parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/28391 | 3 | I came across this following way of defining connection and curvature which is not so obviously equivalent to the definitions as familiar in Riemannian Geometry books like say by Jost.
If $E$ is a complex vector bundle over a manifold $M$ then one defines the space of vector valued $p$-differential forms on them as $\Omega^p(M,E) = \Gamma ( \wedge ^p (T^\*M) \otimes E) $
The connection be defined as the map , $\nabla: \Gamma(E) \rightarrow \Omega^1(M,E)$ satisfying $\nabla(fX) = (df)X+f\nabla X$ where $f \in C^{\infty}(M)$ and $X \in \Gamma(E)$
{It might help if someone can make explicit as to what exactly is $(df)X$"?
To make sense this has to be an element of $\Omega^1(M,E)$}
Now one defines curvature of the connection as the map, $$R = \nabla \circ \nabla : \Gamma(E) \rightarrow \Omega^2(M,E)$$
Now it is not clear to me as to how this makes $ R \in \Omega^2(M,End(E))$ ?
And how does this match up with the more familiar form as,
$R(X,Y) = \nabla \_X \nabla \_Y - \nabla \_Y \nabla \_X - \nabla \_{[X,Y]}$
One can extend the definition of connection to be a map $\nabla: \Omega^p(M,E) \rightarrow \Omega^{p+1}(M,E)$
such that for $\omega \in \Omega^p(M)$ and $X \in \Gamma(E)$ one has,
$$\nabla(\omega X) = (d\omega)X + (-1)^{deg \omega}\omega \wedge \nabla X$$
{Here again it would be helpful if someone can can explain what is the meaning of $(d\omega)X$"?
To make sense it has to be an element of $\Omega^{p+1}(M,E)$}
This looks very much like the de-Rham derivative of vector valued differential forms but the notation looks uneasy.
| https://mathoverflow.net/users/2678 | Definition of the curvature tensor | You need to write $\omega X$ as $\omega \otimes X$.
If you prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial - i.e. $R(X,Y)(fS) = fR(X,Y)S$ where $S\in\Gamma(E)$, $f\in\mathcal{C}^\infty(M)$ and $X,Y\in TM$, then it follows that this mapping is indeed an element of ${\Omega} ^2(M,End(TM))$, because then the value of $R(X,Y)S$ at a point $m\in M$ depends only on $X\_m,Y\_m$ and $S\_m$ and not on their derivatives (even though we have defined $R$ via differentiation). The desired identification then follows from linear algebra $Hom(\Lambda^2 V \otimes E,E) \simeq Hom(\Lambda^2 V, \otimes E^\* \otimes E)$ & $End(E) \simeq E^\*\otimes E$.
The classical formula for curvature follows directly from the definition of the action of $\nabla$ on $\Omega^p(M,E)$.
| 2 | https://mathoverflow.net/users/6818 | 28393 | 18,553 |
https://mathoverflow.net/questions/28386 | 14 | **The question** Consider a topological space $X$ and a family of sheaves (of abelian groups, say) $\; \mathcal F\_i \;(i\in I)$ on $X$. Is it true that
$$H^\*(X,\prod \limits\_{i \in I} \mathcal F\_i)=\prod \limits\_{i \in I} H^\*(X,\mathcal F\_i) \;?$$
According to Godement's and to Bredon's monographs this is correct if the family of sheaves is locally finite (In particular if $I$ is finite). [Bredon also mentions in an exercise that equality holds for spaces in which every point has a smallest open neighbourhood.]
What about the general case?
**A variant** Same question for $\check{C}$ech cohomology: is it true that
$$\check{H}^\*(X,\prod \limits\_{i \in I} \mathcal F\_i)=\prod \limits\_{i \in I} \check{H}^\*(X,\mathcal F\_i) \;?$$
(Of course, $\check{C}$ech cohomology often coincides with derived functor cohomology but still the question should be considered independently)
**A prayer** Godement's book *Topologie algébrique et théorie des faisceaux* was published in 1960 and is still, with Bredon's, the most complete book on the subject. I certainly appreciate the privilege of working in a field where a book released half a century ago is still relevant: programmers and molecular biologists are not so lucky. Still I feel that a new treatise is due, in which naïve/foundational questions like the above would be addressed, and which would take the research and shifts in emphasis of half a century into account: one book on sheaf theory every 50 years does not seem an unreasonable frequency. So might I humbly suggest to one or several of the awesome specialists on MathOverflow to write one? I am sure I'm not the only participant here whose eternal gratitude they would earn.
| https://mathoverflow.net/users/450 | The cohomology of a product of sheaves and a plea. | The answer to the first question is almost always no, see *Roos, Jan-Erik(S-STOC)
Derived functors of inverse limits revisited. (English summary)
J. London Math. Soc. (2) 73 (2006), no. 1, 65--83.* .
**Addendum**: The crucial point is that infinite products are not exact. The most precise counterexample statement is Cor 1.11 combined with Prop 1.6 which identifies the stalks of the higher derived functors of the product with what you are interested in. Formally, it doesn't give a counter example for a single $X$ but Cor 1.11 shows that for any paracompact space with positive cohomological dimension there is some open subset for which your question has a negative answer. It seems clear that one could examples for specific $X$.
| 8 | https://mathoverflow.net/users/4008 | 28398 | 18,557 |
https://mathoverflow.net/questions/28357 | 8 | **Motivation and background** This question is motivated by the problem of classifying the (two-sided) closed ideals of the Banach algebra $\mathcal{B}(L\_\infty)$ of all (bounded, linear) operators on $L\_\infty$ $(=L\_\infty[0,1])$. As far as I am aware, the only known nontrivial ideals in $\mathcal{B}(L\_\infty)$ are the ideals $\mathcal{K}(L\_\infty)$ of compact operators and $\mathcal{W}(L\_\infty)$ of weakly compact operators. Most of the other well-known closed operator ideals not containing the identity operator of $L\_\infty$ seem to coincide with one of these two operator ideals on $L\_\infty$. Let me mention explicitly the following further relevant pieces of background information:
* Any nontrivial closed ideal $\mathcal{J}$ of $\mathcal{B}(L\_\infty)$ must satisfy $\mathcal{K}(L\_\infty)\subseteq \mathcal{J} \subseteq \mathcal{W}(L\_\infty)$.
* We have $\mathcal{S}(L\_\infty) = \mathcal{W}(L\_\infty)$, where $\mathcal{S}$ denotes the (closed) operator ideal of strictly singular operators. Moreover, the ideal of operators $L\_\infty \longrightarrow L\_\infty$ that factor through $L\_2$ $(=L\_2[0,1])$ is a norm dense subset of $\mathcal{W}(L\_\infty)$. With regards to this latter fact, we have that if $\mathcal{J}$ is any closed ideal of $\mathcal{B}(L\_\infty)$ satisfying $\mathcal{K}(L\_\infty)\subsetneq \mathcal{J} \subsetneq \mathcal{W}(L\_\infty)$, then there exists an element of $\mathcal{W}(L\_\infty)\setminus \mathcal{J}$ that factors through $L\_2$.
One possible avenue towards discovering more closed ideals in $\mathcal{B}(L\_\infty)$ is to consider products of closed operator ideals, and my question below concerns but one approach along these lines. We recall now that for operator ideals $\mathcal{A}$ and $\mathcal{B}$, their product $\mathcal{B}\circ\mathcal{A}$ is the class of all operators of the form $BA$, where $A\in\mathcal{A}$, $B\in\mathcal{B}$ and the codomain of $A$ coincides with the domain of $B$ (so that the composition is defined). It is well-known that $\mathcal{B}\circ\mathcal{A}$ is an operator ideal and that $\mathcal{B}\circ\mathcal{A}$ is a closed operator ideal whenever $\mathcal{A}$ and $\mathcal{B}$ are, the latter fact being due to Stefan Heinrich. For $n$ a natural number, one may define the power of an operator ideal $\mathcal{A}^n$ as the product $\mathcal{A}\circ \ldots \circ \mathcal{A}$ with $n$ factors, and $\mathcal{A}^n$ is closed for all $n$ whenever $\mathcal{A}$ is closed. Moreover, we may define $\mathcal{A}^\infty:= \bigcap\_{n\in\mathbb{N}}\mathcal{A}^n$, with $\mathcal{A}^\infty$ being a closed operator ideal whenever $\mathcal{A}$ is.
It is well-known that every operator from $L\_\infty$ to $L\_2$ $(=L\_2[0,1])$ is strictly singular. However, the answer to the following question is not so clear to me:
**Question:** *Is $\mathcal{B} (L\_\infty, L\_2) = \mathcal{S}^\infty (L\_\infty, L\_2)$? If no, what is the least $n$ for which $\mathcal{B} (L\_\infty, L\_2) \neq \mathcal{S}^n (L\_\infty, L\_2)$?*
**Further comments:** A negative answer to my question above will yield through easy arguments that $\mathcal{K}(L\_\infty) \subsetneq \mathcal{S}^\infty (L\_\infty) \subsetneq \mathcal{W}(L\_\infty)$ (note that $\mathcal{K}(L\_\infty) \subsetneq \mathcal{S}^\infty (L\_\infty)$ in any case since there are noncompact operators $L\_\infty \longrightarrow L\_\infty$ that have the formal inclusion operator $\ell\_2 \hookrightarrow \ell\_\infty$ as a factor, and this inclusion operator belongs to $\mathcal{S}^\infty$).
It seems to me that the best chance of obtaining an element of $\mathcal{B} (L\_\infty, L\_2) \setminus \mathcal{S}^\infty (L\_\infty, L\_2)$ would be to consider a surjective element of $\mathcal{B} (L\_\infty, L\_2)$, for instance the adjoint of an isomorphic embedding of $L\_2$ into $L\_1$. A further possibility along these lines would be to factor such an embedding as the product of $n$ operators whose adjoints are strictly singular, for each $n\in \mathbb{N}$. However, it seems to me that such an approach will require a far deeper knowledge of the subspace structure of $L\_1$ than I have at the present time. Interpolation methods may work too, but I am also too ignorant of that theory to know whether it's a genuinely feasible approach.
| https://mathoverflow.net/users/848 | Factoring operators $L_\infty \longrightarrow L_2$ as the composition of $n$ strictly singular operators, $n\in \mathbb{N}$ | I think this is an outline of a proof, Phil. It is enough to factor $I\_{\infty,2}$ as the product of strictly singular operators, where $I\_{\infty,2}$ is the identity from $L\_\infty(\mu)$ to $L\_2(\mu)$ with $\mu$ a probability (since every operator from $L\_\infty$ to $L\_2$ is $2$-summing). I guess it can be assumed that $\mu$ is a separable measure (if not, I think using Maharam's theorem works) and certainly it is enough to look at $\mu$ purely non atomic. But then $I\_{\infty,2}$ can be regarded as the identity from $L\_\infty([0,1])$ to $L\_2([0,1])$ since every separable non atomic probability space is measure isometric to $[0,1]$ with Lebesgue measure. Now $I\_{\infty,2}$ factors through $L\_\Phi([0,1])$ for every fast growing Orlicz function $\Phi$. Consider $\Phi\_p(t)=e^{t^p} -1$ for $p>$. Rodin and Semenov showed that the identity from $L\_{\Phi\_p}$ to $L\_2$
is strictly singular and the injection from $L\_\infty$ to $L\_\Phi$ is strictly singular
always. This takes care of the case $n=2$.
Schechtman suggests that to do general $n$ to show that the injection from $L\_{\Phi\_p}$ to $L\_{\Phi\_r}$ is strictly singular for
$2<r<p$. At least this map is not an isomorphism on the span of the Rademachers.
| 6 | https://mathoverflow.net/users/2554 | 28405 | 18,562 |
https://mathoverflow.net/questions/28389 | 3 | A-The addition of the Grothendieck Universe Axiom (for every set x, there exists a set y that is a universe and contains x as member element) to ZFC (ZFC+GU) is considered as giving an almost good solution permitting the insertion of Category theory inside Set theory. It is known that we obtain an equivalent theory (ZFC+TA) by replacing GU with the Axiom A of Tarski (for every set x, there exists a set y that is an A Tarski set and contains x as a member element), or also (ZFC+IN) by replacing GU by the axiom IN (for every ordinal x, there exists a strongly inaccessible ordinal y containing x as a member element.
1-In each three cases, we can prove (using foundation) that "there exists a proper class of (GU sets, TA sets, inaccessible cardinals)" is a theorem. But is it possible to replace the axioms GU, TA and IN by the formulation with proper classes (this seems possible for IN and GU ) ?
B-If we replace ZFC by ZF, the situation seems much more involved.
As it is provable that AC is a consequence of TA, ZF+TA is equivalent to ZFC+TA.
As it is provable that some GU sets cannot be well-ordered, AC is not a consequence of ZF+GU, that cannot be equivalent to ZFC+GU.
For the third case, this is depending of how we define an inaccessible without AC; but if we take the "correct one", Ac is not derivable from ZF+IN'.
These question are very thoroughly presented in a message of R. Solovay to FOM "AC and stongly inaccessible cardinals" (29/02/2008). But, in fact, the power set axiom and the Infinity axiom can also be derived from the Tarski a axiom. So that one could think that the theory:
Extensionality+Replacement+AT+Union+Foundation is equivalent to ZFC+GU.
But when you try to developp such a theory, it seems that you are obliged to also introduce the Pair Axiom before introducing AT that needs a definition of functions.
2-Is it in fact possible to dispense of the axiom of the pair within this theory ?
C-Tarski's A axiom is given inside his paper (auf deutsch) "Über unerreichbare Kardinalzahlen", Fund Math 1938, page 84.
3-On the same page, Tarski gives another axiom, named A'with four conditions (as in the case of A) and writes ""Übrigens sind vershiedene âquivalente Unformung dieses Axioms [A] bekannt. Man kann Z. B. Bedingungen A-1-A4 beziehungsweise durch folgenden Bedingungen [A'1-A'4] ersetzen (und zwar jede Bedingung unhabhängig von denen anderen).
Does anyone completely understand what is exactly meant here by Tarski, and how is this proved ?
Gérard LANG
| https://mathoverflow.net/users/30395 | About Grothendieck Universe and Tarski's A and A' Axioms | The answer to question 1 is affirmative for GU and IN but
negative for TA. That is, the proper class formulation of
TA is not equivalent to TA, unless both are inconsistent.
Asserting that every ordinal is below an inaccessible
cardinal is clearly equivalent in ZF to asserting that the
inaccessible cardinals form a proper class. And since the
Grothendieck universes are exactly the sets $V\_\kappa$ for
an inaccessible cardinal $\kappa$, the theory GU over ZFC
is equivalent to the assertion that there are a proper
class of universes. So those cases are relatively straightforward, as you had guessed.
But the case of the Tarski axiom is different, since his
universes are not transitive sets. In fact, if there is a
single inaccessible cardinal $\kappa$, one can already form
a proper class of Tarski sets. To see this, suppose that
$\kappa$ is inaccessible and $x$ is any set. Build a Tarski
set $U$ as follows:
* $U\_0=\{\{x\}\}$,
* $U\_{\alpha=1}=P(U\_\alpha)$ and
* $U\_\lambda=\bigcup\_\alpha
U\_\alpha$ at limit ordinals.
That is, we perform the usual cumulative
hiearchy, but starting with object {x} instead of nothing.
The resulting set $U=U\_\kappa$ contains {x} as an element,
has size $\kappa$ and is closed under subsets, power sets
and small unions, so it is a Tarski universe. (Note that if
$y\in U\_{\alpha+1}$, then all subsets of $y$ are also in
$U\_{\alpha+1}$, and so $P(y)$ is added on the next step.
And {x} has only two subsets, which are both in $U$.) But
for sufficiently large $x$, the resulting $U$ will not be
transitive, since $x$ itself will never be added as an
element. Furthermore, since $x$ was arbitrary, from a
single inaccessible cardinal we can build a proper class of
Tarski universes. And so the proper class formulation of TA
will not be equivalent to TA, unless both theories are
inconsistent, because if the existence of an inaccessible
cardinal is consistent with ZFC, then it is consistent that
there is exactly one such cardinal, by chopping the
universe off at the second one. The resulting model will
satisfy the proper class formulation of TA but not TA
itself.
| 6 | https://mathoverflow.net/users/1946 | 28406 | 18,563 |
https://mathoverflow.net/questions/28379 | 4 | From time to time I find myself wishing to calculate basic statistics on words in the English language. For example, today I found myself wanting a graph of the number of English words vs. their length.
Admittedly, such queries usually arise for me in the context of conversational/recreational purposes, but with the obvious links to cryptography, computational game theory (Scrabble AI etc), and statistics, I think the following question easily falls within the purview of mathematical research.
**What good quality resources exist for performing statistical/structural analysis on the set of English words?**
One answer to this would be "get a decent word list and write an appropriate program", but, firstly, I don't know what the best word list is and where to get it , and, secondly, in a post-Wolfram Alpha world, I am compelled to search for something higher level, that I can consult from time to time with little set-up needed. For example Mathematica seems to have an elaborate "WordData" package, though I am somewhat unsure of how exhaustive the data set is, given the following excerpt from the Wolfram site:
"Total number of words and phrases in WordData:
In[1]:= Length[WordData[All]]
Out[1]= 149191"
If anyone has first-hand experience with this package, or even better (as a Maple user), a similar or better resource that is standalone, (or implemented in Maple), then it would be great to hear about it.
| https://mathoverflow.net/users/3623 | Good quality data/packages for statistical/structure analysis of words in the English language | I am very sure this kind ofn things is done in the free (free as in freedom, not as in beer!)
statistical language R. .. See www.r-project.org, and then repeat your question on
the r-help mailing list, where you are sure to get informed answers about how to do it in R.
That it can be done tere, and is done there, is a certainty.
( as for the meta-discussion where this belongs, it belongs on mathoverflow! Statistics is math, not programming.)
| 1 | https://mathoverflow.net/users/6494 | 28408 | 18,565 |
https://mathoverflow.net/questions/28394 | 3 | Suppose we have a topological group $G$, then the multiplication map $\mu$ and the diagonal map $\Delta$ provide the cohomology $H^\ast(G;R)$ (with Pontryagin coproduct and cup coproduct) and homology $H\_\ast(G;R)$ (with Pontryagin product and diagonal coproduct) of $G$ with the structure of a Hopf algebra (assuming vanishing of the Tor-terms in the Kunneth theorem). If both satisfy some mild conditions, these are in fact dual. A nice historical explanation of this can be found in Cartier's "Primer on Hopf Algebras" (<http://inc.web.ihes.fr/prepub/PREPRINTS/2006/M/M-06-40.pdf>). My general question is the following: when we know they are in fact isomorphic as Hopf algebras?
The Samelson theorem says that if $G$ is a Lie group and $R$ is a field of characteristic zero, then $H^\ast(G;R)$ is commutative, associative and coassociative and an exterior algebra on primitive generators of odd degree. I think this implies that $H^\ast(G;R)$ is isomorphic to $H\_\ast(G;R)$ as a Hopf algebra. Is this correct? To what extent does Samelson's result extend to non-zero characteristic?
My goal is to limit the amount of calculation when I do calculations of the homology and cohomology of Lie groups.
| https://mathoverflow.net/users/798 | When are the homology and cohomology Hopf algebras of topological groups equal? | The mod $2$ cohomology of $\mathrm{SO}\_n$ is not an exterior algebra as soon
as $n\geq3$ (there is a unique non-zero element in degree $1$ whose square is
non-zero, think of the case $\mathrm{SO}\_{3}$ which is $3$-dimensional real
projective space) while the homology ring is an exterior algebra.
Another example is $K(\mathbb Z,2)$ (aka $\mathbb C\mathbb P^\infty$), its
integral cohomology ring is a polynomial ring on a degree $2$-generator while
its homology ring is the free divided power algebra on a degree $2$-generator.
| 7 | https://mathoverflow.net/users/4008 | 28413 | 18,570 |
https://mathoverflow.net/questions/28418 | 4 | The title says it all. Despite heavy googling I have not been able to find anything. What I am interested in, is theory (maybe modelling), not for the moment finite difference methods as approximations to partial differential equations! Books, papers, webpages, ....
| https://mathoverflow.net/users/6494 | Reference/Introduction to partial difference(NOT differential!) equations | [Partial Difference Equations](http://books.google.com/books?id=1klnDGelHGEC) by Sui Sun Cheng? A good start (in the context of reading the classics) is the paper by [Courant-Friedrichs-Lewy](http://www.amath.washington.edu/~narc/win08/papers/courant-friedrichs-lewy.pdf). Also useful is a list with [more books](http://www.ericweisstein.com/encyclopedias/books/FiniteDifferenceEquations.html).
For the non-partial case, I must admit I have a soft spot for [Milne-Thompson](http://books.google.com/books?id=NXRLajGtgjEC&printsec=frontcover&dq=%22The+Calculus+of+Finite+Differences%22&source=bl&ots=Z84L819X7z&sig=frrl3VuChAoRCmtDZcd032TcZr0&hl=en&ei=YB0ZTI3xJImDnQfO_uC1Cg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBIQ6AEwAA#v=onepage&q&f=false)'s book.
| 4 | https://mathoverflow.net/users/3993 | 28426 | 18,577 |
https://mathoverflow.net/questions/28421 | 22 | It is well known that if $M, \Omega$ is a symplectic manifold then the Poisson bracket gives $C^\infty(M)$ the structure of a Lie algebra. The only way I have seen this proven is via a calculation in canonical coordinates, which I found rather unsatisfying. So I decided to try to prove it just by playing around with differential forms. I got quite far, but something isn't working out and I am hoping someone can help. Forgive me in advance for all the symbols.
Here is the setup. Given $f \in C^\infty(M)$, let $X\_f$ denote the unique vector field which satisfies $\Omega(X\_f, Y) = df(Y) = Y(f)$ for every vector field $Y$. We define the Poisson bracket of two functions $f$ and $g$ to be the smooth function $\{f, g \} = \Omega(X\_f, X\_g)$. I can show that the Poisson bracket is alternating and bilinear, but the Jacobi identity is giving me trouble. Here is what I have.
To start, let's try to get a handle on $\{ \{f, g \}, h\}$. Applying the definition, this is given by $d(\Omega(X\_f, X\_g))X\_h$. So let's try to find an expression for $d(\Omega(X,Y))Z$ for arbitrary vector fields $X, Y, Z$.
Write $\Omega(X,Y) = i(Y)i(X)\Omega$ where $i(V)$ is the interior product by the vector field $V$. Applying Cartan's formula twice and using the fact that $\Omega$ is closed, we obtain the formula
$$d(\Omega(X,Y)) = (L\_Y i(X) - i(Y) L\_X) \Omega$$
where $L\_V$ is the Lie derivative with respect to the vector field $V$. Using the identity $L\_V i(W) - i(W) L\_V = i([V,W])$, we get:
$$(L\_Y i(X) - i(Y) L\_X) = L\_Y i(X) - L\_X i(Y) + i([X,Y])$$
Now we plug in the vector field $Z$. We get $(L\_Y i(X) \Omega)(Z) = Y(\Omega(X,Z)) - \Omega(X,[Y,Z])$ by the definition of the Lie derivative, and clearly $(i([X,Y])\Omega)(Z) = \Omega([X,Y],Z)$. Putting it all together:
$$d(\Omega(X,Y))Z = Y(\Omega(X,Z)) - X(\Omega(Y,Z)) + \Omega(Y, [X,Z]) - \Omega(X, [Y,Z]) + \Omega([X,Y], Z)$$
This simplifies dramatically in the case $X = X\_f, Y = X\_g, Z = X\_h$. The difference of the first two terms simplifies to $[X\_f, X\_g](h)$, and we get:
$$
\begin{align}
\{\{f, g\}, h\} &= [ X\_f, X\_g ](h) + [ X\_f, X\_h ](g) - [ X\_g, X\_h ](f) - [ X\_f, X\_g ](h)\\
&= [ X\_f, X\_h ](g) - [ X\_g, X\_h ](f)
\end{align}$$
However, this final expression does not satisfy the Jacobi identity. It looks at first glance as though I just made a sign error somewhere; if the minus sign in the last expression were a plus sign, then the Jacobi identity would follow immediately. I have checked all of my signs as thoroughly as I can, and additionally I included all of my steps to demonstrate that if a different sign is inserted at any point in the argument then one obtains an equation in which the left hand side is alternating in two of its variables but the right hand side is not. Can anybody help?
| https://mathoverflow.net/users/4362 | The Jacobi Identity for the Poisson Bracket | The Jacobi identity for the Poisson bracket does indeed follow from the fact that $d\Omega =0$.
I claim that (twice) the Jacobi identity for functions $f,g,h$ is precisely
$$d\Omega(X\_f,X\_g,X\_h) = 0.$$
To see this, simply expand $d\Omega$.
You will find six terms of two kinds:
* three terms of the form
$$X\_f \Omega(X\_g,X\_h) = X\_f \lbrace g,h \rbrace$$
* and three terms of the form
$$\Omega([X\_f,X\_g],X\_h).$$
To deal with the first kind of terms, notice that from the definition of $X\_f$, for any function $g$,
$$X\_f g = \lbrace g, f \rbrace.$$
This means that
$$X\_f \Omega(X\_g,X\_h) = \lbrace \lbrace g,h \rbrace, f \rbrace.$$
To deal with the second kind of terms, notice that
$$\iota\_{[X\_f,X\_g]}\Omega = [L\_{X\_f},\iota\_{X\_g}]\Omega,$$
but since $d\Omega=0$,
$$L\_{X\_f}\Omega = d \iota\_{X\_f}\Omega = 0,$$
and hence
$$\iota\_{[X\_f,X\_g]}\Omega = d \iota\_{X\_f}\iota\_{X\_g}\Omega = d\lbrace g,f\rbrace,$$
whence
$$\Omega([X\_f,X\_g],X\_h) = d\lbrace g,f\rbrace (X\_h) = \lbrace\lbrace g,f\rbrace, h \rbrace.$$
Adding it all up you get twice the Jacobi identity.
| 27 | https://mathoverflow.net/users/394 | 28431 | 18,581 |
https://mathoverflow.net/questions/28415 | 18 | Let $X, Y$ be normed vector spaces, where $X$ is infinite dimensional. Does there exist a linear map $T : X \rightarrow Y$ and a subset $D$ of $X$ such that $D$ is dense in $X$, $T$ is bounded in $D$ (i.e. $\sup \_{x\in D, x \neq 0} \frac{\|Tx\|}{\|x\|}<\infty $), but $T$ is not bounded in $X$?
| https://mathoverflow.net/users/4928 | Unbounded operator bounded in a dense subset | Matthew's answer reminded me of a fact that makes this easy: if $X$ is a normed space (say, over $\mathbb{R}$) and $f : X \to \mathbb{R}$ is a linear functional, then its kernel $\ker f$ is either closed or dense in $X$, depending on whether or not $f$ is continuous (i.e. bounded). The proof is trivial: $\ker f$ is a subspace of $X$ of codimension 1. Its closure is a subspace that contains it, so must either be $\ker f$ or $X$. And of course, a linear functional is continuous iff its kernel is closed. This is Proposition III.5.2-3 in Conway's *A Course in Functional Analysis*.
So let $f$ be an unbounded linear functional on $X$ (which one can always construct as in Matthew's example), and take $D = \ker f$. $D$ is dense by the above fact, and $f$ is identically zero on $D$.
| 28 | https://mathoverflow.net/users/4832 | 28434 | 18,584 |
https://mathoverflow.net/questions/28435 | 11 | I am writing a small thing in which I am required to use manually formatted \bibitem entries rather than a BibTeX file. This takes a lot of time, and I probably get some of the formatting wrong. Is there a way of producing such entries automatically, for example from MathSciNet or from the BibTeX items created by MathSciNet? For arXiv preprints, there is a great tool from the Courant Research Centre ([here](http://www.crcg.de/arXivToBibTeX/)) which can produce BibTeX as well as bibitem entries, and there should be something similar for MathSciNet.
| https://mathoverflow.net/users/349 | Automatically extract a bibitem (not BibTeX!) from MathSciNet? | Hi Andreas. When you run BibTeX on a .bib file then it produces a .bbl file which more-or-less contains your bibliography in \bibitem format. So I recommend exporting everything BibTex style from MathSciNet, creating a .bib file, running BibTeX, and then making any remaining changes to the resulting .bbl file (which you can then just copy and paste into your .tex file).
I suppose some people even know how to write BibTeX style files, so that the resulting .bbl file contains *exactly* the formatting you want.
Despite my advice, when I need \bibitem bibliographies, I usually just copy citations from the MathSciNet clipboard (in Ascii format) and manually insert the additional formatting. I'm not sure if this is more, or less, lazy than the .bbl business...
| 10 | https://mathoverflow.net/users/5830 | 28439 | 18,586 |
https://mathoverflow.net/questions/28428 | 25 | I recently learned of the result by Carleson and Hunt (1968) which states that if $f \in L^p$ for $p > 1$, then the Fourier series of $f$ converges to $f$ pointwise-a.e. Also, [Wikipedia](http://en.wikipedia.org/wiki/Convergence_of_fourier_series#Norm_convergence) informs me that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in $L^p$. Either of these results implies that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in measure.
My first question is about the $p = 1$ case. That is:
>
> If $f \in L^1$, will the Fourier series of $f$ converge to $f$ in measure?
>
>
>
---
I also recently learned that there exist functions $f \in L^1$ whose Fourier series diverge (pointwise) everywhere. Moreover, such a Fourier series may converge (Galstyan 1985) or diverge (Kolmogorov?) in the $L^1$ metric.
My second question is similar:
>
> Do there exist functions $f \in L^1$ whose Fourier series converge pointwise a.e., yet diverge in the $L^1$ metric?
>
>
>
---
(Notes: Here, I mean the Fourier series with respect to the standard trigonometric system. I am also referring only to the Lebesgue measure on [0,1]. Of course, if anyone knows any more general results, that would be great, too.)
| https://mathoverflow.net/users/2318 | Convergence of Fourier Series of $L^1$ Functions | The answer to your first question is no. There is an $L^1$ function with Fourier series not converging in measure.
In the Kolmogorov example of an $L^1$ function $f$ with a.e. divergent Fourier series, there is in fact a set of positive measure $E$ and a subsequence $n\_k$ such that for all $x$ in $E$, the absolute values of the partial sums $S\_{n\_k}$ of the Fourier series goes to infinity with $k$.
$$\forall x\in E,\ \ |S\_{n\_k}f(x)|\rightarrow \infty$$
This can be checked from the construction of $f$ in the original article of Kolmogorov, in its selected [works](http://books.google.fr/books?hl=fr&lr=&id=ikN59GkYJKIC&oi=fnd&pg=PR13&dq=diverge+partout+authornbsp%3Akolmogorov&ots=2OvYAnZNwa&sig=BXuHcDly4I-RQM1rRSuUCmWSZKY#v=onepage&q&f=false).
If $S\_nf$ converges in measure, then $S\_{n\_k}f$ must also converges in measure. This implies that there is a subsequence $n\_{k\_l}$ such that $S\_{n\_{k\_l}}f(x)$ converges a.e. $x$, a contradiction.
| 17 | https://mathoverflow.net/users/6129 | 28445 | 18,591 |
https://mathoverflow.net/questions/28422 | 13 | Let $W$ be a Coxeter group and let $P\_W(q) = \sum\_{w \in W} q^{\ell(w)}$ be its Poincare series. When $W$ is the Weyl group of a simple algebraic group $G$ (hence $W$ is finite), $P\_W(q)$ is the generating function describing the cells in the Bruhat decomposition of the flag variety $G/B$, $B$ a Borel.
What happens when $W$ is infinite, e.g. when $W$ is an affine Weyl group or a hyperbolic Coxeter group? Can $P\_W$ still be associated to an infinite-dimensional "flag variety"?
| https://mathoverflow.net/users/290 | Does the Poincare series of a Coxeter group always describe a "flag variety"? | I think that Shrawan Kumar's book "Kac-Moody groups, their flag varieties, and representation theory" will contain the flag varieties (which are really ind-varieties in the non-finite case) that you are looking for.
A crystallographic Coxeter group is one of the form $\langle s\_1,\ldots,s\_n| s\_i^2=(s\_i s\_j)^{m\_{ij}}=1\rangle$ where each mij is equal to 2,3,4,6 or infinity. Such Coxeter groups are precisely the Weyl groups associated to arbitrary Kac-Moody algebras. In this case, there is a corresponding Kac-Moody group, together with an associated flag variety and Schubert cells, which seems to me to be what you want.
It is this geometry that is the starting point of the geometric interpretation of Kazhdan-Lusztig polynomials in the crystallographic case. As is to be expected, the finite case is easier than the affine case, which again is easier than the arbitrary KM case.
| 6 | https://mathoverflow.net/users/425 | 28447 | 18,592 |
https://mathoverflow.net/questions/28446 | 2 | Let $S$ be a subset of $\mathbb{R}^n$. I would like to call $S$
1. a Lipschitz(1) hypersurface if for every $x\in S$ there is a hyperplane $H$ so that the orthogonal projection onto $H$ is a bi-Lipschitz map from a neighbourhood of $x$ in $S$ onto an open subset of $H$, and
2. a Lipschitz(2) hypersurface if for every $x\in S$ there is a bi-Lipschitz map $\psi$ from $B\times(-1,1)$ onto a neighbourhood of $x$ in $\mathbb{R}^n$ so that $\psi^{-1}(S)=B\times\{0\}$, where $B$ is an open subset of $\mathbb{R}^{n-1}$.
>
> It seems clear enough (\*) that Lipschitz(1) implies Lipschitz(2). But is the converse true? And if not, what is a simple counterexample?
>
>
>
I have come across the notion of regions with Lipschitz boundaries in a number of papers on boundary value problems for PDEs. But every such paper seems to take the notion of Lipschitz-ness for granted.
(\*) If $S$ is Lipschitz(1), then after a rotation of the axes, it locally looks like the graph of a Lipschitz function $\gamma\colon\mathbb{R}^{n-1}\to\mathbb{R}$. Put $\psi(x,t)=(x,\gamma(x)-t)$ to obtain the Lipschitz(2) property.
| https://mathoverflow.net/users/802 | Are these two notions of Lipschitz hypersurface equivalent? | Counterexample in $\mathbb R^3$: Let $C$ be the cube $max(|x|,|y|,|z|)\le 1$. Let $S$ be the intersection of $C$ with $z=0$. Choose a piecewise linear (PL) homeomorphism from the boundary of $C$ to itself such that the boundary of $S$ goes to a very zigzaggy set. Extend to a PL (therefore Lipschitz) homeomorphism from $C$ to itself by coning to the center point $(0,0,0)$. The image of $S$ (minus its boundary) will be Lipschitz(2) but not Lipschitz(1); there will be no good direction such that the projection to a hyperplane is one to one near the center.
| 7 | https://mathoverflow.net/users/6666 | 28449 | 18,594 |
https://mathoverflow.net/questions/28438 | 20 | Mathematics is not typically considered (by mathematicians) to be a solo sport; on the contrary, some amount of mathematical interaction with others is often deemed crucial. Courses are the student's main source of mathematical interaction. Even a slow course, or a course which covers material which one already knows to some level, can be highly stimulating. However, there are usually a few months in the year when mathematics slows down socially; in the summer, one might not be taking any courses, for example. In this case, one might find themselves reduced to learning alone, with books.
It is generally acknowledged that learning from people is much easier than learning from books. It has been said that Grothendieck never really read a math book, and that instead he just soaked it up from others (though this is certainly an exaggeration). But when the opportunity does not arise to do/learn math with/from others, what can be done to maximize one's efficiency? Which process of learning does social interaction facilitate?
Please share your personal self-teaching techniques!
| https://mathoverflow.net/users/6779 | Mathematics and autodidactism | I think the crucial distinction here is not between books and people (after all, books are written by people!) or between physical contact and electronic contact, but between *non-interaction* and *interaction*. In my view, the crucial benefit that interaction provides is *the ability to have your questions answered and your ideas critiqued*. MathOverflow, of course, demonstrates the value of being able to ask a focused question and have an expert reply to it. In a non-interactive setting, you are limited to whatever answers have already been written down somewhere (plus your own ingenuity in answering your own questions).
Having your own ideas critiqued is also important. You may have a good idea (or a bad idea!) but not recognize it as such. An expert can often give you a quick assessment of your idea; you cannot get this without interaction. (Of course, expert assessment is a double-edged sword because sometimes the expert is wrong and you might have been better off without the incorrect feedback!) Even if you have a good idea that you recognize as good, you may have difficulty articulating it properly until you are forced to communicate it to someone else and get them to understand it. Some people are naturally gifted at expressing their ideas clearly and logically, but most people need to be taught communication skills interactively. And even the best mathematicians benefit from the exercise of teaching others what they know (or think they know!). The process of explaining something to someone else is of great value in clarifying your own understanding.
Finally, as for improving your efficiency if you have limited access to interaction, I would try to *write down, as succinctly as possible, the questions and ideas that you want to get feedback on.* Then you can efficiently send off a batch of questions and ideas and get feedback in a batch.
| 18 | https://mathoverflow.net/users/3106 | 28454 | 18,598 |
https://mathoverflow.net/questions/10666 | 20 | My question is about [nonstandard analysis](http://en.wikipedia.org/wiki/Non-standard_analysis), and the diverse possibilities for the choice of the nonstandard model R\*. Although one hears talk of *the* nonstandard reals R\*, there are of course many non-isomorphic possibilities for R\*. My question is, what kind of structure theorems are there for the isomorphism types of these models?
**Background.** In nonstandard analysis, one considers the real numbers R, together with whatever structure on the reals is deemed relevant, and constructs a nonstandard version R\*, which will have infinitesimal and infinite elements useful for many purposes. In addition, there will be a nonstandard version of whatever structure was placed on the original model. The amazing thing is that there is a *Transfer Principle*, which states that any first order property about the original structure true in the reals, is also true of the nonstandard reals R\* with its structure. In ordinary model-theoretic language, the Transfer Principle is just the assertion that the structure (R,...) is an elementary substructure of the nonstandard reals (R\*,...). Let us be generous here, and consider as the standard reals the structure with the reals as the underlying set, and having all possible functions and predicates on R, of every finite arity. (I guess it is also common to consider higher type analogues, where one iterates the power set ω many times, or even ORD many times, but let us leave that alone for now.)
The collection I am interested in is the collection of all possible nontrivial elementary extensions of this structure. Any such extension R\* will have the useful infinitesimal and infinite elements that motivate nonstandard analysis. It is an exercise in elementary mathematical logic to find such models R\* as ultrapowers or as a consequence of the Compactness theorem in model theory.
Since there will be extensions of any desired cardinality above the continuum, there are many non-isomorphic versions of R\*. Even when we consider R\* of size continuum, the models arising via ultrapowers will presumably exhibit some saturation properties, whereas it seems we could also construct non-saturated examples.
So my question is: what kind of structure theorems are there for the class of all nonstandard models R\*? How many isomorphism types are there for models of size continuum? How much or little of the isomorphism type of a structure is determined by the isomorphism type of the ordered field structure of R\*, or even by the order structure of R\*?
| https://mathoverflow.net/users/1946 | Isomorphism types or structure theory for nonstandard analysis | Under a not unreasonable assumption about cardinal arithmetic, namely $2^{<c}=c$ (which follows from the continuum hypothesis, or Martin's Axiom, or the cardinal characteristic equation t=c), the number of non-isomorphic possibilities for \*R of cardinality c is exactly 2^c. To see this, the first step is to deduce, from $2^{<c} = c$, that there is a family X of 2^c functions from R to R such that any two of them agree at strictly fewer than c places. (Proof: Consider the complete binary tree of height (the initial ordinal of cardinality) c. By assumption, it has only c nodes, so label the nodes by real numbers in a one-to-one fashion. Then each of the 2^c paths through the tree determines a function f:c \to R, and any two of these functions agree only at those ordinals $\alpha\in c$ below the level where the associated paths branch apart. Compose with your favorite bijection R\to c and you get the claimed maps g:R \to R.) Now consider any non-standard model \*R of R (where, as in the question, R is viewed as a structure with all possible functions and predicates) of cardinality c, and consider any element z in \*R. If we apply to z all the functions \*g for g in X, we get what appear to be 2^c elements of \*R. But \*R was assumed to have cardinality only c, so lots of these elements must coincide. That is, we have some (in fact many) g and g' in X such that \*g(z) = \*g'(z). We arranged X so that, in R, g and g' agree only on a set A of size $<c$, and now we have (by elementarity) that z is in \*A. It follows that the 1-type realized by z, i.e., the set of all subsets B of R such that z is in \*B, is completely determined by the following information: A and the collection of subsets B of A such that z is in \*B. The number of possibilities for A is $c^{<c} = 2^{<c} = c$ by our cardinal arithmetic assumption, and for each A there are only c possibilities for B and therefore only 2^c possibilities for the type of z. The same goes for the n-types realized by n-tuples of elements of \*R; there are only 2^c n-types for any finite n. (Proof for n-types: Either repeat the preceding argument for n-tuples, or use that the structures have pairing functions so you can reduce n-types to 1-types.) Finally, since any \*R of size c is isomorphic to one with universe c, its isomorphism type is determined if we know, for each finite tuple (of which there are c), the type that it realizes (of which there are 2^c), so the number of non-isomorphic models is at most (2^c)^c = 2^c.
To get from "at most" to "exactly" it suffices to observe that (1) every non-principal ultrafilter U on the set N of natural numbers produces a \*R of the desired sort as an ultrapower, (2) that two such ultrapowers are isomorphic if and only if the ultrafilters producing them are isomorphic (via a permutation of N), and (3) that there are 2^c non-isomorphic ultrafilters on N.
If we drop the assumption that $2^{<c}=c$, then I don't have a complete answer, but here's some partial information. Let \kappa be the first cardinal with 2^\kappa > c; so we're now considering the situation where \kappa < c. For each element z of any \*R as above, let m(z) be the smallest cardinal of any set A of reals with z in \*A. The argument above generalizes to show that m(z) is never \kappa and that if m(z) is always < \kappa then we get the same number 2^c of possibilities for \*R as above. The difficulty is that m(z) might now be strictly larger than \kappa. In this case, the 1-type realized by z would amount to an ultrafilter U on m(z) > \kappa such that its image, under any map m(z) \to \kappa, concentrates on a set of size < \kappa. Furthermore, U could not be regular (i.e., (\omega,m(z))-regular in the sense defined by Keisler long ago). It is (I believe) known that either of these properties of U implies the existence of inner models with large cardinals (but I don't remember how large). If all this is right, then it would not be possible to prove the consistency, relative to only ZFC, of the existence of more than 2^c non-isomorphic \*R's.
Finally, Joel asked about a structure theory for such \*R's. Quite generally, without constraining the cardinality of \*R to be only c, one can describe such models as direct limits of ultrapowers of R with respect to ultrafilters on R. The embeddings involved in such a direct system are the elementary embeddings given by Rudin-Keisler order relations between the ultrafilters. (For the large cardinal folks here: This is just like what happens in the "ultrapowers" with respect to extenders, except that here we don't have any well-foundedness.) And this last paragraph has nothing particularly to do with R; the analog holds for elementary extensions of any structure of the form (S, all predicates and functions on S) for any set S.
| 9 | https://mathoverflow.net/users/6794 | 28457 | 18,601 |
https://mathoverflow.net/questions/28463 | 2 | The [Wikipedia article on numerical differentiation](http://en.wikipedia.org/wiki/Numerical_differentiation#Practical_considerations) mentions the formula
$$
h=\sqrt \epsilon \times x
$$
where $\epsilon$ is the machine epsilon (approx. $2.2\times 10^{-16}$ for 64-bit IEEE 754 doubles), to calculate the optimum "small number" $h$ to be used in differentiation, such as
$$
\frac{f(x+h)-f(x)}{h}
$$
But what if $x$ is zero? Then $h$ will be zero too, and division by zero is certainly not a way to do numerical differentiation. Is the article wrong? Is it otherwise correct, except that near zero (how near?) some small enough constant (how small?) should be used?
| https://mathoverflow.net/users/6861 | Optimum small number for numerical differentiation | If you click through to the reference given for the Wikipedia piece, you'll find an answer. The formula given there is to take $h$ to be roughly $\sqrt{\epsilon\_f}x\_c$, where $\epsilon\_f$ isn't necessarily "machine epsilon," but more to the point, where $x\_c$ isn't necessarily $x$.
| 4 | https://mathoverflow.net/users/3684 | 28465 | 18,606 |
https://mathoverflow.net/questions/28469 | 14 | Let $L\diagup K$ be a Galois extension of fields satisfying $\left[L:K\right] < \infty$. Let $B$ be a finite-dimensional (as a $K$-vector space) $K$-algebra. Then, the Galois group $G$ of $L\diagup K$ acts on the $L$-algebra $L\otimes\_K B$ (although not by $L$-linear homomorphisms), thus also on its unit group $\left(L\otimes\_K B\right)^{\times}$. Is it true that $H^1\left(G,\left(L\otimes\_K B\right)^{\times}\right)$ is the one-element set?
For $B=K$, this would be Hilbert 90. More generally, for $B$ being a matrix algebra over $K$, this would be an extension of Hilbert 90 Milne claims to hold in his CFT. My main reason for generalizing to arbitrary $B$ is to prove the following fact, known as **Noether-Deuring theorem**:
Let $A$ be a $K$-algebra, and let $U$ and $V$ be two finite-dimensional representations of $A$ over $K$. Then, $U$ and $V$ are isomorphic representations if and only if the representations $L\otimes\_K U$ and $L\otimes\_K V$ are isomorphic representations of the algebra $L\otimes\_K A$. Note that this holds not only for Galois extensions $L\diagup K$ but for arbitrary field extensions $L\diagup K$, but the (Galois) case of finite fields is the hardest. This generalizes the fact that two matrices over some field are similar if and only if they are similar over a field extension, which, in turn, is a particular case of ["Conjugacy rank" of two matrices over field extension](https://mathoverflow.net/questions/9162/conjugacy-rank-of-two-matrices-over-field-extension) .
| https://mathoverflow.net/users/2530 | Hilbert 90 for algebras | It's actually easier to go the other way around. Finite dimensional modules over
an algebra $A$ fulfils the Krull-Remak-Schmidt theorem of being isomorphic to a
direct sum of indecomposable modules with the indecomposable factors unique up
to isomorphism. If now $L\bigotimes\_KU$ and $L\bigotimes\_KV$ are isomorphic as
$L\bigotimes\_KA$-modules they are also isomorphic as $A$-modules but as
$A$-modules they are isomorpic to $U^n$ resp. $V^n$ where $n=[L:K]$ and by the
KRS-theorem this implies that $U$ and $V$ are isomorphic. The generalised Thm 90
now follows as the cohomology set classifies $B$-modules whose scalar extension
to $L$ are free of rank $1$.
| 17 | https://mathoverflow.net/users/4008 | 28474 | 18,612 |
https://mathoverflow.net/questions/28481 | 2 | Let $K$ be an algebraic function field of one variable. Then we can define its genus. On the other hand, it can also be seen as a scheme, so we can define the arithmetic and geometric genus. Could anyone please tell me the relation between these definitions?
| https://mathoverflow.net/users/3849 | on the genus of a function field | The definitions coincide, with some caveats: basically for a curve, there is a single notion of genus, which applies equally to smooth curves over algebraically closed fields, and to their function fields; and also over the complex numbers to the associated Riemann surface as two-dimensional manifold. See <http://en.wikipedia.org/wiki/Genus_%28mathematics%29> . On the other hand care is needed for curves that are allowed to be singular, or fields that are not algebraically closed, what definition is in use.
| 1 | https://mathoverflow.net/users/6153 | 28482 | 18,617 |
https://mathoverflow.net/questions/28459 | 9 | Background
----------
I need to solve polynomials in multiple variables using [Horner's scheme](http://en.wikipedia.org/wiki/Horner_scheme) in Fortran90/95. The main reason for doing this is the increased efficiency and accuracy that occurs when using Horner's scheme to evaluate polynomials.
I currently have an implementation of Horner's scheme for univariate/single variable polynomials. However, developing a function to evaluate multivariate polynomials using Horner's scheme is proving to be beyond me.
>
> An example bivariate polynomial would be: $12x^2y^2+8x^2y+6xy^2+4xy+2x+2y$ which would factorised to $x(x(y(12y+8))+y(6y+4)+2)+2y$ and then evaluated for particular values of x & y.
>
>
>
Research
--------
I've done my research and found a number of papers such as:
staff.ustc.edu.cn/~xinmao/ISSAC05/pages/bulletins/articles/147/hornercorrected.pdf
citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.40.8637&rep=rep1&type=pdf
www.is.titech.ac.jp/~kojima/articles/B-433.pdf
Problem
-------
However, I'm not a mathematician or computer scientist, so I'm having trouble with the mathematics used to convey the algorithms and ideas.
As far as I can tell the basic strategy is to turn a multivariate polynomial into separate univariate polynomials and compute it that way.
Can anyone help me? If anyone could help me turn the algorithms into pseudo-code that I can implement into Fortran myself, I would be very grateful.
| https://mathoverflow.net/users/6860 | How to implement Horner’s scheme for multivariate polynomials? | The paper you cite, "On the multivariate Horner scheme" (Pena, Sauer) has an explicit algorithm specified on p.3. The remaining challenge is to penetrate the notation and conventions in the paper
laid out in the first three pages far enough to turn their algorithm presentation into code.
It also seems that this paper (just reading the abstract) specifies an explicit algorithm:
"Evaluation of Multivariate Polynomials and Their Derivatives,"
J. Carnicer and M. Gasca,
*Mathematics of Computation*, Vol. 54, No. 189 (Jan., 1990), pp. 231-243.
[ResearchGate link to full text](https://www.researchgate.net/publication/258986835_Evaluation_of_Multivariate_Polynomials_and_Their_Derivatives).
>
> **Abstract**. An extension of Horner's algorithm to the evaluation of m-variate polynomials and their derivatives is obtained. The schemes of computation are represented by trees because this type of graph describes exactly in which order the computations must be done. Some examples of algorithms for one and two variables are given.
>
>
>
| 7 | https://mathoverflow.net/users/6094 | 28490 | 18,620 |
https://mathoverflow.net/questions/28485 | 20 | I recently came across an interesting phenomenon which confused me slightly, concerning integral points on varieties.
For example, consider $X = \mathbb{A}\_{\mathbb{Z}}^{n+1} \setminus \{0\}$, affine $n$-space over $\mathbb{Z}$ with the origin removed. Naively, one would guess that $X(\mathbb{Z})$ is the set of integers $\{ (x\_0,x\_1,\ldots,x\_n) \in \mathbb{Z}^{n+1} \setminus \{0\}\}$.
However, some work that I have been doing on recently with universal torsors has in fact led me to believe that $X(\mathbb{Z})$ should equal $\mathbb{P}^n(\mathbb{Z})$, at least modulo the action of $\mathbb{G}\_m$. That is $X(\mathbb{Z})$ is actually the set of integers $\{ (x\_0,x\_1,\ldots,x\_n) \in \mathbb{Z}^{n+1}\setminus \{0\}\}$ such that $\gcd(x\_0,x\_1,\ldots,x\_n)=1$.
Is there a simple explanation for why this is the case?
Thanks!
Dan
| https://mathoverflow.net/users/5101 | Integral points on varieties | Your belief is correct. A $\mathbb{Z}$-point has to reduce to an $\mathbb{F}\_p$-point for all $p$, which kills examples with gcd > 1.
If you want to make this precise, try writing down an explicit description of X by patching affine pieces. All the essential ideas are already there in $\mathbb{A}\_{\mathbb{Z}}^1 \backslash 0$: this is the spectrum of $\mathbb{Z}[X, Y] / (XY - 1)$.
| 14 | https://mathoverflow.net/users/2481 | 28491 | 18,621 |
https://mathoverflow.net/questions/28453 | 9 | I understand that a generic $G$-polynomial $f(t\_1,...,t\_n)[X]$ over field $k$ has Galois group $G$ over $k(t\_1,...,t\_n)$. And basically any $G$ extension of $k$ should be generated by a realization of $f$.(even a bit stronger but that is not the point here).
Now as much as I understand, our motivation for hunting these polynomials is that in real (constructive) life, we would like to plug random elements of $k$ into $t\_1,...,t\_n$ and get a $G$-extension. However, it's obvious that the definition doesn't guarantee it. For example as a trivial failure, we know that $X^n + t\_1X^{n-1} + \cdots + t\_n$ is generic for $S\_n$, but not all values for $t\_1, ..., t\_n$ (basically all polynomials) lead to an $S\_n$-extension.
So, basically, my question is this: what is the constructive value of the definition of generic polynomial. Is there any (although I know I'm saying nonsense) high probabilistic/statistic success rate in getting a $G$-extension when a random realization is chosen. Is there some kind of definition of "odd" that says those times that we don't get a $G$-extension are somehow odd and not normal?
| https://mathoverflow.net/users/6776 | Why do generic polynomials work in reality? | Adding unto Boyarsky's answer: Stephen Cohen has given [quantative bounds](http://www.ams.org/mathscinet-getitem?mr=628276) for how often generic polynomials work. If I've skimmed his paper correctly, when the coefficients are integers chosen from the interval $[-N, N]$, the probability that the Galois group comes out wrong is $O(N^{-1/2} \log N)$, with an explicitly computable constant which depends on the group and the precise parameterization being used.
| 6 | https://mathoverflow.net/users/297 | 28494 | 18,623 |
https://mathoverflow.net/questions/28437 | 2 | Hi,
I have this math modeling problem that I need help with. If I have 3 data sources, each being updated at different frequencies, what would be the best way to rank them so the less frequent sources don't get pushed down too quickly? In other words, I want the less likely source to be "weighed down" in its spot. Also, the frequencies of the data source are not consistent.
I was trying to think of the problem in some sort of a bayesian update/inference/learning way but I couldn't properly describe it. Something like "what is the probability that data source A will be in rank position 1" but I'm not sure if that's the right track. I have limited probability experience. Anyone have any thoughts on this?
I don't need the exact solution or anything, just want to be pointed in the direction. I don't have a breadth of mathematical knowledge, but I should be proficient enough to figure it out if anyone can provide articles, papers, examples, etc.
Thanks for the help.
| https://mathoverflow.net/users/6855 | Ranking sources at variable(random) frequencies | You can model the arrivals as a poisson distribution with different arrival rates λi for each one of your sources.
$f(n\_i;\lambda\_i) = \dfrac{\lambda\_i^{n\_i} e^{-\lambda\_i}} { n\_i!}$
You can then assume that the arrival rates are random draws from a gamma distribution which would let you pool information across all your data sources.
$\lambda\_i$ ~ $\Gamma(k,\theta)$.
Assume suitable distributions for $k$ and $\theta$. You can then construct the posterior distribution for $\lambda\_i$, $k$ and $\theta$. The gamma distribution will help you 'shrink' your estimates for $\lambda\_i$. In other words, if the frequencies of arrival is too low then it will be nudged a bit higher and if it is too high it will be nudged a bit lower depending on the data you have. Basically, the pooling of information will enable you to arrive at more robust estimates of the arrival rates.
Estimate these parameters conditional on the data you observe using standard Bayesian methods. Once you have the estimates you can find out the probability that source $i$ is ranked the highest etc from the estimates. Do note that if the observed arrivals are of order of magnitude different from one another then you would be better of simply ordering on the actual arrival numbers as the bayesian computation above would not give any different results.
Google the following terms to get some traction: hierarchical bayesian models, gibbs sampler, mcmc sampler, conjugate priors etc.
| 1 | https://mathoverflow.net/users/4660 | 28498 | 18,626 |
https://mathoverflow.net/questions/28486 | 7 | In any presheaf topos, there exists an object called *Lawvere's segment*, which can be described as the presheaf $L:A^{op}\to Set$ such that for each object $a\in A$, $L(a)=\{x\hookrightarrow\ h\_a: x\in Ob([A^{op},Set])\}$.
That is, $L$ assigns to each object $a\in Ob(A)$ the set of monomorphisms with target $h\_a(\cdot):=Hom(\cdot,a)$
in $[A^{op},Set]$. Since all objects in $[A^{op},Set]$ can be given as colimits of the representable functors $h\_a$ for $a\in A$, we can actually give a characterization of the functor represented by $L$ as the "sub-object assigning functor".
Lawvere's segment is useful, because it naturally has the structure of a cylinder functor (given by taking the cartesian product with a presheaf $X$). In fact, it is a theorem of Cisinski (very closely related to the specialization Jeff Smith's theorem to presheaf toposes) that every accessible localizer on a presheaf category admits the structure of a closed model category generated by the homotopical data (donnée homotopique is the term used by Cisinski, so this is my rough translation) of some *set* (as opposed to proper class) of arrows $S\subset Arr([A^{op},Set])$, some cellular model $\mathcal{M}$, and the canonical cylinder given by Lawvere's segment.
Then my question: How can we describe Lawvere's segment in $Set\_\Delta$ explicitly, that is to say, geometrically?
Edit: A cellular model M on a presheaf topos is a *set* of monomorphisms $M$ such that $llp(rlp(M))$ is the class of all monomorphisms. Note that $llp$ and $rlp$ give the class of arrows with the left lifting property (resp. right lifting property) to the class of arrows given in the argument.
| https://mathoverflow.net/users/1353 | An explicit description of Lawvere's segment in the category of simplicial sets | I've never heard this called 'Lawvere's segment' before, but your $L$ is the subobject classifier in the presheaf topos $[A^{\mathrm{op}},\mathrm{Set}]$. In presheaf toposes generally, the subobject classifier $\Omega$ is the presheaf that sends an object $a$ to the set of all sieves on $a$, i.e. the set of subobjects of $\hom(-,a)$. This means in this case that simplicial subsets $S' \subset S$ are in bijection with simplicial maps $S \to L$.
Off the top of my head, I can't be exactly sure what $L$ looks like, but I'd guess it's the constant simplicial set with $L\_n = 2$ (the set of truth values) and all maps identities. Then the characteristic map $\chi\_{S'} \colon S \to L$ will be the usual $s \mapsto 1$ if $s \in S'$ and $0$ otherwise, while the simplicial subset corresponding to some $\phi \colon S \to L$ will be given by the fibres $\phi\_n^{-1}(1)$ in each degree. $S'$'s being a *simplicial* subset should correspond to $\phi$'s being a simplicial map.
I could easily be wrong about that last paragraph, though. There might be more about this in Mac Lane--Moerdijk.
Edit: I was wrong -- see the comments below.
| 5 | https://mathoverflow.net/users/4262 | 28499 | 18,627 |
https://mathoverflow.net/questions/27721 | 3 | Hi,
We have a real, non-singular and symmetric matrix M of size n by n, with diagonal elements 0's. Its eigenvalues and eigenvectors are computed.
Now we wish to change its diagonal elements arbitrarily to maximize the multiplicy of an eigenvalue ( unnecessary to be the same value with old one).
For example, let n be even, M=ones(n)-diag(ones(n,1)), then
n-1 is the largest. The modification is unnecessary.
$\lambda$1>$\lambda$2=$\lambda$3=...=$\lambda$n
Is the powerful convex programming the proper approach to solve this problem?
I try to creat a more complicated example M of interger entries ( small n ), but it seems to be hard...
To transform a model to convex programming is not easy either... :)
Thank you for your help
Zhi Ming
| https://mathoverflow.net/users/6679 | Maximize the multiplicity of an eigenvalue | Let $W = M + D$, where $M$ is the original $n \times n$ matrix and $D$ is the added diagonal matrix that we want to determine.
$W$ is symmetric, thus diagonalizable by an adjoint action of the orthogonal group.
Larger multiplicities in the eigenvalues of $W$ imply smaller dimensions of the adjoint orbits.
For example, if we have an eigenvalue of multiplicity $n-1$, then the adjoint orbit will be
$O(n)/(O(n-1) \times O(1)) = S^{n-1}/Z\_2 = RP^{n-1}$ of dimension $n-1$;
while, if all the eigenvalues are distinct, the adjoint orbit is the real flag manifold $Fl\_{\mathbb{R}}^n = O(n)/(Z\_2)^n$ of dimension $ \frac{n(n-1)}{2}$. (Of course, in the case of scalar multiple of the unit matrix, the adjoint orbit is just a single point).
Thus in order to achieve large multiplicities, we need to minimize the dimension of the adjoint orbit.
This problem can be reduced to a problem of matrix rank minimization as follows:
Let $\{l\_i \}\_{i=1,...,n(n-1)/2}$, be a set of generators of the Lie algebra of $O(n)$ normalized according to:
$\textrm{tr}(l\_i l\_j) = \delta\_{ij}$. The dimension of the adjoint orbit equals the rank of the Gram matrix $C$ whose elements are given by, $C\_{ij} = \textrm{tr}([l\_i, W][l\_j, W])$. The problem is thus reduced to the minimization of the rank of the Gram matrix whose elements depend linearly on the added diagonal matrix elements.
One of the possible methods to solve this problem is through a convex programming
heuristic approach for the solution of matrix rank minimization,
based on replacing the rank by the nuclear norm (the sum of the singular values), as explained in the following [lecture notes](http://temple.birs.ca/~10w5007/parrilo.pdf) by: P.A. Parillo. The nuclear norm is a convex envelope of the rank which may explain why this method works well in practice in general.
| 1 | https://mathoverflow.net/users/1059 | 28500 | 18,628 |
https://mathoverflow.net/questions/28523 | 2 | Let $f$ be a Hilbert polynomial, and $X := Hilb\_h(P^d\_{F\_p})$ a Hilbert scheme defined over $F\_p$. Then there is an absolute Frobenius map $F: X \to X$. I'm even interested in the case $f \equiv 1$, so $X = P^d$.
Which of the following is a sensical question?
1. What is the family over $X$ induced by pulling back the universal family along $F$? Is there a reasonable way to think about it, that makes it clear that it's not just again the universal family?
2. If that family is just the universal family again, then in what sense is the Hilbert scheme universal? (As it seems I've obtained the same family from two different maps, which I thought wasn't supposed to happen.)
| https://mathoverflow.net/users/391 | What is the family derived from the absolute Frobenius on the Hilbert scheme? | Indeed since the map on the base is not an isomorphism (apart from silly cases such as when $X$ is empty: the Frobenius endomporphism of a locally finite type $\mathbb{F}\_p$-scheme is an isomorphism if and only if the scheme is etale), the pullback cannot be universal. Functorially, $F$ carries a family of closed subschemes of projective space to the family defined (locally over the base) by the equations with coefficients raised to the $p$-power. Concretely, a hypersurface with some coefficient of 1 and another not a $p$-power in the base ring does not occur in the pullback family.
| 7 | https://mathoverflow.net/users/6773 | 28529 | 18,641 |
https://mathoverflow.net/questions/28530 | 2 | Let $A$ and $B$ be two $4\times 4$ matrices. Using Newton's identities, one can prove that if
$$\det(A) = \det(B)\quad \text{and}\quad \mathrm{tr}(A^i) = \mathrm{tr}(B^i)$$ for $i=1,2,3$, then $A$ and $B$ have the same characteristic polynomial, thus the same eigenvalues.
I'm interested in pairs of matrices $A$ and $B$ that satisfy all those equations except the last one, i.e.
$$\det(A)=\det(B)$$
$$\mathrm{tr}(A)=\mathrm{tr}(B)$$
$$\mathrm{tr}(A^2)=\mathrm{tr}(B^2)$$
but $\mathrm{tr}(A^3) \neq \mathrm{tr}(B^3)$.
Does anyone know how to generate such matrices? Have they ever been studied? A reference would be nice.
| https://mathoverflow.net/users/1162 | On matrices that almost have the same eigenvalues | Such matrices will have a characteristic polynomial $z^4+a\_3z^3+a\_2z^2+a\_1z+a\_0$ with the same $a\_3$, $a\_2$, $a\_0$ but distinct $a\_1$. You can generate a plenty of diagonal such matrices by picking roots of such two polynomials.
I cannot vouch that they were not studied but I am pretty certain that nothing groundbreaking came out of such studies.
| 6 | https://mathoverflow.net/users/5301 | 28531 | 18,642 |
https://mathoverflow.net/questions/28525 | 3 | Construction
------------
Suppose I have a density matrix $\rho$ which is proportional to a projector $P$ formed by tensoring together $N$ small projectors $P^{(i)}$ of rank 2:
$P^{(i)} = |a\rangle\_i\langle a| + |b\rangle\_i\langle b|$
$P=P^{(1)}\otimes \cdots \otimes P^{(N)} = \bigotimes\_{i=1}^N P^{(i)}$
$\rho = P/2^n \quad , \quad \mathrm{Tr} \rho = 1 $.
(I'm a physicists, and I don't really know how to express $P^{(i)}$ explicitly without using bra-ket notation.) The small projectors $P^{(i)}$ act on their own (potentially very large) Hilbert space $H^{(i)}$ spanned by, say, $\{|a\rangle\_i, |b\rangle\_i, |c\rangle\_i,...\}$, and $P$ acts on $H = \bigotimes\_{i=1}^N H^{(i)}$.
Then since the $2^N$ non-zero eigenvalues of $\rho$ are all the same, the entropy is obviously
$H[\rho] = -\sum\_{s} \lambda\_s \ln \lambda\_s = N \ln 2 $.
Now, suppose I split the density matrix in half, and operate on each half with either a *local* set of unitaries (all identical except that the act on different $H^{(i)}$) or their inverses. That is, suppose
$U = \bigotimes\_{i=1}^N U^{(i)}$
$U^{(i)} |a\rangle\_i = |a+\rangle\_i \quad , \quad {U^{(i)}}^{-1} |a\rangle\_i = |a-\rangle\_i \quad , \quad U^{(i)} |b\rangle\_i = |b+\rangle\_i \quad , \quad {U^{(i)}}^{-1} |b\rangle\_i = |b-\rangle\_i $
with a new density matrix $\eta$:
$\eta = (U P U^{-1} + U^{-1} P U)/2^{N+1}$.
Assume we know the inner products
$\langle a-|b+\rangle \quad, \quad \langle a+|b-\rangle \quad, \quad \langle a+|a-\rangle \quad, \quad \langle b+|b-\rangle \quad ,$
(which are the same for all $i$ because the $U^{(i)}$ are all identical).
Question
--------
Is it possible to calculate the entropy $H[\eta]$?
If useful, it's fine to assume that $U^{(i)}$ is infitesimal so that
$\langle a-|b+\rangle \sim 0 \quad, \quad \langle a+|b-\rangle \sim 0 \quad, \quad \langle a+|a-\rangle \sim 1 \quad, \quad \langle b+|b-\rangle \sim 1 \quad .$
Simpler case
------------
It might be helpful to see the solution to the (much) simpler case where the $P^{(i)}$ are rank 1. In this case, we can write
$P^{(i)} = |a\rangle\_i\langle a|$
$P = |a\rangle\_1\langle a| \otimes \cdots \otimes |a\rangle\_N\langle a| = |A\rangle\langle A|$
$\rho = P \quad , \quad \mathrm{Tr} \rho = 1$
$\eta = (U P U^{-1} + U^{-1} P U)/2 $
$\quad = (|A+\rangle\langle A+| + |A-\rangle\langle A-|)/2$
and get that the two non-zero eigenvalues are
$\lambda\_\pm = [1 \pm |\langle A+|A-\rangle\_i|]/2$
where
$\langle A+|A-\rangle = \prod\_{i=1}^N \langle a+|a-\rangle\_i = \langle a+|a-\rangle^N$.
Elaboration of Answer
---------------------
Steve Flammia graciously provided the answer below, which hinges on the relationship between the eigenvalues of the two operators $P+Q$ and $PQP$, where $P$ and $Q$ are orthogonal projectors. (The eigenvalues of $PQP$ are easy to compute because the matrix is diagonal in the same basis as $P$ and the matrix elements are expressed in terms of the known inner products.) As an exercise for myself, and to help out anyone else reading this question, the relationship is proved here.
Consider two orthogonal projectors $P$ and $Q$ with respective images $M$ and $N$. We can assume the intersections
$ \quad M \cap N \quad , \quad M \cap N^\perp \quad , \quad M^\perp \cap N \quad , \quad M^\perp \cap N^\perp$
are all {0}, i.e. $M$ and $N$ are in "generic position". (If not, then the Hilbert space can be decomposed into $H=H' \oplus H\_0$, where $M$ and $N$ leave both $H'$ and $H\_0$ invariant, $M$ and $N$ are simultaneously diagonal on $H\_0$, and the above intersections are trivial on $H'$. We then restrict our attention $H'$.) As shown by Halmos [1], for any pair of orthogonal projectors $P$ and $Q$ of equal rank we can decompose the Hilbert space into two subspaces of equal dimension, $H=K \oplus K$, in which
$P = \pmatrix{1& 0 \\\ 0& 0}$
$Q = \pmatrix{C^2 & CS \\\ CS & S^2}$
where $C$ and $S$ are positive matrices on $K$ which commute and for which $C^2 + S^2 =1$. If we let $c\_i$ be the eigenvalues of $C$, then the non-zero eigenvalues of
$PQP = \pmatrix{C^2& 0 \\\ 0& 0}$
are $c\_i^2$ and the eigenvalues of
$P+Q=\pmatrix{1 + C^2& CS \\\ CS & S^2}$
can be shown to be $1 \pm c\_i$.
References
----------
[1] Halmos, P.R. ["Two Subspaces"](http://www.jstor.org/stable/1995288). Transactions of the American Mathematical Society, Vol. 144 (Oct., 1969), pp. 381-389.
| https://mathoverflow.net/users/5789 | What is the entropy of a density matrix which is the sum of two unitarily equivalent projectors? | First observation, you can unitarily rotate the first term without changing the spectrum so you can just consider a single unitary with out loss of generality. Call the rotated term $Q$.
Next, it helps to know about the canonical form for two projectors. Given two projectors P and Q in general position, you can always compute the eigenvalues of P+Q as follows. They are simply $1 \pm x\_j$ where the $x\_j$ are the positive square roots of the eigenvalues of the operator $PQP$ (or $QPQ$... it doesn't matter).
For your problem, because of the tensor product structure, this is actually quite easy to do. The eigenvalues $x\_j$ will be all possible products of the local eigenvalues for $P^{(i)} Q^{(i)} P^{(i)}$, with a square root. The entropy is then a sum over these configurations, suitably normalized.
| 2 | https://mathoverflow.net/users/1171 | 28537 | 18,647 |
https://mathoverflow.net/questions/28524 | 0 | Let $X$ be a complex normal projective variety, let $|L|$ be a non empty linear series on $X$ and let $b(|L|)$ be its base ideal.
Suppose $f:X'\rightarrow X$ is a log resolution of the ideal $b(|L|)$.
Is $f$ a log resolution of the linear series $|L|$ (even if $X$ is not smooth)?
If it is do you have a proof or a reference for this?
| https://mathoverflow.net/users/6430 | Log resolutions of linear series | I don't think so. Suppose for example that $X$ is smooth, and the base locus of $|L|$ is set-theoretically a divisor with normal crossing, but it has an embedded component. In this case $X$ itself will be a log-resolution of the base locus, but not of the linear system $|L|$.
| 1 | https://mathoverflow.net/users/4790 | 28546 | 18,652 |
https://mathoverflow.net/questions/23427 | 32 | Just yesterday I heard of the notion of a fundamental group of a topos, so I looked it up on the [nLab](http://ncatlab.org/nlab/show/fundamental+group+of+a+topos), where the following nice definition is given:
If $T$ is a Grothendieck topos arising as category of sheaves on a site $X$, then there is the notion of locally constant, locally finite objects in $T$ (which I presume just means that there is a cover $(U\_i)$ in $X$ such that each restriction to $U\_i$ is constant and finite). If $C$ is the subcategory of $T$ consisting of all the locally constant, locally finite objects of $T$, and if $F:C\rightarrow FinSets$ is a functor ("fiber functor"), satisfying certain unnamed properties which should imply prorepresentability, then one defines $\pi\_1(T,F)=Aut(F)$.
Now, if $X\_{et}$ is the small étale site of a connected scheme $X$, then it is well known the category of locally constant, locally finite sheaves on $X$ is equivalent to the category of finite étale coverings of $X$, and with the appropriate notion of fiber functor it surely follows that the étale fundamental group and the fundamental group of the topos on $X\_{et}$ coincide.
Similarly, as the nlab entry mentions, if $X$ is a nice topological space, locally finite, locally constant sheaves correspond to finite covering spaces (via the "éspace étalé"), and we should recover the profinite completion of the usual topological fundamental group.
Before I come to my main question: Did I manage to summarize this correctly, or is there something wrong with the above?
>
> My question:
>
>
> Has the fundamental group of other topoi been studied, and in what context or disguise might we already know them? For example, what is known about the fundamental group of the category of fppf sheaves over a scheme $X$?
>
>
>
| https://mathoverflow.net/users/259 | Fundamental groups of topoi | The profinite fundamental group of $X\_{fppf}$ as you define it is again the etale fundamental group of X. More precisely, the functor (of points)
$f : X\_{et} \to \mathrm{Sh}\_{fppf}(X)$
is fully faithful and has essential image the locally finite constant sheaves (image clearly contained there, as finite etale maps are even etale locally finite constant, let alone fppf locally so). Proof in 3 steps:
1. It is fully faithful by Yoneda (note also well-defined by fppf descent for morphsisms).
2. Both sides are fppf sheaves (stacks) in $X$, by classical fppf descent.
3. Combining 1 and 2, it suffices to show that a sheaf we want to hit is just fppf locally hit, which is obvious since locally it's finite constant.
Note that the same proof also works for $X\_{et}$ or anything in between -- once your topology splits finite etale maps it doesn't really matter what it is. So we usually just work with the minimal one, the small etale topology. As Mike Artin said to me apropos of something like this, "Why pack a suitcase when you're just going around the corner?"
| 33 | https://mathoverflow.net/users/3931 | 28555 | 18,660 |
https://mathoverflow.net/questions/28532 | 24 | Suppose I want to understand classical mechanics.
Why should I be interested in arbitrary poisson manifolds and not just in symplectic ones?
What are examples of systems best described by non symplectic poisson manifolds?
| https://mathoverflow.net/users/2837 | Classical mechanics motivation for poisson manifolds? | For many reasons and purposes, it is the Poisson bracket, not the symplectic form, that plays a primary role.
* Equations of motion and, more generally, the evolution of observables have an easy form:
$$ \frac{\partial f}{\partial t}=\{H,f\}.$$
* Conserved quantities form a Poisson subalgebra:
$$\{H,F\}=\{H,G\}=0 \implies \{H,\{F,G\}\}=0. $$
* Since symmetries (Hamiltonian group actions) are Poisson by nature, the moment map is defined in the Poisson setting:
$$ M\ni P\mapsto (X \mapsto H\_X(P)).$$
Unlike in the symplectic case, *both* steps in the Hamiltonian reduction, restriction to the level set and factorization by the action of the stabilizer, are naturally carried out in the Poisson category, even for singular reduction.
* Quasiclassical approximation in quantum mechanics (and conversely, quantization of classical mechanical systems) is expressed via the Poisson bracket:
$$[\hat{F},\hat{G}]=ih\widehat{\{F,G\}} +O(h^2). $$
On the other hand, many natural systems have a degenerate Poisson bracket and/or are infinite-dimensional.
* The phase space of various tops is the dual space $\mathfrak{g}^\*$ of a Lie algebra. This is a universal example of a linear Poisson structure. Symplectic leaves are the coadjoint orbits and the Poisson center is given by the (classical) "Casimir functions".
* Classical integrable systems such as KdV admit a bi-Hamiltonian structure (i.e. a pair of compatible Poisson brackets). This has no analogues in symplectic theory.
* Some of these structures are obtained by a reduction from a linear Poisson structure on a suitable infinite-dimensional Lie algebra (loop algebra, algebra of matrix differential or pseudo-differential operators, etc).
Finally, a related practical consideration: even if you are interested in studying a symplectic manifold, frequently this is best accomplished by embedding it as a symplectic leaf into a simpler Poisson manifold, *whether or not it has direct physical meaning.* In particular, this applies to flag manifolds of semisimple Lie groups, which are topologically complicated objects, but can be identified with coadjoint orbits, thus embedded into a vector space with a linear Poisson structure.
| 30 | https://mathoverflow.net/users/5740 | 28562 | 18,662 |
https://mathoverflow.net/questions/28569 | 27 | Let $G$ be a complex connected Lie group, $B$ a Borel subgroup and $W$ the Weyl group. The Bruhat decomposition allows us to write $G$ as a union $\bigcup\_{w \in W} BwB$ of cells given by double cosets.
One way I have seen to obtain cell decompositions of manifolds is using Morse theory. Is there a way to prove the Bruhat decomposition using Morse theory for a certain well-chosen function $f$? More generally, are there heuristics when obtaining a certain cell decomposition through Morse theory is likely to work?
| https://mathoverflow.net/users/798 | Is there a Morse theory proof of the Bruhat decomposition? | Here's a partial answer. What I'm about to say is taken from Section 2.4 of Chriss and Ginzburg's *Representation Theory and Complex Geometry*. The references I use will be from this book. There aren't complete proofs there, but there are references to complete proofs.
The Bruhat decomposition on $G$ (we'll assume $G$ is reductive, the general case follows from this) can be seen as the preimage of the Bruhat decomposition of its flag variety $G/B$. It's easier to work with $G/B$ since it's a projective variety.
Here $G/B$ has a torus (= ${\bf C}^\*$) action with finitely many fixed points $\{w\_1, \dots, w\_n\}$, so one has a Bialynicki-Birula cell (= affine space) decomposition given by the attracting sets
$X\_i = \{ x \in G/B \mid \lim\_{t \to 0} t \* x = w\_i \}$.
[Normally, one thinks of $G/B$ as having finitely many fixed points with respect to $({\bf C}^\* )^r$, where $r$ is the rank of $G$, but we can take ${\bf C}^\* $ to be a subgroup in general position]
In our case, the fixed points are indexed by the Weyl group, and the attracting sets will be the Bruhat cells.
From this, one can construct a Morse function $H$ (Lemma 2.4.17) whose Morse decomposition coincides with the B-B cell decomposition (Corollary 2.4.24).
So even though it's a very special case, one case where cell decompositions come from Morse functions is when the manifold is a projective algebraic variety with finitely many torus fixed points.
| 33 | https://mathoverflow.net/users/321 | 28575 | 18,673 |
https://mathoverflow.net/questions/28560 | 9 | Let $X$ be the set of $k\times k$ matrix with entries in $\mathbb{C}$, and let $M\in X$. The group $GL(k,\mathbb{C})$ acts on $X$ by conjugation, and according to the Jordan decomposition theorem (see e.g [wikipedia](http://en.wikipedia.org/wiki/Jordan_form)) somewhere in the orbit containing $M$ is a block diagonal matrix with non-zero entries only on the diagonal and superdiagonal.
Suppose now we consider $k\times k$ matrices whose entries lie in the polynomial ring $\mathbb{C}[z\_{1},z\_{2}, \ldots ,z\_{n}]$ and we study the action by conjugation of $GL(k,\mathbb{C}[z\_{1},z\_{2}, \ldots ,z\_{n}])$. Then the Jordan decomposition theorem, as formulated above, clearly no longer holds. For example consider the matrix:
$ M=\left(
{\begin{array}{cc}
0 & 1 \\\
z^{p}\_{1} & 0
\end{array}}
\right)$,
where in the above $p$ denotes a positive integer. If $p $ is odd, then $M$ cannot be diagonalized since the ring $\mathbb{C}[z\_{1},z\_{2}, \ldots ,z\_{n}]$ does not contain the eigenvalues of $M$. On the other hand, if $p$ is even we still cannot diagonalize $M$ since when $z\_{1}=0 $, $M$ is not diagonalizable.
My question is then what, if anything, remains of the Jordan decomposition in this case? Or equivalently given a $k\times k$ matrix $M$ with entries in $\mathbb{C}[z\_{1},z\_{2}, \ldots ,z\_{n}]$ are there any particularly simple matrices related to $M$ via conjugation by an element of $GL(k,\mathbb{C}[z\_{1},z\_{2}, \ldots ,z\_{n}])$?
| https://mathoverflow.net/users/5124 | Jordan Form Over a Polynomial Ring | The short answer is "no". It is not too difficult to construct \*invariants, but the *canonical forms* are hard.**a** What follows is not a full answer, but a useful way to think about the question.
The problem of classifying $k\times k$ matrices over a commutative ring $R$ up to conjugacy is equivalent to the problem of classifying all $R[\lambda]$-module structures on $R^k$, the free $R$-module of rank $k,$ up to isomorphism. Given $A\in M\_k(R)$, let the variable $\lambda$ act on $R^k$ via $A$ and conversely, given an $R[\lambda]$-module structure on $R^k,$ the action of $\lambda$ is $R$-linear, hence yields a matrix. It's a good exercise to see that conjugacy of matrices $\leftrightarrow$ isomorphism of modules.
Now, when $R$ is a field, $R[\lambda]$ is a principal ideal domain, and all finitely-generated modules can be completely classified using the theory of elementary divisors.**b** However, if $R$ is even a bit more complicated, such as $\mathbb{Z}$ or $K[X]$, the question involves modules over a ring of Krull dimension $2$ or larger, and one cannot hope for an explicit easy solution (except $k=1$). Already for $n=1, R=K[x]$ we are looking at the classification of $K[x,y]$-modules. See van der Waerden for classical treatment.
---
Footnotes
**a** For example, the characteristic polynomial and the Fitting invariants of the matrix in **b**.
**b** In the context of the conjugacy problem, you can replace a $k\times k$ matrix $A$ over $R$ modulo conjugation by $GL\_k(R)$ with a $k\times k$ matrix $A-\lambda I\_k$ over $R[\lambda]$ modulo left and right multiplication by $GL\_k(R).$ The "elementary" part refers to the fact that when $R=K$ is a field, the general linear group is generated by elementary transformations, and "divisors" refers to the form of the answer, where the canonical form is diagonal and $d\_i$ divides the entries in rows $1$ through $i$. Neither fact is true for a more general $R.$
| 13 | https://mathoverflow.net/users/5740 | 28588 | 18,683 |
https://mathoverflow.net/questions/28526 | 42 | Suppose that you graduate with a good PhD in mathematics, but don't necessarily want to go into academia, with the post-doc years that this entails. Are there any other options for continuing to do "real math" professionally?
For example, how about working at the NSA? I don't know much of what is done there -- is it research mathematics? Are there other similar organizations? Perhaps corporations that contract with the federal government? Companies like RSA?
Other areas of industry? Is there research mathematics done in any sort of financial or tech company?
I've made this a community wiki, since there aren't any right answers...
| https://mathoverflow.net/users/3028 | "Industry"/Government jobs for mathematicians | I have worked in academia, at the research center of a telecommunications company (Tellabs), and at two different [FFRDCs](http://en.wikipedia.org/wiki/List_of_federally_funded_research_and_development_centers) (MIT Lincoln Laboratory and IDA). At all of the non-academic jobs, I have done "real math," published papers, attended conferences, given talks, etc. So it is certainly possible to continue doing "real math" outside of academia.
You should be aware, however, that in almost any non-academic job, there is pressure on you to produce results that are "useful" for the company or the government. The amount of such pressure varies, but it always exists, because ultimately that is the main justification for your paycheck. In academia, the corresponding fact is that in almost any academic job, there is pressure on you to teach, since that is usually the justification for a significant portion of your salary. Finding a non-academic job where there is no pressure on you to do anything "useful" is akin to finding an academic job where you have no teaching responsibilities.
Certain high-tech companies and certain FFRDC's recognize that a good way to attract top talent is to give their employees the freedom to pursue their own research interests, whatever that may be. All the non-academic jobs I had were like this. They actively encouraged me to spend some amount of my time doing "real math" regardless of whether the results were of any "use." How much time? Well, if the company was doing well, and if I was doing a good job of producing "useful" results that they liked, then they would give me more freedom. But if the company was doing poorly then they would start to squeeze. During the telecom industry meltdown in the late 1990s, Tellabs eventually eliminated its research center entirely, along with my job; Bell Labs (more famously) suffered a similar fate.
So far I have been drawing a dichotomy between "what the company finds useful" and "real math," and maybe you don't find that satisfactory. After all, if you're sufficiently motivated, you can do "real math" on your own time regardless of what your "day job" is. Maybe what you want is a job where *providing what is useful to the company involves doing real math*. This is a taller order; for example, at Lincoln Labs I found that there was almost no real math involved in the work they wanted me to do, and I eventually left that job for that reason even though it was a great job in almost every other respect. However, it is still possible to find such jobs, depending on what area of math you are interested in. If you are interested in large cardinals and are hoping for a job where your theorems about large cardinals will be "useful" then you are probably out of luck. However, if your interests lean towards areas with known relevance to computer science or various branches of engineering then your chances are much better. The NSA scores pretty well in this regard since it is no secret that number theory and various other branches of so-called "pure" mathematics are relevant to cryptology.
In summary, jobs where you do "real math" do exist. When considering such a job, though, you should first ask yourself, *will I enjoy producing what this company considers to be "useful" results?* If the answer is no, then you will probably not be happy at the job even if they give you some freedom to do "real math." However, if the answer is yes, *and* the company gives you some amount of freedom to do "real math," then it will probably be an excellent fit for you.
| 37 | https://mathoverflow.net/users/3106 | 28589 | 18,684 |
https://mathoverflow.net/questions/28553 | 16 | On the way to defining Cartier divisors on a scheme $X$, one sheafifies a presheaf **base-presheaf** of rings $\mathcal{K}'(U)=Frac(\mathcal{O}(U))$ **on open affines $U$** to get a sheaf $\mathcal{K}$ of "meromorphic functions".1
(**ETA**: See [Georges Elencwajg's answer](https://mathoverflow.net/questions/28553/extra-principal-cartier-divisors-on-non-noetherian-rings/28566#28566) for Kleiman's article on why $Frac(\mathcal{O}(U))$ doesn't define an actual presheaf. The correct base-free way to make a presheaf $\mathcal{K}'$ is to let $S(U)$ be the elements of
$\mathcal{O}(U)$ which are "stalk-wise regular", i.e. non-zerodivisors in $\mathcal{O}\_p$ for every $p\in U$, and define
$\mathcal{K}'(U)=\mathcal{O}(U)[S(U)^{-1}]$. This agrees with the base-presheaf above on affines.)
Have you ever wondered what this sheaf does on affine opens? That's how I usually grasp what a sheaf "really is", but Hartshorne's *Algebraic Geometry* (Definition 6.11-, p. 141) doesn't tell us. The answer is non-trivial, but turns out to be nice for lots of nice rings. Q. Liu's *Algebraic Geometry and Arithmetic Curves* shows that:
1. If $A$ is Noetherian, or reduced with finitely many mimimal primes (e.g. a domain), then
$\mathcal{K}(Spec(A))=Frac(A)$. (Follows from Ch.7 Remark 1.14.)
- If $A$ is any ring, then $Frac(A)$ is a *subring* of $\mathcal{K}(Spec(A))$. (Follows from Ch.7 Lemma 1.12b.)
So for $A$ non-Noetherian, we could be getting some extra elements, and presumably, they could be units. In other words, we could have **principal Cartier divisors that don't come from $Frac(A)$**.
Is there an example where this happens?
---
**Follow-up:** Thanks to [BCnrd's proof below](https://mathoverflow.net/questions/28553/extra-principal-cartier-divisors-on-non-noetherian-rings/28591#28591), the answer is "no": even though $\mathcal{K}(Spec(A))$ *can* be strictly larger than $Frac(A)$, it can't contain additional *units*, so there are no such extra principal divisors!
---
Footnotes:
1 Here "Frac" means inverting the non-zero divisors of the ring; I'm not assuming anything is a domain.
| https://mathoverflow.net/users/84526 | Extra principal Cartier divisors on non-Noetherian rings? (answered: no!) | In the setup in the question, it should really say "we could have *invertible* meromorphic functions on Spec($A$) that don't come Frac($A)^{\times}$", since those are what give rise to "extra principal Cartier divisors". This is what I will prove cannot happen. The argument is a correction on an earlier attempt which had a bone-headed error. [Kleiman's construction from Georges' answer is *not invertible*, so no inconsistency. Kleiman makes some unfortunate typos -- his $\oplus k(Q)$ should be $\prod k(Q)$, and more seriously the $t$ at the end of his construction should be $\tau$, for example -- but not a big nuisance.]
For anyone curious about general background on meromorphic functions on arbitrary schemes, see EGA IV$\_4$, sec. 20, esp. 20.1.3, 20.1.4. (There is a little subtle error: in (20.1.3), $\Gamma(U,\mathcal{S})$ should consist of *locally* regular sections of $O\_X$; this is the issue in the Kleiman reference mentioned by Georges. The content of EGA works just fine upon making that little correction. There are more hilarious errors elsewhere in IV$\_4$, all correctable, such as fractions with infinite numerator and denominator, but that's a story for another day.) Also, 20.2.12 there is the result cited from Qing Liu's book in the setup for the question.
The first step in the proof is the observation that for any scheme $X$, the ring $M(X)$ of meromorphic functions is naturally identified with the direct limit of the modules Hom($J, O\_X)$ as $J$ varies through quasi-coherent ideals which contain a regular section of $O\_X$ Zariski-locally on $X$. Basically, such $J$ are precisely the quasi-coherent "ideals of denominators" of global meromorphic functions. This description of $M(X)$ is left to the reader as an exercise, or see section 2 of the paper "Moishezon spaces in rigid-analytic geometry" on my webpage for the solution, given there in the rigid-analytic case but by methods which are perfectly general.
Now working on Spec($A$), a global meromorphic function "is" an $A$-linear map $f:J \rightarrow A$ for an ideal $J$ that contains a non-zero-divisor Zariski-locally on $A$.
Assume $f$ is an *invertible* meromorphic function: there are finitely many $s\_i \in J$ and a finite open cover {$U\_i$} of Spec($A$) (yes, same index set) so that $s\_i$ and $f(s\_i)$ are non-zero-divisors on $U\_i$; we may and do assume each $U\_i$ is quasi-compact. Let $S$ be the non-zero-divisors in $A$. Hypotheses are preserved by $S$-localizing, and it suffices to solve after such localization (exercise). So without loss of generality each element of $A$ is either a zero-divisor or a unit. If $J=A$ then $f(x)=ax$ for some $a \in A$, so $a s\_i=f(s\_i)$ on each $U\_i$, so all $a|\_{U\_i}$ are regular, so $a$ is not a zero-divisor in $A$, so $a$ is a unit in $A$ (due to the special properties we have arranged for $A$). Hence, it suffices to show $J=A$.
Since the zero scheme $V({\rm{Ann}}(s\_i))$ is disjoint from $U\_i$ (as $s\_i|\_ {U\_i}$ is a regular section), the closed sets $V({\rm{Ann}}(s\_i))$ and $V({\rm{Ann}}(s\_2))$ have
intersection disjoint from $U\_1 \cup U\_2$. In other words, the quasi-coherent ideals ${\rm{Ann}}(s\_1)$ and ${\rm{Ann}}(s\_2)$ generate the unit ideal over $U\_1 \cup U\_2$.
A quasi-coherent sheaf is generated by global sections over any quasi-affine scheme, such as $U\_1 \cup U\_2$ (a quasi-compact open in an affine scheme), so we get $a\_1 \in {\rm{Ann}}(s\_1)$ and $a\_2 \in {\rm{Ann}}(s\_2)$ such that $a\_1 + a\_2 = 1$ on $U\_1 \cup U\_2$. Multiplying both sides by $s\_1 s\_2$, we get that $s\_1 s\_2 = 0$ on $U\_1 \cup U\_2$. But
$s\_1$ is a regular section over $U\_1$, so $s\_2|\_ {U\_1} = 0$. But $s\_2|\_ {U\_2}$ is a regular section, so we conclude that $U\_1$ and $U\_2$ are *disjoint*. This argument shows that the $U\_i$ are *pairwise disjoint*.
Thus, {$U\_i$} is a finite disjoint open cover of Spec($A$), so in fact each $U\_i = {\rm{Spec}}(A\_i)$ with $A = \prod A\_i$. But recall that in $A$ every non-unit is a zero-divisor. It follows that the same holds for each $A\_i$ (by inserting 1's in the other factor rings), so each regular section $s\_i|\_ {U\_i} \in A\_i$ is a unit. But the preceding argument likewise shows that $s\_i|\_ {U\_j} = 0$ in $A\_j$ for $j \ne i$, so each $s\_i \in A$ has a unit component along the $i$th factor and vanishing component along the other factors. Hence, the $s\_i$ generate 1, so $J = A$. QED
| 21 | https://mathoverflow.net/users/3927 | 28591 | 18,685 |
https://mathoverflow.net/questions/28590 | 11 | I derived this definition by searching for a representation of a family of sets. I am quite sure that someone should have thought to this before, because it seems to be quite straightforward given a family of sets. However, I did not find anything suitable on google or on wikipedia.
Let a family of sets, say $A\_1, \ldots, A\_n$, be given. To avoid misunderstanding I will call them *modules*. This family induces a unique partition on the union set $A = \bigcup\_{i=1}^{n} A\_i$ in the following way: I call *building block* a maximal subset $B$ of $A$ such that do not exist 2 different modules $A\_i$ and $A\_j$ with:
$B \cap A\_i \not= \emptyset$,
$B \cap A\_j \not= \emptyset$ and
$B\not\subseteq A\_i \cap A\_j$.
For example, if my family consists of 2 different overlapping modules $A\_1$ and $A\_2$, I can partition the set $A = A\_1 \cup A\_2$ as:
(elements in $A\_1 \cap A\_2$);
(elements in $A\_1$ but not in $A\_2$);
(elements in $A\_2$ but not in $A\_1$).
I know that in logic there is something similar, but I am searching for something in set theory. Moreover, I want to underline the dependency of this uniquely derived partition from the family of sets I am given.
See also <http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics#Given_a_family_of_sets.2C_I_can_generate_a_partition_of_their_union_set.2C_by_looking_at_their_overlapping>...
Thanks to all!
A newcomer
| https://mathoverflow.net/users/6882 | I am searching for the name of a partition (if it already exists) | Your building blocks are known as the *atoms* in the [Boolean algebra](http://en.wikipedia.org/wiki/Boolean_algebras) or [field of sets](http://en.wikipedia.org/wiki/Field_of_sets) generated by the $A\_i$. Each building block will consist of points that have the same pattern of answers for membership in the various $A\_i$. To see this, observe first that by maximality any building block will respect this equivalence and therefore be a union of such atoms. Conversely, if a set has points with two different patterns of answers, then it will contain points from two $A\_i$ without being contained in their intersection. (Specifically, if $x,y\in B$ and $x\in A\_i$ but $y\notin A\_i$, then pick $j$ such that $y\in A\_j$, and observe that $B$ meets both $A\_i$ and $A\_j$, buit is not contained in $A\_i\cap A\_j$.) So the building blocks are the atoms.
If your family is finite as you indicated, then the Boolean algebra generated by the $A\_i$ consists precisely of the unions of blocks. This is the representation theorem showing that every finite Boolean algebra is isomorphic to a finite power set---the power set of the atoms.
In the infinite case, however, the Boolean algebra generated by the $A\_i$ may be atomless---it may have no atoms at all, and this is a fascinating case. Nevertheless, your blocks still form a partition, and are precisely the atoms in the infinitary-generated field of sets, still determined by my argument above by the patttern-of-answers to membership in the $A\_i$.
| 19 | https://mathoverflow.net/users/1946 | 28592 | 18,686 |
https://mathoverflow.net/questions/28595 | 6 | My question is a doubt I had in the last point to the first answer to this MO question - ["Algebraic" topologies like the Zariski topology?](https://mathoverflow.net/questions/14314/algebraic-topologies-like-the-zariski-topology)
Can one associate a Riemann surface to any arbitrary field extension? The statement there says its true, it is the (equivalence class of) valuations which fix the base field. The result seems too fascinating to be true. Are there any extra hypotheses?
I am aware of the result that the valuations of C(x)/C form the Riemann sphere (counting the exponent of any (x-a) in each rational function gives a correspondence between points on the complex plane (minus infinity) and valuations.)
Any references?
| https://mathoverflow.net/users/2720 | Does there exist a Riemann surface corresponding to every field extension? Any other hypothesis needed? | Zariski introduced an abstract notion of Riemann surface associated to, for example, a finitely generated field extension $K/k$. It's a topological space whose points are equivalence classes of valuations of $K$ that are trivial on $k$, or equivalently valuation rings satisfying $k\subset R\_v\subset K$. If $A$ is a finitely generated $k$-algebra inside $K$ then those $R\_v$ which contain $A$ form an open set.
In the case of a (finitely generated and) transcendence degree 1 extension all of these valuation rings are the familiar DVRs -- local Dedekind domains -- and they serve to identify the points in the unique complete nonsingular curve with this function field. (There is also the trivial valuation with $R\_v=K$, which corresponds to the generic point of that curve.)
In higher dimensions there are lots of complete varieties to contend with -- you can keep blowing up. Also there are more possibilities for valuations. Most of the valuation rings are not Noetherian. A curve in a surface gives you a discrete valuation ring, consisting of those rational functions which can meaningfully be restricted to rational functions on the curve: those which do not have a pole there. A point on a curve on a surface gives you a valuation whose ring consists of those functions which do not have a pole all along the curve, and which when restricted to the curve do not have a pole at the given point. The value group is $\mathbb Z\times \mathbb Z$ lexicographically ordered. A point on a transcendental curve in a complex surface, or more generally a formal (power series) curve in a surface gives you a valuation by looking at the order of vanishing; the value group is a subgroup of $\mathbb R$.
This space of valuations has something of the flavor of Zariski's space of prime ideals in a ring: it is compact but not Hausdorff, for example. It can be thought of as the inverse limit, over all complete surfaces $S$ with this function field, of the space (Zariski topology) of points $S$.
| 12 | https://mathoverflow.net/users/6666 | 28600 | 18,692 |
https://mathoverflow.net/questions/24234 | 2 | For personal research, I'm doing some analysis on collected data and trying to develop relationships between two variables where the data is collected through a data logger. I'm hypothesising that a = alpha \* b for any point in my data logger.
Anyways I plotted the 5,000+ x,y points below, and the only relationship I see is a cloud or airplane wing shaped one where it is linear at the bottom and and a curved line on the top. Doing log scales didn't help either.
**EDIT** So I'm not allowed to submit image tags as a new user, but [here's the graph](http://koopics.com/ask_math_chart.jpg)
How can I improve my data analysis and is there a tool / cheat sheet that recommend a statistical analysis for a type of data.
It'll be nice if you can help find a solution; however, my goal is to learn about statistical analysis tools, that'll help me better understand the data and solve the problem.
| https://mathoverflow.net/users/152 | Statistical Data Analysis | To find the relationship between the data try this tool - to all of my knowledge this is the best one available (at least I am very excited about it)
<http://ccsl.mae.cornell.edu/eureqa>
| 2 | https://mathoverflow.net/users/1047 | 28603 | 18,695 |
https://mathoverflow.net/questions/28615 | 4 | Given a regular Tetrahedron *A* (i.e. each edge of *A* has same length), is it possible to split *A* into several smaller regular tetrahedra of equal size? I.e. smaller tetrahedra should completely fill volume of *A*, and they should not overlap.
This can be done in 2D with a triangle and square, and it can be done in 3D with cube (i.e. you can split cube into several smaller cubes of equal size). But I see no way to do same thing in 3D with tetrahedron.
If this can be done, how (how smaller tetrahedra should be positioned)?
If this cannot be done, is there a proof that this is impossible?
P.S. I'm not a mathematician, and this is not a homework, but I'd like to know how/if this can be done.
| https://mathoverflow.net/users/6883 | Tetrahedron splitting/subdivision | The answer is: No. There is a somehwat rambling discussion [here](http://answers.google.com/answers/threadview/id/497054.html). Let $B$ be a smaller tetrahedron that is jammed into the apex of $A$. It fills the solid angle there completely.
Let $e$ be a base edge of $B$. Then one cannot fill the neighborhood of $e$ by gluing in
further regular tetrahedra along it. One way to see this is that the dihedral angle of the tetrahedron is $\delta = \cos^{-1}(1/3) \approx 70.5^\circ$, and the dihedral angle along $e$ to be filled
is $\pi - \delta \approx 109.5^\circ$, which cannot be formed from copies of $\delta$.
| 6 | https://mathoverflow.net/users/6094 | 28620 | 18,707 |
https://mathoverflow.net/questions/28622 | 8 | The Zoll surfaces have the property that all of their geodesics are closed.
If one futher stipulates that all geodesics are also *simple*, i.e., non-self-intersecting,
does this leave only the sphere?
Apologies for the simplicity of this question, but I am not finding an answer in the literature,
and I suspect many just know this off the top of their head. Thanks!
| https://mathoverflow.net/users/6094 | Surfaces all of whose geodesics are both closed and simple | From Guillemin's "The Radon transform on Zoll surfaces", it follows that there are deformations of $S^2$ which keep all geodesics closed AND simple.
| 12 | https://mathoverflow.net/users/1441 | 28627 | 18,712 |
https://mathoverflow.net/questions/28647 | 54 | Is it possible to partition $\mathbb R^3$ into unit circles?
| https://mathoverflow.net/users/3375 | Is it possible to partition $\mathbb R^3$ into unit circles? | The construction is based on a well ordering of $R^3$ into the least ordinal of cardinality continuum. Let $\phi$ be that ordinal and let $R^3=\{p\_\alpha:\alpha<\phi\}$ be an enumeration of the points of space. We define a unit circle $C\_\alpha$ containing $p\_\alpha$ by transfinite recursion on $\alpha$, for some $\alpha$ we do nothing. Here is the recursion step. Assume we have reached step $\alpha$ and some circles $\{C\_\beta:\beta<\alpha\}$ have been determined. If some of them contains (=covers) $p\_\alpha$, we do nothing. Otherwise, we choose a unit circle containing $p\_\alpha$ that misses all the earlier circles. For that, we first choose a plane through $p\_\alpha$ that is distinct from the planes of the earlier circles. This is possible, as there are continuum many planes through $p\_\alpha$ and less than continuum many planes which are the planes of those earlier circles. Let $K$ be the plane chosen. The earlier circles intersect $K$ in less than continuum many points, so it suffices to find, in $K$, a unit circle going through $p\_\alpha$ which misses certain less than continuum many points. That is easy: there are continuum many unit circles in $K$ that pass through $p\_\alpha$ and each of the bad points disqualifies only 2 of them.
| 69 | https://mathoverflow.net/users/6647 | 28650 | 18,727 |
https://mathoverflow.net/questions/28649 | 5 | Let be $G=(V,E)$, where $V=\{1,\ldots,n\}$ and $E=\{\{i,j\}\subset V;|i-j|\leq k\}$ and $k<n$.
For which values of $k\geq 2$, can we count explicitly the number of Hamiltonian paths in $G$ ?
| https://mathoverflow.net/users/2386 | How many Hamiltonians Paths there are in almost regular graph ? | S. Kitaev defines Path schemes $P(n,M)$ as graphs with vertex set $\{1,2,\dots,n\}$ and edges $(i,j)$ iff $|i-j|\in M$. Hamiltonian graphs on path schemes were mentioned in "On uniquely k-determined permutations" by S. Avgustinovich and S. Kitaev. The formula is not simple even in the case where $M=\{1,2\}$ ([here](http://oeis.org/A069241)), but I guess it depends on what kind of formula you are looking for.
| 5 | https://mathoverflow.net/users/2384 | 28654 | 18,730 |
https://mathoverflow.net/questions/28656 | 23 | Every student of set theory knows that the early axiomatization of the theory
had to deal with spectacular paradoxes such as Russel's, Burali-Forti's etc.
This is why the (self-contradictory) unlimited abstraction axiom ($\lbrace x | \phi(x) \rbrace$
is a set for any formula $\phi$) was replaced by the limited abstraction
axiom ($\lbrace x \in y | \phi(x) \rbrace$ is a set for any formula $\phi$ and any
set $y$).
Now this always struck me as being guesswork ("if this axiom system does not work,
let us just toy with it until we get something that looks consistent "). Besides, it is not
the only way to counter those "set theory paradoxes" -there's also
Neumann-Bernays-Godel classes.
So my (admittedly vague) question is : is there a way to explain e.g. Russel's paradox
that does better than just saying, "if you change the axioms this paradox disappears ?" Clearly, I'm looking for an intuitive heuristic, not a technical exact answer.
EDIT June 19 : as pointed out in several answers, the view expressed above is historically false and unfair to the early axiomatizers of ZFC. The main point is that ZFC can be motivated independently from the paradoxes, and "might have been put forth even if naive set theory had been consistent" as explained in the reference by George Boolos provided in one of the answers.
| https://mathoverflow.net/users/2389 | Intuitive and/or philosophical explanation for set theory paradoxes | George Boolos has a number of very readable (to the non-expert like me) essays on this subject. Try "The Iterative Conception of Set" in [Logic, Logic and Logic](http://books.google.com/books?id=2BvlvetSrlgC&lpg=PP1&ots=sWb27wzVgM&dq=boolos%20logic%20logic%20harvard&pg=PP1#v=onepage&q=boolos%20logic%20logic%20harvard&f=false). He tries to find a way to look at the axioms of ZF set theory from a perspective that makes them look natural and not simply contrived to avoid paradoxes. I don't know anything about how Boolos's views are seen by other Set Theorists.
| 5 | https://mathoverflow.net/users/1233 | 28661 | 18,733 |
https://mathoverflow.net/questions/28673 | 0 | Here's something that I'd like to use in my thesis.. but Im feeling too lazy to write a proof of it, I feel pretty sure this is correct though. I have a feeling that this can be found in a book on category theory. So maybe someone can point me to a reference (I have only used Adamék, Herrlich and Strecker so far).
Conjecture:
Short Statement: pullback of inverse limits is the inverse limit of pullbacks
Long Statment:
Let $I$ be a directed set and let $\mathbf D,\mathbf E: I \rightarrow \mathcal C$ be two diagrams to a complete category. Let $C$ be an object in $\mathcal C$ and suppose that we have natural sinks $\mathbf Di \rightarrow C$ and $\mathbf Ei\rightarrow C$ (Mac Lane calls these "cones"). Let $A$ and $B$ be the inverse limit of $\mathbf D$ and $\mathbf E$ respectively. We get another diagram that goes to the pullback namely $\mathbf D \times\_C \mathbf E : I \rightarrow \mathcal C \times\_C \mathcal C$. The claim is that the inverse limit of $\mathbf D \times\_C \mathbf E$ is actually the fiber product (or pullback, however you want to call it) $A \times\_C B$ (where $A\rightarrow C$ and $B\rightarrow C$ are canonical map that results from the inverse limit and the natural sinks).
| https://mathoverflow.net/users/1245 | Need a reference for cones and limits that does this... | Yes -- filtered/directed colimits commute with finite limits. See Mac Lane, *Categories for the Working Mathematician*, theorem IX.2.1.
Edit: Oh, I thought you meant colimits, but it seems you meant limits. But limits always commute with each other (CWM IX.2).
| 3 | https://mathoverflow.net/users/4262 | 28675 | 18,741 |
https://mathoverflow.net/questions/28683 | 6 | The word problem (from wikipedia).
>
> Given a semi-Thue system T: = (Σ,R)
> and two words , can u be transformed
> into v by applying rules from R?
>
>
>
This problem is undecidable, but with a certain restriction, it is decidable.
The Restriction:
All the rules in R are of the form A->B where A and B are string of the same length.
What is the computational complexity of this problem?
| https://mathoverflow.net/users/6886 | Computational complexity of the word problem for semi-Thue systems with certain restrictions | The problem is at least NP-hard. Indeed, it is at least PSPACE-hard.
The reason the original semi-Thue rewrite system is undecidable is that
it reduces the halting problem. Given any Turing machine
program $e$ and input $x$, one sets up a rewrite system acting
on strings that code information about the Turing
computation. Thus, the string should display the contents
of the tape, the head position within that information and
the current state. The rewrite rules correspond to
the update procedure of the program $e$ as the computation
proceeds. For the most part, these rewrite rules are
length-preserving. If the head is in the middle of the
currently stored information of the tape, then it moves one
way or the other and the tape is updated, and changing this
information does not make the representing string any longer. One exception to this, however, is when the head moves off to
either end of the represented tape, in effect using more tape for the first time.
In this case, the rewrite transformation rules will have in effect to add an
extra symbol to represent that new cell that was just
encountered. (Finally, the rewrite rules should include some rules that propagate a halting configuration to some informative output string.)
My main observation is now that, therefore, if we know in
advance how much space the computation will require, then
we can set up the rewrite rules with length-preserving transformations, so that the exceptional case is not needed.
Specifically, suppose that we have an NP algorithm $e$ with
known polynomial bound $p$ and input $x$. Thus, $x$ is
accepted if and only if there is some $y$ such that $e$
accepts $(x,y)$, and this computation will in any event complete in time $p(|x|)$. Let $u$ be a string
representing an initial Turing machine set up with $x$ on
the input tape, $0$'s on the work tape and wildcard symbols
on the witness tape, where the length of these tapes as
represented in $u$ is $p(|x|)$. Now, produce a semi-Thue rewrite system
whose rules first of all allow the wildcard symbols to
assume any specific values (this will produce a potential
witness $y$ on the witness tape). Next, the system
also has rewrite rules as above carrying out the instructions of
program $e$ in the manner of the paragraph above. These are very local length-preserving rewrite transformation rules that
correspond to the operation of $e$, and each rule has to look at only a small portion of the represented information, since the Turing machine operation is completely local. Since we know that the
computation will end before the ends of the string are met,
we do not need the extra non-length-preserving rules that
add extra symbols corresponding to the head moving off the
represented portion of the tape, since we know this will not happen. Finally, add rules that
have the effect that if the *accept* state is realized,
then this information is simply copied to every symbol.
Thus, I claim the original input $x$ is accepted by $e$
(with respect to some unknown $y$) if and only if this semi-Thue
rewrite system transforms $u$ to the *all accept* string. Thus, I
have reduced the given NP problem to your restricted semi-Thue
problem. The reduction is polynomial time, since we can
write down the transformation rules I described above in polynomial time from
$e$ and $x$. And so your problem is at least NP-hard.
A similar argument works with PSPACE, without needing any wildcards, so it is also PSPACE hard.
It isn't clear to me, however, whether your problem is actually itself in PSPACE, since although the transformed strings themselves don't take much space, we have somehow to keep track of all possible ways to apply the rewrite rules, and this would seem naively to take exponential space. Certainly your problem is in EXPSPACE, since we could make a list of all possible strings of length |u|, and then just check off which ones are accessible from u by iteratively applying the rules until we have computed the closure of u under those rules, and finally checking if v was obtained. (This argument also shows, crudely, that your problem is at worst double exponential time.)
| 10 | https://mathoverflow.net/users/1946 | 28697 | 18,758 |
https://mathoverflow.net/questions/28462 | 20 | Question: Why do so few $n\equiv 3 \bmod 8$ have an odd number of representations in the form $$n=x\_0^2 + x\_1^2 + \dots + x\_{10}^2$$ with $x\_i \geq 0$?
Note that $x\_i\geq 0$ spoils the symmetry enough that this is not the usual "number of representations as a sum of squares" question. The question may be ill-posed, too: I do not know that there are few such $n$, but computations suggest that their counting function is $\sim cx/\log x$.
The question I'd like to answer is this: which $n$ have an odd number of representations of the form $$n=x\_0^2+2x\_1^2+4x\_2^2+ \dots = \sum\_{i=0}^\infty 2^i x\_i^2,$$ where $x\_i \geq 0$? Does the set of such $n$ have 0 density? The motivation for this question is that it is nontrivially the same as [this question](https://mathoverflow.net/questions/26839/how-thick-is-the-reciprocal-of-the-squares).
If $n$ is even and has an odd number of reps, then (nice exercise) $n$ has the form $2k^2$. If $n\equiv 1\pmod4$ and has an odd number of reps, then (challenging but elementary) $n$ has a special sort of factorization, and there are very few such $n$.
The question I'm asking here is for $n\equiv 3 \bmod 8$. Reducing modulo 8 reveals that $x\_2$ must be even. If it isn't a multiple of 4, then we can pair off the two reps:
$$(x\_0,x\_1,x\_2,x\_3,x\_4,x\_5,\dots) \leftrightarrow (x\_0,x\_1,2x\_4,x\_3,x\_2/2,x\_5,\dots).$$
If $x\_2$ is a multiple of 4 but not 8, then we can make a similar pairing with $x\_2$ and $x\_5$, and so on. This only leaves the situation when $x\_2$ and $x\_4,x\_5,\dots$ are all zero.
This leaves us at: Which $n\equiv 3\bmod 8$ have an odd number of representations in the form:
$$n = x\_0^2 + 2x\_1^2 + 8 x\_3^2 \; , \; \; \mbox{with} \; \; x\_0, x\_1, x\_3 \geq 0 \;?$$
Some play with the parities of binomial coefficients reduces this to the question I led with.
| https://mathoverflow.net/users/935 | Why are there usually an even number of representations as a sum of 11 squares | Throughout $N>0,$ and $N \equiv 3 \pmod 8.$ Let $I$ be the number of ordered triples $(a,d,e) \;\mbox{with} \; a,d,e \geq 0,$ such that
$$a^2+2 d^2+8 e^2=N.$$ I'll use a result of Gauss on sums of 3 squares to show that if there are 3 or more primes whose exponent in the prime factorization of $N$ is odd, then $I$ is even. As a consequence those $N$ for which $I$ is odd form a set of density 0; in fact the number of such $N < x$ for positive real $x$ is
$$ O \left( \frac{x \; \log \log x}{\log x} \right). $$
Let $ R = R(N) $ be the number of triples $ (a,b,c) \; \mbox{with} \; a,b,c > 0 $ and
$$ a^2+b^2+c^2=N,$$ and let $r(N)$ be the number of such
triples with the $ \gcd(a,b,c) = 1.$ Then $R$ is the sum of
the $r(N/k^2),$ the sum running over all $k>0$
for which $k^2 | N.$ Now in Disquisitiones, Gauss shows that if
$N>3, \mbox{then} \; r(N)/3$ is the number of classes (under
proper equivalence) of positive primary binary forms of discriminant $-
N.$ (Or if you prefer, the number of
classes of invertible ideals in the quadratic order of discriminant $-
N$). Now these classes form a group, and
Gauss uses genus theory to show that the order of this group is divisible by $2^{M-1}$ where $M$ is the
number of primes that divide $N.$ So if 3 or more primes have odd exponent in the prime factorization of $N,$
then all these primes divide $N/k^2,$ the corresponding group has order divisible by $2^{3-1}=4,$ so 4 divides
each $r(N/k^2),$ and 4 divides $R.$ $$ $$
Now let $S=S(N)$ be the number of pairs $(a,d) \; \mbox{with} \; a,d > 0$ and
$$a^2+2 d^2=N,$$ and $s(N)$ be the number of such pairs
with $ \gcd(a,d) = 1.$ Then $S$ is the sum of the $s(N/
k^2).$ Using the fact that
$\mathbb{Z} \left[ \sqrt{-2} \right]$ is a UFD
we can calculate $s(N/k^2);$ it is zero when some prime $p \equiv 5,7 \pmod 8$ divides $N/k^2.$ When this doesn't happen there are 3 or more
primes $q \equiv 1,3 \pmod 8$ dividing $N/k^2,$ so 4 divides each $s(N/
k^2)$ and
4 divides $S$ as well as $R.$ We conclude the proof by showing that $$2I=R+S.$$
Suppose $N \equiv 3 \pmod 8$ and $a^2+b^2+c^2=N,$ with $a,b,c>0.$ Of course $a,b, c$ are odd. If
$b \equiv c \pmod 4,$ let $d=(b+c)/2$ and $e = | (b-c)/4 |.$ Otherwise let $d = | (b-c)/2 |$
and $e=(b+c)/4.$ Then $$a^2+2 d^2+8 e^2=a^2+b^2+c^2=N.$$ Furthermore $(a,b,c)$ and $(a,c,b)$ map to the same $(a,d,e).$ The fiber of the map $(a,b,c) \mapsto (a,d,e)$ has 1 element when $e=0$ and 2 elements otherwise. So $2I=R+S.$ $$ $$
If $N = p q$ where $p$ and $q$ are primes congruent to 5 and 7 $ \pmod 8$
respectively, with $ (q | p ) = -1$ it can be
shown that $R \equiv 2 \pmod 4,$ so that $I$ is odd. This should
allow one to get a lower bound for
the number of $ N < x $ with $I$ odd that's a constant multiple of the upper
bound mentioned above. But
whether the number is asymptotic to a constant multiple of $x \; \log \log(x)/
\log (x)$ as Jagy's calculations suggest
isn't clear.
| 14 | https://mathoverflow.net/users/6214 | 28711 | 18,770 |
https://mathoverflow.net/questions/28669 | 15 | I have a real-valued function $f$ on the unit disk $D$ that is fairly well behaved (real-analytic everywhere) and would like to find the integral $\int\_D f(x,y)dxdy$ numerically. After much searching, the best methods I've been able to find have been the simple quadrature rules in Abramowitz and Stegun that sample $f$ at up to 21 points. What work has been done since? In particular, I'm interested in rules that allow sampling at more than 21 points. One reference indicates that finding optimal quadrature rules is a hard problem, but it seems to me that something better must have been published in the last 50 years.
A couple of references suggest integrating over various domains by triangulating them and using numerical integrals over the triangles. Is this the preferred method for a disk?
(I'm trying to improve code that has been implemented using quasi-monte carlo. It seems to me that we could do much better using the knowledge that $f$ is real-analytic and probably well approximated by polynomials.)
Update: I can't easily say exactly what the functions $f$ are as they're the messy result of a chain of computations. I can say that qualitatively it's like a gaussian with a central hump, fast decay, though not exact rotational symmetry. I do have a pretty good handle on how big the hump is and where it's centred. All variations on this might happen and some are easy to dispense with: eg. the hump might be situated well outside the disk so I know the integral is nearly zero. Or the hump may be very wide in which case the integrand is almost constant. Sometimes the hump is contained well within the disk in which case I can switch to more efficient quadrature over the (approximate) support of $f$ rather than the disk. But having said all that, I'd still like to see some general gaussian quadratures rules for the disk that would apply to integrating any function over the disk that is well approximated by a polynomial.
Update2: After much web searching I found some [Fortran code](http://people.sc.fsu.edu/~burkardt/f_src/stroud/stroud.html) to do what I want (and more) and a reference to a book by Arthur Stroud, *Approximate Calculation of Multiple Integrals*. It seems as that this work from 1971 is the state of the art.
| https://mathoverflow.net/users/1233 | Numerical integration over 2D disk | See the [Encyclopedia of Cubature Formulas](http://nines.cs.kuleuven.be/research/ecf/). The site is password protected, but the maintainer will give a password to anyone who asks.
| 6 | https://mathoverflow.net/users/136 | 28713 | 18,771 |
https://mathoverflow.net/questions/28717 | 25 | Has any work been done on Singmaster's conjecture since Singmaster's work?
The conjecture says there is a finite upper bound on how many times a number other than 1 can occur as a binomial coefficient.
Wikipedia's article on it, written mostly by me, says that
* It is known that infinitely many numbers appear exactly 3 times.
* It is unknown whether any number appears an odd number of times where the odd number is bigger than 3.
* It is known that infinitely many numbers appear 2 times, 4 times, and 6 times.
* One number is known to appear 8 times. No one knows whether there are any others nor whether any number appears more than 8 times.
* Singmaster reported that Paul Erdős told him the conjecture is probably true but would probably be very hard to prove.
| https://mathoverflow.net/users/6316 | Singmaster's conjecture | There is an upper bound of $O\left(\frac{(\log n)(\log \log \log n)}{(\log \log n)^3}\right)$ due to Daniel Kane: see "[Improved bounds on the number of ways of expressing *t* as a binomial coefficient](http://www.emis.de/journals/INTEGERS/papers/h53/h53.pdf)," Integers 7 (2007), #A53 for details.
| 26 | https://mathoverflow.net/users/428 | 28718 | 18,775 |
https://mathoverflow.net/questions/28593 | 2 | Let $A$ be a complex torus of (complex) dimension 2 and $X$ the associated Kummer variety $A/\sigma$, where $\sigma(x)=-x$. I would like to compute the cohomology of $X$ with $\mathbb{Z}$ coefficients. My initial instinct was to use Mayer-Vietoris, but the exact sequence involves the cohomology of the quadratic cone minus a point which is also proving to be difficult for me. My hope is that as in the case with $\mathbb{Q}$ coefficients
$$
H^1(X,\mathbb{Z})=H^3(X,\mathbb{Z})=0,\qquad H^0(X,\mathbb{Z})=H^4(X,\mathbb{Z})=\mathbb{Z}\quad\text{ and }\quad
H^2(X,\mathbb{Z})=\wedge^2H^1(A)\\
$$
Any tips as to how to compute $H^i(X,\mathbb{Z})$ or, equivalently, places to find tips in the literature would be very helpful. Thank you.
| https://mathoverflow.net/users/6310 | Betti Cohomology of singular Kummer Surface | I missed that the question concerned the singular Kummer surface (which I think
historically was what was what was called the Kummer surface but our current fixation on
non-singularity has changed that) so one needs a few more steps than Barth,
Peters, van de Ven: Compact complex surfaces (which will be my reference below).
Let $\pi\colon\tilde X\rightarrow X$ be the minimal resolution of singularities
and consider the Leray spectral sequence for $\pi$. We have $\pi\_\ast\mathbb
Z=\mathbb Z$ and $R^2\pi\_\ast\mathbb Z$ the skyscraper sheaf with one $\mathbb
Z$ at each of the 16 singular points. The Leray s.s. thus gives that
$H^i(X,\mathbb Z)=H^i(\tilde X,\mathbb Z)$ for $i\neq2,3$ and hence
$H^i(X,\mathbb Z)=\mathbb Z$ for $i=0,4$ and $H^1(X,\mathbb Z)=0$ as well as a
short exact sequence
$$
0\rightarrow H^2(X,\mathbb Z)\rightarrow H^2(\tilde X,\mathbb Z)\rightarrow
\bigoplus\_{v\in V}\mathbb Zv\rightarrow H^3(X,\mathbb Z)\rightarrow0,
$$
where $V$ is the set of singular points. Now, it is easy to see that $H^2(\tilde
X,\mathbb Z)\rightarrow \mathbb Zv$ is given by $f\mapsto \deg(f\_{E\_v})$, where
$E\_v:=\pi^{-1}(v)$. We have $\deg(f\_{E\_v})=\langle e\_v,f\rangle$, where $e\_v\in
H^2(\tilde X,\mathbb Z)$ is the fundamental class of $E\_v$. Hence, we get to
begin with that $H^2(X,\mathbb Z)$ is the orthogonal complement in $H^2(\tilde
X,\mathbb Z)$ of the $e\_v$. By Cor. 5.6 (of BPV) this can be identified with
$H^2(A,\mathbb Z)$. On the other hand, the image of $H^2(\tilde X,\mathbb Z)$ in
$\bigoplus\_{v\in V}\mathbb Zv$ contains the linear functions given by the $e\_v$
and $e\_v(v')=-2\delta\_{v,v'}$ so that we may consider the image of $H^2(\tilde
X,\mathbb Z)$ in $\bigoplus\_{v\in V}\mathbb Z/2v$. By the fact that the cup
product pairing on $H^2(\tilde
X,\mathbb Z)$ is perfect (by Poincaré duality) and by Prop. 5.5 we get that this
image is dual to the subspace of affine functions of $\bigoplus\_{v\in V}\mathbb
Z/2v$ (where $V$ is identified by the kernel of multiplication by $2$ in $A$)
and hence we get an identification of $H^3(X,\mathbb Z)$ with the dual of the
$\mathbb Z/2$-space of affine functions of $V$, in particular it has dimension
$5$.
**Remark**: It is interesting to note that while the quotient $A/\sigma$ as a
topological space does not use the complex structure of $A$ it still seems
easier to use it (in a very weak form, the blowing up only uses that a conical
neighbourhood has a certain form) as we consider the complex blow up of the
singular points. Indeed, the use of Mayer-Vietoris tried by the poser does look
more difficult (of course that would also use the local form of the singularity
but somehow in a less complex fashion).
| 5 | https://mathoverflow.net/users/4008 | 28724 | 18,780 |
https://mathoverflow.net/questions/28707 | 3 | If a set is equipped with a dense, complete order then the corresponding topological space is connected - and hence, so are its continuous images, even in unordered spaces. What happens if we remove the completeness condition on the order, and consider a general densely ordered space X?
I know that f is continuous from X to another ordered space, then the induced order on f(X) need not be dense. But is there some weaker property of densely ordered spaces which is preserved under continuous maps? If so, can such a property be generalized to unordered spaces in the same sort of way that "densely and completely ordered" generalizes to "connected"?
Edit: as discussed below, [here](https://mathoverflow.net/questions/29067) is the followup question asked by Noah, as well as another one added by me.
| https://mathoverflow.net/users/4336 | What, if anything, can be said about continuous images of densely ordered spaces? | Let me observe that Noah's excellent answer generalizes to solve the full case.
**Theorem.** Every topological space is a continuous image of a dense linear order.
Proof. Let $\kappa$ be any ordinal number. Let $P$ be the linear order $\mathbb{Q}\times\kappa$, under the lexical order, which is obtained by replacing each ordinal less than $\kappa$ with a copy of the rational order $\mathbb{Q}$. This is a dense linear order, resembling the well-known *long line*, but with rationals and with $\kappa$ instead of the unit interval and $\omega\_1$. You could call it the $\kappa$-long rational line. For each $\alpha\lt\kappa$, let $Q\_\alpha=\mathbb{Q}\times\{\alpha\}$ be the copy of $\mathbb{Q}$ surrounding $\alpha$ in $P$. This is an open set, and they partition $P$ just as in Noah's argument.
Now, following Noah, let $X$ be a discrete space with $\kappa$ many points, and let $f:P\to X$ map $Q\_\alpha$ to the $\alpha$-th point of $X$. This map is continuous, since the pre-image of every point is open. Thus, we have mapped $P$ continuously to the discrete space of size $\kappa$, and by composition, we can now map $P$ to any space of size $\kappa$. QED
The argument shows that every infinite topological space is a continuous image of a dense linear order of the same size.
Note that this proof uses the Axiom of Choice. But full AC doesn't seem required, since we didn't really need that $X$ was well-orderable, but rather only that it was linearly orderable. That is, if $X$ is a linearly orderable set, then by replacing each point with a copy of the rationals, we get a dense linear order, with a locally constant map back to $X$. So the discrete topology on $X$ is the continuous image of a dense linear order, and therefore any topology on $X$ is the continuous image of a dense linear order. (The assertion that every set admits a linear order is a weak choice principle.)
| 8 | https://mathoverflow.net/users/1946 | 28751 | 18,798 |
https://mathoverflow.net/questions/28757 | 3 | Is there standard notation for
(1) exterior algebras
(2) free graded commutative algebras
(3) divided polynomial algebras ?
I've seen (and used) $\Lambda$, $\Gamma$, $\Delta$ etc. used for some or all of these things, and I have no idea if there is a consensus about which notation goes with which algebra.
| https://mathoverflow.net/users/3634 | Notation for algebras | It's pretty standard to use $\bigwedge(V)$ or $\Lambda(V)$ for the exterior
algebra on a vector space $V$ and $\bigwedge^k(V)$ or $\Lambda^k(V)$
for the $k$-th graded part. For symmetric algebras $S(V)$ or $\mathrm{Sym}(V)$
etc. are frequent notations with again $S^k(V)$ or $\mathrm{Sym}^k(V)$
for the graded parts. I wouldn't say divided polynomial algebras come up often enough
to have a standard notation; you could probably use the above notations for
exterior or symmetric algebras without further comment and expect to be understood,
but whatever notation you choose for divided polynomial algebras, you'd probably
have to explain it.
| 5 | https://mathoverflow.net/users/4213 | 28759 | 18,803 |
https://mathoverflow.net/questions/28743 | 5 | After getting stuck with the
[previous positivity](https://mathoverflow.net/questions/28374)
(it probably sounds too [complex](https://mathoverflow.net/questions/28612/do-names-given-to-math-concepts-have-a-role-in-common-mistakes-by-students/28614#28614)),
I would like to give a version of the problem which is of most interest to me.
Consider a sequence of real numbers
$a\_1,a\_2,\dots,a\_n,\dots$ with absolute values bounded above by the first term $a\_1=a>0$,
which satisfies, for all $n=1,2,\dots$,
$$
|A\_n|\le A \qquad\text{where}\quad A\_n=a\_1+a\_2+\dots+a\_n.
$$
In addition, assume that *infinitely many terms of the sequence are nonzero*.
These settings and [Dirichlet's convergence test](http://en.wikipedia.org/wiki/Dirichlet%27s_test)
guarantee that the series
$$
\sum\_{n=1}^\infty\frac{a\_n}n
$$
converges.
Assume, in addition, that
$$
\max\_{1\le k\le n}A\_k+\min\_{1\le k\le n}A\_k\ge0
\qquad\text{for all}\quad n=1,2,\dots.
$$
The problem is to show that
$$
\sum\_{n=1}^\infty\frac{a\_n}n>0
$$
and to provide, in terms of $a$ and $A$, a lower (strictly positive) bound for the series.
(The latter is optional, as I am not sure that such a bound exists.)
| https://mathoverflow.net/users/4953 | A plausible positivity | The sum $\sum a\_n/n$ can be negative. Below I construct a finite sequence; one can always add a negligibly small tail to get infinitely many non-zeroes.
Begin with $a\_1=1$ and $a\_2=-1$.
This gives $A\_2=0$ and the partial sum of the main series is $1-1/2=1/2$.
Then, repeat 100 times the following procedure:
Pick an integer $k$ larger than the length of the sequence so far.
Extend $(a\_n)$ by zeroes up to $n=10k-1$.
Then set $a\_n=1$ for all $n$ from $10k$ to $11k-1$ and $a\_n=-1$ for all $n$ from $11k$ to $13k-1$. The $k$ ones contribute less than $1/10k$ each to the main series $\sum a\_n/n$, and this is less than $1/10$ in total. The $2k$ negative ones contribute absolute value at least $1/13k$ each, this sums up to at least $2/13$ of negative amount. So the partial sum of the main series went down by at least $2/13-1/10>1/20$.
But we have $A\_n=-k$ now (for $n=13k-1$). To fix this, extend $(a\_n)$ by a huge amount of zeroes, followed by $k$ ones, so that the contribution of these ones to the main sum is less than $1/100$.
Now we extended the sequence so that the last $A\_n$ is zero again but the partial sum of the main series went down by at least $1/30$. Choose the next $k$ and repeat (finitely many times!).
[Edit] Since the sequence is finite, one can define $A=\max|A\_n|$ to satisfy the condition $|A\_n|\le A$. The $\max+\min$ condition is immediate from the construction.
| 8 | https://mathoverflow.net/users/4354 | 28760 | 18,804 |
https://mathoverflow.net/questions/28766 | 2 | Would anybody be able to share a Mathematica/Matlab/other script for calculating Onsager's exact solution for the magnetisation of the 2d Ising model? I would be most grateful of one in order to test my MC simulations of the system.
| https://mathoverflow.net/users/6915 | Mathematica/Matlab/other for calculating Onsager's exact solution to the 2d Ising model | Not that it's directly relevant, but I have code for the generator matrix of a 1D Glauber-Ising model that could probably be reworked into 2D...
---
```
function y = glauber1d(symb,n,varargin);
% produces the generator matrix etc for a 1D Glauber-Ising model of n spins
% call as either glauber1d(1,n) for a less complete symbolic result, or
% glauber1d(0,5,[a,mu,H,kT,J]) for a more complete numerical result--i.e.,
% symb is a flag indicating whether or not to use symbolic calculations
% (this requires the symbolic toolbox in order to work)
% a (Glauber's alpha) is the spin flip rate, depends on the coupling
% between the GI system and the bath;
% mu is the magnetic moment associated with the spins;
% H is the magnetic field strength;
% kT is (well, you know);
% J is the exchange energy
if symb % SYMBOLICS
syms a b g real;
else % NUMERICS
args = varargin{1};
a = args(1);
mu = args(2);
H = args(3);
kT = args(4);
J = args(5);
b = tanh(mu*H/kT); % Glauber's beta (NOT 1/kT)
g = tanh(2*J/kT); % Glauber's gamma
end
% produce an array with rows equal to spin configurations
temp = dec2bin(0:((2^n)-1),n);
for j = 1:2^n
for k = 1:n
s(j,k) = 2*str2num(temp(j,k))-1;
end
end
% obtain spin flip rates
for j = 1:2^n
for k = 1:n
km = mod(k-2,n)+1;
kp = mod(k,n)+1;
temp = (g/2) * (b - s(j,k)) * (s(j,km) + s(j,kp));
w(j,k) = (a/2) * (1 - b*s(j,k) + temp);
end
end
% generator matrix
if symb
Q = sym(zeros(2^n));
else
Q = zeros(2^n);
end
for j1 = 1:2^n
for j2 = 1:2^n
if sum(abs( s(j1,:) - s(j2,:) )) == 2 % single spin flip
% now find out which spin gets flipped
k0 = find( s(j1,:) - s(j2,:) );
Q(j1,j2) = w(j1,k0);
end
end
end
if symb
Q = simplify( Q - diag(sum(Q,2)) );
else
Q = Q - diag(sum(Q,2));
end
% invariant distribution p (if you want it)
if 2^n - 1 - rank(Q)
'error'
y = 0;
return;
else
p0 = null(Q')';
end
if symb, simplify(p0); end
sp0 = sum(p0);
if symb, simplify(sp0); end
p = p0 / sp0; % invariant distribution
y = Q;
```
| 4 | https://mathoverflow.net/users/1847 | 28768 | 18,808 |
https://mathoverflow.net/questions/28779 | 1 | Hi,
I would like to do this:
```
fit <- test( measured_values, fitted_values )
```
Where:
* the return value from the `test` function is: **0 < fit < 1**.
* **measured\_values** are the observed data.
* **fitted\_values** are the data for the curve produced by GAM for the **measured\_values**.
What test can I use to compare the data sets that will result in a number between 0 and 1, where 0 indicates the measured values and the fitted values are not a good fit and 1 indicates the fitted values fit perfectly to the measured values?
For example, consider the data plotted here: <https://i.stack.imgur.com/wcrey.jpg>
The fit, produced by GAM, is fairly close to ideal. However, the standard correlations (shown in the bottom left) do not accurately indicate the goodness of fit.
Thank you.
| https://mathoverflow.net/users/5908 | Goodness-of-fit test for Generalized Additive Model | I meant only what is usually meant and you're being completely cryptic.
A wikipedia article titled "generalized additive model" says you've got
$$
g(\operatorname{E}(Y))=\beta\_0 + f\_1(x\_1) + f\_2(x\_2)+ \cdots + f\_m(x\_m)
$$
and then says "The functions $f\_i(x\_i)$ may be fit using parametric or non-parametric means, thus providing the potential for better fits to data than other methods. The method hence is very general". In other words, there's a lot you haven't said!! And then "a typical GAM might use a scatterplot smoothing function such as a locally weighted mean for $f\_1(x\_1)$, and then use a factor model for $f\_2(x\_2)$". So again, you're not telling us how you got this fit? E.g. are you using least squares? Or maximum likelihood? Are you picking the function $f\_1$ from some parametrized or otherwise well-behave class of functions? If so, WHICH ONE? And what function are you using for $g$? You're making me guess what you mean. Your graph looked as if your value of $m$ was probably 1. Which makes me begin to wonder why you're calling it "additive". I'm guessing the graph you linked to was not brought down from Heaven by an archangel. Can you say how you got it?
| 2 | https://mathoverflow.net/users/6316 | 28786 | 18,820 |
https://mathoverflow.net/questions/28788 | 147 | A while back I saw posted on someone's office door a statement attributed to some famous person, saying that it is an instance of the callousness of youth to think that a theorem is trivial because its proof is trivial.
I don't remember who said that, and the person whose door it was posted on didn't remember either.
This leads to two questions:
1. Who was it? And where do I find it in print—something citable? (Let's call that one question.)
2. What are examples of nontrivial theorems whose proofs are trivial? Here's a wild guess: let's say for example a theorem of Euclidean geometry has a trivial proof but doesn't hold in non-Euclidean spaces and its holding or not in a particular space has far-reaching consequences not all of which will be understood within the next 200 years. Could that be an example of what this was about? Or am I just missing the point?
| https://mathoverflow.net/users/6316 | Nontrivial theorems with trivial proofs | Bertrand Russell proved that the general set-formation principle known as the Comprehension Principle, which asserts that for any property $\varphi$ one may form the set $\lbrace\ x \mid \varphi(x)\ \rbrace$ of all objects having that property, is simply inconsistent.
This theorem, also known as the [Russell Paradox](https://en.wikipedia.org/wiki/Russell%27s_paradox), was certainly not obvious at the time, as Frege was famously completing his major treatise on the foundation of mathematics, based principally on what we now call naive set theory, using the Comprehension Principle. It is Russell's theorem that showed that this naive set theory is contradictory.
Nevertheless, the proof of Russell's theorem is trivial: Let $R$ be the set of all sets $x$ such that $x\notin x$. Thus, $R\in R$ if and only if $R\notin R$, a contradiction.
So the proof is trivial, but the theorem was shocking and led to a variety of developments in the foundations of mathematics, from which ultimately the modern ZFC conceptions arose. Frege abandoned his work in this area.
| 145 | https://mathoverflow.net/users/1946 | 28791 | 18,824 |
https://mathoverflow.net/questions/28784 | 8 | Let $A$ be an algebra over an algebraically closed field $k.$ Recall that if $A$ is
a finitely generated module over its center, and if its center is a finitely generated
algebra over $k,$ then by the Schur's lemma all simple $A$-modules are finite dimensional
over $k.$
Motivated by the above, I would like an example of a $k$-algebra $A,$ such that:
1) $A$
has a simple module of infinitie dimension over $k,$
2) $A$ contains
a commutative finitely generated subalgebra over which $A$ is a finitely generated
left and right module.
Thanks in advance.
| https://mathoverflow.net/users/6277 | Example of an algebra finite over a commutative subalgebra with infinite dimensional simple modules | Doc, this is a stinker. Your condition (2) forces your algebra to be finitely generated PI, and every little hare knows that simple modules over such algebras are finite-dimensional. See 13.4.9 and 13.10.3 of McConnell-Robson...
| 5 | https://mathoverflow.net/users/5301 | 28816 | 18,842 |
https://mathoverflow.net/questions/28610 | 18 | Suppose we have a ($n-1$ dimensional) unit sphere centered at the origin: $$ \sum\_{i=1}^{n}{x\_i}^2 = 1$$
Given some some $d \in [0,1]$, what is the probability that a randomly selected point on the sphere, $ (x\_1,x\_2,x\_3,...,x\_n)$, has coordinates such that $$|x\_i| \leq d$$ for all $i$?
This is equivalent to finding the intersection of the $(n-1)$-hypersphere with the $n$-hypercube of side $2d$ centered at origin, and then taking the ratio of that $(n-1)$-volume over the $(n-1)$-volume of the $(n-1)$-hypersphere.
As there are closed-form formulas for the volume of a hypersphere, the problem reduces to finding the $(n-1)$-volume of the aforementioned intersection.
All attempts I've made to solve this intersection problem have led me to a series of nested integrals, where one or both limits of each integral depend on the coordinate outside that integral, and I know of no way to evaluate it. For example, using [hyperspherical coordinates](https://en.wikipedia.org/wiki/N-sphere#Hyperspherical_coordinates), I have obtained the following integral: $$ 2^n n! \int\_{\phi\_{n-1}=\tan^{-1}\frac{\sqrt{1-(n-1)d^2}}{d}}^{\tan^{-1}1} \int\_{\phi\_{n-2}=\tan^{-1}\frac{\sqrt{1-(n-2)d^2}}{d}}^{\tan^{-1}\frac{1}{\cos\phi\_{n-1}}}\ldots\int\_{\phi\_1=\tan^{-1}\frac{\sqrt{1-d^2}}{d}}^{\tan^{-1}\frac{1}{\cos\phi\_2}} d\_{S^{n-1}}V$$
where
$$d\_{S^{n-1}}V = \sin^{n-2}(\phi\_1)\sin^{n-3}(\phi\_2)\cdots \sin(\phi\_{n-2})\ d\phi\_1 \ d\phi\_2\ldots d\phi\_{n-1}$$ is the volume element of the $(n-1)$–sphere. But this is pretty useless as I can see no way of integrating this accurately for high dimensions (in the thousands, say).
Using cartesian coordinates, the problem can be restated as evaluating: $$ \int\_{\sum\_{i=1}^{n-1}{x\_i}^2\leq1, |x\_i| \leq d} \frac{1}{\sqrt{1-\sum\_{i=1}^{n-1}{x\_i}^2}}dx\_1 dx\_2 \ldots dx\_{n-1}$$ which, as far as I know, is un-integrable.
I would greatly appreciate any attempt at estimating this probability (giving an upper bound, say) and how it depends on $n$ and $d$. Or, given a particular probability and fixed $d$, to find $n$ which satisfies that probability.
Edit: This question leads to two questions that are slightly more general:
1. I think part of the difficulty is that neither spherical nor Cartesian coordinates work very well for this problem, because we're trying to find the intersection between a region that is best expressed in spherical coordinates (the sphere) and another that is best expressed in Cartesian coordinates (the cube). Are there other problems that are similar to this? And how are their solutions usually formulated?
2. Also, the problem with the integral is that the limits of each of the nested integrals is a function of the "outer" variable. Is there any general method of solving these kinds of integrals?
| https://mathoverflow.net/users/5768 | Probability of a point on a unit sphere lying within a cube | Denote the median of $\max\_{i=1,\dots,n}|x\_i|$ on the sphere by $M\_n$. It is known that the ratio between $M\_n$ and $\sqrt{\log n/n}$ is universally bounded and bounded away from zero. If you take $d=M\_n$ then the quantity you are looking for is exactly $1/2$. It is also known that the $\infty$-norm ($\max\_{i=1,\dots,n}|x\_i|$) is ``well concentrated" on the sphere meaning in particular that for any $\epsilon>0$, if you take $d<(1-\epsilon)M\_n$ the probability you're interested in tends to zero and if you take $d>(1+\epsilon)M\_n$, it tends to 1. Quite precise estimates are known.
The way these things are estimated is by relating the uniform distribution on the sphere to the distribution of a standard Gaussian vector:
If $g\_1,\dots,g\_n$ are independent standard Gaussian variables then the distribution of
$$
\frac{(g\_1,\dots,g\_n)}{(\sum g\_i^2)^{1/2}}
$$ is equal to the uniform distribution on the sphere. So the quantity you're looking for is the probability that $\max\_{i=1,\dots,n}|g\_i|\le d (\sum g\_i^2)^{1/2}$.
Since $(\sum g\_i^2)^{1/2}$ is very well concentrated near the constant $\sqrt n$, this is asymptotically the same as the probability that $\max\_{i=1,\dots,n}|g\_i|\le d \sqrt n$.
For general reference in a much more general setting you can look here: Milman, Schechtman, Asymptotic theory of finite-dimensional normed spaces. For a finer treatment of the special case of the $\infty$-norm, look here:
G. Schechtman: [The random version of Dvoretzky's theorem in $\ell\_n^\infty$](http://www.weizmann.ac.il/math/gideon/sites/math.gideon/files/uploads/ranDv.pdf).
| 16 | https://mathoverflow.net/users/6921 | 28818 | 18,843 |
https://mathoverflow.net/questions/28736 | 8 | Let G and H be locally compact groups, and let $\theta:G\rightarrow H$ be a continuous group homomorphism. This induces a \*-homomorphism $\pi:C^b(H) \rightarrow C^b(G)$ between the spaces of bounded continuous functions on H and G.
If $\theta$ is an injection with closed range, then as locally compact groups are normal, you can use the Tietze extension theorem to show that $\pi$ is a surjection.
Conversely, if $\pi$ surjects, then $\theta$ must be an injection. Need $\theta(G)$ be closed in H??
(If G and H are just locally compact spaces, and $\theta$ just a continuous map, then no: you could let G be non-compact and $H=\beta G$ the Stone-Cech compactification, with $\theta$ being the canonical inclusion. The resulting map $\pi$ is just $C^b(H) = C(\beta G) \rightarrow C^b(G) = C(\beta G)$, which is the identity, once suitably interpreted. Of course, here $\theta$ has open range, and in a topological group, an open subgroup is closed, so maybe there's hope... hence my question).
**More thoughts:** As in my comment, we can extend $\theta$ to a map $\tilde\theta:\beta G\rightarrow\beta H$ between the Stone-Cech compactifications: this induces the map $\pi:C(\beta H)\rightarrow C(\beta G)$. As these are compact, it follows that $\pi$ is surjective if and only if $\tilde\theta$ is injective. By replacing $H$ with the closure of $\theta(G)$, we may suppose that $\theta$ has dense range: this forces $\tilde\theta$ to be a bijection, and hence a homeomorphism. So is it possible for $\theta$ to be an injection with dense range, and $\tilde\theta$ a homeomorphism, but without $\theta$ being onto? For example, certainly H cannot be compact, as then $\beta G$ would be a topological group, which is possible only if $G$ is compact (I think).
| https://mathoverflow.net/users/406 | Group homomorphisms and maps between function spaces | The answer is yes, at least if the group $G$ is metrizable $\iff$ $G$ is Hausdorff and has countable basis of neighborhoods of the identity element $e$. This follows from the following general statement.
>
> **Proposition.** Let $G$ and $G'$ be topological groups, with $G$ locally compact and metrizable and $f:G\to G'$ be a continuous homomorphism such that
>
> (a) $f$ is a bijection; and
>
> (b) the induced map $f^\*:C^b(G')\to C^b(G)$ of the spaces of bounded continuous functions is surjective.
>
> Then $f$ is a homeomorphism.
>
Let $G'=\theta(G)\subset H$ with the subspace topology, $f$ be the same map as $\theta$, but with codomain $G'$. Then $G'$ is also locally compact, therefore, it is closed in $H.$
**Proof.** Let $d:G\to\mathbb{R}$ be the distance to $e$. Without loss of generality, $d$ may be assumed to be bounded. Consider the function $d':G'\to \mathbb{R}, d'(y)=d(f^{-1}(y)).$ Then
(1) $f^{\*}d'=d$;
(2) by (a), $f^{\*}$ is injective, so $d'$ is the only pre-image of $d$ under $f^\*$; and
(3) by (b), $d'$ is continuous.
The open ball in $B(e,r)\subset G$ consists of all $x\in G$ such that $d(x)<r$, so $$f(B(e,r))=\{y\in G':d(f^{-1}(y))<r\}=d'^{-1}((-\infty,r))$$
is open in $G'$ by (3). Since open balls form a neighborhood basis of $e$, the map $f$ is a homeomorphism. $\square$
| 2 | https://mathoverflow.net/users/5740 | 28820 | 18,845 |
https://mathoverflow.net/questions/28800 | 2 | It seems that the ratio of those authors allowing a field to be a Dedekind domain to those who do not is almost 50 - 50. Why such a bewildering lack of consensus for such an elementary notion?
| https://mathoverflow.net/users/5292 | 0 dimensional Dedekind domain? | Historically Emmy Noether's paper introducing the concept "Dedekind domain" certainly included fields (see e.g. Kleiner's book on the history of abstract algebra, which gives axioms). She was characterising the scope of unique factorisation into prime ideals. Now, a question that might actually be answered is "what happened after the late 1920s in commutative algebra to change this?" To which there is a fairly clear answer, implied by Goodwillie's comment: geometric concepts now mean more than those derived from algebraic number theory. "Dimension 1" seems more a propos or right-thinking than "dimension at most 1".
| 5 | https://mathoverflow.net/users/6153 | 28840 | 18,858 |
https://mathoverflow.net/questions/28843 | 7 | Suppose I know what the category of free algebras for a particular monad look like. Can I then describe what the category of Eilenberg–Moore algebras look like? E.g. Suppose that I have a good handle on a category $D$ and that I have two functors $F$ and $G$, with $G:D \to C$ and $F$ left-adjoint to $G$ and $F$ essentially surjective- this guarantees that this adjunction exhibits $D$ as the Kleisli-category for the monad $T:=GF$ on $C$ (see [Characterization of Kleisli adjunctions](https://mathoverflow.net/questions/26075/characterization-of-kleisli-adjunctions)). Is there a way to exhibit the Eilenberg–Moore category $C^T$ as "generalized objects of $D$"? This is somewhat vague of a question I realize, but I'm not sure how to make it more precise. (Feel free to help me do so).
EDIT: To be slightly more specific, in "Toposes, Triples, and Theories" it is said that the Eilenberg-Moore category "is in efect all the possible quotients of objects in Kleisli's category". Can anyone make this precise in a way that answers my question??
| https://mathoverflow.net/users/4528 | Eilenberg–Moore algebras in terms of Kleisli ones | One nice result is Street's theorem 14 in *The formal theory of monads*, generalized in *Elementary cosmoi*, which says that $C^T$ is isomorphic to the full subcategory of $[(C\_T)^{\mathrm{op}}, \mathrm{Set}]$ containing those presheaves that become representable when precomposed with the inclusion $C \to C\_T$. That is, $C^T$ is the pullback of $[F^{\mathrm{op}},\mathrm{Set}]$ along the Yoneda embedding. So at least informally you can think of T-algebras as being represented by certain [formal colimits](http://ncatlab.org/nlab/show/free+cocompletion) of free algebras.
I think it's standard, too, that every algebra has a canonical presentation -- its (2-truncated) bar resolution -- as the coequalizer of maps between free objects. Can't think of a good reference at the moment, but Mac Lane has a section on the bar construction, and there's some good stuff on nlab. Maybe try Kelly & Power, *Adjunctions whose counits are coequalizers, and presentations of finitary enriched monads*, JPAA 89 (1993).
I don't know if there's a relationship between these two ideas, but I'd be very interested to find out! (Can't do it myself, I'm supposed to be on holiday.)
| 11 | https://mathoverflow.net/users/4262 | 28858 | 18,868 |
https://mathoverflow.net/questions/28826 | 57 | While reading the answer to another Mathoverflow question, which mentioned the Poisson summation formula, I felt a question of my own coming on. This is something I've wanted to know for a long time. In fact, I've even asked people, who have probably given me perfectly good answers, but somehow their answers have never stuck in my brain. The question is simple: the Poisson summation formula is incredibly useful to many people, but why is that? When you first see it, it looks like a piece of magic, but then suddenly you start spotting that people keep saying "By Poisson summation" and expecting you to fill in the details. In that respect, it's a bit like the phrase "By compactness," but the important difference for me is that I *can* fill in the details of compactness arguments.
What I would like to know is this. What is the "trigger" that makes people think, "Ah, Poisson summation should be useful here"? And is there some very simple example of how it is applied, with the property that once you understand that example, you basically understand how to apply it in general? (Perhaps two or three examples are needed -- that would obviously be OK too.) And can one give a general description of the circumstances where it is useful? (Anyone familiar with the Tricki will see that I am basically asking for a Tricki article on the formula. But I don't mind something incomplete or less polished.)
For reference, here is a related (but different) question about the Poisson summation formula: [Truth of the Poisson summation formula](https://mathoverflow.net/questions/14568/truth-of-the-poisson-summation-formula)
| https://mathoverflow.net/users/1459 | How does one use the Poisson summation formula? | The existing answers list some important situations where Poisson Summation plays a role, the application to proving the functional equation of $\theta$ and hence of $\zeta$ being my personal favourite. My best answer to Tim's question as he actually asked it might be: why not have it in mind to try using it whenever you have a discrete sum that you are having trouble estimating, especially if you fancy your chances of understanding the Fourier transform of the summands. You'll end up with a different sum and it might be a lot easier to understand, and you might even be able to approximate your first sum by an integral (the term $\hat{f}(0)$ in the Poisson summation formula).
To explain a little more with an example, there's a whole theory concerned with the estimation of exponential sums $\sum\_{n \leq N} e^{2\pi i \phi(n)}$. There are two processes called A and B that can be used to turn a sum like this into something you might be better positioned to understand. Process A is basically Weyl/van der Corput differencing (Cauchy-Schwarz) and process B is essentially Poisson summation. It's not a very straightforward task to put together a theory of how these processes interact, and how they may best be combined to estimate your sum, and in fact this is in general something of an art. The *10 lectures* book by Montgomery contains a nice exposition, and there's a whole LMS lecture note volume by Graham and Kolesnik if you want more details.
I want to share a perhaps slightly obscure paper of Roberts and Sargos (Three-dimensional exponential sums with monomials, Journal fur die reine und angewandte Mathematik (Crelle) 591), in which they use Poisson Summation in the form of Process B mentioned above *arbitrarily many times* to establish the following rather simple-to-state result: the number of quadruples $x\_1,x\_2,x\_3,x\_4$ in $[X, 2X)$ with
$$|1/x\_1 + 1/x\_2 - 1/x\_3 - 1/x\_4| \leq 1/X^3$$
is $X^{2 + o(1)}$. In other words, the quantities $1/a + 1/b$ tend to avoid one another to pretty much the same extent as random numbers of the same size. Very very roughly speaking (I don't really understand the argument in depth) the proof involves looking at exponential sums $\sum\_x e^{2\pi i m/x}$, and it is these that are transformed repeatedly using Poisson summation followed by other modifications (it being reasonably pointless to try and apply Poisson sum twice in succession).
| 33 | https://mathoverflow.net/users/5575 | 28867 | 18,874 |
https://mathoverflow.net/questions/28850 | 1 | It seems an easy problem but I couldn't prove it.
Let $M$ be a manifold with boundary and $N\subset \mathrm{int}(M)$ is a strong deformation retract of $M$.
Then I wonder whether $M-N$ is homotopic equivalent to the boundary of $M$ or not.
| https://mathoverflow.net/users/6569 | Is the complement of a strong deformation retract of a manifold M homotopic equivalent with the boundary of M? | Take a compact contactible manifold $C$ whose boundary is not a homotopy sphere sphere (those are no so easy to construct, but it has been done long ago). Then removing a point (or equivalently a small ball) from the interior of $C$ gives a manifold that is homotopy equivalent (by excison in homology) to the boundary of the small ball, which is a sphere. This is a desired counterexample.
EDIT: References to how to built contactible manifolds can by found e.g. [in my paper pp.8-9.](http://arxiv.org/abs/math/0302221).
| 6 | https://mathoverflow.net/users/1573 | 28870 | 18,876 |
https://mathoverflow.net/questions/28865 | 6 | In Paul Halmos' Measure Theory book, section 53, he defines a content on a locally compact Hausdorff space to be a set function, $\lambda$ that is additive, subadditive, monotone, and $0\le\lambda(C)<\infty$ for all $C$ compact. The "Borel sets" he considers(section 52) in this book are the smallest $\sigma$-ring generated by the class of **compact sets**. The difference between $\sigma$-ring and $\sigma$-field is that the superset need not be contained. It can be shown, and is given as an exercise by Halmos, that the smallest $\sigma$-ring generated by compact sets is the same as the smallest $\sigma$-ring generated by open **bounded** sets. A bounded set is one that is contained in a compact set. Halmos then shows that a content induces a **Borel Regular** measure. He introduces the inner content and an outer measure in order to do so. Regular is the important word for this question. In fact, later in this chapter he proves the Riesz Representation Theorem, proving the existence of a **Regular** measure induced from positive linear functional of the continuous functions of compact support. In the next chapter, Halmos provides an existence proof of the Haar Measure by creating a left-invariant content, which then induces a Regular measure.
The issue I am having is that in almost every other source(books, wikipedia), the Haar Measure --and the Reisz-Rep measure which is related because one can prove the existence via a left or right invariant functional--is a Radon measure. A Radon measure is outer regular on Borel sets, finite on compact sets, and inner regular on open sets and Borel sets with finite measure. **I do realize that in the other books, the Borel sets are the smallest $\sigma$-algebra generated by open sets.**
The questions I have are:
1) Is the Borel $\sigma$-ring generated by compact sets in a locally compact Hausdorff space the same as the Borel $\sigma$-field generated by the open sets. I think the answer is no. One reason is a $\sigma$-ring and and the other is a $\sigma$-field. I don't even think that the $\sigma$-ring generated by compact sets is the same as the $\sigma$-ring generated by open sets. I think the best you can say is bounded open sets(mentioned above). Perhaps they are the same when the space is separable. Is this true? Is the $\sigma$-algebra generated by compact sets the same as the $\sigma$-algebra generated by the open sets the same.
2)Although this questions is more about basic measure theory than the Haar measure, in the end I want to show that the measure Halmos created is the same as the one constructed in most other books. In order to do this, I need to go from a measure defined on the $\sigma$-ring generated on compact sets to the $\sigma$-field generated by open sets. How do I do this?
Thank you in advance for all the help.
| https://mathoverflow.net/users/2048 | Haar Measure Existence/A problem with Borel sets and regularity. | The situation is indeed a delicate one and one needs to carefully check the conventions before transferring a result from one context to another. The situation is summarized in Royden's *Real Analysis*, though with just a few examples.
For a given space $X$, the main players are:
* $\mathcal{B}a$ — the σ-algebra generated by the compact Gδ sets. (The Baire algebra.)
* $\mathcal{B}c$ — the smallest σ-algebra with respect to which all continuous real-valued functions are measurable.
* $\mathcal{B}k$ — the σ-algebra generated by the compact sets.
* $\mathcal{B}o$ — the σ-algebra generated by the closed sets. (The Borel algebra.)
* $\mathcal{S}$ — the smallest σ-ring containing the compact sets.
* $\mathcal{R}$ — the smallest σ-ring containing the compact Gδ sets.
In general, we have the inclusions
$$\mathcal{B}a \subseteq \mathcal{B}k \subseteq \mathcal{B}o\quad\mbox{and}\quad\mathcal{B}a \subseteq \mathcal{B}c \subseteq \mathcal{B}o,$$
but $\mathcal{B}c$ and $\mathcal{B}k$ are not necessarily related. Moreover, $\mathcal{B}a = \mathcal{B}c \cap \mathcal{B}k$ and $\mathcal{B}o$ is generated by $\mathcal{B}c \cup \mathcal{B}k$. The σ-rings $\mathcal{S}$ and $\mathcal{R}$ consist of the σ-bounded elements of $\mathcal{B}a$ and $\mathcal{B}o$, respectively.
* When $X$ is σ-compact and locally compact, then
$$\mathcal{R} = \mathcal{B}a = \mathcal{B}c
\quad\mbox{and}\quad\mathcal{S} = \mathcal{B}k = \mathcal{B}o.$$
A compact example showing the strict inequality is $\beta\mathbb{N}$.
* When $X$ is metrizable, more generally when closed sets are Gδ, we have
$$\mathcal{R} = \mathcal{S} \subseteq \mathcal{B}a = \mathcal{B}k \subseteq \mathcal{B}c = \mathcal{B}o.$$
An example showing strict inequality is the space of irrational numbers.
* When $X$ is locally compact and separable, then we have
$$\mathcal{R} = \mathcal{S} = \mathcal{B}a = \mathcal{B}k = \mathcal{B}c = \mathcal{B}o.$$
That said, the uniqueness (up to normalization) of Haar measure and it's (inner) regularity guarantee that all constructions will agree on their common domain of definition. Proving this from scratch does require some work depending on where you start and end.
| 4 | https://mathoverflow.net/users/2000 | 28874 | 18,879 |
https://mathoverflow.net/questions/28856 | 3 | What is the motivic cohomology $H^{p,q}(\mathbf{P}^n,\mathbf{Z})$ of projective space? By the projective bundle formula, one has
$H^{p,q}(\mathbf{P}^n,\mathbf{Z})$ =
$\oplus\_{i=0}^n\mathrm{Hom}\_\mathbf{DM}(\mathbf{Z}(i)[2i],\mathbf{Z}(q)[p])$
=
$\oplus\_{i=0}^n\mathrm{Hom}\_\mathbf{DM}(\mathbf{Z}, \mathbf{Z}(q-i)[p-2i])$
Is this equal to $\mathbf{Z}$ for (p,q) = (i,2i), $i=0, \ldots, n$ and $0$ else?
| https://mathoverflow.net/users/nan | The motivic cohomology of projective space | Motivic cohomology is an absolute invariant not a geometric one. The projective bundle formula is purely geometric. It reduces the computation of the motivic cohomology of the projective space to that of the base:
$$
H^{p,q}(\mathbb{P}^n\_k) = \bigoplus\_{i=0}^n H^{p-2i,q-i}(Spec(k))
$$
In general $H^{\bullet,\bullet}(Spec(k))$ is highly non trivial. For example, $H^{1,1}(Spec(k)) = \mathbb{H}^1(Spec(k),\mathbb{G}\_m[-1]) = k^\times$ so you will always have $H^{1,1}(\mathbb{P}^n\_k) \neq 0$ (except for $k = \mathbb{F}\_2$).
| 10 | https://mathoverflow.net/users/1985 | 28876 | 18,881 |
https://mathoverflow.net/questions/28869 | 7 | Kenneth Kunen in his “The Foundations of Mathematics” writes:
1. ‘Set theory is the study of models of ZFC’ (p. 7)
2. ‘Set theory is the theory of everything’ (p. 14)
With (1) Kunen is pointing to a change in the intended use of the axioms of ZFC: ‘there are two different uses of the word “axioms”: as “*statements of faith*” and as “*definitional axioms*”.’ (p. 6).
With (2) he means ‘set theory is *all*-important. That is
* *All* abstract mathematical concepts are set-theoretic.
* *All* concrete mathematical objects are specific sets.’ (p. 14)
According to (1), to be a set is to be any of the *individuals* of the *universe* of a particular model of ZFC, just like being a numeral (standard or not) is being any of the individuals of the universe of a particular model of PA (here I am using Shoenfield’s terminology in “Mathematical Logic”, p. 18).
But, according to (2), models are sets too, as any other objects dealt with in the metatheory.
What's more, models of set theory are defined in terms of *relative interpretations* of set theory into itself, a syntactical concept. (See Kunen’s “Set Theory. An Introduction to Independence Proofs”, p. 141), which makes the whole thing a bit more confusing.
The view of the axioms of set theory as "*definitional axioms*" is appealing. And more in regard of (2) since then they pretend to define all that there is. The study of models of set theory has an intrinsic interest, but why reduce the study of set theory to it? Or stated another way, why abandoning the old view?
I would like to know if set theorists do stick to one view or another or shift comfortably between both at need, and the reasons they have to do so.
| https://mathoverflow.net/users/6466 | How to think like a set (or a model) theorist. | I highly recommend reading Andrej Bauer's [excellent answer](https://mathoverflow.net/questions/23060/set-theory-and-model-theory/23077#23077) to an earlier question along the same lines. To summarize the situation in a few words: the fact that set theory is anterior to model theory does not preclude the model theoretic study of set theoretic structures. The reason for that is the same as why we can study the ring **Z** even though integers are conceptually anterior to abstract algebra.
---
I will attempt to answer your revised question as I understand it.
First, it is important to note that set theorists only rarely work with general models of ZFC, so it would be unfair to reduce set theory to the model theory of ZFC. Indeed, set theorists almost exclusively work with well-founded models of (fragments of) ZFC. In fact, the models that set theorists consider are usually of the form (M,∈) where M is a set and ∈ is the true membership relation. While model theoretic ideas play a very important role in set theory, this is very different from how a model theorist would approach models of a given theory.
Second, the language commonly used by set theorists is designed to accomodate multiple philosophical views and focus on the mathematical ideas. When dealing with forcing, set theorist often use a lingo popularized by Baumgartner. In this lingo, one talks of the generic extension V[G] as an object of the same level and equally tangible as the set theoretic universe V. This is easily translated in the multiverse view, but it is just as easy to think of this as no more than a linguistic convenience: set theorists who prefer working with countable transitive models à la Kunen can mentally replace V by such a model M and interpret the words as those of an inhabitant of M; set theorists who prefer working with boolean valued models à la Jech can simply interpret the generic filter G as a convenient abstraction; set theorists who prefer the formal approach can translate the lingo into purely syntactic arguments; etc. In my case, none of these translations ever take place, but I do hold the possibilities in a safe place for comfort.
In conclusion, my advice is to focus on the mathematics and use models as tools just like any other mathematical object.
| 6 | https://mathoverflow.net/users/2000 | 28881 | 18,883 |
https://mathoverflow.net/questions/28878 | 14 | What is Borel-de Siebenthal theory?
| https://mathoverflow.net/users/6772 | What is Borel-de Siebenthal theory? | I'm not sure the term "theory" is appropriate here, but the joint paper by Borel and de Siebenthal has had considerable influence in Lie theory over the years: MR0032659 (11,326d)
Borel, A.; De Siebenthal, J.,
Les sous-groupes fermés de rang maximum des groupes de Lie clos.
Comment. Math. Helv. 23, (1949). 200--221. (There was a short Comptes Rendus announcement in 1948.) This is found near the start of the Springer four-volume collected papers of Borel, with a couple of minor corrections appended.
To quote from the review by P.A. Smith: "Let $G$ be a compact Lie group, $G'$ a closed connected subgroup having the same rank as $G$. Let $Z'$ be the center of $G'$. The main object of this paper is to show that $G'$ is a connected component of the normalizator of $Z'$ in $G$." The proof involves
"a necessary and sufficient condition that a subsystem of root vectors of $G$ be the root vectors of a closed subgroup of $G$" and the subgroups of this type are found explicitly for all simple $G$. [Here $G$ is always assumed to be connected.]
The result on subsystems of root systems carries over in a natural way to the study of semisimple complex Lie (or algebraic) groups and their Lie algebras,
for example the determination of subalgebras of maximal rank in the latter.
| 9 | https://mathoverflow.net/users/4231 | 28884 | 18,884 |
https://mathoverflow.net/questions/24552 | 46 | Brian Conrad indicated a while ago that many of the results proven in AG using universes can be proven without them by being very careful ([link](https://mathoverflow.net/questions/15897/in-what-topology-dm-stacks-are-stacks/15910#15910)). I'm wondering if there are any results in AG that actually depend on the existence of universes (and what some of the more interesting ones are).
I'm of course aware of the result that as long as we require that the classes of objects and arrows are sets (this is the only valid approach from Bourbaki's perspective), for every category C, there exists a universe U such that the U-Yoneda lemma holds for U-Psh(C) (this relative approach makes proper classes pointless because every universe allows us to model a higher level of "largeness"), but this is really the only striking application of universes that I know of (and the only result I'm aware of where it's clear that they are necessary for the result).
| https://mathoverflow.net/users/1353 | What interesting/nontrivial results in Algebraic geometry require the existence of universes? | My belief is that no result in algebraic geometry that does
not explicitly engage the universe concept will fully
require the use of universes. Indeed, I shall advance an
argument that no such results actually need anything beyond
ZFC, and indeed, that they need much less than this. (But please note, I answer as a set theorist rather than an algebraic geometer.)
My reason is that there are several hierarchies of weakened
universe concepts, which appear to be sufficient to carry
out all the arguments that I have heard using universes,
but which are set-theoretically strictly weaker hypotheses.
A universe, as you know, is known in set theory as
$H\_\kappa$, the set of all
hereditarily-size-less-than-$\kappa$ sets, for some
inaccessible cardinal $\kappa$. Every such universe also
has the form $V\_\kappa$, in the cumulative Levy hierarchy,
because $H\_\kappa=V\_\kappa$ for any beth-fixed point, which
includes all inaccessible cardinals. Thus, the Universe
Axiom, asserting that every set is in a universe, is
equivalent to the large cardinal assertion that there are
unboundedly many inaccessible cardinals.
This hypothesis is relatively low in the large cardinal
hierarchy, and so from this perspective, it is relatively
mild to just go ahead and use universes. In consistency
strength, for example, it is strictly weaker than the
existence of a single Mahlo cardinal, which is considered
to be very weak large-cardinal theoretically, and much
stronger hypotheses are routinely considered in set theory.
Nevertheless, these hypotheses do definitely exceed ZFC in
strength, unless ZFC is already inconsistent, and so your
question is a good one. It follows the pattern in set
theory of inquiring the exact large cardinal strength of a
given hypothesis.
The weaker universe concepts that I propose to use in
replacement of universes are the following, where I take
the liberty of introducing some new terminology.
* A *weak universe* is some $V\_\alpha$ which models ZFC.
The Weak Universe axiom is the assertion that every set is
in a weak universe.
This axiom is strictly weaker than the Universe Axiom,
since in fact, every universe is already a model of it.
Namely, if $\kappa$ is inaccessible, then there are
unboundedly many $\alpha\lt\kappa$ with $V\_\alpha$ elementary in $V\_\kappa$, by the Lowenheim-Skolem theorem, and so
$V\_\kappa$ satisfies the Weak Universe Axiom by itself.
From what I have seen, it appears that most of the applications of
universes in algebraic geometry could be carried out with
weak universes, if one is somewhat more careful about how
one treats universes. The difference is that when using
weak universes, one must pay attention to whether a given
construction is definable inside the universe or not, in
order to know whether the top level $\kappa$ of the weak
universe, which may now be singular (and this is the
difference), is reached.
* Let us say that a *very weak universe* is simply a
transitive set model of ZFC. (In set theory, one would want just to call these *universes*, but here that word is taken to have the meaning above; so we could call them *set-theoretic universes*.) The Very Weak Universe Axiom (or Set-theoretic Universe Axiom)
is the assertion that every set is an element of a very
weak universe.
The difference between a very weak universe and a weak
universe is that a very weak universe $M$ may be wrong
about power sets, even though it satisfies its own version
of the Powerset Axiom. Set theorists are very attentive to
such very weak universes, and pay attention in a
set-theoretic construction to which model of set theory it
is undertaken. If the algebraic geometers were to give
similar attention to this point, thereby turning themselves
into set theorists, I believe that all of their arguments
using universes could be essentially replaced with very
weak universes. Another important point is that while
universes are always linearly ordered by inclusion, this is
no longer true for very weak universes.
Now, even the Very Weak Universe Axiom transcends ZFC in
consistency strength, because it clearly implies Con(ZFC).
So let me now describe how one might provide an even
greater reduction in the strength of the hypothesis, and
capture a use of universes within ZFC itself.
The key is to realize that algebraic geometry does not
really use the full force of ZFC. (Please take this with
some skepticism, given my comparatively little exposure to
algebraic geometry.) It seems to me unlikely, for example,
that one needs the full Replacement Axiom in order to carry
out the principal goal constructions of algebraic geometry.
Let me suppose that these arguments can be carried out in
some finite fragment $ZFC\_0$ of $ZFC$, for example, $ZFC$
restricted to formulas of complexity $\Sigma\_N$ for some
definite number $N$, such as $100$ or so. In this case, let
me define that a *good-enough-universe* is $V\_\kappa$,
provided that this satisfies $ZFC\_0$. All such good-enough
universes have $V\_\kappa=H\_\kappa$, just as with universes,
since these will be beth-fixed points. The Good-enough
Universe Axiom is the assertion that every set is a member
of a good-enough universe.
Now, my claim is first, that this Good-enough Universe
Axiom is sufficient to carry out most or even all of the
applications of universes in algebraic geometry, provided
that one is sufficiently attentive to the set-theoretic
issues, and second, that this axiom is simply a theorem of
ZFC. Indeed, one can get more, that the various good-enough
universes agree with each other on truth.
**Theorem.** There is a definable closed unbounded class
of cardinals $\kappa$ such that every $V\_\kappa$ is a
good-enough universe and furthermore, whenever
$\kappa\lt\lambda$ in $C$, then $\Sigma\_N$-truth in
$V\_\kappa$ agrees with $\Sigma\_N$-truth in $V\_\lambda$, and
moreover, agrees with $\Sigma\_N$ truth in the full universe
$V$.
This theorem is exactly an instance of the Levy Reflection
Theorem.
OK, so if I am right, then the algebraic geometers can
carry out their universe arguments by paying a lot of
careful attention to the set-theoretic complexity of their
constructions, and using good-enough universes instead of
universes.
But should they do this? For most purposes, I don't think
so. The main purpose of universes is as a simplifying
device of convenience to stratify the full universe by
levels, which can be fruitfully compared by local notions
of large and small. This makes for a very convenient
theory, having numerous local concepts of large and small.
I can imagine, however, a case where one has used the
Universe theory to prove an elementary result, such as
Fermat's Last Theorem, and one wants to know what are the
optimal hypotheses for the proof. The question would be
whether the extra universe hypotheses are required or not.
The thrust of my answer here is that such a question will
be answered by replacing the universe concepts that are
used in the proof with any of the various weakened universe
concepts that I have mentioned above, and thereby realizing
the theorem as a theorem of ZFC or much less.
| 52 | https://mathoverflow.net/users/1946 | 28913 | 18,901 |
https://mathoverflow.net/questions/27960 | 10 | A set $M$ is called *amenable* if it is transitive and satisfies the following conditions:
1. For all $x,y\in M$, $\{x,y\}\in M$
2. For all $x\in M$, $\bigcup x \in M$
3. $\omega \in M$
4. For all $x,y \in M$, $x\times y \in M$
5. ($\Sigma\_0$ comprehension) Whenever $\Phi$ is a $\Sigma\_0$ formula of one free variable with parameters from $M$, then for all $x\in M$, $\{z\in x | \Phi(z)\}\in M$
Although the definition of an amenable set does not include replacement, some very limited amount of replacement follows from the axioms given. For example, for all $x,y\in M$, it must be that $\{\{z,w\}|z\in x, w\in y\}\in M$ and $\{\{z\}|z\in x\}\in M$. So just how limited is the replacement in amenable sets? In particular,
>
> If $M$ is an amenable set and $x\in M$, does it follow that $\{\bigcup z | z \in x\} \in M$?
>
>
>
| https://mathoverflow.net/users/6649 | When does replacement (accidentally) hold in amenable sets? | I think the following is a counterexample to your specific question. Let AH be the set of those $x$ such that (1) each element of $TC\{x\}$ has cardinality at most $\aleph\_\omega$ and (2) all but finitely many elements of $TC\{x\}$have cardinality strictly smaller than $\aleph\_\omega$. (By $TC\{x\}$, I mean the transitive closure of the singleton, so it contains $x$, all its members, all their members, etc.) If $y\in x$ then $TC\{y\}$ is a subset of $TC\{x\}$, so AH is transitive. AH contains $\omega$ and is easily seen to be closed under pairing, union, and (binary) Cartesian product. Furthermore, it satisfies not only $\Sigma\_0$-comprehension but full comprehension, because if $y$ is a subset of $x$ then $TC\{y\}$ is a subset of $TC\{x\}\cup\{y\}$. So AH is amenable.
For each natural number $n$, let $A\_n=\{\{n\}\times\aleph\_k:k\in\omega\}$, and notice that AH contains not only each of the $A\_n$'s but also the set $X$ consisting of all the $A\_n$'s. The union of any particular $A\_n$ is $\{n\}\times\aleph\_\omega$, which is in AH and has cardinality $\aleph\_\omega$, but the set of all these unions is not in AH because it has these infinitely many elements of size $\aleph\_\omega$. Summary: $X$ is in AH but $\{\bigcup z:z\in X\}$ is not.
Comment: If one modifies the definition of AH by requiring all elements of $TC\{x\}$ to have size strictly below $\aleph\_\omega$, one gets the standard example of a model of all the ZFC axioms except the axiom of union. By allowing, in the definition of AH, finitely many exceptions of size $\aleph\_\omega$ one revives the axiom of union and in particular one lets each of the sets $\bigcup A\_n$ into AH (but just barely) but not the collection of all of them.
| 7 | https://mathoverflow.net/users/6794 | 28918 | 18,903 |
https://mathoverflow.net/questions/28871 | 7 | Question:
---------
---
Let $D:A\to (X\downarrow C)$ be a $\kappa$-good $S$-tree rooted at $X$ for a collection of morphisms $S$ in $C$, where $\kappa$ is a fixed uncountable regular cardinal. Then according to the proof of Lemma A.1.5.8 of *Higher Topos Theory* by Lurie, for any $\kappa$-small downward-closed $B\subseteq A$, the colimit of the restricted diagram, $\varinjlim D|\_B$ is $\kappa$-compact in $(X\downarrow C)$.
Why is this true? (It is stated without proof.)
Definitions:
------------
---
For your convenience, here are the definitions:
Recall that an object $X$ in $C$ is called $\kappa$-compact if $h^X(\cdot):=\hom(X,\cdot)$ preserves all $\kappa$-filtered colimits (where $\kappa$-filtered means "$<\kappa$"-filtered, since the terminology is different depending on the source).
Recall that an $S$-tree rooted at $X$ for a collection of morphisms $S$ in $C$ consists of the following data:
* An object $X$ in C (the root)
* A partially ordered set $A$ whose order structure is well-founded (the index)
* A diagram $D:A\to (X\downarrow C)$ such that given any element $\alpha\in A$, the canonical map $$\varinjlim D|\_{\{\beta:\beta<\alpha\}}\to D(\alpha)$$ is the pushout of some map $U\_\alpha\to V\_\alpha\in S$.
We say that an $S$-tree is $\kappa$-good if for all of the morphisms $U\_\alpha\to V\_\alpha$ above, $U\_\alpha$ and $V\_\alpha$ are $\kappa$-compact, and such that for any $\alpha\in A$, the subset $\{\beta: \beta < \alpha \}\subseteq A$ is $\kappa$-small.
**Edit**: It's easy to reduce the proof to showing that $D(\alpha)$ is $\kappa$-compact, since projective limits of diagrams $B\to Set$ are $|Arr(B)|$-accessible (and therefore $\kappa$-accessible since $B$ is $\kappa$-small), we perform the computation for $I$ a $\kappa$-filtered poset, and $F:I\to C$, assuming that $D(\alpha)$ is $\kappa$-compact for all $\alpha\in B$:
$$\begin{matrix}\
\varinjlim\_I Hom\_C(\varinjlim\_B D, F)&\cong&\varinjlim\_I\varprojlim\_{B^{op}} Hom\_C(D,F)\\
&\cong& \varprojlim\_{B^{op}}\varinjlim\_I Hom\_C(D,F)\\
&\cong& \varprojlim\_{B^{op}} Hom\_C(D,\varinjlim\_IF)\\
&\cong& Hom\_C(\varinjlim\_B D,\varinjlim\_IF)
\end{matrix}$$
**Edit 2**: I think the above reduction actually won't work, since it doesn't use the hypothesis that B is downward-closed.
| https://mathoverflow.net/users/1353 | K-good trees and K-compactness of colimits over K-small downwards-closed subposets (500 point bounty if answered by Midnight EST)) | Does this work?
To prove $D(\alpha)$ is $\kappa$-compact for all $\alpha$ in $A$, assume otherwise, that there exists some counterexample. Then, by the fact $A$ is well-ordered, there is a minimal counterexample (i.e., there is a minimal element $\alpha$ in the set of $\gamma \in A$ such that $D(\gamma)$ is not $\kappa$-compact). This means $D\_\beta$ is $\kappa$-compact for all $\beta \lt \alpha$. Since $\{\beta: \beta \lt \alpha\}$ has cardinality less than $\kappa$, we have that
$$colim\_{\beta: \beta \lt \alpha} D(\beta)$$
is $\kappa$-compact. Now, given a diagram of the form
$$V\_\alpha \leftarrow U\_\alpha \to colim\_{\{\beta: \beta \lt \alpha\}} D(\beta)$$
in the category of $\kappa$-compact objects, its pushout is also $\kappa$-compact. But the hypothesis is that $D(\alpha)$ is the pushout for some such diagram, so $D(\alpha)$ is $\kappa$-compact, and we have reached a contradiction.
So $D(\alpha)$ is $\kappa$-small for all $\alpha \in A$. It follows that $colim\_{\beta \in B} D(\beta)$ is $\kappa$-compact for any subposet $B \subseteq A$ whenever this is a $\kappa$-small colimit. (The restriction to downward-closed $B$ is not much loss of generality, because if $B \subseteq A$ is full, then the colimit over such a $B$ is isomorphic to the colimit over its downward closure, since $B$ is cofinal in its downward closure.)
| 5 | https://mathoverflow.net/users/2926 | 28922 | 18,906 |
https://mathoverflow.net/questions/28892 | 16 | I was searching on MathSciNet recently for a certain paper by two mathematicians. As I often do, I just typed in the names of the two authors, figuring that would give me a short enough list. My strategy was rather dramatically unsuccessful in this case: the two mathematicians I listed have written 80 papers together!
So this motivates my (rather frivolous, to be sure) question: which pair of mathematicians has the most joint papers?
A good meta-question would be: can MathSciNet search for this sort of thing automatically? The best technique I could come up with was to think of one mathematician that was both prolific and collaborative, go to their "profile" page on MathSciNet (a relatively new feature), where their most frequent collaborators are listed, alphabetically, but with the wordle-esque feature that the font size is proportional to the number of joint papers.
Trying this, to my surpise I couldn't beat the 80 joint papers I've already found. Erdos' most frequent collaborator was Sarkozy: 62 papers (and conversely Sarkozy's most frequent collaborator was Erdos). Ronald Graham's most frequent collaborator is Fan Chung: 76 papers (and conversely).
I would also be interested to hear about triples, quadruples and so forth, down to the point where there is no small set of winners.
---
**Addendum**: All right, multiple people seem to want to know. The 80 collaboration pair I stumbled upon is Blair Spearman and Kenneth Williams.
| https://mathoverflow.net/users/1149 | Which pair of mathematicians has the most joint papers? | We get 135 matches for "Author=(Jimbo, Michio and Miwa, Tetsuji)" in mathscinet.
| 21 | https://mathoverflow.net/users/36665 | 28933 | 18,912 |
https://mathoverflow.net/questions/28945 | 16 | There are infinite graphs which contain all finite graphs as induced subgraphs, e.g. the Rado graph or the coprimeness graph on the naturals.
>
> Are there infinite groups which
> contain all finite groups as
> subgroups?
>
>
>
| https://mathoverflow.net/users/2672 | Infinite groups which contain all finite groups as subgroups | Yes, plenty. The group only has to contain all finite permutation groups. Perhaps the most
straightforward example would be the permutations of a countable set. That is bijections
which fix all but a finite set.
| 66 | https://mathoverflow.net/users/3992 | 28946 | 18,917 |
https://mathoverflow.net/questions/28941 | 3 | Can we find finite metabelian (ie with derived length 2) groups with arbitrarily many distinct degrees of irreducible complex characters?
If we cannot, can we somehow find a bound of the form $|cd(G)|\leq f(dl(G))$ for some "interesting" function $f$ (linear would be very cool for instance).
The motivation is that we of course have the bound $dl(G)\leq 2|cd(G)|$ for solvable groups (and conjectured to actually be $dl(G)\leq |cd(G)|$), so I was wondering if a bound in the other direction also existed.
| https://mathoverflow.net/users/4614 | Are there finite metabelian groups with arbitrarily many character degrees? | Sure, you just take direct products of metabelian groups with different character degrees: cd(G×H) = cd(G)×cd(H) and (G×H)′ = G′×H′.
I suggest taking G(p) = AGL(1,p) = Hol(p) to be the normalizer of a Sylow p-subgroup in the symmetric group of degree p, for each prime p, but there are lots of examples. For instance:
* G(3) = Sym(3) has character degrees {1,2},
* G(5) = F20 has character degrees {1,4},
* G(7) = F42 has character degrees {1,6},
* and G(3) × G(5) × G(7) has character degrees { 1, 2, 4, 6, 8, 12, 24, 48 }.
| 3 | https://mathoverflow.net/users/3710 | 28950 | 18,918 |
https://mathoverflow.net/questions/28747 | 4 | Two short questions:
* Is there any work classifying the lattice of subcategories of an arbitrary (sufficiently small) category $\mathcal{C}$, similar to the way that the set of subsets of set $\mathcal{S}$ is isomorphic to the functions $\mathcal{S}\to\mathbf{2}$, where $\mathbf{2}$ is a two point set?
* Is there standard notation denoting the lattice of subcategories of some category?
The definitions found in [nLab](http://ncatlab.org/nlab/show/subcategory) are phrased in terms of functors going into $\mathcal{C}$, but the definition for sets talks about functions out of the set $\mathcal{S}$. Why are things done differently? That is, rather than characterising subcategories in terms of functors into $\mathcal{C}$, why not characterise them in terms of functors out of $\mathcal{C}$? Something like:
>
> The lattice of subcategories of $\mathcal{C}$ is isomorphic to the functor category $\mathcal{SO}^\mathcal{C}$, for some "subobject classifier" $\mathcal{SO}$.
>
>
>
| https://mathoverflow.net/users/2620 | Lattice of subcategories: subobject classifier in Cat | In his comment on Finn Lawler's response, Mike Shulman points out that Cat has no subobject classifier in the 2-categorical sense. It also fails to have a subobject classifier in the 1-categorical sense. To see this, note that if Cat had a subobject classifier, then it would be a topos, as it is cartesian closed. However, Cat is not a topos since, e.g., it is not locally cartesian closed (cf. Johnstone's "Elephant", A1.5).
| 6 | https://mathoverflow.net/users/6485 | 28960 | 18,926 |
https://mathoverflow.net/questions/28947 | 76 | The recent question about the most prolific collaboration interested me. How about this question in the opposite direction, then: can anyone beat, amongst contemporary mathematicians, the example of Christopher Hooley, who has written 91 papers and has yet to coauthor a single one (at least if one discounts an obituary written in 1986)?
| https://mathoverflow.net/users/5575 | Least collaborative mathematician | Lucien Godeaux wrote more than 600 papers and not one of them is a joint paper. He cowrote a textbook in projective geometry. Mathscinet records only 15 citations to all these papers! But there is something called Godeaux surfaces which is mentioned in the literature. This is about the weirdest example I know.
<http://www.ams.org/mathscinet/search/author.html?mrauthid=241534>
| 80 | https://mathoverflow.net/users/2290 | 28973 | 18,934 |
https://mathoverflow.net/questions/28832 | 6 | Dear all.
Let
$$
f(x) = \sum\_{k \in \mathbb{Z}} \hat{f}(k) \exp(2\pi \mathrm{i} kx)
$$
be a function given by usual fourier series.
Since my original question hasn't got any answer yet, and I came across another related question, I am just adding it. Denote by $T\_n$ the set of all trigonometric polynomials of degree $n$, that is $g\in T\_n$ if
$$
g(x) = \sum\_{k=-n}^{n} \hat{g}(k) \exp(2\pi \mathrm{i} kx).
$$
So now what is $\min\_{g \in T\_n} \|f - g\|\_{\infty}$ and what is the optimal $g$?
Since the Fourier series of a continuous function must not converge, I expect that the answer isn't $g(x) = \sum\_{k=-n}^{n} \hat{f}(k) \exp(2\pi \mathrm{i} kx)$ but something else. However, the other choice the Fejer kernel
$$
g(x) = \sum\_{k=-n}^{n} \frac{n - |k|}{n} \hat{f}(k) \exp(2\pi \mathrm{i} kx)
$$
seems to give worse estimates on $\min\_{g \in T\_n} \|f - g\|\_{\infty}$ once $\hat{f} \in \ell^2$.
Thanks,
Helge
**Original question**:
I am interested in the question of how well one can approximate $f$ by functions that are analytic in some strip. The naive approach yields for example that if one sets
$$
f\_R(x) = \sum\_{|k|\leq R} \hat{f}(k) \exp(2\pi \mathrm{i} kx)
$$
and assumes $f \in C^{n+1}$ then $f\_R(x)$ has an extension to a strip of width $\frac{n \log(k)}{2\pi k}$ on which $f\_R$ is bounded by $\|\hat{f}\|\_{\ell^1}$.
This seems like a pretty natural question so I expect it to be well studied, but I don't know where... Does anybody has references? I am also interested in stronger regularity assumptions than $C^n$...
| https://mathoverflow.net/users/3983 | Approximation by analytic functions | The answer to the modified question is given by [Jackson-type theorems](http://en.wikipedia.org/wiki/Jackson%27s_inequality).
The classic book by N.I. Akhiezer which is quoted in the Wikipedia article contains a number of specialised results on optimal approximation by trigonometric polynomials.
A typical optimal result improves the approximation by finite Fourier sums by a logarithmic factor.
>
> **Theorem.** Let $f$ be a periodic function on $\mathbb R$ of class $C^{m}$. Then for any $n\in\mathbb N$
> $$\inf\limits\_{g \in T\_n} \|f - g\|\_{L^\infty}\leq \frac{K\_m}{n^m}\|f^{(m)}\|\_{L^\infty},$$
> where the constant $K\_m$ is sharp (and can be written in a closed form).
>
>
>
A result of S. Bernstein says roughly that the order of approximation $n^{-m}$ cannot be improved.
To find the trigonometric polynomial $g$ which minimizes $\|f - g\|\_{L^\infty}$ for a given $f$ is a difficult problem. I am not sure if it has been solved.
| 2 | https://mathoverflow.net/users/5371 | 28975 | 18,935 |
https://mathoverflow.net/questions/28974 | 7 | Let $K$ be a number field and $\mathfrak{p}$ be a place of good reduction. It is easy to see that the reduction map on prime-to-$p$ torsion $A(K)[p'] \hookrightarrow A\_{\mathfrak{p}}(\kappa(\mathfrak{p}))$ is injective.
But if $p > e(\mathfrak{p}/p) + 1$, the reduction map is even injective on $p$-torsion. This can be seen by showing that the formal group has no $p$-torsion.
Now I ask if one can show this without using the formal group. By a theorem of Raynaud in "Schémas en groupes de type $(p, \ldots, p)$", the inequality for $p$ implies that for a $p^n$-torsion commutative finite flat group scheme the generic fibre can be spread out uniquely over the special fibre and $\mathrm{Hom}(G,H) = \mathrm{Hom}(G(\bar{K}),H(\bar{K}))$. Can this be used somehow?
| https://mathoverflow.net/users/6960 | $p$-torsion in the Mordell-Weil group of Abelian varieties injecting in reduction | One has the finite flat group scheme $\mathbb Z/p$ over $\mathcal O\_{K\_{\mathfrak p}}$
(I write $K\_{\mathfrak p}$ for the $\mathfrak p$-adic completion of $K$, and
$\mathcal O\_{K\_{\mathfrak p}}$ for its integer ring),
as well as
the finite flat group scheme $A[p]$. Giving a $p$-torsion point over $K\_{\mathfrak p}$ (and hence in particular over $K$) is the same as giving a closed embedding
on generic fibres:
$(\mathbb Z/p)\\_{/ K\_{\mathfrak p}} \hookrightarrow A[p]\\_{/K\_{\mathfrak p}}.$
Raynaud's results imply that this extends to a closed embedding over $\mathcal O\_K$:
$\mathbb Z/p \hookrightarrow A[p],$
which is another way of saying the that the non-zero $p$-torsion point has non-zero
reduction.
Just to see concretely what can happen in the situation when $e \geq p-1$, suppose
that $K = \mathbb Q$ and $p = 2$. Then we could have a map
$(\mathbb Z/2)\\_{/\mathbb Q\_2} \hookrightarrow A[2]\_{/\mathbb Q\_2}$
which extends to a closed immersion
$\mu\_2 \hookrightarrow A[2].$
This would correspond to having a 2-torsion point in the kernel of the reduction map.
(Note that $\mu\_2$ has a non-trivial point in char. zero, which collapses down to
the identity in char. two.)
[Added in response to unkwown's comment:]
The point is that one can form the scheme-theoretic closure in $A[p]$ of the image
of $(\mathbb Z/p)\\_K,$ which is *some* finite flat subgroup scheme over $\mathcal O\_K$
which is embedding as a closed subgroup scheme of $A[p]$ (by construction: we formed it as a scheme-theoretic closure). And it
has $(\mathbb Z/p)\\_K$ as its generic fibre (again by construction).
Now when $e < p-1$, Raynaud's results show that this finite flat group scheme has no
choice but to be $\mathbb Z/p$, and so we get a copy of $\mathbb Z/p$ embedding into $A[p]$,
extending the original embedding of generic fibres. Thus the order $p$ point in $A[p](K)$
reduces mod $p$ to an order $p$ point.
But if e.g. $p = 2$, then this scheme-theoretic closure could be $\mu\_2$. Now the non-trivial point ($-1$) of $\mu\_2(K)$ specializes to the trivial point in char. 2,
and so when we have a copy of $\mu\_2$ inside $A[2]$, the non-trivial point of $\mu\_2(K)$
lies in the kernel of the reduction mod 2 map.
| 7 | https://mathoverflow.net/users/2874 | 28981 | 18,939 |
https://mathoverflow.net/questions/28967 | 16 | While doing some research on polytopes I came to the following question. Maybe it's already somewhere but anyway I'll post it here.
Let $X\subset \mathbb{R}^3$ be such that, for every plane $P$, $P\cap X$ is simply connected. Is $X$ convex?
I'm not sure, but I think maybe it's necesary to assume some well behaving like local simple connectedness. Anyway I think this is true with the apropiate asumptions. I would not be surprised if it was true just as stated.
Probably this is true even in greater dimensions.
| https://mathoverflow.net/users/4619 | A characterization of convexity | How about some tomography? This should work if $X$ is open. Assume $X\subset \mathbb R^3$ is nonempty and for every plane $H$ the intersection $H\cap X$ is either contractible or empty. (Note that an open subset of the plane is contractible if it is ((nonempty,) connected, and) simply connected.)
Claim 1: $X$ is contractible.
Proof: Consider the space of all pairs $(x,H)$, $x\in X$ and $H$ a plane containing $x$.
Fiber this by $(x,H)\mapsto x$. It's a trivial bundle over $X$ with fiber $P^2$.
Fiber the same thing by $(x,H)\mapsto H$. The nonempty fibers are contractible, so the domain is homotopy equivalent to the image, which in turn fibers over $P^2$ with contractible fibers.
So $X\times P^2$ has the same homotopy type as $P^2$, and upon further inspection $X$ is contractible.
Claim 2: $X$ is convex.
Proof: Let $L$ be any line whose intersection with $X$ is nonempty, and play the same game again with pairs $(x,H)$, but now the plane $H$ is constrained to contain $L$. Call the space of all such pairs $Y$. On the one hand, $Y$ is equivalent to the circle $P^1$ by the same kind of argument as before. On the other hand, $Y$ is the blowup of the $3$-manifold $X$ along the $1$-manifold $X\cap L$. The complement $Y'$ of $(X\cap L)\times P^1$ in $Y$ is the same as the complement of $X\cap L$ in $X$, so it is connected. Therefore all of the group $H\_1(Y,Y')$ comes from $H\_1(Y)$. But the latter group is of rank $1$ while the former is isomorphic to $H\_0( (X\cap L)\times P^1)$. (I am using mod $2$ coefficients.) It follows that $X\cap L$ is connected.
Maybe this generalizes to $\mathbb R^n$. The hypothesis I am thinking of is that every nonempty hyperplane section is contractible, or equivalently $(n-2)$-connected. I don't believe that every hyperplane section simply connected is enough if $n>3$
| 8 | https://mathoverflow.net/users/6666 | 28987 | 18,942 |
https://mathoverflow.net/questions/28744 | 5 | In their book "Theory of sets" Bourbaki suggested a general theory of isomorphism.
(See also <http://www.tau.ac.il/~corry/publications/articles/pdf/bourbaki-structures.pdf> )
The example of an untransportable relation (i.e. formula) in the book involves 2 principal base sets.
Are there examples of untrasportable formulas when we allow only one principal base set?
| https://mathoverflow.net/users/5761 | Bourbaki theory of isomorphism, examples of untransportable formulas | An example of untrasportable sentence, when there is only one principal base set X, may be the following one:
All elements of the set X are finite sets,
Because, by definition, the truth value of a transportable sentence must be preserved under all bijections from the set X. Obviously, there exists a bijection from X to a set Y, where not all elements of Y are finite sets.
A simpler example is "the set X contains the empty set".
There is a paper "Sentences of type theory: the only sentences preserved under isomorphisms" by Victoria Marshall and Rolando Chuaqui - see The Journal of symbolic Logic, vol 56, #3, Sep 1991.
| 3 | https://mathoverflow.net/users/5761 | 28989 | 18,944 |
https://mathoverflow.net/questions/28986 | 2 | Okay, we know that
$$ \frac{sin(x)}{x} = \prod\_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\cdot\pi^2}\Big) $$ .
Is there some known (trigonometric(?)) function that is equal to the following infinite product?
$$ \prod\_{n=1}^{\infty} \Big(1-\frac{x}{n\cdot\pi}\Big) $$
I'd be happy as well if someone could provide me with a function that is equal to a similar divergent infinite product (a function, for example, that is equal to 'my' inifite product, only $\pi=1$, or $x=x^2$, or something in that direction).
Thanks in advance,
Max Muller
| https://mathoverflow.net/users/93724 | Closed form of divergent infinite product? | It's a **divergent** infinite product. You might as well ask for the sum of
$$\sum\_{n=1}^\infty\frac{x}{n\pi}.$$
You can "cure" the divergence by multipliying each term by a suitable factor, so
for instance
$$f(x)=\prod\_{n=1}^\infty e^{x/n\pi}\left(1-\frac{x}{n\pi}\right)$$
does converge (as the $n$-th term is like $\exp(x^2/2n^2\pi^2)$). You can
express this in terms of the gamma function which satisfies
$$\frac1{\Gamma(x)}=x e^{\gamma x}\prod\_{n=1}^\infty
e^{-x/n}\left(1+\frac{x}{n}\right).$$
By using the identity
$$f(x)f(-x)=\prod\_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right)$$
one can deduce the identity
$$\Gamma(x)\Gamma(1-x)=\frac\pi{\sin\pi x}.$$
| 7 | https://mathoverflow.net/users/4213 | 28990 | 18,945 |
https://mathoverflow.net/questions/28980 | 3 | I need a reference for the proof that the complex orthogonal group
$SO\_{2n+1}($ℂ$) = \{A\in SL\_{2n+1}($ℂ$): A^TA = Id\}$ is simple in a group theoretical sense (if it is true).
How about the simplicity of $SO\_{2n+1}(K)$ in general (i.e. $K$ an arbitrary infinite field)?
It there any criterion? It seems that if $K$ is "big", then $SO\_{2n+1}(K)$ is not simple, e.g. if $K$ has a valuation, then (I think) one can find somehow a proper normal subgroup.
| https://mathoverflow.net/users/6923 | Simplicity of (complex) orthogonal groups | The structure of classical groups goes back a long way and has been treated in a number of books, but in varying generality (arbitrary fields, various commutative rings, etc.). One older source in French is J.A. Dieudonne's concise Springer Ergebnisse volume *La geometrie des groupes classiques* (1963). A probably more readable modern textbook with limited aims is Larry Grove's *Classical Groups and Geometric Algebra* (AMS, 2002), just cited by Skip. There is also Emil Artin's old book *Geometric Algebra* and a much larger book by Hahn-O'Meara oriented more to algebraic K-theory. Anyway your group is simple both as an algebraic and as an abstract group (special orthogonal groups in odd dimension are also adjoint groups). There's no real need to get into algebraic groups, BN-pairs, or the like, though this is the "correct" general setting as Tits showed.
Actually, simplicity of various classical groups is proved sometimes in graduate algebra textbooks (which I don't have at hand). It depends how far you want to go. Over more general fields, especially of characteristic 2, a little more care is needed but these groups are still simple or very close to it even over most finite fields.
| 4 | https://mathoverflow.net/users/4231 | 28994 | 18,948 |
https://mathoverflow.net/questions/28999 | 43 | [This recent MO
question](https://mathoverflow.net/questions/28945/infinite-groups-which-contain-all-finite-groups-as-subgroups),
answered now several times over, inquired whether an
infinite group can contain every finite group as a
subgroup. The answer is yes by a variety of means.
So let us raise the stakes: Is there a countable group
containing (a copy of) every countable group as a subgroup?
The countable random graph, after all, which inspired the
original question, contains copies of all countable graphs,
not merely all finite graphs. Is this possible with groups?
What seems to be needed is a highly saturated countable group.
* An easier requirement would insist that
the group contains merely all finitely generated groups as
subgroups, or merely all countable abelian groups.
(Reducing to a countable family, however, trivializes the question via the direct sum.)
* A harder requirement would find the subgroups in particularly nice ways: as direct summands or as normal subgroups.
* Another strengthened requirement would insist on an
amalgamation property: whenever
$H\_0\lt H\_1$ are finitely generated, then every copy of $H\_0$ in the universal group $G$
extends to a copy of $H\_1$ in $G$. This property implies
that $G$ is universal for all countable groups, by adding
one generator at a time. This would generalize the
saturation property of the random graph.
* If there is a universal countable group, can one find a
finitely generated such group, or a finitely presented
such group? (This would lose amalgamation, of course.)
* Moving higher, for which cardinals $\kappa$
is there a universal group of size $\kappa$? That is, when is there
a group of size
$\kappa$ containing as a subgroup a copy of every group of size
$\kappa$?
* Moving lower, what is the minimum size of a finite group
containing all groups of finite size at most $n$ as subgroups?
Clearly, $n!$ suffices. Can one do better?
| https://mathoverflow.net/users/1946 | Is there a universal countable group? (a countable group containing every countable group as a subgroup) | There isn't a countable group which contains a copy of every countable group
as a subgroup. This follows from the fact that there are uncountably many
2-generator groups up to isomorphism.
The first example of such a family was discovered by B.H. Neumann. A clear
account of his construction can be found in de la Harpe's book on geometric group theory.
| 56 | https://mathoverflow.net/users/4706 | 29001 | 18,951 |
https://mathoverflow.net/questions/29007 | 4 | This is not a math question as much as a process question. For the first time in my (very short) career, I find myself doing one of those messy calculations, where each 'line' of the calculation can spread over a page or three. Essentially all of the calculation is trivial if I'm willing to write some reasonable inequalities in places, and all of these choices are obvious as they're being made, but I am having a rough time keeping the assumptions on term sign required for the various inequalities and the calculations themselves even close to organized, and the copying errors are a nightmare.
Does anybody have any good suggestions on how to stay organized for this sort of trivial-in-theory but messy-in-practice calculation? What do you actually *do* in these situations? This is especially directed at people in areas like statistical physics or mathematical statistics where these sorts of things show up frequently, and there must be some way of dealing with them. Clever renaming of variables, latexing as you go, good use of Maple...?
| https://mathoverflow.net/users/6966 | Medium-Sized Calculations and Organization | Mathematica can be very useful for this kind of thing. If you're good enough at it you can force it to go through calculations pretty much step-by-step if you need it too. You can also export its output to LaTeX, which is very nice and saved me a lot of work on various physics homeworks!
For more complicated things, you can often still make mathematica do it, but it's often better (and can be more insightful) to try to restructure things into smaller blocks that are easier to handle. You can do this through renaming things, or lemmas, or whatever else. In physics, physical intuition about things like conserved or almost conserved quantities can be really useful to use here, since they can tell you about things you might otherwise miss.
| 2 | https://mathoverflow.net/users/3329 | 29013 | 18,959 |
https://mathoverflow.net/questions/28997 | 51 | For many standard, well-understood theorems the proofs have been streamlined to the point where you just need to understand the proof once and you remember the general idea forever. At this point I have learned three different proofs of the Birkhoff ergodic theorem on three separate occasions and yet I still could probably not explain any of them to a friend or even sit down and recover all of the details. The problem seems to be that they all depend crucially on some frustrating little combinatorial trick, each of which was apparently invented just to service this one result. Has anyone seen a more natural approach that I might actually be able to remember? Note that I'm not necessarily looking for a short proof (those are often the worst offenders) - I'm looking for an argument that will make me feel like I could have invented it if I were given enough time.
| https://mathoverflow.net/users/4362 | Does anyone know an intuitive proof of the Birkhoff ergodic theorem? | I don't know whether this helps or whether you've already seen this before, but this made a lot more intuitive sense to me than the combinatorial approach in Halmos's book.
The key point in the proof is to prove the maximal ergodic theorem. This states that if $M\_T$ is the maximal operator $M\_T f= \sup\_{n >0} \frac{1}{n+1} (f + Tf + \dots + T^n f)$, then $\int\_{M\_T f>0} f \geq 0$. Here $T$ is the associated map on functions coming from the measure-preserving transformation.
This is a weak-type inequality, and the fact that one considers the maximal operator isn't terribly surprising given how they arise in a) the proof of the Lebesgue differentation theorem (namely, via the Hardy-Littlewood maximal operator $Mf(x) = \sup\_{t >0} \frac{1}{2t} \int\_{x-t}^{x+t} |f(r)| dr$. b) In the theory of singular integrals, one can define a maximal operator in the same way and prove that it is $L^p$-bounded for $1 < p < \infty$ and weak-$L^1$ bounded in suitably nice homogeneous cases (e.g. the Hilbert transform). One of the consequences of this is, for instance, that the Hilbert transform can be computed a.e. via the Cauchy principal value of the usual integral. c) I'm pretty sure the boundedness of the maximal operator of the partial sums of Fourier series is used in the proof of the Carleson-Hunt theorem. So using maximal operators (and, in particular, weak bounds on them) to establish convergence is fairly standard. Once the maximal inequality has been established, it isn't usually very hard to get the pointwise convergence result, and the ergodic theorem is no exception.
The maximal ergodic theorem actually generalizes to the case where $T$ is an operator of $L^1$-norm at most 1, and thinking of it in a more general sense might meet the criteria of your question. In particular, let $T$ be as just mentioned, and consider $M\_T$ described in the analogous way. Or rather, consider $M\_T'f = \sup\_{n \geq 0} \sum\_{i=0}^n T^if$. Clearly $M\_T'f >0$ iff $M\_Tf >0$. Moreover, $M\_T'$ has the crucial property that $T M\_T' f + f = M\_T' f$ whenever $M\_Tf>0$.
Therefore,
$\int\_{M\_T'f>0} f = \int\_{M\_T'f>0} M\_T'f - \int\_{M\_T'f>0} TM\_T'f.$ The first part is in fact $||M\_T'f||\_1$ because the modified maximal operator is always nonnegative. The second part is at most $||T M\_T'f|| \leq ||M\_T'f||$ by the norm condition. Hence the difference is nonnegative.
Perhaps this will be useful: let $M$ be an operator (not necessarily linear) sending functions to nonnegative functions such that $(T-I)Mf = f$ wherever $Mf>0$, for $T$ an operator of $L^1$-norm at most 1). Then $\int\_{Mf>0} f \geq 0$. The proof is the same.
| 25 | https://mathoverflow.net/users/344 | 29014 | 18,960 |
https://mathoverflow.net/questions/28879 | 7 | A von Dyck group is a group with presentation $< a,b | a^m=b^n=(ab)^p=1 >$ with m,n,p natural numbers. Is it known which of these groups are solvable and which are not? Is there a reference for this? Thanks.
| https://mathoverflow.net/users/3804 | von dyck groups and solvability | You might try Generators and Relations for Discrete Groups by Coxeter and Moser.
Specifically for 1/m + 1/n + 1/p = 1 there are only 3 cases up to permutation, (2,3,6), (2,4,4) and (3,3,3). Map a and b to an appropriate root of unity to get a homomorphism onto C\_6, C\_4, or C\_3, respectively. The kernel of the map is in all three cases isomorphic to Z^2.
| 5 | https://mathoverflow.net/users/6787 | 29032 | 18,974 |
https://mathoverflow.net/questions/28992 | 14 | I am just reading about Iwasawa theory about Coates and Sujatha's book on Iwasawa Theory. I was wondering that since Iwasawa thought about the whole theory from the analogy of curves over finite fields, so what should be the analog of the module $U\_\infty$/$C\_\infty$ in the curve case (if there is any) where $U\_\infty$ is the inverse system of local units and $C\_\infty$ is the cyclotomic units.
| https://mathoverflow.net/users/2081 | A question about Iwasawa Theory | There is a very close analogy but to unravel it requires some work.
So take $X$ a smooth curve over $\mathbb F\_{\ell}$ (more generally you could take $X$ a scheme over $\mathbb F\_{\ell}$) and let $\mathscr F$ be a smooth sheaf of $\mathbb Q\_{p}$-vector spaces on $X$ (you could be much more general in your choice of coefficient ring, and indeed, I think you might need to consider more general coefficient rings in order to really grasp the analogy, but that will do for the moment). Moreover, we will assume for simplicity that $\mathscr F$ comes from a motive over $X$, a sentence which will remain vague but aims to convey the idea that $\mathscr F$ has geometric origin.
Then the cohomology complex $R\Gamma(X,\mathscr F)$ is a perfect complex so it has a determinant $D$. This complex fits in a exact triangle
$$ R\Gamma(X,\mathscr F)\rightarrow R\Gamma(X\otimes\bar{\mathbb F}\_{\ell},\mathscr F)\rightarrow R\Gamma(X\otimes\bar{\mathbb F}\_{\ell},\mathscr F)$$
Here the very important fact to understand in order to grasp the analogy is that the second arrow is given by $Fr(\ell)-1$. This exact triangle induces an isomorphism $f$ of $D$ with $\mathbb Q\_{p}$.
There is conjecturally another such isomorphism. Assume that the action of the Frobenius $Fr(\ell)$ acts semi-simply on $H^{i}(\bar{X},\mathscr F)$ for all $i$ (this is widely believed under our hypothesis on $\mathscr F$). Then degeneracy of a the spectral sequence $H^{i}(\mathbb F\_{\ell},H^{j}(X\otimes\bar{\mathbb F}\_{\ell},\mathscr F))$ gives an isomorphism $g$ between $D$ and $\mathbb Q\_{p}$.
Now, consider $gf^{-1}(1)$. This happens to be the residue at 1 of the zeta function of $X$.
What has all this to do with units in number fields? Change setting a bit and take $X\_{n}=\operatorname{Spec}\mathbb Z[1/p,\zeta\_{p^{n}}]$ and $\mathscr F=\mathbb Z(1)$. We would like to carry the same procedure as above but we can't, because we are lacking crucially the exact triangle involved in the definition of the isomorphism $f$. Nonetheless, there is a significantly more sophisticate way to construct a suitable $f\_{n}$ for all $n$ and it turns out that this construction will crucially involve $U\_{\infty}/C\_{\infty}$.
So to sum up, the analog of $U\_{\infty}/C\_{\infty}$ for curves over finite fields is none other than the Frobenius morphism $Fr(\ell)-1$. You may know that the cyclotomic units form an Euler system, that is to say that they satisfy relations involving corestriction and the characteristic polynomial of the Frobenius morphisms. This fact is I believe what led K.Kato to describe the analogy above.
You can read about all this in much much greater details in the contribution of Kato in the volume Arithmetic Algebraic Geometry (Springer Lecture Notes 1553).
| 9 | https://mathoverflow.net/users/2284 | 29059 | 18,990 |
https://mathoverflow.net/questions/29033 | -1 | Define $$x\_{k+1}(t)=\frac{3x^4\_k(t)+6(1-t)x\_k^2(t)-(1-t)^2}{8x\_k^3(t)},$$
with $x\_0(t)=1$. It is not difficult to see $x\_k(t)$ converges to $\sqrt{1-t}$, whose (Maclaurin expansion) has negative coefficients unless the first one. Let
$$x\_k(t)=\sum\limits\_{i =0}^{\infty}c\_{k,i}t^{i}, \mid t\mid<1$$
Is it true $c\_{k,i}\le0$ for all $k\ge 1, i\ge 1$?
| https://mathoverflow.net/users/6858 | A classical analysis problem | This is not the first time I am fighting with positivity (nonnegativity).
But this problem looks not natural enough for a standard technique, and
it seems to me that the resulting sequence $x\_k(t)$ is always between
$\sqrt{1-t}$ and $1$, in the sense that the expansions of $x\_k(t)-\sqrt{1-t}$
and $1-x\_k(t)$ have nonnegative coefficients only.
I can only give a partial solution. After the change
$$
z\_k(t)=1-\frac{x\_k(t)}{\sqrt{1-t}}
$$
the recursion for $x\_k(t)$ translates into
$$
z\_{k+1}=\frac{z\_k^3(4-3z\_k)}{8(1-z\_k)^3}
\quad\text{for}\quad k=0,1,2,\dots
$$
with the initial data
$$
z\_0=1-\frac1{\sqrt{1-t}}=-\frac12t+O(t^2).
$$
The recursion implies $z\_{k+1}=z\_k^3/2+O(z\_k^4)$, therefore
by induction on $k$ we obtain
$$
z\_k=-\frac{t^{3^k}}{2^{(3^{k+1}-1)/2}}+\text{higher terms}.
$$
This implies that
$$
x\_k(t)=\sqrt{1-t}-\sqrt{1-t}z\_k(t)
$$
agrees with $\sqrt{1-t}$ up to the term $t^{3^k}$. In particular,
as the expansion of $\sqrt{1-t}$ involves negative coefficients besides
the constant term, we conclude that $c\_{k,i}<0$ for $i=1,2,\dots,3^k$
in the expansion
$$
x\_k(t)=1+\sum\_{k=1}^\infty c\_{k,i}t^i.
$$
| 3 | https://mathoverflow.net/users/4953 | 29075 | 18,997 |
https://mathoverflow.net/questions/29073 | 5 | Let $G$ be a linear algebraic group over an algebraically closed field $k$, and $T$ a maximal torus of $G$.
Suppose we have two cocharacter $\mu, \mu' : \mathbb{G}\_m \to T$, which are conjugate under $G$ i.e. there exists $g \in G$ such that $\mu'(z) = g\mu(z)g^{-1}$.
**Question.** Can we always choose $g \in G$ such that $\mu'(z) = g\mu(z)g^{-1}$ and such that $g$ normalizes $T$?
If $G = \mathrm{GL}\_n$ then this is true: suppose that $T$ is the diagonal torus. Then a cocharacter of $T$ is just determined by its weights on the basis vectors of the natural representation of $G$ (i.e. integers $k\_i$ such that $\mu(z)e\_i = z^{k\_i} e\_i$).
Since $\mu'$, $\mu$ are $G$-conjugate, the weights of $\mu$ are a permutation of the weights of $\mu'$. We can choose an element of $G$ which induces this permutation on the basis vectors.
For general $G$, we can embed $G$ in $\mathrm{GL}\_n$ for some $G$, such that $T$ is the intersection of $G$ with the diagonal torus. The weights of $\mu$ on the basis vectors of the corresponding representation are a permutation of the weights of $\mu'$. The question is whether we can choose this permutation (since the weights need not be distinct, there may be several choices for the permutation) so that it is induced by $N\_G(T)$.
For my application, to show that the number of $G$-conjugates of $\mu$ in $T$ is finite, it is enough to know this for $\mathrm{GL}\_n$, but it would be nice to know if it is true in general.
| https://mathoverflow.net/users/1046 | Conjugate cocharacters in a maximal torus | Yes. That $\mu'(z) = g \mu(z) g^{-1}$ means that $g T g^{-1}$ centralizes the image of $\mu'$. Thus, $T$ and $g T g^{-1}$ are maximal tori in the centralizer
of the image of $\mu'$, so there exists $h$ in this centralizer such that $g T g^{-1} = h T h^{-1}$. Now replace $g$ by $h^{-1} g$. It doesn't seem that you need G to be semisimple or reductive for this, but I may be missing something here?
| 7 | https://mathoverflow.net/users/6982 | 29081 | 19,000 |
https://mathoverflow.net/questions/29074 | 9 | The [separation axioms](http://ncatlab.org/nlab/show/separation+axioms) have exploded a little since the original list of four! Amongst them can be found "completely regular" spaces and "perfectly normal" spaces. The former is well-known: a point can be separated from a disjoint closed subset by a continuous real-valued function. The latter may be less well-known, but is fairly simple: given two disjoint closed sets then there is a continuous real-valued function such that the first closed set is the preimage of {0} and the second closed set is the preimage of {1}. This is equivalent to the condition that every closed set be the set of zeros of some continuous real-valued function.
Urysohn's lemma says that there's little point in looking for a *completely normal* space: in a normal space, any disjoint closed sets can be separated by a continuous real-valued function. Similarly, if we imposed the obvious exactness condition from perfectly normal onto completely regular then we'd actually end up with *T6* (perfectly normal and *T0*).
However, there's room in the middle for something else and that's what I'm interested in: spaces which are completely regular and are such that the function separating a point from a closed set can be chosen so that the point is the preimage of its value. No assumption is made on the interaction of the closed set and the function (beyond that already given by the "completely regular" condition).
So my question is simple:
>
> Has anyone encountered this notion before, and if so, where?
>
>
>
If it helps, I think that this condition is equivalent to the space being completely regular and singleton sets being *Gδ* sets (intersection of a countable number of open sets). Note that this isn't the same as first countable.
---
**Motivation**: As one might expect, my motivation comes from [Froelicher spaces](http://ncatlab.org/nlab/show/Froelicher+space). Any Froelicher space comes equipped with two topologies: generated by either the curves or the functions. I'm trying to find conditions under which they are equal. Since the identity is continuous from the curvaceous topology to the functional topology, theorems like "a continuous bijection from a compact space to a Hausdorff space is a homeomorphism" spring to mind. Hausdorff is no problem - I can easily assume that both topologies are Hausdorff. In another context, I decided that I quite liked it when the curvaceous topology was *sequentially* compact. I think that I can prove that a continuous bijection from a sequentially compact space to a "perfectly regular" space is a homeomorphism, so it's a bit like a weakening of one condition at the expense of strengthening the other.
For a purely topological motivation, consider the following. In a completely regular space, one can test whether or not a net converges to a particular point by looking at its image under all continuous functions. In a "perfectly regular" space, one can test whether a **convergent** net converges to a particular point by using just one function and that function depends only on the proposed limit and not on the net.
So for both of these reasons, it's a nice condition to have and I wondered if it was known about so that, hopefully, I could build on others' work rather than having to invent it myself (hey, I've invented this really cool round thing!).
| https://mathoverflow.net/users/45 | Is there a notion of a "perfectly regular" topological space? | An answer is: completely regular plus countable pseudocharacter, the latter means that points are $G\_\delta$-sets. In completely regular spaces a point is a $G\_\delta$-set iff it is the zero-set of a continuous function, the proof is just like that for arbitrary closed $G\_\delta$-sets in normal spaces: if $\lbrace x\rbrace=\bigcap\_u U\_n$ then pick continuous functions $f\_n:X\to[0,1]$ that are zero at $x$ and $1$ outside $U\_n$; then $\sum\_n2^{-n}f\_n$ has $\lbrace x\rbrace$ as its zero-set.
See Engeling's *General Topology*, proof of Corollary 1.5.12.
| 8 | https://mathoverflow.net/users/5903 | 29093 | 19,004 |
https://mathoverflow.net/questions/29088 | 11 | Martin Gardner kept voluminous correspondence with amateur and professional mathematicians worldwide throughout his career. His files are a treasure trove of information about all areas of recreational mathematics. Does anybody here know whether any portion of those files will be made available to the public for research purposes?
| https://mathoverflow.net/users/3106 | What is happening to Martin Gardner's files? | According to a simple Google search, the papers were donated to Stanford.
<http://www.oac.cdlib.org/data/13030/6s/kt6s20356s/files/kt6s20356s.pdf>
| 18 | https://mathoverflow.net/users/454 | 29094 | 19,005 |
https://mathoverflow.net/questions/29102 | 7 | In the book "A = B" by Petkovesk, Wilf, and Zeilberger, [(downloadable here)](http://www.math.upenn.edu/~wilf/Downld.html), the authors provide several algorithmic methods for finding closed forms or recurrences for sums involving e.g. binomial coefficients. Even more exciting, their methods provide seemingly short certificates for the truth of these computer-verified claims. In particular, the WZ method prints a single rational function as such a certificate.
In more detail, here is the broad outline of the WZ method, where I directly quote from page 25 of "A = B":
>
> 1. Suppose that you wish to prove an identity of the form $\sum\_k t(n, k) = rhs(n)$,
> and let’s assume, for now, that for each n it is true that the summand $t(n, k)$
> vanishes for all $k$ outside of some finite interval.
> 2. Divide through by the right hand side, so the identity that you wish to prove
> now reads as $\sum\_k F (n, k) = 1$, where $F (n, k) = t(n, k)/rhs(n)$.
> 3. Let $R(n, k)$ be the rational function that the WZ method provides as the proof
> of your identity (this is described in Chapter 7 of "A=B"). Define a
> new function $G(n, k) = R(n, k)F (n, k).$
> 4. You will now observe that the equation
> $$F (n + 1, k) − F (n, k) = G(n, k + 1) − G(n, k)$$
> is true. Sum that equation over all integers $k$, and note that the right side
> telescopes to 0. The result is that
> $$\sum\_k F (n + 1, k) = \sum\_k F (n, k),$$
> hence we have shown that $\sum\_k F (n, k)$ is independent of $n$, i.e., is constant.
> 5. Verify that the constant is $1$ by checking that $F (0, k) = 1$.
>
>
>
What I want to know is:
>
> Are there known bounds on the length of these certificates $R(n, k)$, in terms of the length of the description of the combinatorial sum in question? If so, what are they?
>
>
>
| https://mathoverflow.net/users/6950 | How long are the certificates produced by the Zeilberger and WZ methods for solving combinatorial sums (A=B)? | Already for the case of Gosper summation (single-variable), it is known that things can get exponentially larger than the input, because the 'answer' fundamentally depends on the *dispersion* of the input term. You will find much more comprehensive answers in the papers of [Sergei Abramov](http://www.ccas.ru/sabramov/publications.htm) as well as those of [Marko Petkovsek](http://www.fmf.uni-lj.si/~petkovsek/papers.html) (especially so in their joint papers!)
The *disperson* of a polynomial $p(x)$ is the largest integer $n$ such that $p(x)$ and $p(x+n)$ have a non-trivial gcd. A good understanding of how dispersion enters the picture can be gotten from the paper [Shiftless decomposition and polynomial-time rational summation](http://dx.doi.org/10.1145/860854.860887).
| 6 | https://mathoverflow.net/users/3993 | 29106 | 19,008 |
https://mathoverflow.net/questions/29115 | 4 | I just read a proof and, after struggling some time with a mental leap, I think that it uses tacitly the following:
Let $\kappa$ be a regular cardinal, $\theta > \kappa$ a regular cardinal too then:
$ S \subset \kappa$ is stationary if and only if
$\forall \mathcal{A} = (H(\theta), \in, <,..) \exists M \prec \mathcal{A}, |M| < \kappa,$ such that $sup(M \cap \kappa) \in S$.
Now my questions are:
1. Is this statement above even true? (I think so as I have a proof, but this doesn't have to mean anything)
2. It appears to me that the latter part of this characterization is a quite strong assumption as $\mathcal{A}$ might contain a lot of additional information, so is there a possibility to weaken it? Or could you mention any similar statements to the one above?
Thank you
EDIT: I accepted the answer of Philip, simply because he has lower points. Francois answer would have deserved it too.
| https://mathoverflow.net/users/4753 | A characterization of stationarity? | (I first wanted to give an answer, but I was not quick enough. I then wanted to add a small comment and found out after 20 minutes that I had insufficient reputation.)
The comment was regarding 2) of oktan's original query: having $H(\theta)$
in the structure is overkill: it suffices to have $( \kappa, <, \in, C)$. (One does not need the structure to be able to express $C$ is closed.'')
| 10 | https://mathoverflow.net/users/6942 | 29128 | 19,019 |
https://mathoverflow.net/questions/29123 | 1 | Can a harmonic function defined on the upper half-plain (or any domain which is unbounded) be extended to the point at infinity. If so, under what condition. What happens to the mean value property then ? Do we still get a integral representation of some sort. Please suggest a reference.
Thank you.
| https://mathoverflow.net/users/6766 | Extension of harmonic function at infinity | One has this type of representations for Herglotz functions under certain growth conditions.
Herglotz means here that the function maps the upper half plane into the upper half plane.
This can for example be found in the spectral theory book by Teschl ( <http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/index.html> ). Section 3.4. seems to be the relevant one.
Of course these theorems need a growth condition. For that book for example $|F(z)| \leq \frac{M}{im(z)}$ for some constant $M > 0$. But I think one can extend this to some moderate growth at infinity. For example $|F(z)| = O(\sqrt{|z|})$ as $z\to\infty$ is ok.
| 1 | https://mathoverflow.net/users/3983 | 29132 | 19,023 |
https://mathoverflow.net/questions/29104 | 26 | As a person who has been spending significant time to learn mathematics, I have to admit that I sometimes find the fact uncovered by Godel very upsetting: we never can know that our axiom system is consistent. The consistency of ZFC can only be proved in a larger system, whose consistency is unknown.
That means proofs are not like as I once used to believe: a certificate that a counterexample for a statement can not be found. For example, in spite of the proof of Wiles, it is conceivable that someday someone can come up with integers a,b and c and n>2 such that a^n + b^n = c^n, which would mean that our axiom system happened to be inconsistent.
I would like to learn about the reasons that, in spite of Godel's thoerem, mathematicians (or you) think that proofs are still very valuable. Why do they worry less and less each day about Godel's theorem (edit: or do they)?
I would also appreciate references written for non-experts addressing this question.
| https://mathoverflow.net/users/1229 | Why are proofs so valuable, although we do not know that our axiom system is consistent? | If you like, you can view proofs of a statement in some formal system (e.g. ZFC) as a certificate that a counterexample cannot be found without demonstrating the inconsistency of ZFC, which would be a major mathematical event, and probably one of far greater significance than whether one's given statement was true or false.
In practice, a given proof is not going to be closely tied to a single formal system such as ZFC, but will be robust enough that it can follow from any number of reasonable sets of axioms, including those much weaker than ZFC. Only one of these sets of axioms then needs to be consistent in order to guarantee that no counterexample would ever be found, and this is about as close to an ironclad guarantee as one can ever hope for.
But ultimately, mathematicians are not really after proofs, despite appearances; they are after *understanding*. This is discussed quite well in Thurston's article "[On proof and progress in mathematics](http://arxiv.org/abs/math/9404236)".
| 68 | https://mathoverflow.net/users/766 | 29133 | 19,024 |
https://mathoverflow.net/questions/29138 | 4 | If $E$ is a complex vector bundle over a manifold $M$ then one defines the space of vector valued $p$-differential forms on them as $\Omega^p(M,E) = \Gamma ( \wedge ^p (T^\*M) \otimes E) $
The connection be defined as the map , $\nabla: \Gamma(E) \rightarrow \Omega^1(M,E)$ satisfying $\nabla(fX) = df\otimes X+f\nabla X$ where $f \in C^{\infty}(M)$ and $X \in \Gamma(E)$
* Can't the connection be thought of as an element of $\Omega^1(M,End(E))$?
Difference of two connections though not a connection is such an element.
Curvature of the connection is defined as the map, $$R = \nabla \circ \nabla : \Gamma(E) \rightarrow \Omega^2(M,E)$$
Corresponding definition of "trace" is as a map $Tr: \Omega^p(M,End(E))\rightarrow \Omega^p(M)$ satisfying some natural conditions.
One can show that if $A,B \in \Omega^p(M,End(E))$ then $Tr[A,B]=0$
The crucial relationship from which a version of the Chern-Weil Theorem becomes almost immediately obvious is this,
$$dTr[A] = Tr[[\nabla,A]]$$
The argument for this begins by choosing a different connection say $\nabla'$ and seeing that, $$Tr[[\nabla',A]] = Tr[[\nabla,A]]$$
Hence the RHS of the desired equation is independent of the connection chosen and hence it can be evaluated for any connection to get the LHS. Using the fact that the a bundle is locally trivial one can choose the "trivial" connection and this should apparently yield $dTr[A]$
* Here I can't see what is a "trivial connection". It would help if someone can write that down in local trivializing coordinates. And how for that does the evaluation of the Tr give a de-Rham derivative of the Tr. The appearance of the de-Rham derivative on the LHS looks very mysterious and thats what makes the Chern-Weil Theorem click! It would help if someone can give the intuition behind this.
One defines the Chern form as $det(I + \frac{\sqrt{-1}}{2\pi}R)$
+ In this definition $I$ is the identity automorphism of $E$ but $R$ takes values in $\Omega^2(M,End(E))$. Then how is the "+" defined?
What is a good reference for this approach to Chern-Weil Theory in the language of connections and curvature?
I found the notation of Kobayashi's lectures very old to relate to and Weiping's book is too terse and Milnor-Stasheff's book does it in the language of cohomology.
| https://mathoverflow.net/users/2678 | Proving the basic identity which implies the Chern-Weil theorem | (1) No, a connection is not a section of $\Omega^1(M,\mathrm{End}(E))$: a section would act tensorially and not satisfy the Leibniz rule. The connection is $\mathbb{C}$ linear and not $C^\infty(M,\mathbb{C})$ linear.
(2) Since you have a vector bundle over some manifold, by definition there is some complex vector space $V$ such that around a neighbourhood of some point $p$ in the base manifold $M$, $E$ splits locally as $U\times V$ (a bundle is by definition locally trivial), where $U$ is a domain in $\mathbb{R}^n$. A section of $E$ can be locally expressed as a $V$-valued function on $U$. The trivial connection is just given by component-wise partial derivation. (In coordinates, we can pick a basis of $V$ and use the standard coordinate on $\mathbb{R}^n$, then you just have a map from a domain in Euclidean space to Euclidean space, given by a collection of functions. The trivial connection acts on each of the functions like the exterior derivative.)
(3) The Chern form takes value in $H^\*(M,\mathbb{R})$... if you accept that you can add objects in $H^2$ and $H^4$, why not the expression in the determinant? If you worry about such things, perhaps you'd be happier with the definition that the k'th Chern class is given by $(i / 2\pi)^k \sigma\_k(\Omega)$, where $\sigma\_k$ is the k'th symmetric polynomial acting on $\Omega$.
(4) Another book that may be useful is Morita's *Geometry of Differential Forms*.
| 2 | https://mathoverflow.net/users/3948 | 29158 | 19,040 |
https://mathoverflow.net/questions/27805 | 31 | Given a solution $S$ of the Navier-Stokes equations, is there a way to make mathematically precise a statement like: "$S$ is [turbulent](https://en.wikipedia.org/wiki/Turbulence) in the spacetime region $U$"?
And if such a definition exists, are there any known exact solutions of Navier-Stokes exhibiting turbulence?
Reading Wikipedia makes me also want to ask the following related questions:
1. Does a single [vortex](https://en.wikipedia.org/wiki/Vortex) count as turbulence?
2. Is the appearance of vortices a necessary feature of turbulence?
3. What's the difference between a vortex and an [eddy](https://en.wikipedia.org/wiki/Eddy_(fluid_dynamics))?
Pointers to (mathematically rigorous) literature are much appreciated.
| https://mathoverflow.net/users/745 | Is there a mathematically precise definition of turbulence for solutions of Navier-Stokes? | There is probably no universally accepted mathematical definition of turbulence. (By the way, is there a physical one?) Moreover, the prevailing definitions seem to be highly volatile and time-dependent themselves.
A few notable examples.
* In the Ptolemaic [Landau–Hopf theory](https://en.wikipedia.org/wiki/Landau%E2%80%93Hopf_theory_of_turbulence) turbulence is understood as a cascade of bifurcations from unstable equilibriums via periodic solutions ([the Hopf bifurcation](https://en.wikipedia.org/wiki/Hopf_bifurcation)) to quasiperiodic solutions with arbitrarily large frequency basis.
* According to [Arnold and Khesin](https://books.google.co.uk/books?id=LJapQwAACAAJ&dq=arnold%20topological%20methods&hl=en&ei=dvEhTO_SDY_9_AaNrogq&sa=X&oi=book_result&ct=result), in the 1960's most specialists in PDEs regarded the lack of global existence and uniqueness theorems for solutions of the 3D Navier–Stokes equation as the explanation of turbulence.
* Kolmogorov suggested to study *minimal attractors* of the Navier-Stokes equations and formulated several conjectures as plausible explanations of turbulence. The weakest one says that the maximum of the dimensions of minimal attractors of the Navier–Stokes equations grows along with the Reynolds number Re.
* In 1970 Ruelle and Takens formulated the conjecture that turbulence is the appearance
of *global attractors* with sensitive dependence of motion on the initial conditions in the phase space of the Navier–Stokes equations ([link](https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-20/issue-3/On-the-nature-of-turbulence/cmp/1103857186.full)). In spite of the vast popularity of their paper, even the existence of such attractors is still unknown.
**Edit 1.** Concerning explicit solutions to the Navier-Stokes equations, I don't think any of them really exhibit turbulence features. The thing is that the nonlinear term $v\cdot \nabla v$ is equal to $0$ for most known classical explicit solutions. In other words, these solutions
actually solve the linear Stokes equation and don't "see" the nonlinearity of the full Navier-Stokes system. This is probably not what one would expect from a truly turbulent flow.
**Edit 2.** As for the quick reference, you may find helpful the short [survey on turbulence theories](http://www.labma.ufrj.br/%7Errosa/dvifiles/encyturb.pdf) by Ricardo Rosa. ([Wayback Machine](https://web.archive.org/web/20170329151350/http://www.labma.ufrj.br/%7Errosa/dvifiles/encyturb.pdf)) It appears as an article in the [Encyclopedia of Mathematical Physics](http://www.sciencedirect.com/science?_ob=RefWorkIndexURL&_idxType=AU&_cdi=27008&_refWorkId=327&_acct=C000015798&_version=1&_userid=273788&md5=b04306d338adb2412b2379b81f51f3d0).
| 19 | https://mathoverflow.net/users/5371 | 29159 | 19,041 |
https://mathoverflow.net/questions/29136 | 7 | The theta function of a lattice is defined to be
$$ \vartheta\_\Lambda = \sum\_{v\in\Lambda} q^{{\Vert v\Vert}^2}$$
which yields as a coefficient of *qk* the number of vectors of norm-squared *k*.
On the other hand, the Jacobi theta function is given by
$$ \vartheta(u,q) = \sum\_{n=-\infty}^\infty u^{2n}q^{n^2}$$
and we have the obvious fact that if $\Lambda = \mathbb{Z}$ with its usual intersection form, then $\vartheta(1,q)$ is the theta function for that lattice.
We also have the fact that $\vartheta\_{\Lambda\_1 \oplus \Lambda\_2} = \vartheta\_{\Lambda\_1}\vartheta\_{\Lambda\_2}$, and so we can decompose our theta functions into products of theta functions of primitive lattices.
Combining these facts, it is not entirely ridiculous to hope that there is some way to write, for a lattice of rank *k*, a ``theta function'' of the form
$$\vartheta(u\_1, \ldots, u\_k, q)$$
such that $\vartheta(1, \ldots, 1, q)$ is the ordinary theta function of the lattice. In some sense, the *u*-variables keep track of the basis elements of the lattice which immediately raises the question as to well-definedness of such an idea; it is worth noting that for the lattice $\Lambda = \bigoplus\_i\mathbb{Z}$ that this definition does make sense.
So is there any literature on such objects? Do they make sense for lattices which are not just sums of copies of ℤ? Do they have nice relations akin to those of normal theta functions?
| https://mathoverflow.net/users/1703 | Is there any literature on multivariable theta functions? | There are three ways to view theta functions
1. as classical homomorphic functions in
vector z and/or period matrix T
2. as matrix coefficients of a representation of the
Heisenberg and/or Metaplectic grp
3. as sections of Line bundles on the Abelian variety
and/or moduli space of the abelian variety
Ram Murty's [Theta functions - from the classical to the modern](http://books.google.com/books?id=ZXvCc0zLtKUC) discusses Weil's representation-theoretic interpretation of theta functions. See chapter 3 by Hoffstein on *Eisenstein series and theta functions on the metaplectic group*. It is the connection to the [metaplectic group](http://en.wikipedia.org/wiki/Metaplectic_group) which gives rise to the functional equation of the multivariable theta function, which you will also find in the chapter on the Metaplectic group in vol 3 of Mumford's *Tata Lectures*
Bellman's [Brief introduction to Theta functions](http://rads.stackoverflow.com/amzn/click/0030103606) Section 61 alludes to theta functions in several complex variables.
You may also want to search for material in books on Abelian varieties. For example, Baker's [Abelian functions Chapter X](http://books.google.com/books?id=q4iHCNXXX_UC&lpg=PP1&pg=PR9#v=onepage&q&f=false) develops the theory based on the period matrix. Also Murty's book on [Abelian varieties](http://books.google.com/books?id=1bJKqHqWgp4C&lpg=PA3&dq=Murty%2520Abelian%2520Varieties&pg=PA3#v=onepage&q&f=false) and Polishchuk's [Abelian Varieties, Theta Functions and the Fourier Transform](http://rads.stackoverflow.com/amzn/click/0521808049)
Tyurin's [Quantization, Classical and Quantum Field Theory, and Theta Functions](http://rads.stackoverflow.com/amzn/click/0821832409) might also be a useful reference, which I haven't browsed.
See also: [Springer Encyclopedia of Math entry on theta functions](http://eom.springer.de/t/t092600.htm)
| 7 | https://mathoverflow.net/users/5372 | 29165 | 19,045 |
https://mathoverflow.net/questions/29149 | 5 | Given a regular tessellation, i.e. either a platonic solid (a tessellation of the sphere), the tessellation of the euclidean plane by squares or by regular hexagons, or a regular tessellation of the hyperbolic plane.
One can consider its isometry group $G$. It acts on the set of all faces $F$. I want to define a symmetric coloring of the tessellation as a surjective map from $c:F\rightarrow C$ to a finite set of colors $C$, such that for each group element $G$ there is a permutation $p\_g$ of the colors, such that $c(gx)=p\_g\circ c(x)$. ($p:G\rightarrow $Sym$(C)$ is a group homomorphism).
Examples for such colorings are the trivial coloring $c:F\rightarrow \{1\}$ or the coloring of the plane as an infinite chessboard.
The only nontrivial symmetric colorings of the tetrahedron, is the one, that assigns a different color to each face. For the other platonic solids there are also those colorings that assign the same colors only to opposite faces.
So my question is: Does every regular tessellation of the hyperbolic plane admit a nontrivial symmetric coloring?
I wanted to write a computer program that visualizes those tessellations, but I didn't find a good strategy which colors should be used. So I came up with this question.
| https://mathoverflow.net/users/3969 | Symmetric colorings of regular tessellations | The answer is yes. Moreover, for every two different faces $A$ and $B$ there is a symmetric coloring assigning different colors to $A$ and $B$.
The isometry group $G$ is residually finite, hence here is a normal finite index subgroup $H$ of $G$ that contains no elements (except the identity) sending $A$ to itself or to $B$. Assign a unique color to each orbit of $H$.
The coloring symmetry condition is essentially he following: if $f\in G$ and faces $X$ and $Y$ are of the same color, then so are $f(X)$ and $f(Y)$. Since $X$ and $Y$ are of the same color, there exists $h\in H$ such that $h(X)=Y$. Since $H$ is normal, $h\_1:=f^{-1}hf\in H$. But $h\_1(F(X))=F(Y)$, hence $F(X)$ and $F(Y)$ are of the same color.
| 5 | https://mathoverflow.net/users/4354 | 29168 | 19,046 |
https://mathoverflow.net/questions/29160 | 1 | **Background**
This can be generalised, but let me be fairly concrete. Let $X$ be a simply-connected riemannian manifold and let $G$ denote the Lie group of isometries, assumed nontrivial. Let $F < G$ be a finite subgroup acting freely and consider the smooth quotient $X/F$ with the induced riemannian structure.
The normaliser $N(F) < G$ still acts on $X/F$ isometrically with $F < N(F)$ acting trivially. So $X/F$ inherits an isometric action of the group $N(F)/F$.
Now let $E < N(F)/F$ be a finite subgroup acting freely on $X/F$ and consider the quotient $(X/F)/E$. This is a smooth manifold, locally isometric to $X$ and hence isometric to $X/D$ for some freely-acting subgroup $D<G$.
**Question**
How is $D$ related to $E$ and $F$? I would expect $D$ to be an extension of $E$ by $F$. Is it? And if so, but does it split? And if not, is there a name for this construction?
| https://mathoverflow.net/users/394 | A question about iterated quotients in riemannian geometry | The group $D$ is the preimage of $E$ in $N(F)$, so it is as you expect. The finiteness hypothesis can be weakened, which is important for many applications. Things become clearer if one thinks categorically in terms of the universal properties.
Say an arbitrary group $G$ acts freely and properly discontinuously and isometrically on $X$ and $H$ is a normal subgroup of $G$. Then $G/H$ acts freely and properly discontinuously and isometrically on $X/H$ with $X \rightarrow (X/H)(G/H)$ a $G$-invariant map. The induced map $f:X/G \rightarrow (X/H)(G/H)$ is an isomorphism. Indeed, both sides composed back with the natural map from $X$ satisfy the same universal property, and $f$ respects the maps from $X$, so $f$ is an isomorphism. QED
In fact, with more work this can all be done more generally with $G$ and Lie group and $H$ a closed normal Lie subgroup, under suitable "niceness" hypotheses for the orbit maps (which are satisfied in the above situation): see Proposition 13 in section 1.6 of Chapter III of Bourbaki LIE.
| 5 | https://mathoverflow.net/users/6773 | 29171 | 19,048 |
https://mathoverflow.net/questions/29118 | 26 | I am trying to prove $\sum\limits\_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$. This inequality has been verified by computer for $k\le40$.
Some clues that might work (kindly provided by Doron Zeilberger) are as follows:
1. Let $Ef(x):=f(x-1)$, let $P\_k(E):=\sum\_{j=0}^{k-1}(-1)^{(j+1)}\binom{2k+1}{j}E^j$;
2. These satisfy the inhomogeneous recurrence $P\_k(E)-(1-E)^2P\_{k-1}(E) =$ some binomial in $E$;
3. The original sum can be expressed as $P\_k(E)x^{(2k-2)} |\_{x=k} $;
4. Try to derive a recurrence for $P\_k(E)x^{(2k-2)}$ before plugging in
$x=k$ and somehow use induction, possibly having to prove a more general statement to facilitate the induction.
Unfortunately I do not know how to find a recurrence such as suggested by clue 4.
| https://mathoverflow.net/users/6989 | Is the sum $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0?$ | Your expression is the difference of two central Eulerian numbers ,
$$A(k):=\sum\_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2}{2k+1 \choose j}=\left \langle {2k-2\atop k-2} \right \rangle-\left \langle {2k-2\atop k-3} \right \rangle$$
as you can easily deduce from their closed formula. The positivity of $A(k)$ is just due to the fact that the Eulerian numbers $\left \langle {n\atop j}\right \rangle$ are increasing for $1\leq j\leq n/2$ (like the binomial coefficients); this fact has a clear combinatorial explanation also.
See e.g.
<http://en.wikipedia.org/wiki/Eulerian_number>
<http://www.oeis.org/A008292>
**[edit]**: although by now all details have been very clearly explained by Victor Protsak, I wish to add a general remark, should you find yourself in an analogous situation again. A healthy approach in such cases is adding variables, following the motto "more variables = simpler dependence" (like when one passes from quadratic to bilinear). In the present case, you may consider
$$A(k):=a(k,\ 2k-2,\ 2k+1)$$
where you define
$$a(k,n,m):=\sum\_{j=0}^{k-1}(-1)^{j+1}(k-j)^{n}{m \choose j}$$
in which it is more apparent the action of the iterated difference operator, or, in the formalism of generating series, the Cauchy product structure:
$$\sum\_{k=0}^\infty a(k,n,m)x^k=-\sum\_{j=0}^\infty j^nx^j\, \sum\_{j=0}^\infty(-1)^j{m \choose j} x^j =-(1-x)^m\sum\_{j=0}^\infty j^nx^j. $$
The series
$$\sum\_{j=0}^\infty j^nx^j$$
is now quite a simpler object to investigate, and in fact it is well-known to whoever played with power series in childhood. It sums to a rational function
$$(1-x)^{-n-1}x\sum\_{k=0}^{n}\left \langle {n\atop k}\right \rangle x^k$$
that defines the Eulerian polynomial of order $n$ as numerator, and the Eulerian numbers as coefficients. In your case, $m=n+3$, meaning that you are still applying a discrete difference twice (in fact just once, due to the symmetric relations; check Victor's answer).
| 41 | https://mathoverflow.net/users/6101 | 29179 | 19,051 |
https://mathoverflow.net/questions/29166 | 7 | I am reading about the L-functions of elliptic curves and I was thinking about the root number as the product of local root numbers. So my question is how to think about the local root numbers geometrically or arithmetically. I have also read that even though the functional equation is conjectural (in different cases) but root number is a well-defined concept. Is it somehow related to the l-adic representation attached to my objects?
| https://mathoverflow.net/users/2081 | Local root number | Yes, if $X$ is a variety over an extension $K$ of $\mathbb Q\_p$, then the $\ell$-adic cohomology spaces
$H^i(X,\mathbb Q\_{\ell})$ are $\ell$-adic representations of $G\_{K}$,
which give rise to Weil--Deligne representations. (See Tate's Corvallis article,
for example.) The resulting Weil--Deligne representation is conjectured to be
independent of the choice of $\ell$ (as long as $\ell \neq p$). This is known
when $X$ has a smooth proper model over the ring of integers of $K$ (although
from the point of view of root numbers this case is not so interesting; in
this good reduction situation the associated Weil--Deligne representation
is unramified, so the local root number is 1, for the right choice of additive
character). It is also known when $X$ is an elliptic curve (in which case
$i = 1$ is the interesting choice; this gives the contragredient of the
Tate module).
Actually computing root numbers is a non-trivial business, especially for
instances of very bad reduction at small primes. For examples, see recent
work by Mazur and Rubin, and by the Dokchitsers.
[Added in response to Arijit's comment below:]
One of the most interesting applications of root numbers is when you can prove that the product of the local root numbers (which is the global root number) of an elliptic curve is -1. Then BSD predicts that there will be a rational point of infinite order.
In the Heegner point situation studied by Gross--Zagier, one is in this context (the global root number of the elliptic curve over a quad. imag. field is -1), and if the order of vanishing is precisely one, they produce a point of infinite order. In general we don't have the technology yet to control Mordell--Weil groups, but can control Selmer groups (which morally should be the same thing, since, following Shafarevic and Tate, one conjectures that Sha has finite order, and hence that the Selmer rank and Mordell--Weil rank coincide).
In the work of Mazur and Rubin, and also of Nekovar, and of the Dokchitsers, the goal is,
under assumptions which makes the global root number -1, to produce the predicted rank in
the Selmer group. (They succeed in doing in this in many cases!) Perhaps developing some understanding of this question, and some sense of what people are doing about it, will help give you motivation. After all, it's pretty surprising that just by computing a sign,
you can predict whether a Diophantine equation will have an interesting solution, and even
more surprising that you can prove non-trivial results in this direction!
| 12 | https://mathoverflow.net/users/2874 | 29181 | 19,052 |
https://mathoverflow.net/questions/29161 | 4 | If we want to define a sheaf F on a topological space X and we have a basis B for the topology of X, what we can do is to define objects and restrictions for guys in B, check that they satisfy the "B-sheaf axioms" and then use
Theorem 1: the B-sheaf extends uninquely to the whole of X.
I was wondering if there's a similar thing for more general sites, and actually not just for sheaves on a given site but for stacks.
The question I'm really interested in is the following:
If one has a fibred category over Schemes (say Schemes over some fixed field with the fppf topology) and one wants to check that descent is effective, would it be sufficient to check it on some subcategory of schemes (using perhaps some vague analogue of Theorem 1)?
Thanks.
EDIT
For example one might want to construct the stack M of coherent sheaves on some scheme X.
One way to do it is to define the functor which associates with each scheme S the groupoid of coherent sheaves on $S\times X$ flat over S
$$ M(S) = \{ E \in Coh\ S\times X,\ E \text{ flat over S} \}.$$
Let's say I want to use a different characterization of $M(S)$, perhaps using Lemma 3.31 on page 82 of Huybrechts' Fourier-Mukai [book](http://books.google.com/books?id=Wq7Kr-7rmbIC&lpg=PP1&dq=huybrechts%20fourier&pg=PA82#v=onepage&q=flat&f=false).
My ignorance prevents me from knowing if the that lemma is valid for a general scheme S (no matter how nice my X might be).
This is why I'd like to work over some nice subcategory of schemes (where the lemma is valid) and then extend.
The stack I'd be interested in defining would be a stack of perverse sheaves on X, where the matters would be a bit worse.
| https://mathoverflow.net/users/3701 | Does the concept of a basis for a topology on a category exist? | Let $S$ be your Grothendieck site. What you want is a subcategory $j:B \hookrightarrow S$ such that the Grothendieck topology of $S$ restricts to $B$ in the sense that every covering sieve of $b \in B$ can be refined by one coming from a family $(U\_i \to b)\_i$ with each $U\_i \in B$, AND such that $j^\*:Sh(B) \to Sh(S)$ is an equivalence of topoi. This is exactly what makes Theorem 1 work.
Concretely, you want the topology to restrict and every $s$ in $S$ to have a covering family $(b\_i \to s)$ with $b\_i \in B$, so that you can then say:
$$F(s):=\varprojlim \left(\prod\_{i}F(b\_i)\rightrightarrows \prod\_{i,j}F(b\_i \times\_{s} b\_j)\right).$$
However, you need to make sure this doesn't depend on the covering family you chose.
Suppose only that every $s$ in $S$ to have a covering family $(b\_i \to s)$ with $b \in B$ and that the Grothendieck topology on $S$ restricts to $B$ in the sense described above. Then, since the Grothendieck topology is subcanonical, $s$ is a colimit of $b\_i$s, hence $s \mapsto Hom(blank,s)$ embeds $S$ into $Sh(B)$ (note that the left-Kan extension of this embedding is precisely $j^\*$, which is literally restriction).
I claim $j^\*$ is fully-faithful. This is essentially because $Hom(j^\*F,j^\*G)$ for two sheaves on $S$ determines $Hom(F,G)$ since the value of $F(s)$ is determined by the value of $F$ on $b\_i$s by the cover $(b\_i \to s)$, by descent.
Now, if $F$ is a $B$-sheaf (i.e. an element of $Sh(B)$), then $j\_\*F(s)=Hom(j^\*s,F).$ Hence, $$j\_\*j^\*(F)(s)\cong Hom(s,j^\*j\_\*F)\cong Hom(j^\*s,j^\*F)\cong Hom(s,F)\cong F(s),$$
by Yoneda, adjointness, and full an faithfulness.
Note also that $j^\*j\_\*(G) \cong G$ for all $G \in Sh(B)$ pretty much by definition. Hence the adjoint pair $j\_\*,j^\*$ is an equivalence.
So, what does this mean concretely? You need to find a subcategory $B$ of schemes such that
1.)every cover in the fppf topology of an element of $b \in B$ can be refined by one with domains in $B$ (at least you need to be able to find a family of morphisms whose SIEVE is in the topology GENERATED by the fppf pretopology)
2.) Every scheme can be covered by elements of $B$.
I'll leave it to you to find such a subcategory, as I don't know much AG.
P.S., everything I said will hold for stacks as well.
EDIT: Condition 2.) implies condition 1.), so this becomes simpler:
You just need a category subcategory $B$ of schemes such that every scheme can be covered by elements of $B$.
| 5 | https://mathoverflow.net/users/4528 | 29182 | 19,053 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.