$. In the local Langlands correspondence, $\pi\_\theta$ lines up with the representation of the Weil group of $K$ induced from $\theta$. All the supercuspidals of $\text{GL}\_2(K)$ arise by induction from an open compact-mod-center subgroup, but the precise construction of these is a little more subtle than the above example.
| 7 | https://mathoverflow.net/users/271 | 8163 | 5,594 |
https://mathoverflow.net/questions/8167 | 5 | If x is an element of a field K and n is a positive integer, we have both a symbol and a name for a root of the polynomial t^n - x = 0: we denote it by x^{1/n} and call it an nth root of x.
Of course nth roots play a vital role in field theory, e.g. in the characterization of solvable extensions in characteristic 0. However, in characteristic p > 0, the extraction of a p-power root is a much different business: it gives rise to purely inseparable extensions, not composition factors of solvable Galois extensions.
To repair the characterization of solvable extensions in characteristic p as those being attainable as a tower of "radical" extensions, one needs to include the operation of taking roots of Artin-Schreier polynomials: t^q - t - x = 0, for q = p^a a power of the characteristic.
Finally my question: do we have a name for an element t solving the equation t^q - t = x and/or a special notation for it? I do not know one. Similarly, whereas classically we often speak of x as being "an nth power", in this case I find myself writing "x is in the image of the Artin-Schreier isogeny \rho". Is there something better than this?
| https://mathoverflow.net/users/1149 | Notation/name for "Artin-Schreier roots"? | Google Scholar finds three papers with the phrase "Artin-Schreier root" or "Artin-Schreier roots" (with quotes). The papers are by Jing Yu, Thomas Scanlon, and Spencer Bloch + Helene Esnault. This is not all that many, but maybe enough for some kind of standard. Various people, probably enough to call it standard, also use the notation $\wp(x)$ (Weirstrass p, or \wp in amslatex) for the Artin-Schreier polynomial. So you could use the notation $\wp^{-1}(x)$ for an Artin-Schreier root of $x$. ~~(The notation is unambiguous, because which polynomial you take is determined by the characteristic of the field containing $x$.)~~ The notation is often used to mean the first Artin-Schreier polynomial with $q=p$, which is unambiguous but only in a lame sense. (As Pete was too polite to emphasize in his comment, I should have read the question more carefully.)
---
If you're bothered by the length of the phrase "Artin-Schreier root" and you want to be a real joker with terminology, you could call it an "asroot". (The current use of this word is a Unix command which means to execute something as root user. You could pronounce it the same except with the emphasis on the first syllable.) Why not? Gromov got away with the barbarous term "a-T-menable". And who came up with "rng" and "rig"?
| 3 | https://mathoverflow.net/users/1450 | 8169 | 5,598 |
https://mathoverflow.net/questions/8160 | 3 | Let $A$ be the set of all quadruples $(a\_0,a\_1,a\_2,a\_3) \in {\mathbb Q}^4$ such that
the polynomial $P=X^4+a\_3X^3+a\_2X^2+a\_1X+a\_0$ is irreducible and if $z$ is any root
of $P$, then ${\mathbb Q}(z)$ contains $\sqrt{2}$. Is there a nontrivial polynomial relation
$R(a\_0,a\_1,a\_2,a\_3)=0$ satisified by all $(a\_0,a\_1,a\_2,a\_3) \in A$ ?
| https://mathoverflow.net/users/2389 | Expressing field inclusions by polynomial equalities on coefficients | If there was a nontrivial polynomial relation between the coefficients, it would be true for a dense subset (~~reducibility is a nowhere dense condition~~ see comment below) of all polynomials of the form $(x^2+(\alpha +\beta\sqrt{2})x+\gamma+\delta\sqrt{2})(x^2+(\alpha -\beta\sqrt{2})x+\gamma-\delta\sqrt{2})$ with rational $\alpha,\beta,\gamma,\delta$, which would mean the same relation would be true for all real $\alpha,\beta,\gamma,\delta$ as well. But all quartic polynomials are of the form above with real $\alpha,\beta,\gamma,\delta$, so there are no nontrivial relations.
| 1 | https://mathoverflow.net/users/2368 | 8171 | 5,600 |
https://mathoverflow.net/questions/8189 | 5 | What's the cardinality of a single equivalence class of Cauchy sequences in ℚ?
To clarify, I'm not asking for the cardinality of the real numbers, but for the cardinality of the set of Cauchy Sequences that are equivalent to any single real number.
| https://mathoverflow.net/users/2394 | Cardinality of Equivalence Classes of Cauchy Sequences | It's the same as the size of the real numbers. Here's a rough sketch of the proof.
For each element of (0,1) (which has the same cardinality as the reals), I'm going to construct a distinct sequence of rationals that converges to 0.
Think of an element of (0,1) in binary, so as an infinite sequence of 0s and 1s. For definiteness, we assume the sequence is never eventually constant 1 (this just gives a well defined bijection between (0,1) and the sequences, since, e.g., .1000000... = .0111111... ). Given such a sequence of 0s 1s $a\_1, a\_2, ...$, create the rational sequence $(-1^{a\_1}(1), -1^{a\_2} (1/2), -1^{a\_3} (1/3), ..., -1^{a\_n} (1/n),...)$.
Convince yourself all these sequences of rational numbers are distinct.
Thus, the size of the reals is less than or equal to the number of rational sequences converging to 0. For a bound in the other direction, note that the collection of ALL sequences of rationals = $\prod\_{\mathbb{N}} \mathbb{Q}$ and $|\prod\_{\mathbb{N}} \mathbb{Q}| = |\mathbb{Q}|^{|\mathbb{N}|} = |\mathbb{N}|^{|\mathbb{N}|} \leq |2^{\mathbb{N}}|^{|\mathbb{N}|} = |2^{\mathbb{N}}| =$ the size of the reals.
By the Cantor-Schroeder-Bernstein theorem, the set off all rational sequences converging to 0 has the same cardinality as the reals.
| 9 | https://mathoverflow.net/users/1708 | 8200 | 5,618 |
https://mathoverflow.net/questions/8190 | 16 | Is there an notion of elliptic curve over the field with one element? As I learned from a [previous question](https://mathoverflow.net/questions/430/homological-algebra-for-commutative-monoids), there are several different versions of what the field with one element and what schemes over it should be (see for example this [article](http://arxiv.org/abs/0909.0069) by Javier López Peña and Oliver Lorscheid). What I want to know is whether there a good notion of elliptic curve over $F\_{un}$? What about modular forms?
| https://mathoverflow.net/users/184 | Elliptic Curves over F_1? | In a strict sense, elliptic curves over the rationals (say) are not defined over $F\_1$ since their reduction modulo $p$ varies with $p$, e.g. they have places of bad reduction. However, CM elliptic curves have some of the properties that one would associate with objects defined over $F\_1$. For example, their L-function looks a bit like the twist of a constant elliptic curve over a function field, except that the role played by the character is actually played by a Hecke character. So, in a sense, a CM elliptic curve is the twist of a curve over $F\_1$ by a Hecke character, although I have never tried to fully formalize this notion. Another way to view this is perhaps through J. Borger's viewpoint in which a variety over $F\_1$ is a variety in which all Frobeniuses lift. That sort of happens for CM elliptic curves.
| 17 | https://mathoverflow.net/users/2290 | 8206 | 5,621 |
https://mathoverflow.net/questions/8202 | 1 | It seems well-known that the system of conics given by $\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1$ for $c>0$ fixed and $a \in (0,c)\cup(c,\infty)$ varying is orthogonal: whenever two of these curves intersect, they do so at a right angle. Does anyone know a good *elementary* proof of this? I.e. no complex analysis, no physics... something the ancients would've appreciated. I am looking to explain this to a sharp 13-year-old...
| https://mathoverflow.net/users/1464 | Systems of conics | As long as you can get an elementary proof of the fact that one of the families consists of ellipses with foci $A=(-c,0)$ and $B=(c,0)$ and the other consists of hyperbolas with the same foci, you can say that for any intersection point $P$ the angle between the lines $PA$ and $PB$ is dissected by the tangent to either curve - otherwise moving on the tangent would cause a first-order error in the sum $|PA|+|PB|$ or the difference $|PA|-|PB|$, respectively. Hence the two tangents are just the two angle bisectors of a pair of lines, and are thus orthogonal.
| 4 | https://mathoverflow.net/users/2368 | 8209 | 5,624 |
https://mathoverflow.net/questions/6640 | 6 | What should be $\text{Spec } \mathbb{Z}[\sqrt{D}] \times\_{\mathbb{F}\_1} \text{Spec } \overline{\mathbb{F}}\_{1}$?
Sure, there's more than one definition.
I'm looking for any answer that uses at least one definition of scheme over $\mathbb{F}\_1$.
This really is more a question of opinion.
What do you think this should be?
Some monoid that has something to do with $\text{Spec }\mathbb{Z}[\sqrt{D}][\mathbb{Q}/\mathbb{Z}]$ would be my guess (where the second brackets mean group ring).
This interests me from the point of view that, say, hyperelliptic curves over a finite field come (geometrically) from the group scheme of a quadratic extension of $\overline{\mathbb{F}}\_p [t]$. In this case the frobenius acts on ideal classes, and satisfies a quadratic equation.
But, from what I understand, the natural analogue of frobenius in the arithmetic case, is like taking any positive power, and taking limits to 0 (or something of the sort).
Would this satisfy some kind of equation on, say, $\text{Pic(Spec } \mathbb{Z}[\sqrt{D}] \times\_{\mathbb{F}\_1} \text{Spec } \overline{\mathbb{F}}\_{1}\text{)}$?
(for whatever definition of Pic that should be natural here)
I've searched for information on $\mathbb{F}\_1$, but most just talk about making $\text{Spec }\mathbb{Z}$ into a curve, getting zeta functions to be Riemann's, etc.
Instead, I want to ask questions that are not just about proving the Riemann hypothesis, like the one above.
| https://mathoverflow.net/users/2024 | What should Spec Z[\sqrt{D}] x_{F_1} Spec \bar{F_1} be? | Sorry I didn't reply before, I somehow didn't read the question till now.
I think your question is a bit misguided. The main problem I see with it is: what is $\text{Spec} \mathbb{Z}[\sqrt{D}]$ over F1? If you think of it as the M\_0 scheme given by $\mathbb{Z}[\sqrt{D}]$ **as a multiplicative monoid**, then it is something huge (since that monoid is not even finitely generated, for starters), so none of the current notions can effectively deal with it (so far one can mostly only control schemes of finite type). If you want to make better sense of the question, one should ask: is it possible to find an algebra $A$ over F1 such that its base extension to $\mathbb{Z}$ gives $A\otimes\_{\mathbb{F}\_1} \mathbb{Z} = \mathbb{Z}[\sqrt{D}]$?
If you find an answer to this question, say in CC setting where it boils down to finding a monoid (with zero) such that the reduced semigroup ring gives you back your original ring,
then you might consider an approximation to your problem (the tensor with the abelian closure of F1) by taking some kind of multivariable-cyclotomic completion of your monoid, as in the papers by [Manin](http://arxiv.org/abs/0809.1564) and [Marcolli](http://arxiv.org/abs/0901.3167).
| 3 | https://mathoverflow.net/users/914 | 8215 | 5,629 |
https://mathoverflow.net/questions/8178 | 3 | I am trying to understand divisors reading through Griffith and Harris but it is difficult to come up with any particular interesting example. I have browsed through Hartshone's book but everything is expressed in terms of schemes, and I believe it is still possible to find some toy example to carry with me without having to learn what a scheme is first. Does anyone know any reference or book which has some exercise or example on this? In particular I would like to see examples of linear systems of divisors and how given a linear system of dimension $n$ I can choose a pencil inside it.
Apologies if this is not the place to write this, it is my first post. I have searched through the questions and have not find anything similar.
| https://mathoverflow.net/users/1887 | Examples of divisors on an analytical manifold | Hartshorne is the reference where you can find the following example which might be useful.
I what follow everything is with **multiplicity**. Now Alberto pointed out above the case of the divisor over $\mathbb{P}^1$ associated to its "tangent bundle": Two points over the sphere counted with multiplicity (from here though, it is not hard to believe that the Chern class of such a bundle is going to be 2). Notice that these two points are given by zeros of polynomials of degree two defined over the sphere. I think nothing stops you taking now polynomial of degree 3, 4 and so on. Then what we get are nothing but 3, 4 points over the sphere: **Divisors** of degree 3, 4 and so on. We can do something similar over all the curves (Riemann Surfaces) and what we get are divisors: **points with labels**. Such labels are the multiplicities. Chapter IV Hartshorne. or Klaus-Hulek: Elementary Algebraic Geometry.
Now, let's take a look at **divisors over the surface $\mathbb{P}^2$: they are algebraic curves** (Riemann Surfaces). Do not get confused please by the name Surface here. Applying the same argument as before, a divisor of degree two is going to be the zero locus of polynomials of degree 2: conics. Same for degree three (cubics), four (quartics), and so on and so forth. For instance, in degree two we might have the divisor $C=([x:y:z]\in \mathbb{P}^2|\ \ x^2+y^2=z^2)$. Deshomogenizing with $H=[z=1]$ you get a perfect polynomial $x^2+y^2=1$ which defines the intersection $H\cap C$. This is how your global divisor $C$ looks like locally.
Now taking a family of divisors of degree two, the conics, it is well known that the space of embeddings of conics in $\mathbb{P}^2$ is (the linear system) $\mathbb{P}^5$. We get this by considering the coefficients in the equation $ax^2+by^2+cz^2+dxy+exz+fyz=0$ as coordinates in $\mathbb{P}^5$. Notice that we get the following map out of the previous considerations, $$\phi:\mathbb{P}^2\rightarrow \mathbb{P}^5$$ given by $[x:y:z]\mapsto [x^2:y^2:z^2:xy:xz:yz]$. Here pencils are a subfamily of conics in the complete linear system given above with a certain property (find out which one). However, we can consider the following subfamily of conics: all those conics passing through a fixed point in $\mathbb{P}^2$. This is nothing but a hyperplane $H$ in $\mathbb{P}^5$. We can even consider $\phi(\mathbb{P}^2)\cap H$. This is going to be a divisor on $\mathbb{P}^2\cong \phi(\mathbb{P}^2)$. Guess which one?. Hartshorne II section 7.
One can apply the the ideas with zero locus of polynomials of degree three: **Divisors** of degree 3 in $\mathbb{P}^2$. These were given the name of **elliptic curves**. (did someone say that in considering such curves, we find the divisor associated to the canonical bundle of $\mathbb{P}^2$?). We can go on with the degree and getting divisors on the projective plane of higher degree. **These were only examples of divisors on $\mathbb{P}^2$**. Notice that all of them have a nontrivial topology and geometry. This fact is not a coincidence and the book of HG argues in this direction in Chapter zero.
| 3 | https://mathoverflow.net/users/1547 | 8222 | 5,634 |
https://mathoverflow.net/questions/8079 | 3 | I have a tangled web of ideas about natural transformations, vector spaces, equivalence classes, local coordinates, etc. in my head that I'm trying to unravel. So here are some of the questions I thought of:
* Vector spaces: Why is it that in linear algebra we always calculate with basis but when we think of a definition we always try to make sure it is basis independent? What kinds of symmetries are we trying to preserve by this?
* Categories: The obvious condition for natural transformations isn't so obvious to me and again I think the definition is the way it is because there is again some kind of symmetry that a natural transformation is preserving. What are the symmetries that a natural transformation is preserving?
* Equivalence classes: Almost the same thing as in the vector space scenario. We do things with representatives but we make sure our definitions are true regardless of this choice. Ok, this one is kinda silly because we are trying to preserve $\cong$ so there isn't anything too complicated happening here.
* Differential geometry: We again calculate with local coordinates but our definitions shouldn't depend on them. What symmetries are we trying to preserve in differential geometry?
So it seems to me we always break symmetry, whatever that means, when we want to work with something concrete but we try to make sure our calculations are preserved by these symmetries. Feel free to correct me and add your answers to some of my questions.
**Edit**: Dmitri gives a good answer for the definition of natural transformation in terms of the exponential and its adjoint.
| https://mathoverflow.net/users/nan | Broken Symmetry | Your examples about vector spaces and differential geometry do not make any sense to me.
One does not need coordinates or bases to prove statements in linear algebra and differential geometry.
Personally, I always use coordinate- and basis-free proofs.
For me the reason to avoid coordinates and bases is that we lose geometric intuition whenever we use them.
See my manifesto on this matter here: [When to pick a basis?](https://mathoverflow.net/questions/4648/when-to-pick-a-basis/4900#4900)
One way to make the definition of natural transformation more natural is to consider the category A
with exactly two objects and one non-trivial morphism between them.
Then the set of morphisms of an arbitrary category C is the set of functors Fun(A, C).
If we want the set of functors Fun(C, D) between two categories C and D to be a category,
then the set of morphisms of this category is Fun(A, Fun(C, D)).
But it is natural to assume that we have the standard adjunction Fun(A, Fun(C, D)) = Fun(A × C, D).
Unraveling the definition of functor from A × C to D yields precisely the usual axioms for natural transformations.
| 3 | https://mathoverflow.net/users/402 | 8229 | 5,637 |
https://mathoverflow.net/questions/8203 | 3 | Let S be a bounded semilattice without maximal elements. Can we always construct an atomless boolean algebra B, containing S as a subsemilattice, such that S is cofinal in B-{1}? That is, for every x≠1 from B there is y∈S such that x≤y.
More detailed explanation since apparently my terminology was rather ambiguous:
S is a partially ordered set with the least element 0, and every pair of elements from S has an infimum (greatest lower bound). S has no maximal elements, that is for every x∈S there is y∈S greater than x.
If B is an atomless boolean algebra, B-{1} would definitely satisfy the above conditions.
The question is, if it is possible for any semilattice S satisfying the above conditions to cofinally embed it into an atomless boolean algebra B. That is, for any x∈B there is y∈S such that y is greater than x in B.
| https://mathoverflow.net/users/200 | Semilattices in atomless boolean algebras | The answer to the original question is that no, in fact we can never do this.
**Theorem.** No nontrivial Boolean algebra has a cofinal
subset of B-{1} that is a join-semilattice. Indeed, B cannot have a cofinal subset of B-{1} that is even upward directed.
Proof. Suppose that B is a nontrivial Boolean algebra and S is cofinal subset of B-{1}. Let b be any element of B that is neither 0 nor 1. Thus, -b is also not 1, so both b and -b must have upper bounds in S. If S is a join-semilattice, or even only upward directed, than there will be a single element of S above both b and -b. But the only such upper bound in B is 1, which is not in S. QED
This is true whether or not B is atomic. In the atomic case, it is particularly easy to see, since the co-atoms have no upper bound in B-{1}.
There is no need in the question to insist that the embedding of S into B is a semilattice embedding, since even just an order-preserving map would give an upward directed image, which the theorem rules out.
The dual version of the question, turning the lattice upside down, is the corresponding fact that no nontrivial Boolean algebra has a filter that is also dense.
(Note: I didn't know what sense of "bounded" you meant in
the first part of the question, so I ignored that...Does this matter?)
---
The situation with the revised version of the question is somewhat better, since as the questioner points out, there are now some examples of the phenomenon. But nevertheless, there are also counterexamples, as I explain, so the answer is still no.
The reason there are counterexamples is that the new conditions on S do not rule out the possibility that S is upward directed. Thus, the same obstacle from above still occurs. For example, any linearly ordered S is a meet-semilattice and a join-semilattice, so it can never be cofinal in a Boolean algebra. More generally, if you take any S you have in mind, and put a copy of the natural numbers on top of it (above all elements), then this will still satisfy your conditions, but since it is also upward directed, it cannot embed cofinally in a Boolean algebra.
| 5 | https://mathoverflow.net/users/1946 | 8231 | 5,639 |
https://mathoverflow.net/questions/8232 | 17 | The Koebe–Andreev–Thurston theorem gives a characterization of planar graphs in terms of disjoint circles being tangent. For every planar graph $G$ there is a disk packing whose graph is $G$. What happens when disks are replaced by closed balls? By closed balls of higher dimension? I have already asked one question about this here:
[Graphs of Tangent Spheres](https://mathoverflow.net/questions/8031/graphs-of-tangent-spheres)
The question I want to ask here is what is known about the chromatic numbers of these graphs? I have updated the numbers and changed the arguments in the following based on some of the answers.
Assume the chromatic number is 14 or more and we have the smallest such graph that is colorable with 14 or more colors. Take one of the smallest closed balls then since the kissing number for three dimensions is 12 there are at most 12 closed balls tangent to this closed ball. Remove this closed ball then the remaining graph can be colored in 13 or less colors. Color it with 13 colors. Then add the closed ball back in since it is tangent to only 12 closed balls it can be given one of the thirteen colors so we have the entire graph can be colored with thirteen colors which gives a contradiction so the chromatic number must be 13 or less. We have an lower bound of 6 from a spindle constructed according to David Eppstein's answer. Can we improve on the 6 to 13 range?
We have the lower bound is a quadratic function and we have an upper bound that is exponential. Which of these two is right?
Is there a case where closed balls of different sizes raise the chromatic number from closed balls the same size?
Finally based on the existing chromatic numbers I am wondering if it is possible to answer this question. Is there a dimension where the chromatic number of the unit distance graph is different from the chromatic number of the graphs in that dimension of tangent closed balls. The unit distance graph is the set of all points in the $n$-dimensional space with two points connected if their distance is one. For dimension two the chromatic number is known to be in the range from 4 to 7. For dimension three the range is 6 to 15. For the graphs of tangent disks we have a chromatic number of 4 and for closed balls a range from 6 to 13. So the possibility that the chromatic numbers of the two types of graphs are the same has not yet been eliminated. So the specific question is what is known and what can be proved about the chromatic number of the graphs of tangent closed balls?
| https://mathoverflow.net/users/1098 | Chromatic number of graphs of tangent closed balls | It's easy to form sets of five mutually-tangent spheres (say, three equal spheres with centers on an equilateral triangle, and two more spheres with their centers on the line perpendicular to the triangle through its centroid). Based on this, I think it should be possible to construct a set of spheres analogous to the Moser spindle [http://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson\_problem] that requires six colors: spheres a, b, and c, where a and b have four mutual neighbors that are all adjacent to each other, a and c have another four mutual neighbors that are all adjacent to each other, neither a and b nor a and c are adjacent, but b and c are adjacent.
I have no idea how tight this lower bound might be, but it's at least better than four.
| 9 | https://mathoverflow.net/users/440 | 8235 | 5,642 |
https://mathoverflow.net/questions/8228 | 3 | Hi,
I apologize for the basic questions. I am looking for good references on the following problems:
1) What is known about the Dehn function of $SL\_n(\mathbb{Z})$?
2) What is known about the Dehn function of mapping class groups?
Thanks!!
| https://mathoverflow.net/users/2409 | How do Dehn functions of special linear and mapping class groups behave? | 1) For $SL\_n(\mathbb{Z})$, it depends on the dimension.
a. For $n=2$, the group is virtually free, so the Dehn function is linear.
b. For $n=3$, a theorem of Thurston-Epstein says that it is exponential (this can be found in the book "Word Processing in Groups").
c. For $n>3$, Thurston conjectured that it should be quadratic. This was recently proven for $n>4$ by [Robert Young](http://www.ihes.fr/~rjyoung/). He hasn't yet written up the proof, but he has a preprint [here](http://arxiv.org/abs/0903.2495) proving that it is at most quartic.
2) For the mapping class group, Lee Mosher proved that it is automatic (see his paper "Mapping class groups are automatic" in the Annals in 1995; a survey of the proof can be found in his beautiful paper "A user's guide to the mapping class group: once-punctured surfaces"). As shown in "Word processing in groups", this implies that its Dehn function is quadratic.
| 8 | https://mathoverflow.net/users/317 | 8237 | 5,643 |
https://mathoverflow.net/questions/8223 | 3 | Here reducible means that the mapping class for the fiber is a reducible auto-homeomorph in the sense of Nielsen-Thruston. So,
could anyone give me a hint to classify them?
In contrast, do you agree that -in the sense of connected sum- all these bundles are irreducible?
| https://mathoverflow.net/users/2196 | Reducible 3d torus bundles | Yes - surface bundles over the circle are irreducible (\*) as long as the fiber is not a two-sphere. This follows from the fact that the universal cover of such a surface bundle is homeomorphic to $\mathbb{R}^3$ and Proposition 1.6 of Hatcher's three-manifold [notes](http://www.math.cornell.edu/~hatcher/3M/3Mdownloads.html).
(\*) in the sense of connect sum.
EDIT: To answer the question posed by Juan in the comment. A orientation preserving homeomorphism $h$ of $T^2$ is reducible (that is, preserves the isotopy class of some essential multicurve) if and only if $h$ is a power of a Dehn twist or is the power of a Dehn twist followed by the hyperelliptic element. Here is a "cut-and-paste" proof: if $h$ preserves a multicurve then it preserves a curve, say $\alpha$. Now, $h$ either preserves or reverses the orientation of $\alpha$. If the latter case replace $h$ by $h$ followed by the hyperelliptic, to reduce to the former case. Isotope $h$ so that $\alpha$ is fixed pointwise. Note that, as $h$ preserves orientation of $T^2$, the sides of $\alpha$ are preserved as well. Thus $h$ restricts to a homeomorphism of the annulus, fixing the boundary pointwise. By the classification of mapping classes in the annulus, $h$ is a Dehn twist.
As a bit of an advertisement: Farb and Margalit have written a [primer](http://www.math.utah.edu/~margalit/primer/) on the maping class group. You can find a discussion of the mapping class groups of the disk, annulus, and torus in Section 2.4, on the "Alexander Method". (In particular they give the usual proof that the group of orientation oreserving classes on $T^2$ is $SL(2,Z)$. They also give as an exercise the characterization of Dehn twists.)
I'll end by pointing out that there is not a contradiction between my answer and Hatcher's. If the monodromy is reducible then the resulting torus bundle is Seifert fibered and in fact has Nil geometry.
| 5 | https://mathoverflow.net/users/1650 | 8241 | 5,647 |
https://mathoverflow.net/questions/8225 | 7 | One of the canonical examples used by Barr & Wells in order to motivate the use of topoi is that we can construct a theory for fuzzy logic and fuzzy set theory as set-valued sheaves on a poset (Heyting algebra) of confidence values for the fuzziness. Doing this constructs a fuzzy theory where both the *membership* and *equality* relations have more truth values than *just* true and false.
How would one construct a ternary approach using this mindset? In other words, is there an easy way to see a sheaf on a poset with three values as a fuzzy set theory for the truth values (true, maybe, false)? Or is this formulation even the wrong approach? Do I want a different Heyting algebra, so that the subobject classifier ends up having three elements?
| https://mathoverflow.net/users/102 | Encoding fuzzy logic with the topos of set-valued sheaves | First I must warn you that there is a difference between fuzzy logics and topos theory. There are some categories of fuzzy sets which are almost toposes, but not quite - they form a quasitopos, which is like a topos, but epi + mono need not imply iso. There is a construction of such a quasitopos in Johnstone's Sketches of an Elephant - Vol 1 A2.6.4(e).
Now for the three valued logic I think a good example is a time-like logic. Suppose you have a fixed point T in time. This gives you two regions of time - before T and after T. Our logic will have three truth values - always true, true after T but not before, and never true. Note that we don't have a case "true before T, but not after", since once something is true, it is always true from that time on. Like knowledge of mathematical theorems (assuming there are no mistakes!).
The topos with this logic is the arrow category of set: Set$^\to$. Objects consist of Set functions $A \to B$, and morphism consist of pairs of set functions forming a commutative square.
For other three valued logics look at different Heyting algebras, but pay close attention to the implication operation, as it is a vital part of topos logic. For the true, false, maybe case I am not sure on how to construct a Heyting algebra which reflects this logic.
| 7 | https://mathoverflow.net/users/1709 | 8249 | 5,652 |
https://mathoverflow.net/questions/8244 | 6 | Is there a name for those categories where objects posses a given structure and every bijective morphism determines an isomorphism between the corresponding objects?
Examples of categories of that type abound: **Gr**, **Set**, ...
An specific example of a category where the constraint doesn't hold is given by **Top**: a morphism there is a continuous function between topological spaces. Now, it is easy to give *here* a concrete example of a bijective morphism between [0,1) and $\mathbb{S}^{1}$ that fails to be an isomorphism of topological spaces (in point of fact, much more is known in this case, right?).
| https://mathoverflow.net/users/1593 | What is the name of the following categorical property? | The comments on the question point out that it's not really well-posed: the property "bijective" isn't defined for morphisms of an arbitrary category.
However, for maps between sets, "bijective" means "injective and surjective". A common way to interpret "injective" in an arbitrary category is "monic", and a common way to interpret "surjective" in an arbitrary category is "epic". So we might interpret "bijective" as "monic and epic".
Then JHS's question becomes: is there a name for categories in which every morphism that is both monic and epic is an isomorphism? And the answer is yes: [balanced](http://ncatlab.org/nlab/show/balanced+category).
It's not a particularly inspired choice of name, nor does it seem to be a particularly important concept. But the terminology is quite old and well-established, in its own small way.
Incidentally, you don't *have* to interpret "injective" and "surjective" in the ways suggested. You could, for instance, interpret "surjective" as "regular epic", and indeed that's often a sensible thing to do. But then the question becomes trivial, since any morphism that's both monic and regular epic is automatically an isomorphism.
| 35 | https://mathoverflow.net/users/586 | 8250 | 5,653 |
https://mathoverflow.net/questions/8216 | 13 | Most of the group theory that is taught in introductory graduate classes is of the form $$(\mbox{number theory} + \mbox{ group actions} + \mbox{ orbit-stabilizer thm}) + \mbox{group axioms} \Rightarrow \mbox{theorems}$$ So what is the equivalent of "**(number theory + group actions + orbit-stabilizer thm)**" in the more advanced parts of group theory?
**To clarify based on some comments**: The techniques I learned in a graduate group theory class were just the orbit-stabilizer thm + some number theory + Lagrange's thm. Adding in some more constructions like semi-direct products allows one to make some inroads into some less elementary parts of group theory, i.e. we get some classification theorems for groups of small order with the help of Sylow theorems which is really just clever number theory + orbit-stabilizer theorem. So I would like to expand my toolbox a little bit by seeing what other tools are used in more advanced group theory but are still applicable to the elementary parts of group theory like classifying groups of small order.
**Tidbits collected from comments**: Kevin McGerty makes some excellent points about the extension of the theory from actions on sets which allow number theoretic arguments to actions on vector spaces which increase the sophistication and depth of the theory. The move from mere sets to vector spaces allows the use of linear algebra as another tool which in turn allows some tools from homological algebra to enter into the game.
| https://mathoverflow.net/users/nan | less elementary group theory | As someone who has spent far too much time thinking about classifying groups of small order, one thing you'd want to do is understand some more results about p-groups. You already probably learned one of the main results, every p-group has a nontrivial center, which is proved by the number theory type arguments you discussed. One of the key ideas in understanding p-groups is the Burnside basis theorem that says that any generating set for $G/\Phi(G)$ is a generating set for G. Here $\Phi(G)$ is the Frattini subgroup which is generated by all commutators and all pth powers. You can think of this as a version of Nakayma's lemma for p-groups and think that p-groups are similar to commutative rings.
| 17 | https://mathoverflow.net/users/22 | 8283 | 5,675 |
https://mathoverflow.net/questions/8307 | 11 | Basically intuitionistic logic is classical logic minus the law of the excluded middle, i.e. $\neg A\vee A$ is not necessarily valid for all formulas. So I would take this to mean that classical logic allows one to prove more theorems but apparently this view is too naive because yesterday I read the following in a book on proof theory:
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> Given a formula *C*, there is a translation giving a formula *C*\* such that *C* and *C*\* are classically equivalent and *C*\* is intuitionistically derivable if *C* is classically derivable. … The translation gives an interpretation of classical logic in intuitionistic logic.
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How am I supposed to understand the above statement? Is it saying that every theorem that uses the law of excluded middle in its proof can be done without the law of excluded middle or is it saying something more subtle?
Edit: Thanks to the answers I received the interpretation of the quote is in fact much more subtle than I was thinking. The double negation translation allows us to prove $\neg\neg C=C^\*$ if we can classically prove $C$ but in intuitionistic logic $\neg\neg C$ and $C$ are not necessarily equivalent. I think intuitively this means that there is a subtle semantic difference between $C$ and $\neg\neg C$. In intuitionistic logic a single classical theorem is potentially two intuitionistic theorems since knowing $\neg\neg C$ does not tell us much about $C$. So my naive interpretation was quite wrong. There are potentially more theorems in intuitionistic logic than there are in classical logic.
Edit 2: The last sentence is not precise and Reid Barton has an excellent comment on what I was actually thinking.
| https://mathoverflow.net/users/nan | intuitionistic interpretation of classical logic | Note that the statements are *classically* equivalent; the intuitionistically derivable statement may be intuitionistically weaker.
For example, the classical ‘law of the excluded middle’ $\lnot A \vee A$ is classically equivalent to $\top$, which is certainly intuitionistically derivable, even though the law of the excluded middle is not.
Does that help to understand the meaning of the statement?
EDIT: fqpc points out that the symbol $\top$ might be non-standard in this context. It probably comes from too much recent reading of the computer-science literature, where it's used for ‘top’ (the type of which all other types are a subtype; as opposed to ‘bottom’, $\bot$, which is a subtype of all other types)—and, indeed, that's the TeX code for it.
I thought it was common usage as a sort of visual pun on ‘true’ or ‘tautology’—an abbreviation for the empty statement.
| 6 | https://mathoverflow.net/users/2383 | 8311 | 5,695 |
https://mathoverflow.net/questions/8309 | 2 | Let $k$ be a positive integer greater than $1$ and suppose that $F \in \mathbb{Z}[x\_{1}, \ldots, x\_{k}]$.
Can we always find a natural number $n(k)$ and $f\_{1}, \ldots f\_{n(k)} \in \mathbb{Z}[x]$ such that
$\displaystyle F\Big(\bigoplus\_{j=1}^{k} \mathbb{Z}\Big) = \bigcup\_{j=1}^{n(k)} f\_{j}(\mathbb{Z})$ ?
Think this question is really cool. What do you guys think?
| https://mathoverflow.net/users/1593 | Decomposition result for multivariate polynomial | Not really. Take $F(x,y,z,t)=x^2+y^2+z^2+t^2$. Then the image consists of all non-negative integers (Lagrange 4-square theorem). On the other hand, any linear polynomial will give you negatives in the image and every polynomial of degree 2 and higher will give you a zero density set.
| 5 | https://mathoverflow.net/users/1131 | 8312 | 5,696 |
https://mathoverflow.net/questions/8263 | 10 | I have recently become interested in game theory by way of John Conway's on Numbers and Games. Having virtually no prior knowledge of game theory, what is the best place to start?
| https://mathoverflow.net/users/2421 | Just starting with [combinatorial] game theory | *Winning Ways for your Mathematical Plays* (in four volumes) has an enormous amount of stuff about combinatorial games. But most of it you probably won't be interested in for a while. There are a few quickly diverging directions one could study in combinatorial games. Here are some that come to mind immediately, and a possible list of topics to study in each:
1) Impartial games. Read a bit of *On Numbers and Games* so that you know how to read the notation and understand game equivalence and addition. Then learn the winning strategy for Nim, read the relevant bits of Chapter 3 and all of Chapter 4 of *Winning Ways*. After that, if you like the infinite theory, Lenstra has a paper called "On the algebraic closure of two" which is really nice. If you like the finite theory, learn about nim multiplication from *ONAG*, and then read Conway and Sloane's paper "Lexicographic codes: error-correcting codes from game theory." I think this part of the theory is the most interesting.
2) (Surreal) numbers. Again, learn how to read the notation and about game equivalence and addition. (You will need this for everything.) Then read the first part of *ONAG*. Then, perhaps learn about real-closed fields in general; you can make most of real analysis work over the Field of surreal numbers. (A Field is something like a field, but it has a proper class of objects instead of a set.)
3) Weird games, for example from Hackenbush and Domineering. Read Volume 1 of *Winning Ways*. The stuff on thermography and all-small games is quite interesting.
| 10 | https://mathoverflow.net/users/2046 | 8318 | 5,700 |
https://mathoverflow.net/questions/8236 | 19 | What are the axioms of four dimensional Origami.
If standard Origami is considered three dimensional, it has points, lines, surfaces and folds to create a three dimensional form from the folded surface. This standard origami has seven axioms which have been proved complete.
The question I have is whether these same axioms apply to four dimensional origami which has lines, surfaces, shapes, and produces shapes folded along surfaces in the fourth dimension. Or are there more axioms in four dimensional Origami.
I cannot find this question approached in the literature. Four Dimensional Origami does not seem to be a field of study yet.
| https://mathoverflow.net/users/2410 | Four Dimensional Origami Axioms | In the $n$-dimensional origami question, you start with a generic set of hyperplanes and their intersections, which can then be some collection of $k$-dimensional planes. An "axiom" is a set of incidence constraints that determines a unique reflection hyperplane, or conceivably a reflection hyperplane that is an isolated solution even if it is not unique. The space of available hyperplanes is also $n$-dimensional. Each incidence constraint has a codimension. A set of independent constraints makes an axiom when their codimensions add to $n$.
It is easy to write down a simple incidence constraint on a reflection $R$ and compute its codimension. If $n=2$, then the simple constraints are as follows:
1. $R(L) = L$ for a line $L$,
2. $R(x) = x$ for a point $x$,
3. $R(x) \in L$,
4. $R(L\_1) = L\_2$, and
5. $R(x\_1) = x\_2$
The first three constraints have codimension 1 and the last two have codimension 2. Formally there are 8 ways that to combine these constraints to make axioms. However, 1 cannot be used twice in Euclidean geometry, so that leaves 7 others. These are the seven axioms listed on Robert Lang's page. As it happens, Huzita missed combining 1 and 3. If you could have hyperbolic origami, you would have 8 axioms.
You can go through the same reasoning in $n=3$ dimensions. The simple constraints can again be written down:
1. $R(x) = x$ for a point $x$ (1)
2. $R$ fixes a line $L$ pointwise (2)
3. $R(L) = L$ by reflecting it (2)
4. $R(P) = P$ for a plane $P$ (1)
5. $R(x) \in L$ (2)
6. $R(x) \in P$ (1)
7. $R(L) \subset P$ (2)
8. $R(x\_1) = x\_2$ (3)
9. $R(L\_1) \cap L\_2 \ne \emptyset$ (1)
10. $R(P\_1) = P\_2$ (3)
I've written the codimensions of these constraints in parentheses. As before, you can combine these constraints. Certain pairs, such as 3 and 4, can't combine. You also can't use 4 three times. If you go through all of this properly, it is not all that hard to make a list of axioms that resembles the ones in 2 dimensions. (Possibly I made a mistake in this list or missed something, but it is not hard to go through this properly.)
However, there is a possible subtlety that I don't know how to address. Namely, suppose that you do make some complicated configuration using combinations of these constraints, at first. Can you create a configuration with the property that the codimensions don't simply add? For instance, ordinarily you can't use condition 7 twice to define the reflection $R$, because the total codimension is 4, which is too large. However, if the lines and planes in this condition are related to each other, is the true codimension sometimes 3? I would guess that you can make this happen. If so, then potentially you'd have to add "unstable" construction axioms to the list. But then it is not clear whether an unstable construction axiom is actually needed, or whether an unstable axiom can always be replaced by a sequence of stable axioms.
| 14 | https://mathoverflow.net/users/1450 | 8321 | 5,703 |
https://mathoverflow.net/questions/8324 | 43 | All definitions I've seen for the statement "$E,F$ are linearly disjoint extensions of $k$" are only meaningful when $E,F$ are given as **subfields of a larger field**, say $K$. I am happy with the equivalence of the various definitions I've seen in this case. Lang's *Algebra* VIII.3-4 and (thanks to Pete) Zariski & Samuel's *Commutative Algebra 1* II.15-16 have good coverage of this.
**"Ambient" definitions** of linear disjointness:
[Wikipedia](http://en.wikipedia.org/wiki/Tensor_product_of_fields#Compositum_of_fields) says it means *the map $E\otimes\_k F\to E.F$ is injective*, where $E.F$ denotes their compositum in $K$, the smallest subfield of $K$ containing them both.
An equivalent (and asymmetric) condition is that any subset of $E$ which is linearly independent over $k$ is *also linearly independent over $F$* (hence the name); this all happens inside $K$.
However, I often see the term used for field extensions which are **NOT subfields of a larger one**, even when the field extensions are not algebraic (so there is no tacit assumption that they live in the algebraic closure). Some examples of these situations are given below.
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> **Question:** What is the definition of "linearly disjoint" for field extensions which are not specified inside a larger field?
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**ANSWER:** (After reading the helpful responses of Pete L. Clark, Hagen Knaf, Greg Kuperberg, and JS Milne -- thanks guys! -- I now have a satisfying and fairly exhaustive analysis of the situation.)
There are two possible notions of abstract linear disjointness for two field extensions $E,F$ of $k$ (**proofs below**):
**(1)** "*Somewhere* linearly disjoint", meaning
"*There exists* an extension $K$ with maps $E,F\to K$, the images of $E,F$ are linearly disjoint in $K$."
This is equivalent to the tensor product $E\otimes\_k F$ being a **domain**.
**(2)** "*Everywhere* linearly disjoint", meaning
"*For any* extension $K$ with maps $E,F\to K$, the images of $E,F$ are linearly disjoint in $K$."
This is equivalent to the tensor product $E\otimes\_k F$ being a **field.**
Results:
**(A)** If *either* $E$ or $F$ is algebraic, then (1) and (2) are equivalent.
**(B)** If *neither* $E$ nor $F$ is algebraic, then (2) is impossible.
Depending on when theorems would read correctly, I'm not sure which of these should be the "right" definition... (1) applies in more situations, but (2) is a good hypothesis for implicitly ruling out pairs of transcendental extensions. So I'm just going to remember both of them :)
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**PROOFS:** (for future frustratees of linear disjointness!)
**(1)** Linear disjointness in some field $K$, by the Wikipedia defintion above, means the tensor product injects to $K$, making it a domain. Conversely, if the tensor product is a domain, then $E,F$ are linearly disjoint in its field of fractions.
**(2)** If the tensor product $E\otimes\_k F$ is a field, since any map from a field is injective, by the Wikipedia definition above, $E,F$ are linearly disjoint in any $K$. Conversely, if $E \otimes\_k F$ is *not* a field, then it has a non-trivial maximal ideal $m$, with quotient field say $K$, and then since $E\otimes\_k F\to K$ has non-trivial kernel $m$, by definition $E,F$ are not linearly disjoint in $K$.
**(A)** Any two field extensions have *some* common extension (take a quotient of their tensor product by any maximal ideal), so (2) always implies (1).
Now let us first show that (1) implies (2) supposing $E/k$ is a *finite* extension. By hypothesis the tensor product $E\otimes\_k F$ is a domain, and finite-dimensional as a $F$-vector space, and a finite dimensional domain over a field is automatically a field: multiplication by an element is injective, hence surjective by finite dimensionality over $F$, so it has an inverse map, and the image of $1$ under this map is an inverse for the element. Hence (1) implies (2) when $E/k$ is finite.
Finally, supposing (1) and only that $E/k$ is algebraic, we can write $E$ as a union of its finite sub-extensions $E\_\lambda/k$. Since tensoring with fields is exact, $E\_\lambda\otimes\_k F$ naturally includes in
$E\otimes\_k F$, making it a domain and hence a field by the previous argument. Then $E\otimes\_k F$ is a union of fields, making it a field, proving (1) implies (2).
**(B)** Now this is easy. Let $t\_1\in E$, $t\_2\in F$ be transcendental elements. Identify $k(t)=k(t\_1)=k(t\_2)$ by $t\mapsto t\_1 \mapsto t\_2$, making $E$,$F$ extensions of $k(t)$. Let $K$ be a common extension of $E,F$ over $k(s)$ (any quotient of $E\otimes\_{k(s)} F$ by a maximal ideal will do). Then $E,F$ are not linearly disjoint in $K$ because their intersection is not $k$: for example the set { $1,t$ }$\subseteq E$ is linearly independent over $k$ but not over $F$, so they are not linearly disjoint by the equivalent definition at the top.
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**Examples** in literature of linear disjointness referring to abstract field extensions:
* Eisenbud, *Commutative Algebra*, Theorem A.13 (p.564 in my edition) says, in characteristic $p$,
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> "$K$ is separable over $k$ iff $k^{1/p^{\infty}}$ is linearly disjoint from $K$."
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* Liu, *Algebraic Geometry and Arithmetic Curves*, Corollary 2.3 (c) (p. 91) says, for an integral algebraic variety $X$ over a field $k$ with function field $K(X)$,
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> "$X$ is geometrically integral iff $K(X)$ and $\overline{k}$ are linearly disjoint over $k$.
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(**Follow-up:** Since in both these situations, one extension is algebraic, the two definitions summarized in the answer above are equivalent, so everything is fantastic.)
**Old edit:** My first guess was (and still is) to say that the tensor product is a domain...
| https://mathoverflow.net/users/84526 | What does "linearly disjoint" mean for abstract field extensions? | The reasonable meaning following example (1) seems to be that $E \otimes\_k F$ is a field. If so, then it is isomorphic to every compositum. If not, then there exists a compositum within which they are not linearly disjoint.
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I am not (yet) getting voter support, but I stand my ground! :-)
First, clearly if $E \otimes\_k F$ is a field, then it is isomorphic to every compositum.
Second, if $E \otimes\_k F$ is not a field, then there exists a compositum in which $E$ and $F$ are not linearly disjoint. It has a non-trivial quotient field, and that field can serve as a compositum. As Pete Clark points out, there is a difference between the case that $E \otimes\_k F$ is an integral domain and the case that it has zero divisors. (And Pete is right that I forgot about this distinction.) In the former case, there exists a compositum in which they are linearly disjoint, namely the fraction field of $E \otimes\_k F$. In the latter case, $E$ and $F$ are not linearly disjoint in any compositum.
If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly disjoint, when $E \otimes\_k F$ is an integral domain, and strongly linearly disjoint, when it is a field. Which you think is the more important condition is up to you. In Andrew's examples, $E$ and $F$ aren't both transcendental, so the distinction is moot.
(I needed to think about this issue in *[Finite, connected, semisimple, rigid tensor categories are linear](https://arxiv.org/abs/math/0209256)*.)
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Actually, the previous isn't the whole story. If $E$ and $F$ are both transcendental, then they are extensions of purely transcendental extensions $E'$ and $F'$. $E'$ and $F'$ are only weakly linearly disjoint, and therefore $E$ and $F$ are too. So the distinction is always moot. Pete and Andrew's intuition was more correct all along. The correct statement is that when $E$ and $F$ are both transcendental, linearly disjoint extensions have different behavior.
| 26 | https://mathoverflow.net/users/1450 | 8327 | 5,706 |
https://mathoverflow.net/questions/6840 | 10 | For any prime p, one has the Frobenius homomorphism Fp defined on rings of characteristic p.
Is there any kind of object, say U, with a "universal Frobenius map" F such that for any prime p and any ring R of characteristic p we can view the Frobenius Fp over R as "the" base change of F from U to R?
I have the following picture in mind: In some sense it should be possible to view the category of Z-algebras as a sheaf of categories over Spec Z such that the fibre over Spec Fp is just the category of F\_p-algebras. A natural transformation f of the identity functor on the category of Z-algebras should restrict to a natural transformation fp of the identity functor on the category of Fp-algebras. In this naive picture one cannot expect the existence of an f such that fp is the Frobenius on Fp-algebras for all primes p. But is there way to make this picture work?
Another possible way to answer my question could be the following: Is there a classifying topos of, say, algebras with a Frobenius action? By this I mean the following: Is there a topos E with a fixed ring object R and an algebra A over it and an R-linear endomorphism f of A such that for any other topos E' with similar data R', A' there is a unique morphism of topoi E' -> E that pulls back R, A to R', A' and such that f is pulled back to the Frobenius fp of A' in case R' is of prime characteristic.
(Feel free to modify my two pictures to make them work.)
| https://mathoverflow.net/users/1841 | Does a universal Frobenius map exist? | I don't really get the categorical picture of what you are asking, but it feels something very similar to the relation between finite fields of characteristic $p$ and the ring (of characteristic 0) of $p$-typical Witt vectors.
You might want to have a look at Borger and Wieland work on [pleythistic algebras](http://wwwmaths.anu.edu.au/~borger/papers/03/paper03.html). The "lifting of all Frobeniuses at the same time" gives you an structure of $\Lambda$-ring, so the paper [The basic geometry of Witt vectors](http://wwwmaths.anu.edu.au/~borger/papers/05/paper05.html) by Borger might also be useful.
| 3 | https://mathoverflow.net/users/914 | 8334 | 5,712 |
https://mathoverflow.net/questions/8331 | 4 | What linear algebraic quantities can be calculated precisely for a nonsingular matrix whose entries are only approximately known (say, entries in the matrix are all huge numbers, known up to an accuracy of plus or minus some small number)? Clearly not the determinant or the trace, but probably the signature, and maybe some sort of twisted signatures? What is a reference for this sort of stuff? (numerical linear algebra, my guess for the name of such a field, seems to mean something else).
| https://mathoverflow.net/users/2051 | Approximately known matrix | SVD is stable, and in some sense incorporates all the stable data you can have, so the answer is: "anything you can see on the SVD". Specifically you can easily see the signature (assuming the matrix is far enough from being singular).
| 7 | https://mathoverflow.net/users/404 | 8336 | 5,714 |
https://mathoverflow.net/questions/8285 | 6 | Does anyone have an opinion on Alain Badiou's use of set theory? Is there anything interesting mathematically there? Also could anyone shed any light on the comment in the Wikipedia article [link text](http://en.wikipedia.org/wiki/Alain_Badiou) that says:
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> This effort leads him, in Being and Event, to combine rigorous mathematical formulae with his readings of poets such as Mallarmé and Hölderlin and religious thinkers such as Pascal.
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| https://mathoverflow.net/users/1977 | Badiou and Mathematics | Badiou's got some mathematical training; reading back and forth between the relevant sections of Goldblatt's *Topoi* and Badiou's account of $\Omega$-sets in *Logics of Worlds*, for example, you can see that the one tracks the other closely. It's not just blind quotation, followed by hand-wavy inference-drawing either: you could actually learn about $\Omega$-sets from Badiou's presentation of them alone, and not be too horribly surprised or confused when you came to read the technical presentation in Goldblatt (this was, in fact, the order in which I did it).
On the axiom of choice and "infinite liberty":
The AoC says that given a set $\{A,B,C,\ldots\}$ none of whose members are the empty set, there exists a set $\{x\in A,y\in B,z\in C,\ldots\}$ which takes one element from each of the first set's members. The point here is that the AoC "freely" chooses an element from each set rather than (for example) identifying a "least" element and choosing that: even when there's no rule that can tell you which element should be chosen, the AoC says that a set exists representing some choice.
The AoC only has any work to do in situations where no rule can be found (for example, no-one knows of a rule that will well-order the reals, but the AoC entails that a real can be chosen, then another from the remaining reals, then another etc. - so "axiomatically" a well-ordering of the reals exists, provided one accepts AoC) - hence it represents, in this sense, the possibility of a *predicatively undetermined* choice. That's the "infinite liberty" he's on about. It is nowhere asserted that AoC "proves" that such a liberty exists, but rather that introducing AoC into ZF makes such a liberty thinkable within the confines of its axiomatic system (this is in line with Badiou's general program of treating mathematics as "ontology", as a means for systematically demarcating what is thinkable of "being as such").
In terms of "interest to mathematicians": Badiou's early text *The Concept of Model* is a good philosophical introduction to model theory, and his *Number and Numbers* is an interesting and accessible guide to the philosophy of number, covering Frege, Peano, Cantor, Dedekind and Conway (surreal numbers).
| 12 | https://mathoverflow.net/users/2451 | 8344 | 5,718 |
https://mathoverflow.net/questions/8182 | 13 | I am asking for some sort of generalization to Perron's criterion which is not dependent on the index of the "large" coefficient. (the criterion says that for a polynomial $x^n+\sum\_{k=0}^{n-1} a\_kx^k\in \mathbb{Z}[x]$ if the condition $|a\_{n-1}|>1+|a\_0|+\cdots+|a\_{n-2}|$ and $a\_0\neq 0$ holds then it is irreducible.)
This would answer a second question about the existence of n+1-tuples $(a\_0,\dots,a\_n)$ of integers for which $\sum\_{k=0}^n a\_{\sigma(k)}x^k$ is always irreducible for any permutation $\sigma$.
What happens if we restrict $|a\_i| \le O(n)$ ? $ |a\_i| \le O(\log n)$ ?
| https://mathoverflow.net/users/2384 | Is a polynomial with 1 very large coefficient irreducible? | OK, about your second question. Let's consider the polynomial $x^n+2(x^{n-1}+\ldots+x^2+x)+4$. I claim that this polynomial is almost good for your purposes: if we permute all coefficients except for the leading one, it remains irreducible. Proof: if the constant term becomes 2 after the permutation, use Eisenstein, if not, look at the Newton polygon of this polynomial mod 2 - you can observe that if it is irreducible, it has to have a linear factor, which is easily impossible.
[If we allow to permute all coefficients, I would expect that something like $9x^n+6(x^{n-1}+\ldots+x^2+x)+4$ would work for nearly the same reasons...]
| 4 | https://mathoverflow.net/users/1306 | 8348 | 5,720 |
https://mathoverflow.net/questions/8339 | 8 | Suppose I have an category additive category C (i.e. the hom sets are enriched in abelian groups and there are finite direct sums). Suppose further that C has cokernels. Then I can make C tensored over finitely presentable Abelian groups by the following ad hoc construction:
First define $\mathbb{Z}^n \otimes X := \oplus\_n X$. Now given a finitely presentable abelian group A, choose a presentation, i.e. realize $A$ as the cokernel of $f:\mathbb{Z}^r \to \mathbb{Z}^g$. Define $A \otimes X$ as the cokernel of the induced map:
$\mathbb{Z}^r \otimes X \to \mathbb{Z}^g \otimes X$
My questions: Is there a way to do the same thing which feels more canonical and less ad hoc? Under what conditions will C be tensored over all abelian groups?
| https://mathoverflow.net/users/184 | Tensored Over Abelian Groups? | Given an object $X$ in an additive category $C$ and an abelian group $A$, define the object $A\otimes X$ in $C$ by the rule $Hom\_C(A\otimes X,\:Y) = Hom\_{Ab}(A,Hom\_C(X,Y))$, where $Ab$ denotes the category of abelian groups. If arbitrary direct sums and cokernels (arbitrary colimits, in other words) exist in an additive category $C$, the tensor product $A\otimes X$ exists in $C$ for any abelian group $A$ and any $X\in C$. It can be constructed just as you describe in your question, except that finite direct sums should be replaced with infinite direct sums.
| 15 | https://mathoverflow.net/users/2106 | 8349 | 5,721 |
https://mathoverflow.net/questions/7914 | 4 | I am looking for an example of a smooth surface $X$ with a fixed very ample $\mathcal O\_X(1)$ such that $H^1(\mathcal O(k))=0$ for all $k$
(such thing is called an ACM surface, I think) and a ~~globally generated~~ line bundle $L$ such that $L$ is torsion in $Pic(X)$ and $H^1(L) \neq 0$.
Does such surface exist? How can I construct one if it does exist? What if one ask for even nicer surface, such as arithmetically Gorenstein? If not, then I am willing to drop smooth or globally generated, but would like to keep the torsion condition.
More motivations(thanks Andrew): Such a line bundle would give a cyclic cover of $X$ which is not ACM, which would be of interest to me. I suppose one can think of this as a special counter example to a weaker (CM) version of purity of branch locus.
To the best of my knowledge this is not a homework question (: But I do not know much geometry, so may be some one can tell me where to find an answer. Thanks.
EDIT: Removed the global generation condition, by Dmitri's answer. I realized I did not really need it that much.
| https://mathoverflow.net/users/2083 | Torsion line bundles with non-vanishing cohomology on smooth ACM surfaces | Let us show that a globaly generated torsion line bundle $L$ on a (compact) complex surface is trivial. Ideed, a globally generated line bundle has at least one section, say $s$. Let us take it. If $s$ has no zeros, then $L$ is trivial. But if $s$ vanishes somewhere then any positive power $L^n$ has a section $s^n$ that vanishes at the same points. So any power of $L$ is not trivial, i.e. $L$ is not a torsion bundle, contradiction.
Notice that we did not use the fact that the surface is smooth. And we also did not use the fact that we work with a surface...
| 4 | https://mathoverflow.net/users/943 | 8355 | 5,726 |
https://mathoverflow.net/questions/8340 | 5 | I am trying to read the Hovey-Shipley-Smith article as defining the stable model structure on symmetric spectra as a left Bousfield localization (as explained on [nLab](http://ncatlab.org/nlab/show/Bousfield+localization+of+model+categories)) of the projective level model structure on symmetric spectra, which has level equivalences as weak equivalences.
Almost all ingredients are there in the article. All I have left to show is that the injective Omega-spectra are indeed the S-local objects, where S is the class of stable equivalences. By defintion any map in S induces a weak equivalence of simplicial hom-sets $Map\_{Sp^\Sigma}(f,E)$ for E a injective Omega-spectrum. Conversely, lemma 3.1.5 and example 3.1.10 conspire to tell you that if a symmetric spectrum is S-local and injective, it is an Omega-spectrum. So, what remains: is any S-local symmetric spectrum injective?
| https://mathoverflow.net/users/798 | Are injective Omega-spectra the S-local objects of symmetric spectra for some class S? | You have to realize it has been a long time since we wrote that paper. But I'll give it my best shot.
I think we intentionally chose the injective Omega-spectra because they are "extra fibrant", so to speak. That is, I think S-local spectra don't have to be injective, just Omega-spectra.
The injective Omega-spectra should be the fibrant objects in a different model structure. There should be an injective level structure, which I guess we did not construct, where the cofibrations are monomorphisms and the weak equivalences are level equivalences. The fibrant objects would then be the injective spectra. The injective Omega-spectra are then the fibrant objects in the left Bousfield localization of this category with respect to the stable equivalences.
| 6 | https://mathoverflow.net/users/1698 | 8360 | 5,730 |
https://mathoverflow.net/questions/8351 | 14 | Hi Everyone!
I'm wondering if anyone knows of a reference for learning global class field theory using the original analytic proofs developed in the 1920s and 1930s. Almost every book I can find either does local class field theory first or uses ideles/cohomology to prove global class field theory. This is not how it was historically done - the ideal-theoretical formulation of class field theory was proven first, using more elementary analytic methods. So I'm wondering if anyone knows of any resources which would teach these proofs.
I'm currently about to take a class which follows global class field theory in this way, and our teacher says he does not know of any textbook for this, so I'm wondering if anyone here would know.
| https://mathoverflow.net/users/1355 | Reference for Learning Global Class Field Theory Using the Original Analytic Proofs? | As far as textbooks your best bet is Janusz's "Algebraic Number Fields."
Also I tried to collect a lot of this stuff in my [senior thesis](http://math.berkeley.edu/~nsnyder/thesismain.pdf). The list of references there should also be very useful. For example, I use Hecke's original approach to abelian L-functions instead of Tate's thesis which I learned from the last section of Neukirch's big book (which is otherwise a very modern book), there's Hilbert's original proof of lifting of the Frobenius from his Zahlbericht (which appears in translation and I highly recommend), and a proof of Kronecker-Weber following the original approach appears both in Mollin's "Algebraic Number Theory" and in Hilbert's Zahlbericht.
As an added bonus for non-German readers I translated (caveat, I didn't know any German at the time and relied heavily on dictionaries) Artin's beautiful paper on L-functions in the appendix which is one of the key original sources here.
| 12 | https://mathoverflow.net/users/22 | 8364 | 5,732 |
https://mathoverflow.net/questions/8361 | 5 | which HNN-extensions are free products? this question is related with another still unsolved about Nielsen-Thruston-reducibility and connected-sum-irreducibility of 3d-torus- bundles...
| https://mathoverflow.net/users/2196 | HNN extensions which are free products | This might help.
**Lemma** If $A$ does not split freely and $C$ is a non-trivial subgroup of $A$ then the HNN extension $G=A\*\_C$ does not split freely.
The proof uses Bass--Serre theory---see Serre's book *Trees* from 1980.
*Proof.* Let $T$ be the Bass--Serre tree of a free splitting of $G$. Because $A$ does not split freely, $A$ stabilizes some unique vertex $v$. But $C$ is non-trivial, so $C$ also stabilizes a unique vertex, which must be $v$. Therefore, $G$ stabilizes $v$, which means the free splitting was trivial. *QED*
A similar argument shows the following.
**Lemma** If $ A\*\_C $ splits non-trivially as an amalgamated free product $ A' \*\_{C'} B'$ then either $A$ splits over $C'$ or $C$ is conjugate into $C'$.
| 13 | https://mathoverflow.net/users/1463 | 8365 | 5,733 |
https://mathoverflow.net/questions/8345 | 2 | Can any body give me a reference of the result about primitive root mod p for a class of prime number p.
The result that I am looking for is something along this line:
$2$ is a primitive root mod $p$ for all prime $p$ of the form $p=4q+1$ where $q$ is also a prime.
Thanks in advance.
| https://mathoverflow.net/users/808 | Reference of primitive root mod p | Take a look at "A criterion on primitive roots modulo p" by H.Park, J.Park, D.Kim.
There is a collection of various criteria including the above for small primes to appear as primitive roots. I hope it helps, or are you looking for something more general?
| 2 | https://mathoverflow.net/users/2384 | 8366 | 5,734 |
https://mathoverflow.net/questions/8374 | 14 | Using the game of [Battleship](http://en.wikipedia.org/wiki/Battleship_(game)) as an example, is there a general solution for determining the number of arrangements of a given set of 1xN rectangles on a X by Y grid?
**Example:** In Battleship, each player has a 10x10 grid on which they must place each of the following rectangular 1xN ships which may not overlap or be placed diagonally:
* 1x5-square "carrier"
* 1x4-square "battleship"
* 1x3-square "submarine"
* 1x3-square "cruiser"
* 1x2-square "destroyer"
**A)** How would one calculate the number of arrangements for this example that use each "ship" exactly once?
**B)** What is the general solution for a grid of height Y and width X, and a given set of 1xN "ships"?
| https://mathoverflow.net/users/2455 | Battleship Permutations | 1) Consider a fixed list of boats in the limit of a large grid. This is a model of a battle in the open Pacific, you know; it's not supposed to be Pearl Harbor. Then the inclusion-exclusion formula is much faster than a back-tracking search. Count all of the arrangements of ships, then add or subtract a term for each pattern of intersection when the ships have had an accident. Each term is a product of squares. Indeed, it is an explicit polynomial in the grid size.
2) As Gjergji says, Kasteleyn's method is only for perfect matchings. Even if the ships were all PT boats, you would not have a determinant formula unless they covered the whole grid. If you truly do want to count saturated traffic jams of PT boats, Kasteleyn himself evaluated the determinant for rectangles and established a trigonometric product formula.
3) If anyone wants to define a generalized $q$-Battleship with an interpretation over $\mathbb{F}\_q$ — no, thank you! (You can make the product formula for each term work, but don't bother.)
| 16 | https://mathoverflow.net/users/1450 | 8380 | 5,742 |
https://mathoverflow.net/questions/8376 | 7 | Let $G$ be a group scheme (for instance, over $k$ a field of characteristic 0).
Let $e$ be its unit.
I denote by $O\_G$ the structural sheaf of $G$.
Let $D\_e : O\_{G,e} \to k$ a derivation.
I would like to get directly (ie, without any consideration about the cotangent bundle, or some canonical isomorphisms...) a derivation $D : O\_G\to O\_G$ that extends $D\_e$, and which is compatible with the action of $G$. That is, I would like to get this derivation by the mean of the multiplication map : $m : G \times G \to G$, etc., etc.
I have guessed this question would not be difficult, and would only be a matter of technics, but I can't manage to do it.
| https://mathoverflow.net/users/2330 | A technical question about derivations of sheaves on group schemes | Your derivation at the origin is a map $\mathrm{Spec} k[\epsilon] / \epsilon^2 \rightarrow G$ whose restriction to $\epsilon = 0$ is the inclusion of the origin. This induces a map
$$G \times\_{\operatorname{Spec} k} \operatorname{Spec} k[\epsilon]/\epsilon^2 \rightarrow G \times G \rightarrow G$$
where the first map is the product with the map described above and the second is the multiplication map. Dually, the composed map gives a map of sheaves of algebras
$$\mathcal{O}\_G \rightarrow \mathcal{O}\_G[\epsilon] / \epsilon^2$$
which is the same thing as a derivation from $\mathcal{O}\_G$ to itself.
| 9 | https://mathoverflow.net/users/32 | 8386 | 5,748 |
https://mathoverflow.net/questions/8367 | 7 | Can anyone describe (or give a reference for) the 2d superspace formulation of N=(2,2) SUSY in Euclidean signature?
I'm reading Hori's excellent introduction to QFT in the book 'Mirror symmetry', and my question is basically Ex. 12.1.1. page 273. What I imagine the answer is is a super version of the usual story of differential forms on complex manifolds, i.e. we complexify, find square roots of the $\partial\_z$ and $\partial\_{\bar{z}}$ operators, then find a subalgebra of 'chiral' fields analogous to the subalgebra of holomorphic forms.
| https://mathoverflow.net/users/2454 | N=(2,2) supersymmetry in two-dimensional Euclidean space | You imagine well. Hori is talking about $\mathbb{R}^{2|2}$, which is arguably the simplest *super Riemann surface*.
There is lots on this subject, mostly in the Physics literature, which I'm hesitant to recommend. In the Mathematics literature, you might wish to read Deligne and Freed's [Supersolutions](http://arxiv.org/abs/hep-th/9901094) and in particular section 2.6.
| 5 | https://mathoverflow.net/users/394 | 8387 | 5,749 |
https://mathoverflow.net/questions/7058 | 8 | In *Proofs and Types*, Girard discusses coherent (or coherence) spaces, which is defined as a set family which is closed downward ($a\in A,b\subseteq a\Rightarrow b\in A$), and binary complete (If $M\subseteq A$ and $\forall a\_1,a\_2\in M (a\_1\cup a\_2\in A)$, then $\cup M\in A$)
It was informally related to topological spaces. Anyway, I have a couple pretty general questions:
Are they particularly useful outside of type theory? Perhaps more specifically, do coherent spaces show up in topology?
The last one raises up a philosophical question I've been pondering: Why is it that some structures seem to show up all over the place, while others that seem like they "should" be more or less equivalently useful don't seem to show up much at all? An example would be matroids versus topologies. I feel, morally, that matroids should be more useful than they seem to be.
The last question probably doesn't have any sort of solid answer, but it would be nice to hear some thought from people with a stronger background.
Cheers and thanks,
Cory
*Edit*:
After thinking about this some more, it has occurred to me that coherent spaces are a sort of "dual" to ultrafilters. I really don't have the background to be terribly formal, but, let me try to explain:
Let $(X,C)$ be a coherent space, and call the elements of $C$ "open" (I think the analogy is justified, because adding $X$ to $C$ makes it a topology), then the closed sets form an ultrafilter. The one problem is that the closure under intersection is a bit strong (the set of closed sets is closed under *arbitrary* intersections).
On the other hand, if $(X,U)$ is an ultrafilter, the set of complements of open sets *almost* forms a coherent space-- but the conditions on unions is just a little too weak.
So, my next question is: Has this link been explored at all? Is there even anything there to explore?
Thanks again.
| https://mathoverflow.net/users/2143 | Coherent spaces | An analogue of your coherence spaces are used extensively in set theory, particularly with the method of forcing, the set-theoretic technique often used to prove statements independent of ZFC. But there is a variation, in that the coherence clause is weakened to cover only some M, such as M of a certain size.
For example, in the standard forcing argument to add κ many generic Cohen reals, one considers the partial order consisting of all finite partial functions from (ω x κ) to {0,1}. This collection of partial functions satisfies your properties, if we restrict to finite M, since the union of a set of partial functions is a function if and only if these functions are coherent, in the sense that any two of them agree. If F is a maximal filter on this partial order, then the union of F is a function fully from (ω x κ) to {0,1}. If the filter is what is known as V-generic, then the slices of this function adds κ many new Cohen reals. If κ is at least ω2, then one can argue that CH fails in the resulting forcing extension.
Many forcing notions have the form of partial functions from one set to another, restricted by size or by other features, and so they also satisfy the corresponding restricted version of your Coherence space.
| 2 | https://mathoverflow.net/users/1946 | 8389 | 5,750 |
https://mathoverflow.net/questions/7350 | 21 | <http://en.wikipedia.org/wiki/Axiom_of_dependent_choice>
Is DC sufficient for the understanding of objects that are countable in some suitable sense?
For example, is DC sufficient for the full development of the theory of von Neumann algebras on a separable Hilbert space?
| https://mathoverflow.net/users/2206 | Is Dependent Choice all we really need? | Let me adopt an extreme interpretation of your question, in order to prove an affirmative answer.
Yes, in the arena of the countable, DC suffices.
To see why, let me first explain what I mean. If one wants to consider only countable sets, then the natural set-theoretic context is HC, the class of hereditarily countable sets. These are the sets that are countable, and all members are countable, and members-of-members and so on. The class HC is the land of all-countable set theory, the land of the fundamentally countable. I am not proposing that you want to remain within HC. Rather, you want to consider the properties of the objects in HC that are expressable there and what you can prove about them in ZF+DC, versus ZFC.
Since hereditarily countable sets can be coded easily by real numbers, it turns out that the structure HC is mutually interpreted in the usual structure of the reals R and vice versa. That is, the structure HC is essentially equivalent in a highly concrete way to the usual structure of the real numbers. And so questions about the fundamentally countable are equivalent to questions in the usual projective hierarchy of descriptive set theory. Thus, in this interpretation, your question is really asking whether one needs ever needs AC as opposed to DC in order to prove a projective statement.
**Question.** Are the projective consequences of ZFC the same as those of ZF+DC?
The answer is Yes, by the following theorem.
**Theorem.** A projective statement is provable in ZFC if and only if it is provable in ZF+DC.
**Proof.** It suffices to show that for every model W of ZF+DC there is a model of ZFC with the same reals. The reason this suffices is that in this case, any projective statement failing in a model of ZF+DC will also fail in a model of ZFC.
So, suppose that W is a model of ZF+DC. Let L(R) be the inner model of W obtained by constructing relative to reals. This is also a model of ZF+DC. Consider the forcing notion P in L(R) consisting of well-ordered countable sequences of real numbers in L(R), ordered by end-extension. Let G subset P be L(R)-generic for P. Since P is countably closed in L(R), it follows using DC that the forcing extension L(R)[G] has no additional real numbers. (The usual argument that countably closed forcing adds no new $\omega$-sequences makes essential use of DC, and indeed, DC is actually equivalent to that assertion.) And since G adds a well-ordering of these reals in order type $\omega\_1$, it follows that L(R)[G] has a well-ordering of the reals. From this, it follows that L(R)[G] is actually a model of ZFC, since it has the form L[A], where A is a subset of $\omega\_1$ enumerating these reals in the order that G lists them. Thus, we have provided a model of ZFC, namely L(R)[G], with the same reals as W. It follows that W and L(R)[G] have the same projective truth, since this truth is obtained by quantifying only over this common set of reals. **QED**
This phenomenon generalizes by the same argument beyond projective statements, to statements of the form "L(R) satisfies $\phi$". The point is that ZFC and ZF+DC prove exactly the same truths for L(R), since any model W of ZF+DC has the same L(R) as the model L(R)[G] in the proof above, and this satisfies ZFC.
The conclusion is that if you wish to study mathematical objects that are essentially countable only in the weak sense that they exist in L(R) — and this includes many highly uncountable objects — then the features about them in L(R) that you can prove in ZFC are exactly the same as the features you can prove in ZF+DC. So this interpretation is not so extreme after all...
| 23 | https://mathoverflow.net/users/1946 | 8392 | 5,752 |
https://mathoverflow.net/questions/8396 | 31 | Not sure if this is appropriate to Math Overflow, but I think there's some way to make this precise, even if I'm not sure how to do it myself.
Say I have a nasty ODE, nonlinear, maybe extremely singular. It showed up naturally mathematically (I'm actually thinking of Painleve VI, which comes from isomonodromy representations) but I've got a bit of a physicist inside me, so here's the question. Can I construct, in every case, a physical system modeled by this equation? Maybe even just some weird system of coupled harmonic oscillators, something. There are a few physical systems whose models are well understood, and I'm basically asking if there's a construction that takes an ODE and constructs some combination of these systems that it controls the dynamics of.
Any input would be helpful, even if it's just "No." though in that case, a reason would be nice.
| https://mathoverflow.net/users/622 | Does every ODE comes from something in physics? | It is possible to solve a large class of ODEs by means of analog computers. Each of the pieces of the differential equation corresponds to an electronic component and if you wire them up the right way you get a circuit described by the ODE. [Wikipedia](https://en.wikipedia.org/wiki/Analog_computer) has lots of information on the subject and a link like [this](http://www.analogmuseum.org/library/handson.pdf) one gives explicit examples of circuits. It's not hard to build circuits for things like the [Lorenz equation](https://en.wikipedia.org/wiki/Lorenz_attractor) and see a nice Lorenz attractors on an [oscilloscope display](https://www.youtube.com/watch?v=6HWfdRZo1I4).
| 18 | https://mathoverflow.net/users/1233 | 8398 | 5,756 |
https://mathoverflow.net/questions/8407 | 6 | Is there a generalized method to find the projective closure of an affine curve? For example, I read that the projective closure of $y^2 = x^3−x+1$ in $\mathbb{P}^2$ is $y^2z = x^3−xz^2+z^3$.
If I want to find the the closure of another affine curve, what method should I employ? I can't seem to find an adequate description in any text.
Thanks
| https://mathoverflow.net/users/2473 | Projective closure of affine curve | When the curve is a plane curve of degree $d$, the formula is simple (and in fact, this works for any hypersurface) you take $f(x,y)=0$ and replace it by $z^d f(x/z,y/z)=0$. This will be homogeneous of degree $d$, and when $z\neq 0$, you recover your curve.
Now, if you have a more general affine variety, given by $\langle f\_1,\ldots,f\_k\rangle$ in $k[x\_1,\ldots,x\_n]$, then you first compute a [Groebner Basis](http://en.wikipedia.org/wiki/Gr%25C3%25B6bner_basis), which you can learn how to do using Cox, Little and O'Shea's "Ideals, Varieties and Algorithms" (assumes no background whatsoever) and then you do this same trick with each polynomial in the Groebner basis.
Why can't you just use any basis? Look at the twisted cubic. $t\mapsto (t,t^2,t^3)$. This is cut out, in affine space, by $y-x^2$ and $z-x^3$, though $y-x^2$ and $xy-z$ are better to use. But still, $yw-x^2$ and $xy-zw$ don't give the twisted cubic! Bezout tells us that they give something of degree four containing the cubic, and it's not a hypersurface, so you get the twisted cubic plus a line. To get the cubic itself, you have to use a third quadratic polynomial that you get in the ideal, $y^2-xz$, which is not in the ideal given by the homogenized generators. These three, however, form a Groebner basis for the ideal, and then homogenization gives the homogeneous ideal of the projective twisted cubic.
| 26 | https://mathoverflow.net/users/622 | 8410 | 5,763 |
https://mathoverflow.net/questions/8415 | 24 | The number of conjugacy classes in $S\_n$ is given by the number of partitions of $n$. Do other families of finite groups have a highly combinatorial structure to their number of conjugacy classes? For example, how much is known about conjugacy classes in $A\_n$?
| https://mathoverflow.net/users/960 | Combinatorial Techniques for Counting Conjugacy Classes | Since $A\_n$ has index two in $S\_n$, every conjugacy class in $S\_n$ either is a conjugacy class in $A\_n$, or it splits into two conjugacy classes, or it misses $A\_n$ if it is an odd permutation. Which happens when is a nice undergraduate exercise in group theory. (And you are a nice undergraduate. :-) )
The pair $A\_n \subset S\_n$ is typical for this question in finite group theory. You want the conjugacy classes of a finite simple group $G$, but the answer is a little simpler for a slightly larger group $G'$ that involves $G$. Another example is $\text{GL}(n,q)$. It involves the finite simple group $L(n,q)$, but the conjugacy classes are easier to describe in $\text{GL}(n,q)$ . They are described by their Jordan canonical form, with the twist that you may have to pass to a field extension of $\mathbb{F}\_q$ to obtain the eigenvalues.
The group $\text{GL}(n,q)$ is even more typical. It is a Chevallay group, which means a finite group analogue of a Lie group. All of the infinite sequences of finite simple groups other than $A\_n$ and $C\_p$ are Chevallay groups. You expect a canonical form that looks something like Jordan canonical form, although it can be rather more complicated.
If $G$ is far from simple, i.e., if it has some interesting composition series, then one approach to its conjugacy classes is to chase them down from the conjugacy classes of its composition factors, together with the structure of the extensions. The answer doesn't have to be very tidy.
I suppose that finite Coxeter groups give you some exceptions where you do get a tidier answer, just because they all resemble $S\_n$ to varying degrees. But I don't know a crisp answer to all cases of this side question. The infinite sequences of finite Coxeter groups consist only of permutation groups, signed permutation groups, and dihedral groups. (And Cartesian products of these.) In the case of signed permutation groups, the answer looks just like $S\_n$, except that cycles can also have odd or even total sign. There is also the type $D\_n$ Coxeter group of signed permutation matrices with an even number of minus signs; the answer is just slightly different from all of the signed permutation matrices, which is type $B\_n$. The crisp answer that I don't have would be a uniform description that includes the exceptional finite Coxeter groups, such as $E\_8$.
| 31 | https://mathoverflow.net/users/1450 | 8418 | 5,768 |
https://mathoverflow.net/questions/8395 | 4 | So, today I started learning the definition of a quiver variety, and wanted to make sure I'm understanding things right, so first, my setup:
I've been looking at the simplest case that didn't look completely trivial: two vertices with one directed edge. Now, my understanding is that then we have two vector spaces $V$ and $W$ of dimensions $n$ and $m$, and the quiver variety is $\hom(V,W)\oplus \hom(W,V)/GL(V)\times GL(W)$, with the action of each group on the domain or codomain, as applicable.
Now, a naive dimension count (this is one place that I may be very, very wrong) is that this is a $2nm-n^2-m^2=-(m-n)^2$ dimensional space. So presumably if $m=n$, the GIT quotient is a single point.
Now, what if $n\neq m$? Presumably, we don't really get anything for the GIT quotient, either a point or the empty set (I don't know GIT very well, so perhaps no stable points?)
Finally, does this object have interesting geometry as a stack? It seems obvious that this should be a smooth Artin stack (if my intuition for them is even vaguely accurate) which just happens to be negative dimensional, but what kinds of properties does it have?
| https://mathoverflow.net/users/622 | Near Trivial Quiver Varieties | First, one important point: people who study quiver varieties seem not to usually take stack quotients, but rather GIT quotients (though there ARE very good reasons to do this if you like geometric representation theory) which leads them to come up different dimension formulae from you. The reason is that you are think of a fine moduli space, whose stack dimension is the geometric dimension of the underlying variety minus the dimension of the automorphism group of the object, and modules over an algebra ALWAYS have automorphisms (if nothing else, they have multiplication by constants).
Now, the answer proper: You seem to have mixed up two of the more popular notions of a quiver variety (leading your confusing dialogue with Greg in the comments), and Kevin seems to have mixed in a third, which may or may not actually be the one you had in mind (all of which are, of course, closely related).
If you have two vertices and one arrow, then there are two things you can do.
You can take the moduli space of quiver representations of the path algebra of that quiver, which is given by $\mathrm{Hom}(V,W)/GL(V)\times GL(W)$. This has a very small dimension $(nm-n^2-m^2)$ and should probably be thought of as finitely many points (indeed, this quiver only has finitely many representations of any given dimension. The indecomposibles look like $k\to 0$, $0\to k$ and $k\to k$), all of which have a bunch of automorphisms.
Now it sounds like what you intended to do was take the "hyperkählerization" of this quiver variety. What you should do for this is take the cotangent bundle of $\mathrm{Hom}(V,W)$, but before you mod out, you have to impose a moment map condition. The reason this is a good idea is that you want a resulting variety which is a holomorphic symplectic result (just like the cotangent bundle), which you can also think of as hyperkähler by picking a hermitian metric on $\mathrm{Hom}(V,W)$. This moment map condition is basically that both possible compositions of maps along your arrows are 0 (note: I think this is not a flat map, so I think if you want to really think properly about this story, you should probably take the derived fiber). Then take the quotient of that.
What Kevin is referring to is probably the most popular definition of quiver varieties for geometric representation theory. This is yet another definition, where he interpreted one of your vertices as a shadow vertex. This extra layer of confusion results from some (IMHO poor) notational choices of Nakajima.
| 7 | https://mathoverflow.net/users/66 | 8420 | 5,770 |
https://mathoverflow.net/questions/7016 | 27 | Assume a convex figure $F\subset \mathbb R^2$ satisfies the following property: if $f:F\to \mathbb R^2$ is a distance-non-increasing map then its image $f(F)$ is congruent to a subset of $F$.
Is it true that $F$ is a round disk?
**Comments:**
* It is easy to see that the **round disk** has this property.
* One can **reformulate** the property: if for some set $G\subset\mathbb R^2$ there is a distance-non-contracting map $G\to F$ then there is a distance-preserving map $G\to F$. (The equivalence follows from [Kirszbraun theorem](http://en.wikipedia.org/wiki/Kirszbraun_theorem))
* No bad map is known for the following figure: intersection of two discs say unit disc with center at (0,0) and a disc with radius 1.99 and center at (0,1) --- see comments of Martin M. W. below. (That might be a **counterexample**.)
* Some figures as **Reuleaux triangle** are bad (see the comments below)
* The construction with **two folds along parallel lines** (see below) gives the following: If $F$ is good then for any point $x\in \partial F$ the restriction of $dist\_x$ to $\partial F$ does not have local minima except $x$. (This property holds for any shape $C^2$-close to a round disc.)
* This problem was meant to be an exercise for school students, but I was not able to solve it :). It appears in print in 2008 (in Russian), see problem #5 in [Плоское оригами и длинный рубль.](http://arxiv.org/abs/1004.0545)
* One answer is **accepted, BUT** it only provides a solution for unbounded figures.
| https://mathoverflow.net/users/1441 | When shorter means smaller? | Here's a few thoughts on the question:
First, although this is probably obvious to everyone who's posted already, the region F must be bounded (if it is not $\mathbb{R}^2$). If not, then since it's convex, it must contain an infinite ray, and F must be contained in a half-space since it is convex. Take the projection onto the ray, then map the ray by arclength onto a spiral that can't live in any half-space, for example. This map then cannot be contained inside of F and is clearly length-decreasing.
So assume F is bounded. Then I think we may assume F is compact, by taking its closure. Any 1-Lipschitz map from F to $\mathbb{R}^2$ will extend to a 1-Lipschitz map from of the closure, and if the image lies in F, then its closure will lie in the closure. This probably doesn't help at all.
**Edit: as Anton Petrunin has pointed out, the following argument is bogus:**
Now, the space of 1-Lipschitz maps to $\mathbb{R}^2$ is convex (and I think it is complete in the sup topology). Also, it's an easy exercise to see that any convex combination of 1-Lipschitz maps lying in isometric copies of F also lie in isometric copies of F (convex combinations of isometries give conformal affine maps with dilatation $\leq 1$).
So to prove the claim for a given region F, we need "only" prove that extremal maps, i.e. ones which are not convex combinations of other maps, are contained in an isometric copy of F. I'm not sure if this helps, but there might be some literature on the convex structure of 1-Lipschitz maps which one could possibly exploit.
| 3 | https://mathoverflow.net/users/1345 | 8424 | 5,773 |
https://mathoverflow.net/questions/8414 | 2 | The question is pretty self-explanatory; we are dealing with the standard symplectic structure on ℝ4.
Some background: I'm reading the thesis "Lagrangian Unknottedness of Tori in Certain Symplectic 4-manifolds" by Alexander Ivrii, which proves that all embedded Lagrangian tori in ℝ4 are smoothly isotopic (and, in fact, Lagrangian isotopic). It uses lots of pseudoholomorphic curves. Obviously, if the question of this post is answered, together with the paper it will imply that all embedded tori in ℝ4 are smoothly isotopic (in other words, there are no torus knots in ℝ4).
I am told that this is, in fact, true, but that every proof that is known uses symplectic topology and Lagrangian tori. However, I have no idea how to do the question from the title, whether it's easy or hard, or whether it involves any pseudoholomorphic curves.
| https://mathoverflow.net/users/2467 | Why (and whether) is any smooth embedded torus in R^4 isotopic to an embedded Lagrangian torus? | Whoever told you that any embedded torus in R4 is isotopic to a Lagrangian torus was sorely mistaken. Luttinger (JDG 1995) observed the following: The manifolds obtained by doing certain Dehn-type surgeries on a Lagrangian torus in R4 admit symplectic structures. But the result X of the surgery is then always minimal and has the complement of a compact set symplectomorphic to the complement of a compact set in R4, whence a result of Gromov shows that X is actually symplectomorphic to R4. In particular X is simply connected. But for many knot types of tori in R4 (such as those mentioned by Ryan Budney) the results of these surgeries would in some cases not be simply connected, leading to the conclusion that these knot types must not admit Lagrangian representatives.
| 13 | https://mathoverflow.net/users/424 | 8425 | 5,774 |
https://mathoverflow.net/questions/8428 | 5 | We can embed $S^2\times I$ into $\mathbb{R}^3$ by taking a compact 3-ball and removing an open 3-ball from its interior. Taking the boundary gives an embedding $i: S^2\sqcup S^2\hookrightarrow\mathbb{R}^3$, as "a sphere contained inside another sphere". Now it's intuitively clear that this embedding is not ambient-isotopic to the embedding $j$ given by putting these two spheres "side-by-side" in $\mathbb{R}^3$. That is, there is no isotopy $F\_t:\mathbb{R}^3\to \mathbb{R}^3$ with $F\_0=1\_{\mathbb{R}^3}$ and $F\_1\circ i=j$. At least, this looks visually obvious. Assuming my intuition isn't betraying me and this isn't false, what's an elegant way to prove this?
Also, (why) does there (not) exist an embedding $S^2\times I\hookrightarrow\mathbb{R}^3$ whose boundary *is* ambient-isotopic to the "side-by-side" embedding? How about for $S^{n-1}\times I\hookrightarrow\mathbb{R}^n$?
| https://mathoverflow.net/users/1182 | Freeing a sphere from within a sphere | **For your first question**, look at complements (a "fundamental" technique in analyzing ambient isotopies/ambient homeomorphisms; e.g. the "[knot group](http://en.wikipedia.org/wiki/Knot_group)"):
Let the union of the initial two spheres be $S$, and the union of the final two spheres be $T$. An isotopy on $\mathbb{R}^3$ taking $S$ to $T$ in particular ends with a homeomorphism from $\mathbb{R}^3$ to itself taking $S$ to $T$, hence producing a homeomorphism from $U=\mathbb{R}^3\setminus S$ to $V=\mathbb{R}^3\setminus T$. But $U$ has only one of its three connected components contractible (the inside of the inner sphere), whereas $V$ has two of its three components contractible (the insides of the two spheres), a contradicting that they be homeomorphic.
$\Big($One way to prove the other components are not contractible is that they have non-trivial second homotopy and homology groups, as exhibited by the elements represented by the spheres themselves.$\Big)$
**For your third (and second) question**, in generality, here's an argument that there is no embedding from $M=S^{n-1}\times I$ to $\mathbb{R}^n$ with boundary ambient isotopic to the "side-by-side" embedding of
$S\_0\sqcup S\_1$. By any letter $S$ I denote a copy of $S^{n-1}$. For comfort of the imagination, think $n=2$:
**(1)** Up to ambient isotopy, there are only three embeddings of $S\_0\sqcup S\_1$ in $\mathbb{R}^n$ : the "avocado$^+$ type" embeddings $A^+$ with $S\_0$ inside $S\_1$, the "avocado$^-$ type" embeddings $A^-$ with $S\_1$ inside $S\_0$, and the "side-by-side type" embeddings $B$ with $S\_0$ and $S\_1$ "next to" each other.
$\Big($That they fall into these three types is just a case analysis; that any two embeddings in the same "type" are ambient isotopic is also easy: (a) first translate and dilate until the $S\_1$ embeddings match up, and then being in the same case means the two $S\_2's$ are embedding in the same component of the compliment of $S\_1$, so (b) you can move them around in that component until they match up, too.$\Big)$
**(2)** $A^\pm$ are not ambient isotopic to $B$, by counting contractible components of their complements, as in the answer to your first question: $A^\pm$ has one, $B$ has two.
**(3)** An embedding from $M=S\times I$ to $R^n$ cannot have $\partial M = B$ up to ambient isotopy.
$\Big($ Why? Here's one proof, which I found kind of fun: Suppose $\partial M = B$, i.e. is in the "side-by-side" arrangement. Since $M$ is connected, $int(M)=M\setminus \partial M$ must lie outside both spheres of $\partial M$, since if it lies inside one, it can't touch the other. Let $x$ be the center of of the $S\_1$ component of $\partial M$. Now $M$ is compact, so it has a point $p$ which is of maximal distance from $x$. Such a $p$ must be a boundary point of
$M$, since at an interior point, we could move slightly in a direction away from $x$ and stay inside $M$. But $p$ can't be on $S\_1$ or $S\_2$ either, since their outer sides are "padded" by the open set $int(M)$... this a Euclidean geometry argument which I'll stop rigorizing here. So $p$ is nowhere, a contradiction.$\Big)$
**(4)** By process of elimination, $\partial M$ must be ambient isotopic to $A^\pm$, which by (2) is *not* ambient isotopic to the "side-by-side" embedding $B$, which is what you wanted.
| 9 | https://mathoverflow.net/users/84526 | 8430 | 5,776 |
https://mathoverflow.net/questions/8426 | 4 | Using Vandermonde's identity we know:
$\sum\_{i=0}^k \binom{k}{i}\binom{n-k}{n/2-i} = \binom{n}{n/2}$.
I'm interested in how close the alternating sum is to 0 when k << n. I.e.,
$\sum\_{i=0}^k (-1)^i\binom{k}{i}\binom{n-k}{n/2-i}$.
| https://mathoverflow.net/users/2476 | alternating sums of terms of the Vandermonde identity | So, you are interested in $f(n,k)=\sum\_{i=0}^k (-1)^i\binom{k}{i}\binom{2n-k}{n-i}$.
Simple manipulations show $f(n,k)=\frac{k!(2n-k)!}{(n!)^2}\left[\sum\_{i=0}^n (-1)^i \binom{n}{i}\binom{n}{k-i}\right]$
Now the second factor counts the coefficient of $x^k$ in $(1-x^2)^n$ and therefore if $k$ is odd $f=0$ otherwise $f=(-1)^{\frac{k}{2}}\frac{k!(2n-k)!}{n!(k/2)!(n-k/2)!}$ which is far from zero...
EDIT: On a different note I see the result is a signed generalized Catalan number of degree 2 (I was not aware they satisfied such simple identities). Since usually providing combinatorial interpretations for generalized Catalan numbers is not easy, may I ask in what combinatorial context did you face the above calculation?
| 5 | https://mathoverflow.net/users/2384 | 8431 | 5,777 |
https://mathoverflow.net/questions/8445 | 25 | **EDIT (Harry):** Since this question in its original form was poorly stated (asked about topology rather than graph theory), but we have a list of Topology books in the answers, I guess you should go ahead and post with regard to that topic, rather than graph theory, which the questioner can ask again in another topic.
EDIT (David): The original question was asking for places to learn topology with an eye towards applying it to computer science (artificial neural networks in particular)
| https://mathoverflow.net/users/2482 | Learning Topology | 1. A self study course I can recommend for topology is **Topology** by **JR Munkres** followed by **Algebraic Topology** by **A Hatcher** (freely and legally available online, courtesy of the author!). But that is if you want to be able to really do the math in all its glorious detail. **Basic Topology** by **MA Armstrong** is a shortcut and a very good one at that.
2. The closest I can get to what you are asking for here is [Network Topology](http://en.wikipedia.org/wiki/Network_topology%20%22Network%20Topology%22). Is that what you mean? In that case you should be probably be looking at topological graph theory. Wikipedia also tells me that something called Computational Topology exists, but that is probably not what you are looking for.
Hope that helps!
| 20 | https://mathoverflow.net/users/262 | 8452 | 5,789 |
https://mathoverflow.net/questions/8451 | 12 | Let $A \to B$ be a ring extension.
What is the definition of $B/A$ étale ?
When $A$ is a field, do we get a nice characterization ?
| https://mathoverflow.net/users/2330 | Definition of étale for rings | You say that a ring homomorphism $\phi: A \to B$ is **étale** (resp. **smooth**, **unramified**), or that $B$ is étale (resp. smooth, unramified) over $A$ is the following two conditions are satisfied:
* $A \to B$ is **formally étale** (resp. **formally smooth**, **formally unramified**): for every square-zero extension of $A$-algebras $R' \to R$ (meaning that the kernel $I$ satisfies $I^2 = 0$) the natural map $$\mathrm{Hom}\_A(B, R') \to \mathrm{Hom}\_{A}(B, R)$$ is bijective (resp. surjective, injective).
* $B$ is **essentially of finite presentation** over $A$: $A \to B$ factors as $A \to C \to B$, where $A \to C$ is of finite presentation and $C \to B$ is $C$-isomorphic to a localization morphism $C \to S^{-1}C$ for some multiplicatively closed subset $S \subset C$.
The second condition is just a finiteness condition; the meat of the concept is in the first one. Formal smoothness is often referred to as the infinitesimal lifting property. Geometrically speaking, it says that if the affine scheme $\mathrm{Spec} \ B$ is smooth over $\mathrm{Spec} \ A$, then any map from $\mathrm{Spec} \ R$ to $\mathrm{Spec} \ B$ lifts to any square-zero (and hence any infinitesimal) deformation $\mathrm{Spec} \ R'$. Moreover, if $\mathrm{Spec} \ B$ is étale over $\mathrm{Spec} \ A$ this lifting is unique.
Differential-geometrically, unramifiedness, smoothness and étaleness correspond to the tangent map of $\mathrm{Spec} \ \phi$ being injective, surjective and bijective, respectively. In particular, étale is the generalization to the algebraic case of the concept of local isomorphism.
There are two references you might want to consult. The first one, in which you can read all about the formal properties of these morphisms, is Iversen's "[Generic Local Structure of the Morphisms in Commutative Algebra](https://doi.org/10.1007/BFb0060790)" (Springer LNM 310). The second one, Hartshorne's book "[Deformation Theory](https://doi.org/10.1007/978-1-4419-1596-2)" (Graduate Texts in Mathematics 257), will give you a lot of information about the geometry; section 4 of chapter 1 talks about the infinitesimal lifting property.
EDIT: The [EGA](https://en.wikipedia.org/wiki/%C3%89l%C3%A9ments_de_g%C3%A9om%C3%A9trie_alg%C3%A9brique) definition of étale morphism of rings is slightly different from the above, in the sense that it requires finite presentation, not just locally of finite presentation: see the comments below.
| 18 | https://mathoverflow.net/users/1797 | 8455 | 5,792 |
https://mathoverflow.net/questions/8460 | 22 | In Weibel's *An Introduction to Homological Algebra*, the Chevalley-Eilenberg complex of a Lie algebra $g$ is defined as $\Lambda^\*(g) \otimes Ug$ where $Ug$ is the universal enveloping algebra of $g$. The differential here has degree -1.
I have been told that the Chevalley-Eilenberg complex for $g$ is
$C^\*(g) = \text{Sym}(g^\*[-1])$,
the free graded commutative algebra on the vector space dual of $g$ placed in degree 1. The bracket $[,]$ is a map $\Lambda^2 g \to g$ so its dual $d : = [,]\* \colon g^\* \to \Lambda^2 g^\*$ is a map from $C^1(g) \to C^2(g)$. Since $C^\*(g)$ is free, this defines a derivation, also called $d$, from $C^\*(g)$ to itself. This derivation satisfies $d^2 = 0$ precisely because $[,]$ satisfies the Jacobi identity.
Finally, Kontsevich and Soibelman in *Deformation Theory I* leave it as an exercise to construct the Chevalley-Eilenberg complex in analogy to the way that the Hochschild complex is constructed for an associative algebra by considering formal deformations.
The first is a chain complex, the second a cochain complex, and what do either have to do with formal deformations of $g$?
| https://mathoverflow.net/users/1676 | Chevalley Eilenberg complex definitions? | The first complex, from Weibel, is a projective resolution of the trivial $\mathfrak g$-module $k$ as a $\mathcal U(\mathfrak g)$-module; I am sure Weibel says so!
Your second complex is obtained from the first by applying the functor $\hom\_{\mathcal U(\mathfrak g)}(\mathord-,k)$, where $k$ is the trivial $\mathfrak g$-module. It therefore computes $\mathrm{Ext}\_{\mathcal U(\mathfrak g)}(k,k)$, also known as $H^\bullet(\mathfrak g,k)$, the Lie algebra cohomology of $\mathfrak g$ with trivial coefficients.
The connection with deformation theory is explained at length in Gerstenhaber, Murray; Schack, Samuel D. Algebraic cohomology and deformation theory. Deformation theory of algebras and structures and applications (Il Ciocco, 1986), 11--264, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 247, Kluwer Acad. Publ., Dordrecht, 1988.
In particular neither of your two complexes 'computes' deformations: you need to take the projective resolution $\mathcal U(\mathfrak g)\otimes \Lambda^\bullet \mathfrak g$, apply the functor $\hom\_{\mathcal U(\mathfrak g)}(\mathord-,\mathfrak g)$, where $\mathfrak g$ is the adjoint $\mathfrak g$-module, and compute cohomology to get $H^\bullet(\mathfrak g,\mathfrak g)$, the Lie algebra cohomology with coefficients in the adjoint representation. Then $H^2(\mathfrak g,\mathfrak g)$ classifies infinitesimal deformations, $H^3(\mathfrak g,\mathfrak g)$ is the target for obstructions to extending partial deformations, and so on, exactly along the usual yoga of formal deformation theory à la Gerstenhaber.
By the way, the original paper [Chevalley, Claude; Eilenberg, Samuel
Cohomology theory of Lie groups and Lie algebras.
Trans. Amer. Math. Soc. 63, (1948). 85--124.] serves as an incredibly readable introduction
to Lie algebra cohomology.
| 21 | https://mathoverflow.net/users/1409 | 8461 | 5,797 |
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