nt.number theory - Stark's conjecture and p-adic L-functions

Not long back I asked a question about the existence of p-adic L-functions for number fields that are not totally real; and I was told that when the number field concerned has a nontrivial totally real or CM subfield, then there is a construction due to various people including Coates-Sinnott and Katz.

But my favourite number field at the moment is K = $mathbb{Q}(sqrt[3]{2})$, and sadly K contains no totally real or CM subfield, so for trivial reasons $L(n, chi) = 0$ for every Groessencharacter $chi$ of $K$ and every $n le 0$. So in this case the above constructions just give zero. When I learnt this, I thought "that can't be the whole story, what about higher derivatives at 0"? Asking around, I was told about Stark's conjectures, which apparently predict that the leading term at $s = 0$ of the L-function of any GC of K should be the product of an explicit transcendental regulator and an algebraic number (which, if I've understood this right, should lie in the field $mathbb{Q}$(values of $chi$).)

My question is this: assuming Stark's conjecture, can we construct a distribution on the Galois group of the maximal unramified-outside-p abelian extension of K whose evaluation at any locally constant character of this group gives the algebraic part of the leading term at 0 of the L-series of the corresponding Groessencharacter?

pr.probability - most general way to generate pairwise independent random variables?

I'm sure that Gil's answer is wise and that it is a good idea to look at Alon and Spencer's book. Here also is a quick summary of what is going on.

Suppose that $X_1,ldots,X_n$ are random variables, and suppose for simplicity that they take finitely many values. Suppose that you prescribe the distribution of each $X_i$, and suppose also that you want the random variables to be pairwise independent or $k$-wise independent. Then the constraints on the joint distribution are a finite list of equalities and inequalities. The solution set is a polytope whose dimension is fairly predictable, and the fully independent distribution is always in the interior of this polytope. If you are interested in $k$-wise independent distributions that are far from $k+1$-wise independent, then it can be difficult to determine what is achievable because the polytope is complicated. (The vertices are a particularly interesting and non-trivial class: $k$-wise independent distributions with small support. These are called "weighted orthogonal arrays".) However, if you're just intersted in examples, it is much easier to write down a small deviation of the fully independent distribution. The deviation just satisfies linear equations.

For example, suppose that $X,Y,Z$ are three unbiased Bernoulli random variables (coin flips) that take values $0$ and $1$. Then there are 8 probabilities $p_{ijk}$, one for each outcome $(X,Y,Z) = (i,j,k)$. Then you can set
$$p_{ijk} = frac18 + (-1)^{ijk}epsilon. qquadqquadqquad text{(1)}$$
to get a pairwise independent but not independent distribution. In this simple example, there is a 1-dimensional space of deviations and it is easy to compute how far you can vary the independent solution. (Up to $|epsilon| = frac18$.) In larger cases, the variations can be multidimensional and the polytope of deviations can be more complicated.

Addendum: If I have not made a mistake, all deviations for any finite list of discrete random variables are linear combinations of those of the form (1). More precisely, given discrete random variables $X_1,ldots,X_n$, let $f_i$ be some function of the value of $X_i$ which is 1 for one value, $-1$ for another value, and $0$ otherwise. Then you can make deviations proportional to $prod f_{i_j}$ as long as there are at least $k+1$ factors. It looks like all deviations are a linear combination of those of this form.

pr.probability - Can you explain a step in an expectation maximization algorithm in a Nature article?

These numbers are the normalized likelihoods that the results given in the 10 toss vector
are obtained from the current distributions the coin A (or respectively B).

I'll work out the first two rows for illustration:

The guessed Bernoulli parameter for type A is 0.6 and for type B is 0.5.
According to the binomial distribution formula,
the unnormalized likelihood for obtaining 5H 5T are
From A:

L_A = C(10,5)(0.6)^5(0.4)^5

where C(10,5) is the binomial coefficient 10!/5!5!

Similarly from B we obtain:

L_B = C(10,5)(0.5)^5(0.5)^5

The normalized likelihoods are obtained as

For A: L_A/(L_A+L_B) = 0.4491

For B: L_B/(L_A+L_B) = 0.5509

For the second case 9H 1T

L_A = C(10,9)(0.6)^9(0.4)^1

L_B = C(10,9)(0.5)^9(0.5)^9

The normalized likelihoods:

For A: L_A/(L_A+L_B) = 0.8050

For B: L_B/(L_A+L_B) = 0.1950

fa.functional analysis - Is the category of Banach spaces with contractions an algebraic theory?

I think this is some kind of infinitary algebraic theory, but that it is not a monadic adjunction. That is, if you take the "closed unit ball functor" $B$ from ${bf Ban}_1$ to ${bf Set}$ and the "free Banach space functor" $L: {bf Set} to {bf Ban}_1$, then $L$ is left adjoint to $B$ but this adjunction is not monadic (IIRC, and I often don't).

See, for instance, the first few pages of this paper by Pelletier and Rosicky.

You say something about a closed structure on ${bf Ban}_1$, if I understand this right then this is symmetric monoidal with the tensor being the projective tensor product of Banach spaces. That seems to be well known but little-used, although IMHO having this kind of perspective takes some of the tedium/clutter out of certain computations/constructions in my corner of functional analysis.

I think the ball functor from (Hilbert spaces & contractions) to Set doesn't have a left adjoint, but that's more of a guess than an intuition. Certainly the `natural' attempt to build a left adjoint falls over.

As for putting a closed structure on Hilb .... well, the fact that the natural norm on B(H) is not Hilbertian suggests to me that this won't work. (Put another way, the natural Hilbertian tensor product would dualise to only considering Hilbert-Schmidt class maps between your Hilbert spaces, which in infinite dimensions rules out the identity morphism.)

pr.probability - measurable sets not depending on even coordinates

Let $Asubset{0,1}^omega$ be a measurable set (w.r.t. the usual borel sigma algebra) which does not depend on any even coordinate (that is, if $xin A$ and $x$ and $y$ agree except on a finite number of even coordinates, then $yin A$).

Is it true that $A$ belongs to the sigma-algebra generated by all the odd coordinates + the tail sigma algebra?

To clarify: the tail sigma algebra consists of all the events which do not depend on any coordinate.

It seems to me that this should be some easy/well known measure theory fact/counterexample, but perhaps I'm wrong? Suggestions on where to look for an answer would be welcome.

Note that it is well known the this statement is false if the ground space would be $[0,1]^omega$.

computability theory - {transcendental numbers} {computable transcendental numbers}

Note: Answer is pending update per attached comments.

The difference, stated informally, is that that the non-computable transcendentals in their k-base digit representation are entirely random and non compressible. A computable transcendental, such as e, can be represented by a finite algorithmic description, such as a series expansion, which is a form of compression. For the non-computable numbers no such shorter representation exist. Their shortest computational description is their own infinite digit sequence.
You can read more about computational complexity here:
http://en.wikipedia.org/wiki/Kolmogorov_complexity.

There is a wealth of similar numbers to the Ω class of numbers. In general it is "easy" to come up with new definitions for such numbers. These all belong to the countably infinite set of non-computable definable numbers.

To make matters worse, what is left are the non-definable (and therefore also non-computable) numbers. They are the numbers that cannot be described in any way what-so-ever, other than by just iterating through their infinite non-compressible digit sequence. The set of all non-definable numbers is uncountable.

How many dimensions I need to embed a graph?

As Charles points out, you can always embed a graph in three dimensions. The interesting question is how complicated a surface one needs to embed a graph into. The number of handles one has to attach to a spehere in order for a graph to become embeddable is called the genus of the graph, see graph embedding on Wikipedia, which offers other useful information.