star trek - Is there a galactic map showing the homeworlds of the various humanoid species?

I don't think either would qualify as canon, but there have been two published books of maps of the Star Trek universe that might at least be seen as semi-official: Star Trek Stellar Cartography, and the earlier Star Trek Star Charts (which is out of print and pricey to get used, though somewhat cheaper used copies can be found on the amazon.co.uk page). The Memory Alpha site has an article about Star Trek Star Charts here which mentions that people who worked on the show like Michael Okuda and Rick Sternbach were consulted:

Star Trek: Star Charts was written and illustrated by Geoffrey Mandel,
with the help of Doug Drexler, Tim Earls, Larry Nemecek, and Christian
Rühl (see below). André Bormanis, Michael Okuda, Rick Sternbach, and
Timo Saloniemi gave technical advice.

There was also an older book called Star Trek Maps from 1980, but that obviously wouldn't include anything from later series like TNG.

reference request - Trying to find a 1949 Russian Paper on Transportation Theory

I am a research student in transportation theory. I have difficulty in obtaining this paper:

L. V. KANTOROVICH and M. K. GAVURIN, "The application of mathematical methods
in problems of freight flow analysis", in V. V. Zvonkov, ed., Collection of Problems
Concerned with Increasing the Effectiveness of Transports, Publication of the Akademii
Nauk SSSR, Moscow-Leningrad, 1949, pp. 110-138. (Russian)

I am wondering if anyone knows how can I can find this specific paper, and if anyone knows how I gain access to Russian journals?

Thank you.

a song of ice and fire - Does each season of Game of Thrones roughly equal a book?

This is how it has broken down so far (more or less):

• Season 1 (2011): A Game of Thrones


• Season 2 (2012): A Clash of Kings


• Season 3 (2013): The first half of A Storm of Swords (aka Steel and Snow)


• Season 4 (2014): The second half of A Storm of Swords (aka Blood and Gold)


• Season 5 (2015): Both A Feast for Crows and A Dance with Dragons

Guesswork from the news (see update):

• Season 6 (2016?): The Winds of Winter (unpublished)


• Season 7 (2017?): A Dream of Spring (unpublished) ?

Seasons 1 and 2 were a book each, but it was changed for Season 3, which became the first half of Storm of Swords, and Season 4 became the second half. The producers then, somewhat surprisingly, managed to squeeze the next two books, A Feast for Crows and A Dance with Dragons, into Season 5, by removing lots of what many people saw as extraneous plot lines.

It's worth noting that there has has been some overlap. Certain events have drifted into different seasons, so the above isn't always strictly true -- just mostly true.

Either way, this means that Season 6 will take most of its story from The Winds of Winter (although there may still be unused things from A Feast for Crows and A Dance with Dragons that appear in it).

Update: It appears George RR Martin has confirmed my suspicions in a recent Vanity Fair article.
George R.R. Martin Has a Detailed Plan For Keeping the Game of Thrones TV Show From Catching Up To Him

Update 2: It's been confirmed that the cast have been signed on for a seventh season, and that the creators have hinted about ending the show in the seventh or eighth season.

Which algebraic groups are generated by (lifts of) reflections?

The Cartan-Dieudonne theorem
states that each element $g in O(V)$, where $V$ is a quadratic space of dimension $n$ over a field of characteristic $neq 2$, can be written as a product of $leq n$ reflections.

Something similar is true for $SL_n(k)$ for $char k neq 2$: each element $g$ can be written as a product of elements of order $4.$ Indeed, it suffices to prove this for $n=2$. Then $s_t:=begin{pmatrix} 0 & t \ -frac 1t & 0 end{pmatrix}$ is of order $4$ . Let $h(t):=s_ts_{-1}=begin{pmatrix} t & 0 \ 0 & frac 1t end{pmatrix}$, then each element in $U$, the group of upper triangular matrices, is a product of two conjugates of elements $h(t),h(t')$ (provided $|k|geq 4$). Similarly for the lower triangular matrices $V$, and then $SL_2$ is generated by these subgroups.

Since a simply connected semisimple $k$-split group is generated by $SL_2$'s, the same argument applies.

What can we say about other algebraic groups? Clearly unipotent groups (in characteristic 0) don't have elements of finite order, so nothing there.

What about (the k-rational points of) anisotropic groups?

Story Identification: Novel/Short Story set in space with a mesh-like satellite

I remember reading this at my local library whilst I waited for my brother. I think it was in a book with a few other short stories.

What I can remember:

• Some sort of mesh/net/grid shaped satellite orbiting earth, and it was malfunctioning. I'm pretty sure it was a communication satellite of some form.


• It has two protagonists in their teens. I don't know how or why they were in space, but they were.
  
  • They fix the satellite against some guy's orders to stay away from it.
  
  
  • I think they 'live' in space, i.e. they haven't come for a trip from Earth, they actually live on a space station nearby the mesh (I remember this because they travel to it from wherever they are).
  
  


• There were some bad guy(s) causing the malfunction.

It was a pretty worn book, so I'm thinking it was written earlier than the 90s (I read it in 98/99).

ca.analysis and odes - continuous selection of a multivalued function?

The title is probably a bit too broad. I frequently encountered the following situation: suppose I need to select a solution to a linear equation from a compact set. Can I make this selection continuous?

Formally, let $S subset mathbb{R}^n$ be a compact set. Let $A$ be a $k times n$ matrix ($k < n$), which we view as a linear function $A: mathbb{R}^n to mathbb{R}^k$. Let $T = A(S)$ be the range of $A$ on $S$. Is there a continuous function $g: T to S$, such that $A g(y) = y$?

To construct $g$, we only need to pick a value from the solution set $A^{-1}({y}) cap S$, which is compact. The question is: can we choose it in a continuous way? It is easy to see that we can choose $g$ to be Borel measurable, say, choose $g$ to be the one with the minimum Euclidean norm from the solution set.

Fourier transform for dummies

One of the main uses of Fourier transforms is to diagonalize convolutions. In fact, many of the most useful properties of the Fourier transform can be summarized in the sentence "the Fourier transform is a unitary change of basis for functions (or distributions) that diagonalizes all convolution operators." I've been ambiguous about the domain of the functions and the inner product. The domain is an abelian group, and the inner product is the L² inner product with respect to Haar measure. (There are more general definitions of the Fourier transform, but I won't attempt to deal with those.)

I think a good way to motivate the definition of convolution (and thus eventually of the Fourier transform) starts with probability theory. Let's say we have an abelian group (G, +, -, 0) and two independent random variables X and Y that take values in G, and we are interested in the value of X + Y. For simplicity, let's assume G = {x₁, ..., x_n} is finite. For example, X and Y could be (possibly biased) six-sided dice, which we can roll to get two independent elements of Z/6Z. The sum of the die rolls mod 6 gives another element of the group.

For x ∈ G, let f(x) be the probability P(X = x), and let g(x) = P(Y = x). What we care about is h(x) := P(X + Y = x). We can compute this as a sum of joint probabilities:

h(x) = P(X + Y = x) = Σ_y+z=xP(X = y & Y = z)

However, since X and Y are independent, P(X = y & Y = z) = P(X = y)P(Y = z) = f(y)g(z), so the sum is actually

h(x) = Σ_y+z=xf(y)g(z) = Σ_y∈Gf(y)g(x-y).

This is called the convolution of f and g and denoted by f*g. In words, the convolution of two probability distributions is the probability distribution of the sum of two independent random variables having those respective distributions. From that, one can deduce easily that convolution satisfies nice properties: commutativity, associativity, and the existence of an identity. Moreover, convolution has the same relationship to addition and scalar multiplication as pointwise multiplication does (namely, bilinearity). In the finite setting, there's also an obvious L² inner product on distributions, with respect to which, for each f, the transformation g -> f * g is normal. Since such transformations also commute, recalling a big theorem from finite-dimensional linear algebra, we know there's an orthonormal basis with respect to which all of them are diagonal. It's not difficult to deduce then that in such a basis, convolution must be represented by coordinatewise multiplication. That basis is the Fourier basis, and the process of obtaining the coordinates in the Fourier basis from coordinates in the standard basis (the values f(x) for x ∈ G) is the Fourier transform. Since both bases are orthonormal, that transformation is unitary.

If G is infinite, then much of the above has to be modified, but a lot of it still works. (Most importantly, for now, the intuition works.) For example, if G = Rⁿ, then the sum Σ_y∈Gf(y)g(x-y) must be replaced by the integral ∫_y∈Gf(y)g(x-y)dy to define convolution, or even more generally, by Haar integration over G. The Fourier "basis" still has the important property of representing convolution by "coordinatewise" (or pointwise) multiplication and therefore of diagonalizing all convolution operators.

The fact that the Fourier transform diagonalizes convolutions has more implications than may appear at first. Sometimes, as above, the operation of convolution is itself of interest, but sometimes one of the arguments (say f) is fixed, and we want to study the transformation T(g) := f*g as a linear transformation of g. A lot of common operators fall into this category. For example:

• Translation: T(g)(x) = g(x-a) for some fixed a. This is convolution with a "unit mass" at a.


• Differentiation: T(g)(x) = g'(x). This is convolution with the derivative of a negative unit mass at 0.


• Indefinite integration (say on R): T(g)(x) = ∫^x_-infinityg(t)dt. This is convolution with the Heaviside step function.

In the Fourier basis, all of those are therefore represented by pointwise multiplication by an appropriate function (namely the Fourier transform of the respective convolution kernel). That makes Fourier analysis very useful, for example, in studying differential operators.

nt.number theory - Polynomial with prime powers values

Actually you can prove a lot more.

Theorem

For any non-constant polynomial $p(x)inmathbb{Z}[x]$ and any positive integer $k$ there is an integer $n$ such that $p(n)$ is divisible by at least $k$ distinct primes.

Proof

If we prove that there exist integers $n_1,ldots,n_k$ and distinct primes $p_1,ldots,p_k$ such that $p(n_i)equiv 0 bmod{p_i}$ then we are done, because there exists an $n$ such that $nequiv n_ibmod{p_i}$ by the Chinese Remainder Theorem, and any such $n$ satisfies $f(n)equiv f(n_i)equiv 0 bmod{p_i}$, as desired.

Now by contradiction suppose that $p$ is only divisible by the primes $p_1,ldots,p_l$, $lle k-1$. Since we have that $p(0)neq 0$, let $p(0)=pm p_1^{alpha_1}ldots p_l^{alpha_l}$, $xequiv 0bmod{p_1^{alpha_1+1}ldots p_l^{alpha_l+1}}$.

Then $p(x)equiv p(0)bmod{p_i^{alpha_i+1}}$, $1leq lleq k-1$, so that the greatest power of $p_i$ that divides $p(x)$ is $p_i^{alpha_i}$.

But by hypothesis $p(x)$ is only divisible by the $p_i$, so we conclude that $p(x)=pm p(0)$. Using the pigeonhole principle and the fact that a non-constant polynomial can only assume a value a finite amount of times this is a contradiction, as desired.

transformers - Is Marissa Faireborn related to Flint?

In the G1 Transformers cartoon, we meet Marissa Faireborn, a member of the Earth Defense Command. We learn that she has family in the military.

Both the Transformers and GI Joe are made by Hasbro, and both TV shows were made by the same animation studios.

Marissa Faireborn shares a last name with Dashiell Faireborn, aka Flint, one of the members of GI Joe. Flint is often depicted as having a romantic relationship with Lady Jaye, another member of the team.

Has there been any official statement linking the characters?

What does 'And Another Thing' add to The Hitchhiker's Guide to the Galaxy?

The reason given for "And Another Thing" being written was a much-quoted (in press releases and interviews) mention by Douglas Adams that:

I suspect at some point in the future I will write a sixth Hitchhiker
book … I would love to finish Hitchhiker on a slightly more upbeat
note. Five seems to be a wrong kind of number, six is a better kind of
number.

And Another Thing does not really add anything to the existing books. While quite a readable story, it is distinctly non-Adamsian in turn of phrase and structure, which was always part of the enjoyment in reading a H2G2 book. It is in effect merely a different author's book, written in the H2G2 universe, which to be honest was the only way Colfer could have come out of writing this book without a fan backlash.

Overall, it's a perfectly adequate and readable book, but it doesn't add anything to the previous stories and is largely self-contained.

transwarp - How did the duplicate Voyager make it as far as the real Voyager?

The episode Course: Oblivion takes place in stardate 52586.3:

"Captain's log, stardate 52586.3. We've had a lot to celebrate lately – Tom and B'Elanna's wedding, Ensign Harper's new baby, and the continued health of our enhanced warp drive, which has taken us within striking distance of home."

The episode Dark Frontier, where the Transwarp coil was acquired, takes place in 52619.2:

"Captain's Log, Stardate 52619.2. We got another 20,000 light years out of the transwarp coil before it gave out. I figure we're a good fifteen years closer to home."

So actually, the real Voyager travelled the extra 20,000 light years after the events in Course: Oblivion.

Interestingly though, both production and release order were first Dark Frontier and then Course: Oblivion.

lord of the rings - Why did Bilbo physically age faster than Gollum?

If I understand your question, you're asking not about the change from Smeagol to Gollum, but the change from "Ian Holm" Bilbo to "Ian Holm in old-person makeup" Bilbo after he gave up the ring, while Gollum didn't rapidly age after losing it.

As far as I know, there is no officially stated reason within the canon. However, there are several things that Bilbo did that Gollum did not, which might account for the difference:

Bilbo didn't use it as much

It's possible that the Ring's preservative powers build up over time, or affect you on a deeper level the more you use it. Gollum used it a lot for centuries, Bilbo used it a little for decades. This especially makes sense because the Ring's other effects, such as its addictive nature, are definitely shown to increase with repeated use.

If this is the case, Gollum was simply still "warm" from the Ring, while Bilbo "cooled off" faster. If given enough time, Gollum may have experienced rapid aging as well.

Bilbo gave it up willingly

Unlike Gollum, Bilbo gave up the Ring (mostly) of his own free will. This may have contributed to a greater "detaching" from the Ring's influence, while Gollum, who was still obsessed and seeking the Ring with every thought and action, may have stayed more under its influence. Put another way: greater control by the Ring could mean greater effect by the Ring.

Bilbo went to Rivendell

If there's anywhere in Middle Earth that will help wash away the influence of Evil, it's Rivendell. Similar to the first possibility, it's possible that whatever "residual charge" from the Ring is preventing Gollum from aging would normally have the same effect on Bilbo, but the ambient magic of Rivendell helped wash it away, returning Bilbo to his non-Ring state. Normally the Elves try to slow down time, but in this particular instance, their power may have sped it back up to where it was supposed to be.

---

There's really no way to know which, or which combination, of these possibilities is actually in play. (And yes, out-of-universe you wouldn't be wrong to say it's just a casting/filmmaking issue). But all three are based on extrapolations from things we see in the canon, so they're probably as close as we'll come to an "official" answer.

at.algebraic topology - Does there exists a (possibly homological) characterization of the Jordan curve property in all dimensions?

More precisely, let $M$ be a subspace $mathbb R^n$ with the following properties:

• $M$ is a topological manifold of dimension $n-1$.


• M is compact.

Does there exist a homological characterization of when the following happens:

• $mathbb R^n backslash M$ has two components, the bounded one being "inside" and the other one "outside". Both are $n$-dimensional manifolds.

If the above is not possible, is there a different formulation of the question which would allow a nice characterization?

The motivation of this question is of course the realization that the solution for $n = 3$ seems to be that $M$ is an oriented surface.

ag.algebraic geometry - are closed subfunctors complimentary to open subfunctors?

I apologize if the following question has already been asked and settled. I couldn't find any thread.

Say, $mathcal{C} = (Sch/k)$, the category of schemes over $k$ (a field). Let $mathcal{F} in mathcal{C}^{wedge}$, be an object of $mathcal{C}^{wedge}$ - the category of contravariant functors from $mathcal{C}$ to $(Sets)$. One has the set of points:

$$ |mathcal{F}| := lim_{to} mathcal{F} (K), $$

the limit taken over fields $K/k$. Given a subfunctor $mathcal{G} hookrightarrow mathcal{F}$ one gets a subset $|mathcal{G}| subset |mathcal{F}|$ (ie. a canonical map from $|mathcal{G}| to |mathcal{F}|$ that is injective). In particular, $|mathcal{U}|$ for the open subfunctors $mathcal{U} hookrightarrow mathcal{F}$ form a topology on $|mathcal{F}|$.

Question: Given a closed subset $Z subset |mathcal{F}|$ does there exist a closed subfunctor (possibly non-unique)
$mathcal{Z} hookrightarrow mathcal{F}$ so that $Z = |mathcal{Z}|$ (as subsets of $|mathcal{F}|$)?

In some sense, are open subfunctors and closed subfunctors really "complimentary"?

gn.general topology - Notion of finite dimensional simplicial space

OK I checked, how the adjoint functors looks like. Given any $Delta|_N $ simplicial space $X$. To define $L(X)$, we have to extend $X$ to the whole category $Delta$. I am just telling, what $L(X)$ does on $[N+1]$. Then you keep extending the functor in the same way:

$L(X)([N+1]):=(0,ldots,N)times X([N])/sim$, where the equivalence relation is given by
$(j,s_k(x))sim (k+1,s_j(x))$ for $0le jle kle N,xin X[N-1]$. The $i$-th degeneracy map is induced by the inclusion of the i-th summand. Using the relations in $Delta$ one can also define the face maps.

The right adjoint functor is given by
$M(X)([N+1]):= ( (x_0,ldots,x_{N+1})|partial_ix_j=partial_{j-1}x_imbox{ for } 0 le i < j le N+1 )subset prod_{i=0}^{N+1}X[N]$. The face maps are just the projections and one can define the degeneracy maps using the relations in $Delta$.

So let $X$ be a $Delta$-space. The natural transformation is given by
$L(R(X))([N+1])rightarrow M(R(X))([N+1])qquad (i,x)mapsto (partial_0 s_i(x),ldots,partial_{N+1} s_i(x))$.

Using the relations in $Delta$ one can show, that this map is injective. So the remaining question is, whether this map is an open map (considered as a map onto the image).

lord of the rings - What did Denethor do in the secret room?

I just discovered canon support for DVK's answer that Denethor was looking into his Palantir.

Starting with the words of Gandalf:

"Though the Stewards deemed that it was a secret kept only by themselves, long ago I guessed that here in the White Tower, one at least of the Seven Seeing Stones was preserved. In the days of his wisdom Denethor would not presume to use it to challenge Sauron, knowing the limits of his own strength. But his wisdom failed; and I fear that as the peril of his realm grew he looked in the Stone and was deceived: far too often, I guess, since Boromir departed. He was too great to be subdued to the will of the Dark Power, he saw nonetheless only those things which that Power permitted him to see. The knowledge which he obtained was, doubtless, often of service to him; yet the vision of the great might of Mordor that was shown to him fed the despair of his heart until it overthrew his mind."

"Now I understand what seemed to strange to me!" said Pippin, shuddering at his memories as he spoke. "The Lord went away from the room where Faramir lay; and it was only when he returned that I first thought he was changed, old and broken."

"It was in the very hour that Faramir was brought to the Tower that many of us saw a strange light in the topmost chamber," said Beregron. "But we have seen that light before, and it has long been rumoured in the City that the Lord would at times wrestle in thought with his Enemy."

"Alas! then I have guessed rightly," said Gandalf.
_{The Lord of the Rings Book 5 Chapter 7: The Pyre of Denethor}

And so has DVK; when Denethor withdrew to the secret room under the Tower, he was looking into the Palantir, and in it he saw primarily the forces of Sauron amassing (as well as the ships that had been captured by Aragorn, though Denethor thought they belonged to the Enemy still).

lord of the rings - Why is Orthanc so indestructible?

It's indestructible because of magic.

Two Towers, Flotsam and Jetsam: "Many of the Ents were hurling themselves against the Orthanc-rock; but that defeated them. It is very smooth and hard. Some wizardry is in it, perhaps, older and stronger than Saruman's"

It was built by the guys in the second age who had access to more powerful magic than they do in the time of LOTR. If they could enchant Palantir to communicate instantly over vast distances, then why not a tower to be really strong?

As a side note, it is strong for the same reason the walls of Gondor are so strong. I don't have a quote for this but they are nigh indestructible.

RotK, The Siege of Gondor: "For the main wall of the City was of great height and marvellous thickness, built ere the power and craft of Númenor waned in exile; and its outward face was like to the Tower of Orthanc, hard and dark and smooth, unconquerable by steel or fire, unbreakable except by some convulsion that would rend the very earth on which it stood."

What do gerbes and complex powers of line bundles have to do with each other?

If L is any line bundle on a space (scheme, whatever) X, A is any (additive) abelian group, and a an element of A, there is a natural construction of an A-gerbe $L^a$ as follows. By definition, $L^a$ should be a "sheaf of categories", or stack (not algebraic) on X, and here are its categories of sections. Identify L with its total space, which is a $mathbb{G}_m$-bundle on X, and for any open set U in X, let $L^a(U)$ be the category of all A-torsors on $L|_U$ whose monodromy about each fiber of $L|_U to U$ is a.

One can check that this really is a gerbe: it is locally nonempty, since if L is trivial over U, you can write $L|_U = mathbb{G}_m times U$ and then pull back the unique A-torsor on $mathbb{G}_m$ with monodromy a. It has a natural action of A-torsors on X, given by pulling up along the bundle map $L to X$ and tensoring. And this action is free and transitive, since the difference of two a-monodromic torsors on $L|_U$ has trivial monodromy on each fiber and therefore descends to X.

Why do I call this $L^a$? Suppose that $L = mathcal{O}_X(D)$ for a divisor D, where for simplicity let's say that D is irreducible of degree n; then L gets a natural trivialization on $U = X setminus D$ having a pole of order n along D. As shown above, this induces a trivialization $phi$ of $L^a$ on U, and if we pick a small open set V intersecting D and such that D is actually defined by an equation f of degree n, then we get a second (noncanonical) trivialization $psi$ of $L^a$ on V. You can check that the difference $psi^{-1} phi$, which is an automorphism of the trivial gerbe on $U cap V$, is in fact described by the A-torsor $mathcal{T} = f^{-1}(mathcal{L}_a)$, where $f colon U cap V to mathbb{G}_m$ and $mathcal{L}_a$ is the A-torsor of monodromy a. Since f has degree n, $mathcal{T}$ has monodromy na about D. Thus, it is only reasonable to say that the natural trivialization $phi$ has a pole of order na, which is consistent with the behavior of the trivialization of L itself on U, when raising to integer powers.

What does this have to do with twisting of differential operators? Suppose we have some kind of sheaves (D-modules, locally constant sheaves, perverse sheaves; technically, they should form a stack admitting an action of A-torsors). On the one hand, one could mimic the above construction of $L^a$ to describe a-monodromic sheaves on L, and this is what is often called twisting. On the other hand, there is a natural way to directly twist sheaves by the gerbe $L^a$ without mentioning L at all (that is, you can twist by any A-gerbe). The procedure is as follows: a twisted sheaf is the assignment, to every open set U in X, of a collection of sheaves on U parameterized by the sections of $L^a(U)$, and compatible with tensoring by A-torsors. Of course, since if $L^a(U)$ is nonempty this is the same as giving just one sheaf, this is sort of overkill, but the choice of just one such sheaf is noncanonical whereas this description is canonical. These collections should be compatible with the restriction functors $L^a(U) to L^a(V)$ when $V subset U$. It is an exercise to reader to check that this is the same as the other definition of twisting :)

Man, you asked the right question at the right time. My thesis is all about this stuff.

classification of Nilpotent Leibniz Algebra

Two more papers by Albeverio, S., Omirov, B.A., Rakhimov, Isamiddin S.:

Varieties of nilpotent complex Leibniz algebras of dimensions less then five, Comm. Algebra 33 (2005), N5, 1575-1585 DOI:10.1081/AGB-200061038

Classification of 4-dimensional nilpotent complex Leibniz algebras, Extr. Math. 21 (2006), No. 3, 197-210 http://www.unex.es/extracta/Vol-21-3/21J3Albe.pdf (also arXiv:math/0611831)

There are probably some other papers on the subject by Ayupov, Omirov, and their collaborators.

dg.differential geometry - A reference for smooth structures on R^n

You can handle the case of $n leq 3$ one at a time, and so the question really is about $n geq 5$. Two important names in this regard are Kirby and Siebenmann. The Wikipedia article on the Hauptvermutung is a good place to start.

If M is an $n$-dimensional topological manifold (and $n geq 5$), then $M$ admits a PL structure if and only if a special cohomology class, the Kirby-Siebenmann class, in $H^4(M; mathbb{Z}_2)$ vanishes. If this class vanishes, then the different PL structures are parametrized up to concordance by $H^3(M; mathbb{Z})$. (Note: The Wikipedia article on the Hauptvermutung assumes that $M$ is compact, but I don't believe that this is a necessary assumption.)

So what does this say about $M = mathbb{R}^n$? Well, we already know that $mathbb{R}^n$ has a PL structure, and since $H^3(mathbb{R}^n; mathbb{Z}_2)=0$, it follows that this structure is unique up to concordance. Since concordance implies diffeomorphism, and since every smooth structure gives us a PL structure, it follows that there can be only one smooth structure on $mathbb{R}^n$ up to diffeomorphism.

Here are the main references (you can find them both here):

1. Kirby and Siebenmann, On the triangulation of manifolds and the Hauptvermutung. Bull. Amer. Math. Soc. 75 1969 742--749.




2. Kirby and Siebenmann, Foundational essays on topological manifolds, smoothings, and triangulations. Annals of Mathematics Studies 88 (1977). (I did some MathSciNet investigating, and the relevant essays are IV and V.)

This expository article by Rudyak, which I found through Wikipedia, also seems interesting.

Finally, I learned all of this from Scorpan's wonderful book, "The Wild World of 4-Manifolds".

reference request - Database of Steiner triple systems

The first answer is not a database of the Steiner triple systems, but rather how many there are up to isomorphism. It is known that they there are 2 of order 13, 80 of order 15, and 11084874829 of order 19. The last number was computed by Kaski and Ostergard, and I suppose that the best approach is to ask them for their data. It's more or less the end of the story, because it is easy to make larger Steiner triple systems, but impossible to compile them into a complete database.

Actually this paper describes a compressed 39-gigabyte file with the Steiner triple systems of order 19, and says that it is available by e-mail request from three of the authors (including Kaski and Ostergard).

It looks like a number of people have the Steiner triple systems of order 15, but I didn't find a paper that simply lists them.

Update 1: It seems that everyone works from the paper "Small Steiner triple systems and their properties", by Mathon, Phelps, and Rosa. This paper is basically an encyclopedia of the 80 Steiner triple systems of order 15 and many of their properties. It also introduces a somewhat standard numbering. The thing to do at this point would be to transcribe the data in this widely cited paper into a file. Google seems to indicate that no such file has been posted to the web.

Update 2: It was done! See file data/steiner.tbl in this GAP package by Nagy and Vojtechovsky. For some reason, a Steiner triple system of order $n$ is also called a Steiner loop of order $n+1$, and that is the terminology that they use. They copied the data from Colbourn and Rosa, Triple Systems, which presumably is the same as in Mathon, Phelps, and Rosa.

How I found it: I Googled one of the hexadecimal strings used to describe one of the STS(15)s. None of the Google's heuristics worked for me, so instead I used the old-fashioned trick of searching for a very specific keyword. It also shows up in a LaTeX PHD thesis, but the GAP file, which has almost the same syntax as Python, is better.

lie algebras - Weyl group Invariants

What are the generators of $mathbb C[V^m]^W$, where $W$ is the Weyl group
of type $E_6, E_7, E_8$, V^m denote 'm' (m > 1) copies of the Cartan subalgebra
and the action is the diagonal action?

Is there any reference where I can find the generators explicitly?

ac.commutative algebra - Determining if a ring satisfies Serre's condition S_{n}

I think there is a neat answer to this question.

Lemma: Let $S$ be regular local of dimension $d$, $M$ a f.g $S$-module. Then:
$$text{depth}(M)geq n Longleftrightarrow text{Ext}^i(M,S)=0 text{for} i>d-n$$

Proof: LHS is equivalent to $e= text{pd}_SM leq d-n$. By using a minimal free resolution of $M$ to compute Ext, one sees that $text{Ext}^i(M,S)$ is not $0$ for $i=e$ (Nakayama's lemma) and $0$ for $i>e$.

Now let $A=k[x_1,cdots,x_d]$, $R=A/I$. Here is the main:

Claim: $R text{is} (S_n) Longleftrightarrow text{dim}(text{Ext}_A^i(R,A))leq d-n-i forall i>d- text{dim}(R)$

(of course, we only need to check for values of $i$ up to $d$, as $A$ has finite global dimension $d$).

Proof: By Lemma one needs to check that for all $pin text{Spec} A$:
$$text{Ext}_{A_p}^i(M_p,A_p)=0 text{for} i>text{dim}(A_p) -min{n,text{dim}(R_p)}=max{text{dim}(A_p)-n,text{dim}(A_p)-text{dim}(R_p)}$$

This condition is equivalent to the fact that for all $i>0$ and each $p$ in the support of $text{Ext}_A^i(R,A)$ we must have $ileq max{text{dim}(A_p)-n,d-text{dim}(R)}$. Note if $i < d-text{dim}(R)$, $text{Ext}^i(R,A)=0$, so the claim follows.

You can compute both Ext and dimension with Macaulay 2.

How do we study the theory of reductive groups?

Sit at a table with the books of Borel, Humphreys, and Springer. Bounce around between them: if a proof in one makes no sense, it may be clearer in the other. For example, Springer's book develops everything needed about root systems from scratch, and has lots of nice exercises relate to that stuff. On the other hand, Borel is better about systematically allowing general ground fields from early on (so one doesn't have to redo the proofs all over again upon discovering that it is a good idea to allow ground fields like $mathbf{R}$, $mathbf{Q}$, $mathbf{F}_ p$, and $mathbf{F} _p(t)$). Pay attention to the power of inductive arguments with centralizers and normalizers (especially of tori).

Unfortunately, none makes good use of schemes, which clarifies and simplifies many things related to tangent space calculations, quotients, and positive characteristic. (For example, the definition of central isogeny in Borel's book looks a bit funny, and if done via schemes becomes more natural, though ultimately equivalent to what Borel does.) So if some proofs feel unnecessarily complicated, it may be due to lack of adequate technique in algebraic geometry. (Everyone has to choose their own poison!) Waterhouse's book has nothing serious to say about reductive groups, but the theory of finite group schemes that he discusses (including Cartier duality and structure in the infinitesimal case) is very helpful for a deeper understanding isogenies between reductive groups in positive characteristic. The exposes in SGA3 on quotients and Grothendieck topologies (etale, fppf, etc.) are helpful a lot too (some of which is also developed in the book "Neron Models"). Galois cohomology is also useful when working with rational points of quotients.

topological groups - Dense cyclic subgroup

First, it is clear the group has to be abelian. Now, if you assume that $G$ is locally compact, then by the classification you can decompose $G$ as $G={mathbb R}^n times H$ where
$H$ has a compact open subgroup. Clearly, there can be no ${mathbb R}^n$ factor, so $G$ has a
compact open subgroup. Now, suppose $G$ is itself compact and topologically generated by $g$. Then any character $chi$ in the dual of $G$ vanishing on $g$ will be identically zero. So, the map
$chi mapsto chi (g)$ is injective, hence the dual is a subgroup of $U(1)$. Conversely, you can also see that if $Gamma$ is a subgroup of $U(1)$ (considered with the discrete topology) then the dual of $Gamma$ has a dense cyclic subgroup. By taking various subgroups you can, for instance, get the $p$-adic integers, or the n-torus.

co.combinatorics - Multiplication of (0,1) matrices

Yes, you can imagine a three columns graph, each column has n points. The resultant matrix (AB)_ij= # of path from i-th point in the first column to j-th column in the last column.

Actually, if we assume the Word RAM computational model, the above interpretation leads to an O(n^3/log^2 n) time algorithm, which is better than O(n^3).

Simple proof that these graphs are perfect

I recently came across a family of infinite graphs (in the context of two-dimensional convexity) that don't have induced 4-paths (paths with 4 vertices).
Note that the complement of a 4-path is again a 4-path.
Clearly, every induced $n+1$-cycle contains an induced $n$-path.
Hence, by the Strong Perfect Graph Theorem of Chudnowski, Robertson, Seymour, and Thomas,
graphs without induced 4-paths are perfect.

Can anyone provide a simple proof of that fact?
Having no induced 4-paths seems like a very strong condition.

qa.quantum algebra - Why do my quantum group books avoid homotopical language?

The question is rather rambling and it is more about not so well-defined appetites (do you have a more conrete motivation?).

There is one thing which however makes full sense and deserves the consideration. Namely it has been asked what about higher categorical analogues of (noncommutative noncocommutative) Hopf algebras. This is not a trivial subject, because it is easier to do resolutions of operads than more general properads. Anyway the infinity-bialgebras are much easier than the Hopf counterpart. There is important work of Umble and Saneblidze in this direction (cf. arxiv/0709.3436). The motivating examples are however rather different than quantum groups, coming from rational homotopy theory, I think.

Similarly, there is no free Hopf algebra in an obvious sense what makes difficult to naturally interpret deformation complexes for Hopf algebras (there is a notion called free Hopf algebra, concerning something else). Boris Shoikhet, aided with some help from Kontsevich, as well as Martin Markl have looked into this.

Another relevant issue is to include various higher function algebras on higher categorical groups, enveloping algebras of higher Lie algebras (cf. baranovsky (pdf) or arxiv 0706.1396 version), usual quantum groups, examples of secondary Steenrod algebra of Bauese etc. into a unique natural higher Hopf setting. I have not seen that.

The author of the question might also be interested in a monoidal bicategorical approach to general Hopf algebroids by Street and Day.

ac.commutative algebra - Limit of a series of singularities

The $A_infty$ and $D_infty$ plane curve singularities have defining equations $x^2=0$ and $x^2y=0$. These equations are "clearly" natural limiting cases of the equations for $A_n$ singularities $x^2 + y^{n+1}=0$ and $D_n$ singularities $x^2y+y^{n-1}=0$ as $n to infty$, since large powers are small in the adic topology. So we're tempted to say that $A_infty$ and $D_infty$ are "limits" of the "series of singularities" ${A_n}$ and ${D_n}$. This was already observed by Arnol'd in 1981, who wrote "Although the series undoubtedly exist, it is not at all clear what a series of singularities is."

Have there been any attempts since Arnol'd to make sense out of the phrases in quotes in the previous paragraph? That is:

Are there precise definitions of a "series of singularities", and of the "limit" of a series of singularities, under which $lim_{nto infty} A_n = A_infty$ and $lim_{nto infty} D_n = D_infty$?

---

If the answer is Yes, here's another desideratum: does the notion of "limit" extend to modules/sheaves over the singularities? My motivation here is that the $A_n$ and $D_n$ are (almost) precisely the equicharacteristic hypersurfaces with finite Cohen-Macaulay type (i.e. only finitely many indecomposable MCM modules), while $A_infty$ and $D_infty$ are precisely the ones with countable or bounded CM type. I'd really like some statement that each MCM module over the "limit" "comes from" a module "at some finite stage".

ag.algebraic geometry - D-modules, deRham spaces and microlocalization

Given a variety (or scheme, or stack, or presheaf on the category of rings), some geometers, myself included, like to study D-modules. The usual definition of a D-module is as sheaves of modules over a sheaf of differential operators, but for spaces that aren't smooth in some sense, this definition doesn't work that well, and you want to use a different definition. My overall question is how to reinterpret microlocalization in this alternative definition.

deRham spaces

This definition is that a D-module on $X$ is a quasi-coherent sheaf on a new space $X_{dR}$, the deRham space of $X$. It's easiest to define this is in terms of its functor of points: a map of Spec R to $X_{dR}$ is by definition a map of Spec $R/J_R$ to $X$ where $J_R$ is the nilpotent radical of $R$. So this is not a topological space, but it is a sheaf on the big Zariski site, and I can make sense of a quasi-0-coherent sheaf on one of those. For more details, you can see the notes of Jacob Lurie on these.

More informally $X_{dR}$ is $X$ "with all infinitesimally close points identified." A sheaf on this space is like a D-module in that a D-module is a sheaf with a connection, i.e. where the fibers of infinitesimally close points are identified. You'll note, I say "space" here, since I want to be vague about what this object is. It's very hard from being a scheme, but I believe it is a (EDIT: not actually algebraic!) stack.

microlocalization

Now, one of the lovely things about D-modules is that they have a secret life on the cotangent bundle of X. You might think a D-module is a sheaf on X, but this is not the whole picture: there is also a microlocal version of things.

The sheaf of functions on $T^*X$ has a quantization $mathcal{O}^h$; this is a non-commutative algebra over $mathbb{C}[[h]]$ such that $mathcal{O}^h/hmathcal{O}^hcong mathcal{O}_{T^*X}$, defined using Moyal quantization.

There's a ring map $p^{-1}mathcal{D}to mathcal{O}^h[h^{-1}]$, and thus a functor from D-modules to sheaves of $mathcal{O}^h[h^{-1}]$-modules on $T^*X$ given by $mathcal{O}^h[h^{-1}]otimes_{p^{-1}mathcal{D}}mathcal{M}$, called microlocalization, because it makes D-modules even more local than they were before. This is an equivalence between D-modules and $mathbb{C}^*$-equivariant $mathcal{O}^h[h^{-1}]$-modules.

Given an $mathbb{C}^*$ invariant open subset $U$ of $T^*X$, one can look at $mathcal{O}^h[h^{-1}]$-modules on $U$, and obtain a microlocalized category of D-modules, which has all kinds of interesting geometry one couldn't see before. I'm particularly interested in the semi-stable points for the action of some group $G$ on $X$ (extended to $T^*X$).

my question:

Now, I'm something of a convert to derived algebraic geometry, so it feels intuitive to me that anything one has to say about D-modules should be sayable using deRham spaces. On the other hand, I have no idea how microlocalization can be phrased in this way. Do any of you out in MathOverflowLand?

graph theory - Efficient way of determining isomorphism

The other answers have treated the case of finite graphs, but there is something interesting to say about infinite graphs as well.

For example, we might consider the case of computable graphs. A graph is computable if it has a computable edge relation on the vertex set of natural numbers. It is quite natural to inquire of isomorphic computable graphs, whether they have a computable isomorphism.

Question. If $G$ and $H$ are computable graphs and isomorphic, must there be a computable isomorphism?

I hope you will find it interesting to learn that the answer is No. To see this, let me describe two isomorphic graphs, which will both be trees of height 2, having infinitely many leaves on level $1$ and infinitely many leaves on level $2$, but no splitting except at the root. On the one hand, we can easily build a computable such graph $G$, by using $0$ as the root, the remaining even numbers as the level $1$ nodes, and giving exactly the multiples of $4$ a successor on level $2$, using the odd numbers. Next, let me describe another computable presentation $H$ of this graph. We again use $0$ as the root and the remaining even numbers $2n$ as the level one nodes. But this time, we give $2n$ a successor, using the $k^{rm th}$ odd number in the construction, only if program $n$ halts on input $0$ in less than time $k$. The graphs $G$ and $H$ are certainly isomorphic, because of their trivial form, but there can be no computable isomorphism between them, since any isomorphism would have to send the node $2n$ in $H$ either to a multiple of $4$ or not, and this would reveal whether it gets a successor in $G$, which would in turn tells us whether the program $n$ halts or not, solving the halting problem. So there can be no such computable isomorphism. QED

The argument generalizes to oracles. That is, for any Turing degree $d$, there are isomorphic graphs that are computable from oracle $d$, but no isomorphism between them is computable from $d$. Thus, in the infinite case, one answer to your question is that you cannot compute the isomorphism of isomorphic graphs $G$ and $H$ just knowing $G$ and $H$. The isomorphism may simply have a greater Turing degree.

More generally, this article by Csima, Khoussainov and Liu investigates the class of computably categorical graphs, the graphs $G$ that have the property that any two computable presentations of $G$ should be computably isomorphic.

I discuss the issues of computable categoricity in this MO answer, which asked whether there could be equivalence without us being able to calculate it, a question for which the subject of computable model theory provides numerous answers.

And there is more to say when one moves to uncountable graphs...

ac.commutative algebra - Sums of two squares in (certain) integral domains

While giving the first of eight lectures on introductory model theory and its applications yesterday, I stated Hilbert's 17th problem (or rather, Artin's Theorem): if $f in mathbb{R}[t_1,ldots,t_n]$ is positive semidefinite -- i.e., non-negative when evaluated at every $x = (x_1,ldots,x_n) in mathbb{R}^n$ -- then it is a sum of squares of rational functions. One naturally asks (i) must $f$ be a sum of squares of polynomials, and (ii) do we know how many rational functions are necessary? The general answers here are no (Motzkin) and no more than $2^n$ (Pfister). Then I mentioned that the case of $n=1$ is a very nice exercise, because one can prove in this case that indeed $f(t)$ is positive semidefinite iff it is a sum of two (and not necessarily one, clearly) squares of polynomials. Finally I muttered that this was a sort of function field analogue of Fermat's Two Squares Theorem (F2ST).

So I thought about how to prove this result, and I was able to come up with a proof that follows the same recipe as the Gaussian integers proof of F2ST. Then I realized that the key step of the proof was that a monic irreducible quadratic polynomial over $mathbb{R}$ is a sum of two squares, which can be shown by...completing the square.

But then today I went back to the general setup of a "Gaussian integers" proof, and I came up with the following definition and theorem.

Definition: An integral domain $R$ is imaginary if $-1$ is a square in its fraction field; otherwise it is nonimaginary. (In fact I will mostly be considering Dedekind domains, hence integrally closed, and in this case if $-1$ is a square in the fraction field it's already a square in $R$, so no need to worry much about that distinction.) Note that nonimaginary is a much weaker condition than the fraction field being formally real.

(Definition: An element $f$ in a domain $R$ is a sum of two squares up to a unit if there exist $a,b in R$ and $u in R^{times}$ such that $f = u(a^2+b^2)$.)

Theorem: Let $R$ be a nonimaginary domain such that $R[i]$ ($= R[t]/(t^2+1)$) is a PID.
a) Let $p$ be a prime element of $R$ (i.e., $pR$ is a prime ideal). Then $p$ is a sum of two squares up to a unit iff the residue field $R/(p)$ is imaginary.
b) Suppose moreover that $R$ is a PID. Then a nonzero element $f$ of $R$ is a sum of two squares up to a unit iff $operatorname{ord}_p(f)$ is even for each prime element $p$ of $R$ such that $R/(p)$ is nonimaginary.

[Proof: Introduce the "Gaussian" ring $R[i]$ and the norm map $N: R[i] rightarrow R$. Follow your nose, referring back to the proof of F2ST as needed.]

Corollaries: 1) F2ST. 2) Artin-Pfister for $n = 1$. 3) A characterization of sums of two squares in a polynomial ring over a nonimaginary finite field (a 1967 theorem of Leahey).

4) Let $p equiv 3,7 pmod{20}$ be a prime number. Then $p$ is a sum of two squares up to a unit in $mathbb{Z}[sqrt{-5}]$ but is not (by F2ST) a sum of two squares in $mathbb{Z}$.

Finally the questions:

Have you seen anything like this result before?

I haven't, explicitly, but somehow I feel subconsciously that I may have. It's hard to believe that this is something new under the sun.

What do you make of the strange situation in which $R$ is not a PID but $R[i]$ is?

Note that one might think this impossible, but $R = mathbb{R}[x,y]/(x^2+y^2-1)$ is an example. [Reference: Theorem 12 of http://math.uga.edu/~pete/ellipticded.pdf.] Do you have any idea about how one might go about producing more such examples, e.g. with $R$ the ring of integers of a number field (or a localization thereof)?

---

Addendum: As I commented on below, a good answer to the first question seems to be the paper

MR0578805 (81h:10028)
Choi, M. D.; Lam, T. Y.; Reznick, B.; Rosenberg, A.
Sums of squares in some integral domains.
J. Algebra 65 (1980), no. 1, 234--256.

In this paper, they prove the theorem above with slightly different hypotheses: $R$ is a nonimaginary UFD such that $R[i]$ is also a UFD. Looking back at my proof, the only reason I assumed PID was not to worry about the distinction between $R/pR$ and its fraction field. Just now I went back to check that everything works okay with PID replaced by UFD. So the second question becomes more important: what are some examples to exploit the fact that $R[i]$, but not $R$, needs to be a UFD?

reference request - Representation theory over Z

See Curtis–Reiner's textbook on the Representation Theory of Finite Groups and Associative Algebras ^{(MR 144979)}, Theorem 74.3, page 507, and especially the introduction starting on page 493.

The result for cyclic groups of prime order, and for order 4 was originally done in:

• Diederichsen, Fritz-Erdmann.
  "Über die Ausreduktion ganzzahliger Gruppendarstellungen bei arithmetischer Äquivalenz"
  Abh. Math. Sem. Hansischen Univ. 13, (1940). 357–412. MR2133.

However, Reiner has written quite a few nice papers on similar subjects. One of his earlier ones is on the same topic:

• Reiner, Irving. "Integral representations of cyclic groups of prime order."
  Proc. Amer. Math. Soc. 8 (1957), 142–146.
  MR83493
  doi:10.2307/2032829

One can also consult texts on things called "crystallographic groups", "space groups", and "p-adic space groups". Plesken has written several nice books using this sort of thing. These give infinite families of nicely related finite groups and of course help crystallographers.

Be careful to distinguish these sorts of representations from ZG-modules. ZG-modules are basically incomprehensible, so instead lots of people focus on ZG-lattices, where the underlying module is projective. This means the idea of using matrices still makes some sense. There is a lot of literature on modules over group rings over nice rings (like Z or Dedekind domains), but a fair amount of it is not applicable to questions about GL(n,Z).

Roughly speaking, even for G=1, ZG modules are too difficult to understand, and adding a G just makes it worse. Another common tack is to look at $hat {mathbb{Z}}_p$ modules, p-adic modules. Again the results are nicest for lattices, but things do not get so bad near so fast there. Reiner's Maximal Orders textbook describes some of the beautiful and well-behaved things you can see there.

co.combinatorics - Trying to sum a series (related to catalan numbers perhaps)

edit: the preceding answer suggests that my browser didn't display the dots, i.e. you really meant the series, not the sequence... Sorry, the below doesn't answer the question.

---

Does the following look right? (might this be homework?)

(1) -> f n == reduce(+, [binomial(2*i, i)*2^i/3^(2*i) for i in 0..n])
                                                                   Type: Void
(2) -> guess([f n for n in 0..20], maxLevel==2)
   Compiling function f with type NonNegativeInteger -> Fraction(
      Integer)

                  s   - 1
                   21      8p   + 12
                   ++-++     20
                 4  | |    ---------
                    | |    9p   + 18
         n - 1    p  = 0     20
          --+      20
   (2)  [ >      ------------------- + 1]
          --+             9
         s  = 0
          21
                                              Type: List(Expression(Integer))
(3) -> guessPRec [f n for n in 0..20]

   (3)
   [
     [f(n): (9n + 18)f(n + 2) + (- 17n - 30)f(n + 1) + (8n + 12)f(n)= 0,
                     13
      f(0)= 1, f(1)= --]
                      9
     ]
                                              Type: List(Expression(Integer))

In general, it's often a good idea to generalise, i.e., introduce more parameters:

(4) -> f n == reduce(+, [binomial(2*i, i)*x^i/y^(2*i) for i in 0..n])
   Compiled code for f has been cleared.
   1 old definition(s) deleted for function or rule f
                                                                   Type: Void
(5) -> guess([f n for n in 0..20], maxLevel==2)
   Compiling function f with type NonNegativeInteger -> Fraction(
      Polynomial(Integer))

                   s   - 1
                    21      (4p   + 6)x
                    ++-++      20
                 2x  | |    -----------
                     | |              2
         n - 1     p  = 0   (p   + 2)y
          --+       20        20
   (5)  [ >      ---------------------- + 1]
          --+               2
         s  = 0            y
          21
                                              Type: List(Expression(Integer))
(6) -> guessPRec [f n for n in 0..20]

   (6)
   [
     [
       f(n):
                   2                      2
           (n + 2)y f(n + 2) + ((- n - 2)y  + (- 4n - 6)x)f(n + 1)
         +
           (4n + 6)x f(n)
           =
           0
       ,
                      2
                     y  + 2x
      f(0)= 1, f(1)= -------]
                         2
                        y
     ]
                                              Type: List(Expression(Integer))

ca.analysis and odes - Simultaneous Equations Involving Power Sums

Actually Darsh gave an almost full solution. Let me fill in the minor technical details.

1) We need the following quantitative form of the inverse function theorem. Suppose that $F:mathbb R^nto mathbb R^n$. Assume also that $|DF(X)^{-1}|le C_1$, that $max_{Yin B(X, delta)}|D^2F(Y)|le C_2$, and that $C_1C_2deltalefrac 12$. Then $F(B(X,delta))supset B(F(X),frac{delta}{2C_1})$.

2) Take $n=2ell-1$ and consider the mapping $F:mathbb R^nto mathbb R^n$ given by $F(y_1,dots,y_n)_k=sum_{j=1}^n y_j^k$ where $k=1,2,dots,n$. Take $X=(x_1,dots,x_n)$ where $x_j=frac{n+j}{n}$ for $j=1,dots,n$.

3) Note that in $B(X,1)$, we have $|D^2F|le A^n$ for some absolute $A>1$.

4) Note also that $DF(X)^*$ is the linear operator that maps the vector $(c_1,dots,c_n)$ to the vector $p(x_1),ldots,p(x_n)$ consisting of the values of the polynomial $p(x)=sum_{k=1}^n c_k kx^{k-1}$. The inverse operator is given by the standard interpolation formula, which allows us to estimate its norm by $B^n$ with some absolute $B>1$.

5) Thus, taking $delta=2^{-1}(AB)^{-n}$, we conclude that the image of the ball $B(X,delta)$ contains a ball of radius $fracdelta{2A^n}ge C^{-n}$ with some absolute $C>2$.

6) In particular, it contains two points with the difference $(0,0,dots,0,D^{-ell},D^{-(ell+1)},dots,D^{-(2ell-1)})$ with some absolute $D>1$, which is equivalent to what we need.

co.combinatorics - Recursions which define polynomials

I have no answer to your question, but some related examples.
Consider the q-Fibonacci polynomials defined by f(0, x, s)=0, f(1, x, s)=1 and f(n, x, s)=x f(n-1, x, s)+q^(n-2) s f(n-2, x, s).
Then the subsequences f(k n, x, s) satisfy a homogeneous recursion with rational coefficients which for k>2 are not polynomials (see e.g. my paper in arXiv 0806.0805).

More precisely
f(k, x, q^k s) f(k n, x, s) – f(2 k, x, s) f(k (n-1), x, q^k s)
+(-1)^k q^(k(3k-1)/2) s^k f(k, x, s) f(k (n-2), x, q^(2k) s)= 0
or equivalently

f(k, x, q^(n-2k) s) f(k n, x, s) – f(2 k, x, q^(k (n-2)) s) f(k (n-1), x, s)
+(-1)^k q^(-k (3k+1)/2) q^(k^2 n) s^k f(k, x, q^(k (n-1)) s) f(k (n-2), x, s)= 0.

Analogous results are true for powers of q-Fibonacci polynomials.

ac.commutative algebra - sum of radical ideals

Let $A$ be a commutative ring and endow the closed subsets of $mathrm{Spec}(A)$ with the Grothendieck topology of finite covers. One may ask if the presheaf $V mapsto A/I(V)$ is a sheaf. This is not true in general and is related (but not equivalent) to the following pure algebraic question:

In which commutative rings $A$ are the radical ideals closed under sum?

The property can be checked locally. It holds in dimension $0$, and also for integral domains of dimension 1. It doesn't hold for the $2$-dimensional ring $k[x,y]$ (consider $(x^2 + y)+(y) = (x^2,y)$), nor for the 1-dimensional ring $bigl(k[x,y]/(x^2 y + y^2)bigr)_{(x,y)}$.

Are there other interesting examples/counterexamples or approaches for a general classification? I think the property has some algebro-geometric interpretation: All intersections of closed subschemes are transversal. See also SE/322872.

ct.category theory - What theorem constructs an initial object for this category? (Formerly "Integrability by abstract nonsense")

The category you described can be written as a lax limit of a diagram in the 2-category of categories. The diagram in question consists of accessible categories and accessible functors, so its limit is again accessible by a theorem of Makkai and Paré. It is obvious from the construction that the category has all small limits (they are computed as in the category of Banach spaces and nonexpanding maps), so it is in fact locally presentable. This means that it has all colimits and in particular an initial object.

Here are some more details and references. Let $mathbf{Ban}_1$ be the category of Banach spaces and nonexpanding maps. This category is locally $aleph_1$-presentable (see e.g. Borceux, Handbook of Categorical algebra, Volume II, 5.2.2.e). Let

$F colon mathbf{Ban}_1 rightarrow mathbf{Ban}_1$

be the functor which sends a Banach space $X$ to $mathbb{R}+Xoplus X$, where + stands for the coproduct. A morphism f of Banach spaces gets sent to $mathrm{id}_{mathbb{R}}+foplus f$. Let $U colon mathbf{Ban}_1 rightarrow mathbf{Set}$ be the functor which sends a Banach space to its underlying set. Since $aleph_1$-filtered colimits are computed as in the category of sets (see e.g. Borceux, Handbook of Categorical algebra, Volume II, 5.2.2.e) we know that the composite UF preserves $aleph_1$-filtered colimits. By the open mapping theorem it follows that $F$ preserves $aleph_1$-filtered colimits.

By the theorem of Makkai and Paré (see e.g. Adámek, Rosický, Locally Presentable and accessible categories, Theorem 2.77), the inserter $mathcal{C}$ of $F$ and the identity functor on $mathbf{Ban}_1$ is again an accessible category. The objects of $mathcal{C}$ are triples $(X,xi,u)$ where $xicolon Xoplus X rightarrow X$ and $ucolon mathbb{R} rightarrow X$ are nonexpanding (i.e., u corresponds to an element of $X$ of norm less than or equal one). The morphisms are morphisms of Banach spaces which are compatible with $u$ and $xi$. Thus $mathcal{C}$ is almost the category we are interested in; the only thing that's missing is the requirement that $xi(u,u)=u$.

Let

$G colon mathcal{C} rightarrow mathbf{Ban_1}$

be the functor which sends every object to $mathbb{R}$ and every morphism to $mathrm{id}_{mathbb{R}}$. This is clearly a functor which preserves $aleph_1$-filtered colimits. Let

$H colon mathcal{C} rightarrow mathbf{Ban_1}$

be the functor which sends $(X,xi,u)$ to $X$ and a morphism to itself; this is again a functor which preserves $aleph_1$-filtered colimits. There are two natural transformations $alpha,beta colon G Rightarrow H$, whose component at $(X,xi,u)$ is given by $u$ and $xi(u,u)$ respectively. The equifier of $G$ and $H$ is precisely the category we are looking for, and from our construction we can conclude that it is accessible.

The obvious forgetful functor to Banach spaces creates limits, so this category is complete. Thus it is locally presentable, and therefore also cocomplete. In particular, it follows that our category has an initial object.

special functions - Finding a recursion for a sum of Legendre polynomials

(this is a variation on Darij's answer)

Note that "holonomic" generating functions $f(x)$, i.e. generating functions satisfying a linear differential equation with polynomial coefficients (in $x$) enjoy rich closure properties.
For a (very brief) summary, see Table 1 of a paper on guessing (sorry, I couldn't find a better reference, but there should be one...). Moreover, all these closure properties can be made "effective": if we know the order and the degree of the coefficients of the diffential equation for $f(x)$ and for $g(x)$, we can (easily) compute bounds for order and coefficient degree of the differential equation for $f+g$, $fcdot g$, etc.

Moreover, it turns out that the (Taylor) coefficient sequence of a holonomic generating function satisfies a linear recurrence with polynomial coefficients, i.e., is P-recursive, and the P-recursive sequences are precisely the coefficient sequences of holonomic generating functions.

Thus, to find such a recurrence (or differential equation), it is often easiest to use gfun for Maple by Bruno Salvy and Paul Zimmermann, or the Mathematica equivalent by Mallinger (I think). Unfortunately, the FriCAS package described in 1 only does the guessing part, i.e., you would need to compute the bounds yourself and then check. In the case at hand the results are below.

However, you also asked, whether the solution is hypergeometric. To check this, you simply feed the recurrence you found below into Petkovsek's program Hyper, it will tell you whether there is a hypergeometric solution. (most probably no, because if the degree of the coefficient polynomials of the hypergeometric solution are not huge, guessPRec and guessHolo would have found it.)

In case you are dealing with orthogonal polynomials, the Askey Wilson scheme by Koekoek and Swarttouw is a great reference to find the basic information.

(1) -> guessHolo(cons(x, [(legendreP(n, x)-(1-1/n)*legendreP(n-2, x)) for n in 2..30]), variableName==t)

   (1)
   [
     [
         n
       [t ]f(t):
                 4     2  2     5     3           4     2  ,,
           ((- 2t  - 2t )x  + (t  + 6t  + t)x - 2t  - 2t )f  (t)

         + 
                 3       2      4      2           3       ,
           ((- 3t  - 7t)x  + (3t  + 15t  + 2)x - 8t  - 2t)f (t)

         + 
              2      2     3            2
           ((t  - 3)x  + (t  + 3t)x - 4t  + 2)f(t)
           =
           0
       ,
                    2           2 3      2       3 4      3 2     3
                3t x  - 2t   15t x  - 13t x   35t x  - 39t x  + 6t       4
      f(t)= x + ---------- + -------------- + --------------------- + O(t )]
                     2              6                   8
     ]
                                              Type: List(Expression(Integer))
(2) -> guessPRec cons(x, [(legendreP(n, x)-(1-1/n)*legendreP(n-2, x)) for n in 2..30])

   (2)
   [
     [
       f(n):
                 4      3      2             2     4      3      2
             ((4n  + 28n  + 67n  + 63n + 18)x  - 4n  - 28n  - 68n  - 68n - 24)
          *
             f(n + 2)
         + 
                    4      3       2              3
               (- 8n  - 52n  - 118n  - 107n - 30)x
             + 
                  4      3       2
               (8n  + 52n  + 120n  + 114n + 36)x
          *
             f(n + 1)
         + 
             4      3      2             2     4      3      2
         ((4n  + 24n  + 51n  + 46n + 15)x  - 4n  - 24n  - 52n  - 48n - 16)f(n)
           =
           0
       ,
                       2
                     3x  - 2
      f(0)= x, f(1)= -------]
                        2
     ]
                                              Type: List(Expression(Integer))

algorithms - Determining the space complexity of van Emde Boas trees

We call S(u) the space complexity of the vEB tree holding elements in the range 0 to u-1, and suppose without loss of generality that u is of the form 2^2k.

It's easy to get the recurrence S(u²) = (1+u) S(u) + Θ(u). (In Wikipedia's article the last term is O(1), but it's wrong because we must count the space for the array.)

Van Emde Boas gave in [1] the trivial bound S(u) = O(u log log u), and later in [2] he found a clever way to combine the data structure with other one so that to reach space complexity O(u), while mantaining the O(log log u) time bounds.

But modern references present the original data structure and state without proof that the space complexity is O(u). For instance, the very recent 3rd edition of "Introduction to algorithms" by Cormen et al. leaves it as an exercise.

I tried with some friends to [dis]prove the O(u) bound without luck.

[1] http://www.springerlink.com/content/h63507n460256241/

[2] http://www.cs.ust.hk/mjg_lib/Library/VAN77.PDF

co.combinatorics - Generalizations of Planar Graphs

A well ordering, $leq$, on a set $S$ is a WELL-QUASI-ORDERING if and only if every sequence $x_iin S$ there exists some $i$ and $j$ natural numbers with $i < j$ with $x_ileq x_j$. (See wikipedia article at bottom)

Robertson-Seymour Theorem: The set $S=Graphs/isomorphism$ are well-quasi-ordered under contraction.

The corollary of this theorem is that any property $P$ of graphs which is closed under the relation of contraction (meaning if $P(G_2)$ and $G_1leq G_2$ then $P(G_1)$) is characterized by a finite set of excluded minors (Which is explained below). An example of such a $P$ is planarity, or linkless embeddability of a graph into R^3. i.e. Every contraction of a planar graph is planar.

Suppose $P$ is a property closed under $leq$.

***If $Bleq G$ and $B$ is not $P(B)$, Then not $P(G)$.

The idea is to characterize $P$ by a collection of bad $B$'s. The finiteness of the set of excluded minors comes from the well-quasi ordering and doesn't use the idea of a graph:
Assuming we have well-quasi-ordered set $(S,leq)$. One can prove that every property $P$ which is closed under the relation is characterized by a Finite set of excluded minors.That is, there exists some $X=lbrace x_1,ldots,x_nrbrace subset S $ such that for all $sin S$, $$ mbox{ not } P(s) iff exists i, x_i leq s $$.

The existence of a finite set $X$ is implicitly in 12.5 of Diestel (link at bottom, see the corollary of Graph Minor Theorem in Diestel). First convince yourself that there exists a set of $B's$ (not necessarily finite) as in * that characterize property $P$. Then consider the smallest such set of $B$'s and using the property of well-quasi ordering show that is is finite. Note that as in the second wikipedia article, we can say any set of elements $Asubset S$ such that for all $a,bin A$ we have $a nleq b$ must be a finite set (provided $leq$ is a well-quasi-ordering).

Actual work done showing stuff in a topological direction has be done by Eran Nevo http://www.math.cornell.edu/~eranevo/

I suspect that Matroids have a well-quasi-ordering and that there is work being done toward proving an analogous theorem for them.

I have a limit on links:

en.wikipedia.org/wiki/Well-quasi-ordering

en.wikipedia.org/wiki/Robertson-Seymour_theorem

diestel-graph-theory.com/GrTh.html

Can an abelian variety be represented as the cohomology of some other object?

Question

Given an abelian variety $V$ and an integer $n$, is there a natural abelian category with a natural object $X$ and natural coefficients $F$ so that $Vsimeq H^n (X,F)$?

Motivation

Studying abelian varieties is awesome. Studying objects in long exact sequences is awesome. How do (somewhat forcefully) combine these two? I mean without taking cohomology of the variety like everyone else does...

Possible answers

The abelian variety is a $G$-module, where $G=Gal(bar{k}/k)$, $k$ the field over which the variety is defined. So, maybe there is an interesting $G$-module that answers the above? The cases of abelian varieties over number fields and finite fields are the most interesting, so $G$ is assumed to be interesting as well (i.e. not trivial).

Maybe it arises as the $n$-th cohomology of some interesting sheaf of some interesting related variety?

Transforming a Diophantine equation to an elliptic curve

Nearly 10 years ago, I gave a talk at Wesleyan, and a gentleman named Roy Lisker asked me the same question: Fix an integral solution $(x, y, z)$ and make the substitution

$$u = 3 frac {n^2 z - 12 x}z qquad v = 108 frac {2 x y - n x z + z^2}{z^2}$$

Then $(u, v)$ is a rational point on the elliptic curve $E_n: v^2 = u^3 + A u + B$ where $A = 27 n (24 - n^3)$ and $B = 54 (216 - 36 n^3 + n^6)$. (It actually turns out that $E_n$ is an elliptic curve whenever $n$ is different from 3, but I’ll discuss this case separately.)

Let me say a little about the structure of this curve for the experts:

This curve has the “obvious” rational point $T=(3 n^2, 108)$ which has order 3, considering the group structure of $E_n$.  It actually turns out that these three multiples correspond to the cases $x = 0$ and $z = 0$, so if such an integral solution $(x, y, z)$ exists then the rational solution $(u, v)$ must correspond to a point on $E_n$ not of order 3. (Of course, I don’t care about the cyclic permutation $x to y to z to x$.)

In the following table I’m computing the Mordell-Weil group of the rational points on the elliptic curve i.e. the group structure of the set of rational solutions $(u, v)$:

$$ begin{matrix}
n & E_n(mathbb Q) \ \
1 & Z_3 \
2 & Z_3 \
3 & text{Not an elliptic curve} \
4 & Z_3 \
5 & Z_6 \
6 & Z_3 oplus mathbb Z \
7 & Z_3 \
8 & Z_3 \
9 & Z_3 oplus mathbb Z \
end{matrix} $$

Hence when $n  =$ 1, 2, 4, 7 or 8 we find no integral solutions $(x, y, z)$.  When $n = 5$, there are only six rational points on $E_n$, namely the multiples of $(u,v) = (3, 756)$ which all yield just one positive integral point $(x,y,z) = (2,4,1)$.

Something fascinating happens when $n = 6$... The rank is positive (the rank is actually 1) so there are infinitely many rational points $(u, v)$.  But we must be careful: not all rational points $(u, v)$ yield positive integral points $(x, y, z)$.  Clearly, we can scale $z$ large enough to always choose $x$ and $y$ to be integral, but we might not have $x$ and $y$ to both be positive.  You’ll note that $x > 0$ if only if $u < 3 n^2$, so we only want rational points in a certain region of the graph.  Since the rank is 1, this part of the graph is dense with rational points! Let me give some explicit numbers.  The torsion part of $E_n( mathbb Q)$ is generated by $T = (75, 108)$ and the free part is generated by $(u,v) = (-108, 2052)$.  By considering various multiples of this point we get a lot of positive integral -- yet unwieldy! --  points $(x,y,z)$ such that $x/y + y/z + z/x = 6$:

$$begin{aligned} (x,y,z) & = (12, 9, 2), \
& = (17415354475, 90655886250, 19286662788) \
& = (260786531732120217365431085802, 1768882504220886840084123089612, 1111094560658606608142550260961) \
& = (64559574486549980317349907710368345747664977687333438285188, 70633079277185536037357392627802552360212921466330995726803, 313818303038935967800629401307879557072745299086647462868546) end{aligned} $$

I’ll just mention in passing that when $n = 9$ the elliptic curve $E_n$ also has rank 1.  The generator $(u,v) = (54, 4266)$ corresponds to the positive integral point $(x,y,z) = (63, 98, 12)$ on $x/y + y/z = z/x = 9$.

What about $n = 3$?  The curve $E_n$ becomes $v^2 = (u – 18) (u + 9)^2$.  This gives two possibilities: either $u = -9$ or $u geq 18$.  The first corresponds to $x = z$ while the second corresponds to $(z/x) geq 4$.  By cyclically permuting $x$, $y$, and $z$ we find similarly that either $x = y = z$ or $x/y + y/z + z/x geq 6$.  The latter case cannot happen by assumption so $x = y = z$ is the only possibility i.e. $(x,y,z) = (1,1,1)$ is the only solution to $x/y + y/z + z/x = 3$.

soft question - Real analysis has no applications?

As it happens, I just finished teaching a quarter of undergraduate real analysis. I am inclined to rephrase Pugh's statement into a form that I would agree with. If you view analysis broadly as both the theorems of analysis and methods of calculation (calculus), then obviously it has a ton of applications. However, I much prefer to teach undergraduate real analysis as pure mathematics, more particularly as an introduction to rigorous mathematics and proofs. This is partly as a corrective (or at least a complement) to the mostly applied and algorithmic interpretation of calculus that most American students see first.

Some mathematicians think, and I've often been tempted to think, that it's a bad thing to do analysis twice, first as algorithmic and applied calculus and second as rigorous analysis. It can seem wrong not to have the rigor up-front. Now that I have seen what BC Calculus is like in a high school, I no longer think that it is a bad thing. Obviously I still think that the pure interpretation is important. On the other hand, both interpretations together is also fine by me. I notice that in France, calculus courses and analysis courses are both called "analyse mathématique". I think that they might separate rigorous and non-rigorous calculus a bit less than in the US, and it could be partly because of the name.

In fact, it took me a long time to realize how certain non-rigorous explanations guide good rigorous analysis. For instance, the easy way to derive the Jacobian factor in a multivariate integral is to "draw" an infinitesimal parallelepiped and find its volume. That's not rigorous by itself, but it is related to an important rigorous construction, the exterior algebra of differential forms.

Finally, I agree that Pugh's book is great. As the saying goes, you shouldn't judge it by its cover. :-)

moduli spaces - Why is the Hodge class of bar{M_g} big and nef?

Some multiple of lambda is defined on the coarse moduli space and this is the pullback of an ample bundle on bar{A_g}, the Satake-Baily-Borel compactification of A_g. Since bar{M_g} maps birationally onto its image in bar{A_g}, it follows that lambda is nef and big, in fact also semi-ample (some multiple is base point free) on the coarse moduli space.

(The map to bar{A_g } contracts the boundary divisor corresponding to irreducible nodal curves so lambda is not ample.)

ra.rings and algebras - Commutative subalgebras of M_n

Although this is not an answer to the original question, readers might be interested to know of a cute open problem in the context of commutative subalgebras of $M_n(mathbb{C})$. Let $A$ be such an algebra; we focus on the minimal number of generators of $A$ as a $mathbb{C}$ algebra vis-a-vis the dimension of $A$ as a $mathbb{C}$ space. If $A$ is generated by just one element, then by Cayley-Hamilton, the dimension of $A$ as a $mathbb{C}$ space is bounded above by $n$. It is an old theorem of Gerstenhaber as also Motzkin and Taussky-Todd that if $A$ can be generated by two elements, then too, the dimension of $A$ is bounded above by $n$. OTOH, if $A$ needs at least four generators, there are examples to show that the dimension of $A$ can be greater than $n$. So here is the open question: are there commutative subalgebras $A$ that can be generated by three elements for which the dimension of $A$ as a $mathbb{C}$ space is greater than $n$?

The work on this question has led naturally to study the variety of commuting triples of matrices (a subvariety of affine 3n^2 space over $mathbb{C}$). This variety has been proved to be reducible (by Guralnick) for $nge 32$ (since sharpened to $nge 29$ by others); one set of enquiries focuses on determining irreducibility of this variety for smaller $n$. I believe it is known now that for $nle 7$, this variety is irreducible. This translates to the result that for $nle 7$, such an $A$ as above must indeed have dimension bounded by $n$.

Given the reducibility for general $n$ of this variety, another set of enquiries focuses on studying specific subvarieties of this commuting triples variety. For instance, studying the varieties of commuting triples $(M_1, M_2, M_3)$ where $M_3$ is fixed to be a matrix in which each eigenvalue appears in at most two blocks (aka "$2$-regular") leads naturally to generalized tangent varieties over determinantal varieties. On the other hand, fixing $M_3$ to be the nilpotent Jordan block of size $k$ repeated $m$ times along the diagonal (so $n = mk$) leads to generalized tangent varieties over the variety of commuting pairs of $mtimes m$ matrices.

Interesting stuff, this!

big list - Undergraduate Level Math Books

I would also recommend the book entitled Analysis on Manifolds by James Munkres. I think that this is a good undergraduate textbook in mathematics for any student wishing to pursue multivariable calculus in greater depth. My only complaint is that Munkres often chooses to include details which can be seen easily after a little bit of thought. Perhaps this can be viewed as an effort to show the student how to "properly do analysis": doing analysis, just like doing any other branch of mathematics, requires you to carefully apply definitions and theorems, and it is important for the student to appreciate this early in his/her mathematical learning.

That said, the book is an excellent text overall for "advanced calculus". The student will need to be familiar with single variable analysis and perhaps some linear algebra. (Even a rudimentary knowledge of linear algebra will do since Munkres develops most of the necessary theory from scratch.)

Roughly speaking, the book splits into two parts. The first part covers most of the results students see when doing multivariable calculus that are stated "without proof" in their texts. For example, "the equality of the mixed partials", "double integrals can be done in any order", "a bounded function is Riemann integral if and only if it is continuous almost everywhere", "the change of variables theorem" etc., are (very) imprecise forms of some of the results Munkres establishes.

In the second part of the book, manifolds and their theory are introduced. Thus, for example, a rudimentary introduction to tensors is given, and this is supplemented by the basic theory of differential forms, the De Rham groups (of the punctured plane), Stokes' theorem etc.

I think that the exposition could be tightened: if you actually pick up the book and really make an effort to read it, it is quite possible to finish the first half of the book in the space of a week (that is, approximately 200 pages in a week) simply because certain topics are explained in more detail (at least in my opinion) than necessary. (One example is Munkres' proof of the linearity, monotonicity, additivity etc. of the Riemann integral. This is proved in three contexts separately: the case of the integral over a rectangle, that over a bounded set, and that of improper integrals, when essentially the proofs can be left as relatively easy exercises in some cases.)

As the above comments suggest, I think that this is an excellent book for undergraduate students, but perhaps less so for graduate students. (Spivak's Calculus on Manifolds is good for both undergraduate and graduate students, in my opinion, but some people may suggest that it is too hard for undergraduates.) And after reading this book, you should have more than enough preparation to read more advanced texts such as William Boothby's An Introduction to Differentiable Manifolds and Riemannian Geometry.

soft question - Are there any (interesting) consequences of the irrationality of π?

I am not sure how appropriate this question is for MO. If it is not, I apologize in advance but I could not resist asking it and if by any chance I get some interesting answers, it will for sure be very useful to keep my students excited about mathematics and physics as September arrives.

We all know very well that $pi$ (the ratio of the circumference of a circle to its diameter in Euclidean space) is irrational and even transcendental. These are some of the famous results in all mathematics.

So I was wondering what will go wrong if $pi$ was just an integer number?

Are
there important theorems that are based on the fact that it is actually irrational
and/or transcendental?

sheaf theory - Cohomology of sheaves in different Grothendieck topologies

Suppose I have a sheaf $mathcal{F}$ on the (small) étale site over $X$. By restriction, $mathcal{F}$ is also a sheaf on $X$ (with the Zariski topology). When is it that the sheaf cohomologies (i.e. derived functors of the global section functors) agree in these two sites?

For example, in SGA 4 (Chapter VII, p355), Grothendieck proves that the above cohomologies agree for quasicoherent sheaves. I do not really understand the proof however, and would appreciate any elaboration on this. It uses the Leray spectral sequence, and it seems that the key ingredient is that the higher direct images $R^q f_*(mathcal{F})$ are $0$, where $f_*$ is the direct image functor induced by the inclusion of the Zariski site in the étale one).
Question: Are there any precise conditions on $mathcal{F}$ that guarantee that the two cohomologies agree? Is the above condition on $R^q f_*(mathcal{F})$ necessary/sufficient, and when does it hold?

I should give an example of failure: constant sheaves seem like the obvious choice, and here the cohomologies don't agree even in the case of a point (Zariski cohomology is zero, étale cohomology reduces to Galois cohomology).

Maybe an easier question would be obtained by instead starting with a sheaf $mathcal{G}$ on $X$ (with the Zariski topology).
Subquestion: Under what conditions can you extend $mathcal{G}$ to a sheaf on the (small) étale site? (I'm thinking about extending by inverse image, i.e. for an étale $f colon U to X$, setting $mathcal{G}(U) =f^*(mathcal{G})(U)$, but maybe there are other possibilities)
(Possibly) Easier question: Supposing that $mathcal{G}$ extends, are there any precise conditions on $mathcal{G}$ that guarantee that the Zariski cohomology and the étale cohomology of $mathcal{G}$ agree?

Here again I have in mind the case of quasicoherent sheaves, where everything works out nicely provided that you extend in the right way: starting with a quasicoherent sheaf on the Zariski topology, you need to make sure your sheaf ends up being quasicoherent for the étale topology. You need to take some tensor products to ensure this, as Scott points out in the comments.

(Of course, I have specialised this question to a particular example of sites/toposes, ie Zariski and étale. If you know how this works in more general cases, please share that too!)

co.combinatorics - Computing Bruhat Order Covering Relations

To put this in context: I am in the process of developing a package for Macaulay 2 (a commutative algebra software,) called "Permutations", which will add permutations as a type of combinatorial object that M2 will handle, hopefully integrating nicely with the (still in development) Posets package, the (still in development) Graphs package, and various current M2 functions.

One of the first things that I'd wanted to put in was functions which compute the poset of the Bruhat order on $S_n$ and compute when one permutation covers another in the Bruhat order. This was all coded and "working" - but has been producing a poset which is decidedly NOT the desired poset (among other things, the graph of the Hasse diagram isn't regular.) I'd like to ask whether the (probably somewhat naive) algorithm I was using to check covering relations seems reasonable (so the problem is just in the coding of it, not in the theory behind it) or not.

To see if $Pleq R$ in the Bruhat order:

Given a pair of permutations P and R, compute their lengths (the number of simple transpositions in their decomposition, or [as implemented right now] the sum of entries in their inversion vectors.) If length(P)=length(R)+1, then we compute
$(P^{-1})*R$.

If R covers P in the Bruhat order, then length$((P^{-1})*R)=1.$

Am I missing some subtlety of the Bruhat order? I thought one permutation covered enough exactly when they differed by a single, simple transposition. This seemed to capture that - but is giving me an incorrect poset.

Coding error or theory error? I'd love to hear it.

fa.functional analysis - Stone-Weierstrass for cones

A version of the Stone-Weierstrass Theorem asserts: If A is a linear subspace of C(K), the set of continuous functions on a compact space, and if A is a subalgebra that contains the constant functions and separates points, then A is dense in C(K) relative to the uniform (or sup-norm) topology. I am looking for a version for cones along the lines: if A is a subcone of X, itself a cone in C(K), if A is closed with respect to products, and if it contains constants and separates points in K, then A is ``dense'' in X. An example, would be the statement that the set of nondecreasing polynomials on [0,1] is dense in the set of nondecreasing continuous functions on [0,1]. (Is this true?)

I would appreciate references to such results, or to counterexamples.

ac.commutative algebra - Alternative proof of unique factorization for ideals in a Dedekind ring

I'm writing some commutative algebra notes, but I'm facing a difficulty in organizing the order of the topics. I'd like to have the topics about factorization before speaking of integral closure. This is fine, as long as I talk of UFD and primary decomposition.

The problem is that a topic worth mentioning is the factorization theory for ideals in a Dedekind ring. Now, there are a few ways to define a Dedekind ring, but I guess one of the most natural is a Noetherian domain, integrally closed, of dimension 1.

At this point I haven't yet introduced the concept of dimension, nor integral closure. It is easy not to speak of dimension, and just say that every prime ideal is maximal. I'm also fine in writing out explicitly what integrally closed means.

The real problem is to get a proof of unique factorization for ideals without using anything about integral closure, apart from direct arguments. For instance, I'd be fine in saying: "...so this element satisfies this monic equation, hence it is in A." Less so in saying "...so this element lies in a ring which is finitely generated as an A-module, hence it is integral. Since it is in the field of fractions of A, is must belong to A."

The only missing step in proving that ideals in a Dedekind ring satisfy unique factorization is the fact that primary ideals are prime powers.

Is there a direct proof of this fact which does not rely on anything about integrally closed domains, apart from the definition?

I should make clear that other standard techniques are available at this point: localization, Noetherian and Artinian stuff, primary decomposition, symbolic powers and so on.

I should also say that changing the order of the topics woule a major headache. I have thought out for long the order, and this is the only point where I get things in the wrong order. If possible, I would like to leave it as it is.

Edit (added in response to KConrad comment).

The steps which are easy are the following. Since $A$ is Noetherian, a primary decomposition exists. Since every prime is maximal, there are no embedded primes, so all primary components are unique. Finally, using again that every prime is maximal, all primary components are coprime, so intersections become products. So the only step where one uses integrality is the proof that the primary ideals are actually prime powers.

For the definition, there is no need to speak about integral closure, let alone proving that the integral closure is a ring. The integral domain $A$ is said to be integrally closed if every element $x$ of the quotient field of $A$, which satisfies a monic equation $x^n + a_{n-1} x^{n-1} + cdots + a_0$ with coefficients in $A$, is itself in $A$.

As for the examples, the compromise for now is to list some number rings, with the promise that it will be shown in a later section that these are actually Dedekind rings. Of course I'm not happy with this solution. But I'm also not happy with putting an aside on integral closure in the middle of a section about factorization and primary decomposition; even less so because there IS a later section on integral closure.

I cannot even reverse the two, because in the section of integral closure I want to be able to speak about the integral closures of $mathbb{Z}$, so I need the factorization theory for Dedekind rings.

ag.algebraic geometry - Degree of divisors and degrees of the corresponding maps to projective space

Here's how I think about it.

Let's assume we are in the case that $dimvarphi(X)=dim X$. Then $varphi : Xto varphi(X)$ is an generic finite map. Let $d$ be the degree of this map which is defined as the degree of field extension $[k(X):k(varphi(X))]$. The degree of $varphi(X)$ is given by $varphi(X)cdot H^{dim X}$ where $H$ is a general hyperplane of $mathbb{P}^n$. Pulling $H$ back to $X$, we get $D$. Then, by projection formula, $D^{dim X} = Xcdot D^{dim X}=dcdot(varphi(X)cdot H^{dim X})$. In the case that $X$ is a curve, $D^{dim X}$ is noting but the degree of $D$. So, the degree of $D$ equals that the degree of image times the degree of the map.

However, in higher dimension, $D^{dim X}$ may not be the degree of $D$. For example, $D$ is a irreducible degree 2 curve in $mathbb{P}^2$. The degree of $D$ is 2 which is not equal to $Dcdot D=4$ by Bézout's theorem.

Edit: I think in higher dimension, to define the degree of a divisor $D$, we need to choose a very ample divisor $A$ at first and then define the degree as the intersection number $Dcdot A^{dim D}$.

at.algebraic topology - A specific branched cover of S^2 as a subgroup of Pi_1

In standard topological terms, the exact sequence that relates homotopy groups of the base $B$, fiber $F$ and total space $E$ of topological fibration gives

$$pi_1(F) to pi_1(E) to pi_1(B) to pi_0(F),$$

that is,

$$ 0to pi_1(T-4) to pi_1(S^2-4) to mathbb Z_2.$$

The middle map works by taking a loop above and pushing it to the base; the right map works by taking a loop on $S^2-4$, lifting it as a path, and taking 0 or 1 depending on whether the resulting path is closed or not.

I believe the OP described the fundamental group of the torus as $left< a, b, c, d, e, f| [a, b]cdef = 1right>$ where $a, b$ are two circles of the torus and $c, d, e, f$ are four loops around the holes. For $S^2-4$ my suggestion would be to use $left< C, D, E, F| CDEF = 1right> = left< C, D, Eright>$ where $c$ is over $C$ etc (rather then $g, h, x, w$).

Now, since the loop around $c$ on torus has to wind twice around it when projected to the sphere (think about complex $zmapsto z^2$ map) it's easy to see that $cmapsto C^2$, $dmapsto D^2$, $emapsto E^2$, $fmapsto F^2$.

What about $a$ and $b$? Careful observer should note that it's a bit tricky to define the loops. E.g, if you move $a$ parallel to itself, you'll get a new $a'$ which would differ by something like $cd$ depending on which points are where and depending on how you draw the basepoints on the loops.

For the exact calculations one should fix the torus to be $mathbb Rtimes mathbb R/mathbb Ztimesmathbb Z$ so that fixed points of $zmapsto z$ are the vertices $c = (0, 0)$, $d = (1/2, 0)$, $e = (0, 1/2)$, $f = (1/2, 1/2)$. Moreover, you should now select some basepoint and draw the cycles around $c, d, e, f$ so that $cdef = 1 $.

Unfortunately, from the picture it's not easy to say where some simple loops like horizontal or vertical go. While in homology they seem to be $C+D$ and $D+F$, one has to draw them really carefully with the basepoints. I couldn't do that, but here's something different instead.

I tried to exhibit some expressions $a, b$, which may be not exactly the cycles above, but which nevertheless satisfy $[a, b]cdef mapsto 1$. In other words, these $a, b$ will be generators, but different ones.

I was able to make $a = CDEC^{-1}, b = CCDC^{-1}$ work:

$$CDEC^{-1}CCDC^{-1}(CDEC^{-1})^{-1}(CCDC^{-1})^{-1}CCDDEE(CDECDE)^{-1} = $$

$$ = CDEC^{-1}CCDC^{-1}CE^{-1}D^{-1}C^{-1}CD^{-1}C^{-1}C^{-1}CCDDEE(CDECDE)^{-1} = $$

$$ = CDEC^{-1}CCDE^{-1}D^{-1}D^{-1}DDEE(CDECDE)^{-1} = $$

$$ = CDECDE (CDECDE)^{-1} = 1 $$

I think this is more-or-less the explicit map you're asking for!

Finally, note that the map $ left< C, D, E, F| CDEF = 1right>
to mathbb Z_2$ is given by counting all the letters modulo 2 (consistent because $F = 1 = 3 = (CDE)^{-1}$), so the image of the map discussed above should contain exactly expressions with even number of letters.