TITLE: Proving a statement about polynomials.
QUESTION [8 upvotes]: I wanted to prove the following statement: in case of a polynomial of an even degree, there exists some $y\in\mathbb{R}$ for which there is NO $x\in\mathbb{R}$ such that $f(x)=y$.
The proof I initially came up with was as following. I thought of splitting the proof into two cases: with positive and negative leading coefficient. Then, we can can prove that the polynomial with a positive leading coefficient attains a global minimum, and hence our statement follows for every $y<y_{\min}$. Similarly, we can prove that each polynomial with a negative leading coefficient attains a global maximum, and hence the statement follows for every $y>y_{\max}$.
However, is there an alternative (or perhaps a simpler) way to prove this statement?

REPLY [0 votes]: Here's an approach (though I haven't rigorously checked its validity), which is a bit more 'complicated' than your approach.
Suppose $f$ is any even polynomial and we assume that it is surjective on $\mathbb{R}$, seeking a (handwavey) contradiction.
By our assumption, $f$ always has a root in $\mathbb{R}$, and in fact it must come in pairs (as a result of the complex conjugate root theorem), hence there must be at least 2 roots.
By looking at roots of $f(x)-y$ we can deduce that for any $y \in \mathbb{R}$, there are at least two values of $x$ for which $f(x)=y$.
Now I believe we can then find a contradiction, though I haven't rigorously validated the following direction:
Without loss of generality, suppose
 $$\lim_{x \to {\infty^{+}}}f(x) = \lim_{x \to {\infty^{-}}}f(x) = \infty^+$$.
(We may assume this as $f$ is an even polynomial)
Then perhaps we could somehow prove that this means there must be some $x=a \in \mathbb{R}$ such that $\lim_{x \to a} f(x) = \infty^-$ because every element in the image is hit at least twice. Hence, if this is true, we have found the contradiction we've been looking for, as the limit to a finite point of a real valued polynomial is finite.
As I've said, I missed many details, not as nearly as simple as your original approach, but I've included it for comparison.


TITLE: Find a primitive element for the extension $Q(3^{1/4}, i)/Q$
QUESTION [1 upvotes]: Find a primitive element for the extension $Q(3^{1/4}, i)/Q$
So, I was guessing the primitive element is $3^{1/4}+i$, and I don't have any trouble to show that $Q(3^{1/4}+i)$ is subset of $Q(3^{1/4}, i)$.
Next, we need to show $Q(3^{1/4}, i)$ is subset of $Q(3^{1/4}+i)$, which means we need to show $3^{1/4}\in{Q(3^{1/4}+i)}$ and $i\in{Q(3^{1/4}+i)}$.
I have no clue how to do it. Can any one help me on this question?

REPLY [0 votes]: Old question, but I just had to deal with it. I presume there also is an elegant solution, but I haven’t found one so far.
One’s strategies are to find some $α ∈ ℚ(\sqrt[4] 3, \mathrm i)$ and then do one of the following:

Show that $\sqrt[4] 3, \mathrm i ∈ ℚ(α)$.
Compute that $[ℚ(\sqrt[4] 3, \mathrm i) : ℚ] = [ℚ(α) : ℚ]$.

One can also do a combination of both. Often, no precise computation is needed, but a convergence of general observations may suffice. For instance, to find the primitive element of a simple extension $L / K$, one could pick an element $α ∈ L$ such that $K(α) ≠ K(α^2)$ and $[K(α^2): K] ≥ 3$, but $[L : K] \mid 8$ , which already entails $L = K(α)$.
Here, one can

pick $α = \sqrt[4] 3 + \mathrm i$ and $β = \sqrt[4] 3 - \mathrm i$,
note that $α + β = 2\sqrt[4] 3$ and $α - β = 2\mathrm i$,
conclude that $ℚ(\sqrt[4] 3, \mathrm i) = ℚ(α, β)$,
observe that $α^2 = \sqrt 3 + 2\sqrt[4] 3 \mathrm i - 1 ∈ ℚ(\sqrt[4] 3 \mathrm i)$, so $ℚ(α^2) ⊆ ℚ(\sqrt[4] 3 \mathrm i)$,
note that the only proper intermediate field of $ℚ(\sqrt[4] 3\mathrm i) / ℚ$ is $ℚ(\sqrt 3)$,
to find that $\sqrt 3 ∈ ℚ(α^2) ⊆ ℚ(α)$,
to finally conclude that $ℚ(α) = ℚ(α, β) = ℚ(\sqrt[4] 3, \mathrm i)$ since $αβ = \sqrt 3 + 1$.

To show that $ℚ(\sqrt[4] 3\mathrm i) / ℚ$ has exactly one proper subfield, observe that $ℚ(\sqrt[4] 3\mathrm i) \cong ℚ(\sqrt[4] 3)$, and since $[ℚ(\sqrt[4] 3) : ℚ] = 4$, any proper subfield $F$ of this extension has $[ℚ(\sqrt[4] 3) : F] = 2$ and, splitting $X^4 - 3$, one can see that the only real divisor of $X^4 - 3$ of degree $2$ which has $\sqrt[4] 3$ as a root is $X^2 - \sqrt 3$, so it’s the only possible candidate for aminimal polynomial of $\sqrt[4] 3$ over $F$, proving $\sqrt 3 ∈ F$, hence $F = ℚ(\sqrt 3)$.


TITLE: Markov Chain Monte Carlo thermolization time estimation (not by eye)
QUESTION [0 upvotes]: For a given MCMC algorithm, there are two important time(=step) scale.

$\tau_{thermolization}$ also known as burn-in time, intialization time.
$\tau_{indenpendent}$ the time scale to make $X_i$ and $X_{i+\tau_{indenpendent}}$  independent from each other

$\tau_{indenpendent}$ can be extract from the auto-correlaton function of certain random variable $$C(t):=\langle X_i X_{i+t}\rangle -\langle X_i \rangle^2 \sim e^{-t/\tau_{indenpendent}}$$
For  $\tau_{thermolization}$, I don't see any explicit definition of it. Usually, I would plot a curve of $X_i$ , and then truncate at a location where the series looks "steady".
My goal is to write a program, which gives me estimation of $\tau_{thermolization}$ and $\tau_{indenpendent}$ before a major Monte Carlo simulation.
So I need an expression or algorithm for $\tau_{thermolization}$ , not by eye judgement.

REPLY [0 votes]: Coming from the molecular simulation community, where MCMC is heavily used along with molecular dynamics, to sample complex configurations of interacting molecules, I agree that this aspect is often glossed over and left as a "rule of thumb". However I'm aware of at least one paper where an attempt has been made, by John Chodera, to tackle it objectively: https://www.biorxiv.org/content/early/2015/07/04/021659 which was also published in J Chem Theo Comp, 12, 1799 (2016). The author also provides a piece of Python software to implement the calculation automatically. Maybe this is what you need.
I won't try to reproduce the mathematics here, but the basic idea seems to be to use the procedure for estimating statistical errors in correlated sequences of data - which involves estimating the correlation time (or the statistical inefficiency), the interval between effectively independent samples, as you mention in your question - and applying it to the interval $(t_0,t_{\text{max}})$ which covers the period between the (proposed) end of the equilibration period and the end of the whole dataset. This calculation behaves in a predictable way if the dataset is at equilibrium; the method systematically carries it out as a function of $t_0$, reducing it from an initial high value, and attempts to detect the value of $t_0$ at which deviations from this expected behaviour are seen.
Anyway, hopefully that paper will be helpful.


TITLE: Is it possible to construct a $p$-adic numbers out of $\mathbb{Z}[[x]]$ like we can for the real numbers?
QUESTION [1 upvotes]: I saw this question earlier today about the $p$-adic numbers generally and am trying to figure out how to visualize them or construct an algebraic entity with the right structure.

There's a straightforward way to build the real numbers out of formal power series in $\mathbb{Z}[[x]]$. Is there a way to do a similar thing for the $p$-adic numbers?
The intuition is that we are considering generalizations of infinite decimals $123456.123123\cdots$, but we only have a single decimal place to the left of the decimal point, and we allow any integer to appear in any decimal place.
I'm defining a notion of normal form to capture the fact that some-but-not-all formal power series converge when evaluated at $\frac{1}{10}$, without circularly assuming that those sums converge to a real number. The normal form for a given real isn't unique, $1.3$ and $1.299999\cdots$ are the same number. Note that in the example below $a_0$ is unconstrained.
$$ \sum_{i=0}^{\infty} a_ix^i \;\; \text{is in normal form} \;\; \textit{if and only if} \;\; \text{$0 \le a_i \le 9$ for all $i \ge 1$} $$
Let's define a relation $R$ as follows
$$ \left(\left(\sum a_i x^i \right), \left(\sum b_i x^i\right) \right) \in R \\ \textit{if and only if} \\ \text{there exists a $k$ such that $10a_k + a_{k+1} = 10b_k + b_{k+1}$} \\ \text{and all indices besides $k$ and $k+1$ are the same. } $$
Let $\simeq$ be the transitive, reflexive, symmetric closure of $R$.
If we consider only the formal power series in $\mathbb{Z}[[X]]$ that are equivalent to a power series in normal form, and then take equivalence classes mod $\simeq$, then we have a field isomorphic to the real numbers $\mathbb{R}$.
We can also define orderings on this structure.
$$ \sum a_i x^i \;\; \text{is non-negative} \;\; \textit{if} \;\;  \text{$\sum a_i x^i$ is in normal form and $a_0 \ge 0$} $$
$$ \sum a_i x^i \;\; \text{is non-negative} \;\; \textit{if and only if} \;\;  \text{$\sum a_i x^i$ is equivalent to a non-negative series in normal form} $$
And then the definition of $\sum a_i x^i \le \sum b_i x ^i$.
$$ \sum a_i x^i \le \sum b_i x ^i \;\; \textit{if and only if} \;\; \sum (b_i-a_i)x^i \;\; \text{is non-negative} $$

Is there a way to do something similar for the $p$-adic numbers?

REPLY [2 votes]: "Decimal expansion" happens to be the/a correct way to view $p$-adics, and power series in particular are a good way in. Intuitively, one can think about $\mathbb Z_p$ as series which look like
$$\sum_{k=0}^\infty a_kp^k,$$
where $a_k\in[0,p)\cap\mathbb Z.$ The $a_k,$ then, are more or less digits in base-$p.$ This should look familiar: if you put $p=\frac1{10}$ into these series, you will recover $\mathbb R.$
However, the caveat here is that we have allowed this to be an infinite series when it certainly does not converge in any meaningful way in $\mathbb R.$ This is not a problem: we don't live in $\mathbb R$ anymore because we live in $\mathbb Z_p,$ and convergence is different in this universe.
Let's be a bit more formal. As suggested, let's take two power series $a(x),b(x)\in\mathbb Z[[x]]$ and ask when they should give the same element of $\mathbb Z_p.$ For concreteness, we'll write
$$a(x)=\sum_{k=0}^\infty a_kx^k,\qquad b(x)=\sum_{k=0}^\infty b_kx^k.$$
To test equality, what we want to do is test$\pmod{p^N}$ for increasingly large $N.$ That is, if
$$\sum_{k=0}^\infty a_kp^k\equiv\sum_{k=0}^\infty b_kp^k\pmod{p^N}$$
as $N\to\infty,$ then we'll have the corresponding elements of $\mathbb Z_p$ be equal. However, these series are only defined formally, so how should we check$\pmod{p^N}$? Well, there's no reason to check elements past $k\ge N$ because they don't add any modular information. So we might as well check if
$$\sum_{k=0}^Na_kp^k\equiv\sum_{k=0}^Nb_kp^k\pmod{p^N}$$
as $N\to\infty.$ Now that we have finite sums, this notion of equality potentially makes more sense and can be made rigorous.
Anyways, the point is that, yes, we can think about $\mathbb Z_p$ as power series in $p.$ This can even be made rigorous, for $\mathbb Z_p\cong\mathbb Z[[x]]/(x-p).$


TITLE: Showing that the sum of any 1-cycle and its opposite orientation is a 1-boundary
QUESTION [0 upvotes]: Suppose that $\phi$ is a $1-cycle$ in a topological space $X$, and that $\phi^*:= \phi r$ is its opposite orientation where $r(t,1-t)=r(1-t,t)$. How should one show that there is a $2-chain$ whose boundary is $\phi+\phi^*$?
I know a proof for a more general fact (given any 1-simplex, the sum of it and its opposite orientation is a 1-boundary), but for this simple case is there any simple solution?

REPLY [0 votes]: I am going to assume that you are asking about the singular chain complex of $X$, so in this answer a "simplex" means a "singular simplex in the topological space $X$".
Suppose at first that one knows how to find a solution for each 1-simplex $\sigma$, i.e. one knows how to find $\tau$ so that $\partial\tau = \sigma + \sigma^*$.
Consider any $1$-chain $\phi$ (not just a 1-cycle). We may write $\phi$ as a finite linear combination of 1-simplices,
$$\phi = \sum_i a_i \sigma_i
$$
where each $\sigma_i$ is a 1-simplex and each $a_i$ is a coefficient. Using your proof for a 1-simplex, it follows that for each $i$ there is a 2-chain $\tau_i$ such that $\partial \tau_i = \sigma_i + \sigma^*_i$. From this we obtain
$$\phi + \phi^* = \sum_i a_i \sigma_i + \sum_i a_i \sigma^*_i = \sum_i a_i (\sigma_i + \sigma^*_i) = \partial \left( \sum_i a_i \tau_i \right)
$$
and so $\sum_i a_i \tau_i$ is the required 2-chain.
So to complete the answer, I'll describe $\tau$ in the special case of a single 1-simplex $\sigma$ (which does not have to be a cycle). I'll follow the standard notations in many algebraic topology books, for example Hatcher's book. By definition, $\sigma$ is a continuous function $\sigma : \Delta^1 \to X$. Let $c : \Delta^2 \to \Delta^1$ be the affine map that maps the first and third vertices of $\Delta^2$ to the first vertex of $\Delta^1$, and maps the second vertex of $\Delta^2$ to the second vertex of $\Delta^1$. Let $d : \Delta^2 \to \Delta^1$ be the constant map which take $\Delta^2$ to the first vertex of $\Delta^1$. Then we just take $\tau = \sigma \circ c + \sigma \circ d$. A short computation shows that $\partial \tau = \sigma + \sigma^*$.


TITLE: Help proving operator inequality
QUESTION [0 upvotes]: Given $P \geq 0$, I need to show that $2Tr(P^{5/2}) \leq Tr(P^3) + Tr(P^2)$. It's trivial to show that the RHS is the trace of a positive operator, but I'm at a loss on how to actually prove this inequality...

REPLY [2 votes]: We have $P^3 + P^2 - 2 P^{\frac{5}{2}} = (P^{\frac{3}{2}} - P)^{2} \geq 0$, so the trace is also non-negative, and this is the announced inequality.


TITLE: Is the density of probability of the Brownian bridge known?
QUESTION [6 upvotes]: The Brownian bridge is a stochastic process defined as a Brownian motion with
the condition that it comes back to the origin at time $t=1$. A Brownian bridge $X$ can be obtained from a Brownian motion using
$$X_t=B_t-tB_1.\tag1$$
This definition extends to any final time $T$ and any final point $Y$ in any dimension $d$ using the same definition
$$X^{d,Y}_t=B^{d}_t-\frac tT(B^d_T-Y),$$
where $B^d$ is a $d$ dimensional Brownian motion.
I was wondering if the probability density of $X$ was known in the literature and if so, if someone could provide me with a reference.

REPLY [6 votes]: Edited to remove useless material
So silly of me, the answer is straightforward.
Consider a Brownian bridge $X$ in $d$ dimensions, starting at $x=0$ at time $t=0$ and reaching its final point $R$ at time $t>0$. Then the caracteristic function is determined by the equation (1) in the Question :
$$ \mathbb{E}[X_u]=\frac utR, \qquad \mathbb{E}[X_u^2]-\mathbb{E}[X_u]^2=\frac{u(t-u)}t.$$
It follows that the density probability of $X_u$ is a Gaussian distribution centered on $\frac utR$ with variance $\frac{u(t-u)}t$ or explicitely
$$p_{X_u}(x)=\left(\frac{t}{2\pi u(t-u)}\right)^{d/2}\exp\left(-\frac{t\left(\frac utR-x\right)^2}{2u(t-u)}\right).$$


TITLE: Ways to add Aronszajn trees which are neither Souslin nor special
QUESTION [11 upvotes]: By an Aronszajn tree, I mean a tree of height $\omega_1$ with countable levels and no branch. Such a tree is Souslin if it has no uncountable antichains and special if it can be written as the countable union of antichains. It is well known that there is both a ccc forcing and a $\sigma$-closed forcing for adding Souslin trees and, given any Aronszajn tree $T$ (Souslin or otherwise) there is a ccc forcing for specializing $T$. Therefore we can add Souslin trees, and then specialize them, giving us new special Aronszajn trees. In this post I have a variety of questions about forcings which add Aronszajn trees that are neither special nor Souslin. The most basic of these is as follows.
Question 1: What are some of the most common ways of forcing to add an Aronszajn tree which is neither Souslin nor special? In this post Mohammad Golshani cites a theorem that immediately implies that such trees exist under certain $\diamondsuit$ hypotheses. Therefore forcings to add diamonds suffice but I'm curious whether there is a more explicit forcing where the generic is actually the tree.
My next question is about killing Souslin-ness (or whatever it's called) without specializing.
Question 2: Suppose $T$ is Souslin. Does there exist a forcing notion $\mathbb P$ such that in $V^\mathbb P$, $T$ is still Aronszajn but no longer Souslin and also not special?
I'm also curious about these questions in the case of fat Aronszajn trees: Aronszajn trees with the requirement of countable levels removed. The same ccc specializing forcing works to make such trees special so under MA all such trees are special. I'm curious whether non-special fat Aronszajn trees can be built by hand using forcing.
Question 3: What forcings will add a fat, non-special Aronszajn tree?

REPLY [3 votes]: Regarding your comment before Question 3, you mean all such trees with size $<\mathfrak{m}$ are special. In particular, it doesn't say anything about trees of size $\geq 2^{\omega}$. Indeed, $T(\mathbb{R})$ where elements are bounded well ordered subsets of $\mathbb{R}$ ordered by end-extension is not special. It definitely has no uncountable branches.
For question 1, you can first add a Suslin tree (there are various ways of doing this), of which the most straightfoward one is to use countable well-pruned trees with end-extension (due to Jech). For the second step (which also gives yet another answer to the second question), you can pick $S\subset \omega_1$ that is stationary and co-stationary, and use a Shelah style forcing $Q(T,S)$ to make $T$ $S$-st-special (this means there exists a regressive function $f$ whose domain is $T\restriction S$ such that $t<t'\in T\restriction S$ implies $f(t)\neq f(t')$), but not special (this is implied by the property of $Q(T,S)$ called $(T,S)$-preserving). Observe that if a tree is $S$-st-special, then there must not be any uncountable branch.
You can find more details in Chapter IX of Shelah's proper forcing book.


TITLE: why every absolutely convex absorbing and closed set is a zero neighborhood?
QUESTION [1 upvotes]: I need to prove that
"If $E=\lim_{n} E_n $, the inductive limit of Banach spaces $E_n$ and $A$ is absolutely convex, absorbing and closed set in $E$, then $id_n^{-1}(A)$ is a zero neighborhood in $E_n$ for all $n$.
Here, $id_n^{-1}$ identity map from $E$ to $E_n$ which is continuous linear map".

REPLY [0 votes]: In a locally convex inductive limit $E=\lim E_n$
(with linking maps $i_n:E_n\to E$) an absolutely convex set $A$ is a $0$-neighbourhood if and only if all preimages $i_n^{-1}(A)$ are $0$-neighbourhoods in $E_n$.
If $E_n$ are Banach and $A$ is closed and absorbing than also $i_n^{-1}(A)$ is closed (continuity of $i_n$) and absorbing and then Baire's theorem (or the fact that Banach spaces are barrelled) implies that $i_n^{-1}(A)$ is a $0$-neighbourhood in $E_n$.
This argument neither uses the injectivity of $i_n$ nor that you have a countable inductive limit.


TITLE: Serre's formula for $\Delta^{1/3}$
QUESTION [8 upvotes]: Let
$$E\quad\colon \quad y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$
be an elliptic curve, and let $b_4 = a_1a_3+2a_4$.
Serre in Propriétés galoisiennes des points d'ordre fini des courbes elliptiques  (Section 5.3, b), p. 305) gives the following formula
$$\Delta^{1/3}=b_4-3(x_i x_j+x_kx_l),\label{1}\tag{1}$$
where the $x_i$ are the $x$-coordinates of points of $E$ of order $3$, and $\{1,2,3,4\}=\{i,j\}\cup\{k,l\}$. This formula can be rewritten as
\begin{equation*}
 j^{-1/3}=\frac{\frac{1}{2}\frac{27j}{j-12^3}+3(x_ix_j+x_kx_l)}{\frac{27j}{j-12^3}}.\label{2}\tag{2}
 \end{equation*}
The relation \eqref{2} is a relation between modular forms, therefore its proof comes for free.
Questions:

Is there an algebraic reason why \eqref{1} should be true?
Is there an algebraic proof of \eqref{1}?
Are there similar formulas for $\Delta^{1/d}$ where $d$ is a divisor of $24$?
Are there similar formulas for $\Delta^{1/d}$ if $d$ is not  a divisor of $24$?
What is the algebraic significance of the number $24$ in this context ($\Delta$ is the $24$-th power of the Dedekind eta function, but what does this mean algebraically)?

REPLY [5 votes]: Regarding Q5, the modular form $\Delta$ has weight 12, so it is a section of the line bundle $\omega^{\otimes 12}$ on the modular curve $Y$. Here $\omega$ is the Hodge bundle, which can be defined as $\omega = \pi_* \Omega^1_{E/Y}$ where $\Omega^1_{E/Y}$ is the sheaf of relative Kähler differentials on the universal elliptic curve $\pi : E \to Y$ (note that here $Y=Y(1)$, so there is no universal elliptic curve, to get around this problem we must treat $Y(1)$ as an algebraic stack).
If $d$ is a divisor of $12$ then $\Delta^{1/d} = \eta^{24/d}$ will be a cusp form of weight $12/d$ (I don't recall the precise level), so everything is algebraic. If you want to get to $\eta$ (in other words $d=24$) then you need half-integral weight modular forms. The interpretation of such modular forms uses the metaplectic cover of $\mathrm{SL}_2(\mathbb{R})$, which is not an algebraic group anymore, only a Lie group (see David Loeffler's answer for more details). The point is that the Hodge bundle $\omega$ is not the square of a line bundle, but it becomes a square once you pull back to the metaplectic group.


TITLE: Show that $F(x)=\int_0^x f(t)\ dt.$ is differentiable. if $f(x)=\begin{cases} \cos(1/x) & x\neq 0\\ 0 & x=0\end{cases}.$ bounded,continuous except x=0
QUESTION [0 upvotes]: The function
$f(x)=\begin{cases} \cos(1/x) & x\neq 0\\ 0 & x=0\end{cases}.$
is bounded, and is continuous except at 0. Hence it is integrable. Show that the function
$F(x)=\int_0^x f(t)\ dt.$
is differentiable.
I have tried proving that by using the derivative of the function $g(x)=\begin{cases} x^2\sin(1/x) & x\neq 0\\ 0 & x=0\end{cases}$
I took the derivative and got $g'(x)=2x\sin(1/x)-cos(1/x)$, and I couldn't proceed from there.

REPLY [2 votes]: Nice approach. The function$$\begin{array}{rccc}h\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ if }x=0\end{cases}\end{array}$$is continuous, and therefore integrable on any interval $[a,b]$. And $f=2h-g'$. Therefore\begin{align}F(x)&=\int_0^xf(t)\,\mathrm dt\\&=2\int_0^xh(t)\,\mathrm dt-\int_0^xg'(t)\,\mathrm dt\\&=2\int_0^xh(t)\,\mathrm dt-g(x).\end{align}So, $F$ is the difference between two differentiable functions, and therefore it is differentiable.


TITLE: Can a prime of form $4m+1$ be expressed as sum of 2 squares in more than one way?
QUESTION [3 upvotes]: It is easy to find dozens of proofs (some of which are elementary or easy to follow) that every prime of the form $4m+1$ can be expressed as the sum of two squares.

How can one prove that every prime of the form $4m+1$ can be expressed as the sum of two squares $p = a^2 + b^2 | a,b \in \Bbb Z^+, a>b$ in only one way?

This is easy to verify for primes up to quite a large value but I can't find a proof.

REPLY [2 votes]: Denote by $p$ a prime number of the form $4m+1$. Thus $$p\equiv 1\pmod 4$$
Assume that $p={a^2}+{b^2}={c^2}+{d^2}$ for some $a,b,c,d\in\mathbb {N}$ such that, w.l.o.g. $a>b$ and $c>d$. Our aim is now to show by contradiction, that this follows to $a=c$ and $b=d$:
We now have $$(ad-bc)(ad+bc)={a^2}{d^2}-{b^2}{c^2}=\bigl(p-{b^2\bigr)}{d^2}-{b^2}\bigl(p-{d^2\bigr)}=p\bigl({d^2}-{b^2}\bigr)\equiv 0\pmod p$$
This implies (see Euclid's lemma) that either $$p|(ad-bc)$$
or $$p|(ad+bc)$$
Let's suppose $p\mid(ad+bc)$. Hence ${a^2},{b^2},{c^2},{d^2}<p$. Thus $a,b,c,d<\sqrt{p}$, which implies that $$0<ad+bc<2p$$
That leads to $ad+bc=p$. However $${p^2}=\Bigl({a^2}+{b^2}\Bigr)\Bigl({c^2}+{d^2}\Bigr)={(ad+bc)^2}+{(ac-bd)^2}={p^2}+{(ac-bd)^2}$$ $$\Rightarrow ac-bd=0$$
which, nevertheless, contradicts the assumption that $a>b$ and $c>d$. This shows, that $p\nmid ({ad+bc})$.
This implies $p\mid ({ad-bc})$. Analugously $$a,b,c,d<\sqrt{p}$$ which implies that $$-p<ad-bc<p \Rightarrow ad=bc$$ Hence $a\mid {bc}$. Note however, that since $p={a^2}+{b^2}$, $a$ and $b$ have to be coprime, so $$a\mid{c}$$
Let $c=xa$. Since $ad=bc$, it follows that $d=xb$
We finally have $$p={a^2}+{b^2}={c^2}+{d^2}={x^2}\bigl({a^2}+{b^2}\bigr)\Rightarrow x=1$$
$$\Rightarrow c=a\quad d=b$$


TITLE: Group action and semidirect product
QUESTION [3 upvotes]: Let $m,n \in \mathbb N$, $k$ a field, $X=(k^{n\times m},+)$, and we consider the groups $GL(n,k)$ and $GL(m,k)$. Let $K:=GL(n,k) \times GL(m,k)$. We define
\begin{align*}
K\times X &\to X\\
((A,B),M) &\mapsto AMB^{-1}
\end{align*}
Note that this defines an action of $K$ on $X$. With this action we can consider $X \rtimes K$. Let
$$
G=\left\{\begin{pmatrix}
        A & M \\
        0 & B \\
        \end{pmatrix} :  A \in GL(n,k), B \in GL(m,k), M \in X \right\} \, .
$$
Show that $G \cong X \rtimes K$.
I've tried to define the following map
\begin{align*}
\psi:G &\to X \rtimes K\\
\begin{pmatrix}
        A & M \\
        0 & B \\
        \end{pmatrix} &\mapsto (M,(A,B))
\end{align*}
but this is not a morphism since $$\begin{pmatrix}
        A & M \\
        0 & B \\
        \end{pmatrix}\begin{pmatrix}
        C & M' \\
        0 & D \\
        \end{pmatrix}=\begin{pmatrix}
        AC & AM'+MD \\
        0 & BD \\
        \end{pmatrix},$$ so $$\psi\left(\begin{pmatrix}
        A & M \\
        0 & B \\
        \end{pmatrix}\begin{pmatrix}
        C & M' \\
        0 & D \\
        \end{pmatrix}\right)=(AM'+MD,(AC,BD)),$$ and  $$\psi\left(\begin{pmatrix}
        A & M \\
        0 & B \\
        \end{pmatrix}\right)\psi\left(\begin{pmatrix}
        C & M' \\
        0 & D \\
        \end{pmatrix}\right)=(M,(A,B)).(M',(C,D))=(M+AM'B^{-1},(AC,BD))$$
Could someone suggest me an appropriate isomorphism between these two groups? Thanks in advance.

REPLY [2 votes]: First we identify the subgroups of $G$ corresponding to $X$ and $K$:
\begin{align*}
X &\cong  \left\{
\begin{pmatrix}
1 & M\\
0 & 1
\end{pmatrix}
: M \in X \right\} =:\tilde{X}\\
\quad
K &\cong \left\{
\begin{pmatrix}
A & 0\\
0 & B
\end{pmatrix}
: A \in GL_n(k), B \in GL_m(k)\right\} =: \tilde{K}
\end{align*}
where $1$ is the identity matrix of the appropriate size.
Note that
\begin{align*}
\begin{pmatrix}
A & M\\
0 & B
\end{pmatrix}
=
\begin{pmatrix}
1 & M B^{-1}\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
A & 0\\
0 & B
\end{pmatrix} \in \tilde{X} \tilde{K} \, .
\end{align*}
This suggests the map
\begin{align*}
\varphi : G & \to X \rtimes K\\
\begin{pmatrix}
A & M\\
0 & B
\end{pmatrix} & \mapsto (MB^{-1}, (A,B)) \, .
\end{align*}
Can you show that this map is a homomorphism?


TITLE: Anybody to solve this definite integral?
QUESTION [0 upvotes]: Can someone help me with this definite integral?
$$ \int _{-\pi}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^x}  dx $$

REPLY [10 votes]: Split the integral into two parts:
$$I = \int _{-\pi}^{0}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^x}  dx + \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^x}  dx$$
Applying the substitution $x \to -x$ on the first integral:
$$I = \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^{-x}}  dx + \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^x}  dx$$
Multiply top and bottom of first integral by $2^x$:
$$I = \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{2^x}{1+2^x}  dx + \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^x}  dx$$
Sum the two integrals:
$$I= \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)} dx$$
Use the fact that $\sin(3x) = 3\sin(x)-4\sin^3(x)$ by the triple-angle formula:
$$I= \int _{0}^{\pi}3-4\sin^2(x) dx$$
Using the fact that $\int_{0}^{\pi}\sin^2(x)dx = \frac{\pi}{2}:$
$$I = 3\pi-4\left(\frac{\pi}{2}\right) = \pi$$


TITLE: How are $Var(X)$ and $Var(X^2)$ both positive?
QUESTION [1 upvotes]: Suppose that the distribution of a random variable
  $X$ is symmetric with respect to the point $x = 0$.  If $\mathbb{E}(X^4)>0$ then $Var(X)$ and $Var(X^2)$ are both positive.

How is that true? I am getting $Var(X)=\mathbb{E}(X^2)$ and  $Var(X^2)=\mathbb{E}(X^4)-(\mathbb{E}(X^2))^2$, but do not know why $\mathbb{E}(X^2)>0$ & $\mathbb{E}(X^4)>(\mathbb{E}(X^2))^2.$

REPLY [0 votes]: I think we can prove everything but the fact that $Var(X^2) > 0.$ We are given that $E(X) = 0$ and $E(X^4) > 0$, the question is how to show both Var$(X)$ and Var$(X^2)$ are strictly positive.
Knowing $E(X^4) > 0$ allows us to say that
$$
\begin{align}
\frac{\sum_{i=1}^nx_i^4}{n} &> 0\\
\Rightarrow \sum_{i=1}^nx_i^4 &> 0
\end{align}
$$
Now each $x_i^4$ individually must be greater than or equal to $0$ as the power is even, and there must exist some $x_i \neq 0$ to allow for $\sum_{i=1}^nx_i^4 > 0$. Discard all $x_i = 0$, and call the remaining observations the set ${i^*}$. The remaining $x_{i^*} \neq 0$ so $\sum_{i^*} x_{i^*}^2 > 0$ as well. All that are left are 0 observations which cannot reduce the value, so $E(X^2)$ must also be greater than 0, thus:
$$
\begin{align}
Var(X) &= E(X^2) - E(X)\\
&= E(X^2) > 0\\
&\blacksquare\; Var(X) > 0
\end{align}
$$
Now we get to the tricky part. We can say
$$
\begin{align}
Var(X^2) &= E((X^2)^2) - E(X^2)^2\\
&= E(X^4) - E(X^2)^2
\end{align}
$$.
What is $E(X^4)$? It's
$$
\begin{align}
&=\frac{\sum_{i^* = 1}^{n^*}x_{i^*}^4}{n^*}\\
&=\frac{\sum_{i^* = 1}^{n^*}\left(x_{i^*}^2\right)^2}{n^*}\\
\end{align}
$$
What is $E(X^2)^2$? It's
$$
\begin{align}
&=\left(\frac{\sum_{i^* = 1}^{n^*}x_{i^*}^2}{n^*}\right)^2\\
&=\frac{\left(\sum_{i^* = 1}^{n^*}x_{i^*}^2\right)^2}{\left(n^*\right)^2}
\end{align}
$$
So the denominator of $E(X^2)^2$ is the square of the denominator of $E(X^4)$ which should make $E(X^4) > E(X^2)^2$. However, compare the numerators. We have a case of Jensen's inequality. Since all the terms are strictly non-negative, and some are strictly positive, being that squaring is a convex function, the sum of the squares is less than or equal to the square of the sum. So we have $E(X^4)$ having a lower numerator and $E(X^2)^2$ having a higher denominator, and I'm not sure if we can always say which one dominates.


TITLE: If krull dimension is $0$, then all prime ideals are maximal?
QUESTION [1 upvotes]: Let $R$ be a ring, If we assume '$R$'s all prime ideals are maximal',then we can say $R$'s krull dimension is exactly $0$. This is easy because there are two different prime ideal $p$ and $q$ which satisfies $p⊂q⊂R$, and this contradicts the fact that $p$ is maximal.
But is the converse true?
my attempt:
If there are prime ideal $p$ which is not maximal,
then there are another ideal $q$ which satisfies $p⊂q⊂R$. If I can deduce $q$ is prime ideal, the proof are done.But I cannot show $q$ is a prime ideal.
Any help would be appreciated, thank you.

REPLY [3 votes]: Every ideal is contained in a maximal ideal.

Basically you need to verify that Zorn's lemma applies. If you have a chain of ideals that all contain $p$ and all don't contain $1$ then their union contains $p$ and doesn't contain $1$ (every chain has an upper bound). So the set of proper ideals containing $p$ contains a maximal element and that element is a maximal ideal.

Every maximal ideal is prime.

Which should be familiar.


text:
\begin{document}

\title{Representations of Logmodular Algebras}
\author[V. I. Paulsen]{Vern I. Paulsen}
\address{Department of Mathematics, University of Houston, Houston, TX, 77204}
\email{vern@math.uh.edu}

\author[M. Raghupathi]{Mrinal Raghupathi}
\address{Department of Mathematics, University of Houston, Houston, TX, 77204}
\email{mrinal@math.uh.edu}
\date{\today}
\urladdr{http://www.math.uh.edu/~mrinal}
\thanks{This research was supported in part by NSF grant
  DMS-0600191.}
\subjclass[2000]{Primary 47L55; Secondary 47A67, 47A20}
\keywords{logmodular algebra, completely contractive, 2-summing, semispectral}

\begin{abstract}
  We study the question of whether or not contractive representations
  of logmodular algebras are completely contractive. We prove that a
  2-contractive representation of a logmodular algebra extends to a
  positive map on the enveloping C*-algebra, which we show generalizes
  a result of Foias and Suciu on uniform logmodular algebras. Our
  proof uses non-commutative operator space generalizations of
  classical results on 2-summing maps and semispectral measures. We
  establish some matrix factorization results for uniform logmodular
  algebras.
\end{abstract}

\maketitle

\section{Introduction}

Let $\cal B$ be a unital $C^\ast$-algebra and let $\cal A \subseteq
\cal B$ be a unital subalgebra. By a {\em representation} of $\cal A$
on a Hilbert space $\cal H$ we mean a unital, contractive
homomorphism, $\rho: \cal A \to B(\cal H)$, where $B(\cal H)$ denotes
the $C^\ast$-algebra of bounded operators on $\cal H.$ A
representation of $\cal A$ is said to have a \textit{$\cal
  B$-dilation} if there exists a Hilbert space $\cal K$, an isometry,
$V: \cal H \to \cal K$ and a unital $\ast$-homomorphism, $\pi: \cal B
\to B(\cal K),$ such that $\rho(a)= V^\ast\pi(a)V,$ for every $a \in
\cal A.$ By a famous result of Arveson~\cite{Ar}, $\rho$ has a $\cal
B$-dilation if and only if $\rho$ is a {\em completely contractive}
map. For this reason there is a great deal of interest in results that
imply that contractive representations are automatically completely
contractive.

In particular, the Sz.-Nagy dilation theorem implies that contractive
representations of the disk algebra are completely contractive and it
is proven in~\cite{MM},~\cite{PP} and~\cite{DPP} that contractive
representations of many finite-dimensional CSL algebras, including
algebras of block upper triangular matrices, are completely contractive.

However, it is still unknown if contractive representations of
$H^{\infty}(\bb D)$ or of the upper triangular operators on
$\ell^2(\bb N)$ are automatically completely contractive. It is known
that weak*-continuous contractive representations of both these
algebras are completely contractive. The case of $H^{\infty}(\bb D)$
follows readily from Sz.-Nagy's theorem and the case of nest algebras
is done in~\cite{PPW}.

Both these algebras are examples of {\em logmodular} algebras and it
is still unknown if, more generally, contractive representations of
logmodular algebras are always completely contractive.

If $\rho:\cal{A}\to B(\cal{H})$ is a representation, then we say that
$\rho$ is $R_n$-contractive if $\rho_{1,n}:M_{1,n}(\cal{A})\to
B(\cal{H}^{(n)},\cal{H})$, defined by $\rho_{1,n}((a_1,\ldots,a_n))=
(\rho(a_1),\ldots,\rho(a_n))$, is contractive. The notion of
$C_n$-contractive is defined by using columns instead of rows.

Foias and Suciu~\cite{FS} proved, in modern terminology, that every
$R_2$- and $C_2$-contractive representation of a logmodular uniform
algebra is automatically completely contractive. Thus, by their result
a $R_2$- and $C_2$-contractive representation of $H^{\infty}(\bb D)$
is completely contractive. However, it is still not known if every
$R_2$- and $C_2$-contractive representation of a general logmodular
algebra is automatically completely contractive. In particular, it is
not known if every $R_2$- and $C_2$-contractive representation of the
upper triangular operators on $\ell^2(\bb N)$ is completely
contractive.

In these notes we try to extend the result of Foias and Suciu to
general logmodular algebras. We prove that if $\cal A$ is a logmodular
subalgebra of a $C^\ast$-algebra $\cal B,$ then every representation
of $\cal A$ on $\cal H$ that is both $R_2$- and
$C_2$-contractive extends to a positive map from $\cal B$ to $B(\cal
H)$ and that this positive map is unique.  The Foias-Suciu result then
follows as a corollary.

A result of Pietsch, Pelczy\'{n}ski and Lindenstrauss~\cite{Pi, LP},
on the existence of ``dominating'' measures for 2-summing maps plays a
central role in Foias-Suciu's proof. We give a new proof of this
result using operator space methods and obtain a non-commutative
analogue that we need for our generalization.

The second key element in the Foias-Suciu proof is Mlak's
theory~\cite{Ml} of semispectral measures and we obtain an analogue of
this theory for non-commutative $C^\ast$-algebras.

Recent work of Blecher and Labuschange~\cite{BL,BL2} has shown that
many classical function theoretic results on logmodular algebras
extend to the ``non-commutative'' logmodular algebras. Their results
consider only completely contractive representations and do not
address the above problems, but we are able to use many of their
ideas. For example, they prove that every completely contractive
representation of an arbitrary logmodular algebra extends uniquely to
a two-positive map on the $C^\ast$-algebra. In their case, the
existence of a completely positive extension is automatic by Arveson's
extension theorem, while the issue in their case is the uniqueness. In
our case, the difficulty is the existence of the extension at all,
while the techniques to prove uniqueness are essentially contained in
their paper~\cite{BL}.

One limitation of the theory of logmodular algebras is the lack of
many natural examples. We prove that the only logmodular CSL algebras
of matrices are the block upper triangular matrices. Thus, by a result
of~\cite{PP} every contractive representation of these types of
logmodular algebras is completely contractive. Possibly every
logmodular subalgebra of the algebra of matrices is of this form. This
result also indicates that possibly every logmodular completely
distributive CSL algebra is a nest algebra.

\section{Definitions and Examples}

Let $X$ be a compact Hausdorff space. Let $C(X)$ denote the continuous
complex valued functions on $X$ and let $C(X)_+$ denote the set of
positive elements in $C(X)$. If $S\subset C(X),$ then let
$\cal{S}^{-1}$ denote the set of elements in $\cal{S}$ that are
invertible in $\cal{S}$. A {\em logmodular algebra} on $X$ is a
subalgebra $\cal{A}$ of $C(X)$ such that the set $\{\log \mod{a}\,:\,
a\in \cal{A}^{-1}\}$ is dense in the real-valued functions on $C(X)$,
which we denote $C_\bb{R}(X)$. The following well-known fact is
immediate.

\begin{lemma}
  $\cal{A} \subseteq C(X)$ is logmodular if and only if the set
  $\{\mod{f}^2\,:\, f\in \cal{A}^{-1}\}$ is dense in the set of
  positive invertible functions on $X$.
\end{lemma}

This equivalent characterization of logmodularity motivates the
following definition. Let $\cal{B}$ be a $C^\ast$-algebra and let
$\cal{A}$ be a subalgebra. Following ~\cite{BL}, we say that $\cal{A}$
is {\em logmodular} in $\cal{B}$ if the set $\{a^\ast a\,:\, a\in
\cal{A}^{-1}\}$ is dense in $\cal{B}_+^{-1}$.

From the invertiblity of the $a\in \cal{A}^{-1}$, it follows that the
set $\{aa^\ast\,:\, a\in \cal{A}^{-1}\}$ is also dense in
$\cal{B}_+^{-1}$.  Thus, if $\cal{A}$ is logmodular, then so is the
subalgebra $\cal{A}^\ast$.

Some algebras possess a property called {\em stong logmodularity} or
{\em factorization}. This means that every positive invertible element
in $\cal{B}$ is of the form $a^\ast a$ for some $a\in \cal{A}^{-1}$.

Examples of logmodular algebras include the disk algebra $A(\bb{D})
\subseteq C(\bb T)$, a conseqeunce of the Fej\'{e}r-Riesz theorem on
the factorization of positive trigonometric polynomials. However, the
disk algebra is not a logmodular subalgebra of $C(\bb D^-).$ More
generally, $H^\infty(\bb{D})$ is (strongly) logmodular when viewed as
a subalgebra of $L^\infty(\bb T)$.

An algebra $\cal{A}\subseteq C(X)$ is called \textit{Dirichlet} if and
only if $\Re(\cal{A})$ is uniformly dense in $C_\bb{R}(X)$. This is
equivalent to $\cal{A}+\overline{\cal{A}}$ being uniformly dense in
$C(X)$. Every Dirichlet algebra $\cal{A}$ is logmodular on
$C(X)$. However, the algebra $H^\infty$ is not Dirichlet. We refer the
reader to \cite[Section~3]{KH1} for more about these examples.

If $\bb{A}:=\{z\in\bb{C}\,:\, 0<r \leq |z|\leq 1\}$ is a closed
annulus, then the algebra of functions that are continuous on
$\partial \bb{A}$ and analytic on the interior of $\bb{A}$ provides an
example of an algebra that is not logmodular. However, a deep result
of Agler shows that every representation of this algebra is completely
contractive~\cite{agler}.

We now give some examples of non-commutative logmodular algebras. The
Cholesky decomposition shows that the algebra of upper triangular
matrices $\cal{T}_n$ is logmodular in $M_n$ and more generally, nest
subalgebras for countable nests are logmodular subalgebras of $B(\cal
H)$ since they have factorization by ~\cite{Ar}.

We now prove that among a certain family of subalgebras of matrix
algebras, the upper triangular matrices are essentially the only
logmodular algebras.  We let $M_n$ denote the $n \times n$ matrices
and let $\cal D_n \subseteq M_n$ denote the subalgebra of diagonal
matrices. Every subalgebra $\cal A, \cal D_n \subseteq \cal A
\subseteq M_n$ is known to be the algebra of matrices left invariant
by a {\em commutative lattice of subspaces.} Moreover, every
subalgebra of $M_n$ that is left invariant by a commuting lattice of
subspaces can be seen to be unitarily equivalent to such an
algebra. Such algebras are called {\em CSL algebras.} For further
details on CSL algebras see ~\cite{Da}. While the results below do not
need this characterization, the characterization serves to put the
result in perspective.

We call $\cal A\subseteq M_n$ an algebra of \textit{block upper triangular
matrices} if there are positive integers $n_1,\ldots,n_k$ such that
$n=n_1+\ldots+n_k$, and $\cal{A}$ is the set of matrices of the form
$(A_{i,j})$, where $A_{i,j}\in M_{n_i,n_j}$ and $A_{i,j}=0$ whenever
$i>j$.

\begin{thm} Let $\cal D_n \subseteq \cal A \subseteq M_n.$ Then $\cal
  A$ is a logmodular subalgebra of $M_n$ if and only if after a
  permutation of the basis, $\cal A$ is an algebra of block upper
  triangular matrices.
\end{thm}
\begin{proof} If $\cal A$ is an algebra of block upper triangular
  matrices, then it contains the algebra of upper triangular matrices
  and hence is logmodular.

  Conversely, assume that $\cal A$ is logmodular. Since $\cal D_n
  \subseteq \cal A,$ we easily see that $\cal A$ is the span of the
  matrix units, $\{ E_{i,j} \}$ that belong to $\cal A.$

  By a compactness argument, it follows that every positive, $P \in
  M_n,$ factors as $P=A^*A,$ with $A \in \cal A.$ Let $J$ denote the
  rank one positive matrix with all entries equal to 1 and write
  $J=A^*A$ with $A \in \cal A.$ If we let $R_1,...,R_n$ denote the
  rows of $A$, then $J= R_1^*R_1 + ... + R_n^\ast R_n$ and each
  $R_i^*R_i$ is a rank one positive. From this it follows that for
  each $i,$ either $R_i^*R_i=0$ or $R_i^*R_i$ is a positive multiple
  of $J$.

  Hence, we have a row $k$ such that $R_k^*R_k$ is a positive multiple
  of $J$ and so each entry of this row must be non-zero. Thus, $\{
  E_{k,j}: j=1,...,n \} \subseteq \cal A.$ Re-numbering, we have that
  $\{ E_{1,j}: j=1,...,n \} \subseteq \cal A.$ Now let $S= \{ i :
  E_{i,1} \in \cal A \}.$ For $i \in S$ and any $j$, we have that
  $E_{i,j} = E_{i,1}E_{1,j} \in \cal A.$ If $i \notin S, j \in S,$
  then $E_{i,j} \notin \cal A,$ or else, $E_{i,1} = E_{i,j}E_{j,1} \in
  \cal A.$

  Thus, re-numbering so that $S= \{1,...,m \},$ we have that $E_{i,j}
  \in \cal A,$ for all $1 \le i,j \le m,$ while $E_{i,j} \notin \cal
  A,$ whenever, $m < i$ and $1 \le j \le m,$ that is, $\cal A$
  contains the entire upper left hand block.  Relative to this
  decomposition, setting $p=n-m,$ we have that $M_n = \begin{bmatrix}
    M_m & M_{m,p} \\ M_{p,m} & M_p \end{bmatrix}$ and
  \[\cal A = \left\{ \begin{bmatrix} A & B\\ 0 & C \end{bmatrix} :
    A \in M_m, B \in M_{m,p}, C \in \cal A_1 \right\},\] where $\cal A_1
  \subseteq M_p$ is a subalgebra that contains the diagonal
  subalgebra, $\cal D_p.$

Now if we let $P= \begin{bmatrix} I_m &0 \\0 & Q \end{bmatrix}$ where $Q \in M_p$ is positive, then factoring, $P= A^*A,$ with $A= \begin{bmatrix} X & Y\\ 0 & Z \end{bmatrix} \in \cal A,$ we find that $I_m= X^*X, 0 = X^*Y$ and $Q= Y^*Y+Z^*Z.$ But the first equation implies that $X$ must be a unitary, and hence, $Y=0,$ so that $Q= Z^*Z,$ with $Z \in \cal A_1.$

Thus, $\cal A_1$ is a logmodular subalgebra of $M_p$ and we are done by induction.

\end{proof}

\begin{cor} Let $\cal D_n \subseteq \cal A \subseteq M_n.$ If $\cal A$ is a logmodular subalgebra of $M_n,$ then every contractive representation of $\cal A$ is completely contractive.
\end{cor}
\begin{proof}  In ~\cite{PP} it is proven that every contractive representation of an algebra of block upper triangular matrices is completely contractive.
\end{proof}

We do not know if every logmodular subalgebra of $M_n$ is of this form, but we believe that this is the case.
This result makes it natural to conjecture that every logmodular completely distributive CSL algebra is a nest algebra.

In ~\cite[Proposition~4.3]{BL} it is shown that if $\cal A$ is
logmodular on $ \cal
B,$ then $C^*_e(\cal A)= \cal B,$ where $C^*_e(\cal A)$ denotes Hamana's
boundary algebra, thus $\cal B$ is already the smallest $C^\ast$-algebra that
any completely isometric representation of $\cal A$ can generate.

\section{Noncommutative Analogues of Some Classic Results}

In this section we prove analogues of a classic factorization result for 2-summing maps and obtain a noncommutative analogue of Mlak's~\cite{Ml} theory of semispectral measures.

We first turn our attention to $2$-summing maps and their relation to completely bounded maps. Given a Hilbert space $\cal H,$ we form two operator spaces, {\em row Hilbert space,} $\cal H_r = B(\cal H, \bb C)$ and {\em column Hilbert space,} $\cal H_c = B(\bb C, \cal H).$ For more details on the structure of these operator spaces, see~\cite{Pa}.

Let $X$ be a compact Hausdorff space and let $\cal{V}$ be a subspace of $C(X)$. A map $\psi:\cal V \to \cal{H}$ is called 2-summing if there exists a constant $c$ such that
\[\sum_{j=1}^n \norm{\psi(f_j)}^2_{\cal H} \leq c^2 \norm{\sum_{j=1}^n \mod{f_j}^2}_{\infty}\]
for all $n\geq 1$ and $f_1,\ldots,f_n\in\cal{A}$. The least such constant $c$ is denoted $a_2(\psi)$.

The following result follows from \cite[Proposition~5.11]{Pis}(see
also \cite[Theorem~5.7]{ER}) and the
fact that, $a_2(\psi) = \pi_{2,c}(\psi) = \pi_{2,r}(\psi),$ but we
provide a direct argument.

\begin{lemma}
Let $\cal V \subseteq C(X)$ be a subspace, let $\cal H$ be a Hilbert
space and let $\psi:\cal V \to \cal{H}$ be a linear map. Then the
following are equivalent:
\begin{enumerate}
\item $\psi : \cal V \to \cal H$ is 2-summing,
\item $\psi: \cal V \to \cal H_r$ is completely bounded,
\item $\psi: \cal V \to \cal H_c$ is completely bounded.
\end{enumerate}
Moreover, in this case the 2-summing norm, $a_2(\psi)$ and the two
completely bounded norms are equal.
\end{lemma}

\begin{proof} We only prove the equivalence of (1) and (2) and the
  equality of the 2-summing norm with the completely bounded norm into
  $\cal H_r.$ The proof of the equivalence of (1) and (3) and the other
  equality is identical.

To this end,
let $(h_{i,j})\in M_n(\cal{H}_r)$ and note that
\begin{align*}
\norm{(h_{i,j})}^2&=\norm{(h_{i,j})(h_{i,j})^\ast}=\norm{(\sum_{k=1}^n\inp{h_{i,k}}{h_{j,k}})}\\
&=\sup \left\{\mod{\sum_{i,j,k=1}^n \inp{\lambda_i h_{i,k}}{\mu_j h_{j,k}}}\,:\, \sum_{i=1}^n \mod{\lambda_i}^2=\sum_{j=1}^n \mod{\mu_j}^2=1\right\}
\end{align*}
Now,
\begin{multline}\label{mat1}
\norm{(\psi(f_{i,j}))}^2=\\\sup \left\{\mod{\sum_{i,j,k=1}^n \inp{\lambda_i \psi(f_{i,k})}{\mu_j \psi(f_{j,k})}}\,:\, \sum_{i=1}^n \mod{\lambda_i}^2=\sum_{j=1}^n \mod{\mu_j}^2=1\right\}
\end{multline}
Let $g_k=\sum_{i=1}^n \lambda_i f_{i,k}$ and $h_k=\sum_{j=1}^n \mu_jf_{j,k}$. We have that the quantity in \eqref{mat1} is smaller than
\begin{align*}
\sum_{k=1}^n \mod{\inp{\psi(g_k)}{\psi(h_k)}}&\leq (\sum_{k=1}^n \norm{\psi(g_k)}^2)^{1/2}(\sum_{k=1}^n \norm{\psi(h_k)}^2)^{1/2}\\
& \leq  a_2(\psi)^2\norm{\sum_{i=1}^n \mod{g_k}^2}^{1/2}\norm{\sum_{j=1}^n \mod{h_k}^2}^{1/2}
\end{align*}

Now,
$(g_1,\ldots, g_n)=(\lambda_1,\ldots,\lambda_n) (f_{i,j})$ implies
 \begin{align*}
\sum_{k=1}^n \mod{g_k}^2 &= (g_1,\ldots,g_n)(g_1,\ldots,g_n)^\ast \\
&=(\lambda_1,\ldots,\lambda_n)(f_{i,j})(f_{i,j})^{\ast} (\lambda_1,\ldots,\lambda_n)^\ast\\
& \leq (\lambda_1,\ldots,\lambda_n) \norm{(f_{i,j})}^2 (\lambda_1,\ldots,\lambda_n)^\ast  \leq \norm{(f_{i,j})}^2.
\end{align*}
and similarly we can show $\sum_{k=1}^n \mod{h_k}^2\leq \norm{(f_{i,j})}^2$.

Combining the above inequalities, we get,
$$\norm{(\psi(f_{i,j}))}^2\leq a_2(\psi)^2 \norm{(f_{i,j})}^2.$$
Hence, $\norm{\psi}_{cb} \le a_2(\psi),$ and so every 2-summing map is
completely bounded into the row Hilbert space.

Conversely, if $\psi$ is completely bounded into the row Hilbert
space, then
we have,
$$\sum_{i=1}^n \norm{\psi(f_i)}^2=\norm{(\psi(f_1),\ldots,\psi(f_n))}^2\leq \norm{\psi}^2_{cb}.$$
and so $\norm{\psi}_{cb} \ge a_2(\psi)$. Thus, the map $\psi$ is 2-summing
and the norms agree.
\end{proof}

The result of~\cite{Pi, LP} on the existence of dominating measures is essential to the proof of
Foias-Suciu~\cite{FS}. We now present a non-commutative version of this
result from which the result of~\cite{Pi, LP} follows readily. Note
that if $v \in \cal H_r = B(\cal H, \bb C),$ then $v^*v \in B(\cal H)$
is a rank one positive operator. While if $v \in \cal H_c,$ then $vv^*
\in B(\cal H)$ is a rank one positive operator. The existence of the
state in the following result is related to \cite[Proposition~5.1]{Pis}.

\begin{thm}\label{dominatingstate} Let $\cal B$ be a unital $C^\ast$-algebra, let $\cal X
  \subseteq \cal B$ be an operator subspace and let $\cal H$ be a
  Hilbert space. If $\psi: \cal X \to \cal H_r= B(\cal H, \bb C)$ (respectively, $\cal H_c = B(\bb C, \cal H)$)  is completely
  contractive, then there exists a state, $s: \cal B \to \bb C$ and a
  unital completely positive map $\Phi: \cal B \to B(\cal H)$ such
  that for every $x \in \cal X,$ we have that
$\|\psi(x)\|^2 \le s(xx^*)$ and $\psi(x)^*\psi(x) \le \Phi(x^*x)$(respectively, $\|\psi(x)\|^2 \le s(x^*x)$ and $\psi(x)\psi(x)^* \le \Phi(xx^*)$).
\end{thm}

\begin{proof}
Writing an arbitrary operator $T \in B(\bb C \oplus \cal H)$ as a $2 \times 2$ operator
matrix, we have that $T = \begin{bmatrix} \lambda & r \\ c &
  B \end{bmatrix}$ where $\lambda \in \bb C, r \in \cal H_r, c \in
\cal H_c$ and $B \in B(\cal H).$ That is we identify
\[B(\bb C \oplus \cal H) = \begin{bmatrix} \bb C & \cal H_r \\ \cal H_c & B(\cal H)\end{bmatrix}.\]
Let
\[\cal{S}_\cal{X}:=\left\{\begin{bmatrix}\lambda 1 & x \\ y^* & \mu 1 \end{bmatrix}\, :\,
  \lambda, \mu\in \bb{C}, x,y \in \cal{X} \right\} \subseteq M_2(\cal
B),\]
where $1$ denotes the unit of $\cal B.$
Since $\psi$ is completely contractive, the map $\Psi:
\cal{S}_{\cal{X}} \to B(\bb{C} \oplus \cal{H})$ given by
$\Psi\left(\begin{bmatrix}\lambda & x \\ y^* & \mu\end{bmatrix}\right) =
\begin{bmatrix}\lambda & \psi(x) \\ \psi(y)^* & \mu I_{\cal H}\end{bmatrix}$ is
completely positive by ~\cite[Lemma~8.1]{Pa}. Thus, by Arveson's
extension theorem, we may extend $\Psi$ to a completely positive map
on $M_2(\cal B),$
which we still denote by $\Psi.$ Thus,
$\Psi:M_2(\cal B)\to B(\bb{C}\oplus \cal{H})$. By
~\cite[Theorem~8.3]{Pa}, there exists two completely positive maps
$\phi_1:\cal B \to \bb{C}$ and $\Phi_2:\cal B \to B(\cal{H})$ and a
completely bounded map $u$ such that
\[\Psi\left(\begin{bmatrix}b_{1,1} & b_{1,2} \\ b_{2,1} & b_{2,2}\end{bmatrix} \right)=\begin{bmatrix}\phi_1(b_{1,1}) &
  u(b_{1,2}) \\ u(b_{2,1}^*)^\ast & \Phi_2(b_{2,2})\end{bmatrix}.\]

 For any $x \in \cal{X}$ the matrix $\begin{bmatrix}xx^* & x \\ x^* & 1\end{bmatrix}\geq 0$
which implies that $\begin{bmatrix}\phi_1(xx^*) & \psi(x) \\ \psi(x)^\ast & I\end{bmatrix}$ is positive which yields
$$\norm{\psi(x)}^2\leq \phi_1(xx^*).$$ Since $\phi_1(1)=1,$ we have that $s= \phi_1$ is the desired state.

On the other hand, $\begin{bmatrix}1 & x \\x^* & x^*x\end{bmatrix} \ge 0,$ which implies that
$\begin{bmatrix}1 & \psi(x) \\ \psi(x)^* & \Phi_2(x^*x)\end{bmatrix}$ is positive, from which it follows that $\psi(x)^*\psi(x) \le \Phi_2(x^*x),$ and $\Phi = \Phi_2$ is the desired unital completely positive map.

For the case where $\psi:X \to \cal H_c$ is completely contractive, we define
\[\cal S := \left\{ \begin{bmatrix}\lambda & y^* \\ x & \mu I\end{bmatrix}\, :\, \lambda, \mu \in \bb C, x,y \in \cal X \right\}\]
and $\Psi: \cal S \to B(\bb C \oplus \cal H)$ via $\Psi \left(\begin{bmatrix} \lambda & y^* \\ x & \mu I\end{bmatrix} \right) = \begin{bmatrix}\lambda & \psi(y)^* \\ \psi(x) & \mu I\end{bmatrix}$, and argue as above.
\end{proof}

\begin{cor}[Lindenstrauss-Pelczy\'{n}ski,Pietsch] Let $\cal V
  \subseteq C(X)$ and let $\psi: \cal V \to \cal H$ with $\cal H$ a
  Hilbert space. If $\psi$ is $2$-summing, then there exists a
  positive measure $\mu$ on $X$ with $\mu(X)=1$ such that
  $\norm{\psi(f)}^2\leq  a_2(\psi)^2 \int_X \mod{f}^2 \,d\mu$.
\end{cor}

We now turn our attention to obtaining $C^\ast$-algebraic generalizations of Mlak's theory of semispectral measures~\cite{Ml}.
We shall need the following folklore result about quadratic forms
on Hilbert spaces.

\begin{prop}\label{quadratic} Let $\cal H$ be a complex Hilbert space
  and let $\phi: \cal H \to \bb C$ be a bounded quadratic form, that is:
\begin{enumerate}
\item $\phi(\lambda
  h) = |\lambda|^2 \phi(h),$ for every $\lambda \in \bb C, h \in \cal H,$
\item there exists a constant $C>0,$ with $|\phi(h)| \le C \|h\|^2,$
\item $\phi(h+k) + \phi(h-k) = 2 \phi(h) + 2\phi(k).$
\end{enumerate}
Then there exists a bounded operator, $T \in B(\cal H),$ such
that $\phi(h) = \langle Th,h \rangle,$ for every $h \in \cal H$ and
$\|T\| \le 2C$ and such a $T$ is unique.
\end{prop}
\begin{proof} This result is ~\cite[Exercise~4.18]{Pa}, so we only
  sketch the proof. One defines a two variable function, $\psi: \cal H
  \times \cal H \to \bb C$ by polarization and
  then copies the steps of the simplified proof of the Jordan-von~Neumann theorem
  to verify that $\psi$ is a bounded sesquilinear map.
\end{proof}

Note that if we are given a unital $C^\ast$-algebra $\cal B,$ a Hilbert space $\cal H$ and a positive map $\Phi: \cal B \to B(\cal H),$ then for each vector $h \in \cal H,$ we obtain a positive functional $\gamma(h)$ on $\cal B$ by setting $\gamma(h)(b) = \langle \Phi(b)h,h \rangle.$ The following result characterizes such maps.
We let $\cal B^{\dagger}_+$ denote the set of positive linear functionals on $\cal B.$

\begin{thm}\label{semispectral2} Let $\cal B$ be a $C^\ast$-algebra, let $\cal H$ be a Hilbert space and let $\gamma: \cal H \to \cal B^{\dagger}_+$ be a map. Then there exists a positive linear map, $\Phi: \cal B \to B(\cal H)$ such that $\gamma(h)(b) = \langle \Phi(b)h,h \rangle$ for every $b \in \cal B$ and every $h \in \cal H$ if and only if
\begin{enumerate}
\item $\gamma( \lambda h) = |\lambda|^2 \gamma(h)$ for every $\lambda \in \bb C$ and every $h \in \cal H,$
\item there is a constant $C$ such that $\|\gamma(h)\| \le C \|h\|^2,$ for every $h \in \cal H,$
\item $\gamma(h+k) + \gamma(h-k) = 2 \gamma(h) + 2 \gamma(k),$ for every $h,k \in \cal H.$
\end{enumerate}
Moreover, if $\cal B$ is unital, then $\Phi$ is unital if and only if $\gamma(h)$ is a state for every $\|h\|=1.$
\end{thm}

\begin{proof} For each fixed $b \in \cal B,$ we see that $h \to \gamma(h)(b)$ is a bounded quadratic form. Hence, there exists an operator $\Phi(b) \in B(\cal H)$ such that $\gamma(h)(b) = \langle \Phi(b)h,h \rangle .$  The linearity and positivity of the map $b \to \Phi(b)$ now follows from the fact that each $\gamma(h)$ is linear and positive.

Finally, $\Phi(I)=I$ if and only if $1 = \langle \Phi(I)h,h \rangle = \gamma(h)(I),$ for every unit vector $h$.
\end{proof}

\begin{cor}[Mlak's Dilation Theorem for Semispectral Measures] Let $X$ be a locally compact Hausdorff space, let $\cal B(X)$ be the $\sigma$-algebra of Borel measurable subsets of $X$ and let $\cal H$ be a Hilbert space and assume that for each $h \in \cal H$ we are given a positive, regular Borel measure $\mu_h$ on $X,$ satisfying:
\begin{enumerate}
\item $\mu_{\lambda h} = |\lambda|^2 \mu_h,$ for every $\lambda \in \bb C$ and every $h \in \cal H,$
\item there is a constant $C$ such that $\mu_h(X) \le C \|h\|^2,$ for every $h \in \cal H,$
\item $\mu_{h+k} + \mu_{h-k} = 2 \mu_{h} + 2 \mu_k,$ for every $h,k \in \cal H.$
\end{enumerate}
Then there exists a Hilbert space $\cal K$, a linear map $V: \cal H \to \cal K$ and a regular Borel projection-valued measure, $E: \cal B(X) \to B(\cal K)$ such that for every Borel set, $B \subseteq X,$ and every vector $h \in \cal H,$ we have $\mu_h(B) = \langle E(B)Vh, Vh \rangle.$
\end{cor}
\begin{proof} By Theorem~\ref{semispectral2} there exists a positive map $\Phi: C_0(X) \to B(\cal H),$ with $\langle \Phi(f) h,h \rangle = \int_X f d \mu_h ,$ for every $h \in \cal H.$ By Stinespring's theorem ~\cite{St}, this positive map is completely positive. Therefore the map $\Phi$ dilates, that is, there exists a $\ast$-homomorphism $\pi: C_0(X) \to B( \cal K)$ and a map $V: \cal H \to \cal K$ such that $\Phi(f) = V^*\pi(f)V.$
The result now follows by letting $E$ be the regular Borel projection-valued measure such that $\pi(f) = \int_X f dE.$
\end{proof}

It would be interesting to find a similar characterization of maps $\gamma$ of the form $\gamma(h)(b) = \langle \Phi(b)h,h \rangle$ such that $\Phi: \cal B \to B(\cal H)$ is a {\em completely} positive map that did not reduce to a simple tautology.

\section{Representations of Logmodular Algebras}

In this section we obtain our principal results on representations.

Given operator spaces $X$ and $Y$ and a linear map $\psi:X \to Y,$ we call $\psi$ {\em $R_n$-contractive} provided that
\[\|(\psi(x_1), \ldots, \psi(x_n))\| \le \|(x_1, \ldots, x_n)\|,\]
for every $x_1, \ldots, x_n \in X$ and {\em row contractive} if it is $R_n$-contractive for every $n$. We define {\em $C_n$-contractive} and {\em column contractive}, analogously.

\begin{lemma}\label{colimpliescc}
Let $X$ be an operator space and $\psi:X\to \cal{H}_c$ (respectively, $\cal H_r$) be a linear map. If $\psi$ is column contractive (respectively, row contractive), then $\psi$ is completely contractive.
\end{lemma}

\begin{proof}
Begin by noting that we may identify $M_{m,n}(\cal{H}_c)$ with $B(\bb{C}^n,\cal{H}^{(m)})$. Let $(x_{i,j})\in M_{m,n}(X)$ and let $\lambda=\begin{bmatrix}\lambda_1 \\ \vdots \\\lambda_n\end{bmatrix}\in \bb{C}^n$, with $\sum_{j=1}^n \mod{\lambda_j}^2=1$. Write $(x_{i,j})=[C_1,\ldots,C_n]$, $C_j\in M_{m,1}(X)$.  We have,
\[\norm{(\psi(x_{i,j}))\lambda}=\norm{\psi_{m,1}\left(\sum_{j=1}^n \lambda_j C_j\right)}\leq \norm{\sum_{j=1}^n \lambda_j C_j}=\norm{(x_{i,j})\lambda}\leq \norm{(x_{i,j})}. \]
Hence, $\psi$ is completely bounded and $\norm{\psi}_{cb}=\norm{\psi}_{col}$.

The row case is analogous.
\end{proof}

\begin{thm}\label{tworowimpliescc}
Let $\cal B$ be a $C^\ast$-algebra, let $\cal A \subseteq \cal B$ be logmodular and let $\rho:\cal{A}\to B(\cal{H})$ be a $R_2$-contractive representation, then $\rho$ is row contractive. If $\rho$ is $C_2$-contractive, then $\rho$ is column contractive.
\end{thm}

\begin{proof}
Let $a_1,\ldots,a_n \in \cal{A}$ with $\sum_{j=1}^n a_ja_j^\ast \leq 1$. We will prove our result by induction on $n$. The case $n=2$ is our hypothesis. Now let $\epsilon >0$ and choose $b\in \cal{A}^{-1}$ such that $\sum_{j=1}^{n-1} a_ja_j^\ast < bb^\ast <1+\epsilon -a_na_n^\ast$. This choice being possible since $\cal{A}$ is logmodular. Set $b_j=b^{-1} a_j$ and note that $\sum_{j=1}^{n-1} b_jb_j^\ast=\sum_{j=1}^{n-1} b^{-1}(\sum_{j=1}^{n-1} a_ja_j^\ast)b^{\ast -1}\leq 1$, hence the induction hypothesis yields $\sum_{j=1}^{n-1}\rho(b_j)^\ast \rho(b_j) \leq 1$ or equivalently
$$\sum_{j=1}^{n-1} \rho(b^{-1}a_j) \rho(b^{-1}a_j)^\ast =\rho(b)^{-1} (\sum_{j=1}^{n-1} \rho(a_j)\rho(a_j)^\ast) \rho(b)^{\ast -1}\leq 1.$$

Applying the $R_2$ condition again we get
$$\sum_{j=1}^n \rho(a_j)\rho(a_j)^\ast \leq \rho(b)\rho(b)^\ast +\rho(a_n)\rho(a_n)^\ast \leq 1+\epsilon.$$
Since this is true for all $\epsilon$ we have our result.

The case of $C_2$-contractive is similar.
\end{proof}

\begin{prop}\label{unique} Let $\cal{A} \subseteq \cal{B}$ be logmodular, let $\rho: \cal{A} \to B(\cal H)$ be a unital homomorphism and let $h \in \cal H$ be a unit vector. If $\phi, \psi: \cal B \to \bb C$ are states such that $\|\rho(a)h \|^2 \le \phi(a^*a)$ and $\|\rho(a)^*h\|^2 \le \psi(aa^*)$ for all $a \in \cal A^{-1},$ then $\phi = \psi.$
\end{prop}

\begin{proof}
For any $a \in \cal A^{-1},$ we have
\begin{align*}
1&=\norm{h}^4=\inp{h}{h}^2=\inp{\rho(a)^{-1}\rho(a) h}{h}^2\\
&= \inp{\rho(a)h}{\rho(a^{-1})^\ast h}^2\leq \norm{\rho(a)h}^2\norm{\rho(a^{-1})^\ast h}^2\\
&\leq  \phi(a^\ast a)\psi(a^{-1}(a^\ast)^{-1})=\phi(a^\ast a)\psi((a^\ast a)^{-1}).
\end{align*}
For $x\in \cal{B}_{sa}$ and $t\in \bb{R}$, $e^{tx}$ is a positive invertible element of $\cal{B}$ and so the above inequality gives
$$1\leq \phi(e^{tx})\psi(e^{-tx}).$$
Let $u(t)=\phi(e^{tx})\psi(e^{-tx})$ and note that this is a differentiable function with a minimum at $t=0$. Taking the derivative and setting $t=0$ we get
\[0=u'(0)=\phi(xe^{tx})-\psi(xe^{-tx})|_{t=0}=\phi(x)-\psi(x).\] Since this holds for all selfadjoint elements in $\cal{B}$ and $\phi,\psi$ are selfadjoint maps we get $\phi=\psi$.
\end{proof}

Note that the above result does not guarantee that such a state $\phi$
exists, nor does it guarantee that if $\phi$ exists satisfying,
$\|\rho(a)h \|^2 \le \phi(a^*a)$, then $\|\rho(a)^*\| \le \phi(aa^*).$

\begin{prop}\label{dominatingrep}
Let $\cal{A}\subseteq \cal{B}$ be logmodular and let $\rho:\cal{A}\to
B(\cal{H}) $ be a representation which is $R_2$-contractive and $C_2$-
contractive. If $h\in \cal{H}$ is fixed, then there exists a
positive linear functional $\phi:\cal{B}\to \bb{C}$ such that
$\phi(1)= \|h\|^2$, $\norm{\rho(a)h}^2\leq \phi(a^\ast a)$, $\norm{\rho(a)^\ast h}^2\leq \phi(aa^\ast)$ and $\phi(a)=\inp{\rho(a)h}{h}$ and such a $\phi$ is unique.
\end{prop}

\begin{proof} It will be enough to consider the case where $\|h\|=1$,
  and prove that there exists a state $\phi$ as above, since uniqueness will follow by Proposition~\ref{unique}.
Since the map $a\mapsto \rho(a)$ is $C_2$-contractive, it follows by Theorem~\ref{tworowimpliescc} that $a\mapsto \rho(a)$ is column contractive. Hence, by Lemma~\ref{colimpliescc} the map $a\mapsto \rho(a)h$ is completely contractive as a map from $\cal{A}$ into the column Hilbert space $\cal{H}_c$. By Theorem~\ref{dominatingstate} there exists a state $\phi_1$ such that
$\norm{\rho(a)h}^2\leq \phi_1(a^\ast a).$
Moreover, since $a \mapsto \rho(a)$ is $R_2$-contractive, the map $a^\ast \mapsto \rho(a)^\ast h$ is completely contractive as a map from $\cal{A}^\ast$ to the column Hilbert space $\cal{H}_c$. Applying Theorem~\ref{dominatingstate} again yields a state such that
$\norm{\rho(a)^\ast h}^2\leq \phi_2(aa^\ast)$. From Proposition~\ref{unique} we get that $\phi_1=\phi_2$.

Let $\lambda$ be a complex scalar and consider $\norm{\rho(a+\lambda)h}^2\leq \phi((a+\lambda)^\ast (a+\lambda))$. Expanding both sides and rearranging we get
\[2\Re (\lambda(\inp{\rho(a)h}{h}-\phi(a)))\leq \phi(a^\ast a)-\norm{\rho(a)h}^2.\]
This quantity being positive for all $\lambda\in \bb{C}$ implies that $\inp{\rho(a)h}{h}=\phi(a)$.
 \end{proof}

\begin{lemma}\label{semispectralstate} Let $\cal A \subseteq \cal B$ be logmodular and let
  $\rho: \cal A \to B(\cal H)$ be a representation that is $R_2$-contractive and
  $C_2$-contractive and for each $h \in \cal H$ let $\phi_h$
  denote the unique positive linear functional as obtained in Proposition~\ref{dominatingrep}, then
  for any vectors, $h,k \in \cal H$, $\phi_{h+k} + \phi_{h-k} = 2 \phi_h
  + 2 \phi_k.$
\end{lemma}
\begin{proof}  The proof is identical to the one outlined in~\cite[Lemma~3]{FS}. For any $a \in \left(\cal A\right)^{-1}$ we have
\begin{align*}
&2(\|h\|^2 + \|k\|^2) \\
=& \|h+k\|^2 + \|h-k\|^2 \\
=& \inp{\rho(a)(h+k)}{\rho(a^{-1})^\ast(h+k)}+\inp{\rho(a)(h-k)}{\rho(a^{-1})^\ast (h-k)} \\
\le & \left(\|\rho(a)(h+k)\|^2 + \|\rho(a)(h-k)\|^2\right)^{1/2} \\
&\times \left(\|\rho(a^{-1})^*(h+k)\|^2 + \|\rho(a^{-1})^*(h-k)\|^2\right)^{1/2} \\
\le & \left( \phi_{h+k}(a^*a) + \phi_{h-k}(a^*a)\right)^{1/2}\left(2\phi_h((a^*a)^{-1}) + 2\phi_k((a^*a)^{-1})\right)^{1/2}.
\end{align*}
Thus, for any $x=x^* \in \cal B,$ and $t \in \bb R,$ we have that
$$4(\|h\|^2 + \|k\|^2)^2 \le [ \phi_{h+k}(e^{tx}) +
\phi_{h-k}(e^{tx})][2\phi_h(e^{-tx}) + 2\phi_k(e^{-tx})].$$
Since $t=0$ is the minimum of the function on the right hand side,
differentiating and evaluating at $t=0$, yields
\begin{multline*}
0=\left(\phi_{h+k}(x)+ \phi_{h-k}(x)\right)\left(2 \|h\|^2 + 2 \|k\|^2\right) \\- \left(\|h+k\|^2
+ \|h-k\|^2\right)\left(2\phi_h(x)+ 2\phi_k(x)\right),
\end{multline*}
from which the result follows.
\end{proof}

We can now state and prove our key result.

\begin{thm}\label{uniqueextension} Let $\cal A \subseteq \cal B$ be logmodular. If $\rho:
  \cal A \to B(\cal H)$ is a unital homomorphism which is $R_2$-contractive
  and $C_2$-contractive, then there exists a positive map,
  $\Phi: \cal B \to B(\cal H),$ extending $\rho$ and satisfying,
  $\Phi(a^*a) \ge \rho(a)^*\rho(a),$ $\Phi(aa^*) \ge
  \rho(a)\rho(a)^*$, for every $a \in \cal A$. Such a map is unique.
Moreover, if $\Psi: \cal B \to B(\cal H)$ is any completely positive
map extending $\rho,$ then $\Psi = \Phi.$
\end{thm}
\begin{proof}
Let $\phi_h: \cal B \to \bb C, h \in \cal H,$ be the family of
positive functionals guaranteed by Proposition~\ref{dominatingrep}.
If we fix $b \in \cal B,$ then by Lemma~\ref{semispectralstate} the map $h \mapsto \phi_h(b),$ is easily
seen to be a bounded quadratic form on $\cal H$ and hence there exists
a unique bounded operator, $\Phi(b)$ with $\inp{\Phi(b)h}{h}= \phi_h(b).$
The map $b \mapsto \Phi(b)$ is easily checked to be linear, positive and
to satisfy the other conclusions stated in the theorem.

If $\Psi: \cal B \to B(\cal H)$ is any positive map
satisfying the above conclusions, then for each $h \in \cal H,$
the positive linear maps, $\psi_h(b) = \inp{\Psi(b)h}{h}$
satisfy the conclusions of Proposition~\ref{dominatingrep}, and hence
$\psi_h(b)=\phi_h(b),$ for every $b$ and every $h$.

Finally, if $\Psi$ is a completely positive map extending $\rho$, then
by the Cauchy-Schwarz inequality for 2-positive maps, it will
satisfy $\Psi(a^*a) \ge \rho(a)^*\rho(a)$ and $\Psi(aa^*) \ge
\rho(a)\rho(a)^*,$ and so we will have $\Psi =\Phi,$ by the uniqueness
result.
\end{proof}

Thus, a completely positive extension of $\rho$ exists if and only if
the positive extension that we have constructed is completely
positive. We now re-capture the main result of Foias-Suciu~\cite{FS}.

\begin{cor} Let $X$ be a compact, Hausdorff space, let $\cal A
  \subseteq C(X)$ be logmodular, let $\rho: \cal A \to B(\cal H)$ be a
  unital homomorphism that is  $R_2$-contractive and $C_2$-contractive. Then there exists a Hilbert space $\cal K,$ an isometry
  $V: \cal H \to \cal K,$ and a unital $\ast$-homomorphism $\pi:C(X) \to
  B(\cal K)$ such that $\rho(a) = V^*\pi(a)V,$ for every $a \in \cal
  A.$
If the span of $\pi(C(X))V\cal H$ is dense in $\cal K,$ then this
  representation is unique up to unitary equivalence.
\end{cor}
\begin{proof} By Theorem~\ref{uniqueextension}, there
  exists a positive map, $\Phi: C(X) \to B(\cal H).$ By Stinespring's
  theorem~\cite{St} (see also~\cite{Pa}) such a map is automatically
  completely positive and hence by Stinespring's dilation
  theorem~\cite{St} (see also~\cite{Pa}) the map $\Phi$ has
  a unique dilation of the above type.

Conversely, given any dilation of the above type, if we set $\Phi(b) =
V^*\pi(b)V,$ then $\Phi$ is a positive map satisfying the above
conditions and hence is unique.
\end{proof}

\section{Matrix Factorization in Logmodular Algebras}

In~\cite[Corollary~18.11]{Pa}, a necessary and sufficient condition is
given for an operator algebra to have the property that every
contractive representation is completely contractive in terms of a
certain type of factorization. In particular, it is proven that a
unital operator algebra $\cal A$ has the property that every
contractive representation is completely contractive if and only if
for every $n$ and for every $(a_{i,j}) \in M_n(\cal A)$ with
$\|(a_{i,j})\| < 1,$ there exists some $m,$ scalar matrices, $C_0,
\ldots, C_m$ (of appropriate sizes) all with norm less than one and
diagonal matrices, $D_1, \ldots, D_m$ (of appropriate sizes), whose
diagonal entries are elements of the open unit ball of $\cal A,$ such
that
\[(a_{i,j}) = C_0D_1C_1 \cdots D_mC_m.\]
The term {\em appropriate sizes} means simply that the sizes are such
that the product is defined. Combining this result with the result of
Foias-Suciu, yields some new results about uniform logmodular
algebras.

\begin{thm} Let $\cal A \subseteq C(X)$ be a uniform logmodular
  subalgebra.  Then every contractive representation of $\cal A$ is
  completely contractive if and only if for each $f_1, f_2 \in \cal
  A,$ with $|f_1|^2+|f_2|^2 < 1,$ there exists an $m$ and scalar
  matrices, $C_0, \ldots, C_m$ (of appropriate sizes) all with norm
  less than one and diagonal matrices $D_1, \ldots, D_m$ (of
  appropriate sizes) whose diagonal entries are elements of the open
  unit ball of $\cal A,$ such that,
\[\begin{bmatrix} f_1 \\ f_2 \end{bmatrix} = C_0D_1 C_1 \cdots
D_mC_m.\]
\end{thm}
\begin{proof} If $\rho$ is contractive and $D$ is a diagonal matrix in
  $M_n(\cal A)$, then $\norm{\rho_n(D)}\leq \norm{D}$. If the
  factorization condition is met, then we see, by applying $\rho$ to
  both sides of the above equation, that $\rho$ is
  $C_2$-contractive.

  Note that whenever $\begin{bmatrix} f_1 \\ f_2 \end{bmatrix}$ has a
  factorization as above, then
  \[ \begin{bmatrix} f_1 & f_2 \end{bmatrix} = C_m^tD_m \cdots
  C_1^tD_1C_0^t \]
  where $C^t$ denotes the transpose.  Since $\|C\| = \|C^t\|$ for
  scalar matrices, it follows from the same argument as in the
  previous paragraph that $\rho$ is also $R_2$-contractive. Hence, by
  the result of Foias-Suciu, the representation $\rho$ is completely
  contractive.

  Conversely, if every contractive representation is completely
  contractive, then by the result cited above, every element of the
  unit ball of $M_n(\cal A)$ has such a factorization. Thus, in
  particular, every element of the unit ball of $M_2(\cal A)$ of the
  form $\begin{bmatrix} f_1 & 0\\ f_2 & 0 \end{bmatrix}$ can be
  factorized, from which the factorization in the statement of the
  theorem follows.
\end{proof}

Of course, the columns of size two in the above theorem can be equally
well replaced by rows of size two.

\begin{thm} Let $\cal A \subseteq C(X)$ be a uniform logmodular subalgebra.
Then for every $n$ and every $(f_{i,j})$ in the open unit ball of
$M_n(\cal A),$ there exists an $m,$
 scalar matrices, $C_0,
\ldots, C_m$(of appropriate sizes) all with norm less than one and
block diagonal matrices, $D_1, \ldots, D_m$(of appropriate sizes),  whose
direct summands are elements of the open unit ball of either
$M_{2,1}(\cal A)$ or $M_{1,2}(\cal A),$ such
that
\[(f_{i,j}) = C_0D_1C_1 \cdots D_mC_m.\]
\end{thm}
\begin{proof}  Let $\cal C= \cal A$ as algebras but define a new family of
  matrix norms on $\cal C$ by the above factorization.  That is, we
  declare the open unit ball of $M_n(\cal C)$ to be the set of all matrices
  that can be expressed as products of scalar contractions and
  block diagonal matrices as above of norm less than one. Arguing as in~\cite{BP} or in~\cite[Theorem~18.1]{Pa} one shows that this definition yields a
  family of norms on $M_n(\cal C)$ and that these norms satisfy the
  Blecher-Ruan-Sinclair~\cite{BRS} axioms to form an abstract operator algebra.

The proof will be complete if we can show that the identity map from $\cal A$ to $\cal C$ is a complete isometry.

Clearly, any matrix that is in the unit ball of $M_n(\cal C)$ is also
in the unit ball of $M_n(\cal A).$ This shows that the identity map
from $\cal C$ to $\cal A$ is completely contractive.

Conversely, since we are allowing contractions from $M_{2,1}(\cal A)$
in our definition of the norm, we have that $\norm{\begin{bmatrix}
    f_1\\f_2 \end{bmatrix}}_{M_{2,1}(\cal C)} \le
\norm{\begin{bmatrix} f_1 \\ f_2 \end{bmatrix}}_{M_{2,1}(\cal A)}.$
Thus, the identity map is also an isometry from $M_{2,1}(\cal A)$ to
$M_{2,1}(\cal C),$ that is, the identity map from $\cal A$ to $\cal C$
is $C_2$-contractive. Similarly, this identity map is $R_2$-contractive.

Since $\cal C$ is an abstract operator algebra, by the
representation theorem of~\cite{BRS} (see~\cite[Corollary~16.7]{Pa} for
an alternate proof), it can be represented completely isometrically on a Hilbert space via some map,
$\rho: \cal C \to B(\cal H).$ Regarding, $\rho: \cal A \to B(\cal H),$
we see that $\rho$ is $C_2$-contractive and $R_2$-contractive on the
uniform logmodular algebra $\cal A,$ so by the Foias-Suciu result, it is a completely contractive map.

Thus, the identity map is completely contractive as a map from $\cal
A$ to $\cal C.$ As we saw earlier, the identity map is completely
contractive from $\cal C$ to $\cal A,$ and hence the identity map is a
complete isometry. Thus, the original family of matrix norms on $\cal A$ is equal to the new family.
\end{proof}

The above result can also be used as the basis of an approach to
deciding whether or not the analogue of the Foias-Suciu result is true
for logmodular subalgebras of non-commutative $C^\ast$-algebras. Given a
logmodular subalgebra $\cal A \subseteq \cal B,$ one can form a new
operator algebra $\cal C$ by endowing $\cal A$ with a (possibly) new
operator algebra norm as above, allowing the diagonals to be direct
sums of block diagonal matrices where each block is either a $2 \times
1$ matrix or a $1 \times 2$ matrix over $\cal A.$ The proof of the
above result shows that every $\rho: \cal A \to B(\cal H)$ that is
both $R_2$- and $C_2$-contractive will be completely
contractive if and only if the identity map from $\cal A$ to $\cal C$
is a complete isometry.  Thus, some progress could conceivably be made
by studying the abstract operator algebra $\cal C.$ These
considerations motivate the following problem.

\begin{ques} Let $\cal A \subseteq \cal B$ be a logmodular subalgebra
  of a $C^\ast$-algebra and let $\cal C$ be the abstract operator algebra
  defined in the above paragraph. Is $\cal C$ a logmodular algebra?
\end{ques}

config: arxiv


text:
{\bf Problem.} Find the point where the line passing through $(3,4,1)$ and $(5,1,6)$ intersects the $xy$-plane.
{\bf Level.} Level 3
{\bf Type.} Precalculus
{\bf Solution.} The direction vector the line is $\begin{pmatrix} 5 - 3 \\ 1 - 4 \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix},$ so the line is paramaterized by
\[\begin{pmatrix} 3 \\ 4 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 + 2t \\ 4 - 3t \\ 1 + 5t \end{pmatrix}.\]We want the $z$-coordinate to be 0, so $1 + 5t = 0.$  Then $t = -\frac{1}{5},$ so the point of intersection is $\boxed{\left( \frac{13}{5}, \frac{23}{5}, 0 \right)}.$

set_name: MATH