parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/4562 | 24 | Many people know that there is a (3×3) [nine lemma](http://en.wikipedia.org/wiki/Nine_lemma) in category theory. There is also apparently a sixteen lemma, as used in [a paper on the arXiv](http://arxiv.org/abs/0706.3547) (see page 24). There might be a twenty-five lemma, as it's mentioned satirically on Wikipedia's nine lemma page.
Are the 4×4 and 5×5 lemmas true? Is there an n×n lemma? How about even more generally, if I have an infinity × infinity commutative diagram with all columns and all but one row exact, is the last row exact too? For all of these, if they are true, what are their exact statements, and if they are false, what are counterexamples?
*Note*: There are a few possibilities for what infinity × infinity means -- e.g., it could be **Z**×**Z** indexed or **N**×**N** indexed. Also, in the **N**×**N** case, there are some possibilities on which way arrows point and which row is concluded to be exact.
| https://mathoverflow.net/users/1079 | Is there an infinity × infinity lemma for abelian categories? | Yes, there is an $n\times n$ lemma, and even an $\mathbb N\times\mathbb N$ lemma. The spectral sequence argument that Reid gives works. Another elementary proof uses the salamander lemma, a result of George Bergman's that I [blogged about at SBS](http://sbseminar.wordpress.com/2007/11/13/anton-geraschenko-the-salamander-lemma/). It's exactly the same as the proof of the $3\times 3$ lemma I wrote up there.
Here's a counterexample to the $\mathbb Z\times\mathbb Z$ lemma. If you read about the salamander lemma, you'll understand how I came up with it. All non-zero maps are the identity
$$\require{AMScd}
\begin{CD}
0 @>>> 0 @>>> 0 @>>> 0 @>>> 0\\
@. @VVV @VVV @VVV @VVV\\
0 @>>> 0 @>>> 0 @>>> \mathbb{Z} @>>> 0\\
@. @VVV @VVV @VVV @VVV\\
0 @>>> 0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 \\
@. @VVV @VVV @VVV @VVV\\
0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 @>>> 0
\end{CD}
$$
Extend the diagram by copies of $\mathbb Z$ down and to the left, and put $0$'s everywhere else. All columns are exact, and all rows except one (the one with a single $\mathbb Z$ in it) are exact.
| 18 | https://mathoverflow.net/users/1 | 4575 | 3,065 |
https://mathoverflow.net/questions/4591 | 8 | Can anyone prove that a [Weyl Algebra](http://en.wikipedia.org/wiki/Weyl_algebra) is not isomorphic to a matrix ring over a division ring?
| https://mathoverflow.net/users/494 | Proof a Weyl Algebra isn't isomorphic to a matrix ring over a division ring | Notation: The Weyl algebra is
$$k[x\_1, x\_2, \ldots, x\_n, \partial\_1, \partial\_2, \ldots, \partial\_n]$$
with the obvious relations.
The Weyl algebra doesn't contain any division rings larger than $k$, and it is infinite dimensional over $k$. So, assuming you don't allow infinite matrices, that's a proof.
How to see that it doesn't contain any division ring larger than $k$? I just need to show that any nonconstant differential operator is not invertible. One way to see this is to notice that multiplying differential operators multiplies symbols, and the symbol of a nonconstant differential operator is a nonconstant polynomial.
| 16 | https://mathoverflow.net/users/297 | 4593 | 3,073 |
https://mathoverflow.net/questions/4590 | 15 | Let $A$ be a commutative integral domain, with fraction field $K$. Let $T$ be a torsion-free finitely generated $A$ module, so $T \otimes\_A K$ is a finite dimensional vector space $V$. Let $T^\*$ be the set of $y$ in the dual vector space, $V^\*$, such that $\langle x, y \rangle \in A$ for every $x \in T$.
Under what hypotheses on $A$ can I conclude that $T^\*$ is a free $A$-module? My current conjecture is that this holds whenever $A$ is a [UFD](http://en.wikipedia.org/wiki/Unique_factorization_domain). (Of course, it trivially holds if $A$ is a PID.)
Here are a few ideas of mine. Define the rank of $T$ to be $\dim\_K T \otimes\_A K$. I can show that, if $A$ is a UFD, then $T^\*$ is free for $T$ of rank $1$. For any $T$, we can make a short exact sequence
$$0 \to S \to T \to U \to 0$$
where $S$ is rank $1$ and $U$ is torsion free with rank one less than $T$. So we have
$$0 \to U^\* \to T^\* \to S^\* \to \mathrm{Ext}^1(U,A) \to \cdots$$.
This looks like a good start, but I don't know how to control that Ext group. I suspect that one of you does!
This is motivated by Kevin Buzzard's [question](https://mathoverflow.net/questions/3270/which-rings-are-subrings-of-matrix-rings) about matrix rings.
| https://mathoverflow.net/users/297 | When are dual modules free? | The dual module of a finitely generated module is *reflexive*, that is, $M^{\*\*}=M$, and reflexives are awfully close to projectives. Specifically, if $R$ is a Noetherian domain, then a module is projective if $Ext^i(M,R)=0$ for all $i>0$, and its reflexive if $Ext^i(M,R)=0$ for $i=1,2$.
It is also worth noting that **every** reflexive is the dual of some module, specifically of $M^\*$. Therefore, your question amounts to "for what rings is every reflexive module free?" In this light, its very similar to the question of when every projective module is free.
From the above Ext criterion, its clear that if the global dimension of $R$ is less than or equal to 2, that being reflexive is the same as being projective. I would go so far as to conjecture the converse is true: that if gldim of R is 3 or more, that there is a non-projective module which is reflexive (and hence it is non-free).
If this conjecture is true, then the answer to your question is "rings with global dimension 2 or less, such that every projective is free". Of course, its not immediately clear what these are, but its a start.
| 15 | https://mathoverflow.net/users/750 | 4608 | 3,083 |
https://mathoverflow.net/questions/4612 | 20 | I'm wondering what the landscape looks like for proofs of Hironaka's desingularisation theorem.
>
> Are there many proofs in the literature?
>
>
> Is there a commonly accepted simplest bare-knuckle proof out there
> that might be considered a pleasant read for people outside of
> algebraic geometry?
>
>
> Would that still be Hironaka's original manuscript?
>
>
>
| https://mathoverflow.net/users/1465 | Hironaka desingularisation theorem -- new proofs in literature? | There's an article by Herwig Hauser in the Bulletin of the AMS: *[The Hironaka theorem on resolution of singularities (Or: A proof we always wanted to understand)](https://doi.org/10.1090/S0273-0979-03-00982-0)* (Bull. Amer. Math. Soc. **40** (2003), 323-403 )
which is aimed at giving an accessible account (I must admit I didn't read most of it though).
Mircea Mustata also gave a 5 lecture course at the 2008 Park City summer school. He did not prove every fact he needed, but the lectures contained all of the relevant ideas. If I remember correctly, the main idea was to find the right invariants for varieties (I think it has something to do with multiplier ideals, but I could be wrong) and just show that one can make the invariants "smaller" by choosing the right blow-ups to do at each step. Someone correct me if I am wrong, please. I don't know if those notes can be downloaded anywhere, but they will appear in the summer school proceedings.
As far as I have been informed, the original proof of Hironaka is extremely difficult to understand.
| 17 | https://mathoverflow.net/users/321 | 4616 | 3,086 |
https://mathoverflow.net/questions/4583 | 9 | Sorry if this is easy/well-known, I don't know much algebraic topology and I'm just curious about this question.
One of the easier proofs of the Brouwer fixed-point theorem (we'll say for n = 2 for concreteness in the terminology, although it generalizes) proceeds by establishing Sperner's lemma and noting that a continuous map gives us a Sperner labeling on a triangulation of our disk. All the "real work" in this proof is in establishing Sperner's lemma, which can be done completely combinatorially.
So I know that the [Lefschetz fixed-point theorem](http://en.wikipedia.org/wiki/Lefschetz_fixed-point_theorem) generalizes Brouwer's theorem, and that it applies to more general spaces than Brouwer. Is there a (relatively) simple combinatorial statement, analogous to Sperner, that can be easily shown to imply the Lefschetz theorem, at least on some large class of topological spaces?
| https://mathoverflow.net/users/382 | Analogue of Sperner's lemma for Lefschetz theorem? | Probably there aren't any such combinatorial statements in the literature, the difficulty being that the statement of Lefschetz theorem involves looking at the induced action on the homology.
There is a combinatorial lemma - Tucker lemma - which implies the Borsuk-Ulam theorem. Both the Sperner Lemma and Tucker lemma had many generalizations (Ky Fan contributed many, there is a famous deep extension of Sperner lemma by Shapley), although there is less activity in this line of reaserch these days.
| 7 | https://mathoverflow.net/users/1532 | 4618 | 3,088 |
https://mathoverflow.net/questions/4625 | 5 | Suppose $F$ has discrete Fourier transform $(a\_n)$ where $a\_n=0$ unless $n=2^k$ for some $k > 0$, in which case $a\_n=1/k$ (or $a\_n=1/k^2$ if you want: I'm happy with anything polynomial). What sort of regularity conditions does $F$ have? Is it Holder continuous, or not?
To be explicit:
$$
F(x)=\sum\_{k=1}^\infty k^{-2} \exp(ix2^k)
$$
for example.
More generally, I'm interested in two dimensional (discrete) Fourier transforms: is there a good reference for this sort of thing?
| https://mathoverflow.net/users/406 | Regularity of sparse Fourier transforms | If $0 < \alpha < 1/2$ then a continuous function on the circle is $\operatorname{Lip}\_\alpha$ only if the Fourier coefficients satisfy $a\_n = {\rm O}( n^{-\alpha})$; this is in Katznelson's book (Chapter I, Corollary 4.6) for instance.
[*EDIT (2013-07-10)*: at the time I thought this was "iff" but a comment points out that I misremembered; in any case, for lacunary series such as the one in the question, a lot more is known than in the general case; see e.g. Katznelson Chapter V for the basics.]
So the function you defined above isn't going to be Hölder continuous for any positive exponent, even though it's clearly continuous (absolutely convergent Fourier series).
Off the top of my head, I don't know of any particularly good source for the higher-dimensional stuff.
| 7 | https://mathoverflow.net/users/763 | 4626 | 3,093 |
https://mathoverflow.net/questions/4578 | 24 | There is a well-known topological proof of the fact that subgroups of free groups are free. Many people, myself included, think it is easier and more natural than the purely algebraic proofs which had been given earlier by (IIRC) Nielsen and Schreier. It goes as follows:
1. If $S$ is any set, then the CW-complex $X$ obtained as the wedge of $|S|$ circles is a graph whose fundamental group is isomorphic to $F(S)$, the free group on $S$.
2. If $H$ is a subgroup of $F(S)$, then by covering space theory $H$ is the fundamental group of a covering space $Y$ of $X$.
3. The covering space of any graph is again a graph.
4. Any graph has the homotopy type of a wedge of circles, so the fundamental group of $Y$ is again free.
My question is: to what extent is there an analogous proof of the result with "free" replaced everywhere by "free abelian"?
In the case of a finitely generated free abelian group -- say $G \cong \mathbb{Z}^n$ -- there is at least an evident topological *interpretation*. Namely, we can take $X$ to be the $n$-torus
(product of $n$ copies of $S^1$), and then observe that any covering space of a torus is homeomorphic to a torus of dimension $d$ cross a Euclidean space of dimension $n-d$, hence homotopy equivalent to a torus of rank $d <= n$.
Even in this case though I'd like some assurance that the proof of this topological fact does not use the algebraic fact we're trying to prove. (Is for instance some basic Lie theory relevant here?)
Then, what happens if the free group has arbitrary rank? Can we take $X$ to be a direct limit over a family of finite-dimensional tori? Does the proof go through?
| https://mathoverflow.net/users/1149 | Subgroups of free abelian groups are free: a topological proof? | The "free group" proof rests on proving that that the fundamental group of a graph is free. For the analogue we'd need to essentially prove that the fundamental group of a "torus" (something that looks like a quotient of a vector space by a discrete subgroup) is free abelian. A sketch:
Given a real vector space V, we can put the direct limit topology on it (so that subsets are closed if and only if their intersection with any finite dimensional subspace is closed). This is a contractible topological group.
If A is a free abelian group, then A is a discrete subgroup of the associated real vector space (ℝ ⊗ A) and the quotient space has fundamental group A. Any covering space is a quotient of (ℝ ⊗ A) by a discrete subgroup B of A.
So the question boils down to showing: Any discrete subgroup of a vector space (with the direct limit topology) is free abelian.
Let's say that a partial basis is a set S of elements of B such that
* S is linearly independent, and
* S generates B ∩ Span(S).
Then partial bases are a partial order under containment, and Zorn's lemma implies that there is a maximal element S. I claim that S is a basis of B as a free abelian group.
S is linearly independent by construction, so it generates a free abelian group, and hence it suffices to show that it generates all of B. If b in B is not in S, then it is not in Span(S). Let S' be (S ∪ {b}). Then Span(S')/Span(S) is a 1-dimensional vector space and the image of B ∩ Span(S') must be discrete, because otherwise Span(S') would contain an element (rb + v) for v in Span(S) that we could use to generate a non-discrete subset of B. (If v is a combination of w1...wn in S, then it suffices to check that any subgroup of the finite-dimensional space Span(w1...wn,b) requiring more than n generators is indiscrete.)
Thus any lift of a generator of B ∩ Span(S') would extend to a larger generating set, contradicting maximality.
(My apologies for the comment last night, which this morning looks snarkier than I intended. I'm a fan of using this topological reasoning for free groups myself, because it compartmentalizes the proof into much more understandable pieces. In particular, I don't think I'd really understand a purely algebraic proof that an index n subgroup of a free group on m generators is free on nm - n + 1 generators.)
| 20 | https://mathoverflow.net/users/360 | 4634 | 3,099 |
https://mathoverflow.net/questions/4630 | 9 | Some time ago I was told there's an interesting classical Satake correspondence which I will write as
$$[\mathop{\mathrm{disk}} \Rightarrow G] \,\backslash\, [\mathop{\mathrm{disk}^\times} \Rightarrow G] \,/\, [\mathop{\mathrm{disk}} \Rightarrow G] \,=\, X\_\*/W \,=\, G^\vee\mbox{-reps}$$
(where $G$ is reductive, $\mathrm{disk}$ could be the spectrum of either $\mathbb Z\_p$ or $k[[t]]$ and $\Rightarrow $ denotes algebraic morphism) and its categorified/geometric version (equivariant perverse sheaves on affine grassmanian of $G$ form the tensor category of representations of $G^\vee$).
I think I'm missing the larger context here, though. I don't mean the context of perverse sheaves, geometric Langlands, etc. On the contrary, I feel like I miss any intuition for classical representation theory. Why would statements like this be interesting?
I wasn't able to find anything in [wikipedia](http://en.wikipedia.org/wiki/Satake_correspondence) or [nLab](http://ncatlab.org/nlab/search?_form_key=4209aaac3878c2dab6d077f12cd5c0363e94a24c&query=satake+correspondence).
One thing I know is that the correspondence allows us to construct the Langlands dual group in a natural way. But still, it would be interesting to know if it's a part of larger picture and if there are related results.
>
> **Question:** is there an intuition for Satake correspondence that would make its statement obvious?
>
>
>
| https://mathoverflow.net/users/65 | Explanation for Satake correspondence | What you have written above isn't classical Satake; it's the generalized Bruhat decomposition. Classical Satake is a much more interesting theorem, which says that the Hecke algebra of $G(\mathcal{K})$ over $G(\mathcal{O})$ (the compactly supported $G(\mathcal{O})$ bi-invariant functions on $G(\mathcal{K})$ with convolution multiplication) is isomorphic to the representation *ring* of $G^\vee$.
Why is this interesting? Because the Hecke algebra $L^2[G(\mathcal{O})\backslash G(\mathcal{K})/G(\mathcal{O})]$ is the endomorphism algebra of $L^2(G(\mathcal{K})/G(\mathcal{O}))$, so there's a bijection between $W$-orbits on $T^\vee$ and representations of $G(\mathcal{K})$ appearing in the $L^2$ above.
| 10 | https://mathoverflow.net/users/66 | 4639 | 3,102 |
https://mathoverflow.net/questions/4580 | 17 | I'm beginning to run into work where I have to do a significant amount of learning of math by myself, with a book rather than with a teacher. Now, I do know that doing problems tends to be the best way to learn these things, but my question is a bit different.
How do you pace yourselves when you're learning new mathematics? Or reading a paper, let's say. Any pointers for reading large swathes of mathematics? (I'm running into this problem with my category theory references).
One answer per post.
| https://mathoverflow.net/users/429 | Pacing for learning new material | Here are my advise that are mostly based on experience:
If I start a totally new math, especially in the graduate level. I'd give my self at least 1.5-2 years (especially if it's an area in which a lot of lot of reading is involved.. say algebraic geometry). One of the things I find important, is not getting frustrated that you haven't learned to the level you need to learn. It is indeed frustrating, but when I look back in my PhD years.. I indeed took 2 years of just reading before even being able to start any new ideas of my own. You just don't have enough knowledge to make a ground-breaking mathematics and until you do be patient and learn it and try new ideas and create new examples (out of the book). I personally hardly answer excercises in the book, but created my own questions and tried to answer them first. If you were able to make new ideas and even publish a paper or two during these 1.5-2years then thats a bonus but you shouldn't feel incapable during that time.
The other advise I'd give is the references. Never stick to one or even just two single reference. Especially if they are the references that are difficult to digest. You should collect as many of the references in that area of mathematics as possible. If this is math that people have done already, chances are there are many many references about it that you don't probably know yet. And never read linearly through the references (esp. textbooks). I don't know of any professional mathematicians that has actually finished reading an entire book that he has not written himself. Switch from the different references as much as possible and try to get as much goodies from each as possible. There is no ONE book in homological algebra and different mathematician find different book suitable, you should find one that writes in a style your prefer and every now and then look at the other books as well. There is NO one book in commutative algebra, you can read certain characterization of testing for flatness of modules/algebras in commutative algebra books but you can hardly find ALL of them in ONE single book and some of them don't even have all of the proof.
Third advise, is to collaborate or speak as often as possible with people very knowledgeable in the topic. Attend seminar and conferences in that area of mathematics, even if you don't understand a pea. Chances are you learn something new or you learn about a question you think you find interesting in that area that is unanswered. There are some people who are knowledgeable in some area and make me feel like sh\*t when I speak to them, I tend to avoid them.. but sometimes I mingle nevertheless. For me, true authentic mathematicians must good educators as well, so that if they find someone not knowledgeable in one thing they actually help him become knowledgeable instead of making him feel bad about it.
| 19 | https://mathoverflow.net/users/1245 | 4641 | 3,103 |
https://mathoverflow.net/questions/4640 | 14 | Supervector spaces look a lot like the category of representations of $\mathbb{Z}/2\mathbb{Z}$ - the even part corresponds to the copies of the trivial representation and the odd part corresponds to the copies of the sign representation. This definition gives the usual morphisms, but it does not account for the braiding of the tensor product, which I've asked about [previously](https://mathoverflow.net/questions/4427/what-is-the-conceptual-significance-of-supercommutativity).
Since the monoidal structure on $\text{Rep}(G)$ comes from the comultiplication on the group Hopf algebra, which in this case is $H = \mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$, it seems reasonable to guess that the unusual tensor product structure comes from replacing the usual comultiplication, which is $\Delta g \mapsto g \otimes g$, with something else, or alternately from placing some extra structure on $H$. What extra structure accounts for the braiding?
| https://mathoverflow.net/users/290 | Are supervector spaces the representations of a Hopf algebra? | The answer is yes, but comultiplication is not what you change. The symmetrizer (or braiding as you call it) is given by an $R$-element in $H \otimes H$ that makes $H$ into a [triangular Hopf algebra](http://en.wikipedia.org/wiki/Quasitriangular_Hopf_algebra). (The Wikipedia article says quasitriangular; triangular means that plus that the modified switching map $\widetilde{R}$ is an involution.)
The $R$ element is the correction to the usual switching map $x \otimes y \mapsto y \otimes x$. You can solve for it directly in terms of the usual description of the switching map on supervector spaces, and worry later about what axioms it satisfies. Let $1$ and $a$ be the group elements of $H$ and let $\epsilon$ and $\sigma$ be the dual vectors on $H$ which are the trivial and sign representations. Then
we want
$$(\epsilon \otimes \epsilon)(R) = 1 \quad (\epsilon \otimes \sigma)(R) = 1 \quad (\sigma \otimes \epsilon)(R) = 1 \quad (\sigma \otimes \sigma)(R) = -1,$$
because that is the correction in the modified switching map $x \otimes y \mapsto (-1)^{|x||y|} y \otimes x$. You can solve the linear system of equation to obtain
$$R = \frac{1 \otimes 1 + a \otimes 1 + 1 \otimes a - a \otimes a}{2}.$$
| 22 | https://mathoverflow.net/users/1450 | 4644 | 3,106 |
https://mathoverflow.net/questions/4596 | 19 | It is well-known that
A: The series of the reciprocals of the primes diverges
My question is whether property A is in some sense a truth strongly tied to the nature of the prime numbers.
Property A tells us that the primes are a rather *fat* subset of $\mathbb{N}$. Is there a way to define a topology on $\mathbb{N}$ such that every dense subset of $\mathbb{N}$ (under this topology) corresponds to a *fat* subset of the natural numbers?
What do you think about this?
| https://mathoverflow.net/users/1593 | On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ... | Yes, it's possible. Define the closed sets to be the sets the sum of whose reciprocals converges, together with $\mathbb{N}$. This collection of subsets is closed under arbitrary intersection and finite union, so it does form the closed sets of a topology.
A subset of $\mathbb{N}$ is dense in this topology if its closure is $\mathbb{N}$, in other words, if it is not contained in any smaller closed set -- in other words, if it is not contained in any set the sum of whose reciprocals converges. This is equivalent to the sum of its reciprocals not converging.
| 55 | https://mathoverflow.net/users/468 | 4645 | 3,107 |
https://mathoverflow.net/questions/4648 | 30 | Picking a specific basis is often looked upon with disdain when making statements that are about basis independent quantities. For example, one might define the trace of a matrix to be the sum of the diagonal elements, but many mathematicians would never consider such a definition since it presupposes a choice of basis. For someone working on algorithms, however, this might be a very natural perspective.
What are the advantages and disadvantages to choosing a specific basis? Are there any situations where the "right" proof requires choosing a basis? (I mean a proof with the most clarity and insight -- this is subjective, of course.) What about the opposite situation, where the right proof never picks a basis? Or is it the case that one can very generally argue that any proof done in one manner can be easily translated to the other setting? Are there examples of proofs where the only known proof relies on choosing a basis?
| https://mathoverflow.net/users/1171 | When to pick a basis? | One answer to your question is already hinted at in the question. At the level of algorithms, basis-independent vector spaces don't really exist. If you want to compute a linear map $L:V \to W$, then you're not really computing anything unless both $V$ and $W$ have a basis. This is a useful reminder in our area, quantum computation, that has come up in discussion with one of my students. In that context, a quantum algorithm might compute $L$ as a unitary operator between Hilbert spaces $V$ and $W$. But the Hilbert spaces have to be implemented in qubits, which then imply a computational basis. So again, nothing is being computed unless both Hilbert spaces have distinguished orthonormal bases. The reminder is perhaps more useful quantumly than classically, since serious quantum computers don't yet exist.
On the other hand, when proving a basis-independent theorem, it is almost never enlightening (for me at least) to choose bases for vector spaces. The reason has to do with data typing: It is better to write formulas in such a way that the two sides of an incorrect equation are unlikely to even be of the same type. In algebra, there is a trend towards using bases as sparingly as possible. For instance, there is widespread use of direct sum decompositions and tensor decompositions as a way to have partial bases.
I think that your question about examples of proofs can't have an explicit answer. No basis-independent result needs a basis, and yet all of them do. If you have a reason to break down and choose a basis, it means that the basis-independent formalism is incomplete. On the other hand, anything that is used to build that formalism (like the definition of determinant and trace and the fact that they are basis-independent) needs a basis.
There is an exception to the point about algorithms. A symbolic mathematics package can have a category-theoretic layer in which vector spaces don't have bases. In fact, defining objects in categories is a big part of the interest in modern symbolic math packages such as Magma and SAGE.
| 34 | https://mathoverflow.net/users/1450 | 4649 | 3,110 |
https://mathoverflow.net/questions/4607 | 0 | Can anyone explain what Iwahori order is? All I know is that it is mentioned [here](https://mathoverflow.net/questions/3483/intuitive-example-of-a-jacobson-radical/3490#3490).
| https://mathoverflow.net/users/494 | Explanation and Definition of Iwahori order | To translate Rob's answer above, you take a ring $R$ and ideal $I$, and take the ring of consisting of matrices with coefficients in $R$ such that the image in $R/I$ is upper-triangular. This is a subring of all matrices.
The name "Iwahori order" is borrowed from the name "Iwahori subgroup" which refers to the invertible elements in this ring (usually in the case where R is a valuation ring, and I the valuation ideal).
| 2 | https://mathoverflow.net/users/66 | 4656 | 3,115 |
https://mathoverflow.net/questions/4558 | 6 | [Jonah's question](https://mathoverflow.net/questions/3519/uniformization-theorem-in-higher-dimensions "MO link") makes me wonder: What is with uniformization in algebraic/arithmetic geometry? E.g. [this article by Faltings](http://archive.numdam.org/article/CM_1983__48_2_223_0.pdf "numdam link") seems to be about that, the Shimura-Taniyama statement too, Mochizuki discusses a p-adic version of Fuchsian uniformization of hyperbolic curves. Do you know surveys or expositions of the theme? Or is 'uniformization' a mistaken concept in the context of arithmetic geometry (a remark in Faltings article sounds as if saying that)?
| https://mathoverflow.net/users/451 | Uniformization in algebraic/arithmetic geometry? | Presumably, you want to look at uniformizations of curves, since blow-ups make it difficult to classify covers of higher dimensional varieties. Scheme-theoretically, there doesn't seem to be a good notion of uniformization, but one can get good arithmetic results using analytification.
You already mentioned a couple references. For the complex-analytic view, I think any book on modular forms (e.g., Shimura) should have some treatment. Mochizuki's work in the nonarchimedean setting has a counterpart in the maximally degenerate case of Mumford curves. There is a book by Gekeler and van der Put on uniformizations of these (essentially using the Bruhat-Tits building for SL2). I think Darmon and Dasgupta have done some arithmetic work using this "upper half plane" uniformization. It historically originates from Tate's work around 1960 on rigid analytic spaces (apparently, Grothendieck expressed doubt that such a theory would exist).
| 3 | https://mathoverflow.net/users/121 | 4670 | 3,126 |
https://mathoverflow.net/questions/4653 | 14 | I know of several results to the effect that two triangulated categories are equivalent categories (usually one coming from algebra and one coming from topology). However, it's never been clear to me how to classify the possible different triangulated structures lifting a given "graded" category (a category enriched in graded abelian groups).
For example, how many different triangulated structures can there be with underlying category the derived category of Z-modules? All objects in this category are equivalent to direct sums of Z-modules concentrated in different degrees.
As another example, we have the category of "Z/2-graded abelian groups", which is equivalent to the derived category of differential graded modules over a Laurent polynomial ring on a generator in degree 2. There's another, equivalent-looking, homotopy category of modules over the periodic complex K-theory spectrum KU. These have equivalent underlying categories but I have never been clear on whether the triangulated structures are the same.
| https://mathoverflow.net/users/360 | Classifying triangulated structures on a graded category | Generally speaking a unique lifting does not exist and I believe it is open as to what the possible liftings can be.
As an example of the non-uniqueness consider a slight variant of the particular category you asked about - namely $K^b(\mathbb{Z})$ the homotopy category of bounded complexes of finitely generated abelian groups. Considering this as a graded category it has at least 2 different structures of triangulated category. The usual one and one we denote $K^b(\mathbb{Z})^{-}$ where we declare that
$$X \stackrel{u}{\to} V \stackrel{v}{\to} Y \stackrel{w}{\to} \Sigma X$$
is a triangle if and only if
$$X \stackrel{-u}{\to} V \stackrel{-v}{\to} Y \stackrel{-w}{\to} \Sigma X$$
is a triangle with respect to the usual triangulation. The point that was implicit (before this edit) is that these triangulations do not agree. This also works for $D^b(\mathbb{Z})$, the bounded derived category of finitely generated abelian groups, indeed one just needs to suppose the two categories agreed and consider [TR3] applied to a diagram obtained by mapping
$$\mathbb{Z} \stackrel{3}{\to} \mathbb{Z} \to K(3) \to \Sigma \mathbb{Z}$$
to
$$\mathbb{Z} \stackrel{-3}{\to} \mathbb{Z} \to K(3) \to \Sigma \mathbb{Z}$$
by $(1,-1,h,1)$ where $h$ is some map provided by [TR3] but one can check this is ridiculous.
If you want more lifts of the suspended category with the usual suspension to a triangulated category I feel like the answer should be there aren't any but I don't know a proof or where this is written down if it is true. I think I know how to make some headway on the problem more generally if one rigidifies the situation a little but this is still work in progress.
More generally one can play this game with any unit in the degree zero part of the central ring of our suspended category (in plainer language we can twist by automorphisms of the identity functor commuting with suspension). In [this](http://www.math.ucla.edu/~balmer/research/Pubfile/TriangulationS.pdf) note Balmer shows that this leads to suspended categories with infinitely many triangulations. I believe that it is unknown if there can be other sorts of triangulations at least on an indecomposable category.
I currently can't think of any examples that arise naturally and I am not familiar enough with the one you mention to be able to say at the moment whether or not the triangulations agree. If the suspensions agree then I'd guess they are equivalent up to something of the form above. I'd be interested to know if they weren't.
**Update** Here is the sort of thing I now have in mind for the bounded derived category of finitely generated abelian groups. As a warning I haven't really thought this through too thoroughly.
The statement I think one should be after is that no matter the triangulation there is no choice involved in the isomorphism class of the cones - more precisely if one is given $f\colon X\to Y$ then $\mathrm{cone}(f)$ is up to isomorphism independent of the triangulation. It should follow from this that the sign trick that gives you the second triangulation is all the wiggle room one has.
Denote by $\#X$ the non-negative integer $\lvert \{i\in \mathbb{Z} \;\vert\; H^i(X)\neq 0\} \rvert$, and try to run an induction on this quantity. If $X$ is the suspension of a group and we are given $f\colon X\to Y$ we can desuspend and assume it is in degree zero since this will just rotate the triangle. Complete $f$ to a triangle
$$ X \stackrel{f}{\to} Y \to Z \to \Sigma X$$
Applying $\mathrm{Hom}(\mathbb{Z},-)$ gives an exact sequence of abelian groups which tells us that (where since we are not assuming the standard triangulation the use of cohomological notation is slightly abusive but just denotes the corresponding graded piece) $H^i(Z) \cong H^i(Y)$ for $i\leq -2$ and $i\geq 1$, that $H^0(Z) \cong \mathrm{coker} \; H^0(f)$, and that $H^{-1}(Z)$ is determined up to extension by
$$0 \to H^{-1}(Y) \to H^{-1}(Z) \to \mathrm{ker}\;H^0(f) \to 0$$
but there is still no choice involved here since the element of $\mathrm{Ext}^1$ giving this extension is determined by the "degree 1 part" (what I mean by this is if $X$ has torsion it might map up to the piece of $Y$ in degree 1) of $f$ (I think - this is the bit I didn't really check). Thus the graded pieces of $Z$ are determined up to isomorphism and hence so is $Z$ since everything is a sum of its cohomology.
Now if we assume we have uniqueness for maps out of guys with less than $n$ pieces and let $\#X = n$ and suppose we are given $f\colon X\to Y$ which completes to a triangle with cone $Z$. Write $X$ as $X' \oplus X''$ where $\#X' < n$. Using the octahedral axiom we get a triangle (we can actually cook up 4 in this way but we only need 1)
$$W \to \Sigma^{-1}Z \to X' \to \Sigma W$$
and since $\#X' < n$ the inductive hypothesis implies $Z$ is uniquely determined up to isomorphism.
I'd be interested if anyone has any ideas for an invariant that would detect the difference between these two triangulations on $D^b(\mathbb{Z})$ - they are really almost the same in every way I can think of except that the mapping axiom can fail between the triangles in each of them.
| 10 | https://mathoverflow.net/users/310 | 4674 | 3,129 |
https://mathoverflow.net/questions/4669 | 23 | When people work with infinite sets, there are some who (with good reason) don't like to use the Axiom of Choice. This is defensible, since the axiom is independent of the other axioms of ZF set theory.
When people work with finite sets, there are still some people who don't like to use the "finite Axiom of Choice" -- i.e., they don't like to pick out a distinguished element of a set, or a distinguished isomorphism between a set with $n$ elements and $\{0, 1, \ldots, n-1\}$ (without some algorithm to pick it, that is). This is still an aesthetically defensible position, since oftentimes proofs that proceed that way don't give as much insight as proofs that don't use finite choice. But ZF set theory allows us to do this for finite sets!
Is there a general framework in which we can disallow, if we so choose, the method of distinguished element? I have a hunch that the fact that, even for a finite-dimensional vector space $V$, $V$ and $V^\ast$ aren't naturally isomorphic is the first step towards the "right answer," but I don't really see where to go from there.
(To be clear, as Ilya points out, I'm referring primarily to set theory; I know that/how category theory tells us about the non-naturality of the dual vector space isomorphism, in particular. My question is, is there something that either subsumes this or parallels it for more combinatorial constructions?)
| https://mathoverflow.net/users/382 | Can we disallow finite choice? | You might want topos theory. A topos is something like the category of sets, but the internal logic of a general topos is much weaker than ZF; it need not even be Boolean. An example of a topos is the category of sheaves of sets on a topological space. You can use topos theory to turn voodoo-sounding statements of constructive mathematics into ordinary mathematical statements which you can understand: for example "a nonempty set S might not have any elements" becomes "a sheaf S which is not the empty sheaf might not have any global sections" (or possibly "might not have local sections everywhere"), which is of course true. A canonical reference on this subject is Mac Lane & Moerdijk, Sheaves in Geometry and Logic.
| 22 | https://mathoverflow.net/users/126667 | 4677 | 3,131 |
https://mathoverflow.net/questions/4678 | 4 | Hey,
Is countinng the number of shortest paths in a weighted directed acyclic graph with nonnegative weights #P-complete?
If so, is there a proof I can read somewhere?
Thanks
| https://mathoverflow.net/users/1612 | Number of Shortest paths problem | No, it's easily solved in polynomial time. Suppose you have some designated start vertex s from which you want to count shortest paths. Then, if D(v) denotes the distance from s to v and N(v) denotes the number of shortest paths from s to v, then these two quantities may be computed by a single pass through all the vertices of the DAG, in a topological ordering: if v=s then D(v)=0 and N(v)=1, else D(v) is the minimum of D(w)+lenghth(w,v) over edges from w to v and N(v) is the sum of N(w) over vertices w achieving that minimum value.
The assumption of nonnegative weights is irrelevant.
| 11 | https://mathoverflow.net/users/440 | 4681 | 3,134 |
https://mathoverflow.net/questions/4585 | 7 | Let $f : X \rightarrow Y$ be a finite type morphism of Noetherian schemes. The valuative criterion for properness runs as follows. Suppose that for any DVR $R$ with fraction field $K$ that any $K$-valued point of $X$ lying above an $R$-valued point of $Y$ extends uniquely to an $R$-valued point of $X$. Then $f$ is proper.
Does it suffice to check instead that any $K$-valued point of $X$ lying above an $R$-valued point of $Y$ extends to an $R'$-valued point of $X$, where $R'$ is the integral closure of $R$ in a finite extension $K'$ of $K$?
| https://mathoverflow.net/users/1594 | Valuative criterion for properness | Yes.
I claim that, for any $K'$ point of $X$, if this point extends both to an $R'$ point and an $K$ point, then it extends to a $R$ point. This obviously proves the result.
Let $x'$ be the closed point of $\mathrm{Spec}(R')$. Let $\mathrm{Spec}(A)$ be an affine neighborhood of the image of $x'$. Then the $R'$ point must factor through $\mathrm{Spec}(A)$. The $K$ point also factors through $\mathrm{Spec}(A)$, as it has the same image as the $K'$ point.
We can now transform the question into algebra. The algebraic statement is that we have a map $A \to K'$ which factors through both $K$ and $R'$. But then the image of this map must lie in $K \cap R' = R$. So the map factors through $\mathrm{Spec}(R)$, and our $K$ point extends to an $R$ point.
| 7 | https://mathoverflow.net/users/297 | 4686 | 3,137 |
https://mathoverflow.net/questions/4687 | 3 | According to [Wikipedia](http://en.wikipedia.org/wiki/Regular_representation): If G is a finite group and K is the complex number field, the regular representation is a direct sum of irreducible representations, in number at least the number of conjugacy classes of G.
Can anyone prove this? (Only an honors level student, so please try to keep it as simple as possible)
| https://mathoverflow.net/users/494 | Number of irreducible representations | If I remember correctly, that statement can be proven via the equivalence between the (group) representations of $G$ and the (algebra) representations of the group algebra $K[G]$. If $K=\mathbb{C}$ is the field of complex numbers (or any other field of characteristic 0, or in general if $\textrm{char} K$ does not divide the order of $G$) then by Maschke's Theorem $K[G]$ is semisimple, and moreover it is Artinian (by finite-dimensionality) so by Artin-Wedderburn it decomposes (as a module over itself) as a direct sum of matrix rings over some division ring over $K$. If $K=\mathbb{C}$, there are no finite dimensional division rings over $\mathbb{C}$ so each summand is a ring of matrices over the complex numbers. The irreducible representations of the group are in one to one correspondence with left ideals of this ring, which are well known (look for instance [here](http://books.google.co.uk/books?id=oWzoOWp_tWgC&pg=PA76&lpg=PA76&dq=regular+representation+of+a+group+ring&source=bl&ots=qo3ZDIsInc&sig=TYRFfwKXCKQdjX83IGH9aeVlzvk&hl=en&ei=Bk33SqmPLYKj4Qbc7ozeAw&sa=X&oi=book_result&ct=result&resnum=10&ved=0CDEQ6AEwCQ#v=onepage&q=regular%20representation%20of%20a%20group%20ring&f=false). A simple dimension counting argument should convince you that you have at least one ideal for each conjugacy class.
| 6 | https://mathoverflow.net/users/914 | 4691 | 3,139 |
https://mathoverflow.net/questions/4689 | 9 | There is a theorem of Schwede and Shipley which classifies categories of modules over an A∞ ring spectrum as those stable presentable (∞,1)-categories with a compact generator. Suppose I allow my A∞ rings to "have many objects", that is, I consider categories of the form FunSp(Iop, Sp) where Sp is the category of spectra, I is a small Sp-enriched category (in some appropriate sense) and Funsp denotes the category of Sp-enriched functors. Is there a classification of which stable presentable categories can be obtained in this way? Is it possible that *all* stable presentable categories are of this form?
| https://mathoverflow.net/users/126667 | Stable presentable categories as module categories | According to the abstract of <http://arxiv.org/abs/math/0108143> (Schwede & Shipley, *Classification of Stable Model Categories*), they deal with the case of stable model categories (=stable presentable (∞,1)-categories, I suppose) which have a set of compact generators, and show they are the same as model categories of functors from spectral-enriched categories.
(Edited, in the light of Reid's comment, to include the hypothesis of *compact* generators.)
| 8 | https://mathoverflow.net/users/437 | 4696 | 3,142 |
https://mathoverflow.net/questions/4699 | 8 | I am looking for concrete examples of cancellative, left reversible semigroups. Left reversible semigroups are also called "Ore semigroups". See [this wikipedia page](http://en.wikipedia.org/wiki/Special_classes_of_semigroups) for the definition of a left reversible semigroup. Of course, commutative semigroups are automatically left reversible, and I am looking for non-commutative examples.
Please also mention if these semigroups arise in an interesting setting.
| https://mathoverflow.net/users/1193 | Examples of left reversible semigroups | Some examples and further references (and an interesting setting) are given in this paper by Laca:
<http://arxiv.org/abs/math/9911135>
See section 1.1, pages 2-3.
| 5 | https://mathoverflow.net/users/1119 | 4707 | 3,150 |
https://mathoverflow.net/questions/4695 | 15 | Let S be a finite set of primes in Q. What, if anything, do we know about K3 surfaces over Q with good reduction away from S? (To be more precise, I suppose I mean schemes over Spec Z[1/S] whose geometric fibers are (smooth) K3 surfaces, endowed with polarization of some fixed degree.) Are there only finitely many isomorphism classes, as would be the case for curves of fixed genus? If one doesn't know (or expect) finiteness, does one have an upper bound for the number of such K3 surfaces X/Q of bounded height?
| https://mathoverflow.net/users/431 | K3 surfaces with good reduction away from finitely many places | Some thoughts.
There are no such varieties when S = 1. This is a consequence of a theorem of Fontaine, MR1274493 (Schémas propres et lisses sur Z).
I think that one should only expect finitely many such varieties for any fixed S. Let me give an argument that uses every possible conjecture I know. There may be an unconditional proof, but that would probably require knowing something about K3-surfaces.
I first want to claim that the ramification at primes q|S is "bounded" independently of X. The corresponding fact for elliptic curves will be that the power of the conductor for each q|N is bounded by 2 (if p > 3) or (if p = 2 or 3) by some fixed number I can't remember.
The most obvious argument along these lines is to consider the representation on inertia I`_`q acting on the p-adic etale cohomology groups H^2(X). These correspond to Galois representations with image in GL`_`22(Z`_`p). The argument I have in mind for elliptic curves works directly in this case, providing that one has "independence of p" statement for the Weil-Deligne representations at q (quick hint: the image of wild inertia divides the gcd of the orders of GL`_`22(F\_p) over all primes p). This may require the existence of semi-stable models, which one certainly has for elliptic curves, but I don't know for K3-surfaces.
The next step is to use a Langlands-type conjecture. The p-adic representation V on H^2(X) may be reducible, but at least we know that each irreducible chunk will correspond to an irreducible Galois representation of Q into GL`_`n(Z`_`p) for some n (at most 22). Each of these, conjecturally, will correspond to a cuspidal automorphic form of fixed weight and level divisible only by q|S. Moreover, from the previous paragraph, the level will be *bounded* at q|S. Thus there will only be finitely many representations which can occur as H^2(X) for any K3-surface X/Z[1/S]. (Maybe I am assuming here that the Galois representation acting on H^2(X) is semi-simple --- let us do so, since this is a conjecture of Grothendieck and Serre.)
Finally, I want to deduce from any equality H^2(X) = H^2(X') that X is (essentially) X'.
From the Tate conjecture we deduce the existence of correspondences X~~>X' and X'~~>X over Q whose composition induces an isomorphism on H^2(X) --- and now hopefully some knowledge of the geometry of K3 surfaces is enough to show that these sets of "isogenous" K3 surfaces form a finite set.
---
EDIT:
As Buzzard points out, I obscured the fact in the last paragraph that some more arithmetic may be necessary. What I meant to say is that understanding isogeny classes of K3's over Q will first require understanding isogeny classes over C, and hopefully this second task will be the hard part.
As David points out, the Torelli theorem for K3 will surely be relevant here. I think there can be non-isomorphic isogenous K3s, however. If one takes an isogeny of abelian surfaces A->B then one can presumably promote this to an isogeny of the associated Kummer surfaces.
---
EDIT:
Here is another thought. Deligne proves the Weil conjecture for K3 surfaces:
<http://www.its.caltech.edu/~clyons/DeligneWeilK3trans.pdf>
The philosophy is that there should be an inclusion of motives H^2(X) --> H^1(A) tensor H^1(A) for some abelian variety A (possibly of some huge but uniformly bounded dimension, like 2^19). It may be possible (conjecturally or otherwise) to reduce your question to the analogous statement for A, for which it is known. (Prop 6.5 is relevant here). It may well be possible to show that the variety A is defined over Z[1/2S], for example. I could make this edit more coherent but I'm off to lunch, so treat this as a thought fragment.
| 8 | https://mathoverflow.net/users/nan | 4710 | 3,153 |
https://mathoverflow.net/questions/4589 | 14 | I've seen in college that some functions are not computable.
The proof for that was the case of Halt(x,y) function.
The thing is, the proof used a very artificial (IMHO) case
which is evaluating the function in it's own program number.
In fact the whole idea of the Halt function is quite self-referencial...
I'd like to know if there is a more "normal" function which
cannot be computed.
Can someone show me a function which is non-computable but
does not refer to itself?
I hope I made myself clear... if not, just comment and I'll explain.
Thanks!
| https://mathoverflow.net/users/1592 | Is there a non self-referencing non-computable function? | Check this [blog post](http://xorshammer.com/2008/09/04/a-geometrically-natural-uncomputable-function/).
| 14 | https://mathoverflow.net/users/158 | 4725 | 3,165 |
https://mathoverflow.net/questions/4724 | 15 | How does random noise in the digital world typically look?
Suppose you have a memory of n bits, and suppose that a "random noise" hits the memory in such a way that the probability of each bit being affected is at most t.
What will be the typical behavior of such a random digital noise? Part of the question is **to define "random noise" in the best way possible**, and then answer it for this definition.
The same question can be asked for a memory based on a larger r-letters alphabet.
In particular (as Greg mentioned) I am interested to know: **Will the "random noise" behave typically in a way which is close to be independent on the different bits? or will it be highly correlated?** or pehaps the answer depends on the size of the alphabet and the value of t?
The source of this question is an easy fact about quantum memory which asserts that if you consider a random noise operation acting on a quantum memory with n qubits, and if the probability that every qubit is damaged is a tiny real number t, then typically the noise has the following form: with large probability nothing happens and with tiny probability a large fraction of qubits are harmed. (So in this case, random noise is highly correlated.)
I made one try for the digital (binary) question but I am not sure at all that it is the "correct" definition for what "random noise" is. I considered a random permutation of all the 2^n binary strings of length n subject to the condition that at most t bits are flipped. I will be happy if somebody can think about the case of larger alphabet and also offers the conceptually "correct" framework for this question.
Perhaps the best way to ask this question (following Jason's and Greg's remarks) is this: What is the appropriate way to define a "random" probability distribution on {0,1}^n and on {0,1,...,r}^n.
Given such a definition we the would like to understand the typical behavior of such a random probability distribution with the additional property that the probability for the ith coordinate to be non zero is at most t for every i.
**Updated formulation:** One way to solve the question was proposed by Greg: The space of stochastic maps on a finite probability space is a convex polytope. So you can choose one using standard Euclidean measure. (In our case, the finite probability space is the uniform probability space on {0,1}^n, and more generally, the uniform probability space on {0,1,...,r}^n.)
The question is (stated for the binary case) this: Let t>0 be a small real number. How does a random stochastic map, conditioned on the property that a proportion of t bits are corrupted looks like. Is it also the case, as Greg expects, that (like in the quantum case,) with large probability nothing happens and with small probability many bits are corrupted.
I will be happy to see how to do the calculation precisely or heuristically, or perhaps to find it in the literature.
**Update:** The notion of Arikan's **polar coding** (See e.g. [this paper](http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=05560464)) is somewhat related to this question.
| https://mathoverflow.net/users/1532 | How Does Random Noise Typically Look? | I think people might be misinterpreting the question.
The easy fact that I think Gil has in mind is that if you randomly choose a quantum noise operation with certain properties, then for most choices of that noise operation, although the total error rate will be very low, the errors will be highly correlated. The orthodox interpretation is that this is an artificial way to choose a quantum noise model.
One analogous classical question is as follows: If you randomly choose a stochastic map on the probability space of bit strings, with similar properties, what will it look like?
Gil proposes a randomly chosen permutation subject to the condition that at most t bits are flipped. I am not sure that that is really a good analogy to the quantum noise model that he proposes.
The Poisson process that flips bits is a specific model of random noise on bit strings. It is not a randomly chosen model of noise.
---
Gil has added the more specific question of why a random stochastic map conditioned on few errors will have the "ambush noise" property that he describes for a random unitary operator. There is a more complicated version, in which each bit separately has a bound on error, and a simpler version, where you just demand that at most $tn$ of the $n$ bits are changed. I will look at the simpler version. As a first step, the different inputs to the stochastic map do not communicate in this question. Each bit string, for instance the string of 0s, is sent to another bit string according to a distribution which is uniform on the simplex of all $2^n$ bit strings. So I will concentrate on just what happens to the 0 string.
One remark is that the fate of just the 0 string is actually statistically identical, whether you choose a random stochastic map classically or a random unitary operator quantumly. The first column of a random unitary operator is a random vector in $\mathbb{C}P^{N-1}$ (with $N = 2^n$ for us). The induced map from $\mathbb{C}P^{N-1}$ to the $(N-1)$-simplex of distributions is a toric moment map, and a famous fact (due to Archimedes for $\mathbb{C}P^1$) is that toric moment maps are measure-preserving. The difference between a random stochastic and a random unitary only first appears with pairwise correlations, and when $N$ is large only very-high-order correlations are significant. If Gil has an argument for random unitaries, this remark suggests that it also applies for random stochastics.
More directly, a distribution on bit strings induces a distribution on Hamming weights of bit strings. This is a linear map from a huge simplex $\Delta\_{N-1}$ to the smaller simplex $\Delta\_n$ whose corners are labelled by the Hamming weights $0,1,\ldots,n$. We are interested in the push-forward of uniform measure. Let $p\_0,\ldots,p\_n$ be barycentric coordinates on $\Delta\_n$; $p\_k$ is also just the probability that the random bit string has weight $k$. Then push-forward of uniform measure on $\Delta\_{N-1}$ is proportional to $f(p) \propto \prod\_k p\_k^{\binom{n}{k}}$. So, in this induced measure on $\Delta\_n$, there is an enormous statistical attraction to the corners with middle values of $k$.
Now let's impose the restriction that the total error is at most $tn$, in other words that
$$\sum\_k kp\_k \le tn.$$
This cuts the simplex $\Delta\_n$ by a hyperplane. At this point I'll switch to a rough calculation. If $t \ll \frac12$, and if you maximize the log of the probability density $\log f(p)$, the maximum puts most of the probability in the weights $k \approx n/2$. That's because the corresponding term in $\log f(p)$ is $\binom{n}{k}(\log p\_k)$. The logarithmic dependence on $p\_k$ is outweighed by the sizes of the coefficients. There is a also a geometric factor if your are close to a sharp corner of the cut simplex; however after a logarithm this geometric factor is of order $n$, which is again much smaller than the binomial coefficients.
| 10 | https://mathoverflow.net/users/1450 | 4736 | 3,171 |
https://mathoverflow.net/questions/919 | -3 | I want to prove that the positive powers of two, mod 10m, cycle with period 4\*5m-1. It's simple to prove that the powers of FIVE cycle with this period (2 is a primitive root mod powers of five), but how do you make the leap to powers of TEN?
I'm sure it's something simple -- perhaps related to the Chinese Remainder Theorem -- but I don't see the connection yet.
Thanks for the help.
| https://mathoverflow.net/users/565 | Cycle Length of the Positive Powers of Two Mod Powers of Ten | The answer I like best is based on the proof in the "physics forums" thread linked to in the comments above. I wrote about it in detail here: <http://www.exploringbinary.com/cycle-length-of-powers-of-two-mod-powers-of-ten/>
| -1 | https://mathoverflow.net/users/565 | 4742 | 3,174 |
https://mathoverflow.net/questions/4745 | 12 | $\DeclareMathOperator\GL{GL}$The ordinary Grassmannian of k-planes in n-space is a coset space for $\GL\_n$.
It is $\GL\_n$ mod a maximal parabolic. Here there is a nice basis given by Schubert varieties, which can be indexed by Young diagrams that fit in a $(k)\times(n-k)$ box. The structure constants for the cup product are then given by Littlewood–Richardson numbers.
My question: is there a similarly nice picture for Grassmannians of arbitrary simple groups? Here the ordinary Grassmannian is replaced by $G/P$ where $G$ is a simple group and $P$ is a maximal parabolic. There are still Schubert varieties in this case, but I don't know how to say anything about the cup product.
| https://mathoverflow.net/users/788 | Littlewood–Richardson–Type Rule for Cohomology Ring of Grassmannians | As yet, such a nice rule has only been formulated in the case that $G/P$ is minuscule or co-minuscule. See
* Hugh Thomas, Alexander Yong, *A combinatorial rule for (co)minuscule Schubert calculus*, Adv. Math. **222** (2009), no. 2, 596–620, doi:[10.1016/j.aim.2009.05.008](https://doi.org/10.1016/j.aim.2009.05.008), arXiv:[math/0608276](https://arxiv.org/abs/math/0608276)
for details.
| 15 | https://mathoverflow.net/users/297 | 4747 | 3,177 |
https://mathoverflow.net/questions/4743 | 5 | Hey,
I need to count the number of paths from node $s$ to $t$ in a weighted directed acyclic graph s.t. the total weight of each path is less than or equal to a certain weight $W$. I have an algorithm to do it in $O(nW)$ using dynamic programming. Let $N\_W(s,t)$ denote the number of such paths (with weight less than $W$) from $s$ to $t$.
It seems like doing it in polynomial time would be hard, since for instance if I take the strategy of calculating $N\_W(s,u)$ for $u$ in the paths between $s,t$, then I would have to keep track of how many paths are there from $s\rightarrow p,p\in parents(u)$ for every weight in $\{1,\dots,W\}$ which alone would take $O(nW)$ time and space.
So my question is what is the complexity of this problem?
Thanks
Edit: Changed $O(W)$ to $O(nW)$ for the running time of the dynamic programming approach. Correction thanks to David Eppstein.
| https://mathoverflow.net/users/1612 | Number of paths equal less than equal to a certain length | The problem is and is not NP-hard. If the length of the input is defined by writing the weights in binary, then it is indeed NP-hard and in fact #P-complete. Say that the target weight $W$ is written in base 10. Suppose further that the graph is a string, as in Reid's construction, with two edges from $i$ to $i+1$, one weight 0 and the other with weight some integer in base 10 with only 0s and 1s in its digits. Also, choose all of the weights so that they cannot add together with any carries. Then the $k$th digit $d$ of the target weight $W$ is a Boolean clause, which says that of those edges that have a non-zero $k$th digit, exactly $d$ are used used. It is not hard to build general Boolean circuits with these clauses.
Also, to clarify a couple of points about this: You can convert from weight at most $W$ to weight exactly $W$ by subtracting off weight at most $W-1$; so in this quick argument it is only #P-hard with a Turing reduction and not a Cook reduction. And of course I'm really showing that the knapsack counting problem is #P-hard.
On the other hand, if the length is defined by writing the weights in unary, to favor small integer values of the weights, then the dynamic programming algorithm given in the question is polynomial time.
Which complexity model is more appropriate depends on the context of the problem, which was not given.
| 5 | https://mathoverflow.net/users/1450 | 4755 | 3,183 |
https://mathoverflow.net/questions/4756 | 3 | Is it true that there are no projective curves which are also flag manifolds? If so, why?
| https://mathoverflow.net/users/1648 | Flag Varieties - Projective Curves | The projective line is both a curve of genus $0$ and a flag variety for $SL\_2$. This is the only example. This is true for about a zillion reasons:
Flag varieties are rational (because of the Bruhat decomposition.) Curves, of genus $>0$, are not.
Flag varieties have transitive group actions. Curves of genus $\geq 2$ do not (see my answer [here](https://mathoverflow.net/questions/3296/projective-curves-which-are-principal-bundles/3301#3301).)
If you look at the classification of flag varieties, there are only finitely many of any given dimension. In paritcular, there is only one example of dimension $1$.
| 10 | https://mathoverflow.net/users/297 | 4759 | 3,186 |
https://mathoverflow.net/questions/4765 | 5 | What is the exact relationship between Lie groups and Lie algebras? I know it's not bijective because all commutative Lie groups have isomorphic Lie algebras.
| https://mathoverflow.net/users/1648 | Lie Groups and Lie Algebras | Up to isomorphism, there is one *simply connected* Lie group for every Lie algebra. Indeed, there is also a homomorphism of simply connected Lie groups for every homomorphism of the corresponding Lie algebras so one gets an equivalence of categories this way.
This pans out nicely in yr commutative example: the simply connected abelian groups are just the Lie algebras under addition. All other abelian Lie groups have a non-simply connected torus component and look like $T^k\times\mathbb{R}^{n-k}$.
To get the remaining Lie groups into the picture, recall that the universal cover of any Lie group is a simply connected Lie group.
| 10 | https://mathoverflow.net/users/1143 | 4770 | 3,191 |
https://mathoverflow.net/questions/4782 | 7 | Using the [axioms](http://en.wikipedia.org/wiki/Triangulated_category#Definition) for a triangulated category, is it possible to prove the following:
>
> $A\stackrel{0}{\to}B\to A\oplus B\to$ is a distinguished triangle.
>
>
>
From the first axiom, the map `0:A-->B` extends to its cone, but there is no guarantee I see that the direct sum fits into a triangle. If it does, however, clearly they are (should be?) isomorphic.
I have tried using the universal properties of the direct sum, in that finite coproducts and finite products coincide in additive categories, so that I have two diagrams, and extended each of these diagrams into triangles in every way I can imagine, but I think I'm just getting lost in the plethora of sequences. `0:A-->B` extends to a triangle `A-->B-->X-->` and so I can get things like $X\to A\oplus B\to \Sigma^{-1}X\cong X$, but by moving away from triangles, I have lost notions of exactness (so that this sequence of maps merely commutes..)
I ask this because the proof of Lemma 3.3(2) of [this paper](http://www2.math.uni-paderborn.de/fileadmin/Mathematik/AG-Krause/publications%5Fkrause/support.pdf) seems to use this without reference.
| https://mathoverflow.net/users/707 | Splitting in triangulated categories | So if I understand correctly the question you wanted to ask was:
Is it true that a triangle $$X \stackrel{u}{\to} Y \stackrel{v}{\to} Z \stackrel{w}{\to} \Sigma X$$ is split if and only if one of $u$, $v$, or $w$ is zero. The answer to this is yes.
It is clear (I think I can add details if someone wants) that if the triangle is split then one map must be zero (basically since we have an epi composing to zero). Conversely suppose that $w$ is zero, which is sufficient since we can always just rotate. Now we know by the axioms that $$Z \stackrel{-1}{\to} Z \to 0 \to \Sigma Z$$ and hence $$0 \to Z \stackrel{1}{\to} Z \to 0$$ are triangles (as an exercise check that any sequence of this form given by an isomorphism is necessarily a distinguished triangle), and $$X \stackrel{1}{\to} X \to 0 \to \Sigma X$$ is also a triangle. It is easy to check then that so is the direct sum $$ X \to X\oplus Z \to Z \stackrel{0}{\to} \Sigma X$$. The identity maps on $X$ and $Z$ then induce a map via [TR3] from this to the original triangle (since we have $w=0$ this trivially satisfies the necessary commutativity to apply [TR3]) and since two of the maps are isomorphisms so is the third. Hence any triangle as above with $w=0$ is isomorphic to one obtained by summing triangles on identity maps.
Proof of the claim that the direct sum of triangles is a triangle:
Let $X \to Y \to Z \to \Sigma X$ and $X' \to Y' \to Z' \to \Sigma X'$ be two distinguished triangles. We can complete the map $X\oplus X' \to Y \oplus Y'$ to a triangle $$X\oplus X' \to Y \oplus Y' \to Q \to \Sigma(X\oplus X')$$. We then get two diagrams whose rows are triangles by projecting to the factors $X, Y$ and $X', Y'$ respectively which we can complete to maps of triangles by the mapping axiom [TR3]. The maps $Q\to Z$ and $Q\to Z'$ such obtained induce a map $Q\to Z\oplus Z'$ by the universal property which gives a map from the triangle $X\oplus X' \to Y\oplus Y' \to Q$ to the pretriangle (a pretriangle is one where the maps compose to zero and it plays nicely with homological functors, that is they take it to a long exact sequence, this is clear from the fact that homological functors are additive and it is a sum of distinguished triangles) $X\oplus X' \to Y\oplus Y' \to Z\oplus Z'$. Two of the maps are isomorphims, namely the identities on the terms $X\oplus X'$ and $Y\oplus Y'$ so that the third must be also - this follows from the fact that $Hom(A,-)$ is a homological functor for any $A$, Yoneda, and the 5 lemma. So the triangle in question is isomorphic to a distinguished triangle and hence itself distinguished.
This works in the generality of arbitrary coproducts/products provided the coproducts/products in question exist and we consider a pretriangle to be one which homological/cohomological functors preserving the coproducts/products take to long exact sequences.
I'd recommend reading through the first chapter of Neeman's book Triangulated Categories - this is certainly covered in there as well as a bunch of other facts you might find useful in reading that paper. The reference for this result is Corollary 1.2.7 (it seemed lazy not to check since it is on my shelf) and the proof there is pretty much identical to the one here except that the facts I glossed over are proved earlier.
| 17 | https://mathoverflow.net/users/310 | 4785 | 3,200 |
https://mathoverflow.net/questions/4763 | 9 | Since $\Gamma(N)$ is normal in $\mathrm{SL}(2,\mathbb{Z})$, the quotient group $\mathrm{SL}(2,\mathbb{Z}/N)$ acts on the spaces of cusp forms $S\_k(\Gamma(N))$. How do these spaces decompose into irreducible representations?
I can do the case $N=2$. I'm mostly interested in the case of $N$ a prime.
| https://mathoverflow.net/users/1310 | SL(2,Z/N)-decomposition of space of cusp forms for Gamma(N) | See Theorem 1.0.3 of Jared Weinstein's [phd thesis](http://www.math.ucla.edu/~jared/jswthesis.pdf) (it uses equivariant Riemann Roch).
| 5 | https://mathoverflow.net/users/1253 | 4791 | 3,205 |
https://mathoverflow.net/questions/3131 | 10 | By 'work' I would like the correspondence between fiber functors (to finitely generated projective modules) and algebraic groups to be the same as in the field case.
Specifically, if $A$ is an affine ring, and if $\operatorname{Proj}(A)$ is the category of finitely generated projective A-modules, when can we say that a fiber functor $w:\mathcal{T}\to\operatorname{Proj}(A)$ corresponds to an algebraic group over $A$, where $\mathcal{T}$ is an $A$-linear tensor category.
| https://mathoverflow.net/users/100 | When does Tannakian theory work over affine schemes besides fields? | If I understand your question correctly you are asking whether or not there is a characterization of those A-linear functors C-->Proj(A) which are equivalent to the forgetful functor Rep(G)-->Proj(A), where G is an affine group scheme over A and Rep(G) is the category of representations of G whose underlying A-module is finitely generated projective.
In this case the categories Rep(G) need not be abelian because Proj(A) is in general not abelian, so the classical Tannakian formalism probably won't help you. However, as in the classical case we can rephrase the problem in terms of comodules and Hopf algebras: An affine group scheme G over A is of the form Spec(H) for some Hopf algebra H, and a representation of G is the same as an H-comodule.
The classical characterization of Deligne (found in "Categories Tannakiennes", Grothendieck Festschrift Vol. II) is split up into the following parts:
1) Every faithful exact functor w:C-->Vect(k) is equivalent to a forgetful functor Comod(L(w))-->Vect(k) for some k-coalgebra L(w).
2) If the category C has a symmetric monoidal structure and if w is a strong monoidal functor, then L(w) is a bialgebra.
3) If C is rigid, then L(w) is a Hopf algebra.
For some time now I've been working on a generalization of step 1) to the case of arbitrary rings A, where we replace Vect(k) by Proj(A). I have recently uploaded a paper ([here](http://arxiv.org/abs/0911.0977 "0911.0977")) which contains the following result (see Corollary 9.8):
Let C be an A-linear category (with finite direct sums) and w:C-->A-Mod an A-linear functor whose image is contained in Proj(A). We say that a diagram F:D-->C is w-*rigid* if the colimit of wF:D-->A-Mod is finitely generated and projective. If
a) w reflects isomorphisms,
b) the category el(w) of elements of w is cofiltered (equivalently, w is *flat*),
c) C has colimits of w-rigid diagrams and w preserves them,
then there is a flat coalgebra L(w) such that w:C-->Proj(A) is equivalent to the forgetful functor Comod(L(w))-->A-Mod, where Comod(L(w)) denotes the category of L(w)-comodules whose underlying A-module is finitely generated and projective. Condition a) and c) are necessary conditions. I don't know if b) is a necessary condition.
I have convinced myself that result 2) from above should work at this level of generality, and I think that 3) should not cause any trouble either. In other words, if your category C is a symmetric monoidal category where every object has a dual and if w is a strong monoidal functor satisfying a)-c), then the coalgebra L(w) should be a Hopf algebra, and C is in fact a category of representations of an affine group scheme.
| 8 | https://mathoverflow.net/users/1649 | 4805 | 3,216 |
https://mathoverflow.net/questions/4807 | 16 | Suppose you are a mathematics student who has just graduated and you haven't yet come to graduate school, or maybe you are in your first year of graduate school. Which magazines should you read? I mean general interest magazines, not journals of a specific field, but the kind of magazine that in each volume contain at least a couple of articles that every mathematician could understand (or reasonably try to understand) and be interested in. I don't even know how to describe exactly what I mean. That's why I am asking.
Please post one suggestion per answer.
| https://mathoverflow.net/users/1651 | Which magazines should I read? | Collected answers from Kim Greene and one from Gerald Edgar:
1. I hear Mathematical Intelligencer is good but I have never read it.
2. I have heard that reading things that you don't completely understand is good for mathematicians so I also recommend the Notices of the AMS.
3. *[Plus](http://plus.maths.org/issue52/index.html)* is a math themed magazine. I feel doubtful it would prepare you for graduate school in any way.
4. College Mathematics Journal
5. Mathematics Magazine
6. American Mathematical Monthly
| 15 | https://mathoverflow.net/users/812 | 4809 | 3,219 |
https://mathoverflow.net/questions/4459 | 6 | ### Background
Lagrangian mechanics on $\mathbb R^n$ is usually defined by picking a **Lagrangian** function $L: {\rm T}\mathbb R^n \to \mathbb R$, where ${\rm T}\mathbb R^n = \mathbb R^{2n}$ is the tangent bundle of the configuration space $\mathbb R^n$. Such a function determines the **Euler-Lagrange** equations:
$$ \frac{d}{dt}\left[ \frac{\partial L}{\partial v^i}\bigl( \dot\gamma(t), \gamma(t)\bigr) \right] - \frac{\partial L}{\partial q^i} \bigl( \dot\gamma(t), \gamma(t)\bigr) = 0$$
Here $(v^i,q^i)$ for $i=1,\dots,n$ are the standard coordinates on ${\rm T}\mathbb R^n$, $\gamma: [0,T] \to \mathbb R^n$ is a smooth function, and $\dot\gamma^i(t) = \frac{d\gamma^i}{dt}$. Suppose that the matrix $\frac{\partial^2 L}{\partial v^i\partial v^j}(v,q)$ is invertible for any $(v,q) \in {\rm T}\mathbb R^n$. Then the Euler-Lagrange equations are a nondegenerate second-order differential equation on $\mathbb R^n$. I am interested in the **boundary-value problem** for $L$. Namely, fix $T > 0$ and $q\_1,q\_2 \in \mathbb R^n$; the BVP asks to find the set $C(q\_1,q\_2,T)$ of all paths $\gamma: [0,T] \to \mathbb R^n$ with $\gamma(0) = q\_1$ and $\gamma(t) = q\_2$. Generically, this is a discrete set.
### My question
Suppose that if instead of the Euler-Lagrange equations above, I pick some small parameter $\epsilon$ and consider the differential equation
$$ \frac{d}{dt}\left[ \frac{\partial L}{\partial v^i}\bigl( \dot\gamma(t), \gamma(t)\bigr) \right] - \frac{\partial L}{\partial q^i} \bigl( \dot\gamma(t), \gamma(t)\bigr) = \epsilon\gamma^{(4)}(t)^i $$
where $\gamma^{(4)}(t)^i$ is the $i$th component of the fourth derivative of $\gamma$ with respect to $t$ (the "jounce", a word I just learned from [Wikipedia](http://en.wikipedia.org/wiki/Jounce)). For $\epsilon \neq 0$, the EL equations are a nondegenerate fourth-order differential equation, and so generically solutions to the boundary-value-problem above form a two-dimensional family. To restrict to a discrete set, we should fix more boundary values. Pick $(v\_1,q\_1), (v\_2,q\_2) \in {\rm T}\mathbb R^n$ and $T > 0$, and define $C\_\epsilon(v\_1,q\_1,v\_2,q\_2,T)$ to be the set of solutions $\gamma$ to the $\epsilon$-dependent EL equations with $(\dot\gamma(0),\gamma(0)) = (v\_1,\_1)$ and $(\dot\gamma(T),\gamma(T)) = (v\_2,q\_2)$.
My question is: as $\epsilon \to 0$, in what sense do we have $C\_\epsilon(v\_1,q\_1,v\_2,q\_2,T) \to C(q\_1,q\_2,T)$?
| https://mathoverflow.net/users/78 | What happens to the solutions of a fourth-order boundary-value problem as you turn off the fourth-order coefficient? | I believe that the method of solution to your problem is called the method of "dominant balance", and in this case, "singular dominant balance." If you do a web search for that, you should be able to find the information you need.
This method will you give a perturbative solution to as high a degree as you have the propensity to calculate. You can analyze this solution to answer various questions that you implied in your original question, such as what the decay behavior of the solution is, which continuity and smoothness properties it has, etc...
If you want to study the solutions of a large class of coefficient functions, not just a specific set, you can leave arbitrary constants in a solution "ansatz" and then develop a parameterized family of solutions. Note that the algebraic expressions involved in finding the simple-looking solutions grow exponentially in the number terms which end up simplifying in the end. Computer algebra is needed to find the simplified form of these solutions, lest you go mad and kill many trees.
You may also want to search for "catastrophe theory", which catalogs the types of bifurcations that happen in systems such as you have described. This is a one-dimensional bifurcation problem, which are well-studied.
| 5 | https://mathoverflow.net/users/1653 | 4814 | 3,221 |
https://mathoverflow.net/questions/4812 | 2 | I'm at a sticky spot in my research. Namely, I have a particular fact, and it ought to have a short proof, but the only way I know how to show it is long and drawn out, and I don't like it and worry I might have an error. I'm hoping one of y'all will either see a short proof or respond with "all your questions are answered in [link]". And I'm hoping this isn't too close to "homework question".
I have a linear second-order differential operator $\mathcal D$ on $C^\infty( [0,1], \mathbb R^n)$, where $\mathbb R^n$ has its usual metric, and of the following form:
$$ \mathcal D = \frac{d^2}{dt^2} + B(t) \frac{d}{dt} + C(t) $$
where $B,C$ are $n\times n$ matrix-valued functions on $[0,1]$, $B(t)$ is antisymmetric for each $t$, and $C(t) - C(t)^{\rm T} = B'(t)$, where $C^{\rm T}$ is the transpose of $C$. I happen to know a lot of solutions to $\mathcal D[f] = 0$. In particular, I have two matrix-valued functions $f\\_1(t)$ and $f\\_2(t)$, which satisfy $\mathcal D[f\\_a] = 0$, and also $f\\_1(0) = \delta = f\\_2(1)$ and $f\\_2(0) = 0 = f\\_1(1)$, where $\delta$ is the unit $n\times n$ matrix.
(Incidentally, this implies that the columns of the $f\\_a$ are a basis for the space of solutions of $\mathcal D[f]=0$, so that there are no nonzero solutions with $f(0) = 0 = f(1)$. Indeed, any solution with $\mathcal D[f] = 0$, $f(0) = 0$ is determined by the derivative $f'(0)$, so that there is a linear map $\mathbb R^n \to \mathbb R^n$ sending $v$ to the value $f(1)$ where $f'(0) = v$. But $f\\_2(1) = \delta$, and so $f\\_2'(0)$ is full-rank, and so if $f$ solves the differential equation with $f(0) = 0$, then $f(t) = f\\_2(t)\left(f\\_2'(0)\right)^{-1}f'(0)$.)
Anyhoo, so my question is this. Let $g\\_1(t),g\\_2(t)$ be matrix-valued functions such that:
$$ f\\_1g\\_1 + f\\_2g\\_2 = 0 \text{ and } f\\_1' g\\_1 + f\\_2' g\\_2 = \delta$$
**Prove that $\mathcal D[(g\\_a)^{\rm T}] = 0$.**
For example, when $n=1$, $B(t) = 0$ because there are no antisymmetric $1\times 1$ matrices, and then by Abel's formula the determinant of the matrix $\left(\begin{smallmatrix} f\\_1 & f\\_2 \\\ f\\_1' & f\\_2' \end{smallmatrix}\right)$ is constant. Therefore, $g\\_2$, which is the lower-right corner of the inverse of this matrix, is a constant times $f\\_1$, and $g\\_1$, which is the upper right-hand-corner of the inverse, is a constant times $f\\_2$.
| https://mathoverflow.net/users/78 | In an n-dimensional linear 2nd-order ODE, why is the transpose-inverse to a system of solutions also a solution? | I promised an answer, so I'll sketch it here, but I hope someone can give a better one.
The operator $\mathcal D$ is *self-adjoint* in the following sense. Let $\langle f,g\rangle = \int\_0^1 f(t) \cdot g(t) dt$ be the usual inner-product on $C^\infty([0,1],\mathbb R^n)$. Then if $f(0) = g(0) = 0 = f(1) = g(1)$, we have $\langle \mathcal D[f],g\rangle = \langle f, \mathcal D[g]\rangle$. This follows from integration by parts and the (anti)symmetry of $B,C$.
Define a Green's function to be a matrix-valued function $G(t,s)$ on the square $[0,1]\times [0,1]$ such that $\mathcal D\\_t[\mathcal G] = \delta(t-s)$ (times the identity matrix), and also satisfying $G(0,s) = 0 = G(1,s)$. In particular, $\mathcal G$ is smooth away from the diagonal, and has a corner like $|s-t|$ at the diagonal. $\mathcal G$ is unique if it exists. Because $\mathcal D$ is self-adjoint, switching $G$ is *symmetric* in the sense that $G(t,s)$ is the transpose of $G(s,t)$.
By the usual variation-of-parameters mumbo-jumbo, I can explicitly write down a formula for $G$. Namely, $G(t,s) = f\\_1(t)g\\_1(s) \Theta(t-s) - f\\_2(t)g\\_2(s) \Theta(s-t)$, or something similar. Then use symmetry: $G(s,t) = f\\_1(s)g\\_1(t) \Theta(s-t) - f\\_2(s)g\\_2(t) \Theta(t-s) = (G(t,s))^{\rm T}$, and so $f\\_1(s)g\\_1(t) = (f\\_2(t)g\\_2(s))^{\rm T}$. But $\mathcal D[f\\_1] = 0$, so $(g\\_1)^{\rm T}$ must be a solution as well.
The problems with this argument are:
* Actually going through the variation-of-parameters is tedious and unenlightening.
* I'm not completely sure I believe that $\mathcal G$ is symmetric in the way that I said it is. I mean, it should be, and I believed it until I started doubting myself.
| 1 | https://mathoverflow.net/users/78 | 4815 | 3,222 |
https://mathoverflow.net/questions/616 | 26 | There is a standard way to construct the sheafification of a presheaf on a Grothendieck topology which involves matching families. Details may be found here:
<http://ncatlab.org/nlab/show/matching+family>
In short, there is a functor + sending presheaves to separated presheaves and then separated presheaves to sheaves. So P^++ is always a sheaf.
Gelfand/Manin's Methods of Homological Algebra has a wrong proof that P^+ is a sheaf, and I have seen in several places a proof that P^++ is a sheaf. However, it seems that for any presheaf P I run into, P^+ is already a sheaf.
Does anyone know an example of a presheaf P where P^+ is not a sheaf i.e. where you actually need to apply the functor + twice to get a sheaf?
| https://mathoverflow.net/users/332 | What is an example of a presheaf P where P^+ is not a sheaf, only a separated presheaf? | I think this works:
Consider a topological space consisting of 4 points $A$, $B$, $C$, $D$, where the topology is given by open sets $ABC$, $BCD$, $B$, $C$, $ABCD$, $\emptyset$.
Then let the presheaf $\mathcal{F}$ be given by:
$$\mathcal{F}(ABC)=\mathbb{Z}$$
$$\mathcal{F}(BCD)=\mathbb{Z}$$
$$\mathcal{F}(BC)=\mathbb{Z}$$
$$\mathcal{F}(ABCD)=\mathbb{Z}$$
$$\mathcal{F}(B)=\mathbb{Z}/2\mathbb{Z}$$
$$\mathcal{F}(C)=\mathbb{Z}/2\mathbb{Z}$$
$$\mathcal{F}(\emptyset)=0$$
where all restrictions are what you expect (identity in the case of $\mathbb{Z} \to \mathbb{Z}$ and canonical surjection in the case $\mathbb{Z} \to \mathbb{Z}/2 \mathbb{Z}$).
Then if we we get $\mathcal{F}^+$ is given by:
$$\mathcal{F}^+(ABC)=\mathbb{Z}$$
$$\mathcal{F}^+ (BCD)=\mathbb{Z}$$
$$\mathcal{F}^+ (BC)= \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$$
$$\mathcal{F}^+ (ABCD)=\mathbb{Z}$$
$$\mathcal{F}^+ (B)= \mathbb{Z}/2\mathbb{Z} $$
$$\mathcal{F}^+ (C)=\mathbb{Z}/2\mathbb{Z}$$
$$\mathcal{F}^+ (\emptyset)=0$$
where the map from $\mathcal{F}^+ (BCD)$ to $\mathcal{F}^+ (BC)$ is given by taking the canonical surjection on both copies, and other restrictions are obvious. Then note that if we take 1 over $BCD$ and 3 over $ABC$, these two are compatible over $BC$ but they do not patch.
The key point is that being compatible over a refinement is not the same thing as being compatible. That is, the way the plus construction works is by taking $F^+$ of a space to be some direct limit over open covers of guys on the covers which are compatible on intersections. If we had said instead take direct limit over open covers of guys on the covers which compatible on some refinement of the intersection, then applying just once probably works.
So in our example, 1 and 3, over $ABC$ and $BCD$, in our original presheaf were compatible on a refinement of $BC$ but not on $BC$.
| 23 | https://mathoverflow.net/users/1655 | 4817 | 3,223 |
https://mathoverflow.net/questions/3234 | 4 | It is common to solve PDEs with e.g. Fourier and Laplace Transforms. It is often said that Wavelets are a progression compared to them with many nice features.
My question: Which Ansätze do you know to solve PDEs with Wavelets? Are these solution methods actually superior to the classical Ansätze?
Are there even Ansätze to solve stochastic partial differential equations? I am also especially interested in parabolic equations like e.g. diffusion equations.
| https://mathoverflow.net/users/1047 | Ansätze for solving PDEs with wavelets | The method of choosing a solution Ansatz to an equation and then actually deriving an exact solution is quite common in soliton theory, which is a sub-field of the study of hyperbolic equations. All methods described below, to my knowledge, only work on hyperbolic equations. Sorry, diffusion folks.
You must know properties of your equations to know which Ansatz will yield reasonable or good results. If you know that the tails of the solution die off quickly, you may choose a Gaussian $$ A \exp(-b x^2) $$, or if they die off **very** quickly, a super Gaussian $$ A \exp(-b a(x)^2) $$, where a(x) can be any polynomial. Also, based on the properties of your equation, you may want to multiply these 'basic' Ansatzen by other functions, to represent behavior that is known to be present. For example, if you know that solutions to the equation are not monotonic and/or 'wiggly', then you might want
$$ A \exp(-b x^2) \sin(k x) $$ The latter Ansatz is a two-parameter Ansatz and is the most likely to have a chance of working on a real equation. You may think that $ k $ is a third parameter, but actually, it is determined, usually algebraicly, by $A$ and $b$. Single parameter Ansatzen usually only work on very specific coefficients of equations and are too simple to model real equations.
There are obviously many, many other good Ansatzen, such as soliton solutions
$$ A\ {\rm sech}^n{\left(k x - \omega t\right)} $$ (where $n$ is a positive even integer, and $\omega=\omega\left(k\right)$ is the dispersion relation)
if your equations has symmetry properties. There is a large theory, mostly derived from the work of R. Hirota, of how to derive exact solutions to systems of nonlinear PDE's which have certain symmetry properties or invariants, using the properties of bilinear operators.
Note: Directly translated, the word der Ansatz in German has many meanings, but it most usually is translated as 'approach' or 'basic approach', but it really just means: an educated guess of a solution, with enough degrees of freedom (in the form of parameters) such that the Ansatz is able to solve the equation.
Also, in the above equations, $A$ can be constant, or only a function of $ t $ or a function of both $x$ and $t$, depending on which behavior is being modeled.
| 1 | https://mathoverflow.net/users/1653 | 4828 | 3,230 |
https://mathoverflow.net/questions/4851 | 6 | This might be a silly/obvious question. I know that we have the removable singularity theorem (of Riemann) on the complex line, and we also have the generalization of this to algebraic curves. (Namely: if $U$ is an open subset of a Riemann surface and $a\in U$, and $f \in \mathcal{O}(U-a)$ is bounded in some neighborhood of a, then f can be extended uniquely to a function $F\in \mathcal{O}(U)$.) In particular, if a function on a curve vanishes over a divisor, we can extend over it uniquely.
What do we know about extending of functions over divisors/hyperplanes in higher dimensions?
| https://mathoverflow.net/users/1177 | Divisors, extensions of functions | The very same result holds in arbitrary dimensions: a locally bounded holomorphic function
defined in the complement of a divisor extends. In many textbooks (like Gunning's) this is also called Riemmann's removable singularity Theorem.
If you know more about your divisor you can do even better. For instance, if your ambient is a surface and you have a holomorphic function defined on the complement of a smooth and compact rational curve of self-intersection $-1$ then your function extends to the whole surface since
the curve is contractible and after contraction you obtain a smooth surface. More generally, holomorphic functions defined on the complement of compact contractible divisors can also be extended if, after contraction, you obtain a normal variety.
| 7 | https://mathoverflow.net/users/605 | 4858 | 3,251 |
https://mathoverflow.net/questions/4840 | 4 | I have some statistical data from which I want to graph the means and use the standard deviations as error bars. However this produces a graph with some of the error bars passing below zero. A negative value is silly for this data (mean trip times), so I was wondering what is a sensible way to graph the data.
| https://mathoverflow.net/users/1664 | Is it alright for STD error bars to be below zero? | Your error bars may be giving you a hint to look more closely at the distribution of your data: it may not be symmetric. For example, if your data is essentially log-normal you could work with the logs of your numbers and the problem will automatically go away.
I'm not a fan of error bars. In theory they let you visually do some statistical significance estimates and perhaps give some sense of the underlying data. But there are a lot of subtleties and at least one study has found that even experienced scientists often misinterpret them. This [nice blog post](https://scienceblogs.com/cognitivedaily/2008/07/31/most-researchers-dont-understa-1) discusses some of the issues.
If you do need to summarize the data with a few statistics, I'd argue for [boxplots](https://en.wikipedia.org/wiki/Box_plot) as a better way to represent asymmetric distributions, along with text/captions that highlight important statistical significance conclusions.
| 11 | https://mathoverflow.net/users/1227 | 4861 | 3,254 |
https://mathoverflow.net/questions/4825 | 7 | like serre's thm for ampleness?
| https://mathoverflow.net/users/1657 | Is there a cohomological criterion of nefness? | Well, kind of -- a line bundle is nef if and only if its tensor product with any ample line bundle is ample.
| 3 | https://mathoverflow.net/users/1528 | 4863 | 3,255 |
https://mathoverflow.net/questions/4848 | 13 | I was thinking about the Gelfand-Naimark theorem asserting the isometric \* isomorphism between a commutative C\* algebra (with unit) A and the C\* algebra of continuous complex-valued functions on its spectrum (via the Gelfand transform). Explicitly: let spec(A) denote the spectrum of A and C(X) the algebra of complex continuous functions on X. Then spec and C define contravariant functors from commC\* alg to CompHausTop, which (correct me if i'm wrong) establish an equivalence between the two categories.
Gelfand-Naimark theorem has a non-commutative analogue, which is based on the so-called GNS construction and which shows that every non commutative C\* algebras has a faithful isometric \*-representation on a Hilbert space H. In this case I can't see an analogue of the preceding equivalence of categories, which is equally meaningful. Does it exist?
| https://mathoverflow.net/users/1049 | Gelfand-Naimark from the category-theoretic point of view | The proper analogue is rather based on the characterization of the state space of a unital C\*-algebra found in (sorry about the self-advertisement) E. Alfsen, H. Hanche-Olsen and F.W. Shultz: *State Spaces of C∗-Algebras*, Acta Math. **144** (1980) 267–305. So the category to replace CompHausTop would be the category of state spaces equipped with orientations on their facial 3-balls, and whose morphisms are certain affine maps between these compact convex sets.
In this context, a compact Hausdorff space *X* is represented by the set of probability Baire measures on *X*, which is in particular a Choquet simplex.
| 15 | https://mathoverflow.net/users/802 | 4864 | 3,256 |
https://mathoverflow.net/questions/4835 | 18 | Are there any suggestions for introductory books on wavelets? I want a book, not online material or tutorials.
| https://mathoverflow.net/users/1662 | Introduction to wavelets? | The canonical answer used to be Ingrid Daubechies, *Ten lectures on wavelets* (1992), ISBN 0898712742. It may be somewhat outdated by now, but probably still good.
| 11 | https://mathoverflow.net/users/802 | 4867 | 3,259 |
https://mathoverflow.net/questions/4841 | 75 | (And what's it good for.)
Related MO questions (with some very nice answers): [examples-of-categorification](https://mathoverflow.net/questions/43579/examples-of-categorification); [can-we-categorify-the-equation $(1-t)(1+t+t^2+\dots)=1$?](https://mathoverflow.net/questions/1465/can-we-categorify-the-equation-1-t1-t-t2-1); [categorification-request](https://mathoverflow.net/questions/55721/categorification-request).
| https://mathoverflow.net/users/1532 | What precisely Is "Categorification"? | The Wikipedia answer is one answer that is commonly used: replace sets with categories, replace functions with functors, and replace identities among functions with natural transformations (or isomorphisms) among functors. One hopes for newer deeper results along the way.
In the case of work of Lauda and Khovanov, they often start with an algebra (for example ${\bf C}[x]$ with operators $d (x^n)= n x^{n-1}$ and $x \cdot x^n = x^{n+1}$ subject to the relation $d \circ x = x \circ d +1$) and replace this with a category of projective $R$-modules and functors defined thereupon in such a way that the associated Grothendieck group is isomorphic to the original algebra.
Khovanov's categorification of the Jones polynomial can be thought of in a different way even though, from his point of view, there is a central motivating idea between this paragraph and the preceding one. The Khovanov homology of a knot constructs from the set of $2^n$ Kauffman bracket smoothing of the diagram ($n$ is the crossing number) a homology theory whose graded Euler characteristic is the Jones polynomial. In this case, we can think of taking a polynomial formula and replacing it with a formula that inter-relates certain homology groups.
Crane's original motivation was to define a Hopf category (which he did) as a generalization of a Hopf algebra in order to use this to define invariants of $4$-dimensional manifolds. The story gets a little complicated here, but goes roughly like this. Frobenius algebras give invariants of surfaces via TQFTs. More precisely, a TQFT on the $(1+1)$ cobordism category (e.g. three circles connected by a pair of pants) gives a Frobenius algebra. Hopf algebras give invariants of 3-manifolds. What algebraic structure gives rise to a $4$-dimensional manifold invariant, or a $4$-dimensional TQFT? Crane showed that a Hopf category was the underlying structure.
So a goal from Crane's point of view, would be to construct interesting examples of Hopf categories. Similarly, in my question below, a goal is to give interesting examples of braided monoidal 2-categories with duals.
In the last sense of categorification, we start from a category in which certain equalities hold. For example, a braided monoidal category has a set of axioms that mimic the braid relations. Then we replace those equalities by
$2$-morphisms that are isomorphisms and that satisfy certain coherence conditions. The resulting $2$-category may be structurally similar to another known entity. In this case, $2$-functors (objects to objects, morphisms to morphisms, and $2$-morphisms to $2$-morphisms in which equalities are preserved) can be shown to give invariants.
The most important categorifications in terms of applications to date are
(in my own opinion) the Khovanov homology, Oszvath-Szabo's invariants of knots, and Crane's original insight. The former two items are important since they are giving new and interesting results.
| 26 | https://mathoverflow.net/users/36108 | 4872 | 3,264 |
https://mathoverflow.net/questions/1010 | 11 | For periodic/symmetric tilings, it seems somewhat "obvious" to me that it just comes down to working out the right group of symmetries for each of the relevant shapes/tiles, but its not clear to me if that carries over in any nice algebraic way for more complicated objects such as a [penrose tiling](http://en.wikipedia.org/wiki/Penrose_tiling)
and just following [wikipedia](http://en.wikipedia.org/wiki/Quasicrystal) just leads to the statement that groupoids come into play, but with no references to example constructions! Moreover, at least naively thinking about, it seems any such algebraic approach should naturally also apply to fractals.
1. what references am I somehow not able to find that do a good job talking about this further?
2. is my "intuition" that the mathematical structure for at least some classes of fractals and quasicrystals being equivalent correct?
| https://mathoverflow.net/users/426 | What is the right way to think about / represent general tilings? | Aperiodic tilings can be thought of (in a sometimes useful way) as leaves of laminations; the groupoid in question (as in Emily's answer) is then the holonomy groupoid of the lamination.
There is a standard description of the Penrose tiles in this way; think of an irrational plane (i.e. an $R^2$) in $R^n$ for some $n>2$, and consider the set of 2-dimensional faces of the $Z^n$ lattice in $R^n$ that intersect a (uniform) thickened tubular neighborhood of your plane. Project each such 2-dimensional face perpendicularly down to your plane; the result is an aperiodic tiling. If the irrational plane happens to be chosen with extra symmetries (eg it could be an eigenspace of a finite order element in $GL(n,Z)$) one gets quite a tile set with extra "partial symmetries". The Penrose tiling is of this kind: think of $Z/5Z$ permuting the coordinate axes in $R^5$. This fixes the vector $(1,1,1,1,1)$ and has two perpendicular irrational eigenspaces on which it acts as an order 5 rotation; translates of these eigenspaces give rise to the "standard" Penrose tilings.
The lamination in this case is the "irrational foliation" of the torus $R^5/Z^5$ by planes with slope equal to the slope of the $R^2$ (and one can easily imagine generalizations).
| 14 | https://mathoverflow.net/users/1672 | 4889 | 3,274 |
https://mathoverflow.net/questions/4895 | 33 | I have recently begun to study algebraic geometry, coming from a differential geometry background. It seems that there is a deep link between complex manifolds and complex varieties. For example, one often hears that they are two different ways of looking at the same thing. Can anybody give a precise statement of this relationship?
| https://mathoverflow.net/users/1648 | The Relationship between Complex and Algebraic Geomety | The Wikipedia article is more technical than it should be, and for the reader in a hurry not all that well written. Here is a summary of the main points as best I understand them:
Complex manifolds are analogous to smooth complex algebraic varieties, not to the singular ones. But that discrepancy is surmountable, because you can also have complex analytic varieties which can have similar singularities. Then the first and most important relation is that every complex algebraic variety is a complex analytic variety. Every Zariski open set is analytically open; analytic gluing maps are more general than algebraic ones; and the allowed analytic charts are more general than the allowed algebraic charts. Also every algebraic morphism is an analytic morphism, so you get a morphism between categories.
But the connection is better than that because of the GAGA principle (globally analytic implies globally algebraic). My understanding of GAGA is very sketchy, but I think that the following is correct. Among other consequences of GAGA, a closed analytic subvariety of a proper (equivalent to compact) algebraic variety is algebraic. An analytic isomorphism between two proper algebraic varieties is algebraic. I would suppose that there is a similar principle for compact fibrations as well.
So, ~~if you make compact analytic varieties algebraically, you can't escape from the algebraic class~~ some of the main constructions of complex manifolds do not escape from the algebraic class. (But not all: deformations and infinite group actions can escape.) All projective analytic varieties are algebraic, and in dimension 1 all compact curves are projective. Moreover, there are limited ways for a compact analytic manifold to avoid being projective, by Moishezon's theorem and Kodaira's theorem. In practice, then, most of the complex manifolds that people make are algebraic. Also, most of the analytic calculations on a proper algebraic variety are algebraic: Many global calculations are algebraic by GAGA, and many local calculations are algebraic just by truncating Taylor series.
Contrast all this with real algebraic vs real analytic. It is still true that (the real points of) a smooth real algebraic variety is a real analytic manifold. More strongly than in the complex case, although it is highly non-trivial, every compact real analytic manifold is real algebraic. But the real algebraic structure is massively not unique, even for a circle, and that makes all the difference.
---
The other answerers in this thread, who are more expert in this topic than I am, had more information about why a compact complex manifold might not be a smooth projective variety. Just for clarity, I will restrict attention to the compact, smooth case. Also, you say "proper" rather than "compact" in the algebraic category because every algebraic variety is "compact" in the extremely coarse Zariski topology. An algebraic variety is proper if and only if it is analytically compact. The main use of the word proper is to emphasize that it is more general than projective, which means given by polynomial equations in complex projective space.
There are two very different initial reasons that an analytic complex manifold might not be projective. It might not be Moishezon: A complex $n$-manifold is Moishezon if it has $n$ algebraically independent meromorphic functions. (The number of algebraically independent elements or the transcendence degree of a field is called the Krull dimension. The meromorphic Krull dimension of a compact complex $n$-manifold is at most $n$.) Or it might not be Kähler: A complex $n$-manifold is Kähler if it has a Riemannian metric such that the covariant derivative of the complex structure vanishes. So to summarize what people said about compact complex manifolds (much of which is in the back of Hartshorne's book):
projective ⇒ algebraic ⇒ Moishezon ⟺ bimeromorphically projective
projective ⇒ Kähler ⇒ symplectic ⇒ non-zero $H^2$
algebraic ⇒ non-zero $H^2$ ([exposited](http://sbseminar.wordpress.com/2008/02/14/complex-manifolds-which-are-not-algebraic/) by David Speyer)
Moishezon and Kähler ⟺ projective (Moishezon)
Kähler and integrally symplectic ⟺ projective (Kodaira)
In addition, projective and algebraic structure and the Moishezon property are all unstable with respect to analytic deformation. And bimeromorphic equivalence preserves $\pi\_1$. Taubes found compact complex manifolds that have the wrong $\pi\_1$ to be Kähler; indeed they can have any $\pi\_1$. Voisin found compact Kähler manifolds with the wrong homotopy type to be projective, disproving Kodaira's conjecture that every compact Kähler manifold can be deformed to projective. And, way out in left field, a left-invariant complex structure (LICS) on a compact simple Lie group is a compact complex manifold that has no $H^2$ and can be simply connected too.
Still, despite these beautiful non-projective compact complex manifolds, it's generally easier to study projective examples. It's generally easier to sidestep analysis and do algebra instead.
| 55 | https://mathoverflow.net/users/1450 | 4902 | 3,280 |
https://mathoverflow.net/questions/4766 | 13 | I noticed that the generator of the second stable stem b is the square of the generator of the first stable stem a, in the sense that if take two copies of a and smash product them together you get b out. I'm wondering if there are any ( many) other exmples of this. What are the elements in the stable homotopy of spheres which a squares in the above sense?
| https://mathoverflow.net/users/1643 | squares in stable homotopy | Appendix 3 of Ravenel's Green Book, <http://www.math.rochester.edu/u/faculty/doug/mu.html#repub>, has a chart of stable homotopy groups including much of the multiplicative structure. Figure A3.1 depicts some of this structure visually, while Table A3.3 lists the elements out by name and degree.
The next example of a square after \eta^2 is the element in the sixth stable stem, which is the square of the Hopf map \nu in the third stable stem.
Though stable homotopy is not multiplicatively finitely generated, you can consider Toda brackets, which are a form of higher multiplication in homotopy analogous to Massey products in cohomology, and it is known that the entire stable homotopy groups of spheres are generated by Toda brackets on the Hopf elements 2, \eta, \nu, and \sigma.
| 11 | https://mathoverflow.net/users/288 | 4905 | 3,283 |
https://mathoverflow.net/questions/4897 | 5 | Okay, I'm going to ask a naiive question that surely has an interesting answer. So, a first approximation of defining a (small) weak n-category probably goes something like this. Take a pre-n-category C of all the cells, source and target maps that do the right thing (i.e. are globular), and a composition defined for each r in {0,...,n}.
For C, define a family of coherent sets $(\Sigma\\_1, \Sigma\\_1, \ldots)$ as a family of sets $\Sigma\_r$ of r-cells in C such that
1. $f : a \rightarrow b \in \Sigma\\_r \Rightarrow \exists f' : b \rightarrow a \in \Sigma\\_r$
2. $f, f' : a \rightarrow b \in \Sigma\\_r \Rightarrow \exists \alpha : f \rightarrow f' \in \Sigma\\_{r+1}$
Now, suppose C admits such a family of coherent sets and all r-cells have associators, uniters, and interchangers in $\Sigma\\_{r+1}$, one might be tempted to say C is an $\infty$-category. If for all $r \geq n+1$, $\Sigma\\_r$ is only identites, one might say this defines an n-category.
So, the reason I say "one might be tempted to say" is that, if it were that easy, someone much smarter than me would have done it already. :) So, where does the above recipe fail? Or is this definition unsatisfactory because it doesn't express the structure using a finite generating set of commutative diagrams (cf. Mac Lane's coherence etc.)?
| https://mathoverflow.net/users/800 | Where does the "easy" definition of a weak n-category fail? | If I understand correctly what you're getting at, I think the reason this fails is because for n>2, *not* every diagram of constraints can be expected to commute (even up to higher constraints) in a weak n-category. For example, a braided monoidal category can be regarded as a weak 3-category with one 0-cell and one 1-cell, but then the "double twist" is a constraint isomorphism which is not equal to the identity (also a constraint isomorphism).
One way to get around this is, instead of talking about diagrams of constraints *in* some particular n-category, to talk about "formal" diagrams of constraints, i.e. diagrams of constraints in a *free* n-category. I think that when one makes your idea precise using this corrected approach, one will end up with something very similar to Batanin's higher-operadic definition of weak n-category.
| 9 | https://mathoverflow.net/users/49 | 4906 | 3,284 |
https://mathoverflow.net/questions/4810 | 5 | The Erdős–Stone theorem theory says that the densest graph not containing a graph H (which has chromatic number r) has number of edges equal to $(r-2)/(r-1) {n \choose 2}$ asymptotically.
However, this doesn't say much for bipartite graphs (since r=2). I wanted to know what are the best results known for the densest graphs not containing a particular bipartite graph H. I'm guessing this problem is still open and hasn't been completely resolved.
This problem is easy if H is a forest, since every graph with $|E| > k|V|$ contains every forest on k vertices as a subgraph. For even cycles, I know there is a result of Bondy and Simonovits which says:
"if $|E| \geq 100k|V|^{1+1/k}$ then G contains a $C\_{2l}$ for every $l$ in $[k, n^{1/k}]$."
So can someone point me to the best known results now for bipartite cyclic graphs?
| https://mathoverflow.net/users/1042 | Erdős–Stone theorem type edge density estimates for bipartite graphs? | Let $ex(n, H)$ denote the maximum number of edges of a graph on $n$ vertices not containing a copy of $H$. The exact bounds are difficult already if you forbid *complete* bipartite graphs $K\_{m,n}$.
Erdös, Rényi, Sós (1954) showed that $$ex(n,K\_{2,2}) \sim \frac{1}{2}n^{3/2}.$$
According to the classical Kövári-Sós-Turán Theorem (1954), $$ex(n,K\_{t,s}) \leq c\_{s,t} n^{2-\frac{1}{t}}$$ for $s\geq t\geq 2$,
while a random construction gives the lower bound $$ex(n,K\_{t,s})\geq cn^{2-\frac{s+t-2}{st-1}}.$$
Brown (1966) showed Kövári-Sós-Turán is tight for $s=t=3$: $$ex(n,K\_{3,3}) \geq cn^{2-\frac{1}{3}},$$ and Füredi (1996) proved that the constant in Brown's construction is optimal, giving $$ex(n,K\_{3,3}) \sim \frac{1}{2}n^{2-\frac{1}{3}}.$$
Alon, Kollár, Rónyai, Szabó (1995, 1999) showed that for each $t\geq 2$ there exists $c\_t>0$ such that for all $s\geq(t-1)!+1$,
$$ex(n,K\_{t,s}) \geq c\_tn^{2-\frac{1}{t}},$$ thus matching Kövári-Sós-Turán asymptotically.
I'm not sure if a construction matching Kövári-Sós-Turán has been found for the case $K\_{4,4}$. There must also be more known than what is mentioned here. But as you can see, somewhat more than a just a little more is known, although the gaps in our knowledge are still huge here.
| 9 | https://mathoverflow.net/users/932 | 4923 | 3,295 |
https://mathoverflow.net/questions/4917 | 19 | So looking at Euclid's proof he says
1)take a finite family of primes (F)
2)multiply all the primes and add one
3)this new number has at least 1 new prime factor
So I was wondering about what kind of primes you get by recursively feeding this process into it self.
Since the number you must factor grows exponentially, it's hard to get a lot of numerical evidence for what happens.
I calculated a few:
[2]-> [2,3]-> [2,3,7]->[2,3,7,43]->[2,3,7,43,13,139]->[2,3,7,43,13,139,3263443]
->[2,3,7,43,13,139,3263443,547,607,1033,31051]-> cannot factor 113423713055421844361000443
[5] (x5)-> [5,2,3,31,7,19,37,3343,79,193662529] -> cannot factor 234069798025176583891
Obviously quite a few primes are missing, 5,11,19,etc from the first list, but could show up later.
So my question is does a finite family of primes exist that eventually generates all the primes? I figure this probably doesn't have an easy answer, but any information related to this process would be appreciated, or even why it can't be done.
| https://mathoverflow.net/users/695 | On Euclid's proof of the infinitude of primes and generating primes | Here's the reason why keeping primes with multiplicity makes the answer "no." If $p\_n$ denotes the product of all the numbers you have so far, where $p\_1$ is the product of the primes you start with, then $p\_n = p\_1 ... p\_{n-1} + 1$. But we can rewrite this as $p\_n = p\_{n-1}(p\_{n-1} - 1) + 1 = f(p\_{n-1})$ where $f(x) = x^2 - x + 1$, and it is well-known that any prime divisor of $f(n)$ for an integer $n$ must be $2, 3$, or congruent to $1 \bmod 3$, i.e. the primes $5, 11, 17, ...$ will **never** appear (unless they divide $p\_1$ to start with.)
(Sketch: if $q | x^2 - x + 1$ then $q | x^3 + 1$, hence $x$ has order $6 \bmod q$ or $q = 2, 3$. Since the multiplicative group $\bmod q$ has order $q - 1$, this is possible if and only if $6 | q-1$.)
| 21 | https://mathoverflow.net/users/290 | 4928 | 3,299 |
https://mathoverflow.net/questions/4919 | 4 | I want to make sure I completely understand the isomorphism classes of smooth unitary irreducible finite-dimensional representations of $U(n)$. We have the irreducible defining representation $R$, and we can apply any Young diagram to this, hitting it with a Schur functor, to get a load more irreps. We can also take the complex conjugate $R^\*$ of the defining representation, and hit this with all the Young diagrams, to get another series of irreps. Some of these irreps will actually be the same, if the number of rows in our Young diagrams gets as large as $n$, but I'm not worried about that.
I understand how to tensor together representations that arise from applying Young diagrams to $R$, and decompose this into a direct sum of irreducible representations. But what if I tensor together an irrep coming from a Young diagram applied to $R$, and an irrep coming from a Young diagram applied to $R^\star$? How does this product decompose into a sum of irreps? I've convinced myself it just isn't possible when you're restricted to using the irreps that arise from applying Young diagrams to $R$ and $R^\star$ --- there must be some more of them. For example, consider $U(2)$ and the representation $R \otimes R^\star$, where $R$ is the defining representation. Since $R$ and $R^\star$ are dual, this must decompose as $R \otimes R^\* \simeq 1 \oplus X$, where $X$ is a 3-dimensional self-dual possibly-reducible representation. But I don't think it's possible to build such an $X$ by taking direct sums of the irreps described in the first paragraph.
So, apart from applying Young diagrams and conjugate Young diagrams to the defining representation, how to you get the rest of the of the irreps? In particular, I don't think I know how to build any self-dual irreps, apart from the trivial irrep, but some have to exist by the argument in the previous paragraph.
| https://mathoverflow.net/users/799 | Smooth unitary irreducible finite-dimensional representations of U(n) | The irreps of $U(n)$ are indexed by decreasing $n$-tuples of integers $\lambda\\_1 \geq \lambda\\_2 \geq \cdots \geq \lambda\\_n$. Note that I do NOT impose that $\lambda\_n \geq 0$.
One way to see this is to note that the conjugacy classes of $U(n)$ are $(S^1)^n/S\\_n$, where $S\\_n$ acts by the obvious permutations. In other words, the conjugacy class of a unitary matrix is determined by its eigenvalues up to permutation. For $\theta \in \mathbb{R}^n$, write $[e^{i \theta}]$ for the conjugacy class with eigenvalues $e^{i \theta\\_1}$, ..., $e^{i \theta\\_n}$. Using the Fourier transform, a class function will look like
$$f([e^{i \theta}]) = \sum\_{\omega} a(\omega) e^{i \langle \omega, \theta \rangle}$$
function $a(\omega)$ on $\mathbb{Z}^n$ which unchanged by permuting the input. In the case of a character, this will be a finite sum, because in any representation of $U(n)$ the diagonal matrices will act diagonalizably. I'll write $x\\_j$ for the character $x\\_j(e^{i \theta}) = e^{i \theta\\_j}$. So the character ring of $U(n)$ is the **symmetric Laurent polynomials** in the $x$'s, and you should believe that there is one irrep for each point in $\mathbb{Z}^n/S\\_n$.
Now, how does this relate to the examples you raised? The defining representation has character $x\\_1 + x\\_2 + \cdots x\\_n$. Applying any Schur functor to this will get you another **polynomial** in the $x$'s. So, just using Schur functors, you are only seeing the representations that correspond to polynomials; in other words, to the case that $\lambda\_n \geq 0$. In other (more algebraic) contexts, these are called the *polynomial representations*, and I'm going to use that term here.
For a unitary matrix, the complex conjugate is the same as the inverse transpose; so taking the complex conjugate moves you to the dual representation. So, the complex conjugate of the dual representation has character $x\\_1^{-1}+x\\_2^{-1}+\cdots+x\\_n^{-1}$. Taking Schur functors of this will give you **polynomials in the variables $x\\_i^{-1}$.**
On the $\lambda$'s, taking the dual sends $(\lambda\\_1, \ldots, \lambda\\_n)$ to $(-\lambda\\_n, \ldots, -\lambda\\_1)$.
You are right that there are representations you are not seeing, because there are symmetric Laurent polynomials which are neither polynomials in the original variables nor in the inverse variables. In terms of $\lambda$'s, it is possible to have an $n$-tuple of numbers, some of which are positive and some nonnegative. The simplest example is adjoint rep, which is $V\\_{1,0,\ldots,0,-1} \oplus V\\_{0,0,\ldots, 0}$, and has character
$$(x\\_1 + x\\_2 + \cdots x\\_n) (x\\_1^{-1} + x\\_2^{-1} + \cdots x\\_n^{-1}).$$
Note that none of these representations are much different from the polynomial representations, though. By tensoring with a high enough power of the determinant, you can always get back into the polynomial representations. In other words, if you multiply a Laurent polynomial by a high enough power of $x\\_1 x\\_2 \cdots x\\_n$, it becomes a polynomial.
This sounds complicated, but if you know the $SU(n)$ case well then you'll find this easier. In $SU(n)$, the torus is $T / S\_n$ where $T$ is the subtorus of $(S^1)^n$ where the $n$-components add up to $0$. So you have to remember that the Schur polynomials $1$ and $x\\_1 x\\_2 \cdots x\\_n$ represent the same representation. All of those little details go away in the $U(n)$ case!
| 7 | https://mathoverflow.net/users/297 | 4929 | 3,300 |
https://mathoverflow.net/questions/4912 | 7 | When I was working on my PhD dissertation, I came across a physical situation involving nodes and flows between them. It turned out that I was working with a complete oriented graph $K\_n$ (all nodes are connected to each other), and I needed to calculate the pseudoinverse of its incidence matrix T, i.e. the rectangular matrix N(V) x N(E) where N(E) = n is the number of edges and N(V) is the number of vertices, with matrix element 1 if the edge enters the vertex, -1 if the edge leaves the vertex, and 0 otherwise.
To my surprise the pseudoinverse turns out to be proportional to the transpose of the incidence matrix! Specifically
$T^{+} (K\\_n) = \frac{1}{n} T^{\prime} (K\\_n)$
where the prime denotes transposition.
---
My question is:
A) What other graphs $G$, if any, have this property, i.e. that
$T^+(G) \propto T^\prime(G)$,
or some suitable generalization thereof, and
B) How can I show that this result is invariant of orientation? (I determined empirically is certainly true for all possible orientations of the small complete graphs, and I haven't been able to find a counterexample, but I don't have a proof of this statement yet)
I'm not a professional mathematician, so any thoughts would be welcome.
---
One class of generalizations that is possible (but not so interesting IMO) are disconnected graphs where each subgraph is a complete graph, i.e.
$G = K\_{n\_1} \oplus K\_{n\_2} \oplus \dots \oplus K\_{n\_m}$
In this case one gets
$T^+(G) = \frac{1}{n\_1} T^\prime(K\_{n\_1}) \oplus \frac{1}{n\_2} T^\prime(K\_{n\_2}) \oplus \dots \oplus \frac{1}{n\_m} T^\prime(K\_{n\_m})$
which is not really that exciting, but perhaps could point the way to a more interesting generalization.
---
P.S. There is a short proof of the first fact, which relies on the fact that the complete graph has a Laplacian of the form
$\Delta = n \mathbf{I} - \mathbf{1} \mathbf{1}^\prime$
where $\mathbf{I}$ is the n x n identity matrix and $\mathbf{1}$ is a column vector of ones.
With this fact, together with knowing that the column sums of $T$ are all zero, it is straightforward to show that $T^\prime(K\_n)/n$ satifies all the Moore-Penrose conditions for the pseudoinverse.
---
P.P.S. If anyone is interested in the physical context, [here](http://link.aip.org/link/?JCPSA6/129/214113/1) is where it came from.
| https://mathoverflow.net/users/1674 | Graphs with incidence matrices whose pseudoinverses are proportional to their transposes | Let the vertices be numbered $1,\ldots,n$. For a vertex $i$, let $\mathrm{deg}(i)$ be the total number of (unsigned) edges incident to $i$. If $A$ is the incidence matrix, consider $AA^TA$. An entry of $AA^TA$ is indexed by a vertex $i$ and an edge $e=(j,k)$. If I did this correctly, it's straightforward to show that $[AA^TA]\_{i,(j,k)}$ is:
* $-\mathrm{deg}(i)$ if $i=j$,
* $\mathrm{deg}(i)$ if $i=k$,
* the number of undirected edges $(i,k)$ minus the number of undirected edges $(i,j)$ otherwise.
If $A^T$ is a scalar multiple of the pseudoinverse of $A$, then $AA^TA$ must be a scalar multiple of $A$. It's easy to see from the above that this is equivalent to requiring that each connected component of the graph is complete, each vertex has the same (undirected) degree, and within each component, all edges have the same (undirected) multiplicity. (For example, I think you could have two components $K\\_3$ and $K\\_5$, where there are two edges between each pair of vertices in the $K\\_3$ component and one edge between each pair in the $K\\_5$ component.) None of this depends on the orientations of the edges: flipping an edge of the graph is equivalent to negating a column of $A$, and if $AA^TA$ was a scalar multiple of $A$ before, then it still is now. Since $(AA^TA)^T = A^TAA^T$, if $AA^TA = cA$, then $A^TAA^T = cA^T$, so $(1/c)A^T$ is the Moore-Penrose pseudoinverse of $A$. ($AA^T$ and $A^TA$ are obviously symmetric.) This completely characterizes graphs with your property.
| 5 | https://mathoverflow.net/users/302 | 4935 | 3,304 |
https://mathoverflow.net/questions/4934 | 2 | This question is not urgent; just a matter of curiosity...
It is relatively easy to generate an arbitrary 3D or even 4D rotation matrix using conjugation (i.e. *YXY*−1) of orthogonal rotations. I expect that there are ways to choose the contributing orthogonal angles of rotation in order to get a uniform random distribution of the resulting axis (and angle). (3D rotations are also related to quaternions.)
There's a wonderfully symmetrical 3D rotation matrix presentation, given in the first edition of D. F. Rogers and J. A. Adams book "Mathematical Elements for Computer Graphics". The elements of the matrix are:
u2+(1-u2)c; uv(1-c)+ws; uw(1-c)-vs
uv(1-c)-ws; v2+(1-v2)c; vw(1-c)+us
uw(1-c)+vs; vw(1-c)-us; w2+(1-w2)c
Here, (u,v,w) is a unit vector along the chosen axis of rotation, and s and c are the sine and cosine of the chosen angle of rotation.
[I mention this on my [blog page](http://rhubbarb.wordpress.com/2009/04/07/rotation-matrix/).]
Now, a 4D rotation must be about a "2D-axis", or plane (where a 3D rotation is about a "1D-axis", or line).
**I wonder if there's an equally elegant 4x4 matrix, in terms of a pair of mutually orthogonal unit vectors (defining the plane of rotation) and the sine and cosine of the angle of rotation in 4D.**
Any on-line references regarding anything mentioned here would we gratefully received.
| https://mathoverflow.net/users/1536 | Symmetrical Presentation of 4-Dimensional Rotation Matrix | A 4-d rotation does not have to fix a 2-d axis. For example, (complex) multiplication by $e^{i\theta}$ rotates every vector of unit length in $C^2$ ``the same way''.
| 3 | https://mathoverflow.net/users/1672 | 4936 | 3,305 |
https://mathoverflow.net/questions/4939 | 28 | Apologies for the very simple question, but I can't seem to find a reference one way or the other, and it's been bugging me for a while now.
Is there a compact (Hausdorff, or even T1) (topological) group which is infinite, but has countable cardinality? The "obvious" choices don't work; for instance, $\mathbb{Q}/\mathbb{Z}$ (with the obvious induced topology) is non-compact, and I get the impression that profinite groups are all uncountable (although I might be wrong there). So does someone have an example, or a reference in the case that there are no such groups?
| https://mathoverflow.net/users/382 | Is there a compact group of countably infinite cardinality? | No, there is no countably infinite compact Hausdorff topological group.
Indeed such a group $G$ would have a left-invariant Haar measure $m$ with $m(G)=1$
and all points would have the same measure (since the group acts transitively on itself).
But then, by countable additivity of the measure $m$, the group itself would have measure $m(G)=0$ or $m(G)=\infty$ according as its points all had $m(p)=0$ or $m(p)>0$ . A contradiction in both cases to the fact that $m(G)=1$ .
| 69 | https://mathoverflow.net/users/450 | 4950 | 3,316 |
https://mathoverflow.net/questions/4930 | 11 | This is not an urgent question, but something I've been curious about for quite some time.
Consider a Boolean function in *n* inputs: the truth table for this function has 2*n* rows.
There are uses of such functions in, for example, computer graphics: the so-called ROP3s (ternary raster operations) take three inputs: *D* (destination), *S* (source) and *M* (mask). All three are planes of bits, and the result is returned to the destination plane of bits. Now, this is only really applicable to 2-colour (black & white) or 8-colour displays (regarding the red, green, blue planes separately). A given ROP3 is given a code from 0 to 255, a number which represents the pattern of the output bit, down the rows of the table. Similarly, ROP2s have a value from 0 to 15. ROPs may also be given names, especially when the logical connective of which the truth table is an extension is a simple one, such as AND, XOR or SRC (source).
An expression for any truth table (or ROP) may be found in terms of an expressively-complete set of connectives (usually unary or binary, sometimes nullary too). [Well, I suppose this statement is itself a tautology!] For example, the sets {NOT, AND}, {NOT, OR}, {NAND} are each expressively complete.
One commonly used (redundant) expressively complete set is {NOT, AND, OR}. Two particularly common canonical sets of expressions over this set are the conjunctive normal form (CNF) and disjunctive normal form (DNF). Both of these are rather verbose.
There is also a notion of a minimal expression over a set of connectives, defined variously. The count might be of instances of a variable or of connectives. There might be a bound to the depth or breadth of the expression (perhaps).
The Boolean connectives might be extended to the set ***R***[0,1], for fuzzy logic; that is, the connectives are over ***R***[0,1], with the restriction to {0,1} being the usual Boolean function. There are many ways to do this; it is possible to preserve some but not all the usual properties (e.g. associativity, idempotency) of the connectives. [NOT(x) would usually be interpreted as (1−x); AND(x,y) could be (x\*y) or (min{x,y}), or in many other ways.]
Such extensions may be used, for example, to give a meaning to a ROP3 as applied to 256-level monochrome images (to combine or 'mix' such images) or to planes of full-colour images. (Necessarily, some truncation or 'quantisation' must take place.)
However, two expressions have the same function over {0,1} will generally have different functions over ***R***[0,1]. Rather than choosing some arbitrary expression, it would be an advantage to choose some canonical or minimal expression.
How much is known about this field? Are there any good on-line references? I'm particularly interested in definitions of, theorems about, and algorithms for the generation of minimal or canonical expressions.
| https://mathoverflow.net/users/1536 | Finding minimal or canonical expressions for Boolean truth tables | If you play around with this a bit, you'll probably come to the same conclusion that I did a long time ago, namely that there is indeed a **one true extension** of boolean functions $\{0,1\}^n\to \{0,1\}$ to functions $[0,1]^n\to [0,1]$. There are many simple characterizations of this extension (which tells you that it's important). The most intuitive one for me is to relax boolean logic to probability: take a boolean function of some number of variables, like $$y \text{ or not } (x \text{ and } z).$$ Now suppose we flip possibly biased coins independently to determine the truth values of $x,y,z$. The probability of $x$ being true is $p$, the probability of $y$ being true is $q$, and the probability of $z$ being true is $r$, so we have $0\leq p,q,r \leq 1$. Now consider our expression $y \text{ or not } (x \text{ and } z)$: what is the probability that this will be true? It's some function of $p,q,r$ that takes values in $[0,1]$. If you work it out (which is straightforward), you'll find that the function is $$f(p,q,r) = 1-p(1-q)r.$$ You should check that when you restrict $p,q,r$ to be 0 or 1, this agrees with the original boolean function when the corresponding boolean variables are respectively false or true. One of the other main characterizations of this extension is as the unique multiaffine interpolating polynomial. ("Multiaffine" means that if you set all but one of the variables, say $p$, to constant, then you get a linear function $ap + b$ of $p$ for some constants $a$ and $b$ (which depend on the constants you chose for the other variables)).
Unfortunately, *computing* this extension for a boolean function of many variables is easily seen to be #P-hard, which means it's one of those hard problems that people hate because they can't solve it efficiently but can't prove that they can't. (Indeed, any easily computable canonical representation (extending to [0,1] or otherwise) applicable to all boolean functions would prove P=NP, so your problem as you stated it is pretty much hopeless. This particular canonical extension happens to be powerful enough to be #P-hard.) However, if your boolean expressions have only a few variables, then there is a very easy recursive algorithm to compute the extension. Here it is:
$$f(p\\_1,p\\_2,\ldots,p\\_n) = p\\_1f(1,p\\_2,\ldots,p\\_n) + (1-p\\_1)f(0,p\\_2,\ldots,p\\_n).$$ $f(1,p\\_2,\ldots,p\\_n)$ and $f(0,p\\_2,\ldots,p\\_n)$ are respectively the extensions in which you set the first variable to true and to false, which you can compute recursively. The basis case is when $n=0$, in which case you have a nullary function 0 or 1.
[**Edit:** If your boolean function is actually represented literally as a truth table, then there's no difficulty in computing its canonical extension to real values; actually, the recursive algorithm given above has $O(N)$ complexity, where $N=2^n$ is the number of rows in the truth table. Of course, *producing* a truth table for a function of many variables is time- and space-consuming, but if you already have the truth table, then you've already committed to it.]
There's actually one special case in which the extension is easy to compute for any number of variables, which may or may not help you: if you can write your boolean expression so each variable only appears once, then you can compute the extension in linear time. You use the transformation rules
* $(\text{not } A)\to (1-A)$,
* $(A \text{ and } B)\to (AB)$,
* $(A \text{ or } B)\to (A + B - AB)$
and simply recurse on the expression. My example $y \text{ or not } (x \text{ and } z)$ has this property, so let's use this procedure on it:
$$ y \text{ or } (\text {not } (x \text{ and } z)) \to y + (1-xz) - y(1-xz), $$ which you can see is the same as $1-x(1-y)z$. This, very unfortunately, fails miserably when you have multiple occurrences of any variable (just try $x \text{ and not } x$). However, the situation is just slightly more general than may appear initially: note that I gave transformation rules for AND, OR, and NOT. I did that because it's often convenient to express boolean expressions in terms of those three. You can in fact formulate a corresponding transformation rule for the boolean function $f$ of your choice (which is identical to the canonical extension of $f$: e. g., $1-x$ is the canonical extension of $\text{not }x$ and $x + y - xy$ is the canonical extension of $x\text{ or }y$), and you can apply that transformation to $f(A\\_1,\ldots,A\\_n)$ *provided that no variable appears in more than one subexpression $A\\_1,\ldots, A\\_n$*. For example, exclusive-or might be a useful boolean function, so you can use the transformation rule
* $(A \text{ xor } B)\to A + B - 2AB$.
This way, you can effectively build up a library of your favorite boolean functions and their canonical extensions, which you can use as transformation rules whenever you apply those functions to subexpressions with disjoint variable sets.
Finally, let me give you one "fun" way to compute the canonical extension based on the transformation rules that is tremendously useful when computing a canonical extension by hand: it turns out that you can apply any canonical transformation rule *arbitrarily*, even for arguments with non-disjoint sets of variables, expand everything out into a sum of monomials, and use the additional transformation
* $(x^2) \to x$.
as many times as you can until you're left with a multiaffine polynomial. (In fact, you're free to apply that transformation even without expanding your polynomial: e. g., $(1 -xy)^2 \to (1-xy)$ is perfectly valid.) To illustrate, let's apply that to the exclusive-or function:
$$ x \text{ xor } y = (x \text{ or } y) \text{ and not } (x \text{ and } y).$$
If we blindly apply the transformations for AND, OR, and NOT, we get:
$$ x \text{ xor } y \to (x + y - xy)(1-xy). $$ Expand that out into monomials:
$$ (x + y - xy)(1-xy) = x + y - xy - x^2y - xy^2 + x^2y^2.$$Now replace $A^2$ by $A$ whenever possible:
$$x + y - xy - x^2y - xy^2 + x^2y^2 \to x + y - xy - xy - xy + xy = x + y - 2xy,$$ which agrees with my canonical expression/transformation rule for XOR stated above.
| 11 | https://mathoverflow.net/users/302 | 4951 | 3,317 |
https://mathoverflow.net/questions/4973 | 15 | Let $G$ be a simple group (say $SL\_n$) and let $B$ be a Borel subgroup (say upper triangular matrices). Then all irreducible representations of $G$ are induced from one-dimensional representations of $B$, i.e characters of $B$. However, only some of the weights will induce to non-zero representations. Relative to the standard maximal torus of diagonal matrices, these weights appear to be the anti-dominant weights. However, $B$ contains other tori as well, and no one choice is canonical.
My question is: since induction from $B$ didn't require a choice of torus, how are some characters of $B$ already deemed to be anti-dominant? Where is the symmetry broken?
| https://mathoverflow.net/users/788 | How Does a Borel Subgroup Know Which Weights Are Dominant | One of the sneaky tricks that Lie theorists play on students is that they tell them about Cartan subalgebras, and then at some point, they pull the rug out and say "just joking; really you should think about the **abstract Cartan**." The **abstract Cartan** is a Borel mod its radical. You might say "which Borel?" but it doesn't matter; all these quotients are canonically isomorphic (any two Borels are conjugate, and any two ways of making them conjugate differ by an element of the Borel).
Note, a Cartan subgroup of your Lie algebra isn't canonically isomorphic to the abstract Cartan; you have to choose a Borel containing that Cartan subgroup first. That's where the symmetry is broken.
The abstract Cartan has a canonical notion of positive weight: you look at the action of $B/N$ on the quotient of the Lie algebra $n/[n,n]$, and the weights that appear there are your simple roots (all the positive roots can be obtained by taking the associated graded for the filtration $n\supset [n,n]\supset [n,[n,n]]$). Again, any way of making Borels conjugate carries these weights to the corresponding ones for the other Borel. So once you've picked a Borel, you can use the isomorphism of the abstract Cartan to your chosen one to get a root system on your chosen one.
**EDIT:** As Allen notes below: I learned most of this from the book of Chriss and Ginzburg.
You get fundamental coroots by taking minimal parabolics over your Borel (those obtained by adding one negative simple root). There's a canonical map from the abstract Cartan of $P/[P,P]$ to that of $G$, and there's unique coweight of $P/[P,P]$ (which is $SL\_2$ or $PSL\_2$) so that the action on the unique simple root space has eigenvalue 2. The image of that is the simple coroot.
| 16 | https://mathoverflow.net/users/66 | 4980 | 3,332 |
https://mathoverflow.net/questions/981 | 13 | What can be said about rational self-maps of $\mathbb P^1$
for which all critical points are also fixed points ?
If all but one of the fixed points are critical, there is
a characterization in <http://arxiv.org/abs/math/0411604v1>
( see Corollary 1 and the discussion just after the statement ).
Still assuming that all critical points are fixed: Is it possible to bound the degree of the rational map if all but two of the fixed points are critical ?
I think that the answer is probably no, but I would really
love to hear the contrary.
---
**Motivation.** The question is motivated by a rather specific problem I like to think about from time to time. It concerns the
classification of some special arrangements of lines on the projective plane. More specifically, I would like to classify arrangements
of $3d$ lines(or rather hyperplanes through the origin of $\mathbb C^3$) invariant by degree $d$ homogeneous polynomial vector fields on $\mathbb C^3$. Given one arrangement like that one can produce a degree $d$ rational map having all its critical points fixed.
---
**Update. (08/28/2013)** The paper [On the classification of critically fixed rational maps](http://arxiv.org/abs/1308.5895) by Cordwell, Gilbertson, Nuechterlein, Pilgrim, and Pinella classifies rational maps for which all critical points are fixed. In particular, for every degree $d\ge 3$ there exists a rational such that all but two of the fixed points are critical.
| https://mathoverflow.net/users/605 | Rational maps with all critical points fixed | Consider
$$z \mapsto \frac{(n-2) z^n + n z}{n z^{n-1} + (n-2)}.$$
This has $n+1$ fixed points, at $0$, $\infty$, and the $(n-1)$-st roots of $-1$. The only critical points are the roots of $-1$, each of which is ramified of index $3$. So this is a map with all critical points fixed, and all fixed points but two critical.
I am tempted to leave it at that. But, being a nice person, I will explain how I found this. Moreover, I will show that this is (up to conjugacy), the only degree $n$ map with $n+1$ distinct fixed points, two of which are not critical and the rest of which are critical with multiplicity $2$ (ramification index $3$.)
We can take the two noncritical fixed points to be $0$ and $\infty$. So our map is of the form
$$z \mapsto z - z p/q,$$
where $p$ and $q$ are of degree $n-1$ and relatively prime. The $n-1$ fixed points other than $0$ and $\infty$ are the roots of $p$.
The derivative of this map is
$$\frac{q^2 - zqp' + zpq' - pq}{q^2}.$$
The condition that the fixed points other than $0$ and $\infty$ be critical means $p$ divides the numerator. So $p$ divides $q (q-zp')$ and, as $p$ and $q$ are relatively prime, we conclude that $q - z p' = kp$. Checking degrees, $k$ has degree $0$ and is thus a constant, to be determined later.
Now, we want to impose the stronger condition that every zero of $p$ be doubly a critical point, so the numerator is $\ell p^2$ for some constant $\ell$.
Plugging in $q = kp + z p'$, and simplifying
$$\ell p^2 = p \left( k(k-1) p + 2 k z p' + z^2 p'' \right).$$
Cancelling $p$ from both sides,
$$\ell p = k(k-1) p + 2 k z p' + z^2 p''.$$
Plugging in $z=0$, and noting that $p(0) \neq 0$, we get $\ell = k(k-1)$. So
$$2 kz p' + z^2 p'' =0.$$
The solution to this differential equation is $p = C z^{1-2k} + D$. But we know that $z$ has degree $n-1$, so $1-2k=n-1$ and $k = -(n-2)/2$.
Taking the simplest choices $C=D=1$ and plugging back in gives the above solution. All other solutions are related to this one by rescaling the variable $z$.
---
I was thinking about this very old question again, due to conversations with Sarah Koch. Here is another infinite family of examples which we can write down explicitly: Let $g(x)$ be the degree $n$ [Gegenbauer polynomial](https://mathoverflow.net/questions/292703) satisfying
$$g'' = \frac{n(n-1)}{x^2-1} g.$$
Let $f(x)$ be the Newton recursion
$$f(x) = x-g/g'.$$
Then
$$f' = \frac{g g''}{(g')^2} = \frac{n (n-1) g^2}{(x^2-1) (g')^2} .$$
So every critical point of $f$ is a root of $g$, and hence a fixed point of $f$.
More specifically, $\pm 1$ are critical points of multiplicity $1$ and the other $n-2$ roots of $g$ have multiplicity $2$. The only fixed point which is not critical is the fixed point at $\infty$.
On reading further, I see that this example is in section 11 of the Cordwell et al paper.
---
There has been a lot of research on this question in the last few years. As well as the paper of Cordwell et al linked by the OP, see the Ph. D thesis of [Hlushchanka](https://opus.jacobs-university.de/frontdoor/index/index/docId/750) and [this paper of his](https://arxiv.org/abs/1904.04759).
| 12 | https://mathoverflow.net/users/297 | 4983 | 3,334 |
https://mathoverflow.net/questions/4943 | 9 | I'm working my way through Lang's *Fundamentals of Differential Geometry*, and when he introduces vector bundles, he states that for finite dimensional bundles, the third axiom is redundant. I'm hoping someone can give a counterexample in infinite dimensions.
His axioms (for a $C^p$ bundle) are (1) local triviality, (2) transition maps are Banach space isomorphisms (linear homeomorphisms), and (3) the maps $x\mapsto A\_x$ are $C^p$ (where $A\_x$ is a transition map).
Essentially, I'm looking for a $C^p$ automorphism of $B\times V$ of the form $(x,v)\mapsto(x,A\_xv)$ such that $x\mapsto A\_x$ is not $C^p$. Here, $B$ is open (say the unit ball) in some Banach space, $V$ is another Banach space, and the linear maps are given the operator norm.
All derivatives are Fréchet derivatives.
| https://mathoverflow.net/users/300 | "Vector bundle" with non-smoothly varying transition functions | I suspect that the issue here is actually to do with continuity, not differentiability. Without more details on the exact definition of vector bundle as given, I can't be sure (and I don't have a copy of Lang's book to hand to check). If the definition is a "top down" one, then continuity is certainly an issue. By "top down" then I mean that a vector bundle consists of two smooth manifolds, $E$ and $B$, and a smooth map $p : E \to B$ satisfying the above conditions. (The other approach is to build it up from the transition functions.)
I'll assume that this is so.
Then from local triviality and the fibrewise description, we obtain the transition functions $\psi : U \times F \to U \times F$. Now, from the properties we can find a function $\theta : U \to GL(F)$ with the property that $\psi(x,v) = (x,\theta(x)v)$. There is no difficulty in simply *defining* this function.
The problem is that continuity of $\psi$ (even higher differentiability) is *not* sufficient to guarantee the continuity of $\theta$ **in infinite dimensions**. The map $\theta$ is continuous if $GL(F)$ is given the **weak** topology where $(A\_n) \to A$ if $(A\_nv) \to Av$ and $(A\_n^{-1}v) \to A^{-1}v$ for all $v$. But normally we ask for the strong (norm) topology on $GL(F)$.
In finite dimensions, this topology agrees with the standard topology but in infinite dimensions they are very different. For example, if we take $\ell^2$ and let $P\_n$ be the projection onto the first $n$-coordinates then $(P\_n) \to I$ in the weak topology but not in the strong topology (this is an example in $L(H)$ but can be easily tweaked to give an example in $GL(H)$).
Further reading:
* Topological Vector Spaces, Schaefer. Contains lots about the different topologies.
* A Convenient Setting of Global Analysis, Kriegl and Michor. Contains lots about the intricacies of infinite dimensional differential topology.
**Edit:** I finally remembered the classic example of this: $L^2$-functions on a Lie group. For simplicity, let's take $S^1$. Then $S^1$ acts in the obvious way on $L^2(S^1,\mathbb{C})$ (hereinafter $L^2$). The action $S^1 \times L^2 \to L^2$ is jointly continuous but the associated map $S^1 \to GL(H)$ is most assuredly not. Indeed, if $\lambda \ne \mu \in S^1$ then $\|R\_\lambda - R\_\mu\| = 2$ so the image is discrete. So if we have an $S^1$-principal bundle $P \to B$ then we can form a new space by taking the quotient $P \times\_{S^1} L^2$. This will be locally trivial, and the transition maps will be fibrewise linear as they are of the form $x \mapsto R\_{\lambda(x)} : L^2 \to L^2$ where $x \mapsto \lambda(x)$ are the transition functions of the $S^1$-bundle. But the associated map $x \to GL(H)$ is not continuous and so it won't be a genuine vector bundle in the sense Lang defines.
(What's particularly embarrassing about how long it took me to remember this example is that in a recent paper I go into great detail about the different "levels" one can require for continuity of the action of subgroups of $Diff(S^1)$ on various loop spaces. The paper in question is [this one](http://www.math.ntnu.no/~stacey/Research/Papers/smooth.html) in case anyone's interested.)
| 12 | https://mathoverflow.net/users/45 | 4997 | 3,344 |
https://mathoverflow.net/questions/4994 | 217 | It is not unusual that a single example or a very few shape an entire mathematical discipline. Can you give examples for such examples? (One example, or few, per post, please)
I'd love to learn about further basic or central examples and I think such examples serve as good invitations to various areas. (Which is why a bounty was offered.)
---
Related MO questions: [What-are-your-favorite-instructional-counterexamples](https://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples),
[Cannonical examples of algebraic structures](https://mathoverflow.net/questions/3242/canonical-examples-of-algebraic-structures), [Counterexamples-in-algebra](https://mathoverflow.net/questions/29006/counterexamples-in-algebra), [individual-mathematical-objects-whose-study-amounts-to-a-subdiscipline](https://mathoverflow.net/questions/49082/individual-mathematical-objects-whose-study-amounts-to-a-subdiscipline), [most-intricate-and-most-beautiful-structures-in-mathematics](https://mathoverflow.net/questions/49151/most-intricate-and-most-beautiful-structures-in-mathematics), [counterexamples-in-algebraic-topology](https://mathoverflow.net/questions/55365/counterexamples-in-algebraic-topology), [algebraic-geometry-examples](https://mathoverflow.net/questions/34110/algebraic-geometry-examples), [what-could-be-some-potentially-useful-mathematical-databases](https://mathoverflow.net/questions/68442/what-could-be-some-potentially-useful-mathematical-databases), [what-is-your-favorite-strange-function](https://mathoverflow.net/questions/22189/what-is-your-favorite-strange-function); [Examples of eventual counterexamples](https://mathoverflow.net/questions/15444/examples-of-eventual-counterexamples) ;
---
To make this question and the various examples a more useful source there is a **[designated answer](https://mathoverflow.net/questions/4994/fundamental-examples/37153#37153)** to point out connections between the various examples we collected.
---
In order to make it a more useful source, I list all the answers in categories, and added (for most) a date and (for 2/5) a link to the answer which often offers more details. (~year means approximate year, \*year means a year when an older example becomes central in view of some discovery, year? means that I am not sure if this is the correct year and ? means that I do not know the date. Please edit and correct.) Of course, if you see some important example missing, add it!
**Logic and foundations:** [$\aleph\_\omega$](https://mathoverflow.net/questions/4994/fundamental-examples/6191#6191) (~1890), [Russell's paradox](https://mathoverflow.net/questions/4994/fundamental-examples/6706#6706) (1901),
[Halting problem](https://mathoverflow.net/questions/4994/fundamental-examples/5171#5171) (1936), [Goedel constructible universe L](https://mathoverflow.net/questions/4994/fundamental-examples/5326#5326) (1938), [McKinsey formula](https://mathoverflow.net/questions/4994/fundamental-examples/5185#5185) in modal logic (~1941), [3SAT](https://mathoverflow.net/questions/4994/fundamental-examples/5227#5227) (\*1970), The theory of Algebraically closed fields (ACF) (?),
**Physics:** [Brachistochrone problem](https://mathoverflow.net/questions/4994/fundamental-examples/6539#6539) (1696), Ising model (1925), [The harmonic oscillator,](https://mathoverflow.net/questions/4994/fundamental-examples/4998#4998)(?) Dirac's delta function (1927), [Heisenberg model](https://mathoverflow.net/questions/4994/fundamental-examples/13840#13840) of 1-D chain of spin 1/2 atoms, (~1928), Feynman path integral (1948),
**Real and Complex Analysis:** [Harmonic series](https://mathoverflow.net/questions/4994/fundamental-examples/5245#5245) (14th Cen.) {and Riemann zeta function (1859)}, the Gamma function (1720), li(x), [The elliptic integral that launched Riemann surfaces](https://mathoverflow.net/questions/4994/fundamental-examples/5052#5052) (\*1854?), [Chebyshev polynomials](https://mathoverflow.net/questions/4994/fundamental-examples/8818#8818) (?1854) [punctured open set in C^n](https://mathoverflow.net/questions/4994/fundamental-examples/5006#5006) (Hartog's theorem \*1906 ?)
**Partial differential equations**: [Laplace equation (1773), the heat equation, wave equation,](https://mathoverflow.net/questions/4994/fundamental-examples/5011#5011) [Navier-Stokes equation](https://mathoverflow.net/questions/4994/fundamental-examples/6490#6490) (1822),[KdV equations](https://mathoverflow.net/questions/4994/fundamental-examples/5038#5038) (1877),
**Functional analysis:** Unilateral shift, The spaces $\ell\_p$, $L\_p$ and $C(k)$, [Tsirelson spaces](https://mathoverflow.net/questions/4994/fundamental-examples/5208#5208) (1974), Cuntz algebra,
**Algebra:** Polynomials (ancient?), Z (ancient?) and Z/6Z (Middle Ages?), symmetric and alternating groups (\*1832), Gaussian integers ($Z[\sqrt -1]$) (1832), [$Z[\sqrt(-5)]$](https://mathoverflow.net/questions/4994/fundamental-examples/5042#5042),[$su\_3$ ($su\_2)$](https://mathoverflow.net/questions/4994/fundamental-examples/5002#5002), [full matrix ring over a ring](https://mathoverflow.net/questions/4994/fundamental-examples/5010#5010), $\operatorname{SL}\_2(\mathbb{Z})$ and SU(2), [quaternions](https://mathoverflow.net/questions/4994/fundamental-examples/5201#5201) (1843), [p-adic numbers](https://mathoverflow.net/questions/4994/fundamental-examples/5498#5498) (1897), Young tableaux (1900) and Schur polynomials, cyclotomic fields, Hopf algebras (1941) Fischer-Griess monster (1973), Heisenberg group, ADE-classification (and Dynkin diagrams), Prufer p-groups,
**Number Theory:** conics and pythagorean triples (ancient), [Fermat equation](https://mathoverflow.net/questions/4994/fundamental-examples/5050#5050) (1637), [Riemann zeta function](https://mathoverflow.net/questions/4994/fundamental-examples/5245#5245) (1859) [elliptic curves](https://mathoverflow.net/questions/4994/fundamental-examples/5222#5222), transcendental numbers, Fermat hypersurfaces,
**Probability:** [Normal distribution](https://mathoverflow.net/questions/4994/fundamental-examples/4995#4995) (1733), [Brownian motion](https://mathoverflow.net/questions/4994/fundamental-examples/5319#5319) (1827), The percolation model (1957), The Gaussian Orthogonal Ensemble, the Gaussian Unitary Ensemble, and the Gaussian Symplectic Ensemble, SLE (1999),
**Dynamics:** [Logistic map](https://mathoverflow.net/questions/4994/fundamental-examples/5046#5046) (1845?), [Smale's horseshoe map](https://mathoverflow.net/questions/4994/fundamental-examples/24058#24058)(1960). [Mandelbrot set](https://mathoverflow.net/questions/4994/fundamental-examples/5174#5174) (1978/80) (Julia set), [cat map](https://en.wikipedia.org/wiki/Arnold's_cat_map), (Anosov diffeomorphism)
**Geometry:** [Platonic solids](https://mathoverflow.net/questions/4994/fundamental-examples/5462#5462) (ancient), the Euclidean ball (ancient), The configuration of 27 lines on a cubic surface, The configurations of Desargues and Pappus, construction of regular heptadecagon (\*1796), [Hyperbolic geometry](https://mathoverflow.net/questions/4994/fundamental-examples/6709#6709) (1830), Reuleaux triangle (19th century), [Fano plane](https://mathoverflow.net/questions/4994/fundamental-examples/5016#5016) (early 20th century ??), cyclic polytopes (1902), Delaunay triangulation (1934) [Leech lattice](https://mathoverflow.net/questions/4994/fundamental-examples/5172#5172) (1965), [Penrose tiling](https://mathoverflow.net/questions/4994/fundamental-examples/6586#6586) (1974), [noncommutative torus](https://mathoverflow.net/questions/4994/fundamental-examples/5034#5034), cone of positive semidefinite matrices, the associahedron (1961)
**Topology:** Spheres, [Figure-eight knot](https://mathoverflow.net/questions/4994/fundamental-examples/5052#5052) (ancient), trefoil knot (ancient?) (Borromean rings (ancient?)), the torus (ancient?), Mobius strip (1858), Cantor set (1883), Projective spaces (complex, real, quanterionic..), Poincare dodecahedral sphere (1904), [Homotopy group of spheres](https://mathoverflow.net/questions/4994/fundamental-examples/5019#5019), Alexander polynomial (1923), [Hopf fibration](https://mathoverflow.net/questions/4994/fundamental-examples/5024#5024) (1931), The standard embedding of the torus in R^3 (\*1934 in Morse theory), pseudo-arcs (1948), Discrete metric spaces, Sorgenfrey line, Complex projective space, [the cotangent bundle](https://mathoverflow.net/questions/4994/fundamental-examples/5000#5000) (?), The Grassmannian variety,[homotopy group of spheres](https://mathoverflow.net/questions/4994/fundamental-examples/5019#5019) (\*1951), [Milnor exotic spheres](https://mathoverflow.net/questions/4994/fundamental-examples/5004#5004) (1965)
**Graph theory:** [The seven bridges of Koenigsberg](https://mathoverflow.net/questions/4994/fundamental-examples/24125#24125) (1735), [Petersen Graph](https://mathoverflow.net/questions/4994/fundamental-examples/5049#5049) (1886), two edge-colorings of K\_6 (Ramsey's theorem 1930), K\_33 and K\_5 (Kuratowski's theorem 1930), Tutte graph (1946), Margulis's expanders (1973) and Ramanujan graphs (1986),
**Combinatorics:** tic-tac-toe (ancient Egypt(?)) (The game of nim (ancient China(?))), [Pascal's triangle](https://mathoverflow.net/questions/4994/fundamental-examples/24010#24010) (China and Europe 17th), [Catalan numbers](https://mathoverflow.net/questions/4994/fundamental-examples/5212#5212) (18th century), (Fibonacci sequence (12th century; probably ancient), Kirkman's schoolgirl problem (1850), surreal numbers (1969), alternating sign matrices (1982)
**Algorithms and Computer Science:** Newton Raphson method (17th century), Turing machine (1937), RSA (1977), universal quantum computer (1985)
**Social Science:** [Prisoner's dilemma](https://mathoverflow.net/questions/4994/fundamental-examples/5017#5017) (1950) (and also the chicken game, chain store game, and centipede game), the model of exchange economy, second price auction (1961)
**Statistics:** the Lady Tasting Tea (?1920), Agricultural Field Experiments (Randomized Block Design, Analysis of Variance) (?1920), Neyman-Pearson lemma (?1930), Decision Theory (?1940), the Likelihood Function (?1920), Bootstrapping (?1975)
| https://mathoverflow.net/users/1532 | Fundamental Examples | The [harmonic oscillator](http://en.wikipedia.org/wiki/Harmonic_oscillator) is a fundamental example in both classical and quantum mechanics.
| 66 | https://mathoverflow.net/users/394 | 4998 | 3,345 |
https://mathoverflow.net/questions/4964 | 18 | The basic relationship in algebraic geometry is between a variety and its ring of functions. Arguably a similarly basic relationship in quantum mechanics is between a state space and its algebra of observables. In what sense is algebraic geometry a "classical" (i.e. commutative) phenomenon? How does intuition from quantum mechanics influence how one should think about algebraic geometry and vice versa? What modern areas of research study their interaction?
(This question is loosely inspired by a [similar question](https://mathoverflow.net/questions/2557/examples-of-applications-of-the-borel-weil-bott-theorem) about representation theory.)
| https://mathoverflow.net/users/290 | What is the relationship between algebraic geometry and quantum mechanics? | I also would say it's non-commutative geometry and in particular the notion of spectral triple that is the geometric formalism directly coming out of quantum mechanics.
A sketchy description from this point of view is at [nLab:spectral triple](https://ncatlab.org/nlab/show/spectral+triple).
| 16 | https://mathoverflow.net/users/381 | 5028 | 3,367 |
https://mathoverflow.net/questions/5001 | 12 | In algebraic topology, we define a degree for a map $f: S^n \to S^n$ as where the induced map $f\_\*$ on the $n$-th homology group of $S^n$ sends $1$.
In differential topology, we have a different (same?) notion of degree for $f$. You take a regular value $b \in S^n$, consider $f^{-1} (b)$ (which is finite by the inverse function theorem and some compactness argument), and take the difference between the number of points in the preimage where the Jacobian of $f$ is positive and the number of points in the preimage where the Jacobian of $f$ is negative.
Geometrically, I can see that they are the same, but I couldn't convince myself rigorously. In Prop 2.30 of Hatcher, he mentions that the degree of $f$ is the sum of the local degrees of $f$ at each preimage point, and local degrees are either $\pm 1$. (Local degree is defined in the middle of page 136 in Hatcher.)
So, the final question is, must the sign of the local degree of $x \in f^{-1}(b)$ the same as the sign of the Jacobian of $f$ at $x$?
| https://mathoverflow.net/users/491 | Notions of degree for maps $S^n \to S^n$? | I think what you need is the following lemma (usually called the "Stack of records" lemma):
Consider a smooth proper map of manifolds of the same dimension $f \colon M \to N$ and let $y \in N$ be a regular value of $f$.
Then there exists a neighbourhood $V \subset N$ of $y$ such that $f^{-1}(V) = \cup\\_{i=1}^n U\\_i$ with $U\\_i \cap U\\_j = \emptyset$ for $i \neq j$ and $f|\\_{U\_i} \colon U\\_i \to V$ is a diffeomorphism for all $i$.
Now from this you can just sum up $\pm 1$ according to orientation on each $U\_i$ to get the local degree of $f$ at $y$, and this works for both definitions of degree.
| 7 | https://mathoverflow.net/users/362 | 5030 | 3,369 |
https://mathoverflow.net/questions/4901 | 3 | I have often heard of various statements being independent from the axioms of set theory (typically ZFC). Some examples include
* The continuum hypothesis is probably the most famous
* The independence of the axiom of choice from plain ZF
* My professor told me that the following theorem is independent from the standard axioms: *Theorem*: If $U$ is a regular set then $U\times [0,1]$ is regular.
I'm wondering what a proof of such a statement would look like. What context do you do the proof in? What kind of theoretical framework do you have to build up before you can answer such questions?
In addition to answers I would also be interested in resources that would let me find out more about these ideas. I'm looking for books/web sites that start a relatively elementary level but still build up to dealing with some of the examples I mentioned above.
| https://mathoverflow.net/users/936 | Independence from Set Theory Axioms | It might help to understand look at how the fifth postulate is proved independent of Euclid's other axioms: One constructs a model, such as the Poincare disc, where the axioms can be given new interpretations. So the word LINE now means "arc perpindicular to the boundary of the disc", the word CONGRUENT now means "related by a Mobius transformation fixing the boundary of the disc" and so forth. One then checks that, if you take each axiom and replace the capitalized words by the quoted strings, the axiom remains true, except for the fifth postulate, which becomes false.
Similarly, to show that AC is independent of ZF, one works inside ZFC and builds a model where the axioms of ZF, suitably reinterpreted, stay true but where choice is false. The technical tool used to build that model is called forcing. I don't really understand it, but Timothy Chow's introduction is the closest I have come to doing so.
| 8 | https://mathoverflow.net/users/297 | 5048 | 3,382 |
https://mathoverflow.net/questions/5045 | 11 | Why the word "hidden" present in hidden markov model? What exactly is hidden.
Whatever is hidden in HMM isn't it hidden in normal Markov Models?
| https://mathoverflow.net/users/1692 | What is hidden in Hidden Markov Models? | The unobserved state.
Let's consider a hidden Markov model for my cat's behavior. Bella can be in five states: hungry, tired, playful, cuddly, bored. She can respond to these states with six behaviors: whining, scratching, cuddling, pouncing, sleeping and stalking.
A hidden Markov model would consist of two matrices, one 5x5 and the other 5x6. The 5x5 matrix gives the probabilities that, if she is hungry at time $t$, she will be tired at time $t+1$, and so forth. So we can compute the probability that she is in different emotional states by taking powers of this matrix.
However, we can't observe her emotions -- they are hidden. The 5x6 matrix gives the probability that, if she is hungry at time $t$, she will whine at time $t+1$. (Very close to $1$.) These are the behaviors we observe.
In an ordinary Markov model, there would just be a single 6x6 matrix, which directly described the probability of transitions like whining ---> clawing. As you can see, an ordinary Markov model is less able to reflect the complexity of my cat's inner life.
See the [wikipedia article](http://en.wikipedia.org/wiki/Hidden_markov_model) for much more information.
| 28 | https://mathoverflow.net/users/297 | 5053 | 3,387 |
https://mathoverflow.net/questions/5008 | 4 | Given a graded ring $S$ and a graded S-module $M$ we can carry out a construction in order to get $\tilde{M}$, which is a sheaf over the scheme $\mathrm{Proj}~ S$. With this in view, I have an equivalence of categories between the category of (quasi-finite generate) modules and the category of $q$-coherent sheaves over $\mathrm{Proj}~S$.
On the other hand, given a locally free sheaf $\mathcal{F}$ of rank $n$ over $\mathrm{Proj}~S$ we can get a vector bundle out of it and further we have a 1-1 correspondence between isomorphism classes of free sheaves of rank $n$ and isomorphism classes of vector bundles of rank $n$.
With the last two facts in view, my question is the following. if I start with a finite generated module $M$ then the sheaf $\tilde{M}$ is locally free? If so, I can get from $M$ a locally free sheaf $\tilde{M}$ and from such a sheaf a vector bundle (and perhaps backwards as well). Therefore, Is it the same having a vector bundle over $\mathrm{Proj}~ S$ than a $S$-Module? or what are the limits of such a relation described here among Vector Bundles & $S$-Modules?. By "Is it the same" I mean, We have the same amount of information in such objects.
| https://mathoverflow.net/users/1547 | Modules, Sheaves and Vector bundles | There are (at least) two details you are missing.
(1) This is not an equivalence of categories between finitely generated graded modules and coherent sheaves. If your module is $0$ in all sufficiently large degrees, then the corresponding sheaf will be zero. For example, let $S=k[x,y]$ and $M=S/\langle x,y \rangle$. The sheaf on $\mathbb{P}^1$ corresponding to $M$ is the zero sheaf.
The category you want to work with is the one whose objects are finitely generated graded modules, and where we formally invert any map which is an isomorphism in sufficiently large degree. (Alternative formulation: we formally invert a map $f:M \to N$ if, for any $s$ in the irrelevant ideal, $s^{-1} f: s^{-1} M \to s^{-1} N$ is an isomorphism.)
(2) Not every coherent sheaf is a vector bundle. Correspondingly, not every finitely generated graded module will correspond to a vector bundle. If we were dealing with an affine variety, vector bundles would correspond to locally free modules. (Also called projective modules.) For projective varieties, things are a little trickier: the criterion is to be locally free away from the irrelevant ideal.
Sadly, I believe that there exist examples of modules which are locally free away from the irrelevant ideal, but are not isomorphic (in the above category) to any module which is locally free at the irrelevant ideal. This should be related to my question [here](https://mathoverflow.net/questions/4590/when-are-dual-modules-free). But you can go a long while without paying attention to this detail.
| 7 | https://mathoverflow.net/users/297 | 5056 | 3,390 |
https://mathoverflow.net/questions/5036 | 6 | Gelfand-Naimark structure theorem for $C^\* $ algebras gives a canonical isometric \* isomorphism between any commutative unital $C^\* $ algebra $A$ and the algebra of continuous complex-valued functions on $A$^. This is the spectrum (or structure space) of $A$, i.e. the non-zero multiplicative linear continuous functionals with the topology of pointwise convergence (alias weak\*), which is compact and hausdorff. Apart from the easy case $A = C(X)$, with $X$ compact hausdorff, for which $A$^ is $X$ itself, there are a lot of non trivial and not immediately visible examples of spectra, for example:
If $X$ loc. compact hausdorff $A = C\_b(X)$ (continuous and bounded functions with uniform topology) is a $C^\*$ algebra. If X is non compact then A^ cannot be $X$ and is in fact $\beta X$, the Stone-Cech compactification of $X$.
If $X$ is loc. compact hausdorff and you take $C\_0(X)$, then you get another compactification of $X$.
If instead you simply take $C(X)$ for $X$ compact non-hausdorff you get a natural "hausdorfization" of $X$.
I'm particularly interested in the existence of other constructions which can be described by gelfand theory as above. I mean to associate functorially to each space (in an appropriate subcategory of Top, maybe not full) a $C^\*$ algebra and then to look at its spectra.
A related question: what are the spectra of $L^\infty(R)$, and similar algebras (maybe $L^\infty(G)$, G loc. compact group with haar measure)?
| https://mathoverflow.net/users/1049 | Spectra of $C^*$ algebras | The spectrum of $L^\infty(R)$ is the hyperstonean space associated with the measurable space R.
More information can be found in Takesaki's Theory of Operator Algebras I, Chapter III, Section 1.
| 7 | https://mathoverflow.net/users/402 | 5058 | 3,391 |
https://mathoverflow.net/questions/4988 | 9 | This is a very silly question.
For all regions S contained inside the unit square, what is the infimum of the quantity Perimeter(S)/Area(S)? This ratio being considered is not scale invariant, so it is only the constraint of being contained within the square which implies that this infimum is non-zero.
There are some "obvious" configurations to try, but I do not even know how to use a calculus of variations argument to show that these are local maxima.
Can one do better than $\displaystyle{2 + \sqrt{\pi}}$?
EDIT: One can probably show that an optimal S has $\mathbf{Z}/4\mathbf{Z}$ symmetry. Assume that S consists of line segments along part of boundary, leaving segments of length x at each corner, and then (four quarters of) a shape T in the corner. We can assume that T has volume Ax^2 and perimeter Px. Minimizing the quantity with respect to x, one obtains a minimum of:
$$2 +\sqrt{4 + \frac{(P/2-4)^2}{(A - 4)}},$$
which becomes $2 + \sqrt{\pi}$ when T is the circle, i.e., $A = \pi$ and $P = 2 \pi$.
| https://mathoverflow.net/users/1684 | Minimize Perimeter(S)/Area(S) for S inside the unit square. | Basic observation: the solution will consist of circular arcs joining the sides of the square to each other, and some sides of the square.
Lemma: Consider a line segment of length $\ell$, and real number $p \geq \ell$. Consider all planar regions which contain the line segment in their boundary, and for which the rest of their boundary has length $p$. The largest possible area of such a region is obtained when the nonlinear part of the boundary is a circular arc.
Proof: Otherwise, we could take a circle, draw a chord of length $\ell$ across it cutting off an arc of length $p$, and replace this segment by another shape, to obtain a shape with the same perimeter as the circle and larger area.
Using the basic observation, your problem turns from a problem in the calculus of variations to one in multivariable calculus: mark the 8 points at which your shape leaves the sides of the square, and the radii of the 4 circles which cut the corners. Since, as you observed, the solution has eight-fold symmetry, there are really just two quantities: the length of the four boundary segments and the radius (or the angle) of the arcs cutting the corners.
| 9 | https://mathoverflow.net/users/297 | 5060 | 3,393 |
https://mathoverflow.net/questions/4958 | 17 | Do you know natural examples of triangulated categories (or [presentable] stable $\infty$-categories) which are not compactly generated? (ideally they'd be defined algebraically, but curious to hear any examples.. the ones I know are gotten as opposites of compactly generated categories or by slightly ad hoc geometric constructions)
| https://mathoverflow.net/users/582 | Categories which are not compactly generated | As David says, D(R) is compactly generated. This means Brown representability for COHOMOLOGY is automatically true, but that does NOT mean Brown representability for HOMOLOGY is true, and in fact it is not always true. That is what the Christensen-Keller-Neeman example shows.
Let K(R) be the category of chain complexes over R and chain homotopy classes of chain maps. Then K(Z) is not compactly generated (and K(R) is not either for many R). This is proved in a paper of mine with Christensen.
[Quillen model structures for relative homological algebra](http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=122357)
| 16 | https://mathoverflow.net/users/1698 | 5062 | 3,395 |
https://mathoverflow.net/questions/5057 | 7 | This is a general question about group cohomology. I'm interested in the case when the coefficients are the rational numbers and hence I suppose when my groups are infinite. The question splits into two:
1) Are there any favoured examples that you would recommend a look at? (Recommended references would be just as welcome.)
And the main question:
2) What sort of functors on the category of groups leave the rational cohomology unchanged? In particular is there a projection onto a special subcategory of groups that is in some way the right category to study?
I have a feeling that someone with a good knowledge of rational homotopy theory would be able to answer this question with relative ease.
| https://mathoverflow.net/users/109 | Rational Group Cohomology | Stallings showed that if $f:\Gamma \to \Delta$ is a homomorphism between finitely presented groups where $f\_i:H\_i(\Gamma,Q) \to H\_i(\Delta,Q)$ is bijective for $i\le 1$ and surjective for $i=2$ then $f\otimes Q: \Gamma \otimes Q \to \Delta \otimes Q$ is an isomorphism between the $Q$-unipotent completions. So "taking the $Q$-unipotent completion" is a functor with some interesting properties with respect to rational group (co)homology. I'm not sure if this is the kind of thing you're after ...
Stallings' paper is
MR0175956 (31 #232)
Stallings, John
Homology and central series of groups.
J. Algebra 2 1965 170--181.
| 11 | https://mathoverflow.net/users/1672 | 5064 | 3,396 |
https://mathoverflow.net/questions/5068 | 3 | Hello
I am trying to get a good book that explains the Dolbeault Cohomology, does anyone know of a good one?
| https://mathoverflow.net/users/997 | Dolbeault cohomology | Chern's book "Complex manifolds without potential theory" is a good book, and it does explain Dolbeault Cohomology. But it's a short book, and it explains it concisely. If you need more details, you could also try Griffiths-Harris (but I greatly prefer Chern's book).
Kodaira's book "Complex manifolds and deformations of complex structures" is much more leisurely, and with great attention paid to exposition and detail (it doesn't appeal to some people, but I enjoyed it a lot).
| 7 | https://mathoverflow.net/users/1672 | 5072 | 3,401 |
https://mathoverflow.net/questions/5065 | 26 | This is a question asked out of curiosity, and because I can't understand the Wikipedia page.
I have often been told that PA cannot prove the validity of induction up to $\varepsilon\_0$, which has been expressed to me roughly as the claim that $\varepsilon\_0$ is well-ordered. I understand what ordinals are, and what $\varepsilon\_0$ is. I also understand first order logic and axiom schemes, so I understand how the induction axiom scheme formalizes the notion that $\omega$ is well-ordered.
What I don't understand is how one could formulate the statement that $\varepsilon\_0$ is well-ordered as a first order sentence in arithmetic. Would someone mind spelling this out for me?
| https://mathoverflow.net/users/297 | What is induction up to $\varepsilon_0$? | Here's a more detailed answer:
The [above-mentioned](https://mathoverflow.net/questions/5065/what-is-induction-up-to-varepsilon-0#comment5716_5065) link (recreated below [1]) constructs a recursive relation $E$ on $\omega$, such that $(\omega, E)$ is isomorphic to $(\epsilon\_0, \in )$. Then, induction up to $\epsilon\_0$ is interpreted as $E$-induction, that is, for every predicate $\phi$, if $(\forall x E y \phi(x))\rightarrow \phi(y)$ then $\forall y \phi(y)$.
[1] P. Tosi, *Normal derivability and first-order arithmetic*,
Notre Dame J. Formal Logic 21(2): 449-466 (April 1980) doi:[10.1305/ndjfl/1093883058](https://doi.org/10.1305/ndjfl/1093883058).
| 12 | https://mathoverflow.net/users/1061 | 5076 | 3,404 |
https://mathoverflow.net/questions/5116 | 2 | I am reasonably certain this is the case, but can't find a reference that actually states this, although the Wikipedia article states something close.
| https://mathoverflow.net/users/290 | Is the existence of a well-ordering on R independent of ZF? | It is possible to have all the subsets of R be measurable (Solovay, Robert M. (1970). "A model of set-theory in which every set of reals is Lebesgue measurable". Annals of Mathematics. Second Series 92: 1–56.) which implies the nonexistence of a well ordering of R.
| 5 | https://mathoverflow.net/users/1061 | 5119 | 3,435 |
https://mathoverflow.net/questions/5109 | 2 | A binomial distribution is the distribution of the number of successes of n independent, identical Bernoulli trials. What happens when the trials are dependent and the Bernoulli trials are not identical by which I mean that the probability of success from trial to trial varies? How identical and close to independent do they have to be before we see something that resembles a binomial distribution?
| https://mathoverflow.net/users/812 | When do binomial distributions occur? | You are asking, I think, when a [Central Limit Theorem](http://en.wikipedia.org/wiki/Central_limit_theorem) holds. The simplest form of the CLT is that the binomial distributions Binomial(n,p), suitably rescaled, converge to a normal distribution as n goes to infinity. (This binomial case is usually not called the CLT, but goes under the name of the [de Moivre-Laplace theorem](http://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem).)
Now, a Binomial(n,p) random variable is the sum of n Bernoulli(p) random variables. The usual form of the CLT states that if $S\\_n = X\\_1 + ... + X\\_n$, where the $X\\_i$ are independent and identically distributed with mean μ and standard deviation σ, then $(S\\_n - \mu n)/(\sigma \sqrt{n})$ converges in distribution to the standard normal as n → ∞.
If the $X\\_i$ are in fact dependent, see the link provided by Ori Gurel-Gurevich above.
If the $X\\_i$ are independent but *not* identically distributed, then there are two standard conditions for proving that the rescaled distribution of $S\\_n = X\\_1 + ... + X\\_n$ converges to the standard normal: [Lindeberg's condition](http://en.wikipedia.org/wiki/Lindeberg%27s_condition) and [Lyapunov's condition](http://en.wikipedia.org/wiki/Lyapunov_condition). Both are a bit difficult to understand when you first look at them. But the basic idea behind both of them is that if no one of the summands $X\\_i$ is too large (in variance) compared to the others, then the normal distribution still appears.
| 5 | https://mathoverflow.net/users/143 | 5121 | 3,437 |
https://mathoverflow.net/questions/5117 | 7 | Can anyone point out some good reference to understand how Paul Cohen proved that the continuum hypothesis is independent of ZFC? I know he used the so called forcing technique to construct two different models of ZFC, but I don't quite understand how.
| https://mathoverflow.net/users/1172 | Independence of the continuum hypothesis on ZFC | Raymond Smullyan and Melvin Fitting wrote a long (but very readable) monograph, called "Set theory and the continuum problem" (Oxford Logic Guides, 34. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1996. xiv+288 pp. ISBN: 0-19-852395-5) which starts from the very beginning, introducing the von Neumann-Bernays-Godel "class-set" formalism, establishing all the basic properties, etc. culminating in a complete, self-contained exposition of Cohen's result and the basics of forcing.
There are three parts. The first part is foundational. The second is an exposition of Godel's relative consistency result, which says that the continuum hypothesis is consistent with NBG (or ZFC if you prefer). The third is about forcing, and Cohen's result.
In very, very broad outline, the point of Godel's argument is to exhibit a "model" of class-set theory with as few sets as possible: the only sets are those that are absolutely required to be there by some application of the axioms. Each set therefore carries with it a formula which requires it to exist. There are so few objects in this model that the continuum hypothesis is seen to be true because every element of the continuum that is required to exist is required by some explicit "reason", and the reasons can be enumerated.
The point of forcing is to show that one can build a new model in which there are many more objects, by adding new objects when there is no explicit reason why they can't exist (i.e. they are "forced" to exist) and keeping careful track of how many such objects you can add without reaching a contradiction. Certain tools (eg "compactness") are required to be able to add infinitely many new objects in this way.
| 8 | https://mathoverflow.net/users/1672 | 5125 | 3,440 |
https://mathoverflow.net/questions/4507 | 10 | This question is closely related to [another one I asked recently](https://mathoverflow.net/questions/4459/), and may be thought of as a warm-up to that one.
Consider $\mathbb R^n$ with its usual metric, and pick a one-form $b$ and a function $c$. Let $m$ be a positive constant, and consider the second-order differential equation for a function $q(t)$
$$ m\ddot q = db \cdot \dot q + dc $$
where I have used the metric to identify vectors and covectors, $dc$ is the differential of $c$, and $db$ is the exterior derivative of $b$ (it is contracted with $\dot q$ to yield a covector). In coordinates, and using Einstein's summation convention:
$$ m\ddot q^i = \left(\partial\\_i b\\_j - \partial\\_j b\\_i\right)\dot q^j + \partial\\_i c $$
I am interested in the limit when $m\to 0$. For example, when $m=0$ and $b=0$ (or anyway when $b$ is closed), then the differential equation forces the path $q(t)$ to stay within the set of critical points of $c$ (this set is generically discrete, so that the only solutions are constant). At another (more generic) extreme, $db$ might be nondegenerate, and hence a symplectic form on $\mathbb R^n$. Then the equation $0 = db \cdot \dot q + dc$ is a nondegenerate first-order differential equation, exactly equivalent to Hamilton's equations for the symplectic manifold $(\mathbb R^n,db)$ with Hamiltonian $-c$. There is some gradation when $db$ is nonzero but has nontrivial kernel (as for example must happen if $n$ is odd).
So I basically get what happens when $m=0$. But can we understand the limit $m\to 0$? For example, if $m\neq 0$, then any initial value $(\dot q(0),q(0))$ determines a solution; for fixed initial values, how does this solution vary as $m\to 0$? Alternately, we can try to solve the boundary value problem, in which we prescribe $q(0)$ and $q(1)$. Then what happens to the solutions as $m$ shrinks? Since when $m=0$ we cannot find solutions with arbitrary initial velocity, it is unlikely that anything is particularly well-behaved in the limit, but not impossible.
Very specifically, I would like to know about the asymptotics of the solutions to the boundary and initial-value problems — what do solutions look like when $m$ is a formal variable? But more generally I'm happy with some statements about the regularity in the $m\to 0$ limit.
| https://mathoverflow.net/users/78 | What happens to Newtonian systems as the mass vanishes? | This is not a real answer, but there is a branch of mathematics called "semiclassical analysis" which might be related. For example, consider a degenerate version of the problem above:
$$
(-h^2 \partial\_x^2+V(x))u=0.
$$
Here $h^2=m$ and $V=dc$; we assume that $n=1$. Then the limit as $h\to 0$ is called "semiclassical limit". What should happen (if you prescribe some boundary conditions) is that the possible solutions $u$ should get "microlocalized" near the zero set $\{p=0\}$ of the semiclassical symbol
$$
p(x,\xi)=\xi^2+V(x).
$$
Here a function $u$ is "microlocalized" near a subset $K$ of the cotangent bundle if a certain norm $\|Au\|$ is small for any pseudodifferential operator $A$ with symbol $a$ supported outside of $K$.
A physical explanation of the above would be that our static Schr\"odinger equation governs the behaviour of a single quantum particle under the potential $V$ near the zero energy level; for small values of Planck constant $h$, this should correspond to the motion of a classical particle at this fixed energy level.
There are several sources to read about semiclassical analysis, including
[lecture notes by Evans-Zworski](http://math.berkeley.edu/~zworski/semiclassical.pdf) and [a book by Dimassi and Sjostrand](http://books.google.com/books?id=2wh1Lb4F3oYC&dq=Spectral+analysis+in+the+semiclassical+limit&source=gbs_navlinks_s).
| 2 | https://mathoverflow.net/users/1704 | 5129 | 3,443 |
https://mathoverflow.net/questions/5114 | 12 | Let X be a compact Hausdorff space. Swan's theorem provides an equivalence between the category of (say real) vector bundles on X and the category of finitely generated projective modules over the ring C(X,R) of continuous functions from X to the real numbers. This relates the topological K0 to the algebraic K0 of a ring, i.e. the group completion of the semiring with elements finitely generated projective modules.
Higher topological K-Groups are defined as K0 of suspensions, i.e. Kn(X)=K0(Sn(X)) and in particular Kn(X)=K0(C(Sn(X),R)). Higher algebraic K-groups are defined as certain homotopy groups of a space, given by topological constructions like the +-construction, Q-construction or Waldhausen's S-construction which remind one of the group completion.
Is there a kind of analogous statement to Swan's theorem for higher K-groups? To be more precise, is it possible to construct a functor Rn(-) from well behaved topological spaces to the category of rings such that Kn(X)=Kn(Rn(X))?
| https://mathoverflow.net/users/467 | Relation between higher algebraic K-groups and topological K-groups | I am quite sure that such a construction exists (for complex topological K-theory): Start with a space X, from this produce the C^\*-algebra of continuous complex-valued functions A:=Cont(X,C) - there you already have a ring encoding your space and I am sure this is step one.
The C^ \*-algebra-K-theory of A is the topological K-theory of X. But the C^ \*-K-theory is algebraic K-Theory made homotopy invariant (see Example 2.1.3 and what follows in the Notes of Cortinas you find [here](http://cms.dm.uba.ar/Members/gcorti/workgroup.swisk/index_html.html)). Now I think one could enforce homotopy invariance somehow on a ring level, but can not recall how.
If you take the topological tensor product of A with K:=kernel of the map from the Toeplitz algebra to Cont(S^1,C) (see start of section 2.3 in Cortinas' notes), then algebraic and topological K-theory coincide on the result (Thm 3.4.1.3 in Cortinas' notes) - if this leaves topological K-theory unchanged for C^ \*-algebras coming from spaces then that is what you do. If not then it is something else, which you might find in those notes. Actually I thought it sounded less complicated than what I wrote, something like, pass to infinite matrices over A...
There also is an article comparing algebraic and topological K-theory, from the handbook of K-theory, which might contain what you look for: [here](http://www.math.umd.edu/~jmr/algtopK.pdf)
| 5 | https://mathoverflow.net/users/733 | 5138 | 3,449 |
https://mathoverflow.net/questions/5144 | 8 | Normalize the Alexander polynomial (in $t$) so that the positive and negative exponents are balanced. For example in the Conway normalization, make the substitution $z = t^{1/2} - t^{-1/2}$. The trefoil gives $t^{-1} - 1 + t$.
>
> **Conjecture:** Suppose that $K$ is an alternating knot. Then the sequence of absolute values of the coefficients is unimodal. Specifically, if $\Delta\_K(t) = \sum\_i a\_i t^i$, then $|a\_0| \ge |a\_1| \ge |a\_2| \ge \cdots$.
>
>
>
This is a conjecture due to Murasugi, I believe. Where is it written? Has this been proved or disproved?
| https://mathoverflow.net/users/813 | Is Murasugi's conjecture still open? | Didn't Hosokawa prove (1958, Osaka J. Math) that the Alexander polynomial of a knot can be any integral Laurent polynomial $p(t)$ such that $p(t^{-1}) = p(t)$ and $p(1) = \pm 1$?
If that's right, then according to Hosokawa, $2t^{-2}+t^{-1}-7+t+2t^2$ would be the Alexander polynomial of a knot, contradicting this conjecture.
It's been a while but I think you construct these knots very explicitly using ribbon diagrams -- Rolfsen's knots and links, also Kawauchi's big survey book should have the construction.
| 3 | https://mathoverflow.net/users/1465 | 5151 | 3,454 |
https://mathoverflow.net/questions/5148 | 7 | Let S be the open unit square in R^2: the set of points (x,y) with 0 < x < 1 and 0 < y < 1. Consider an area-preserving smooth map S --> R^2, that is, a map whose Jacobian has determinant 1 at every point. What can the image of S look like?
Can the image of the square have a smooth boundary? I think you can smooth out the two corners on top with a transformation of the form (x,y) --> (x,y+f(x)), but this makes the other two corners sharper.
(I had added, and now removed, some nonsense about Gromov's nonsqueezing theorem, which does not hold in dimension 2, and doesn't say what I thought it said besides.)
| https://mathoverflow.net/users/1048 | What are the possible images of a square under an area-preserving map? | Yes, you can smooth all corners.
Say, in the class of maps $(x,y)\mapsto (f(x),g(x,y))$.
Moreover, one can map open unit square to any domain bounded by convex smooth curve
by a map of the above type.
I'm sure that any simply connected domain of the same area can be also obtained as an image...
| 2 | https://mathoverflow.net/users/1441 | 5154 | 3,456 |
https://mathoverflow.net/questions/4775 | 65 | I've been prodded to ask a question expanding [this one on Ramanujan's constant](https://mathoverflow.net/questions/4741/a-very-very-good-approximation-to-ramanujan-constant-why-closed) $R=\exp(\pi\sqrt{163})$.
Recall that $R$ is very close to an integer; specifically $R=262537412640768744 - \epsilon$ where $\epsilon$ is about $0.75 \times 10^{-12}$. Call the integer here $N$, so $R = N - \epsilon$.
So $R^2 = N^2 - 2N\epsilon + \epsilon^2$. It turns out that $N\epsilon$ is itself nearly an integer, namely $196884$, and so $R^2$ is again an almost-integer. More precisely,
$$j(\tau) = 1/q + 744 + 196884q + 21493760q^2 + O(q^3)$$
where $q = \exp(2\pi i\tau)$. For $\tau = (1+\sqrt{-163})/2$, and hence $q = \exp(-\pi\sqrt{163})$, it's known that the left-hand side is an integer. Squaring both sides,
$$j(\tau)^2 = 1/q^2 + 1488/q + 974304 + 335950912q + O(q^2).$$
To show that $1/q^2$ is nearly an integer, we can rearrange a bit to get
$$j(\tau)^2 - 1/q^2 - 974304 = 1488/q + 335950912q + O(q^2)$$
and we want the left-hand side to be nearly zero. $1488/q$ is nearly an integer since $1/q$ is nearly an integer; since q is small the higher-order terms on the right-hand side are small.
As noted by [Mark Thomas in this question](https://mathoverflow.net/questions/4741/a-very-very-good-approximation-to-ramanujan-constant-why-closed), $R^5$ is also very close to an integer -- but as I pointed out, that integer is not $N^5$. This isn't special to fifth powers. $R$, $R^2$, $R^3$, $R^4$, $R^5$, $R^6$, respectively differ from the nearest integer by less than $10^{-12}$, $10^{-9}$, $10^{-8}$, $10^{-6}$, $10^{-5}$, $10^{-4}$, and $10^{-2}$. But the method of proof outlined above doesn't work for higher powers, since the coefficients of the $q$-expansion of $j(\tau)^5$ (for example) grow too quickly. Is there some explanation for the fact that these higher powers are almost integers?
| https://mathoverflow.net/users/143 | Why are powers of $\exp(\pi\sqrt{163})$ almost integers? | Another take on this:
As David Speyer and FC's answer shows, this question can be answered without any additional deep theory.
However, I'd like to explain a variant on their arguments that puts this in a little more context regarding modular forms. It also means we can use a technique which makes it easier to see how good these approximations are in terms of the growth rate of the coeffients of the j-function.
The important fact here is that any modular function (for SL\_2(Z)) with integer coefficients in its q-expansion takes on integer values at τ = (1+√(-163))/2 (and so q = exp(-π√163)). This fact is in fact a consequence of the integrality of the j-value here, since any such function can be expressed as a polynomial in j with integer coefficients (although similar things are true in other contexts, such as modular functions of higher level, where there is not a canonical generator for the ring of such modular invariants).
This means that, just as we can use the integrality of
$j(\tau) = q^{-1} + 744 + O(q) $
to get an integer approximation to $q^{-1}$, if we have a modular function $f\_n$ with power series of the form
$f\_n(\tau) = q^{-n} + integer + O(q)$,
we can get an integer approximation to $q^{-n}$. How good this approximation is will depend upon the size of the coefficients of the power series for the $O(q)$ part.
Fortunately for us, such a function $f\_n$ always exists (and is unique up to adding integer constants). How can we construct it? One way is to take an appropriate polynomial in $j$, that is, take an appropriate linear combination of $j, j^2, \cdots , j^n$ to get a function with the right principal part. This clearly works, and if one works out the details, it should turn out equivalent to FC's and David's approach.
However, now that we're in the modular forms mindset, we have other tools at our disposal. In particular, another way to create new modular functions is to apply Hecke operators to existing modular functions, such as $j$. This turns out to be an effective way to get modular functions of the type we need, since Hecke operators do predictable things to principal parts of q-series (for example, if $p$ is prime, $T\_p j = q^{-p} + O(1)$). I'll just explain how this works for $n = 5$, although the method should generalize immediately to any prime $n$ (composite $n$ might be a little trickier, but not much).
The theory of Hecke operators tells us that the function $T\_5 j$ defined by
$$(T\_5 j)(z) = j(5 z) + \sum\_{i \ mod \ 5} j (\frac{z + i}{5})$$
is modular, with q-expansion given by
$$(T\_5 j)(\tau) = q^{-5} + \sum\_{n = 0}^{\infty} (5 c\_{5n} + c\_{n/5}) q^n$$.
where the $c\_n$ are the coefficients in
$$j(\tau) = q^{-1} + \sum\_{n = 0}^{\infty} c\_n q^n$$
(and $c\_n = 0$ if $n$ is not an integer).
So if as before we set $q = e^{- \pi \sqrt{163}}$, we find that
$q^{-5} + 6 c\_0 + 5 c\_5 q + 5 c\_{10} q^2 + 5 c\_{15} q^3 + 5 c\_{20} q^4 + (c\_1 + 5 c\_{25}) q^5 + \dots$
is an integer.
Now, $q$ is roughly $4 \cdot 10^{-18}$, and looking up the coefficients for $j(z)$ on [OEIS](http://oeis.org/A000521 "OEIS"), we find that
$q^{-5} + 6 \cdot 744 + 5 \cdot (\sim 3\cdot 10^{11}) (\sim 4\cdot10^{-18}) + \text{clearly smaller terms}$
is an integer. Hence $q^{-5}$ should be off from an integer by roughly $6 \cdot 10^{-6}$. This agrees pretty well with what Wolfram Alpha is giving me (it wouldn't be hard to get more digits here, but I'm feeling back-of-the-envelope right now and will call it a night :-)
| 50 | https://mathoverflow.net/users/422 | 5156 | 3,458 |
https://mathoverflow.net/questions/5162 | 6 | Here's the context for the question: Proposition 4.6 of Freitag and Kiehl's book on etale cohomology shows that a sheaf (of sets) $\mathcal{F}$ (on the site Et(X)) is constructible if and only if it is the coequalizer of an etale equivalence relation $\mathcal{R}\rightrightarrows \mathcal{Y}$, where $\mathcal{R}$ and $\mathcal{Y}$ are representable sheaves. Here an etale equivalence relation is defined exactly as you would expect: $\mathcal{R}\rightarrow \mathcal{Y}\times \mathcal{Y}$ is injective, and for every etale $U\rightarrow X$, $\mathcal{R}(U)\subset \mathcal{Y}(U)\times \mathcal{Y}(U)$ is an equivalence relation (of sets).
Now if the quotient $\mathcal{Y}/\mathcal{R}$ "should" be represented by the quotient $Y/R$ (where $Y$ represents $\mathcal{Y}$ and $R$ represents $\mathcal{R}$), well, it sounds like constructible sheaves should be algebraic spaces, or at least there should be some relationship. On the other hand, I don't think this could be right.
So is the problem that you can't take sheafy quotients like this, or is it something more subtle?
| https://mathoverflow.net/users/88 | Do quotients of representable sheaves represent quotients? | It seems like your interpretation is correct. The bottom of page 39 reads
>
> As we have already seen in § 1, for every sheaf of sets $\mathcal F$ on the scheme $X$ there is a family $X\_\alpha$ of etale $X$-schemes and a surjective sheaf mapping $\coprod \tilde X\_\alpha\to \mathcal F$ (where $\tilde X\_\alpha$ is the sheaf represented by $X\_\alpha$). The sheaf $\mathcal F$ is called constructible if one can get by with a finite family. In this case $\mathcal F$ actually turns out to be the quotient of a representable sheaf by a representable equivalence relation (not separated, in general). Thus one can interpret $\mathcal F$ as an etale algebraic space over $X$ (not necessarily separated) in the sense of M. Artin, Knutson [91].
>
>
>
Since the disjoint union of an arbitrary collection of etale $X$-schemes is an etale $X$-scheme, they must be implicitly assuming each $X\_\alpha$ is finite type or something.
| 5 | https://mathoverflow.net/users/1 | 5169 | 3,466 |
https://mathoverflow.net/questions/3249 | 8 | Suppose you have a homogeneous ideal $I$ inside the algebra $\mathbb{C}[x\_1,...,x\_d]$ of complex polynomials in $d$-variables. Can one find a basis for $I$, say $\{f\_1,...,f\_k\}$, such that every $h \in I$ can be written as
$$h = a\_1 f\_1 + ... + a\_k f\_k $$
where the coefficients appearing in each summand $a\_i f\_i$ are not much bigger then the coefficients appearing in $h$?
More specifically, given that $\{f\_1,...,f\_k\}$ is a Groebner basis for $I$, can one modify the standard division algorithm so that one gets
$h = a\_1 f\_1 + ... + a\_k f\_k$
with controlled terms?
**Added 13.11.09 -** By *controlled* I mean that the **coefficients** of the terms $a\_i f\_i$ are bounded in a non-exponential manner by the coefficients of $h$. There is no problem with **degree** of the $a\_i$'s.
I will share that I found this possible in some special cases, for example when $d=2$ or when $I$ is generated by monomials, and I am now interested in the general question.
Note: My question begins **after** a basis has been found, I am *not* concerned here with the terrible complexity of actually computing a Groebner basis.
**Another note (added 12.11.09): The answers and links that I am getting suggest that this problem has not been considered before. So I re-eamphasize my note from above: assume that a Groebner basis, even a universal Groebner basis, has already been found for the ideal. What can be said about the stability of certain variants of the division algorithm now?**
| https://mathoverflow.net/users/1193 | Is there a stable algorithm for polynomial division (in several variables)? | In Bayer and Mumford's [What Can Be Computed in Algebraic Geometry?](http://arxiv.org/pdf/alg-geom/9304003v1) section 3, you can find a 17 years old survey of known results. The bottom line is that without controlling the Castelnuovo Mumford regularity of the variety in question, there is very little you can do (you follow Buchberger's algorithm step by step and get some doubly exponential bound on the coefficients). The situation when imposing geometric conditions is a mater of current research.
| 2 | https://mathoverflow.net/users/404 | 5170 | 3,467 |
https://mathoverflow.net/questions/5166 | 15 | Reading Ravenel's ["green book"](http://www.math.rochester.edu/u/faculty/doug/mu.html), I wonder about his question on p.15 "that the spectrum MU may be constructed somehow using formal group law theory without using complex manifolds or vector bundles. Perhaps the corresponding infinite loop space is the classifying space for some category defined in terms of formal group laws. Infinite loop space theorists, where are you?". What is the state of things on that now?
| https://mathoverflow.net/users/451 | complex cobordism from formal group laws? | As far as I know, there is still no such interpretation. The closest I've heard is some rumored (but unpublished) work in derived algebraic geometry interpreting MU as some kind of representing object.
Such a construction of MU in terms of formal group data be very welcome (probably even more now than when Ravenel wrote the green book).
EDIT: Some elaboration.
We do know a lot about MU. We know that it has an orientation (Chern classes for vector bundles), and in it's universal for this property. It's not then extremely suprising that we get a formal group law from the tensor product for line bundles, but the fact that MU carries a universal formal group law, and that MU ^ MU carries a universal pair of isomorphic formal group laws, is surprising. At this point it's something we *observe* algebraically. Even Lurie's definition of derived formal group laws, assuming I understand correctly, is geared to construct formal group laws objects in derived algebraic geometry carrying a connection to the formal group law data that we already know is there on the spectrum level, and hence ties it to the story we already knew for MU implicitly.
Some reasons these days we might want to know how to construct MU from formal group law data:
* Selfish, ordinary homotopy-theoretic reasons. It's very useful to be able to construct other spectra with specific connections to formal group law data (like K-theory, TMF, etc) and constructing them is generally very difficult. Things like the Landweber exact functor theorem, the Hopkins-Miller theorem, and Lurie's recent work give us a lot of progress in this direction, but they only apply to restricted circumstances. None of these general methods will construct ordinary integral cohomology, corresponding to the additive formal group law (only rational cohomology). If we understood how to build MU, we might understand how to generalize.
* Equivariant homotopy theory. I would tentatively say that we don't have nearly as good computational and "qualitative" pictures of the equivariant stable categories, because we don't have something like the startling MU-picture that relates it all to some stack like the moduli stack of 1-dimensional formal group laws. If we found MU by \_accident\_ then we don't really know how the analogue should play out in other, more general, stable categories.
* Motivic homotopy theory. Hopkins and Morel found that there is some data to formal group laws appearing in motivic stable homotopy theory via the motivic bordism spectrum MGL. I'm not up with the state of the art here but a better understanding of this connection would be very important too - for understanding MGL itself, but also hopefully for understanding the analogues of chromatic data in these categories related to algebraic geometry.
* (space reserved for connections to other subjects that I've forgotten)
| 17 | https://mathoverflow.net/users/360 | 5193 | 3,486 |
https://mathoverflow.net/questions/4982 | 7 | A Delzant polytope in R^n by definition is a simple, rational, and smooth convex polytope in R^n (Ana Cannas da Silva's book for notions.) Do you guys have any insight of the definition, for example, anything we can say about the shape? They satisfy some rigidity conditions? All related comments are welcome!
| https://mathoverflow.net/users/1468 | look into Delzant Polytope | The standard model of a vertex which satisfies the Delzant condition is the positive "quadrant" $x\_i \geq 0$ of $\mathbb{R}^n$ near the origin. In general a polytope is Delzant if and only if every vertex can be taken to this standard model by some element of $\mathrm{GL}(n, \mathbb Z)$.
The motivation for this definition, coming from symplectic geometry, is the following fact, due to Archimedes. Consider the standard radius-one sphere in 3-space enclosed in a cylinder of radius one. Project the sphere outwards onto the cylinder, orthogonal to the cylinder's axis. Archimedes' theorem says that this map preserves areas. In terms of Delzant polytopes, you should think of the sphere as 2-dimensional toric manifold. The corresponding "polytope" is the interval $[-1,1]$. Archimedes' theorem says that if you take the cylinder $S^1\times [-1,1]$ and collapse the circles $S^1\times\{\pm 1\}$ you obtain the sphere *as a symplectic manifold*. To see how to generalise this to higher dimensions, take the n-fold product of the cylinder $S^1\times [0,\infty)$ and consider the map to $\mathbb{R}^n$ given by forgetting the $S^1$-factors. The image is the positive "quadrant" mentioned in the first paragraph above. Archimedes' theorem tells us that if we collapse the preimage of the boundary of this quadrant in a controlled way we obtain something which admits a smooth symplectic structure. I.e., over the orgin we collapse the whole $T^n$-fibre, more generally, over each coordinate $r$-plane we contract the corresponding orthogonal copy of $T^{n-r}$ inside $T^n$. (To do this recall each $S^1$-factor corresponds to a direction in $\mathbb{R}^n$).
Now the meaning of Delzant's condition should be clear: given a Delzant polytope P we built a symplectic manifold by taking $P \times T^n$ and then collapsing parts of the boundary. To decide how, we map each vertex to the standard model described above and collapse as in that case. Archimedes theorem tells us the resulting object is a smooth symplectic manifold with a torus action.
| 20 | https://mathoverflow.net/users/380 | 5195 | 3,488 |
https://mathoverflow.net/questions/5131 | 9 | Let me start with some background to set the notation before I ask my question.
Let G be a semisimple algebraic group over some algebraically closed field K, and suppose we have fixed a Borel subgroup B and an opposite Borel subgroup B'. Let P be a parabolic subgroup containing B and consider the partial flag variety X = G/P. The closures of the B-orbits on X are the Schubert varieties, and the closures of the B'-orbits on X are the opposite Schubert varieties. Their intersections are Richardson varieties.
As far as I am aware, all of this stuff can be defined over the integers (anyway I'm only interested in the case when G is a symplectic or orthogonal group), so we can ask about how many F\_q-rational points these various varieties have. For a Schubert varieties Z I know how to do this: consider the poset of Schubert varieties contained in Z. It's graded, and specializing q into the rank generating function of this poset gives me the number of points. For opposite Schubert varieties it's similar.
My question is what can one say about Richardson varieties? What I want to say is to just do the same thing: the Schubert varieties correspond to lower intervals in the Bruhat order, and the Richardson varieties correspond to arbitrary intervals. So it's tempting to say just take the rank generating function of this interval and specialize at q. Does this work? What if I assume something like G/P is minuscule? If that's not enough, I really only care about the case when the opposite Schubert variety is replaced by the open B' orbit on X.
A related question: do the Richardson varieties have nice cell decompositions like the Schubert varieties?
My feeling is that something like this is relatively easy and known (I just can't find it), or hopelessly complicated. Please inform me that I'm in the first case.
| https://mathoverflow.net/users/321 | Richardson varieties over finite fields | The intersections of opposite Schubert cells have a very nice decomposition into products of tori and affine spaces due to Deodhar which, of course, induces such a decomposition of the Richardson. This decomposition is defined over $\mathbb{Z}$ (actually it works in any building), so it lets you count points, and the strata are combinatorially described by special subwords of a reduced decomposition of one of the words. I recommend reading the paper of [Marsh and Rietsch.](https://arxiv.org/abs/math/0307017)
| 10 | https://mathoverflow.net/users/66 | 5197 | 3,489 |
https://mathoverflow.net/questions/5159 | 25 | In Mumford's "The red book of varieties and schemes" one of the examples (G on pg 74) is the space Spec $(\prod\_{i=1}^\infty k)$, where $k$ is a field. He mentions that "Logicians assure us that we can prove more theorems if we use these outrageous spaces".
Are the any examples of theorems proved using such spaces, or any references to logicians saying such a thing?
| https://mathoverflow.net/users/1709 | Logic comment in Mumford's Red Book | At (about) the time Mumford was giving his lectures at Harvard, Ax was lecturing on his work with Kochen in which they proved a conjecture of Artin for almost all p by using ultrafilters. This is clearly what Mumford was thinking of. The reference for the Ax-Kochen work is:
MR0184930 Ax, James; Kochen, Simon Diophantine problems over local fields. I. Amer. J. Math. 87 1965 605--630. Ib. 87 1965 631--648.
| 42 | https://mathoverflow.net/users/930 | 5199 | 3,490 |
https://mathoverflow.net/questions/4424 | 7 | I have a fairly specific question. My intuition says the answer is "yes", but there is a natural generalizations in which I take out all the "physics", and then I think the answer is "no".
### Edit number 2: the question without all the background
In response to Andrew's comments, here's the question I want to ask without all the infinite-dimensional preamble:
On $\mathbb R^d$ with its usual metric, pick a differential one-form $b$ and a smooth function $c$, and suppose that each has compact support. Consider the following (nondegenerate, nonlinear, second-order) differential equation for a path $\gamma(t)$:
$$ \ddot \gamma = db \cdot \dot\gamma + dc $$
This is the **Euler-Lagrange equation**, and so I will abbreviate it as (EL). In coordinates, it is:
$$ \ddot \gamma^i = (\partial\\_i b\\_j - \partial\\_j b\\_i) \dot\gamma^j + \partial\\_i c $$
Since (EL) is nondegenerate and $b,c$ have compact support, every solution to (EL) extends to have domain all of $\mathbb R$, and the solutions are in bijection with the tangent bundle ${\rm T}\mathbb R^d = \mathbb R^{2d}$ by identifying $\gamma$ with $(\dot\gamma(0),\gamma(0))$.
For each $(v,q) \in {\rm T}\mathbb R^d$, define a second-order linear differential operator $h\\_{(v,q)}$, given in coordinates by:
$$ h\\_{(v,q)}[\eta]^j(t) = \ddot\eta^j(t) + \bigl(\partial\\_i b\\_j|\\_{\gamma(t)} - \partial\\_j b\\_i|\\_{\gamma(t)}\bigr) \dot\eta^i(t) + \bigl( \partial\\_i \partial\\_k b\\_j|\\_{\gamma(t)} \dot\gamma^k(t) - \partial\\_i\partial\\_j b\\_k|\\_{\gamma(t)} \dot\gamma^k(t) - \partial\\_i\partial\\_j c|\\_{\gamma(t)}\bigr) \eta^j(t) $$
where $\gamma$ is the solution to (EL) with initial conditions $(\dot\gamma(0),\gamma(0)) = (v,q)$.
Let $C = {\rm T}\mathbb R^d \times \mathbb R\\_{\>0}$. For $(v,q,T) \in C$, consider the operator $h\\_{(v,q)}$ as a map
$$ h\\_{(v,q,T)} : \bigl\{ \eta: [0,T] \to \mathbb R^d \text{ s.t. } \eta(0) = 0 = \eta(T) \bigr\} \to \bigl\{ \eta: [0,T] \to \mathbb R^d \bigr\}$$
Define $C' \subseteq C$ to be the set $\{ (v,q,T) \in C \text{ s.t. } \ker h\\_{(v,q,T)} = 0\}$.
Then I have the following questions:
1. Is $C'$ open (with the topology induced from $C$)? I asserted that it was, because the coefficients of the second-order operator depend smoothly, and I think that kernels can only jump in dimension at closed regions. But I'm not 100% sure.
2. Is $C'$ (path) connected? This was my originally-posed question, and it definitely requires that $b,c$ have compact support.
3. Actually, for my research I need that for each $T > 0$, we have $C' \cap {\rm T}\mathbb R^d \times \{T\}$ is path connected. Since $C'$ includes every $\gamma \in C$ with $\gamma([0,T])$ always outside the support of $b,c$, and since this set is path connected if the supports are compact, 3. implies 2., but perhaps 3. is stronger. Also, perhaps 3. does not require that $b,c$ have compact support?
Bonus question: I used the metric exactly once in (EL) and exactly once in (HJ), to compare the folks with raised indices to the ones with lowered indices. Does anything happen if I change the signature of the metric?
The rest is what I wrote before:
### Background and definitions
On $\mathbb R^d$ (with its usual metric), pick a differential one-form $b$ and a smooth function $c$. The tangent bundle $T\mathbb R^d$ is just $\mathbb R^{2d}$; define the **Lagrangian** $L: T\mathbb R^d \to \mathbb R$ by $L(v,q) = \frac12 |v|^2 + b(q)\cdot v + c(q)$, where $v$ is the fiber coordinate on $T\mathbb R^d$, $q$ is the base coordinate on $\mathbb R^d$, and $\cdot$ is the canonical pairing of a one-form with a vector. A **path of length $t$** is a smooth map $\gamma: [0,t] \to \mathbb R^d$; it has a canonical lift $(\dot\gamma,\gamma): [0,t] \to T\mathbb R^d$. The **action** of a path $\gamma$ of length $t$ is the integral $A[\gamma] = \int\_0^t L(\dot\gamma(\tau),\gamma(\tau))d\tau$. By adjusting signs, one can include paths of negative length; a path of length $0$ is a point in $T\mathbb R^d$ and has zero action.
Consider the set $P$ of all paths (of arbitrary length); it is an infinite-dimensional smooth manifold. There are various natural projections from $P$ to finite dimensions. The "initial-value map" $P \to T\mathbb R^d \times \mathbb R$ takes a path $\gamma: [0,t]\to \mathbb R^d$ to the triple $(\dot\gamma(0),\gamma(0),t)$. I will be more interested in the "boundary-value map" $P \to \mathbb R^d \times \mathbb R^d \times \mathbb R$ taking $\gamma \mapsto (\gamma(0),\gamma(t),t)$. The fiber over a point in $\mathbb R^d \times \mathbb R^d \times \mathbb R$ is an affine space modeled on the space of **Dirichlet** paths $\gamma: [0,t] \to \mathbb R^d$ with $\gamma(0) = 0 = \gamma(t)$.
I like to think of the action $A$ as a Morse function on fibers of the boundary-value map. Let $C \subset P$ be the set of **classical** paths, i.e. paths $\gamma$ so that $dA|\_\gamma \cdot \xi = 0$ if $\xi$ is Dirichlet ($dA|\_\gamma$ is the differential of the action at $\gamma$; $\cdot$ is the canonical pairing). Equivalently, $\gamma \in C$ if $\gamma$ satisfies the **Euler-Lagrange equations** $\frac{\partial L}{\partial q}(\dot\gamma,\gamma) = \frac{d}{d\tau}\bigl[ \frac{\partial L}{\partial v}(\dot\gamma,\gamma) \bigr]$. Since the Euler-Lagrange equations are second-order nondegenerate, the initial-value map restricts to a diffeomorphism of $C$ to an open subset of $T\mathbb R^d \times \mathbb R$ containing $T\mathbb R^d \times \{0\}$.
If I really want to think of $A$ as a Morse function, I should require that its critical points (the classical paths) be nondegenerate. Let $\gamma$ be a (classical) path of length $t$, and $V$ the vector space of Dirichlet paths of length $t$. Then the second derivative or Hessian of $A$ is well-defined as a map $H : V \to V^\*$. In fact, the Hessian makes sense as a second-order linear differential operator on the space of all paths of length $t$. Let's say that a classical path is **nondegenerate** if $\ker H = 0$ (or, rather, does not intersect the space $V$ of Dirichlet paths). The set $C'$ of nondegenerate classical paths is an open (I'm pretty sure) subset of $C$.
### My question
Is the space $C'$ (path) connected?
Bonus question: what happens if you change the signature of the metric on $\mathbb R^d$?
### Edit
The answer to my original question is "no". Let $d = 1$, $b = 0$, and $c(q) = \frac12 q^2$. Then a classical path of length $t$ is nondegenerate if and only if $t$ is not an integer multiple of $\pi$. This is a very nongeneric Lagrangian (it is the **harmonic oscillator**, and is exactly solvable). Also, I think with my definitions, paths of length $0$ are always degenerate.
So let me ask a more restricted question. Let's suppose that $b$ and $c$ are only supported in a compact neighborhood. Then classical paths that do not enter this neighborhood are precisely the straight lines, and they are all generic (provided $t \neq 0$). Is it true that the space of classical nondegenerate paths with positive length is connected with the restriction that $b,c$ have compact support?
| https://mathoverflow.net/users/78 | Is the space of nondegenerate classical paths connected? | If you consider a Riemannian manifold with the Lagrangian $L(\nu,q)=|\nu|^2$, where $q$ is a point in the manifold, $\nu$ is a tangent vector, and $|\nu|$ is defined using the metric at the point $q$, then the first variation gives you the geodesic equation and the second variation gives Jacobi fields. This case has been thoroughly studied (for example, it is one of the main topics of the beautiful book "Morse Theory" by J. Milnore). The space of paths $P$ between two fixed points (or the space of loops, which are Dirichlet paths) with the compact/open topology turns out to be homotopy equivalent to a cell complex with dimensions of cells determined by conjugate points (degenerate paths in your case).
In the case of your question, if we put $b=0$, we will get the equation
$$\ddot\eta^i=-R\_{ij}\eta^j,$$
where
$R\_{ij}=\partial\_{ij}c$.
This is exactly the Jacobi equation along a fixed geodesic if we let $R$ be the curvature tensor with respect to the tangent vector of the geodesic. (It may be that in the case $b=0$ we can interpret our problem in the sense of the first paragraph of this answer for a certain Riemannian metric, but I am not sure. If this is true, then the harmonic oscillator should become the round sphere.) We should expect that the curvature will put certain restrictions on when conjugate points occur or do not occur. For example, if $R$ is positive and large, then by Bonnet-Myers Theorem, there exists certain $T\_0$ such that for any $\nu$ and $q$, there exists $T\in (0,T\_0)$ such that $(\nu,q,T)\not\in C$; therefore, $C$ cannot be connected. This is exactly what happens in the case of the harmonic oscillator and this would happen if we restricted our attention to solutions with bounded value of $|\nu|$, because in that case our solution will stay for time $T\_0$ in a fixed compact region, where we could cook up $c$ so that $R$ is very large. However, if $|\nu|$ is too large, then $\gamma$ will escape from our compact region very fast and conjugate points will not have enough time to develop. (In other words, $(\nu,q,T)\in C$ for all $T>0$.) I think that in the nonnegative curvature case, the complement of $C$ should consist of a set of closed hypersurfaces that go to infinity as $|\nu|$ approaches a certain threshold; therefore, $C$ will still not be connected. On the other hand, if $R$ is always negative, then Hadamard-Cartan Theorem gives that all possible points lie in $C$ (thus it is connected).
Here is a particular outcome of the thoughts in the previous paragraph. For each nonnegative integer $k$, consider the set
$$
C\_k=\{(\nu,q,T)\in C\mid N(T'\in (0,T)\mid (\nu,q,T')\not\in C)=k\};
$$
namely, we count how many degenerate points we had before time $T$. (We might need to introduce some multiplicities.) Then each $C\_k$ is open; since $C\_k$ form a nonintersecting cover of $C$, they should represent different connected components. The component $C\_0$ is always nonempty; if we are able to prove that, say, $C\_1$ is nonempty, then $C$ is not connected.
| 2 | https://mathoverflow.net/users/1704 | 5210 | 3,499 |
https://mathoverflow.net/questions/5211 | 39 | Is there any geometric way to understand the exact sequence in Example 8.20.1 in Ch II of Hartshorne (p. 182), or its dual from theorem 8.13?
Here is the sequence:
$0\to O\_{\mathbb{P}^n}\to O\_{\mathbb{P}^n}(1)^{n+1}\to T\_{\mathbb{P}^n}\to 0$
| https://mathoverflow.net/users/1724 | Geometric meaning of the Euler sequence on $\mathbb{P}^n$ (Example 8.20.1 in Ch II of Hartshorne) | Yes! The geometric picture is very nice and very easy. It is explained on pages 408-409 of Griffiths-Harris.
Here is roughly how it works:
Let's work over $\mathbb{C}$ for simplicity. Think of $\mathbb{P}^n$ as being the quotient of $X := \mathbb{C}^{n+1} - 0$ by the action of $\mathbb{C}^\ast$. On $X$ we have the vector fields $d/dx\_i$, where the $x\_i$ are the standard coordinates on $\mathbb{C}^{n+1}$. Check that if $v\_i$ are linear functionals on $\mathbb{C}^{n+1}$, then the vector field $\sum v\_i d/dx\_i$ on $X$ descends to a vector field on $\mathbb{P}^n$. The surjection $\mathcal{O}(1)^{n+1} \to \mathcal{T}$ corresponds to taking $n+1$ linear functionals $v\_i$ and projecting the vector field $\sum v\_i d/dx\_i$ down to $\mathbb{P}^n$.
The kernel $\mathcal{O}$ corresponds to the vector field $E = \sum\_i x\_i d/dx\_i$. Intuitively, $E$ is a "radial" vector field on $X$, and if you pretend that $\mathbb{P}^{n}$ is a "sphere" in $X$, then $E$ is "normal" to this "sphere", so it vanishes when we project it down.
Jonathan Wise gives a nice (and rigorous) explanation of this below.
Aside: I think the reason why this is called the Euler sequence is because the vector field $E$ is known as the Euler vector field. And perhaps the reason why $E$ is called the Euler vector field is because its flow is exponential, and $e = 2.718\dots$ is also known as Euler's number. But I'm not sure, and someone should correct me if I'm wrong about this. Edit: Today somebody told me that the relation $E f = d f$ for $f$ a homogeneous degree $d$ polynomial (as in Charles' answer) was discovered by Euler and is known as "Euler's relation".
| 50 | https://mathoverflow.net/users/83 | 5218 | 3,505 |
https://mathoverflow.net/questions/4927 | 4 | Let k be a finite field, G the k-points of GL\_2, T1, T2 the k-points of the split and non-split tori of G.
Then the G-representations C[G/T1] and C[G/T2] are almost the same.
More precisely, they differ by two copies of a certain irreducible representation (the Steinberg). I might have slightly miscomputed, but the point is that the decomposition of C[G/T1] and C[G/T2] into irreducibles is much more similar than what you might naively expect.
Question: Is there a general phenomenon, of which this is a special case?
Edit added: it seems like a corresponding alternating sum over tori for GL\_3 might be six copies of the Steinberg, see comment below.
| https://mathoverflow.net/users/1253 | Induction from split and non-split tori for GL_2 over a finite field | I don't have much time, but maybe the following can lead you to a solution. I'm sloppy too, writing $G$ both for the algebraic group (over some finite field $k$) and for the set of points $G(k)$.
This is semilar to a special case of a formula of Humphreys on Deligne-Lusztig characters.
("Deligne-Lusztig characters and principal indecomposable modules". J. Algebra 62 (1980), no. 2, 299--303.) The special case says that
$\sum\_{w \in W} R\_{T\_w}(1) = |W| St\_G$,
where $W$ is the Weyl group and $T\_w$ the rational maximal torus defined by $w$ (by twisting a fixed rational maximal torus $T$), and $St\_G$ is the Steinberg character. In
the simplest case, $T$ is split, and then the assignment $w \mapsto T\_w$ induces a bijection between conjugacy classes in the Weyl group and rational conjugacy classes
of rational maximal tori. This is like the multiplicites you found for $GL\_3$. Also
signs come up, since the sign of the "dimension" of a DL representation is related to the parity of the split rank of the
corresponding torus.
[Note that $R\_T(1) = Ind\_B^G(1)$ if $T$ is split and lies in the rational Borel $B$.]
(Humphreys proved it for $G$ simply connected, semisimple, split algebraic groups over finite fields and Jantzen (unpublished)
generalised it to arbitrary reductive groups over finite fields.)
EDIT: $R\_T(\theta)$ is the virtual representation of DL defined by the rational maximal torus $T$ and the character $\theta$ of the finite group $T(k)$.
For $GL\_2$ the formula boils down to $(St\_G+1)+(St\_G-1) = 2St\_G$.
| 5 | https://mathoverflow.net/users/1729 | 5240 | 3,520 |
https://mathoverflow.net/questions/5190 | 7 | Back [here](https://mathoverflow.net/questions/2100/is-there-a-coalgebraic-characterisation-of-the-hyperfinite-ii1-factor) I was asking for a coalgebraic characterisation of the hyperfinite $II\_1$ factor. Recall the latter's construction by forming the inductive limit of a chain of matrix algebras $R \to M\_2(R) \to M\_{2^2}(R) \to ...$, where a matrix is sent to two copies of itself placed in the diagonal blocks, zero elsewhere. Then completion in the weak topology gives the hyperfinite factor.
The trace is halved each step along the chain. Traces for projections in the inductive limit are dyadic rationals in [0, 1], while in the hyperfinite factor they are the whole real interval [0, 1]. Now, the dyadic rationals are the initial algebra for the endofunctor on Bipointed Set, $X \mapsto X \vee X$, identifying the second point of the first copy with the first point of the second copy. The [0, 1] interval is the terminal coalgebra for the same functor and the Dedekind and Cauchy [completion](http://ncatlab.org/nlab/show/completion) of the initial algebra. Perhaps results such as Adamek's [Final Coalgebras are Ideal Completions of Initial Algebras](http://logcom.oxfordjournals.org/cgi/content/abstract/12/2/217) may be extended here.
This put me on the quest of characterising the hyperfinite factor coalgebraically. So now the question: what relevant facts are known about the endofunctor on algebras over the reals: $X \to M\_2(X)$? I believe the hyperfinite factor is a fixed point. Presumably there is a need to be clear over isomorphism versus Morita equivalence. Might it be that the hyperfinite factor is the greatest fixed point up to isomorphism? Maybe I should be looking in the category of a certain kind of algebra.
| https://mathoverflow.net/users/447 | A coalgebraic description of the hyperfinite II_1 revisited | This is an interesting question, but the motivation is a bit misaligned. $C^\*$ algebras are a non-commutative or quantum generalization of compact Hausdorff spaces and von Neumann algebras are a non-commutative or quantum generalization of (not too unreasonable) measurable spaces. However, both of these generalizations are contravariant. Your motivation is a covariant comparison between von Neumann algebras and topological spaces, which is problematic.
The von Neumann algebra or $C^\*$-algebra $M\_2(\mathbb{C})$ is now famously known as a "qubit"; it is a great non-commutative analogue of $\mathbb{C} \oplus \mathbb{C}$, which is of course the complex functional algebra of a classical bit. The endofunctor that you ask about is a geometric product of $X$ and a qubit, and the morphism in your question is a geometric projection back to $X$ with a qubit fiber. So the completion that you ask about is thus a geometric product with a quantum Cantor set. I forget what the fiber is called in the $C^\*$-algebra setting, but I remember that, unlike a classical Cantor set, its isomorphism type depends on the sizes of the matrices. In the von Neumann case, this quantum Cantor set is interpreted as a measurable space, and then it is always the hyperfinite $II\_1$ factor and does not depend on the matrix sizes.
I think that the hyperfinite factor is not the only fixed point of tensoring with a qubit. Let $S$ be any set, let $F$ be the set of functions from $S$ to a bit (or any finite set), and then let $M$ be the von Neumann closure of the local operators on $\ell^2(F)$. Here a local operator is one that affects only finitely many values of $f \in F$. If $S$ is an infinite set of any cardinality, then $M$ goes to itself when you tensor it with a qubit.
| 6 | https://mathoverflow.net/users/1450 | 5246 | 3,525 |
https://mathoverflow.net/questions/3430 | 3 | Weil's proof of the Riemann Hypothesis for projective curves relies upon the following positivity result: Let $\mathbb{F}q$ be the finite field with $q$ elements, $\overline{\mathbb{F}q}$ its closure, and $X$ a projective curve defined over $\overline{\mathbb{F}q}$. Let $D$ be a divisor class on $X \times X$ (where two divisors are related if they differ by a principal divisor), and let $\{p\} \times C$ and $C \times \{p\}$ denote the classes containing these elements. Define an integer valued mapping Tr on the divisor classes by
Tr$(D) = D \bullet (\{p\} \times C) + D \bullet (C \times \{p\}) - D \bullet \Delta$,
where $\bullet$ is the standard intersection product, and $\Delta$ is the diagonal divisor class. Now let $\circ$ be the multiplication induced on divisor classes by composition and let $D^t$ be the divisor given by the composition of $D$ and the permutation of the two factors of an element of $C \times C$. It can be shown that
Tr$(D \circ D^t) \geq 0$,
for all $D$. This result is usually called Castelnuovo Positivity, or Weil Positivity. My questions are as follows:
(1) Does anyone know of a good expository proof of this result available online?
(2) Does the result hold for any projective surfaces which are not curves? More specifically, does it hold for $\mathbb{C}P^2$ or for flag varieties?
(3) Is the failure of this Castelnuovo Positivity for projective varieties in general the sole reason that Weil proof does not generalise to all projective varieties?
| https://mathoverflow.net/users/1095 | Castelnuovo Positivity (Rewrite of: Weil's original proof for FP^2) | I believe the question you meant to ask in (2) is: For $S$ a surface, is there some theorem like Castelnouvo positivity, regarding the $4$-fold $S \times S$? The answer to this question is "There is an analogous theorem, called the Hodge index theorem, but it is more complicated."
Let me explain what the Hodge index theorem says. Let $X$ be a smooth, algebraic variety over $\mathbb{C}$, with a specified projective embedding. For the purposes of the Riemann hypothesis, you would want to be working over a field of finite characteristic instead, but many of the things I want to say are much more subtle, and are conjectures rather than theorems, in finite characteristic. You should think of $X$ as $S \times S$, where $S$ is the variety for which you want to prove the Riemann hypothesis.
The cohomology $H^k(X, \mathbb{C})$ breaks up in the Hodge decomposition $H^k = \bigoplus\\_{p+q=k} H^{p,q}$. For the purposes of the Riemann hypothesis, we only care about $H^{m,m}$, so I'll limit my discussion to that case. Now, our specified projective embedding $X \to \mathbb{P}^N$ gives us a map in cohomology in the other direction. $H^2(\mathbb{P}^N)$ is one dimensional and has a standard choice of generator called the hyperplane class; let $\omega$ be the image of this generator in $H^\*(X)$. It turns out that $\omega$ lands in $H^{1,1}$.
Cupping with $\omega$ maps $H^{(m-1), (m-1)}$ to $H^{m,m}$. The **hard Lefschetz theorem** states that this map is injective when $m \leq \dim X$. Let's assume we're in this case, the other is related to this one by Poincare duality. So, $H^{m,m}$ has a filtration as
$$H^{m,m} \supset \omega H^{(m-1), (m-1)} \supset \omega^2 H^{(m-2), (m-2)} \supset \cdots.$$
Abbreviate this as
$$L^m \supset L^{m-1} \supset \cdots L^1 \supset L^0.$$
Define an inner product on $H^{m,m}$ by
$$\langle f,g \rangle = \int \omega^{\dim X-2m} f g.$$
The **Hodge index theorem** says (in part) that this will be positive definite on $L^0$, negative definite on the orthogonal complement of $L^0$ within $L^1$, positive definite on the orthogonal complement of $L^1$ within $L^2$, and so forth. Let $M^i$ be the orthogonal complement of $L^{i-1}$ in $L^i$. (Not sure of the standard nomenclature here.) The case of $H^{1,1}$ of a surface is particularly easy, because $M^0$ is one-dimensional, spanned by $\omega$, and $M^1$ is the orthognonal complement of $M^0$.
It is relatively easy to prove Castelnuovo positivity from the Hodge index theorem for surfaces; see, for example Hartshorne exercise V.1.9. I have not seen anyone write out an analogue of Castelnuovo positivity for $S \times S$ when $S$ is higher dimensional. However, it is known how to adapt Weyl's proof of the Riemann hypothesis to higher dimensional $S$, if one had an analogue of the Hodge index theorem for $S \times S$ in characteristic $p$. I've been told that a good reference for this is Kleiman's [Algebraic Cycles and the Weil Conjectures](http://www.ams.org/mathscinet-getitem?mr=292838) but I have not read this myself.
Let me explain why it is difficult to extend the Hodge index theorem to finite characteristic. If $X$ is defined in characteristic $p$, then $H^k(X)$ must be interpreted as cohomology with coefficients in $\mathbb{Q}\\_{\ell}$ (or, nowadays, $\mathbb{Q}\\_p$). Since these fields aren't ordered, we can't talk about positive definiteness.
Weil dodges this obstacle by talking about the vector space of algebraic cycles. This is a the $\mathbb{Q}$-vector space spanned by algebraic cycles, which I'll denote $A^{m}$. In characteristic $0$, it is a subspace of $H^{2m}(X, \mathbb{Q}) \cap H^{m,m}(X, \mathbb{C})$. The Hodge conjecture says that this it is precisely this subspace. In any characteristic, we have a map
$$A^m \to H^{2m}.$$
This map is either known or conjectured to be an injection, depending on exactly how you define $A^m$. Let's assume that it is an injection.
The inner product $\langle, \rangle$ is $\mathbb{Q}$-valued on $A^m$, so it makes sense to talk about its signature restricted to subspaces of $A^m$.
I'm going to make a secret switch of notation here, and use $L^i$ and $M^i$ to now refer to constructions in $H^{2m}$ rather than $H^{m,m}$. In the end, we'll be interested in things like $A^m \cap L^i$ which, in characteristic $0$, would live in $H^{m,m}$ anyway. By making this switch, I avoid having to explain how the Hodge decomposition works (and doesn't) in characteristic $p$.
In the case where $X$ is a surface, the generator $\omega$ of $M^0$ lies in $A^m$. One can use this to show that
$$A^m = (A^m \cap M^0) \oplus (A^m \cap M^1).$$
The analogue of the Hodge index theorem then says that $\langle, \rangle$ is positive definite on $A^m \cap M^0$ and negative definite on $A^m \cap M^1$.
In all higher dimensional cases, this falls apart. It is (I believe) not known that $\langle, \rangle$ is nondegenerate on $L^i$, so it is not known that we can define the $M^i$. It is certainly not known that
$$A^m = \bigoplus (A^m \cap M^i).$$
And it is not known that $(-1)^i \langle, \rangle$ restricted to $A^m \cap M^i$ is positive definite.
Grothendieck's [standard conjectures](http://www.ams.org/mathscinet-getitem?mr=268189) assert that all of this works. This is a major, and challenging, field of research.
I'll close by mentioning a challenge that is more suited to a combinatorial algebraic geometer like me. Harry Tamvakis told me that he tried, and failed, to prove the hard Lefschetz and Hodge index theorems for grassmannians by brute force. Here the cohomology ring is given by well known formulas, so the difficulties are all combinatorial. I can't say this is an important problem, but it sounds fun.
| 5 | https://mathoverflow.net/users/297 | 5247 | 3,526 |
https://mathoverflow.net/questions/5243 | 94 | This is related to [another question of mine](https://mathoverflow.net/questions/2748/what-is-the-right-definition-of-a-ring). Suppose you met someone who was well-acquainted with the basic properties of rings, but who had never heard of a module. You tell him that modules generalize ideals and quotients, but he remains unimpressed. How do you convince him that studying modules of a ring is a good way to understand that ring? (In other words, why does one have to work "external" to the ring?) Your answer should also explain why it is a good idea to study a group by studying its representations.
| https://mathoverflow.net/users/290 | Why is it a good idea to study a ring by studying its modules? | In short, I'd tell your friend: *"If you believe a ring can be understood geometrically as functions its spectrum, then modules help you by providing more functions with which to measure and characterize its spectrum."*
Elements of a module over a ring $R$ are like generalized functions on $Spec(R)$. We can talk about the support of a module element, or its vanishing set. More concretely, think of how global sections of a line bundle can act as functions you can use to define map into projective space.
When you glue together a module on open sets of a spectrum or a scheme, you get to glue using maps which are *module* isomorphisms, which are more flexible than the ring isomorphisms required to glue together a scheme. Borrowing intuition from smooth manifold land, the twist in the Moebius band (as a line bundle on the circle) is formed by gluing a copy of the reals to itself via multiplication by $-1$, a module map, not a ring map. This allows us to think of functions like $\cos(\theta/2)$ as being globally defined: as a map to the Moebius band.
In the same vein, when you have a representation $V$ of a group $G$, each element $v\in V$ gives you a nice evaluation map from $G$ into $V$, so lurking everywhere we've got these morphisms from our object of interest into a known object, which are nicely related to each other via the group laws. A fortiori, this certainly doesn't capture the full utility of group representations, but a priori I think it's a decent justification.
| 49 | https://mathoverflow.net/users/84526 | 5250 | 3,528 |
https://mathoverflow.net/questions/5257 | 30 | I think it's pretty intuitive how singular/simplicial cohomology detects "holes" in a space.
>
> How can we directly visualize how and in what sense the Cech cohomology of a cover does this?
>
>
>
In case it's of any interest, here are two examples I've looked at with the constant sheaf $\mathbb{Z}$:
(1) The disk, covered "Venn diagram style" with three open patches $U\_1, U\_2, U\_3$ overlapping near the center (like [this](https://etc.usf.edu/clipart/40500/40529/Pie_01-03a_40529_md.gif), but with overlaps), and
(2) The restriction of this cover to the boundary circle of the disk: three opens $U\_1, U\_2, U\_3$ with 3 double intersections $U\_{12}, U\_{13}, U\_{23}$ and *no triple intersection*.
If you look at the Cech complex in (2), the $H^1=\mathbb{Z}$ "comes from" the fact that you can write down a triple of elements $(1,0,0)$ on $U\_{12}, U\_{13}$,$U\_{23}$ which "would" disagree on the triple overlap in (1), but since it's "missing", $(1,0,0)$ gets counted as a cocycle, which is not a coboundary. Even better, the presentation of this $H^1$ you get from the Cech complex is $\mathbb{Z}^3/\{(a,b,c)=(b,c,a)\}$, which is isomorphic to $\mathbb{Z}$ because you can "rotate" all the coordinates "around the missing intersection" into the first component.
I think a like minded analysis of higher dimensional analogues provides similar intuition. Are there any formulations of the Cech complex to really make precise how this intuition should work? What's going on here?
---
**Follow up:** Following Mariano's answer below, I started reading about [Abstract simplicial complexes and their cohomology](https://planetmath.org/SimplicialComplex), which seem like just what I was looking for. What helped me most were the ideas that
1. The (constant sheaf) Cech cohomology of a cover $\cal{U}$ of $X$ "is" the simplicial cohomology of its [nerve](https://en.wikipedia.org/wiki/Nerve_of_a_covering)
$N(\cal{U})$, an [abstract simplicial complex](https://planetmath.org/SimplicialComplex),
2. The simplicial cohomology of an abstract simplicial complex "is" the singular cohomology of its geometric realization, and
3. The geometric realization of the [nerve of a covering](https://planetmath.org/SimplicialComplex) of $X$ is a "simple approximation" of $X$,
So in this sense, we can say precisely that
>
> Cech (constant sheaf) cohomology on a cover detects holes in a "simple approximation" to $X$ defined by that cover.
>
>
>
In particular, seeing the faces of a simplicial complex encoded as formal wedge products of its vertices [totally made my day](https://etc.usf.edu/clipart/40500/40529/Pie_01-03a_40529_md.gif) :)
| https://mathoverflow.net/users/84526 | Visualizing how Cech cohomology detects holes | The complex which computes Cech cohomology for a covering is the "same" one as the one that computes the cohomology of the nerve of the covering. It is not hard to see that the geometric realization of that nerve is, in some sense, an approximation to the original space.
Since you apparently find it intuitive that simplicial cohomology detects the geometry, then this should convince you that Cech cohomology also does :P
| 23 | https://mathoverflow.net/users/1409 | 5258 | 3,534 |
https://mathoverflow.net/questions/5253 | 6 | I have an acyclic digraph that I would like to draw in a pleasing way, but I am having trouble finding a suitable algorithm that fits my special case. My problem is that I want to fix the x-coordinate of each vertex (with some vertices having the same x-coordinate), and only vary the y. My aesthetic criteria are (in order of importance):
1. Ensure no two vertices are too close together
2. Minimize edge crossings and near misses
3. Make a reasonable use of the entire drawing space
I have tried several (modified) force-directed algorithms, but they haven't met my expectations on at least one of these - usually too many edge crossings.
Has anyone come across a problem like this, or can you point me to some good papers that deal with restrictions like this?
| https://mathoverflow.net/users/1733 | Algorithms for laying out directed graphs? | If the x-coordinates are compatible with the acyclic structure of your DAG (that is, for an edge u->v, the x coordinate of u should always be less than that of v) then this is a standard problem in graph drawing, known as Sugiyama-style layered drawing. (Usually it is the y coordinates that are fixed but that makes no difference.) Some versions of the problem (e.g. finding the exact minimum number of edge crossings) can be NP-hard but effective heuristics are known. See e.g. chapter 9 of Di Battista, Tamassia, and Eades, "Graph Drawing: Algorithms for the Visualization of Graphs", Prentice-Hall, 1999.
Searching Google scholar for "layered graph drawing" should also turn up some more recent references, but be careful that some of them (including mine) which use the term to mean something unrelated.
| 3 | https://mathoverflow.net/users/440 | 5261 | 3,536 |
https://mathoverflow.net/questions/5236 | 6 | Every presheaf (let's say on a topological space) comes with restriction maps. The open sets of a topological space are ordered by inclusion and these inclusions yield the restrictions. Now a sheaf satisfies a gluing condition: that you can glue along elements which coincide on common restrictions.
Every simplicial object (let's say a simplicial set) comes with face maps. The simplex category is ordered by faces & degeneracies and these maps yield simplicial maps. Now a Kan complex satisfies a gluing\* condition: that you can glue along simplices which coincide on common faces.
Is there a deeper theoretical framework to relate these 2 notions? I guess that this is the case, and that it is rather trivial.
Side-question: Can we define "degeneracies" for presheaves?
Ideas?
\*it's not a gluing condition, but "somehow similar" (see answers below)
| https://mathoverflow.net/users/956 | Abstract Relation between Presehaves and Simplicial Sets | I don't really get how you see the Kan Horn filling condition as a gluing condition.
But sheaves and Kan simplicial sets play parallel roles in their categories if you look at it through model category theory: In both situations you have an endofunctor replacing a presheaf by a sheaf, a simplicial set by a Kan set respectively. Both categories have a model structure - that is a bunch of data allowing to handle the formal inversion of morphisms which are called weak equivalences.Both times you have a morphism from the old to the new object which is a weak equivalence, this process is called fibrant replacement and is formalized in the theory of model categories.
In the presheaf case the weak equivalences are those morphisms which become isomorphisms after applying the sheafification functor. If you formally invert these, the resulting category is equivalent to the category of sheaves.
In the simplicial set case the weak equivalences are those maps which induce isomorphisms of homotopy groups after applying geometric realization. If you formally invert those you get a category equivalent to the homotopy category of spaces.
| 3 | https://mathoverflow.net/users/733 | 5263 | 3,537 |
https://mathoverflow.net/questions/5268 | 21 | The [Whitehead tower](http://ncatlab.org/nlab/show/Whitehead+tower) of a (pointed) space is a tower of spaces which successively kills the bottom homotopy groups. The first two spaces can be constructed functorially (at least for suitably nice spaces) as the connected component and the universal cover.
Can the remaining spaces be constructed functorially?
For the dual situation the answer is yes. I.e. for the [Postnikov tower](http://en.wikipedia.org/wiki/Postnikov_system) where we have a tower of spaces where the bottom homotopy groups are intact, but where we have killed off all the higher homotopy groups does have a functorial construction (again for nice spaces). The construction I know passes through simplicial sets. I'm wondering if something similar exists for the Whitehead tower?
| https://mathoverflow.net/users/184 | Functorial Whitehead Tower? | The nth stage of the Whitehead tower of X is the homotopy fiber of the map from X to the nth (or so) stage of its Postnikov tower, so you can use your functorial construction of the Postnikov tower plus a functorial construction of the homotopy fiber (such as the usual one using the path space of the target).
The nth stage of the Whitehead tower of X is also the cofibrant replacement for X in the right Bousfield localization of Top with respect to the object Sn (or so). Since Top is right proper and cellular this localization exists by the result of chapter 5 of Hirschhorn's book on localizations of model categories. You might look there to see how the cofibrant replacement functor is constructed. With some care you should be able to define functorially the maps in the tower as well.
(BTW, the Postnikov tower can similarly be obtained functorially by a left Bousfield localization of Top.)
| 22 | https://mathoverflow.net/users/126667 | 5274 | 3,547 |
https://mathoverflow.net/questions/5283 | 6 | There are certain sequences such as
0, 1, 0, 1, 0, 1, 0, 1, ...
that do not converge, but that may be assigned a generalised limit. Such a sequence is said to *diverge*, although in this case a phrase such as *has an orbit* might be preferable.
One way to generalise a limit is by considering the sequence of accumulated means: given a sequence
a1, a2, a3, a4, ...
the accumulated mean sequence would be
a1, (a1+a2)/2, (a1+a2+a3)/3, (a1+a2+a3+a4)/4, ...
If this sequence has a limit, then the original sequence may be said to have that value as its *generalised limit*. In this way, the example sequence above has the generalised limit of 1/2; this seems natural as the sequence oscillates around this 'mean' value.
Is there a name for this kind of generalised limit? Are there other ways to define such a thing. Do you know of any good on-line references for this?
Thanks.
| https://mathoverflow.net/users/1536 | Are there Generalisations of a Limit (for Just-divergent Sequences)? | Another common technique is [Abel summation](http://en.wikipedia.org/wiki/Divergent_series#Abel_summation), which works a little better than Cesaro summation. [Zeta regularization](http://en.wikipedia.org/wiki/Zeta_function_regularization) is also important in physics.
You might enjoy reading [these posts at The Everything Seminar](http://cornellmath.wordpress.com/2007/07/28/sum-divergent-series-i/) and [this column from John Baez](http://math.ucr.edu/home/baez/week126.html).
| 6 | https://mathoverflow.net/users/290 | 5289 | 3,555 |
https://mathoverflow.net/questions/5279 | 2 | Do you know of any on-line references regarding relationships among the elementary number-theoretic functions?
The sort of thing I'm interested in is as at the Wikipedia page on [Arithmetic Functions](http://en.wikipedia.org/wiki/Arithmetic_function#Relations_among_the_functions).
Are there any others?
Thanks.
| https://mathoverflow.net/users/1536 | References for Relationships amongst the Number-theoretic Functions | There [some](http://www.math.ucla.edu/~cbm/aands/page_826.htm) in Abramowitz-Stegun, [Handbook of Mathematical Functions](http://www.math.ucla.edu/~cbm/aands/).
| 1 | https://mathoverflow.net/users/532 | 5290 | 3,556 |
https://mathoverflow.net/questions/5277 | 9 | Is there a finite dimensional closed manifold $M$ which is a $K(\pi,1)$, whose fundamental group is not word-hyperbolic, but which has a positive simplicial volume (ie "Gromov norm")?
(Added:) The answers of Jim and Richard are both excellent; another example is any closed, irreducible locally symmetric manifold (of non-positive curvature). But these examples are all CAT(0); I wonder if there is an example which is not CAT(0)? (of course then it is hard to see the example is a $K(\pi,1)$ . . .)
| https://mathoverflow.net/users/1672 | Simplicial volume | You can just take the double of a hyperbolic knot complement.
See Soma's paper The Gromov invariant of links, Invent. Math. 64 (1981) 445–454
| 9 | https://mathoverflow.net/users/1335 | 5294 | 3,559 |
https://mathoverflow.net/questions/5286 | 8 | Let $f(s)=\sum\_n a\_n n^{-s}$ be a Dirichlet series whose coefficients satisfy $\lvert a\_n\rvert\leq n^{C}$. Then $f(s)$ converges absolutely in some halfplanes, and is conditionally convergent in (potentially different) halfplane. It might happen sometimes that $f(s)$ admits meromorphic continuation to a larger domain. Consider maximal such domain. Is this domain always a halfplane? For purposes of this question, I take it to mean that $\mathbb{C}$ is a halfplane.
The question is motivated by known results about analytic continuation of $L$-functions. It has vexed me: to what extent the analytic continuation is special to the algebraic world of $L$-functions, and how much its properties are common to the analytic world of Dirichlet series.
| https://mathoverflow.net/users/806 | Is the maximum domain to which a Dirichlet series can be continued always a halfplane? | It should be possible to make a Dirichlet series whose domain of meromorphicity is as screwy as you want. Notice that $\zeta(s - 1 - \alpha) = \sum n^{1+\alpha}/n^s$, so $\zeta(s - 1-\alpha)$ is a Dirichlet series, with pole at $\alpha$. Let $\gamma$ be a curve dividing the complex plane into two pieces, one of which contains all $z$ with $\Re(z)$ sufficiently large; and choose $\alpha\_i$ a sequence of points of $\gamma$ which is dense in $\gamma$. Consider
$$\sum c\_i \zeta(s -1- \alpha\_i)$$
where $c\_i$ is some sequence which goes to zero very fast.
As long as the $c\_i$ go to zero fast enough, this should converge absolutely on away from $\gamma$; and will be represented by a Dirichlet series on the half plane to the right of the rightmost point of $\gamma$. The dense set of poles will ensure that you can't continue past $\gamma$. Details are left to the reader. :)
| 14 | https://mathoverflow.net/users/297 | 5297 | 3,562 |
https://mathoverflow.net/questions/5262 | 14 | I'm trying to get a better handle on the relation between Lie groups and the Manifolds they correspond to. Firstly, is the relationship injective? that is, does each Lie group correspond to a unique manifold? Or are all the manifolds corresponding to a particular group homeomorphic?
Also, what formal form does the relationship take? I can intuitively understand the relationship between, say, $SO(3)$ and $S^2$ by thinking about rotating the sphere into itself, but what how does this generalize to a more general group or manifold.
| https://mathoverflow.net/users/936 | Lie Groups and Manifolds | To add a bit,
There are also many examples of compact manifolds with multiple group structures.
As a quick example, first recall that $SU(2)$ is the collection of all $A \in M\_2(\mathbb{C})$ with $A\overline{A}^t = Id$ and $det(A) = 1$. It is a Lie group (which is actually diffeomorphic to $S^3$.)
The manifold $S^1\times SU(2)$ has (at least) 2 group structures. The first is simply the product. The second is isomorphic to the Lie group $U(2)$, those matrices $A\in M\_2(\mathbb{C})$ such that $ A\overline{A}^t = Id$ (no extra condition no the determinant).
For another example, recall that $SO(n)$ is the Lie group consisting of all $A\in M\_n(\mathbb{R})$ such that $AA^t = Id $. Then $SO(3)\times SU(2)$ is diffeomorphic to $SO(4)$ but the group structures are different.
| 12 | https://mathoverflow.net/users/1708 | 5301 | 3,564 |
https://mathoverflow.net/questions/5282 | 3 | Given $n$ i.i.d. variables $X\_1$ to $X\_n$ with an unknown probability distribution, the sample average is an unbiased estimator for the mean of the distribution. Is there some non-trivial probability distribution for which $\min(X\_1,\ldots,X\_n)$ is an unbiased estimator? (Non-trivial meaning the variables can have more than one potential value).
| https://mathoverflow.net/users/1646 | Is the min function ever an unbiased estimator for the mean? | No. The minimum as always smaller than or equal to the arithmetic mean, and is strictly smaller with positive probability (i.e., when not all the $X\_i$ have the same value). Hence its expected value is strictly smaller than that of the mean.
| 7 | https://mathoverflow.net/users/802 | 5311 | 3,572 |
https://mathoverflow.net/questions/4971 | 12 | Let $F$ be a free group, and $w$ an element of $F$. In any group $G$, a $w$-word is the image of $w$ or $w^{-1}$ under a homomorphism from $F$ to $G$. The subgroup of $G$ generated by $w$-words is denoted $G(w)$.
For any $g \in G(w)$, the $w$-length of $g$, denoted $l(g|w)$, is the minimum number of $w$-words in $G$ whose product is $g$, and the stable $w$-length of $g$, denoted $sl(g|w)$, is the limit $sl(g|w) = lim\_{n \to \infty} l(g^n|w)/n$.
If $w$ is not in the commutator subgroup of $F$, the stable $w$-length of every element in any group is trivial. Otherwise, one has a universal inequality
$$1/2 \le sl\_F(w|w) \le 1$$
(where the subscript $F$ indicates that stable $w$-length is being calculated in the free group $F$ containing $w$ itself.)
The lower bound of $1/2$ is realized e.g. by the word $w=xyx^{-1}y^{-1}$ (i.e. a standard commutator) in $F\_2$ but I don't know how to compute (or even approximate!) $sl(w|w)$ in (essentially) any other case.
What values are achieved by $sl(w|w)$? Are they all rational? Are they dense? Is $1$ ever achieved? Is $1/2$ ever achieved for a word other than $xyx^{-1}y^{-1}$?
(Added:) After reading FC's answer, it is probably worth pointing out that the lower bound $1/2 \le sl\_F(w|w)$ comes from the inequality $scl\_G(g) \le sl\_G(g|w)(scl\_F(w)+1/2)$ for any $g$ in any $G$, so one gets a better lower bound on $sl\_F(w|w)$ if one knows $scl\_F(w)>1/2$ (the estimate $scl\_F(w)\ge 1/2$ is always true). Upper bounds can be established by exhibiting identities (like FC's identity below). Does a blind computer search yield any interesting examples?
| https://mathoverflow.net/users/1672 | Stable w-length | Here are some weak observations that don't quite answer any of your questions. Let $g$ be a positive integer, and consider the free group $F\_{2g}$ generated by $a\_k$ and $b\_k$ for $k = 1$ to $g$. Consider the word:
$$w\_g = [a\_1,b\_1][a\_2,b\_2][a\_3,b\_3] \ldots [a\_g,b\_g].$$
Suppose that $\lambda\_g = sl(w\_g,w\_g)$. I claim that for any $x$ in the commutator of $F\_2$ with $cl(x) = g$, the *stable* commutator length $scl(x)$ of $x$ is $\le g \cdot \lambda\_g$. Suppose otherwise. First of all, note that for large $n$ we can write $w^n\_g$ as the product of (roughly) $n \cdot \lambda\_g \cdot g$ commutators. Since the commutator length of $x$ is $g$, there exists a map from $F\_{2g}$ to $F\_{2}$ such that the image of $w\_g$ is $x$. On the other hand, we see that the image of $w^n\_g$ is $x^n$, and thus the commutator length of $x^n$ is (asymptotically) at most by $n \cdot \lambda\_g \cdot g$, and thus $scl(x) \le \lambda\_g \cdot g$.
Example: $cl([x,y]^3) = 2$ and $scl([x,y]^3) = 3/2$, and thus $\lambda\_3 \ge 3/4$. In general, the fact that $scl([x,y]) = 1/2$ implies that that $\lambda\_g$ tends to one as $g$ increases.
I think one can promote this example to a word in $F\_2$. Consider the characteristic homomorphism $\phi\_n:F\_2 \rightarrow \mathbf{Z} \oplus \mathbf{Z}
\rightarrow \mathbf{Z}/n\mathbf{Z} \oplus \mathbf{Z}/n\mathbf{Z}$. Suppose that $n$ is odd, and write $2g = n^2 + 1$. The kernel of $F\_2$ is free of rank $2g$. Pick generators for $\ker(\phi\_n)$ once and for all, and call them $a\_k$ and $b\_k$ for $k = 1$ to $g$. We may think of $a\_k$ and $b\_k$ as elements in $F\_2$, but also as formal words. Since $\ker(\phi\_n) = F\_{2g}$ is *characteristic*, the formal words $a\_k$ and $b\_k$ always yield elements of $F\_{2g}$ (alternatively, the images of $a\_k$ and $b\_k$ in $\mathbf{Z} \oplus \mathbf{Z}$
are divisible by $n$, and this will be so for any substitution of elements of $F\_2$ for the generators). Let
$$w\_g = [a\_1,b\_1][a\_2,b\_2] \ldots [a\_g,b\_g].$$
The argument proceeds as above. If $sl(w\_g,w\_g) = \mu\_g$, then we can write $w^n\_g$ (for large $n$) as the product of $n \cdot \mu\_g \cdot g$ commutators, each of which is the commutator of a pair of elements of $F\_{2g}$ (by the characteristic property of the words $a\_k$ and $b\_k$ described above). Hence, choosing an appropriate map from $F\_{2g}$ to $F\_2$, we may deduce that for any $x \in F\_2$ with $cl(x) = g$ that $scl(x) \le g \cdot \mu\_g$. Thus we have found words $w\_g$ in $F\_2$ such that $sl(w\_g,w\_g)$ tends to $1$ as $g$ goes to infinity. Of course, this says nothing about whether $sl(w\_g,w\_g)$ actually equals $1$ for any $g$.
Finally, a random other example. If $w = [a,b^2]$, then
$$w^3 = [ab^2a^{-1},b^{-2} ab^2a^{-2}][b^{-2} a b^2,b^4] = [b^{-2} a b^2 a^{-2},(aba^{-1})^2]^{-1}[b^{-2} a b^2,(b^2)^2],$$
so $sl(w,w) \le 2/3$.
I wrote this on a very old computer that was too slow for previewing LaTeX, but hopefully this can still be read.
| 5 | https://mathoverflow.net/users/nan | 5312 | 3,573 |
Subsets and Splits