Firstly, to take supranational organisations. The case for them is not automatically a yes. The UN, WTO, World Bank, yes; FIFA, maybe; the group of People’s Republics of Eastern Europe after World War II, hmmm. You may say that example is a bit far-fetched, but my point is that you should look at the nature of a supranational organisation before embracing it.

You talk of Sovereignty, and I’m going to talk of Democracy, and why people might want decision-making to be taken at the level of the nation state. Not necessarily, is what I say; but for a system to work democratically there has to be a perceived link between the citizen and the decision-maker. Having previously lobbied my UK MP successfully, I assumed that the situation was similar with MEPs. I learned the hard way, being fobbed off by a secretary. And now that MEP’s names don’t appear on our ballot papers – only party loyalties – I could not vote this MEP out without crippling the London’s labour vote, which I am neither willing nor able to do. I am not criticizing the MEP here – I’m sure he is a decent man who works hard – I’m criticizing a system that means he doesn’t need to care what I think.

I also discovered that, as you probably know, the European Parliament is like ours but with no Government present. Imagine how democratic that is! Legislation happens elsewhere, in the European Commission, so our main representative is – what is his name? He’s resigned now anyway, so it doesn’t matter. Again, I’m sure he did the best he could. But democracy it ain’t.

One characteristic of the EU that I really like, and you mention, is the principle of Subsidiarity. Unfortunately, there is another principle which contradicts it – ‘Ever Closer Union’. When these two clash, guess which wins. Subsidiarity produces a flexible diversity, but that’s not what I see the EU doing.

Your discussion about the National Interest, and when you make decisions for your own good and when for the greater good is a bit theoretical for me. But I’d like to point out that if you make decisions that are for the greater good, and in the process alienate sections of your own population, you are heading for disaster. Don’t call a referendum whatever you do! And don’t assume, if you are losing support in many countries across the continent, that it’s because people are getting nastier. It may be something you are doing.

]]>I made a similar post about sports in general: https://blogofmaths.wordpress.com/2016/07/07/a-question-of-sport/

]]>I’ll make some comments addressed solely to you (since I

suspect not many other people will read this post again).

I agree that it would have been helpful to set the problems in a more historical context, but I didn’t do that for the sake of time (the goal I set to myself was to state the theorem precisely, which required most of the talk to consist of definitions and examples, and didn’t allow for a historical overview). I’m still not sure how important the result is compared to other breakthroughs in Thurston’s “program”, and I should be clear, it pales in comparison to Perelman’s resolution of the geometrization conjecture. I think it got some attention since it was “last” of a list of problems of Thurston, which included the geometrization conjecture and the ending lamination conjecture (resolved by Brock-Canary-Minsky). Actually, there are still a few unresolved open-ended questions from this list – there is a nice survey of Thurston’s paper by Otal. http://link.springer.com/article/10.1365%2Fs13291-014-0079-5

I chose to describe Haken manifolds as containing π_1-injective surfaces rather than incompressible surfaces since this is easier to describe (π_1 is taught in any introductory algebraic topology course). It is a non-trivial result that the two notions (for co-orientable surfaces) are equivalent, called the “loop theorem”. Again, I didn’t want to get into these things, because they take some time to describe, even though the loop theorem was one of the most important early results on 3-manifolds of the 20th century (by Pappakyriokopoulos).

I’ll say that the virtual Haken conjecture had some historical importance (in the view of the 3-manifold community) because it would have implied the geometrization conjecture. However, as we know, the implication went the other direction (geometrization was used to prove virtual Haken). By the time geometrization was proved, however, the virtual Haken question was a well-studied open question – Freedman, Lackenby, and Thurston among others had worked on it extensively, and hence it gained some notoriety.

I think the solution I gave (based on the visionary ideas of Wise and his collaborators) will be more important in the future for the applications to cubulated hyperbolic groups, rather than the implications for 3-manifolds (although it does have many applications in 3-manifold topology that were not foreseen when it was conjectured).

The example of the Seifert-Weber dodecahedral space is as you say: taking a hyperbolic dodecahedron with dihedral angles 2π/5 along each edge, the gluing by twists will yield a single vertex with solid angle 4π (one may apply the Poincaré polyhedron theorem to determine the hyperbolic structure). In fact, the cell-decomposition of the Seifert-Weber space is self-dual: put a point in the dodecahedron, an edge through each face, and a pentagon dual to each edge, and the complementary cell structure is isometric.

As for the complements of knots and links (usually in the 3-sphere, so R^3 compactified by a point), these are manifolds for trivial reasons: they are open subsets of a manifold. In fact, the hyperbolic metric is on the open subset obtained by removing the link, and the diameter of the metric is infinite, even though the volume is finite. I like to think of the link as being “pulled” off to infinity like taffy, which gives one a rough idea of what the hyperbolic metric looks like (technically, the end is a warped product of a torus and a ray, with exponential warping function).

]]>For the costs Tim only mentions $10 per submission for the software managing the refereeing process, to be provided by Scholastica. But what about the human side of managing the refereeing process? Some people have to spend time on this. Either this time is free time or it is time paid by their institution, or it means additional costs for the journal to be paid by others.

]]>As to open drug use, the argument against this is simple enough. We can be pretty sure that doing enough drugs will kill people. If we allow some amount of doping, then the same people who today covertly take drugs will covertly exceed the limits, which can only make matters worse. If we don’t place any restrictions, then the same people will try to tread the fine line between maximum performance and killing themselves before winning the race (you’ll easily find with Google studies suggesting a fair fraction of elite athletes would take a drug guaranteed to kill them in 10 years if it would guarantee major wins beforehand). What would you suggest a doctor should do in that situation? Refuse to get involved and let the athlete self-medicate? Or join in and accept responsibility for the occasional accident? Or report someone likely to attempt suicide, which currently a doctor would be required to do in most places?

]]>When talking about the theorem Gowers here wrote that the factorization of 36 is 2 x 2 x 3 x 3. But the composite number 36 by that notation looks like a product of a number and a composite number (2 x 18), a product of a composite number and a composite number (4 x 9), or a product of a composite number and a prime number (12 x 3). A notation like x(2, 2, 3, 3) or (2, 2, 3, 3)x doesn’t get used all that often when talking about the Fundamental Theorem of Arithmetic. And thus, it’s not so obvious, because authors don’t use appropriate notation to theorem all that often when talking about it or make it clear that the theorem implies that for any composite number, there exists of a k-ary product of prime numbers only, where k is some natural number.

]]>$ A=\left( \begin{array}{ccc}

Id & 0 \\

X & Id

\end{array} \right)$

where the rows of $X$ are all the rows with five $1$ and two $0$

like $(1111100)^t,(1111010)^t,(1110110)^t…(1101011)^t…..$

The line separator of $A$ are the line of

$ A=\left( \begin{array}{ccc}

X & Id

\end{array} \right)$

and they have 15 times $1$ and 12 times $0$

So maybe we have change it a litlle bit, I’ve got few idees… but I’m waiting for reactions, it would seem a bit ridiculous to go on, while the comments are not even put in latex^^

]]>That’s a tragedy, really. I think it would very much be worth reading.

Thank you for revealing the common structure (i.e. the prisoner’s dilemma) of such agreements to those of us that didn’t immediately see it (in hindsight, good math often seems so obvious).

The case has occasionally been made that Germany is currently a sweatshop of Europe at least in comparison to some of its immediate neighbours like France: With an ever-growing sector of temporary/lent work, late retirement and very low wages in jobs that do not require a qualification (e.g. working the fields), Germany is making it very hard for a country like France to maintain its desirably worker-friendly status. If France eventually gives in, both German and French workers will end up losing.

(To those who can read German — I’m afraid I could not find a translation) I can also recommend this slightly dated piece: http://www.monde-diplomatique.de/pm/2013/09/13.mondeText.artikel,a0004.idx,0

]]>1) a is prime means there exists no number between 2 and a-1 that divides it

2) there exists a prime factorization of every integer (algo exists for calculating it)

3) let a prime factorization be pi0 * pi1 * … pin

let a different prime factorization be pj0 * pj1 … pjm

pi0 * … pin = pj0 * … pjm

pi0 must divide pj0 * … pjm

thus pi0 must divide some pj, since it itself cannot be written as a product of 2 or more primes

this is a contradiction, so there cannot be a different pj0 … pjm

Let $M\in \mathcal M_n(R)$, with $n>1$, be a lower triangular invertible matrix with all coefficient $0$ or $1$, and suppose that for each integer $0<i\leq j\leq n$

$M_{i,j}=1\Leftrightarrow M_i-M_j\in \left\{0,1\right\}^n$(*)

where $M_i$ and $M_j$ are the line number $i$ and $j$ of $M$

we will say that $M$ is triangular connectable.

If a line in a matrix is such that by removing it we get two equal rows, or a full-0 row : we will say that it is a line-separator:

Sep-FUNC:

In a triangular connectable $n\times n$ matrix without the full-1 line ,there is a line separator $s$ such that $||s||_1\leq n/2$

This does not need the $\wedge$ product which give to matrices of caracteristical fonction of intersection closed Familly a very NOT-VECTORIAL shape, if I can speak this way…but the Sep-FUNC get reed of this Boole ring product and the condition (*) is the vestige of the "inclusion", no more intersection or union…^^

to give a summary of the reason why this is stronger than FUNC, we will say that a version of FUNC still holds in a union closed familly $\mathcal A$ with the same**** size than the ground set $X$, we than see that $\mathcal X=\left\{\mathcal A_x\space|\space x\in X\right\}$ is also closed for the union. and FUNC is equivalent to the fact that there exist a $x\in X$ such that $x$ is aboundant and such that $\mathcal A_x$ is a generator of $\mathcal X$…

****same size up to trivial "lines" and "row" but this is in the upper comment

If we than indexate correctly $X=\left\{x_1…,x_n\right\}$, and $\mathcal A=\left\{A_1,…A_n\right\}$ such that the $Z_2$-matrix, $A$ such that $A_{i,j}=0\Leftrightarrow x_i\in A_j$ we get a anti-caracteristic matrix, and this matrix is a connectable matrix, and the rows of this matrix correspunding to a generators are exactly the separated rows

(of course we have the same statement for lines, please note that I pretend that $M$ is connectable if and only if $M^t$ is connectable, everything is exposed in the upper comment)

I think this conjecture is quite intersting

]]>0) preliminaries and main triangular property

let’s index matrices in $\mathbb N$ with sets $A$ and $B$ with NO ORDER. A classical matrix will then be obtain by giving bijections from $a:[1,|A]\cap \mathbb N|\to A$ and $b:[1,|B|]\cap\mathbb N\to B$

IN BOTH CASES we will note for any matrix $M$, $lin(M)$(resp.$row(M)$) the set of line of $M$ (resp. rows of $M$)

If $M$ is NO-ORDER-indexed we will note that $lin(M)$ (resp.$row(M)$) define a unique NO-ORDERED-indexed matrix, and we will say that $M$ is $line-separated$ ,(resp. $row-separated$)

we will say $separated$ when the matrix is both $line-separated$ and $row-separated$).

we note that $row(lin(M))=lin(row(M)):=M^*$ and if we note $A^*$ and $B^*$ the (non-ordered) sets of indexation of $M^*$, each line $l\in lin(M^*)$ is a element of $\mathbb Z_2^{A^*}$, which is a Boole ring, and we can canonically associate the product $\wedge$ in this ring with the intersection in $\mathcal P(A^*)$

The same can be said with rows, and we can considere what we will call the line-cloture (resp. row-cloture) of a NO-ORDERED-matrix $M$, it will be the $separated$ matrix $L(M)$ (resp. $R(M)$ such that $lin(L(M^*))$ (resp. $row(R(M))$) is closed for $\wedge$ and is the smalest set containing $lin(M^*)$ (resp.$row(M^*)$)

We can easyly see that $R$ and $C$ commute, and I will give a demonstration that if $M$ has the line full-$0$ and not the line full $1$, than $R(L(M))=L(R(M))$ is a NON-ORDERED square matrix, and that you can order the indexed set such that it is a triangular matrix (I will give lter a demonstration and more very interesting properties)

lets finish with terminology!

if we call $l(M)\subset lin(M)\subset L(M)$ the smalest separated NO-ORDRED matrix to have the property $L(l(M))=L(M)$ we will call $l(M)$ the set of Line Generators, the obvious definition handle for rows and we will call $r(M)$ the set of Row-Générators.

$a\in l(M)$ if and only if for any $x,y\in L(M)$, such that $x\ne a\ne y$ we have $x\wedge y\ne a$

another way to see $l(M)$ is that they are the elements $l$ of $L(M)$ such that $L(M)\setminus \left\{l\right\}$ is closed for the product of lines, we will say that a element of $l(M)$ is minimal element if no-one in $lin(M)$ is a divisor (Then for the inclusion it correspond to maximal element, since we chosed the $\wedge$-cloture, corresponding to intersection, if we did the same job with the union (corresponding to “$\vee$”) the definition of our minimal would be the minimal corresponding subset of inclusion)

for example the set if $B$= {$a$,$b$,$c$,$d$,$e$} and $A$= {{$a$},{$b$},{$a,c$},{$a,c,d$},$\emptyset$} $\subset \mathcal P(B)$ can be AFTER having ordered $A$ and $B$… give the ordered matrix (= “usual” matrix) :

$N= \left( \begin{array}{ccc}

1 & 0 & 0 & 0 & 0\\

1 & 1 & 0 & 0 & 0\\

1 & 0 & 1 & 0 & 0\\

1 & 0& 1& 1 & 0 \\

0 & 0 & 0 & 0 & 0

\end{array} \right)$

if $P$ and $Q$ are permutation matrices $PNQ$ will define the same NO-ORDERED-matrix, $M$, in our case $M$ is indexed not by ordered sets but by $A$ and $B$ themselfs!

$M$ is here, line-closed and row-closed ,

$L(M)=lin(M)$ and $R(M)=row(M)$

we can give an ordered representation of $l(M)$ as

$ \left( \begin{array}{ccc}

1 & 1 & 0 & 0 & 0\\

1 & 0 & 1 & 0 & 0\\

1 & 0& 1& 1 & 0 \\

0 & 0 & 0 & 0 & 0

\end{array} \right)$

and we can see that $r(M)=M$ : every row is a generator in this example!

the minimal lines are

$(1,0,1,1,0)$ and $(1,1,0,0,0)$ in $N$

and we can see that they are the one that intersect the unique $1$ coefficient of a row that have only one $1$

this is a caractarisation of minimal lines in a row-closed separated matrix.

to see this suppose that there is a minimal line in a row-closed separated matrix that intersect not such a row, tht meen has $1$ as soon as the minimal line has a $1$ in the corresponding row, and that mean that this line is a divisor of our minimal line, so by minimality , the two lines are equal, but we said the matrix was separated so it is absurd.

We can use this statement to proove by induction on the number of rows (by “removing” a minimal line) that a row-closed separated matrix with row $0$, and without row $1$, is line-closed if and only if it is a NO-ORDERED square matrix (then it has a triangular $n\times n$ representation, of rank $n-1$, because all the diagonal coefficient are $1$ except the commun coefficient of the $0$-row and the $0$-line.

We will call BISTABLE any separated NO-ORDERED matrix which is line-closed and row-closed, and which contain line full $0$ line and row

We can call $\mathbb B$ the set of BISTABLE matrix and $\mathbb B_n$ Bistable matrix of size $n\times n$

If we have a NO-ORDERED matrix $M$, and $l\in lin(M)$ (resp. $r\in row(M)$) we a unique NON-ORDERED matrix $M-l$ (resp. $M-r$ such that $lin(M-l)=lin(M)\setminus \left\{l\right\}$ (resp $row(M-r)=lin(M)\setminus \left\{r\right\}$)

We will also call $\mathbb T$ the (ordered) lower-triangular representation of BISTABLE (NO-ORDERED)-matrix

FUNC is equivalent to the fact that there exist in such a matrix a line-generator with more $0$ than $1$

1)quick definitions and immediate properties

$T$ will be a member of $\mathbb T$ and we will note $T^*$ the corresponding NO-ORDERED matrix in $\mathbb B$ obtained by “forgeting” the indexation (the notation $T^*$ has already been used it for the separated NO-ORDERED matrix of a NO-ORDERED matrix, but as soon as member of $\mathbb B$ are separated, it is not confusing)

we get these property easily by induction and they can themeself be usefull for induction

definition a)

We say that a $x\in lin(T)\cup row(T)$ is a $separator$ when $T^*-x$ is not separated

property a)

separators are generators and generators are separators

definition b)

if $a$ is a line of $T^*$ and $b$ a row of $T^*$, we note $a\wedge b$ the common coeficient of $a$ and $b$. and we will aslo assume that $a\wedge 1=1\wedge a=a\wedge b=b\wedge 1$ and that $0=0\wedge a=a\wedge 0=b\wedge 0=0\wedge b$

we call $T_*=(lin(T^*)\cup row(T^*)\cup \left\{0,1\right\})/\mathcal Z$ the square monoïd

were $\mathcal Z$ is the identificaton $0$=full-0 line=full-0 row

property b)

$(T_*,\wedge)$ is a monoïd

definition c)

if $i\in \mathbb Z_2$ and $a$ is a line (resp. row) we way that $b$ is $i$-perpendicular to $a$ if and only if $b$ is a row (resp. line) and $a\wedge b=i$

We call $a^{\perp}:=\bigwedge_{b\wedge a=1}b$ the product of all rows (resp.line) that are $1$-perpendicular to $a$.

properties c1)

$a\wedge b=a\Leftrightarrow b^{\perp}\wedge a^{\perp}=b^{\perp}$

property c2)

$(a^{\perp})^{\perp}=a$

note1

in terms of intersection-closed separated subset $\mathcal A$ of size $n$ with a groud set $X$ of size $n$ such that $\emptyset \in \mathcal A$ and such that $\mathcal A_{\emptyset}=\emptyset$, the ortogonal of any $x\in X$ is $\mathcal A_x=\left\{a\in \mathcal A\space |\space x\in a\right\}$

and property c1) become $a\subset b\Leftrightarrow \mathcal A_b \subset \mathcal A_a$

note2

in if $a$ is the line nuber $j$ in $T$ than $a^{\perp}$ is the row number $j$

2_very intersting properties of $\mathbb T$ !!

we can give a impressively better result than the main (triangular) property

This result will hold in a bigger set than $\mathbb T$, and we will caracterize this set few interseting equivalent ways!

lets $D$ be a diagonal matrix in $\mathcal M(\mathbb Z_2)$ than

I will say that $P$ a principal matrix of $M$ if $P=(D.M.D)^*$

lets note by $\mathbb P$ the set of all principal matrices of matrices in $\mathbb T$

Very obvioussly members of $\mathbb P$ are lower-triangular matrices.

Quite obvioussly to if $P\in \mathbb P$ and $a\in lin(P)$, we have $(P-a)-a^{\perp}=(P-a^{\perp})-a\in mathbb P$

wait… it is obvious if we call $a^{\perp}$ the row that intesect $a$ on the diagonal… but it mayght not be so clear that the definition and property in c) still work!

in fact it does and $\mathbb P$ is the bigest set of lower-triangular matrix that has those properties , we can say this WITHOUT USING $\wedge$ anymore (!!)

For any lower trangular matrix in $\mathbb R$, $A$ of size $n$, for any lets call $A_i$ the line numer $i$ in $A$, $A^j$ the row number $j$ and $A_i^j$ the $i,j$ coefficient of $A$.

lets note by $\mathbb P^0\subset GL_n(\mathbb R)$ the set of lower-triangular matrices $P$ such that

for any $0<i\leq j\leq n$ integers,

$P_i^j\in \left\{0,1\right\}$ and $ P_i^j=1 \Leftrightarrow P_j-P_j\in \left\{0,1\right\}^n$

if you see $P\in \mathbb P^0$ like matrix with coeficient in $\mathbb Z_2$ you get all the invertible matrix in $\mathbb P$ (we removed the full-0 line and row of $\mathbb P_0$

This is a way to say that $\mathbb P$ is the set of rinagular $\mathbb Z_2$ matrices closed under the $a\mapsto a^{\perp}=\bigwedge_{b\wedge a=1}b$ application where $\wedge$ and $P_*$ can be defined like in definition c) but where $P_*$ is no longer a monoid, because it might not be closed for the $\wedge$

3_connectable lines/rows

let $M$ be the NO-ORDERED matrix indexed by $A$ and $B$ and let $x\in lin(M)$ with $|x|$ the number of $1$ in $x$.

If there exist $a$ bijection from $[1,|A|]\cap \mathbb N\to A$ and $b$ bijection from $[1,|B|]\cap \mathbb N\to B$

such that

1)the matrix $M'=(M_{a(i),b(j)})_{i,j\leq n}$ is trigonal

2) $a^{-1}(x)=|x|$

we say that $x$ is connectable in $M$

(in other terms you have a orderded matrix that represent $M$, which is lower triangular and where $x$ is the line where there is only $1$ under the diagonal, in this diagonal ordered matrix we say that this line is "connected")

a connectable matrix is a matrix where all line are connectable

property e)

$P$ is connectable separated matrix $P^*\in \mathbb P$

(easy by induction)

corollary :

if $P$ is connectable $P^t$ is also connectable

if $T$ is Bistable , then $T$ is connectable

important property f):

if $P\in \mathbb P_n$ and $a\in lin(P)$ and if $mult(a):=\left\{x\in lin(P),\space \exists y\in 2^n, x=a\wedge y\right\}$ then

$|a|=|mult(a)|-1$

(if $P$ is bistable we can write $mult(a)=a\wedge lin(P)$, we note that $MULT(P)=\left\{mult(a),\space a\in lin(P)\right\}$ is a monoide for intersection, and that $a\mapsto mult(a)$ is an isomorphism of monoïd.

note for any monoïd $P,\wedge$, we can also define

MULT'(P) is always a monoïde when the basic monoïd $P$ is idempotent

if we call $divi(a):=\left\{x\in P,\space \exists y\in P, a=x\wedge y\right\}$ we see that $a\mapsto divi(a)$ is an ismorphism of monoïd between $lin(P)$ and $DIVI'(P)=\left\{divi(a),\space a\in P\right\}$ but for the Union (only because of the idempotence $a\wedge a=a$)

$DIVI(P)=(DIVI'(lin(P)), \cup)$ is isomorphic to $(MULT(row(P)),\cap)$

this is given by property c)

4) a week FUNC!

in a bistable if $a$ and $b$ are 0-perpendicular than $|a|+|b|<n$

(it is funny to see that it still holds if $a$ and $b$ are not "perpendicular", but verify $a\wedge b=0$ in $T_*$, (see property and definition b) in paragraph 1))

this a direct corollary of the connectability of a bistable.

we sayd that FUNC is equivalent to the fact that there is a line-generator $a$ in a bistable $B$ such that $|a|<n/2$(**)

this is also equivalent to the fact that there is a line-generator $a$ and a row-generator $b$ such that $|a|<n/2$ and $|b|<n/2$(***)

what would then hapend if we have not (**): we would have

for all genertor line $a\in l(M)$, $ |a|\geq n/2$ , and then each row $r\in row(M)$ that would intersect any generator-line in a $0$ would verify $|r|<n/2$, but all rows are 0-perpendicular to at list one generator line, because if it was not the case, this row would be the full-1 row and as soon as we have the full-0 line this is not possible!

So a contrexemple to FUNC (**) would give us matrix with all rows verifying FUNC,

if we call the equivalent (***) version of FUNC the "and-FUNC", we than have, the $or-FUNC$ :

there is a line-generator $a$ or a row-generator $b$ such that $|a|<n/2$ or $|b|<n/2$

the weeker-FUNC that we HAVE is better than this it say that if the line-generator are not Frankl the row-generators are "uniformally-Frankl", but I think it is not a new result.

Note also that the duality of rows and line, as well as the duality for MULT and DIVI, give, with the property f) the DUAL-FUNC

if $(P,.)$ is a finite commutative idempotent monoïd, and if $g$ is a generator (that mean that $P-\left\{g\right\}$ is still a monoïd, then

$card(g.P)\leq card(P)/2$

(this is exactly the lattice version)

So a intersection closed (or union closed) familly that would not verify the DUAL-lattice-FUNC=FUNC) would verify FUNC uniformilly.

again that is not new but we have it with the tool that I built.

And with this tool I think that we "see" whats happening… therefore it is now time to give a stronger conjecture that the tools ask naturally : we note that bistable matrix are particular connectable matrix, and that the generator are exactly the separators then….

5) Separator-FUNC

If $P\in \mathbb P_n$ there exists $a\in P$ such that $a$ is separator and such that $|a|<n/2$

]]>