co.combinatorics - Hamilton cycle decompositions of the complete graph

Just reporting that I wrote another algorithm for this and found the following values:

3 1
5 6
7 960
9 40037760

I ran this through the superseeker on Sloane and it came up with nothing (so perhaps nobody has counted these before).

Here's my code below (it uses GAP). We generate a (n-1) x n Latin rectangle where each row is an n-cycle and the i-th and (i+(n-1)/2)-th rows are inverses.

EnumerateHamiltonDecompositionsBacktrackingAlgorithm:=function(n,L,step)
  local i,j,k,count,A;
  i:=Int((step-1)/n)+1;
  j:=(step-1) mod n+1;
  count:=0;

  if(n mod 2=0 or n<3) then return fail; fi;
  if(j=1) then A:=[Minimum(Filtered([2..n],i->ForAll([1..n-1],t->L[t][1]<>i)))]; else A:=Filtered([1..n],s->ForAll([1..n-1],t->L[t][j]<>s) and ForAll([1..n],t->L[i][t]<>s)); fi;
  for k in A do
    L[i][j]:=k;
    L[i+(n-1)/2][k]:=j;
    if((j=n and CycleLengths(PermList(L[i]),[1..n])=[n]) or j<n) then
      if(i=(n-1)/2 and j=n) then count:=count+1;
      else count:=count+EnumerateHamiltonDecompositionsBacktrackingAlgorithm(n,L,step+1); fi;
    fi;
    L[i][j]:=0;
    L[i+(n-1)/2][k]:=0;
  od;
  return count;
end;;

EnumerateHamiltonDecompositions:=function(n)
  local L;
  if(n mod 2=0 or n<3) then return fail; fi;
  if(n=3) then return 1; fi;
  L:=List([1..n-1],i->List([1..n],j->0));
  L[1]:=List([1..n],i->i mod n+1);
  L[1+(n-1)/2]:=ListPerm(Inverse(PermList(List([1..n],i->i mod n+1))));
  return Factorial(n-2)*EnumerateHamiltonDecompositionsBacktrackingAlgorithm(n,L,n+1);
end;;

The extra data point comes from assuming that (12..n) is one of the cycles, then multiplying the result by (n-2)!. This is legitimate since each decomposition contains a unique cycle with the edge 12, and by permuting the remaining n-2 edges, we generate a unique decomposition with the cycle (12..n). There are no automorphisms under this group action, so each orbit has cardinality (n-2)!.

fa.functional analysis - L_p norm balls for 12?

The L_1 ball in 2D is shaped like a diamond (L_1 is also known as the Manhattan norm). The L_∞ ball is shaped like a square (L_∞ is also known as the supremum norm). They are similar, i.e. have same shape. The L2 ball is shaped like a circle.

Hypothesis: For all p in the interval (1,2), there is q>2 such that the q-ball and the p-ball are similar. One further hypothesis is that this occurs precisely when p,q are Hölder conjugates.

I wasn't sure how to tag this.

at.algebraic topology - Does homology detect chain homotopy equivalence?

Yes, this is true. Suppose $C_*$ is such a chain complex of free abelian groups.

For each $n$, choose a splitting of the boundary map $C_n to B_{n-1}$, so that $C_n cong Z_n oplus B_{n-1}$. (You can do this because $B_{n-1}$, as a subgroup of a free group, is free.) For all $n$, you then have a sub-chain-complex $cdots to 0 to B_n to Z_n to 0 to cdots$ concentrated in degrees $n$ and $n+1$, and $C_*$ is the direct sum of these chain complexes.

Given two such chain complexes $C_*$ and $D_*$, you get a direct sum decomposition of each, and so it suffices to show that any two complexes $cdots to 0 to R_i to F_i to 0 to cdots$, concentrated in degrees $n$ and $n+1$, which are resolutions of the same module $M$ are chain homotopy equivalent; but this is some variant of the fundamental theorem of homological algebra.

This is special to abelian groups and is false for modules over a general ring.

Algebra / unital associative algebra: better terminology?

If your work requires you to work mostly with unital assosiative $k$-algebras with unital morphisms, define somewhere prominent in your work 'algebra' and 'morphism of algebras' to mean precisely that, and say 'Lie algebras', 'not necessarily associative algebra', 'possibly non unital morphism of algebra', and so on in the meagre 5% of the remaining cases.

nt.number theory - Irrational logs and the harmonic series

Consider the series
$$ S_f = sum_{x=1}^infty frac{f}{x^2+fx}. $$
Goldbach showed that, for integers $f ge 1$,
$$ S_f = 1 + frac12 + frac13 + ldots + frac1f $$
(this follows easily by writing $S_f$ as a telescoping series).
Thus $S_f$ is rational for all natural numbers $f ge 1$.
Goldbach claimed that, for all nonintegral (rational) numbers $f$,
the sum $S_f$ would be irrational.

Euler showed, by using the substitution
$$ frac1k = int_0^1 x^{k-1} dx, $$
that
$$ S_f = int_0^1 frac{1-x^f}{1-x} dx. $$
He evaluated this integral for $f = frac12$ and found
that $S_{1/2} = 2(1 - ln 2)$ (this also follows easily from
Goldbach's series for $S_f$). Thus Goldbach's claim holds for all
$f equiv frac12 bmod 1$ since $S_{f+1} = S_f + frac1{f+1}$.

Here are my questions:

1. The irrationality of $ln 2$ was established by Lambert, who
   proved that $e^r$ is irrational for all rational numbers
   $r ne 0$. Are there any (simple) direct proofs?




2. Has Goldbach's claim about the irrationality of $S_f$ for
   nonintegral rational values of $f$ been settled in other cases?

ag.algebraic geometry - Geometric Intuition for Big Monodromy

In various contexts, I have come across results referred to as "big monodromy." A standard arithmetic example is the open image theorem for the image of Galois action on non-CM elliptic curves. A general setup for such a result in algebraic geometry is:

Given a proper, generically smooth map $pi:X rightarrow S$ of relative dimension d, say S is connected. This gives rise to an $l$-adic representations of the etale fundamental group $pi_1(U)$ where $U$ is smooth locus of $pi$ corresponding to higher pushforward $R^d pi_* Q_l$. One might say it has "big monodromy" if the Zariski closure of the image is as big as it can be given that it has to respect cup-product, etc.

My specific question is what are the geometric consequences of big monodromy? If we know such a result for $pi$, what does that say about the geometry of the fibration or at the very least is there geometric intuition for what it should mean?

I welcome intuition from number theory, algebraic geometry, or complex geometry.

I have also heard that "one should expect big monodromy unless there is a reason not to" (for example, complex multiplication). What are other examples of things which inhibit big monodromy?

linear algebra - Spheres over rational numbers and other fields

Any orthonormal set extends to an orthonormal basis, over any field
of characteristic not $2$. This is a special case of Witt's theorem.

EDIT: In response to Vipul's comment: The proof of Witt's theorem is constructive, and leads to the following recursive algorithm for extending an orthonormal set $lbrace v_1,ldots,v_r rbrace$ to an orthonormal basis.

Let $e_1,ldots,e_n$ be the standard basis of $K^n$, where $e_i$ has $1$ in the $i^{operatorname{th}}$ coordinate and $0$ elsewhere. It suffices to find a sequence of reflections defined over $K$ whose composition maps $v_i$ to $e_i$ for $i=1,ldots,r$, since then the inverse sequence maps $e_1,ldots,e_n$ to an orthonormal basis extending $v_1,ldots,v_r$. In fact, it suffices to find such a sequence mapping just $v_1$ to $e_1$, since after that we are reduced to an $(n-1)$-dimensional problem in $e_1^perp$, and can use recursion.

Case 1: $q(v_1-e_1) ne 0$, where $q$ is the quadratic form. Then reflection in the hyperplane $(v_1-e_1)^perp$ maps $v_1$ to $e_1$.

Case 2: $q(v_1+e_1) ne 0$. Then reflection in $(v_1+e_1)^perp$ maps $v_1$ to $-e_1$, so follow this with reflection in the coordinate hyperplane $e_1^perp$.

Case 3: $q(v_1-e_1)=q(v_1+e_1)=0$. Summing yields $0=2q(v_1)+2q(e_1)=2+2=4$, a contradiction, so this case does not actually arise.

What can be said about the homotopy groups of a CW-complex in terms of its (co)homology?

Try looking up some references on rational homotopy theory. Rational homotopy theory studies the homotopy groups tensor Q, so basically you kill all torsion information. If we focus only on homotopy groups tensor Q, the question you ask becomes easier. As Steven Sam mentions in the comments, the homotopy groups of spheres are really crazy. But the rational homotopy groups of spheres are quite tractable (in fact completely known, by a theorem of Serre) and can be more or less obtained from cohomology, if I recall correctly.

One particularly impressive theorem, of Deligne-Griffiths-Morgan-Sullivan, says that if your space is a compact Kähler manifold (e.g. a smooth complex projective variety), and if you know its rational cohomology ring, then you can compute for instance the ranks of all of its homotopy groups (maybe you need an extra assumption that the space is simply connected or has nilpotent fundamental group).

gr.group theory - Abelianization of a semidirect product

I agree with Ryan and Victor, except that you don't need presentations. The subgroup $[G ltimes H,G ltimes H]$ is generated by $[H,H] cup [G,H] cup [G,G]$, so you can write
$$(G ltimes H)^{ab} = (G ltimes H) / langle [H,H] cup [G,H] cup [G,G] rangle.$$
If you apply the relators $[H,H]$, you get $G ltimes H^{ab}$; then if you apply the relators $[G,H]$, you get $G times (H^{ab})_G$; then finally if you apply $[G,G]$, you get $G^{ab} times (H^{ab})_G$. You can add this as an extra half-paragraph or footnote rather than giving a citation.

I don't think that the referee has the right to demand a longer explanation than this, unless maybe you are writing a textbook.

ct.category theory - What precisely Is "Categorification"?

The Wikipedia answer is one answer that is commonly used: replace sets with categories, replace functions with functors, and replace identities among functions with natural transformations (or isomorphisms) among functors. One hopes for newer deeper results along the way.

In the case of work of Lauda and Khovanov, they often start with an algebra (for example ${bf C}[x]$ with operators $d (x^n)= n x^{n-1}$ and $x cdot x^n = x^{n+1}$ subject to the relation $d circ x = x circ d +1$) and replace this with a category of projective $R$-modules and functors defined thereupon in such a way that the associated Grothendieck group is isomorphic to the original algebra.

Khovanov's categorification of the Jones polynomial can be thought of in a different way even though, from his point of view, there is a central motivating idea between this paragraph and the preceding one. The Khovanov homology of a knot constructs from the set of $2^n$ Kauffman bracket smoothing of the diagram ($n$ is the crossing number) a homology theory whose graded Euler characteristic is the Jones polynomial. In this case, we can think of taking a polynomial formula and replacing it with a formula that inter-relates certain homology groups.

Crane's original motivation was to define a Hopf category (which he did) as a generalization of a Hopf algebra in order to use this to define invariants of $4$-dimensional manifolds. The story gets a little complicated here, but goes roughly like this. Frobenius algebras give invariants of surfaces via TQFTs. More precisely, a TQFT on the $(1+1)$ cobordism category (e.g. three circles connected by a pair of pants) gives a Frobenius algebra. Hopf algebras give invariants of 3-manifolds. What algebraic structure gives rise to a $4$-dimensional manifold invariant, or a $4$-dimensional TQFT? Crane showed that a Hopf category was the underlying structure.

So a goal from Crane's point of view, would be to construct interesting examples of Hopf categories. Similarly, in my question below, a goal is to give interesting examples of braided monoidal 2-categories with duals.

In the last sense of categorification, we start from a category in which certain equalities hold. For example, a braided monoidal category has a set of axioms that mimic the braid relations. Then we replace those equalities by
$2$-morphisms that are isomorphisms and that satisfy certain coherence conditions. The resulting $2$-category may be structurally similar to another known entity. In this case, $2$-functors (objects to objects, morphisms to morphisms, and $2$-morphisms to $2$-morphisms in which equalities are preserved) can be shown to give invariants.

The most important categorifications in terms of applications to date are
(in my own opinion) the Khovanov homology, Oszvath-Szabo's invariants of knots, and Crane's original insight. The former two items are important since they are giving new and interesting results.

mathematics education - effective teaching

I just watched the video -- I didn't know of it and it is certainly very interesting and provides much food for thought.

From the perspective of someone who teaches, I could relate to parts of it: particularly the decreasing ability to understand student's questions. From the perspective of someone who was a student, though, I do remember learning during lectures: perhaps not in all courses, but certainly in some which I still remember fondly. These courses were usually lecture-based; although in one case the lecturer would ask questions to the students all the time: picking a starting person and then moving systematically along the audience. This used to instill the "fear of God" in some people, but it meant one had to be on top of the material. I enjoyed that and, in fact, it boosted my confidence.

Now to answer the question, in the University of Edinburgh (where I am based) we started a few years ago to teach some of the introductory courses incorporating some element of Peer Instruction. I personally have not taught introductory courses for a while, so I cannot say how this is panning out. The School of Physics (I'm in Maths) has been teaching the first-year introductory course using Peer Instruction for some time now and they seem to be very happy with the result.

I wonder whether some variant of this method can work for final-year courses, though.

pr.probability - Is there an introduction to probability theory from a structuralist/categorical perspective?

One can argue that an object of the right category of spaces in measure theory is not a set equipped
with a $sigma$-algebra of measurable sets, but rather a set $S$ equipped with a $sigma$-algebra $M$ of measurable
sets and a $sigma$-ideal $N$ of $M$ consisting of sets of measure $0$.
The reason for this is that you can hardly state any theorem of measure theory or probability
theory without referring to sets of measure $0$.
However, objects of this category contain less data than the usual measured spaces,
because they are not equipped with a measure.
Therefore I prefer to call them measurable spaces.
A morphism of measurable spaces $(S,M,N)to(T,P,Q)$ is a map $Sto T$ such that
the preimage of every element of $P$ is a union of an element of $M$ and a subset of an element of $N$
and the preimage of every element of $Q$ is a subset of an element of $N$.

Irving Segal proved that for a measurable space the following properties are equivalent:

1. The Boolean algebra $M/N$ of equivalence classes of measurable sets is complete;


2. The space of equivalence classes of all bounded (or unbounded) real-valued functions on $S$ is Dedekind-complete;


3. Radon-Nikodym theorem is true for $(S,M,N)$;


4. Riesz theorem is true for $(S,M,N)$;


5. Equivalence classes of bounded functions on $S$ form a von Neumann algebra (aka $W^*$-algebra).


6. $(S,M,N)$ is a coproduct (disjoint union) of points and real lines.

A measurable space that satisfies these conditions is called localizable.

This theorem tells us that if we want to prove anything nontrivial about measurable
spaces, we better restrict ourselves to localizable measurable spaces.
We also have a nice illustration of the claim I made in the first paragraph:
None of these statements would be true without identifying objects that differ
on a set of measure $0$.
For example, take a non-measurable set $G$
and a family of one-element subsets of $G$ indexed by themselves.
This family of measurable sets does not have a supremum in the Boolean algebra
of measurable sets, thus disproving a naïve version of (1).

Another argument for restricting to localizable spaces is the following version of
Gelfand-Neumark theorem:

The category of localizable measurable spaces is equivalent to the category of commutative von Neumann algebras (aka $W^*$-algebras) and their morphisms (normal unital homomorphisms of $^*$-algebras).

I actually prefer to define the category of localizable measurable spaces
as the opposite category of the category of commutative $W^*$-algebras.
The reason for this is that the classical definition of measurable space
exhibits immediate connections only to descriptive set theory (and with additional
effort to Boolean algebras), which are mostly
irrelevant for the central core of mathematics,
whereas the description in terms of operator algebras immediately connects
measure theory to other areas of the central core (noncommutative geometry,
algebraic geometry, complex geometry, differential geometry etc.).
Also it is easier to use in practice. Let me illustrate this statement
with just one example: When we try to define measurable bundles of Hilbert spaces
on a localizable measurable space set-theoretically, we run into all sorts of problems
if the fibers can be non-separable, and I do not know how to fix this problem in the set-theoretic framework.
On the other hand, in the algebraic framework we can simply say that a bundle
of Hilbert spaces is a Hilbert module over the corresponding $W^*$-algebra.

Categorical properties of $W^*$-algebras (hence of localizable measurable spaces)
were investigated by Guichardet.
Electronic version of this paper is available here.
Let me mention some of his results.
The category of localizable measurable spaces admits equalizers and coequalizers,
arbitrary coproducts and hence arbitrary colimits.
It also admits products, although they are quite different from what one might think.
For example, the product of two real lines is not $Bbb R^2$ with the two obvious projections.
The product contains $Bbb R^2$, but it also has a lot of other stuff, for example, the diagonal of $Bbb R^2$,
which is needed to satisfy the universal property for the two identity maps on $Bbb R$.
The more intuitive product of measurable spaces ($Bbb Rtimes Bbb R=Bbb R^2$) corresponds to the spatial
tensor product of von Neumann algebras and forms a part of a symmetric monoidal
structure on the category of measurable spaces.
See Guichardet's paper for other categorical properties (monoidal structures
on measurable spaces, flatness, existence of filtered limits, etc.).

Finally let me mention pushforward and pullback properties of measures on measurable spaces.
I will talk about more general case of $L^p$ spaces instead of just measures (i.e., $L^1$ spaces).
For the sake of convenience let $L_p(M) := L^{1/p}(M)$, where $M$ is a measurable space.
Here $p$ can be an arbitrary complex number with a nonnegative real part.
Note that you don't need a measure on $M$ to define $L_p(M)$.
In particular, $L_0$ is the space of all bounded functions (i.e., the $W^*$-algebra itself),
$L_1$ is the space of finite complex-valued measures (the dual of
$L_0$ in the $sigma$-weak topology), and $L_{1/2}$ is the Hilbert space of half-densities.
I will also talk about extended positive part $E^+L_p$ of $L_p$ for real $p$.
In particular, $E^+L_1$ is the space of all (not necessarily finite) positive measures.

Pushforward for $L_p$ spaces: Suppose we have a morphism of measurable spaces $Mto N$.
If $p=1$, then we have a canonical map $L_{1}(M) to L_{1}(N)$, which just the dual of $L_{0}(N) to L_{0}(M)$
in the $sigma$-weak topology. Geometrically, this is the fiberwise integration map.
If $pneq 1$, then we only have a pushforward map of the extended positive parts:
$E^+L_p(M) to E^+L_p(N)$, which is non-additive unless $p=1$.
Geometrically, this is the fiberwise $L_p$ norm.
Thus $L_1$ is a functor from the category of measurable spaces to the category of Banach spaces
and $E^+L_p$ is a functor to the category of "positive homogeneous $p$-cones".
The pushforward map preserves the trace on $L_1$ and hence
sends a probability measure to a probability measure.

To define pullback of $L_p$ spaces (in particular, $L_1$ spaces) one needs to pass to a different
category of measurable spaces.
In algebraic language, if we have two $W^*$-algebras $A$ and $B$, then
a morphism from $A$ to $B$ is a usual morphism of $W^*$-algebras $f: Ato B$
together with an operator valued weight $T: E^+(B)→E^+(A)$ associated to $f$.
Here $E^+(A)$ denotes the extended positive part of A (think of positive functions on $mathrm{Spec} A$
that can take infinite values).
Geometrically, this is a morphism $mathrm{Spec} f:  mathrm{Spec} B to mathrm{Spec} A $ between the corresponding
measurable spaces and a choice of measure on each fiber of $mathrm{Spec} f$.
Now we have a canonical additive map $E^+L_p(mathrm{Spec} A) to E^+L_p(mathrm{Spec} B)$,
which makes $E^+L_p$ into a contravariant functor from the category of measurable spaces
equipped with a fiberwise measure to the category of “positive homogeneous additive cones”.

If we want to have a pullback of $L_p$ spaces themselves and not just their extended positive parts,
we need to replace operator valued weights in the above definition
by finite complex-valued operator valued weights $T: B to A$ (think of fiberwise complex-valued measure).
Then $L_p$ becomes a functor from the category of measurable spaces to the category
of Banach spaces (if the real part of $p$ is at most $1$) or quasi-Banach spaces (if the real part of $p$ is greater than $1$).
Here $p$ is an arbitrary complex number with a nonnegative real part.
Notice that for $p=0$ we get the original map $f: Ato B$ and in this (and only this) case we don't need $T$.

Finally, if we restrict ourselves to an even smaller subcategory of measurable
spaces equipped with a finite operator valued weight T such that $T(1)=1$
(i.e., T is a conditional expectation; think of fiberwise probability measure), then the pullback map preserves
the trace on $L_1$ and in this case the pullback of a probability measure is a probability measure.

There is also a smooth analog of the theory described above: The category of measurable spaces
and their morphisms is replaced by the category of smooth manifolds and submersions,
$L_p$ spaces are replaced by bundles of $p$-densities, operator valued weights are
replaced by sections of the bundle of relative $1$-densities,
integration map on $1$-densities is defined via Poincaré duality (to avoid circular dependence
on measure theory) etc.
The forgetful functor that sends a smooth manifold to its underlying measurable space
commutes with everything and preserves everything.

ag.algebraic geometry - Stacks and sheaves

Let me see if I understand your example correctly: you are fixing $X$ and $Y$, families
of curves over $S$, and now you are considering the functor which maps an $S$-scheme $T$
to the set of $T$-isomorphisms $f^*X to f^*Y$ (where $f$ is the map from $T$ to $S$).

If I have things straight, then this functor shouldn't be so bad to think about, because it is actually representable, by an Isom scheme. In other words, there is an $S$-scheme
$Isom_S(X,Y)$ whose $T$-valued points, for any $f:T to S$, are precisely the $T$-isomorphisms
from $f^*X$ to $f^*Y$. (One can construct the Isom scheme by looking inside a
certain well-chosen Hilbert scheme.)

One way to think about this geometrically is as follows: one can imagine that two
curves over $k$ (a field) are isomorphic precisely when certain invariants coincide
(e.g. for elliptic curves, the $j$-invariant). (Of course this is a simplification,
and the whole point of the theory of moduli spaces/schemes/stacks is to make it precise,
but it is a helpful intuition.) Now if we have a family $X$ over $S$, these invariants
vary over $S$ to give a collections of functions on $S$ (e.g. a function $j$ in the
genus $1$ case), and similarly with $Y$. Now $X$ and $Y$ will have isomorphic
fibres precisely at those points where the invariants coincide, so if we look
at the subscheme $Z$ of $S$ defined by the coincidence of the invariants,
we expect that $f^*X$ and $f^*Y$ will be isomorphic precisely if the map $f$
factors through $Z$. Thus $Z$ is a rough approximation to the Isom scheme.

It is not precisely the Isom scheme, because curves sometimes have non-trivial
automorphisms, and so even if we know that $X_s$ and $Y_s$ are isomorphic for
some $s in S$, they may be isomorphic in more than one way. So actually the
Isom scheme will be some kind of (possibly ramified) finite cover of $Z$.

Of course, if one pursues this line of intuition much more seriously, one will
recover the notions of moduli stack, coarse moduli space, and so on.

Added: The following additional remark might help:

The families $X$ and $Y$ over $S$ correspond to a map $phi:S to {mathcal M}_g
times {mathcal M}_g$. The stack which maps a $T$-scheme to $Isom_T(f^*X,
f^*Y)$ can then seen to be the fibre product of the map $phi$ and the diagonal
$Delta:{mathcal M}_g to {mathcal M}_g times {mathcal M}_g$.

In the particular case of ${mathcal M}_g$ the fact that this fibre product is representable is part of the condition that ${mathcal M}_g$ be an algebraic stack.

But in general, the construction you describe is the construction of a fibre product
with the diagonal. This might help with the geometric picture, and make the relationship to Mike's answer clearer. (For the latter:note that the path space into $X$ has a natural
projection to $Xtimes X$ (take the two endpoints), and the loop space is the fibre product
of the path space with the diagonal $Xto Xtimes X$.)

set theory - Does every non-empty set admit a group structure (in ZF)?

You cannot in general put a group structure on a set. There is a model of ZF with a set A that has no infinite countable subset and cannot be partitioned into finite sets; such a set has no group structure.

See e.g at http://groups.google.com/group/sci.math/msg/06eba700dfacb6ed

---

Sketch of proof that in standard Cohen model the set $A={a_n:ninomega}$ of adjoined Cohen reals cannot be partitioned into finite sets:

Let $mathbb{P}=Fn(omegatimesomega,2)$ which is the poset we force with. The model is the symmetric submodel whose permutation group on $mathbb{P}$ is all permutations of the form $pi(p)(pi(m),n)=p(m,n)$ where $pi$ varies over all permutations of $omega$, (that is we are extending each $pi$ to a permutation of $mathbb{P}$ which I also refer to as $pi$) and the relevant filter is generated by all the finite support subgroups.

Suppose for contradiction that $pVdash " bigcup_{iin I}dot{A_i}=A$ is a partition into finite pieces"; let $E$ (a finite set) be the support of this partition. Take some $a_{i_0}notin E$ and extend $p$ to a $q$ such that $qVdash ``{a_{i_0},ldots a_{i_n}}$ is the piece of the partition containing $a_{i_0}$". Then pick some $j$ which is not in $E$ nor the domain of $q$ nor equal to any of the $a_{i_0},ldots a_{i_l}$. If $pi$ is a permutation fixing $E$ and each of $a_{i_1},ldots a_{i_n}$ and sending $a_{i_0}$ to $a_j$, it follows that $pi(q) Vdash " {a_j,a_{i_1},ldots a_{i_n}}$ is the piece of the partition containing a_j". But also $q$ and $pi(q)$ are compatible and here we run into trouble, because $q$ forces that $a_{i_0}$ and $a_{i_1}$ are in the same piece of the partition, and $pi(q)$ forces that this is not the case (and they are talking about the same partition we started with because $pi$ fixes $E$). Contradiction.

gt.geometric topology - Triangulations coming from a poset. Or: What conditions are necessary and sufficient for a finite simplicial complex to be the order complex of a poset?

Here are necessary and sufficient conditions for an abstract, finite simplicial complex $mathcal{S}$ to be the order complex of some partially ordered set.

(i) $mathcal{S}$ has no missing faces of cardinality $geq 3$; and

(ii) The graph given by the edges (=$1$-dimensional simplices) of $mathcal{S}$ is a comparability.

[Definitions.
(a) A missing face of $mathcal{S}$ is a subset $M$ of its vertices (=$0$-dimensional simplices) such that
$M not in mathcal{S}$, but all proper subsets $Psubseteq M$ satisfy $Pin mathcal{S}$.
(b) A graph (=undirected graph with no loops nor multiple edges) is a comparability if its
edges can be transitively oriented, meaning that whenever edges ${p, r_1}, {r_1, r_2},ldots, {r_{u−1}, r_u}, {r_u, q}$ are oriented as $(p, r_1), (r_1, r_2),ldots, (r_{u−1}, r_u), (r_u, q)$, then there exists an edge ${p, q}$ oriented as $(p, q)$.]

This characterisation appears with a sketch of proof $-$ which is not hard, anyway $-$ in

M. M. Bayer, Barycentric subdivisions. Pacific J. Math. 135 (1988), no. 1, pp. 1-16.

As Bayer points out, the result was first observed in

R. Stanley, Balanced Cohen-Macaulay complexes, Trans. Amer. Math. Soc, 249 (1979), pp. 139-157.

@Rasmus and @Gwyn: The characterisation might perhaps disappoint you if you were expecting something more topological. However, it's easy to prove that no topological characterisation of order complexes is possible, and therefore a combinatorial condition such as the one on comparabilities must be used. For this, first check that the barycentric subdivision of any simplicial complex indeed is an order complex. Next observe that barycentric subdivision of a simplicial complex does not change the homeomorphism type of the underlying polyhedron of the complex. Finally, conclude that for any topological space $T$ that is homeomorphic to a compact polyhedron, there is an order complex whose underlying polyhedron is homeomorphic to $T$.

I hope this helps.

pr.probability - What is known about the Gaussian measure of the unit ball in a Hilbert Space?

You can't talk about "the" Gaussian measure on an infinite-dimensional Hilbert space, for the same reason that you can't talk about a uniform probability distribution over all integers. It doesn't exist; see Richard's answer. However, there are a lot of non-uniform Gaussian measures on infinite dimensional Hilbert spaces.

Consider the measure on $mathbb{R}^infty$ where the $j$th coordinate is a Gaussian with mean 0 and variance $sigma_j^2$, where $sum_{j=1}^{infty} sigma_j^2 < infty$ (and different coordinates are independent). This is almost surely bounded in the $ell_2$ metric, and any projection onto a finite-dimensional space has a Gaussian distribution. The squared length of a vector drawn from this measure is a sum of squares of Gaussians, and so follows some kind of generalized $chi$-square distribution. If I knew more about generalized $chi$-square distributions, I might be able to tell you what the measure of the unit ball was.

This kind of Gaussian distribution is very important in quantum optics. In fact, in quantum optics, a thermal state is Gaussian, so "the" Gaussian measure actually makes some sense.

lo.logic - What is Realistic Mathematics?

This is more a philosophical question, and therefore hasn't a definite answer.

But if you want to make plausible that AC isn't realistic mathematics, you might reason something like the following:

There is a set of mathematical sentences, that has a direct (so, not indirect yet) relation with physics. Call this set B (of Basic). Personally, I think they are in the following four areas:

• Computation
• Probability
• Geometry
• Topology

Note, I count arithmetic as part of computation, since numbers are not a physical entity, but computation is. But, likely many people will disagree.

Also note, that the sentences in B, might be far simpler than the mathematical sentences suggested in your question or in one of the answers.

Now, we have more complex mathematical sentences, that are still "realistic", if they can be converted or "instantiated" to sentences of B. Call this extended set be E. These more complex sentences capture a higher principle, which can be powerful in the science of physics. Still, there is no direct link with physics, the mathematic sentence first needs to be instantiated, to make a direct link with physics. Example, any sentence with a real number, is not be part of B, because we can not observe real numbers in physics, but they can be part of E.

Consider that there is a sentence s1 ∈ E and s2 ∈ E. Furthermore, that s2 follows from s1, but with a rather difficult proof. Suppose there is a proposed axiom a, such that s1 + a leads more directly to s2 (a simpler proof). However, a ∉ E. So, axiom a is independent from sentences in B.

From above concept it follows that axioms can exist that are "useful", because they make proofs shorter, but have nevertheless no "meaning". I do believe that AC is such axiom.

About CH, I think it is not useful and not having a meaning.

But again, this is more an opinion.

computer science - Reconstructing an ordering of a multiset from its consecutive submultisets

I have some unpublished notes that I think are relevant to this problem, and I am cherry-picking from them. (See also here for a related question.) Because the notes are unpublished I hope it will not be considered poor form to provide a rather lengthy response.

To save myself trouble while picking the relevant parts from them, I will write $q$ in place of $r$ and $k$ in place of $j$. Also, let $B = q^k$ and let $N$ be the length of a generic cyclic word (or necklace) over a $q$-ary alphabet $mathcal{A}_q$.

As Gerhard suggests, the problem can be recast in terms of necklaces and their subwords of length $k$. We introduce the appellation dimgraph for “directed multigraph”. Let $[w]_k$ denote the generalized order $(k – 1)$ de Bruijn dimgraph of the $q$-ary word $w$: i.e., vertices correspond to the $(k – 1)$-tuples, and directed edges correspond to the $k$-tuples in $w$ in the usual way. If $[w′]_k = [w]_k$, we say that $w$ and $w′$ belong to the same order $k$ de Bruijn homology class. It is easy to see that this really is an equivalence relation, which we denote $sim_k$ (suppressing subscripts when there is no risk of confusion). If the size of an equivalence class is greater than unity, it will be necessary to take into account the additional data associated to $p$, but I will not touch on that (or on the formula to determine the multiplicity of a nontrivial equivalence class, though if anyone wants this please leave a comment and I will try to reply with a PDF) here.

For convenience, assume first that $q = 2$ and define the frequency tuple $alpha = alpha(w)$ as the $B$-tuple of natural numbers indicating the frequency of occurrence in $w$ of the (lexicographically ordered) $k$-tuples over $mathcal{A}_q$. For later convenience we take the frequency tuple to be an integral column vector. For example,

$w = 00011 in mathcal{A}_2^5, k = 3 Rightarrow alpha(w) = (alpha_{000},dots,alpha_{111})^T equiv (alpha_0,dots,alpha_7)^T = (1,1,0,1,1,0,1,0)$.

Notice that we start the index at zero: this will be our default convention for all indices relating to $k$-tuples, and context should generally suffice to determine the indexing scheme for a particular object.

The Eulerian property of generalized de Bruijn graphs means that $alpha$ is completely specified by the $B/2 + 1$ components $alpha_0,alpha_1,dots,alpha_{B/2-1},alpha_{B-1}$: conversely, any such assignment completely specifies the corresponding generalized de Bruijn dimgraph—though notice here that we make no implicit requirement that such a dimgraph must as a matter of course be strongly connected (i.e., correspond to a word). We could make other choices for the “free” components: the essential thing is that the two “loop” components $alpha_0$ and $alpha_{B-1}$ must be free, and any other $B/2 – 1$ components will suffice to complete a specification. The reason for this is basically that the Eulerian property (besides its requirement of strong connectivity) provides $B/2$ independent equations in $B$ unknowns.

More explicitly, for k > 1, write the frequency tuple as a column vector:

$alpha = (alpha_0,alpha_{[0]}^{(k)},alpha_{[1]}^{(k)},alpha_{B-1})^T; quad alpha_{[0]}^{(k)} = (alpha_1,dots,alpha_{B/2-1})^T, quad alpha_{[1]}^{(k)} = (alpha_{B/2},dots,alpha_{B-2})^T$.

Now there is an integral matrix—actually there are many integral matrices—satisfying

$alpha_{[1]}^{(k)} = M_{[0 rightarrow 1]}^{(k)} alpha_{[1]}^{(k)}$.

There is an algorithm for computing these matrices for arbitrary $k$: it is just an exercise in index-juggling, however, so we omit the details. The point is that we can dispense with the unnecessary components of frequency tuples for the binary case, although some care must be taken. Also, it turns out to simplify matters considerably if we omit the words of all zeros and all ones from consideration: we will do so without comment when convenient. More generally, it is convenient to require that none of the components of the frequency tuple are zero (which also implies that none of them equal $N$), but we shall be explicit when making such an assumption.

We can treat some of the components of a frequency tuple in the manner just outlined as functions into the natural numbers. In turn, the matrix-tree and BEST theorems can be jointly recast as a recipe for another function—the BEST function—from this space into the natural numbers that gives the cardinality of the de Bruijn equivalence class.

...

With the binary case in hand, we seek to generalize to $k$-tuples over $mathcal{A}_q$. The approach will be somewhat different here. Define ancestor and descendant matrices by

$M_leftarrow := 1_{1 times q} otimes I_{B/q}, quad M_rightarrow := I_{B/q} otimes 1_{1 times q}$.

Now the indegree/outdegree (I/O) or weak Eulerian property reduces to

$M_leftarrow alpha = M_rightarrow alpha$.

Nothing is lost here by forming the augmented ancestor and descendant matrices

$tilde M_leftarrow := delta_1 otimes M_leftarrow + I_B, quad tilde M_rightarrow := delta_1 otimes M_rightarrow + I_B$

and characterizing the I/O property instead by

$tilde M_leftarrow alpha = tilde M_rightarrow alpha$.

The augmented matrices are invertible, and so we can rewrite the I/O property once more as

$M_{(I/O)}alpha equiv left( tilde M_rightarrow^{-1} tilde M_leftarrow -I right)alpha = 0$.

In other words, a proper frequency tuple must belong to the kernel of the consistency matrix on the LHS above. The dimension $d$ of this in/out (I/O) kernel $ker_{I/O}(q,k)$ is just the number of independent variables required to determine a frequency tuple.

This number is $d = B – B/q + 1$. A sketch of a proof is as follows: of the $B$ edges (including any with multiplicity zero) in an order $(k – 1)$ generalized de Bruijn dimgraph, $q$ are loops: the corresponding frequency tuple components obviously need to be treated as free variables. For the remaining edges, we have B/q independent linear equations in $B – q$ variables. This gives a net of $B – B/q$ independent variables for a minimal integral solution to the linear equations. The +1 is accounted for by scaling.

Accordingly, if the word length is fixed to $N$ then evidently there are $B – B/q$ free components.

For small $q, k$ an exact/integral basis for the in/out kernel can be formed easily (e.g., by using the null command in MATLAB with the r option), and since these are the only cases we can realistically address, we will not bother with an algorithm.

We have two comments at this point. The first is that the reader should be convinced (or take it as an exercise to prove) that the I/O kernel contains an integer lattice whose primitive cells are relatively small (a precise formulation and attendant proof are probably rather technical, and certainly uninteresting). The second is that the lattice can be generated by a positive integral basis: the few negative entries above can clearly be made to go away with a few elementary linear operations.

That is, we consider the I/O lattice

$Lambda_{I/O}(q,k) :- ker_{I/O}(q,k) cap mathbb{Z}^B$

and the associated lattice cone

$Lambda_{I/O}^{0+}(q,k) := ker_{I/O}(q,k) cap mathbb{N}^B$.

Except for some degenerate cases corresponding to disconnected generalized de Bruijn dimgraphs (to be addressed in the sequel), points in this cone correspond to valid frequency tuples. For computational purposes we want to be able to efficiently generate the frequency tuple simplices

$Lambda_{I/O}^{(N)}(q,k) := Lambda_{I/O}^{0+}(q,k) cap N Delta_B$

where a dilation of the standard simplex is indicated on the RHS. (It will be an implicit corollary of our construction of these sets that the I/O lattice is unimodular, but we do not use this.)

We first define a bucketspace (“$s$ balls in $r$ buckets”) $X_{r,s} := {alpha in mathbb{N}^r : sum_{j=1}^r alpha_j = s}$. Now it is clear that there exists a $B times d$ integer matrix $M^{(Lambda)}$ s.t.

$M^{(Lambda)}X_{B,N} = Lambda_{I/O}^{(N)}(q,k)$

(where the LHS is interpreted in the obvious way). In fact there may be several such integer matrices—each reflecting a particular choice of basis—but all we need is one. We will call it a bucketspace-I/O lattice matrix.

The problem of its construction appears subtle (but perhaps that is because we have focused on explicit examples). A useful heuristic is to invoke the LLL lattice basis reduction algorithm on an integer basis for the I/O kernel, then use the structure of the de Bruijn graph—in particular, a table of its cycles—to improve on LLL. Note that the cycle enumeration for de Bruijn graphs has only been effected for small values of $k$.

In practice there are some subtle considerations, which we sketch in the context of the two specific cases $q = 2$ and $k = 3, 4$. Since the I/O kernel is the span of the dimgraph cycle space (see the sequel), it makes sense to look for a set of cycles corresponding to a “nice” basis. A natural candidate is a reduced basis of positive lattice vectors with minimal length.

Once the simple cycles of the de Bruijn graph are enumerated, it is easy to produce such a positive basis with a greedy algorithm. Start with the cycle representatives as a proto-basis. Now pick one of the remaining simple cycles of minimal weight uniformly at random: if it is linearly independent of the existing proto-basis, add it. Keep going until a basis is formed.

For $q = 2, k = 3$ the 6 simple cycles and corresponding simple cycle vectors are

$(0) leftrightarrow (1,0,0,0,0,0,0,0)^T$

$(001) leftrightarrow (0,1,1,0,1,0,0,0)^T$

$(0011) leftrightarrow (0,1,0,1,1,0,1,0)^T$

$(01) leftrightarrow (0,0,1,0,0,1,0,0)^T$

$(011) leftrightarrow (0,0,0,1,0,1,1,0)^T$

$(1) leftrightarrow (0,0,0,0,0,0,0,1)^T$

and the greedy algorithm produces the unique minimal positive basis

${(0),(001),(01),(011),(1)}$.

Similarly for $q=2, k=4$ we have the 19 simple cycles

$(0),(0001),(0001011),(00010111),(00011),(0001101),(000111),(00011101),(001)$,
$(001011),(0010111),(0011),(001101),(00111),(0011101),(01),(011),(0111),(1)$

and the greedy algorithm produces either of the two minimal positive bases

${(0),(0001),(001),(001011),(0011),(01),(011),(0111),(1)}$

${(0),(0001),(001),(0011),(001101),(01),(011),(0111),(1)}$

Note that the simple vectors of length 5 are rejected by the greedy algorithm, as is the simple vector $(000111)$ of weight 6.

(Notice that Gordan’s lemma implies that the monoid [commutative semigroup with identity and cancellation] of integral points in a lattice cone is finitely generated. This in turn implies that we can use one of the minimal positive bases to produce frequency tuple simplices if we are willing to pay a large overhead factor over a more efficient [if less well-understood] method outlined below.)

By performing elementary column operations it is possible to find a workable bucketspace-I/O lattice matrix by trial and error (of course an algorithm would be preferable, but the cycle decomposition of de Bruijn graphs is known only for small values of $k$, suggesting that such a goal is too lofty). We find by direct experiment with $N = 16$ that the following matrices work:

The columns of the $k = 4$ matrix can be indicated graphically:

Finally, notice that the method outlined essentially boils down to finding the vertices of a frequency tuple simplex—but this appears to be more easily said than done.

...

Now that the I/O condition has been suitably developed, we move on to the question of (strong) connectivity. Let $B$ (with an abusive resue of notation) denote the incidence matrix of a dimgraph $G$. Without bothering to restate the common definition, we simple provide a relevant example: for $q = 2$ and $k = 4$ we have

Notice by way of passing that we can recast the I/O equations as

$B1 = 0$.

It can be shown that $ker B$ is the cycle space of $G$.

In any event it is a standard result for dimgraphs that

$mbox{rank}(B(G)) = |V(G)| - c(G)$

where on the RHS we indicate the number of vertices minus the number of components. In short, the rank of the incidence matrix allows us to judge whether or not the corresponding dimgraph is (strongly) connected.

Our intent here is that a generalized de Bruijn dimgraph has only a single component by definition: although we have not really produced a proper mathematical definition, this intent should suffice for the remainder. Now the requirement on a frequency tuple is that

$|V(alpha)| - mbox{rank}(B(alpha)) = 1$.

The number of vertices is just the number of $(k – 1)$-tuples making an appearance; the rank can be computed by Gaussian elimination. For $q = 2$, it can quickly be seen that the rank is just the number of nonzero components in $(alpha_1,dots,alpha_{B/2-1})$.

With that in hand, define the tuple canonical ensemble (TCE) by

$hat Lambda_{I/O}^{(N)}(q,k) := Lambda_{I/O}^{(N)}(q,k) cap {alpha: |V(alpha)|-mbox{rank}(B(alpha)) = 1}$.

The TCE gives all the frequency tuples for a given length, and no garbage points, although some care is required in its construction.

In conclusion: check to see if a given point is in the TCE. If it is, then use the matrix-tree and BEST theorems (again, ask me how and I'll provide a PDF if I see the comment) to determine its cardinality. Providing an enumeration and selecting $p$ minimal completes the solution to the problem, although I'm not sure about the value of $p$, complexity, etc.

ca.analysis and odes - Are two probability distributions uniquely constrained by the sum of their p-norms?

Here is a proof that Steve's rescaling gives you all solutions, together with the trivial operation of permuting the components of $A$, $B$, and $C$ if you view them as vectors with positive coeifficients. (If you view them this way, then Steve's notation $||A||_p$ is just the usual $p$-norm.)

I first tried what Alekk tried: You can take the limit as $p to infty$ and eventually obtain certain power series expansions in $1/p$. Or you can take the limit $p to 0$ and obtain certain power series expansions in $p$. The problem with both approaches is that the information in the terms of these expansions is complicated. To help understand the second limit, I observed that the two sides of Steve's equation are analytic in $p$, but it only helped so much.

Then I realized that when you have a complex analytic function of one variable, you can get a lot of information from looking at singularities. So let's look at that. Let
$alpha_k = ln a_k$, so that
$$||A||_p = expleft( frac{ln bigl[exp(alpha_1 p) + exp(alpha_2 p) + cdots + exp(alpha_d p) bigr]}{p} right).$$
The expression inside the logarithm has been called an exponential polynomial in the literature, which I'll call $a(p)$. As indicated, $||A||_p$ has a logarithmic singularity when $a(p) = 0$. $||A||_p$ has another kind of singularity when $p = 0$, but won't matter for anything. Also $a(p)$ is an entire function, which means in particular that it is univalent and has isolated zeroes. Also, none of the zeroes of $a(p)$ are on the real axis. Let $b(p)$ and $c(p)$ be the corresponding exponential polynomials for $B$ and $C$.

Suppose that you follow a loop that starts on the positive real axis, encircles an $m$-fold zero of $a(p)$ at $p_0$, and then retraces to its starting point. Then the value of $||A||_p$, which is non-zero for $p > 0$, gains a factor of $exp(2mpi i/p_0)$. Thus Steve's equation is not consistent unless all three of $a(p)$, $b(p)$ and, $c(p)$ have the same zeroes with the same multiplicity. (Since $exp(2mpi i/p_0)$ cannot have norm 1, geometric sequences with this ratio but with different values of $m$ are linearly independent.)

At this point, the problem is solved by a very interesting paper of Ritt, On the zeros of exponential polynomials. Ritt reviews certain results of Tamarkin, Polya, and Schwengler, which imply in particular that if an exponential polynomial $f(z)$ does not have any zeroes, then it is a monomial $f_alpha exp(alpha z)$. Ritt's own theorem is that if $f(z)$ and $g(z)$ are exponential polynomials, and if the roots of $f(z)$ are all roots of $g(z)$ (with multiplicity), then their ratio is another exponential polynomial. Thus in our situation $a(p)$, $b(p)$, and $c(p)$ are all proportional up to a constant and an exponential factor. Thus, $A$, $B$, and $C$ must be the same vectors up to permutation, repetition, and rescaling of the coordinates. Repetition is an operation that hasn't yet been analyzed. If $A^{oplus n}$ denotes the $n$-fold repetition of $A$, then $||A^{oplus n}||_p = n^{1/p}||A||_p$. Again, since geometric sequences with distinct ratios are linearly independent, Steve's equation is not consistent if $A$, $B$, and $C$ are repetitions of the same vector by different amounts.

The same argument works for the generalized equation
$$x_1||A_1||_p + x_2||A_2||_p + cdots + x_n||A_n||_p = 0.$$
The result is that any such linear dependence trivializes, after rescaling the vectors and permuting their coordinates.

Update (by J.O'Rourke): Greg's paper based on this solution was just published:

"Norms as a function of $p$ are linearly
independent in finite dimensions," Amer. Math. Monthly, Vol. 119, No. 7, Aug-Sep 2012, pp. 601-3
(JSTOR link).

fourier analysis - A sum involving sines

Consider a scaled sine function, $sin(2pi x/2^n)$, for some positive integer $n$. For this, I have the following linear combination.

$$ sum_{x=1}^{2^{n-2}} c_x sin(2pi x/2^n).$$
(The upper limit to the sum is $2^{n-2}$.)

The question is whether there exist $c_x in {0, pm 1, pm 2}$, not all $0$, that make the above expression $0$, for infinitely many $n$?

If it helps, the above came up in a computation concerning the discrete Fourier Transform.