TITLE: Is there evidence of a larger universe?
QUESTION [15 upvotes]: While mapping out the position, distance and movement of galaxies and quasars at the limits of the observable universe, have astronomers ever observed the motion of these object due to gravitational sources that may reside outside the limits of the observable universe?
If so, would this break the principle of information travelling faster than the speed of light?
If not, are these objects in fact the furthest objects there are?

REPLY [12 votes]: No, it would not violate the principle of special relativity, which proscribes information traveling faster than the speed of light. Let's call the farthest object we can see X. Let's call X's neighbor on the far side from us Y. The only way for us to get information about Y is for information to travel to X (at or below the speed of light), which would then influence X's behavior. Then, information about X's behavior would have to travel to us (at or below the speed of light). So, nowhere is information traveling faster than light. The only information we can get from X about Y is already "out of date"* enough that special relativity isn't violated.
*The technical term relativists use is "retarded," but be careful how you use it in everyday conversation.
EDIT: We can make a distinction between the theoretically observable Universe according to relativistic considerations, which is what I was talking about above, and the Universe observable with present technology and practical limitations. Getting sneaky like inferring about Y from X actually does increase the latter meaning of observable Universe.

REPLY [2 votes]: If it's a real phenomenon the Dark Flow is the motion of the entire observable universe relative to the CMB due to attraction to something that was inflated beyond the horizon.  ~~The validity of the theory is currently disputed and probably not resolvable until better data is made available from the Plank satellite's CMB measurements.~~  Even after the Plank data has become available, different research teams continue to dispute if it is a real phenomenon.  The Plank team itself claims to have invalidated it, but the original proposer of the theory claims to still see it in the Plank data.


TITLE: Definition of "finite group of Lie type"?
QUESTION [41 upvotes]: The list of finite simple groups of Lie type has been understood for half a century, modulo some differences in notation (and identifications between some of the very small groups coming from different Lie types).   Call this collection of isomorphism classes of finite groups $\mathcal{S}$.  But it's not clear to me that there is a similar consensus about the meaning of "finite group of Lie type".  Probably most people would place the finite general linear groups on this list.  Maybe also the Weyl group of the $E_8$ root system, which is related in the Atlas of Finite Groups to the simple group denoted $G=\mathrm{O}_8^+(2)$ as the group $2.G.2$.
The term "finite group of Lie type" comes up frequently in the literature (almost 500 times in a MathSciNet search of titles and reviews).   There are two main directions in which such groups are approached, which might possibly lead to different lists.   Often the starting point is a simple algebraic group over a finite field (with no nontrivial proper closed connected normal subgroups), though $\mathrm{GL}_n$ doesn't quite fit here.
(1) Steinberg's efficient organization in terms of finite fixed point groups under endomorphisms, denoted $G_\sigma$, is now often expressed in terms of "Frobenius morphisms" and their fixed points $G^F$.  Basically the groups of interest then come in three flavors: split (Chevalley) groups, quasi-split (Hertzig, Steinberg, Tits) groups in types $A_n (n \geq 2), D_n, E_6$, or Suzuki/Ree groups in types $B_2, G_2, F_4$ with $p = 2, 3, 2$ as defining characteristic.
This leads to a collection $\mathcal{L}_1 \supset \mathcal{S}$ of (isomorphism classes of) Lie-type groups if one is allowed to take derived groups and to factor groups by subgroups of their centers.   For this I certainly want to include general linear and unitary groups, so for type $A_{n-1}$ the collection $\mathcal{L}_1$ includes $\mathrm{GL}_n, \mathrm{SL}_n, \mathrm{PGL}_n, \mathrm{PSL}_n$ and related unitary groups.   Often people specify at the outset a connected reductive group $G$ over a finite field, but it's best to assume that $G$ has a simple derived group for the current purpose.
(2) The Atlas focuses instead on the finite simple groups and then builds character tables for various related groups, as in my Weyl group example above.   Here the simple group $G$ can be enlarged to a group in which $G$ is normal and the quotient is abelian, or can acquire nontrivial central extensions using its Schur multiplier.    Starting with the collection $\mathcal{S}$, one obtains in this way a larger collection $\mathcal{L}_2$ of (isomorphism classes of) "groups of Lie type".   Partly because of questions about general linear groups and the like, I'm left with a question:

How close to being the same are the collections $\mathcal{L}_1$ and $\mathcal{L}_2$? (Is there a natural way to define them to ensure equality?)

REPLY [18 votes]: I feel a bit funny posting this as an answer to someone who has written a book with "finite groups of Lie type" in the title.
But I also find the matter both confusing and interesting (and full of conflicting terminology!), so here's my outside take on this, being a topologist of training:
(My own literature references include the standard sources: Carter, Gorenstein-Lyons-Solomon3..., as well as the recent (recommendable!) book by Malle-Testerman):
I thought the most general definition of a finite group of Lie type was close to your definition $\cal L_1$:
Definition ($\cal L_3$): A finite group of Lie type $G$ is a finite group obtained as ${\mathbf G}^F$, where $\mathbf G$ is a connected reductive algebraic group over an algebraically closed field of characteristic $p$, and $F$ is a Steinberg endomorphism (aka "twisted Frobenius", a composite of a Frobenius map and an automorphism of $\mathbf G$), or a group obtained from such a group ${\mathbf G}^F$ by modding out by a central subgroup, or passing to a normal subgroup with abelian quotient group.
Groups of this form will have many "Lie-like" properties, and in particular admit a canonical BN-pair structure. (Admitting a BN-pair structure would be another possible definition!)
Some authors restrict $\cal L_3$ by assuming $\mathbf G$ is semi-simple, or even simple, and some authors don't allow central quotients or passing to subgroups contained in the commutator subgroup. (One reason to allow the central quotients and subgroups is that otherwise the construction rarely produces simple groups: $|{\mathbf G}^F|$ only depends on the isogeny class of $\mathbf G$, e.g., adjoint and simply connected form have the same order, so ${\mathbf G}^F$ itself will never be simple unless $\mathbf G$ has associated Cartan matrix of determinant $1$.) Sticking with reductive (rather than simple or semi-simple) has the advantage that Levi subgroups in finite groups of Lie type are still finite groups of Lie type, which is useful if you're doing induction.
If you (for some reason) want to avoid tori $\mathbb F_q^{\times}$ and twisted products of non-trivial groups in this definition, but keep $GL_n(\mathbb F_q)$, you can maybe add the assumption to $\mathcal L_3$ that the root datum associated to $\mathbf G$ is indecomposable, and the associated root system not a product of two non-trivial root systems, or something like that? To an outsider this seems artificial?
However, not all classical groups will be in $\mathcal L_3$, e.g., $SO_{2n}^+(\mathbb F_q)$ (with the standard (classical!) definition). The problem arises from that its not connected as an algebraic group when $q$ is a power of two, so one has to pass to an index 2 subgroup, the kernel of the "pseudo/quasi-determinant", instead of the old-fashioned determinant. In particular the Weyl group of $E_8$, which is a central extension $2\cdot SO_8^+(8)$, is also not in $\mathcal L_3$, as Derek alluded to. ($W_{E_8}$ is "even further" freakish than $SO_8^+(2)$ since the central extension is related to that the index two subgroup $Spin^+_8(2)$ has non-generic Schur multiplier $\mathbb Z/2 \times \mathbb Z/2$, not zero as usual.)
I would (perhaps naively) argue that these groups also should not be considered finite groups of Lie type since they (unless I'm mistaken) do not admit the structure of a BN-pair; the structure of their p-radical subgroups is more chaotic (the Borel-Tits theorem fails).
Interpreting your $\mathcal L_2$ to consist of groups $G$ which has a filtration by normal subgroups $1< Z < H < G$, where Z is central in $G$, $G/H$ is abelian, and $H/Z$ is a simple non-abelian subquotient of a group in $\mathcal L_3$ (or equivalently $\mathcal L_1$), we will have $\mathcal L_3 \subseteq \mathcal L_2$.
With these interpretations, we can give the following response to your question:

$\mathcal L_1 \subseteq \mathcal L_3 \subseteq \mathcal L_2$, with $W_{E_8}$ or $SO_{2n}^+(\mathbb F_q)$, $q$ even, being examples of groups in $\mathcal L_2$ not in $\mathcal L_3$.

Note that a group like $GL_n(\mathbb F_p)$ of course would be in $\mathcal L_3$. Whether it is in $\mathcal L_1$ depends a bit on which $\mathbf G$ you are willing to start with.
Now, you may want to limit the class in $\mathcal L_2$ in some way to ensure that it is not larger than $\mathcal L_3$ and avoid the "pathologies". I would be surprised if there was a good way to do this, without referring back to $\mathcal L_3$: The main problem, I think, it that there is no really good "purely group theoretic" way of e.g., specifying that you only allow extensions "on top" by diagonal automorphisms (not field or graph), without getting back into the Lie theory. This is at least what goes wrong in the $SO_{2n}^+(\mathbb F_q)$, $q$ even, case. (Derek also hinted at this.)
You may also, conversely, argue that $\mathcal L_2$ is anyway too small, since it is not closed under iteratively putting things "on top" and "on the bottom"; to be "almost" of Lie type should be a bigger class... The cleanest fix for this is to go all the way and study the groups which are virtually a finite group of Lie type, i.e., has a finite index subgroup which a finite group of Lie type. Now, this class of groups is however of course just the class of virtually trivial groups, aka all finite groups...
This is not as silly as it sounds, since many deep open problems in finite group theory, such as Alperin's conjecture and its cousins, stem from the philosophy that all finite groups behave almost like finite groups of Lie type....


TITLE: Positive Definite iff the determinant of all upper-left submatrices $> 0$ - Always true?
QUESTION [3 upvotes]: Theorem. A $M_{n \times n}$ matrix is positive (negative) definite iff the determinate of all upper-left sub matrices are positive (negative).
However, consider this matrix:
\begin{pmatrix}
        3 & 1 & 2 \\
        2 & 4 & 3 \\
        -1 & -2 & 1 \\
\end{pmatrix}
If I am not mistaken I take the upper-left submatrices and their respective determinant to be
$$\begin{pmatrix}
 3 \\
\end{pmatrix}
\Longrightarrow 3$$
$$\begin{pmatrix}
        3 & 1 \\
        2 & 4
\end{pmatrix}
\Longrightarrow 10$$
$$\begin{pmatrix}
        3 & 1 & 2 \\
        2 & 4 & 3 \\
        -1 & -2 & 1 \\
\end{pmatrix}
\Longrightarrow 25$$
But this matrix isn't strictly positive definite of negative definite. It you find the eigenvalues (I'd suggest using some type of software - can also try to plug in for the Cholesky Decomposition and should given an error) you get eigenvalues with real and complex parts.
The real parts are all $> 0$, but the complex parts are not. Does this mean the matrix is positive definite in the reals, but not strictly positive definite?

REPLY [2 votes]: As was stated in comments, the Sylvester's criterion requires the matrix to be symmetric. A simpler example with a non-symmetric matrix would be
$$
A = \begin{pmatrix} 1 & 4 \\ 0 & 1\end{pmatrix}
$$
which makes it clear that none of the upper-left determinants are influenced by the entry $2$; thus they do not detect its effect on the signature of the matrix.
To determine positive-definiteness of a non-symmetric matrix one can apply Sylvester's criterion to $(A+A^T)/2$, which generates exactly the same quadratic form as $A$ itself.


TITLE: Find the inverse of the equivalence class
QUESTION [1 upvotes]: I have to check if the equivalence class has an inverse(without calculations). If yes, I have to find it.
$$[7] \in \mathbb{Z}_{36}$$
We know that $[a] \in \mathbb{Z}_m$ has an inverse $\Leftrightarrow (a,m)=1$.
In this case, knowing that $(7,36)=1$ we conclude that $[7]$ has an inverse.
$$\text{So there is a } x \text{ such that } [7][x]=[1] \text{ in } \mathbb{Z}_{36}.$$
But how can I find this $x$? Do I have to check all integers that are in $\mathbb{Z}_{36}$, that means all integers in $\{0,1,...,35\}$?

REPLY [1 votes]: Hint $\ {\rm mod}\ 36\!:\,\ 7x\equiv 1\,\Rightarrow\, x\,\equiv\,  \dfrac{1}7 \,\equiv\, \dfrac{-35}7 \,\equiv\, -5.\,$

Or $\bmod 6^2\!:\,\ \dfrac{1}{1+6}= 1-6\ $ by simpler multiples or Hensel lifting or Newton's method.

More generally we can employ the Extended Euclidean Algorithm and related techniques to compute modular inverses. A few worked examples using a handful of methods are  here and here and here.


TITLE: Quasi-coherent envelope of a module
QUESTION [11 upvotes]: Let $X$ be a scheme. It is known that $Qcoh(X)$ is cocomplete, co-wellpowered and has a generating set. The special adjoint functor theorem tells us that then every(!) cocontinuous functor $Qcoh(X) \to A$ has a right-adjoint. Here $A$ is an arbitrary category (which I always assume to be locally small).
a) Is there a nice description of the right-adjoint to the forgetful functor $Qcoh(X) \to Mod(X)$? Here, you may impose finiteness conditions on $X$. This functor may be called a quasi-coherator.
b) Let $f : X \to Y$ be a morphism of schemes. Then $f^* : Qcoh(Y)  \to Qcoh(X)$ is cocontinuous, since $f^* : Mod(Y) \to Mod(X)$ is cocontinuous and the forgetful functor preserves and reflects colimits. In particular, there is a right-adjoint $f_+ : Qcoh(X) \to Qcoh(Y)$. If $f$ is quasi-separated ans quasi-compact, then this is the direct image functor $f_*$. Is there a nice description in general? Note that $f_+$ is the composition $Qcoh(X) \to Mod(X) \to Mod(Y) \to Qcoh(Y)$, where the latter is the quasi-coherator. This is only nice if we have answered a).
c) Since $Mod(X)$ is complete and $Qcoh(X) \to Mod(X)$ has a right adjoint, $Qcoh(X)$ is also complete. Is there a nice description for the products? They are given by taking the quasi-coherator of the product, can we simplify this? I mean, perhaps they turn out to be exact although the products in $Mod(X)$ are not exact?
Answer (after reading the article Leo Alonso has cited)
We have the following description of the quasi-coherator: Let $X$ be a concentrated scheme, i.e. quasi-compact and quasi-separated. If $X$ is separated, say $X = \cup U_i$ with finitely many affines $U_i$ such that the intersections $U_i \cap U_j$ are affine, then the quasi-coherator of a module $M$ on $X$ is the kernel of the obvious map
$\prod_i (u_i)_* \tilde{M(U_i)} \to \prod_{i,j} (u_{i,j})_*  \tilde{M(U_i \cap U_j)}$,
where $u_i : U_i \to X$ and $u_{ij} : U_i \cap U_j \to X$ are the inclusions. If $X$ is just quasi-separated, there is a similar description using the separated case.
The idea is quite simple and can be generalized to every flat ring representation of a finite partial order, which has suprema (for example the dual of the affine subsets of a quasi-compact separated scheme). On an affine part, the quasi-coherator consists of sections of all other affine parts over it, which are compatible in the obvious sense.
If we have no finiteness conditions, the description is basically also valid, but you have to take the quasi-coherators of the products or the direct images, since they don't have to be quasi-coherent. In general there is no nice description. Also in nice special cases, b) and c) have no nice answers (and infinite products are not exact, even in the category of quasi-coherent modules on nice schemes).

REPLY [2 votes]: Let us do the case of an affine scheme $X$ first. This is easy. If $M$ is an $O_X$-module, we define $\tilde{M}$ as the quasicoherent $O_X$-module defined by the global sections $M(X)$. Notice that the restriction gives the canonical map $\tilde{M}->M$.
As step 2, we extend to the case of quasiaffine $X$. Quasiaffinity means that $X-> Spec O_X(X)$ is an open embedding. Equivalently, all its quasicoherent modules are generated by global section. Hence, the coherator functor $M |-> \tilde{M}$ is defined in the same way. BTW, it is clear how the coherator works on maps too.
Finally, we have everything ready as a general scheme admits an affine open cover $X=\cup_i U_i$. The quasiaffine case is useful as double and triple intersection $U_{i,j}$, $U_{i,j,k}$ are all quasiaffine. Given an $O_X$-module $M$, we use its open pieces $M_i=M|_{U_i}$ and gluing maps $\phi_{i,j}$ from ${M_i}|_{U(i,j)}$ to ${M_j}|_{U(i,j)}$,
where $U(i,j)=U_{i,j}$ (because tex translator is finding it difficult to comprehend it too without each formula starting in a new line), that satisfy the cocycle conditions on triple intersection.
Now the coherator is glued from open pieces $\tilde{M}_i$
using isomorphisms $\tilde{\phi}_{i,j}$ which inherit the cocycle conditions. And That's All Folks!


TITLE: Dependence of smoothness of a real-valued function on a manifold
QUESTION [1 upvotes]: Loring Tu's An introduction to manifold, Proposition 6.3 asserts that:
Let $M$ be a smooth $n$-manifold and $f:M \to \Bbb R$ a real valued function on $M$. Then the following are equivalent:
(1) The function $f:M \to \Bbb R$ is $C^{\infty}$
(2) $M$ has an atlas such that for every chart $(U,\varphi)$ in the atlas $f  \circ \varphi ^{-1}$ is $C^{\infty}$
But, I have a trouble with this. Since $M$ is a smooth $n$-manifold, it already has a smooth structure, and condition (1) is saying $f$ is smooth with respect to the structure of $M$. On the other hand, (2) is saying that $M$ has another atlas. Although the atlas in (2) is contained in a unique maximal atlas of $M$, it is not necessarily the same with the already given maximal atlas of $M$, isn't it?

REPLY [3 votes]: The atlas in $(2)$ must be compatible with the smooth structure on $M$.  What $(2)$ is saying is that you don't need to use a maximal atlas, just any atlas (it cuts down on the work needed to check things if there's an atlas with a finite number of charts).
The content in $(2)$ is that if $f$ is smooth with respect to an atlas $\mathcal{A}$, then it is also smooth with respect to the maximal atlas containing $\mathcal{A}$.  The content is the extension to the charts of the maximal atlas not in $\mathcal{A}$.
For example, on a circle (with the standard atlas) you only need to check that a function $f$ is smooth with respect to the upper, lower, left, and right semicircles to conclude that $f$ is smooth instead of checking that its smooth with respect to all charts in a maximal atlas.


TITLE: Generic features, semi-heuristic derivations
QUESTION [1 upvotes]: Einstein was able to make some predictions before GR was fully formulated. e.g. gravitational time dilation.
Such predictions before the full theory are referred to as "semi-heuristic derivations" here: http://motls.blogspot.cz/2012/09/albert-einstein-1911-12-1922-23.html
and have also been called "generic features" by Smolin I think about loop quantum gravity although I can't find the right reference right now.
What such phenomena are predicted today about theories that we don't yet have in full form ?

REPLY [3 votes]: This is the way in which all physical theories get formulated--- you first acquire certainty regarding the behavior of many special cases you have some experimental data or theoretical insight about, then you try to formulate a precise theory which extends these heuristic laws to a precise understanding, and when you succeed in matching the heuristic laws (when they apply) and you can predict everything consistently and correctly, you are done.
In fundamental physics, one has a pretty solid understanding of phenomena which are not quantum gravitational, because we have a precise fundamental theory of relativistic quantum fields. So the most significant things that are not fully understood at the precise level are the class of insights deriving from Hawking radiation and black hole classical behavior. These are semi heuristic, because there are puzzles that are not yet fully resolved within string theory. This class includes:

The near horizon behavior of semiclassical black holes: an observer falling into a black hole sees nothing special when crossing the horizon, and this has not been rigorously demonstrated in string theory, it is only rigorously true classically. Now some people claim that it can't work in quantum gravity, that black holes are "firewalls". I read the arguments, and I find them uncompelling, because they mix assumptions about the semiclassical behavior at late times with full quantum observations on the Hawking radiation which are restricted when you measure the semiclassical state at late times, so the argument smells fishy and doesn't stop smelling fishy even with later clarifications, although understanding what is going wrong is important, and this is at the heuristic level, because we can't reconstruct black hole interiors completely from quantum gravity scattering data.
The related holographic principle and black hole complementarity: this is also heuristic for semiclassical thermal black holes, although it is precise in AdS/CFT, for certain extremal black holes.
Cosmological horizon entropy: the entropy of the cosmological horizon is not even in-principle understood in string theory, and it is a major clue to understanding how to do quantum gravity in deSitter space, because it is understood at the heuristic level--- it's the same as black hole entropy.
Rindler horizons: The behavior of strings on a Rindler background is nontrivial, even though this is just Minkowski space. Quantum fields are completely understood on Rindler, but string theory on Rindler is harder, because you don't have an S-matrix (everything falls into the Rindler horizon).

For string theory, we only have heuristic guidelines regarding compactification and SUSY breaking. These heuristics are summarized in guidelines about what topologies and matter configurations give rise to which gauge fields and representations, and where one should expect to find the standard model. These are relatively well understood, since there are many explored vacua, but there are always surprises.
In cases where we know the fundamental laws without serious doubt, there are still cases where we understand things only at the heuristic level. The following are in QCD:

Confinement and Regge theory: we know the QCD confines to Regge trajectories for mesons and topological soliton Baryons, but this formulation is not mathematically linked to QCD by any rigorous path.
Pomerons: This is related--- we know that high energy scattering is dominated by Pomeron exchange, and this is heuristic only, because we can't relate this to QCD high energy diffractive scattering except in certain regimes which are not completely diffractive.
Chiral perturbation theory: Again related, this is the low-energy approximation to QCD, and the parameters are from assumptions on low energy condensates. I suppose I shouldn't include this, because you can extract chiral data from lattice data in principle exactly, but I had something else in mind for what constitutes a full understanding--- it would mean that any chiral configuration can be mapped to a QCD configuration, and you could do the path integral for QCD in two steps--- as a path integral over long wavelength chiral configurations plus an additional path integeral over QCD in the given chiral background. Nobody did this, although it shouldn't be hard.
Instanton fluid: there is a class of semiheuristic models that give the QCD vacuum as a dense instanton fluid, and this is not 100% understood. It's essentially the same problem as before--- condensates and confinement.
Quark condensates: there are condensate models where we can semi-heuristically calculate the effects on hadrons using the SVZ sum-rules (QCD sum rules), but the condensate values again are not derived from QCD, although they can be measured from experiment. In principle, lattice gives them, but this is not enough--- you want to know the values and effects with more insight.

There are no doubts QCD resolves these questions, but the exact best way to make these heuristic things precise is unclear.
There are condensed matter systems where the understanding is heuristic in the same way--- the most famous and the one I like best is probably:

HighTc: The condensate state of the HighTc superconductor can be described phenomenologically using a D-wave superconducting condensate. Getting this D-wave from the fundamental interactions has not been done in a universally convincing way, although there is no doubt that the fundamental theory will do it.

There are many other such things, this is usually all the open problems people work on. Historically, if you look at any phenomenon that was understood, it was understood heuristically before it was understood precisely, and learning the heuristic stuff is an important prerequisite for getting certainty about the final explanation, because physical theories are evolved by common sense, they don't emerge fully formed from nothing.


text:
\begin{document}

\begin{abstract}
We extend a higher-order sum rule proved by B. Simon to matrix valued measures on the unit circle and their matrix Verblunsky coefficients.
\end{abstract}
\begin{keyword}Sum rules, Szeg{\H o}'s theorem, Verblunsky coefficients,  matrix measures on the unit circle,  relative entropy\end{keyword}

\maketitle

\section{Introduction}
A probability measure $\mu$  on the unit circle $\mathbb T$ with infinite support is characterized by its Verblunsky coefficients $(\alpha_j(\mu))_{j \geq 0}$, elemnts in the interior of the unit disc. They are associated  with the Szeg{\H o} recursion of orthogonal polynomials in $L^2(\mathbb T, d\mu)$. A sum rule is an identity between an entropy-like  functional of this measure and a functional of the sequence of its Verblunsky coefficients (for short, we say "V-coefficients" in the sequel). The most famous is  Szeg{\H o}'s theorem.
\begin{theorem}
Let $d\mu= w(\theta) \frac{d\theta}{2\pi} + d\mu_s$ be the Lebesgue decomposition of a probability measure on $\mathbb T$ and let $(\alpha_n)_{n \geq 0}$ its V-coefficients. Then
\begin{align}
\label{SZ}
\int_0^{2\pi} \log w(\theta)  \frac{d\theta}{2\pi} = \sum_0^\infty\log\left(1-|\alpha_k|^2\right)\,.
\end{align}
where both members can be simultaneously finite or $-\infty$.
\end{theorem}

In his  book \cite{simon05}, B. Simon proved  the following statement (higher-order Szeg{\H o} theorem).

\begin{theorem}[\cite{simon05} Th. 2.8.1]
\label{1.1}
Let $d\mu= w(\theta) \frac{d\theta}{2\pi} + d\mu_s$ be a probability measure on $\mathbb T$ and let $(\alpha_n)_{n \geq 0}$ its V-coefficients. Then
\begin{align}
\notag
\int_0^{2\pi} (1 - \cos \theta) \log w(\theta)  \frac{d\theta}{2\pi} =  &\frac{1}{2}(1 - |1+\alpha_0|^2)  - \frac{1}{2}\sum_0^\infty|\alpha_{k+1} - \alpha_k|^2\\
\label{DH}
&+ \sum_0^\infty \left(\log\left(1-|\alpha_k|^2\right) + |\alpha_k|^2\right)\,,
\end{align}
where both members can be simultaneously finite or $-\infty$.
\end{theorem}
Actually this formula may be written in terms of entropies. For probability measures $\nu$ and $\mu$ on $\mathbb T$, let $\mathcal K (\nu | \mu)$ denote the Kullback-Leibler divergence or relative entropy of $\nu$ with respect to $\mu$:
\begin{equation}
\label{KL}
{\mathcal K}(\nu\, |\, \mu)= \begin{cases}  \ \displaystyle\int_{\mathbb T}\log\frac{d\nu}{d\mu}\!\ d\nu\;\;& \mbox{if}\ \nu\ \hbox{is absolutely continuous with respect to}\ \mu ,\\

   \  \infty  &  \mbox{otherwise.}
\end{cases}
\end{equation}
Usually, $\mu$ is the reference measure. Here the spectral side will involve the reversed Kullback-Leibler divergence, where $\nu$ is the reference measure and $\mu$ is the argument. In this case, we have that $\mathcal{K}(\nu\,  \, | \, \mu)$ is finite if and only if
\begin{align} \label{eq:KL2}
\int_0^{2\pi} \log w(\theta) \, d\nu(\theta) > - \infty ,
\end{align}
where $d\mu = w(\theta)d\nu(\theta) + d\mu_s$ is the Lebesgue decomposition of $\mu$ with respect to $\nu$.
If we denote
\label{defl0}
\begin{align}
d\lambda_0(\theta)  = \frac{d\theta}{2\pi} \ , \ d\lambda_1(\theta)  = (1-\cos\theta) \frac{d\theta}{2\pi}
\end{align}

the sum rule (\ref{SZ}) may be written
\begin{align}
\label{SZsr}
\mathcal K\left(\lambda_0\, |\,  \mu\right) =

 - \sum_0^\infty \log\left(1-|\alpha_k|^2\right)\,,
\end{align}
and
the sum rule (\ref{DH}) may be written
\begin{align}
\notag
\mathcal K\left(\lambda_1\, |\,  \mu\right) = \mathcal K(\lambda_1 \, |\,  \lambda_0)
+ \Re\alpha_0 + \frac{|\alpha_0|^2}{2} +\frac{1}{2}\sum_0^\infty|\alpha_{k+1} - \alpha_k|^2\\
\label{KKone}
 - \sum_0^\infty \left(\log\left(1-|\alpha_k|^2\right)+ |\alpha_k|^2\right)
\end{align}
with
\[ \mathcal K(\lambda_1 \, |\, \lambda_0) =  \int_0^{2\pi}(1-\cos\theta) \log(1-\cos \theta)\frac{d\theta}{2\pi}  = 1 - \log 2\,.\]
In (\ref{KKone}) both sides may be infinite simultaneously, and they are finite if and only if
\begin{align}
\label{gem0}
\sum_k \alpha_k^4 + |\alpha_{k+1}- \alpha_k|^2 < \infty\,.
\end{align}

Actually, it is easy to include (\ref{KKone}) and (\ref{SZsr}) into a family of sum rules depending on a parameter $\g$ such that $|\g| \leq 1$. Let
\begin{align}\label{KKdefg}
d\lambda_\g(\theta) = (1-\g\cos\theta)\ d\lambda_0(\theta)
\end{align}
(called one single nontrivial moment in \cite{simon05} p. 86). Combining (\ref{KKone}) and Szg{\H o}'s formula, we get, as mentioned in  \cite{GNROPUC} Cor. 5.4 :
\begin{align}\notag
\mathcal K(\lambda_\g\, | \ \mu) &=
\mathcal K(\lambda_\g\, [ \ \lambda_0)
 + \g \left(\Re \alpha_0 + \frac{|\alpha_0|^2}{2} + \frac{1}{2}\sum_1^\infty |\alpha_k - \alpha_{k-1}|^2\right)\\
\label{KKg}
&+ \sum_0^\infty -\log (1 - |\alpha_k|^2) -\g|\alpha_k|^2\,,
\end{align}
where
\begin{align}
\mathcal K(\lambda_\g\, | \ \lambda_0) = \int (1-\g \cos \theta)\log (1- \g\cos \theta) \frac{d\theta}{2\pi}  = 1 - \sqrt{1 - \g^2} + \log \frac{1 + \sqrt{1-\g^2}}{2}\,.\end{align}
It may be  called GW sum rule, since $\lambda_\g$ is the equilibrium measure in a random matrix model due to Gross and Witten (\cite{GrossW}).

For $\g = 0$, we recover  (\ref{SZ}) formula and when  $\g=1$, we recover (\ref{KKone}).

Simon's proof of Theorem \ref{1.1} (see Sect. 2.8 in \cite{simon05}) was based on the use of the Szeg{\H o} function
\[D(z) = \exp \int_0^{2\pi} \frac{e^{\ii \theta} +z}{e^{\ii \theta} -z}\log w(\theta) \frac{d\theta}{4\pi}\,,\]
the asymptotics of the orthogonal polynomial and Szeg{\H o}'s theorem.
Later on,  Simon gave another proof of this theorem in Sect.  2.8 of \cite {Simon-newbook}. The new proof  uses a relative Szeg{\H o} function and a step-by-step sum rule provided by the coefficient stripping.

In a series of papers, Gamboa et al. tackled sum rules on the real line and on the unit circle\footnote{See references in \cite{GNROPUC}.},  on a probabilistic way, using large deviations techniques. The main argument is the uniqueness of the rate function when the large deviations of a random measure are considered under two different encodings. In particular, in \cite{GNROPUC}, they (re)proved  Szeg{\H o}'s theorem as a sum rule, stated a new sum rule for the Hua-Pickrell measure, and
asked
for a possible probabilistic proof of the higher-order sum rule quoted above.
Shortly after,
 Simon et al. \cite{BSZ} gave that proof.

It turns out that probabilistic tools are robust enough to be extended to matrix measures, which allowed Gamboa et al. to give  a probabilistic proof of the famous matrix  Szeg{\H o}'s theorem of  Delsarte et al.  \cite{delsarte} involving matrix V-coefficients. With the notations of the following section, this theorem says that if $d\mu = w(\theta) d\lambda_0 + d\mu_s$ is a non-trivial matrix-measure, then\footnote{We use $\dagger$ for matrix adjoint, keeping the notation $^*$ for
reversed polynomials.}
\begin{align}
\int_0^{2\pi}
\log \det w(\theta)  d\lambda_0(\theta) = \sum_0^\infty \log\det (\bdone - \alpha_k\alpha_k^\dagger)\,.
\end{align}

In \cite{GNROPUC} the authors proved also  a matrix version of  the Hua-Pickrell sum rule and

conjectured a matrix version of the GW sum rule (\ref{KKg}).

These considerations open the way to two challenges: analytical proof and probabilistic proof.

The second way seems accessible by combining the machinery of \cite{GNROPUC} and of \cite{BSZ}, i.e. a large deviation for a random measure encoded by its V-coefficients, but it seems more natural to begin with  the first way, which will be done in this note.
Of course, a possible issue comes from the non-commutativity of the product of matrices, but as usual, the story ends well.

We present the notations and main results  in Sect. \ref{mr}. Theorem \ref{main1} is a matrix-version of (\ref{KKg}) and Prop. \ref{4.1} is a  {\it gem} i.e. a condition of finiteness of the entropy. In Sect. 3, we give the proof of the first result,  involving the coefficient stripping method  and a limiting argument.
In Sect. 4 we give the proof of the {\it gem}
. Finally Sect. 5 is devoted to the proofs of intermediate results.

\section{Notations and main result}
\subsection{Notations}
\label{mr}
Let us begin with some introductory elements on matrix measures.  For a more detailed exposition, see  \cite{damanik2008analytic} Sect. 1, \cite{derevy} Sect. 4, \cite{GNROPUC} Sect. 6.

Let $p > 1$ be an integer and let $\mathcal M_p$ be the set of complex $p \times p$ matrix measures $\mu$ on $\mathbb T$ which are Hermitian, nonnegative and normalized by $\mu (\mathbb T) = \bdone$
 (the $p \times p$ identity matrix). A
matrix measure is called quasi-scalar if it may be wriiten ${\bf 1}\cdot \sigma$ with $\sigma$ a probability measure on $\mathbb T$.
A $p \times p$ matrix polynomial is a polynomial with coefficients in $\mathbb C^{p\times p}$.
Given a measure $\mu \in \mathcal M_p$, we define two  inner products on the space of $p \times p$ matrix polynomials by setting
\begin{align*}\langle\langle f, g \rangle\rangle_R = \int f(e^{\ii \theta})^\dagger d\mu(\theta) g(e^{\ii \theta})\\
\langle\langle f, g \rangle\rangle_L = \int g(e^{\ii \theta}) d\mu(\theta) f(e^{\ii \theta})^\dagger
\,.\end{align*}
A sequence of matrix polynomials $(\varphi_j)$ is called right-orthonormal if, and only if,
\[\lal\varphi_i, \varphi_j\rar_R= \delta_{ij}\bdone \, .\]
A matrix measure is called non-trivial if
\[\tr\!\ \langle\langle f, f \rangle\rangle_R
 >0\]
for every non-zero polynomial $f$.
We define the right monic matrix orthogonal polynomials $\Phi_n^R$ by applying the block Gram-Schmidt algorithm  to the sequence $\{\mathbf 1, z\mathbf 1, z^2 \mathbf 1, \dots\}$.  In other words, $\Phi_k^R$ is the unique matrix polynomial
$\Phi_k^R(z) = z^k \mathbf 1 +$ lower order terms, such that $\lal z^j\mathbf 1, \Phi_k^R\rar_R =0$ for $j=0, \dots, k-1$.
The normalized orthogonal polynomials are defined by
\[\varphi_0^R = \bdone\ \  ,\ \  \varphi_k^R = \Phi_k^R\kappa_k^R.\]
Here the sequence of $p\times p$ matrices $(\kappa_k^R)$
satisfies, for all $k$, the condition $\left(\kappa_k^R\right)^{-1}\kappa_{k+1}^R>0_p$
and is such that the sequence $(\varphi_k^R)$ is orthonormal.
We define the sequence of left-orthonormal polynomials $(\varphi_k^L)$ in the same way except that the above condition is replaced by  $\kappa_{k+1}^L \left(\kappa_k^L\right)^{-1}> 0$. The matrix Szeg\H{o} recursion is then
\begin{eqnarray}
\label{SzL}
z\varphi_k^L -\rho_k^L\varphi_{k+1}^L &=& \alpha_k^\dagger(\varphi_k^R)^*\\
\label{SzR}
z\varphi_k^R - \varphi_{k+1}^R\rho_k^R &=& (\varphi_k^L)^* \alpha_k^\dagger\,,
\end{eqnarray}
where for all $k\in\mathbb{N}_0$,
\begin{itemize}
\item $\alpha_k$ belongs to  $\mathbb{B}_p$, the closed unit ball of $\mathbb C^{p\times p}$  defined by
\begin{equation}
\label{defBp}\mathbb{B}_p:=\{M \in \mathbb C^{p\times p}: MM^\dagger \leq \mathbf 1\}\,,\end{equation}
\item $\rho_k^R$ and  $\rho_k^L$ are the so-called defect matrices defined by
\begin{align}
\label{defrho}
\rho^R_k :=  \left(\bdone - \alpha_k\alpha_k^\dagger\right)^{1/2}\  , \  \rho^L_k =    \left(\bdone - \alpha_k^\dagger\alpha_k\right)^{1/2}\,,
\end{align}
\item for a matrix polynomial $P$ with degree $k$, the reversed polynomial $P^*$ is defined by
\[P^*(z) := z^k P(1/\bar z)^\dagger\,.\]
\end{itemize}
Verblunsky's theorem establishes a one-to one correspondance between non-trivial (normalized) matrix measures on $\mathbb T$ and sequences of elements in the interior of $\mathbb B_p$ (Theorem 3.12 in  \cite{damanik2008analytic}).

In an alternative way, these V-coefficients may be introduced as matrix Schur coefficients as follows. Let $F$ be the  Caratheodory (or Herglotz) transform of $\mu$ defined by:
\[F(z) = \int \frac{e^{\ii \theta} + z}{e^{\ii \theta} - z} d\mu(\theta)
 \ , \ z \in \mathbb D = \{ z : |z| < 1\}\,,\]
and $f$ the Schur transform defined by:
\[f(z) = z^{-1}(F(z) -\bdone)(F(z) + \bdone)^{-1}\,,\]
which is equivalent  to
\begin{align}
\label{fF}
F(z) = (\bdone + zf(z))(\bdone-zf(z))^{-1}\,.
\end{align}
The Schur recursion is defined as follows.
At  step $0$ we set
\[\alpha_0 = f(0)\,,\]
which gives the first V-coefficient. We define the defect matrices  (right and left) by
\begin{align}
\label{defect}
\rho_0^R = (\bdone - \alpha_0\alpha_0^\dagger)^{1/2} \ , \ \rho_0^L = (\bdone - \alpha_0^\dagger\alpha_0)^{1/2}\,,
\end{align}
and then, at step $1$ we set

\begin{align}
\label{Schur1}
Sf := f_1 = z^{-1}(\rho_0^R)^{-1}(f(z) - \alpha_0)\left(\bdone- \alpha_0^\dagger f(z)\right)^{-1}\rho_0^L
\end{align}
and the second V-coefficient is
\[\alpha_1 = f_1 (0)\,.\]
The other coefficients are defined with the same algorithm
\[f_{k+1} = Sf_k, \ \alpha_{k+1} = f_{k+1}(0) , ...\,.\]

The following theorem gives the connection between $F$ and the absolutely continuous part of $\mu$.
\begin{theorem}[\cite{damanik2008analytic} Prop. 3.16]
\label{DPS}
For $z \in \mathbb D$, we have
\begin{align}
\label{3.52-o}
\Re\!\ F(z) = (\bdone- \bar zf(z)^\dagger)^{-1}(\bdone - |z|^2f(z)^\dagger f(z)) (\bdone -zf(z))^{-1}\,.
\end{align}
and the non-tangential boundary values $\Re F(e^{\ii \theta})$ and $f(e^{\ii \theta})$ exist for a.e.  $\theta$.

If $\mu$ is a normalized matrix measure with Lebesgue decomposition
\[d\mu(\theta) = w(\theta) d\lambda_0(\theta) + d\mu_s (\theta)\]
(where $w$ is a $p\times p$ matrix),
then for a.e.  $\theta$
\iffalse
\begin{enumerate}
\item
the non-tangential boundary values $\Re F(e^{\ii \theta})$ and $f(e^{\ii \theta})$ exist,
\item
\fi

\[w(\theta) = \Re F (e^{\ii \theta})\,,\]

and for a.e. $\theta$, $\det w(\theta) \no=0$ if and only if $f(e^{\ii \theta})^\dagger f(e^{\ii \theta}) < \bdone$.

\end{theorem}

\subsection{Main result}
When $\Sigma = \bdone \cdot \sigma$ is a pseudo-scalar measure and $d\mu(\theta) = h(\theta) d\sigma(\theta) + d\mu_s (\theta)$, we define the relative entropy
\begin{align}
\mathcal K(\Sigma\!\ | \ \mu) = - \int_\mathbb T \log\det h(\theta) d\sigma(\theta)\,.
\end{align}
We will consider two reference measures:
\begin{align}

d\Lambda_0 (\theta) = \bdone\cdot  d\lambda_0(\theta) \ , \ d\Lambda_\g (\theta) = \bdone\cdot  d\lambda_\g(\theta)\,.
\end{align}

Our main result is the following.

\begin{theorem} For $|\g| \leq 1$, let $d\mu(\theta) = w(\theta) d\lambda_0(\theta) + d\mu_s(\theta)$ be a non-trivial matrix measure, then
\label{main1}
\begin{align}
\label{mainf1}
\int_0^{2\pi}  (1 - \g \cos \theta) \log\det w(\theta) d\lambda_0(\theta)= \sum_0^\infty \log\det (\bdone - \alpha_k\alpha_k^\dagger) - \g T(\alpha_0, \alpha_1, \cdots)
\end{align}
with
\begin{align}
\label{defT}
T(\alpha_0, \alpha_1, \cdots) := \Re\!\ \tr\!\ (\alpha_0 - \sum_0^\infty \alpha_k\alpha_{k+1}^\dagger)\,,
\end{align}

or in an equivalent form
\begin{align}
\label{mainf2}
\mathcal K( \Lambda_\g \, | \  \mu) = \mathcal K(\lambda_\g \, | \ \lambda_0) -\sum_0^\infty \log\det (\bdone - \alpha_k\alpha_k^\dagger) + \g  T(\alpha_0, \alpha_1, \cdots)\,.
\end{align}

In (\ref{mainf2}), both sides, which are nonnegative, may be simultaneously infinite.

\end{theorem}

 It is
exactly  Conjecture 6.11 1. in \cite{GNROPUC}.  For $\g = 0$, we recover of course the matrix Szeg{\H o} formula.

The right hand side may also be written
\begin{align}
\notag
T(\alpha_0, \alpha_1, \cdots) = \Re \tr\!\ \alpha_0
&+ \frac{1}{2} \tr\!\ \alpha_0\alpha_0^\dagger\\
\label{otherT}
&+ \frac{1}{2}\sum_0^\infty \tr\!\ (\alpha_k - \alpha_{k+1})(\alpha_k^\dagger - \alpha_{k+1}^\dagger) - \sum_0^\infty \tr\!\ \alpha_k\alpha_k^\dagger\,.
\end{align}
According to the definition of B. Simon \cite{Simon-newbook}, the {\it gems} are equivalent conditions for the finiteness of entropies.
Like in Corollary 5.4 in \cite{GNROPUC}, we have the following result.

\begin{proposition}
\label{4.1}
\begin{enumerate}
\item If $|\g| < 1$,
\begin{align}
\label{gema}\mathcal K(\Lambda_\g \!\ | \ \mu) < \infty\ \Longleftrightarrow \sum_k \tr\!\ \alpha_k\alpha_k^\dagger < \infty\end{align}
\item
\begin{align}
\label{gemb}
\mathcal K(\Lambda_1\!\ | \mu) < \infty &\Longleftrightarrow \sum_k \tr\!\ (\alpha_k\alpha_k^\dagger)^2  +
\sum_k\tr\!\ (\alpha_{k+1} - \alpha_k)(\alpha_{k+1}^\dagger  - \alpha_k ^\dagger )< \infty\\
\label{gemc}
\mathcal K(\Lambda_{-1}\!\ | \mu) < \infty &\Longleftrightarrow \sum_k \tr\!\ (\alpha_k\alpha_k^\dagger)^2  +
\sum_k\tr\!\ (\alpha_{k+1} + \alpha_k)(\alpha_{k+1}^\dagger  +\alpha_k ^\dagger )< \infty\,.
\end{align}
\end{enumerate}
\end{proposition}

\section{Proof of Theorem \ref{main1}}
We need a preliminary remark to reduce the case $\g <0$ to the case $\g >0$.
\begin{lemma}[Simon \cite{simon05} 3.2.6 and \cite{Simon2} 9.5.28]
If $\mu$ is a non-trivial matrix measure and $\tilde\mu$ is defined by
\[d\tilde\mu (\theta) = \begin{cases} d\mu(\pi + \theta)& \hbox{if} \ \theta \in [0, \pi]\\
d\mu(\theta-\pi )& \hbox{if} \ \theta \in [\pi, 2\pi]
\end{cases}
\]
then
\begin{align}
\label{+pi}
\alpha_k(\tilde \mu) = (-1)^{k+1}\alpha_k(\mu) \ , \ (k \geq 0)\,.
\end{align}
\end{lemma}
If $\g = -\gamma$ with $\gamma > 0$, we have,
\[\int (1 - \g\cos\theta) \log\det w(\theta)d\lambda_0(\theta) = \int (1 - \gamma\cos\theta) \log\det \tilde w(\theta) d\lambda_0(\theta)\,,\]
where $w$ (resp. $\tilde w$) is the a.c. part of $\mu$ (resp. $\tilde \mu$).

If we take for granted the result for $\gamma$, we get
\begin{align}\notag
 \int (1 - \gamma\cos\theta) \log\det \tilde w(\theta) d\lambda_0(\theta) &=\\  \notag =\sum_0^\infty \log\det (\bdone -
\alpha_k(\tilde\mu)\alpha_k^\dagger)(\tilde\mu))& - \gamma T(\alpha_0(\tilde \mu), \alpha_1(\tilde\mu), \cdots)
\end{align}
but, it is straightforward to see that from (\ref{defT}) and (\ref{+pi})
\begin{align}
 T(\alpha_0(\mu), \alpha_1(\mu), \cdots) = - T(\alpha_0(\tilde \mu), \alpha_1(\tilde\mu), \cdots)
\end{align}
so that (\ref{mainf1}) holds true.

From now on, in this section we assume $0 \leq \g\leq 1$.

If $\mu$ is a probability measure on $\mathbb T$ with V-coefficients $(\alpha_j (\mu))_{j \geq 0}$ and if $N$ is some positive integer, we denote by $\mu_N$ the measure whose V-coefficients are shifted:
\[\alpha_j(\mu_N) = \alpha_{j+N}(\mu) \ , \ j \geq 0\,.\]
When $\mu$ has a density $w$ with respect to $\Lambda_0$,
we denote by $w_N$ the density of $\mu_N$.

The key point is the following "recursion" theorem, matrix version of Theorem 2.8.2 in \cite{Simon-newbook}, whose proof is postpone to Sect. \ref{proofs}.
\begin{theorem}
\label{recursion}
If $\det w \not=0$ a.e., we have
\begin{align}
\label{2.29}
\int
\log\det\left(w(\theta)w_1(\theta)^{-1}\right) d\lambda_\g(\theta)

  = \log \det (\bdone - \alpha_0\alpha_0^\dagger) - \g  \Re\!\ \tr\!\ ( \alpha_0 -  \alpha_1  -  \alpha_1\alpha_0\dagger)\,.
\end{align}
\end{theorem}

This implies that $\det w_1 \not= 0$ a.e. and then we may iterate. We get, for $N > 1$

\begin{align}\label{entrel}\int
\log\det \left(w(\theta)w_N(\theta)^{-1}\right) d\lambda_\g(\theta)
= G_N(\mu)
\end{align}
where
\begin{align}
\label{itN}
G_N(\mu) =-\g \Re \tr\!\ (\alpha_N -\alpha_0) + \g \sum_0^{N-1}\Re \tr\!\ \alpha_k
\alpha_{k+1}^\dagger + \sum_0^{N-1}
\log\det (\bdone -\alpha_k \alpha_k^\dagger)\end{align}
In terms of entropy, we have the equivalent form of (\ref{2.29}):
\begin{align}
\label{KKF}
\mathcal K(\Lambda_\g \!\ | \ \mu_N) -  \mathcal K(\Lambda_\g \!\ | \  \mu) = G_N(\mu)\,.
\end{align}
To look for a limit when $N \to \infty$, we need a careful study of

 $G_N(\mu)$
. We have
\begin{align}
\label{3.6}
G_N(\mu) = -\g \Re \tr\!\ (\alpha_N -\alpha_0) + \frac{\g}{2} \tr\!\  (\alpha_N
\alpha_N^\dagger - \alpha_0\alpha_0^\dagger) - \sum_0^{N-1} A_k\,,
\end{align}
with
\begin{align}
\label{defAk}A_k := -\log\det (1 -\alpha_k \alpha_k^\dagger) - \g\!\  \tr\!\  \alpha_k
\alpha_{k}^\dagger +\frac{\g}{2}\!\  \tr\!\ (\alpha_{k+1} - \alpha_k) (\alpha_{k+1}^\dagger - \alpha_k)^\dagger\,.\end{align}

For $\alpha\alpha^\dagger < 1$, we have
\begin{align}\notag
- \log\det (1 - \alpha\alpha^\dagger) = \tr\!\ \alpha\alpha^\dagger + \frac{1}{2}\!\ \tr(\alpha\alpha^\dagger)^2 + R(\alpha)\,,
\end{align}
with
\begin{align}
\label{logdet}
R(\alpha) > 0 \  , \ R(\alpha) =o(\tr\!\ (\alpha\alpha^\dagger)^2)\,.\end{align}
This yields
\begin{align}
\label{cruz}
A_k \geq (1-\g) \tr\!\ \alpha_k\alpha_k^\dagger  + \frac{1}{2}\!\  \tr\!\ (\alpha_k\alpha_k^\dagger)^2+\frac{\g}{2}\!\  \tr\!\ (\alpha_{k+1} - \alpha_k) (\alpha_{k+1}^\dagger - \alpha_k^\dagger)\,.
\end{align}
In particular,  $A_k \geq  0$ for every $k$ (remind that we have assumed $\g \geq 0$), which gives
\[S_N (\mu) := \sum_0^{N-1} A_k \ \uparrow \
S_\infty(\mu) = \sum_0^\infty A_k \leq \infty\,,\]
(this argument of monotonicity is like in Simon \cite{Simon-newbook} Prop. 2.8.6.

The identity (\ref{mainf2}) will be the result of two inequalities.

{\bf A)} The first one uses

  the Bernstein-Szeg{\H o} approximation of
 $\mu$.
We know, from Theorem 3.9 in \cite{damanik2008analytic}, for every $\theta$ and every integer $k$, $\varphi_k^R(e^{\ii \theta})$   is invertible and from Theorem 3.11 of the same article that
 the measure
\begin{align}
d\mu^{(N)}(\theta) = \left[\varphi_{N-1}(e^{\ii \theta})\varphi_{N-1}(e^{\ii \theta})^\dagger\right]^{-1} d\lambda_0(\theta)
\end{align}
satisfies
\begin{align}
\alpha_j(\mu^{(N)}) = \begin{cases}\alpha_j (\mu)\ &\mbox{if} \ 0 \leq j \leq N-1\\
0 &\mbox{if} \ j \geq N\,.
\end{cases}
\end{align}
We have $(\mu^{(N)})_N
= \Lambda_0$.
We may apply  (\ref{KKF}) with $\mu = \mu^{(N)}$, which
 gives
\begin{align*}

\mathcal K(\Lambda_\g \!\ | \  \Lambda_0)
 -  \mathcal K(\Lambda_\g \!\ | \ \mu^{(N)}) = G_N(\mu^{(N)}) = \g \Re\!\ \tr\!\ \alpha_0 - \frac{\g}{2}\!\ \tr\!\ \alpha_0\alpha_0^\dagger - S_N(\mu)
\,.
\end{align*}
Since $\mu^{(N)}$ converges weakly to $\mu$, the lower semicontinuity of
 $\mathcal K (\Lambda_\g \!\ | \ \cdot)$ gives

\begin{align}
\notag
\mathcal K(\Lambda_\g \!\ | \ \Lambda_0) -  \mathcal K(\Lambda_\g \!\ | \  \mu)  \geq \mathcal K(\Lambda_\g \!\ | \  \Lambda_0)
 -  \liminf_N \mathcal K(\Lambda_\g \!\ | \  \mu^{(N)})\\
\label{sens1}
 \geq
\g \Re\!\ \tr\!\ \alpha_0 - \frac{\g}{2}\!\ \tr\!\ \alpha_0\alpha_0^\dagger - S_\infty(\mu)
\geq -\infty\,.
\end{align}

{\bf B)} If $\mathcal K(\Lambda_\g \!\ | \ \mu) = \infty$ the inequality
\begin{align}
\label{reverse}
\mathcal K(\Lambda_\g \!\ | \  \Lambda_0) -  \mathcal K(\Lambda_\g \!\ | \  \mu) \leq \g \Re \tr\!\ \alpha_0 - \frac{\g}{2}\!\ \tr\!\ \alpha_0\alpha_0^\dagger - S_\infty(\mu)
\end{align}
is trivial.

If  $\mathcal K(\Lambda_\g \!\ | \  \mu) < \infty$, then $\det w(\theta) > 0$ a.e. and then from (\ref{KKF}) we have $\det w_N(\theta) > 0$ a.s. too.
We want to let

 $N \to \infty$ in (\ref{KKF}) in order to get
(\ref{reverse}). To begin with,
let us  prove that
\begin{align}
\label{alphazero}
\lim_N \alpha_N(\mu) = 0\,.
\end{align}

From (\ref{KKF}) we deduce
\begin{align*}
G_N(\mu) \leq K(\Lambda_\g \!\ | \  \mu) < \infty\,,
\end{align*}
and then, since
\[-p \leq -\Re \tr\!\ \alpha_N +
\frac{1}{2} \tr\!\ \alpha_N\alpha_N^\dagger \leq \frac{3p}{2}\]
($p$ is the dimension)
we have $S_\infty (\mu) <\infty$.

Let us split the study into two cases:
\begin{enumerate}
\item
if $0\leq \g <1$,  $S_\infty(\mu) < \infty$ implies
\begin{align}
\label{A}
\sum_k  \tr\!\ \alpha_k\alpha_k^\dagger < \infty
\end{align}
hence (\ref{alphazero}) holds true.
\item if $\g =1$, we have
\begin{align}
\label{B}
\sum_k  \tr\!\ (\alpha_k\alpha_k^\dagger)^2+  \tr\!\ (\alpha_{k+1} - \alpha_k) (\alpha_{k+1}^\dagger - \alpha_k)^\dagger < \infty
\end{align}
which in particular implies that (\ref{alphazero}) holds true.
\end{enumerate}

This result has consequences for both sides of (\ref{KKF}). On the one hand, since for every $j$
\[\lim_N \alpha_j(\mu_N) = \lim_N \alpha_{N+j}(\mu) \rightarrow 0\,,\]
the sequence $(\mu_N)$ converges weakly to $\Lambda_0$, so
 using again the semicontinuity, we get
\begin{align}\notag
\mathcal K(\Lambda_\g \!\ | \  \Lambda_0) -  \mathcal K(\Lambda_\g \!\ | \ \mu) \leq \liminf_N \mathcal K(\Lambda_\g \!\ | \  \mu_N)-  \mathcal K(\Lambda_\g \!\ | \  \mu)\,.
\end{align}
On the other hand, from (\ref{3.6})
\begin{align}
\label{liminf}
\lim G_N(\mu)
= \g \Re\!\ \tr\!\ \alpha_0 - \frac{\g}{2}\!\ \tr\!\ \alpha_0\alpha_0^\dagger -\g S_\infty(\mu)\,.
\end{align}
and then  (\ref{reverse}) holds true also in this case.

Gathering (\ref{sens1}) and (\ref{reverse})
 ends the proof of (\ref{mainf2}) hence (\ref{mainf1})  when $0\leq \g \leq 1$.
\qedhere
\medskip

\section{Proof of Proposition \ref{4.1}}

We consider only the case $0 \leq \g\leq 1$, since for $-1< \g <0$ the reduction from $\g <0$ to $\gamma > 0$ as in  the beginning of Sect. 3 leads directly to the result.

We already saw in the above section, that when $\mathcal K(\Lambda_g \!\ | \  \mu) < \infty$ and $0\leq \g \leq 1$, the good conditions  are fulfilled.

Conversely, we consider three cases.

If $0 \leq \g < 1$ and (\ref{A}) is fulfilled, then
\[- \sum_k \log\det \alpha_k\alpha_k^\dagger < \infty\]
and since
\[ \tr\!\ (\alpha_{k+1} - \alpha_k) (\alpha_{k+1}^\dagger - \alpha_k)^\dagger \leq 2( \tr\!\ \alpha_k\alpha_k^\dagger+   \tr\!\ \alpha_{k+1}\alpha_{k+1}^\dagger)\,,\]
the expression $T(\alpha_0, \alpha_1, \cdots)$ in (\ref{otherT}) is well defined and finite, so is the left hand side of (\ref{mainf2}) and then $\mathcal K(\Lambda_g \!\ | \  \mu)$ is finite.

If $\g =1$, condition (\ref{B}), jointly with (\ref{logdet})
entails that
\[\sum_0^\infty -\log\det(
1- \alpha_k\alpha_k^\dagger) - \tr\!\  \alpha_k\alpha_k^\dagger + \frac{1}{2} \tr (\alpha_k - \alpha_{k+1})(\alpha_k^\dagger - \alpha_{k+1}^\dagger) < \infty\]
and then
 gathering
 (\ref{mainf2}) and (\ref{otherT}) show that
 $\mathcal K(\Lambda_g \!\ | \  \mu)$ is finite.

\section{Proofs of intermediate results}
\label{proofs}

\subsection{Proof of Theorem \ref{recursion}}

To compute the LHS of (\ref{2.29}) we need the values of the  Fourier coefficients : \[\int e^{\ii k\theta} \log\det \left(w(\theta) w_1(\theta)^{-1}\right) \frac{d\theta}{2\pi}\ \ \hbox{for} \  k=-1,0, 1\,.\]
The strategy is to approach  $\log\det \left(w(\theta) w_1(\theta)^{-1}\right)$ by a function of $z= re^{\ii \theta}$, sufficiently smooth to apply Cauchy's formula.

In view of Theorem \ref{DPS}, it is natural to approximate $w(\theta)(w_1(\theta))^{-1}$
 by $\Re F(z)(\Re F_1 (z))^{-1}$ with $z = re^{\ii \theta}$. We define the auxiliary matrix function:
\begin{align}
\label{2.6.7n-m}
D_0(z) := (\bdone-zf(z))^{-1}(\bdone-zf_1(z))\left(\rho_0^L\right)^{-1}\left(\bdone - f(z) \alpha_0^\dagger\right)\,.
\end{align}

We need the following formula whose proof  is  postponed in Sect. \ref{proofsl}.
\begin{lemma}
\label{FF1}
\begin{align}
\label{2.6.8-nm}
\det \left(\Re F(z)  \left(\Re F_1(z)\right)^{-1}\right) = \det  (D_0(z)D_0(z)^\dagger) \frac{\det (\bdone - |z|^2f(z)^\dagger f(z))}{\det (\bdone- f(z)^\dagger f(z))}\,.
\end{align}
\end{lemma}

From Theorem \ref{DPS}, for a.e. $\theta$ we have
\begin{align*}
\lim_{r\uparrow 1} \det \left(\Re F(re^{\ii \theta})  \left(\Re F_1(re^{\ii \theta}))\right)^{-1}\right) &= \det \left(w(\theta w_1(\theta)^{-1}\right)\\
\lim_{r\uparrow 1} \frac{\det (\bdone - |r|^2f(re^{\ii \theta})^\dagger f(re^{\ii \theta}))}{\det (\bdone- f(re^{\ii \theta})^\dagger f(re^{\ii \theta}))}&=1\,,
\end{align*}
so that,
\begin{align*}
\det(w(\theta)w_1(\theta)^{-1}) = \lim_{r\uparrow 1} \det(D_0(re^{\ii \theta})D_0(re^{\ii \theta})^\dagger)\,,
\end{align*}
and
 the remaining part of the proof  is based  on the study of $\det  (D_0(z)D_0(z)^\dagger)$. Some properties of $D_0$ are collected in the following lemma, whose proof is also in Sect. \ref{proofsl}.

\begin{lemma}
\label{2.6.2}
The function $\det D_0$ is analytic in $\mathbb D$ and non-vanishing. Moreover
\begin{align}
\label{Hp}
h := 2 \log\det D_0 \in
 H^2(\mathbb D)\,.
\end{align}
\end{lemma}

Since  $h\in H^2(\mathbb D) \subset H^1(\mathbb D)$, we have
\begin{align}
\label{Fh}
\int e^{-\ii \theta}h(e^{\ii \theta})\frac{d\theta}{2\pi} = h'(0)\ , \
 \int h(e^{\ii \theta})\frac{d\theta}{2\pi}= h(0)\ , \
\int e^{\ii \theta}h(e^{\ii \theta})\frac{d\theta}{2\pi} =0\,,
\end{align}
and then
\begin{align}
\label{3Fourier}
\int (1-\g \cos \theta) \Re\!\ h(e^{\ii \theta}) d\lambda_0(\theta) =\Re\!\  h(0) - \frac{\g}{2}\Re\!\ h'(0)\,.
\end{align}
Let us compute $h(0)$ and $h'(0)$. As $|z| \to 0$,
\begin{align}
\label{Taylorff1}
\det (\bdone-zf(z)) = 1- z (\tr\!\ \alpha_0) + O(z^2) \ , \ \det (\bdone -zf_1(z)) = 1 - z(\tr\! \ \alpha_1)  + O(z^2)
\end{align}

 Now, formula (\ref{Schur1}) can be inverted into
\begin{align}
\label{Schurinv}
f(z) = (\rho_0^R)^{-1}(\alpha_0 + zf_1(z))\left(\bdone + z \alpha_0^\dagger f_1(z)\right)^{-1} \rho_0^L\,,
\end{align}
which gives the expansion
\begin{align*}\notag
f(z)
&=  (\rho_0^R)^{-1}\left(\alpha_0 + z(\bdone-\alpha_0\alpha_0^\dagger) f_1 (z) + O(z^2)\right)\rho_0^L\\
&= \alpha_0 +  z (\rho_0^R
\alpha_1\rho_0^L) +  O(z^2)\,,
\end{align*}
so that
\begin{align*}\notag
\bdone - f(z)\alpha_0^\dagger &= \bdone-
\alpha_0
\alpha_0^\dagger - z(\rho_0^R
\alpha_1\rho_0^L\alpha_0^\dagger)   +  O(z^2)
= (\rho_0^R)^{2}- z(\rho_0^R\alpha_1\alpha_0^\dagger \rho_0^R)  +  O(z^2)\\
&= \rho_0^R\left(\bdone - z(\alpha_1\alpha_0^\dagger)  +  O(z^2) \right)\rho_0^R
\end{align*}
and
\begin{align}\notag
\det\left(\bdone - f(z)\alpha_0^\dagger\right) &=\det (\rho_0^R)^2 \det\left(\bdone - z(\alpha_1\alpha_0^\dagger)  +  O(z^2)
 \right)\\
\label{1-falpha}
&= \det (\rho_0^R)^2 \left(\bdone - z\!\ \tr\!\ (\alpha_1\alpha_0^\dagger) +  O(z^2)\right)\,.
\end{align}
Gathering (\ref{2.6.7n-m}), (\ref{Taylorff1})  and (\ref{1-falpha}) and using $\det \rho_0^R =  \det \rho_0^L$ we get
\begin{align*}
\det D_0 (z) =(\det \rho_0^R)
\left(1 -z\!\ \tr\!\ \left( \alpha_0 -  \alpha_1  - \alpha_1\alpha_0^\dagger \right)+ O(z^2)\right)
\end{align*}
Coming back to the definition of $h$, we get
\begin{align*}
h(z) = \log\det (\bdone- \alpha_0\alpha_0^\dagger)
 - 2 z\!\ \tr\!\ \left( \alpha_0 -  \alpha_1  - \alpha_1\alpha_0^\dagger \right)
 + O(z^2)
\end{align*}
and from (\ref{3Fourier})
\begin{align*}
\int(1 - \g \cos\theta) \Re\!\ h(e^{\ii \theta}) d\lambda_0(\theta) = \log\det (\bdone- \alpha_0\alpha_0^\dagger) + \g\Re\!\ \tr\!\ \left( \alpha_0 -  \alpha_1  - \alpha_1\alpha_0^\dagger \right)\,.
\end{align*}
\qedhere

\subsection{Proof of Lemma \ref{FF1}}
\label{proofsl}

To simplify, we omit the variable $z$ if unnecessary.
Applying (\ref{3.52-o}) to $F_1$
\begin{align}
\label{3.52-1}
\Re F_1  = (\bdone- \bar zf_1^\dagger)^{-1}(\bdone - |z|^2f_1^\dagger f_1) (1 -zf)^{-1}
\end{align}
  so we need an expression of $\bdone - |z|^2 f_1f_1^\dagger$ as a function of $f$. From (\ref{Schur1}) we get
\begin{align*} |z|^2f_1(z)^\dagger f_1(z) =&\\  \rho_0^L\left(\bdone- f(z)^\dagger \alpha_0 \right)^{-1}&(f(z)^\dagger - \alpha_0^\dagger)(\rho_0^R)^{-2}(f(z) - \alpha_0)\left(\bdone- \alpha_0^\dagger f(z)\right)^{-1}\rho_0^L
\end{align*}
which,
with the help of the trivial identity
\begin{align}
\notag
\bdone &=   \rho_0^L\left(\bdone- f(z)^\dagger \alpha_0 \right)^{-1}\left(\bdone- f(z)^\dagger \alpha_0 \right)(\rho_0^L)^{-2}\left(\bdone- \alpha_0^\dagger f(z)\right)\left(\bdone- \alpha_0^\dagger f(z)\right)^{-1} \rho_0^L \,,
\end{align}
yields
\begin{align}
\notag
\left(\bdone- f^\dagger \alpha_0 \right)(\rho_0^L)^{-1}
\left(\bdone- |z|^2f_1^\dagger f_1\right)(\rho_0^L)^{-1}\left(\bdone- \alpha_0^\dagger f\right)  &=
\\
\label{5.10} \left(\bdone- f^\dagger \alpha_0 \right)(\rho_0^L)^{-2}\left(\bdone- \alpha_0^\dagger f\right)-
(f^\dagger - \alpha_0^\dagger)(\rho_0^R)^{-2}&(f - \alpha_0)\,.
\end{align}
Now, we use (\ref{defect})
and
\[
 (\rho_0^R)^{-2} = \sum_{n\geq 0} (\alpha_0\alpha_0^\dagger)^n \ , \  (\rho_0^L)^{-2}
 = \sum_{n\geq 0} (\alpha_0^\dagger\alpha_0)^n\]
($\alpha_0$ is a contraction).
Expanding the RHS of (\ref{5.10}) and cancelling terms gives
\[\left(\bdone- f^\dagger \alpha_0 \right)(\rho_0^L)^{-2}\left(\bdone- \alpha_0^\dagger f\right)-
(f^\dagger - \alpha_0^\dagger)(\rho_0^R)^{-2}(f - \alpha_0) = \bdone- f^\dagger f\]
so that
\begin{align}
\label{2.6.6}
\bdone -| z|^2f_1^\dagger f_1 = \rho_0^L\left(\bdone- f^\dagger \alpha_0 \right)^{-1}\left(\bdone- f^\dagger f\right)\left(\bdone- f \alpha_0^\dagger \right)^{-1}\rho_0^L\,.
\end{align}
Plugging into (\ref{3.52-1}) yields
\begin{align*}
&\Re F_1\\ &= (\bdone - \bar zf_1^\dagger)^{-1} \rho_0^L\left(\bdone- f(z)^\dagger \alpha_0 \right)^{-1}\left(\bdone- f(z)^\dagger f(z)\right)\left(\bdone- f(z) \alpha_0^\dagger \right)^{-1}\rho_0^L(\bdone - zf_1)^{-1}
\end{align*}
and
\begin{align}\nonumber
&(\Re F)  \left(\Re F_1\right)^{-1}= (\bdone- \bar zf^\dagger)^{-1}(\bdone - |z|^2f^\dagger f) (\bdone -zf)^{-1}  \\
\notag &\times(\bdone - zf_1)\left(\rho_0^L\right)^{-1}\left(\bdone- f \alpha_0^\dagger \right)\left(\bdone- f^\dagger f\right)^{-1}
\left(\bdone- f^\dagger \alpha_0 \right)\left(\rho_0^L\right)^{-1}(\bdone - \bar z f_1^\dagger)\\
\notag
&= (\bdone- \bar zf^\dagger)^{-1} (\bdone - |z|^2f^\dagger f) D_0\left(\bdone- f^\dagger f\right)^{-1}D_0^\dagger(1-\bar zf^\dagger)\,.
\end{align}
Then, taking determinants
\begin{align}
\notag
\det \left((\Re F)  \left(\Re F_1\right)^{-1}\right) = \det  (D_0D_0^\dagger) \ \frac{\det (\bdone - |z|^2f^\dagger f)}{\det (\bdone- f^\dagger f)}\,,
\end{align}
ends the proof.

\subsection{Proof of Lemma \ref{2.6.2}} We  repeat here the argument of
Theorem 2.6.2 in \cite{Simon-newbook} for the sake of completeness.
For  $z \in \mathbb D$ we have $f(z)f^\dagger(z) < \bdone$ , hence analyticity and non-vanishing are straightforward. Moreover, since  $|\zeta| < 1$ implies $|\arg(1-\zeta)| < \pi/2$,
we conclude from (\ref{2.6.7n-m}) and (\ref{Hp}) that
\[ |\Im h|< 3\pi/2\,.\]
Since $|h|^2 - 2(\Im h)^2$ is harmonic we have
\begin{align}\notag\int |h|^2  d\lambda_0- 2 \int (\Im h)^2 d\lambda_0 = |h(0)|^2 - 2 \left(\Im h(0)\right)^2\,,
\end{align}
and since $h(0) = \log\det(1 - \alpha_0\alpha_0^\dagger)  < 0$, we get
\[\int|h(re^{\ii \theta})|^2 d\lambda_0(\theta) \leq \frac{9\pi^2}{2}+\left(  \log\det(1 - \alpha_0\alpha_0^\dagger)\right)^2\,,\]
which yields
\[\sup_{r <1} \int|h(re^{\ii \theta})|^2 d\lambda_0(\theta)  < \infty\,.\]
\bibliographystyle{plain}
\bibliography{bibil}
\end{document}

config: arxiv


TITLE: Getting nervous refereeing a paper
QUESTION [44 upvotes]: I am refereeing my first paper and I'm quite excited! But inexperienced and I would like to ask an advice to the Maths Community of MO. Let me tell you that I have already read Refereeing a Paper, but it seems that my question is quite different. Roughly:

What is the point after which you get nervous while refereeing a paper?

Specifically:

I've found many English mistakes. (Well, I'm not a native English and so I can understand. So I am not nervous yet)
I've found some maths inaccuracies like "let A be any set"... and then I have discovered that the proof of the first result works only for finite sets. (OK, those are only inaccuracies - I am not nervous yet)
There are many references like "we use the notation of [X]", "this result is proved in [X]", where [X] is a BOOK, without specifying a precise section, or the number of the result... should I get this book and read all to find out the correct references? - just thinking of it, makes me a bit nervous..
(most importantly). There is a mathematical more serious mistake. Something that might be fixed, but not obviously (my definition of obvious is three evenings, in this case). I am not saying that the paper is completely wrong but that now... now I'm getting nervous!

Now, taking into account that the person who asked me to referee this paper told me: be selective, we accept only 20% of submitted papers,

what should you do in these cases? Reject? Ask for a revision? Not getting nervous and try to see if the rest of the piece is good (I'm quite a good guy and I'm doing that at the moment)?

Of course I will talk with the editor, but I also would like to know more opinions that might be helpful in future.
Update: Thank you very much to everybody for the numerous and helpful comments.
Valerio

REPLY [13 votes]: I like the comment of Francis Crick in `What mad pursuit' that one should always start a referee report with praise, but of course not hold back on the detailed criticism.
I have found that a referee can also help the author by asking for clarification in the Introduction of the aims of the paper, what does it accomplish in relation to previous work in the area, and what questions are still open. The author may also not have got the title right in relation to the expectations of readers: some of my early papers had lousy titles which not not convey easily what was in the paper.
It is useful to know that the referee is not responsible for the correctness of the paper: that is the responsibility of the author. Of course, if the referee has doubts on the correctness,  these must be raised.
I am unhappy about remarks such as "there are doubts about the interest of this area", since some papers are of interest precisely because they go against the so-called `mainstream', which, to mix a metaphor, flaps about like a sail in a gusty wind.
The job of referee is very important, and I wish you luck and interest in this.


TITLE: Maximum likelihood question with exponential distribution, simple calculation
QUESTION [1 upvotes]: I cannot seem to get he same answer as the book - book answer is: 0.122

Use the sample, (8.2, 9.1, 10.6, 4.9) to calculate the maximum likelihood estimate for $\lambda$ in the exponential pdf.

$$f_Y(y;\lambda)= \lambda e^{-\lambda y} \;, \quad y \ge 0$$
$$L(\lambda) = \prod_{i=1}^{n}\lambda e^{-\lambda y}$$
$$=\lambda^n e^{-\lambda \sum_{i=1}^{n}y_i}$$
$$\text{Let }  T = \frac{d\big[ln(L(\lambda))\big]}{d\lambda} = \frac{n}{\lambda} - \lambda \sum_{i=1}^{n}y_i$$
so i sum the y's to get 32.8, let T equal zero and n = 4
$$\lambda(32.8) = \frac{4}{\lambda}$$
$$\lambda^2 = \frac{4}{32.8}$$
$$\lambda = \sqrt{\frac{4}{32.8}}$$
$$\lambda = 0.349$$
I have definitely got those four sample values right. I have done this sum three times now. Can anybody see where i'm going wrong?

REPLY [2 votes]: Your derivative is wrong. The sum should not have a $\lambda$ next to it. The correct loglikelihood derivative is
$$\frac{dL}{d\lambda} = \frac{n}{\lambda} - \sum y_i$$
setting equal to zero then yeilds
$$\hat{\lambda} = \frac{n}{\sum y_i} =\frac{1}{\bar{y}}$$
Plugging in your values yields $$\hat{\lambda} = \frac{1}{8.2} \approx 0.122$$


TITLE: Why proximal gradient instead of plain subgradient methods for LASSO?
QUESTION [1 upvotes]: I was thinking to solve LASSO via vanilla subgradient methods. But,I have read people suggesting to use Proximal GD. Can somebody highlight why proximal GD instead of vanilla subgradient methods be used for LASSO?

REPLY [0 votes]: OK, since you liked my comments... :-)

Better performance. That's it.
Longer, but more handwaving, answer: proximal methods allow you to use a lot more information about the function at each iteration. Also, proximal gradient methods take into account a much larger neighborhood around the initial point, enabling longer steps.
Third, practical answer: you should not be implementing your own LASSO solver. There are too many good implementations out there. If you need to port one from one language to another, that's one thing, but to write one from scratch at this point makes no sense unless you have very specific problem structure you need to exploit.


TITLE: What is $\int e^{-\frac{(y-\mathbf{x}^T\mathbf{\theta})^2}{2}}d\mathbf{\theta}$?
QUESTION [0 upvotes]: $y \in \mathbb{R}, \mathbf{x} \in \mathbb{R}^2$ are constants, and $\mathbf{\theta} \in [-5,5]\times[-5,5] \subset \mathbb{R}^2$.
I'm just not familar with vector integrals, any reference to read?
EDIT: I'm sorry, I missed the square. It was actually part of a Gaussian, $\mathcal{N}(y|\mathbf{x},\mathbf{\theta})$, and I want to integrate w.r.t $\mathbf{\theta}$.

REPLY [1 votes]: Writing $x^T\theta = x_1 \theta_1 + x_2 \theta_2$, your integral is separable into multiplicative components involving $\theta_1$ alone and $\theta_2$ alone, so you can just use standard integration by integrating with respect to each variable $\theta_i$ in order, to get the indefinite integral.

REPLY [0 votes]: $x^{T}\theta$ just signifies the inner product between $\theta$ and $x$, so
$$\int_{[-5,5]^{2}} e^{-\frac{y-x^{T}\theta}{2}}\;d\theta=\int_{[-5,5]^{2}} e^{-\frac{y-(x_{1}\theta_{1}+x_{2}\theta_{2})}{2}}\;d\theta=\int_{-5}^{5}e^{\frac{\theta_{1}x_{1}-y}{2}}\;d\theta_{1}\int_{-5}^{5}e^{\frac{\theta_{2}x_{2}-y}{2}}\;d\theta_{2}.$$


TITLE: Poincaré hyperbolic geodesics in half-plane and disc models including outer branch
QUESTION [8 upvotes]: The objective of this post is to state that:
1) the Poincaré  hyperbolic metric results in a solution of complete geodesic circles in both half-plane and disk models.
2) the choice of one or other is quite arbitrary.
3) while for the half plane model negative side arbitrariness provides no new solution or insight, the disc model hyperbolic geodesic proposed outside the horizon/boundary can be validated in full.
EDIT1:
The derivation of geodesics of the two models follows in a straightforward  manner from the metric.
For the half-plane we have in Cartesian coordinates
$$ ds^2 = (dx^2 + dy^2)/y^2  \tag{1} $$
Employing Euler-Lagrange (from calculus of variations approach )(primes wrt x)
$$ s = \int \dfrac{\sqrt{ 1 + y^{'2}}}{ y } dx,\,\tag{2}  $$
the functional is $$ F =\dfrac{\sqrt{ 1 + y^{'2}}} { y } \tag{3}$$
$$ F - y^{'}\dfrac {\partial F }{ \partial y^{'} }  = 1/c  \tag{4}  $$
where c is an arbitrary constant leading  to (since $ y^{'}= \tan\phi,$ $\phi =$ inclination to x-axis),
$$ \cos \phi = \frac{y}{c} \tag{5}  $$
$$ \frac{dy}{dx}= \frac{\sqrt{c^2-y^2}}{y} \tag 6 $$
Let $ y^2= v,\, 2y dy = dv $ Integrating $$-2 \sqrt{c^2-v} = 2x-2h \tag7$$
where $2h$ is an integration constant.
$$ (x-h)^2 + y^2 = c^2 \tag8 $$
which are all all full circles centered on x-axis and lying on either side of x-axis. Argument about distances. Whether we approach from below the x-axis or from above, x-axis is at an infinite distance so lower symmetric half part may have been excluded (by Poincaré). Distance to points on x-axis is infinite whether we take the upper half or the lower, it appears to be entirely arbitrary. I.e, either of them could serve as a valid geodesic model.
But **the disc model has no such direct reflective symmetry, but possesses definitely an "inversion symmetry" arising from same metric  **.
Similarly for the disc in Polar coordinates (primes with respect to $\theta $) we land into a situation for an arbitrarily forced choice of range.
$$ s = \int \dfrac{\sqrt{ r^2 + r^{'2}}}{ (T^2-r^2) } d\theta,\, \tag{9} $$
Where $T$ is a more convenient symbol for the radius of fundamental circle than unity as we shall see..
$$ \dfrac{\sqrt{ r^2 + r^{'2}}}{ (T^2-r^2) } - \dfrac{ r{'}}  {( T^2- r^2)  }  \cdot \dfrac{r^{'}}  {\sqrt{ r^2 + r^{'2}}} = C_1 \tag{10} $$
and after error indicated by $A.\Gamma.$ corrected it leads to
$$ \frac{r^2}{T^2-r^2}\cdot\frac{1}{\sqrt{r^2+r'^2}}=C_1.\tag{11} $$
Using substitutions of (12) and (14) we get Equation (15) that is an ODE of product of segments constant property for an entire circle. At tangential point $T$ where $\psi =0,  \pi $  we have the horizon/boundary circle $r=T.$
which  are orthogonal red trajectories to unit "horizon" circles obtained by integrating this ODE. However they are these orthogonal hyperbolic geodesic circles on either side of the horizon or boundary.

It is well known that the half plane model is a special case of disc model where unit disc radius tends to $ \infty$.
Why Poincaré had deemed the portion of the red circle lying  outside the horizon as not belonging to $ \mathbb H^2 $ .. is to me not clear, so the question here.
EDIT 2:
( .. In response to (since deleted) workaholic's request for showing the above to be hyperbolic geodesics by integration)
Instead of integrating an eccentric circle in polar coordinates (8) can be differentiated to directly obtain pure formula of curvature in polar coordinates in $\mathbb R^2$.
$$ \kappa_g = \dfrac{r^2 + 2 r^{'2 }- r r ^{''}}{{(r^2 +  r^{'2}})^{3/2}} = 2 C_1 $$
Direct integration is needlessly cumbersome and when direct methods are yielding the result readily.
A second way is the following. In the diagram above let
$$ OT =1=T, K=-1/T^2, MCB =\psi, OA =r, BC =a ,..$$
$$ AM = MB = a \sin \psi, C_1 = 1/(2 a)\tag{12}$$
where $a$ is another convenient arbitrary constant.
Note that the central circle has a radius $=T$ which is tangent to all arbitrary circles of radius $a$. If inversion circle/mirroring circle has radius $T,$ the inner hyperbolic geodesics are inversions of outer geodesic circles (and vice-versa)  now brought into picture.
$$ \dfrac{r}{T^2-r^2} \cdot \dfrac{1}{\sqrt{1 + u^2 }} = \dfrac{1}{2 a} \tag{13} $$
where by geometry of differential right triangle
$$ u = \cot \psi = \dfrac{dr} {r d \theta }\tag{14} $$
$$ 2 a r \sin \psi = T^2 - r^2 $$
$$ r ( r+ 2 a \sin \psi) = T^2 \tag{15}$$
$$ OA \cdot OB = OT^2 = T^2 \tag{16}$$
which relation is readily recognized as the Circle property that product $T^2$  of the two cutting segments on circle is square of tangent $T$ or its power.
Alternately in a third method we can get the above into Liouville's form for $2D$ curvature by differentiation with respect to arc and simplifying:
$$ \psi^{'} + \dfrac{\sin \psi}{r} = \dfrac{1}{a}  \tag{17} $$
Yet a fourth way to compute geodesics from Poincaré metric (9) is to take the object function for geodesic
$$ \dfrac{\sqrt{ r^2 + r^{'2}}}{ (T^2-r^2) } \,\tag{18}$$
and apply Euler-Lagrange theorem of calculus of variations
$$ \dfrac{\sqrt{ r^2 + r^{'2}}}{ (T^2-r^2) } - \dfrac{ r{'}}  {( T^2- r^2)  }  \cdot \dfrac{r^{'}}  {\sqrt{ r^2 + r^{'2}}} = 1/2a \tag{19}  $$
where wlog the constant is taken as reciprocal of diameter $2a.$ Simplifying,
$$ \dfrac{r}{ (T^2-r^2) } =\dfrac {\sqrt{ r^2 + r^{'2}}}{2a} \tag {20}$$
Since in differential form we can plug in from (14)
$$ \cot \psi= \dfrac {r^{'}}{r} \tag {21} $$
(20) simplifies to an eccentric circle radius $a$ and power $T$ which is the arbitrary hyperbolic geodesic:
$$ \dfrac{\sin \psi}{r}= \dfrac{1}{2a}\cdot \dfrac{r^2-T^2}{r^2} \tag {22} $$
BTW, the name geodesic has been introduced by Liouville.
External geodesics mentioned here are inversions of internal geodesics with respect the boundary of unit radius horizon circle as mirror of inversion.
$$ r_{outer geodesic} \rightarrow T^2/ r_{inner geodesic} \tag {23} $$
Thus we have circles of arbitrary radius $a$ orthogonally cutting $T$ radius  as valid Poincaré disk model in $ \mathbb H^2 $ on either side of the boundary circle which is hence no truncated geodesic arc in this sense, considered  from Poincaré's defining metric. Either side/segment is equally valid.
All points outside the horizon are at an infinite distance to the horizon along the given external geodesics $TQB$ as well.
The outside hyperbolic geodesics mentioned here are indicated in red to complement with internal purple Poincaré hyperbolic geodesics around a unit (T=1) "boundary" in magenta.
EDIT 3:
Distance of nearest (inner) and farthest (outer) hyperbolic geodesics from any point
For a constant curvature $-1 /T^2$  Poincaré disk (Klein-Beltrami)  model metric hyperbolic distance $ (0,0) - (r,0) $
$$ \frac{4 \,ds^2}{(T^2-r^2)} \tag{24}$$
$$ s=T^2\int_0^r \frac{2 dr}{T^2-r^2}= T \log\frac {T+r}{T-r} =2\,T \tanh^{-1}(r/T) \tag{25} $$

Finding Euclidean distances in the eccentric circle orthogonal to the hyperbolic boundary circle as:
$$r^2-2 a r \sin \psi -T^2 =0,    \quad, r (r-2 a \sin \psi)=T^2 \tag{26}  $$
Roots of quadratic:
$$r_{1,2}= a \sin \psi \pm \sqrt{(a \sin\psi)^2 +T^2} $$
Taking positive radius and plug in from above using $(h^2=a^2+T^2)$
$$r_{1,2}= a \sin \psi \pm \sqrt{h^2-(a \cos\psi)^2},\quad r_1\cdot r_2=T^2  \tag{27} $$
$$\frac{ s_{1,2}}{T}= log \,\frac{T+\sqrt{h^2-(a \cos \psi)^2}+a \sin \psi}{T -\sqrt{h^2-(a \cos \psi)^2}-a \sin \psi} \tag{28} $$
So the metric can be defined to work both ways
$$ \pm \frac{4 \,ds^2}{(T^2-r^2)} \tag{29}$$
Points on the present new outer hyperbolic locus points  $(B,Q)$ correspond to the conventional inner hyperbolic locus points  $(A,P)$ of disc model obtained by mutual inversions.
A diameter of the boundary circle maps to line at infinity in inversion.
Their Euclidean distances are $ (h-a, h, h+a) $ respectively,  for the inner/outer extreme distant hyperbolic points.
Plugging in from (27) these hyperbolic distances are respectively:
$$ T \,log \frac{T+h+a}{T-h-a},\quad \infty,\,T \,log \,\frac{T+h-a}{T-h+a} \tag{30}$$
Cross-Ratio is same for inner and outer hyperbolic arcs
Cross Ratio of a pencil of four concurrent rays depends upon angles enclosed between them...as it is one valid way to define them. Whether the transverse cutting is by a straight line, a circle cutting them inside the tangent points as in case of standard Poincaré  model or cutting them outside of the tangent points as in the model presently being proposed is hence one and the same.
When the Poincaré arc/geodesic shrinks to zero size, then in the limit a circle passing through the pencil point of concurrency on its arc anywhere stands as a valid and simple hyperbolic geodesic. It forms a special case when $T=0,\sin \psi= r/2a$ ... and yet another special case when touching externally..

REPLY [2 votes]: The mistake is after the words "leading to" where the square root in the denominator is missing. It should be
$$
\frac{r^2}{1-r^2}\cdot\frac{1}{\sqrt{r^2+r'^2}}=C_1.
$$
P.S. Now it is much more fun integrating this guy :-)