TITLE: First-order Euclidean Geometry
QUESTION [9 upvotes]: I gather that the axiomatisation of Euclidean Geometry, e.g., by Hilbert, can be used with First-order Logic and has good properties.  Is it fairly complete in the sense, say, that it proves everything Euclid did? E.g., Eudoxus' theory of the reals and the area of the circle....

REPLY [9 votes]: Hilbert's axiomatization is second-order. However, there is a first-order axiomatization by Tarski, which is complete. (And because the theory is recursively axiomatized, it is decidable.)
We cannot develop the full second-order theory of real numbers within the Tarski geometry. For instance, the "plane" $\mathbb{A}^2$, where $\mathbb{A}$ is the field of real algebraic numbers, is a model of Tarski's geometry.  The Eudoxean theory of equality of ratios, which comes so strikingly close to the modern definition of real number, cannot be developed within Tarski's geometry, for the set of integers is not definable in that geometry.
However, for example all of Book I consists of theorems of Tarski's geometry.
For a complete list of Tarski's axioms, with additional information and references, please see this.

REPLY [3 votes]: No, Pythagorean planes, i.e. planes satisfying Hilbert's fourteen axioms up through the Archimedian axiom), form a strictly weaker geometry than Euclidean planes.
Hilbert planes correspond to constructions with straightedge and "segment transporter," which almost but can't quite do the job of a straightedge and a compass. Because of this, I get the impression that you cannot do as much with circles as you can in Euclidean geometry. You'll notice a distinct absence of circles in Hilbert's Foundations of geometry.
Old and New Results... by Greenberg gives a particularly helpful summary of how they are related.


TITLE: Integrating by substitution containing a further factor problem
QUESTION [0 upvotes]: The problem:
$$ {\int } (x+1)(3x+1)^9  dx $$
let u = 3 x +1
3 x = u - 1
$ x = \frac{1}{3} (u-1) $
Hence,
$ x + 1 = \frac{1}{3} (u-1) + 1$
$  = \frac{1}{3} (u+2) $  This line here I do not understand.
The + 1 become a 3? How does this happen?

REPLY [1 votes]: Hint:
$$
x+1=\frac{1}{3}(u-1)+1 \quad \iff \quad x+1=\frac{1}{3}u -\frac{1}{3}+1=\frac{1}{3}u+\frac{2}{3}
$$


TITLE: Fermionic ghost path integral results in $\delta$ function?
QUESTION [2 upvotes]: This is related to a statement in pg 20 of hep-th/9408074 formula (2.39).
Suppose $$\mathcal{L}\sim\frac{i}{\lambda^{\prime}}\bar{\eta}^xg_{ij}U_x{}^i\psi^j+\cdots \tag{2.35}$$where $\bar{\eta}$ to my guess is ghost field as it is non-dynamical and assume $\cdots$ does not contain contribution of $\bar{\eta}$. Consider $\int d\eta e^{\mathcal{L}}$.
The paper says integration of $\eta$ gives $$\left(\frac{-i}{\lambda^{\prime}}\right)^t\delta(g_{ij}U_x{}^i\psi^j)\tag{2.39}.$$
$\textbf{Q:}$ How do I see this does give rise to Dirac delta function? I had expanded the exponential function to perform fermionic integral but this does not give me $\delta(g_{ij}U_x^i\psi^j)$ but $\sim U_x^i\psi^j$. Normally, Faddeev-Popov ghosts involve 2 fermions (ghost and anti-ghost) to perform gauge fixing. What is the argument here?

REPLY [2 votes]: Eq. (2.39) is a $t$-dimensional Grassmann-odd delta function$^1$
$$\prod_{x=1}^t \delta(\frac{i}{\lambda^{\prime}} g_{ij}U_x{}^i\psi^j)~=~\prod_{x=1}^t  \int \! d\eta^x~\exp\left\{ \frac{i}{\lambda^{\prime}}\eta^x g_{ij}U_x{}^i\psi^j\right\}~=~\prod_{x=1}^t \frac{\pm i}{\lambda^{\prime}} g_{ij}U_x{}^i\psi^j .\tag{2.39}$$
Recall that if $f$ is an arbitrary function of a Grassmann-odd variable $\theta$, then Berezin integration yields$^2$
$$\int d\theta~(\pm\theta)~f(\theta)~=~f(0), $$
which formally satisfy the defining property
$$\int d\theta~\delta(\theta)~f(\theta)~=~f(0) $$
of a Dirac delta distribution! So we can identify
$$\delta(\theta)~=~\pm\theta.$$
References:

C. Vafa & E. Witten, arXiv:hep-th/9408074; eqs. (2.35) & (2.39).

--
$^1$ There is a typo in the action (2.35) of Ref. 1: The Grassmann-odd $\bar{\eta}$ variable in the fifth term should be an $\eta$ variable.
$^2$ The $\pm$ sign denotes different Berezin conventions $\int d\theta~\theta~=\pm 1 $.


TITLE: Weak convergence + pointwise convergence on dense subspace of $L^2$
QUESTION [3 upvotes]: Let $H$ be a dense subspace of $L^2(X,\mu)$ ($X$ being a separable and complete metric space with finite measure $\mu$) and $(f_n)_{n\in\mathbb N}$ be a sequence in $H$ that converges to $f\in L^2(X,\mu)$ in the following sense:
$$
\langle f_n - f, h \rangle_{L^2} \xrightarrow{n\to\infty} 0
\quad
\text{for all } h\in H.
$$
In addition, let $f_n$ converge pointwise $\mu$-almost everywhere to some function $g$, i.e. $f_n(x)\xrightarrow{n\to\infty} g(x)$ for $\mu$-almost every $x\in X$.
I need to show that $f$ and $g$ agree $\mu$-almost everywhere, in other words $f_n(x)\xrightarrow{n\to\infty} f(x)$ pointwise for $\mu$-almost every $x\in X$.
I have been struggling to prove this (and also to find counterexamples), but so far I cannot find the right analytical tools. Any help will be very appreciated!
Remark: I know very little about the $f_n$, there may be no upper bound on their norm etc.
Remark: The background is that I am trying to prove a convergence result for conditional mean embeddings in reproducing kernel Hilbert spaces (this is not an homework exercise).

REPLY [3 votes]: Another counterexample for $L^2(-\pi,\pi)$:

Lemma If $Z$ is an inner product space, $x_1,\dots,x_n\in Z$ are independent, and $c_1,\dots,c_n$ are scalars then there exists $x\in Z$ with $\langle x,x_j\rangle=c_j$ for $j=1,\dots, n$.

Proof: By linearity (and symmetry) we can assume $c_1=\dots=c_{n-1}=0$, $c_n=1$. Let $X$ be the span of $x_1,\dots,x_n$ and let $Y$ be the span of $x_1,\dots,x_{n-1}$. Since $Y$ is a proper subspace of $X$ there exists $x\in X$ with $x\ne0$ but $\langle x,x_j\rangle=0$ for $j=1,\dots,n-1$.  Now $x\ne0$ implies that $\langle x, x_n\rangle\ne0$.

Cor. Given $\delta>0$, a positive integer $N$ and complex numbers $c_{-N},\dots,c_N$ there exists $g\in  L^2(-\pi,\pi)$ such that $g$ vanishes off $(0,\delta)$ and $\hat g(k)=c_k$, $-N\le k\le N$.

Now let $F$ be any non-zero element of $L^2(-\pi,\pi)$, and choose $g_n$ so $g_n$ vanishes off $(0,1/n)$ and $$\hat g(k)=-\hat F(k),\quad(|k|\le n).$$Let $$f_n=F+g_n.$$Then $f_n\to F$ almost everywhere, but $$\lim_n\hat f_n(k)=0\quad(k\in\Bbb Z),$$hence $$\lim_n\langle f_n, h\rangle=0$$for every trigonometric polynomial $h$.


TITLE: Has the speed of gravity been measured experimentally? How?
QUESTION [2 upvotes]: In Newtonian physics, changes in gravity propagate instantly. In general and special relativity, gravity propagates at the speed of light, $c$. From reading answers to questions about gravity on this site, the consensus seems to be that the real world fits the relativity answer, $c$.
Numerous experiments have measured the speed of light in various media with increasing precision. Have there been any experimental measurements of the speed of gravity with generally accepted results?
If so, how did they work? Or if not, how might such a measurement be performed and why hasn't it been done yet?

REPLY [1 votes]: When you talk about the speed of gravity, you are talking about the speed of gravitational waves. If something jerks to one side or changes shape in a way that the change in gravity could be detected at a distance, the progress of that change through space is nothing other than a gravitational wave.
They have just recently been detected by LIGO, but I don't think that was enough for a direct measurement of the speed. It would be possible to measure the speed in a straightforward way if a gravitational pulse were observed coming from something that could be detected by other means at a known distance. I don't know if a supernova would be expected to produce detectable gravitational waves, for instance.
So, direct measurement has not happened, but probably will some day. Others can do a better job than me of talking about indirect measurements.


TITLE: Proving statement about sequences
QUESTION [0 upvotes]: I have the following exercise:

Let $a_0 > b_0 > 0,$, with $ a_0, b_0 \in \mathbb{R}$. We define $a_n$
  and $b_n$:
$$ a_{n+1} = \frac{a_n+b_n}{2} $$ $$ b_{n+1} = \sqrt{a_n b_n} $$
Prove the following:

$a_n \ge b_n \forall n$.
$a_n$ is decreasing and $b_n$ is increasing.

Question 2 is simple enough, it's question 1 I'm having trouble with. I tried to use induction: I first show that for $n= 0$ the inequality is true, and then, assuming $a_n \ge b_n$, I want to show that $a_{n+1} \ge b_{n+1}$, i.e., $\frac{a_n+b_n}{2} \ge \sqrt{a_n b_n}$
I have tried for a while but haven't managed to get anything. Does anyone have any suggestions?

REPLY [1 votes]: Hint:
Arithmetic Mean $ \geq $  Geometric Mean

REPLY [1 votes]: It is consequence of the inequality $$\frac{a^2+b^2}{2}\geq ab,\quad\tag{1}$$
(which holds for any $a,b\in\mathbb{R},$ because you can rewrite it as the triviality $(a-b)^2\geq 0.$)
Infact $a_{n+1}=\frac{a_n+b_n}{2}\geq \sqrt{a_nb_n}=b_{n+1}$


TITLE: Notation for defining a set of distinct elements.
QUESTION [2 upvotes]: Suppose I write the following.

Let $X = \{x,y,z\}.$

Then its pretty clear that what I really mean is the following.

Let $x,y$ and $z$ be fixed but arbitrary; suppose they're distinct; and let $X = \{x,y,z\}$.

Not that its really necessary or anything, but if there is an established convention for emphasizing that what I really mean is the second of these, then I'd like to start using it. Something like the following.

Let $X \equiv \{x,y,z\}.$

Does any such notational convention currently exist?

REPLY [5 votes]: I think I've seen the notation $X=\{x,y,z\}_\ne$ used to indicate that $X$ is the set containing the distinct elements $x,y,z$; probably in some old paper by Erdős, Rado, and Hajnal, or some subset thereof. I know for a fact that the notation $\{x_0,x_1,\dots\}_\lt$ has been used to denote the set $\{x_0,x_1,\dots\}$ while expressing the fact that $x_0\lt x_1\lt\dots$; this notation is introduced on p. 428 of P. Erdős amd R. Rado, A partition calculus in set theory, Bull. Amer. Math. Soc. 62 (1956), 427-489. Most likely the notation was Rado's invention. I don't know to what extent it has caught on.
I think Rado's(?) notation has the advantage over your $X\equiv\{x,y,z\}$ in that the meaning is easier to remember or guess; also you can simply write $\{x,y,z\}_\ne$ without necessarily having "$X=$" in front of it.
Update. Quoting from L. Mirsky, Transversal Theory, Academic Press, 1971, p. 2:

We shall use the symbol $\{x_1,\dots,x_k\}_\ne$ to denote the set consisting of the elements $x_1,\dots,x_k$ and at the same time express the fact that these elements are distinct. If the suffix '$\ne$' is not appended to the curly bracket, then no assumption is made about the distinctness of the elements listed.


TITLE: Gradient of the $ {L}_{2, 1} $ Mixed Norm
QUESTION [1 upvotes]: The $(1, 2)$-mixed norm of some matrix $X\in \mathbb{R}^{m \times n}$ is defined as
$$\| X \|_{1,2} = \sum_{j=1}^m \|x_{i,*}\|_2$$
where $x_{i,*}$ denotes the $i^{\text{th}}$ row of $X$. I am trying to find $$\frac{\partial}{\partial X} \|X\|_{1,2}$$ but am confused how to go about this. Does anyone know what this reduces to? I have already consulted the matrix cookbook.

REPLY [1 votes]: For convenience, define the following vector and its associated diagonal matrix.
$$\eqalign{
b &= {\rm diag}(XX^T),\quad B &= {\rm Diag}(b) \doteq {\rm Diag}(XX^T) \\
}$$
Then @frank 's gradient for the $L_{1,2}$ norm can be reformulated as
$$\eqalign{
\frac{\partial\|X\|_{1,2}}{\partial X}
 &= X\odot\Big((X\odot X)\,{\tt\large 11}^T\Big)^{\odot-1/2} \\
 &= X\odot\Big({\rm diag}(XX^T)\,{\tt\large 1}^T\Big)^{\odot-1/2} \\
 &= X\odot\Big(b{\tt\large 1}^T\Big)^{\odot-1/2} \\
 &= X\odot\Big(b^{\odot-1/2}\,{\tt\large 1}^T\Big) \\
 &= B^{-1/2}X \\
 &= {\rm Diag}(XX^T)^{-1/2}X \\
}$$
In this form, it is strikingly similar to the gradient of the Nuclear Norm.
$$\eqalign{
\frac{\partial\|X\|_{*}}{\partial X} &= (XX^T)^{-1/2}X \\
\\
}$$
NB:  Utilizing the above B-variables affords a simpler way to calculate the gradient.
$$\eqalign{
\|X\|_{1,2} &= b^{\odot 1/2}:1 \\
d\,\|X\|_{1,2}
 &= \tfrac{1}{2}b^{\odot-1/2}\odot db:1 \\
 &= \tfrac{1}{2}b^{\odot-1/2}:db \\
 &= \tfrac{1}{2}B^{-1/2}:dB \\
 &= \tfrac{1}{2}B^{-1/2}:2\,{\rm Diag}({\rm sym}(dX\,X^T)) \\
 &=B^{-1/2}X:dX \\
\frac{\partial\|X\|_{1,2}}{\partial X} &=B^{-1/2}X \\
}$$
One more comment. The above norm is row-oriented, but there is an analogous column-oriented norm given by $\|X^T\|_{1,2}$.
Temporarily changing the independent variable to $Y=X^T$, and then back to $X$ yields
$$\eqalign{
d\,\|Y\|_{1,2}
 &= {\rm Diag}(YY^T)^{-1/2}Y:dY \\
 &= {\rm Diag}(X^TX)^{-1/2}X^T:dX^T \\
 &= X\;{\rm Diag}(X^TX)^{-1/2}:dX \\
\frac{\partial\|X^T\|_{1,2}}{\partial X} &= X\;{\rm Diag}(X^TX)^{-1/2} \\
}$$
Once again, there is a similar expression for the gradient of the Nuclear Norm.
$$\eqalign{
\frac{\partial\|X\|_{*}}{\partial X} &= X\,(X^TX)^{-1/2} \\
}$$


TITLE: Finding a basis of the kernels and the images of $f_k$ and $g_k$
QUESTION [0 upvotes]: Let $\ f_k:M_2(\Bbb R)\to\Bbb R[x]_{\le2}, g_k:M_2(\Bbb R)\to\Bbb R^3$ be linear maps such that $f_k\left(\begin{bmatrix} a& b \\ c&d  \end{bmatrix}\right)=ax^2+kbx+d,\ g_k\left(\begin{bmatrix} a& b \\ c&d  \end{bmatrix}\right)=\begin{bmatrix} a-kb \\ 0 \\ k(k-2)(b-c)+d  \end{bmatrix}.$ I'm supposed to find a basis of the kernels and the images of both maps above.
Now, I understand that we should use the fact that $R[x]_{\le2}$ is isomorphic to $\Bbb R^3$, but I'm lost when it comes to the representative matrices. How can it be that multiplied by a 2x2 matrix, they give a 3x1 one?
I think they're not necessary for the kernels, but they should be for the images...

REPLY [1 votes]: Guide:
View $M_2(\mathbb{R})$ as $\mathbb{R}^4$. A matrix is just a matter of representation.
For example find the kernel and image for the following transformation:
$$\hat{f}_k\left(\begin{bmatrix} a\\ b \\ c\\d  \end{bmatrix}\right)=ax^2+kbx+d$$
After you find the conditions for the vector, reshape them back to a matrix.


text:
\section{Proof of lower bounds (Theorem \ref{thm:lower-bound-entrywise-rank-r}
and Theorem \ref{thm:l2-error-lower-bound-rankr})}

\label{sec:Proof-of-lower-bounds}

In this section, we establish the lower bounds claimed in Theorems
\ref{thm:lower-bound-entrywise-rank-r} and \ref{thm:l2-error-lower-bound-rankr}
(which subsume Theorems \ref{thm:lower-bound-entrywise} and \ref{thm:l2-error-lower-bound}
as special cases, respectively). Recall the assumption that $\{E_{i,j,k}\}$
are independent Gaussians. For the sake of notational simplicity,
we shall assume throughout this proof that $\sigma_{i,j,k}^{2}\equiv\sigma_{\min}^{2}$
for all $1\leq i,j,k\leq d$.

\begin{comment}
Clearly, if we reduce the variance of all noise components to $\sigma_{\min}^{2}$,
then the fundamental lower limits on the variance$\,$/$\,$covariance
can only be further decreased. As a result,
\end{comment}

Given that the noise components $\{E_{i,j,k}\mid(i,j,k)\in\Omega,1\leq i\leq j\leq k\leq d\}$
are assumed to be independent Gaussian, the probability density function
(conditional on $\Omega)$ can be computed as
\begin{align*}
f(\bm{T}^{\mathsf{obs}}) & =c\prod_{1\leq i\leq j\leq k\leq d,\,(i,j,k)\in\Omega}\exp\left(-\frac{\big(T_{i,j,k}^{\mathsf{obs}}-\sum_{l=1}^{r}u_{l,i}^{\star}u_{l,j}^{\star}u_{l,k}^{\star}\big)^{2}}{2\sigma_{\min}^{2}}\right)
\end{align*}
for some normalization constant $c>0$. Here, we abuse the notation
$f(\cdot)$ to represent the probability density function whenever
it is clear from the context. Denote by $\mathsf{vec}(\bm{U}^{\star})$
the vectorization of $\bm{U}^{\star}=[\bm{u}_{1}^{\star},\cdots,\bm{u}_{r}^{\star}]$,
namely,
\[
\mathsf{vec}(\bm{U}^{\star}):=\left[\begin{array}{c}
\bm{u}_{1}^{\star}\\
\vdots\\
\bm{u}_{r}^{\star}
\end{array}\right]\in\mathbb{R}^{dr}.
\]
By virtue of the Cram\'er-Rao lower bound, any unbiased estimator
$\widehat{\bm{U}}$ for $\bm{U}^{\star}$ necessarily obeys
\[
\mathsf{Cov}\big[\mathsf{vec}(\widehat{\bm{U}})\big]\succeq(\bm{\mathcal{I}}_{\Omega})^{-1},
\]
where $\bm{\mathcal{I}}_{\Omega}\in\mathbb{R}^{dr\times dr}$ denotes
the corresponding Fisher information matrix (conditional on $\Omega$)
as follows
\begin{align}
\bm{\mathcal{I}}_{\Omega}:=\bm{\mathcal{I}}_{\Omega}(\bm{U}^{\star}) & =\mathbb{E}\Big[\nabla_{\mathsf{vec}(\bm{U}^{\star})}\log f(\bm{T}^{\mathsf{obs}})\big(\nabla_{\mathsf{vec}(\bm{U}^{\star})}\log f(\bm{T}^{\mathsf{obs}})\big)^{\top}\Big]\nonumber \\
 & =\bigg[\mathbb{E}\Big[\nabla_{\bm{u}_{l_{1}}^{\star}}\log f(\bm{T}^{\mathsf{obs}})\big(\nabla_{\bm{u}_{l_{2}}^{\star}}\log f(\bm{T}^{\mathsf{obs}})\big)^{\top}\Big]\bigg]_{1\leq l_{1},l_{2}\leq r}.\label{eq:defn-Fisher-information}
\end{align}

It then suffices to compute the Fisher information matrix. Towards
this end, we start by observing that
\begin{equation}
\frac{\partial T_{i,j,k}^{\star}}{\partial u_{l,s}^{\star}}=\sum_{\tau=1}^{r}\frac{\partial u_{\tau,i}^{\star}u_{\tau,j}^{\star}u_{\tau,k}^{\star}}{\partial u_{l,s}^{\star}}=u_{l,j}^{\star}u_{l,k}^{\star}\ind_{\{i=s\}}+u_{l,i}^{\star}u_{l,k}^{\star}\ind_{\{j=s\}}+u_{l,i}^{\star}u_{l,j}^{\star}\ind_{\{k=s\}},\label{eq:T-ijk-u-derivative}
\end{equation}
and
\begin{align}
\sigma_{\min}^{2}\frac{\partial\log f(\bm{T}^{\mathsf{obs}})}{\partial u_{l,s}^{\star}} & =\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}\Big(T_{i,j,k}^{\mathsf{obs}}-\sum_{\tau=1}^{r}u_{\tau,i}^{\star}u_{\tau,j}^{\star}u_{\tau,k}^{\star}\Big)\Big(u_{l,j}^{\star}u_{l,k}^{\star}\ind_{\{i=s\}}+u_{l,i}^{\star}u_{l,k}^{\star}\ind_{\{j=s\}}+u_{l,i}^{\star}u_{l,j}^{\star}\ind_{\{k=s\}}\Big)\nonumber \\
 & =\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}E_{i,j,k}\Big(u_{l,j}^{\star}u_{l,k}^{\star}\ind_{\{i=s\}}+u_{l,i}^{\star}u_{l,k}^{\star}\ind_{\{j=s\}}+u_{l,i}^{\star}u_{l,j}^{\star}\ind_{\{k=s\}}\Big)\label{eq:log-f-calculation}
\end{align}
for any $1\leq l\leq r$ and $1\leq s\leq d$. In addition, let us
define a collection of vectors $\{\bm{h}_{i,j,k}\}_{1\leq i\leq j\leq k\leq d}$
in $\mathbb{R}^{dr}$ with entries
\begin{equation}
h_{i,j,k}(l,s):=h_{i,j,k}\big((l-1)\times r+s\big):=u_{l,j}^{\star}u_{l,k}^{\star}\ind_{\{i=s\}}+u_{l,i}^{\star}u_{l,k}^{\star}\ind_{\{j=s\}}+u_{l,i}^{\star}u_{l,j}^{\star}\ind_{\{k=s\}}\label{eq:h-entry}
\end{equation}
for any $1\leq l\leq r$ and $1\leq s\leq d$. One can then express
\begin{align}
\bm{\mathcal{I}}_{\Omega} & =\frac{1}{\sigma_{\min}^{4}}\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}\mathbb{E}\left[E_{i,j,k}^{2}\right]\bm{h}_{i,j,k}\bm{h}_{i,j,k}^{\top}=\frac{1}{\sigma_{\min}^{2}}\sum_{1\leq i\leq j\leq k\leq d}\chi_{i,j,k}\bm{h}_{i,j,k}\bm{h}_{i,j,k}^{\top},\label{eq:Fisher-Omega}
\end{align}
where we recall the notation $\chi_{i,j,k}:=\ind_{\{(i,j,k)\in\Omega\}}$.
Let us further define
\begin{equation}
\bm{\mathcal{I}}:=\mathbb{E}_{\Omega}\big[\bm{\mathcal{I}}_{\Omega}\big]=\frac{p}{\sigma_{\min}^{2}}\sum_{1\leq i\leq j\leq k\leq d}\bm{h}_{i,j,k}\bm{h}_{i,j,k}^{\top}\label{eq:Fisher}
\end{equation}
where the expectation is taken over randomness of $\{\chi_{i,j,k}\}_{1\leq i,j,k\leq d}$.
In what follows, we shall compute the spectrum of $\bm{\mathcal{I}}$,
and show that $\mathcal{\bm{\mathcal{I}}}_{\Omega}$ is, with high
probability, sufficiently close to $\bm{\mathcal{I}}$ in the spectral
norm. In addition, denote $(l,s):=(l-1)\times r+d$ for all $1\leq l\leq r,1\leq s\leq d.$

\paragraph{Spectrum of $\bm{\mathcal{I}}$.}

First, it is straightforward to calculate that for any $1\leq s\leq d$
and any $1\leq l_{1},l_{2}\leq d$,
\begin{align*}
\frac{\sigma_{\min}^{2}}{p}\mathcal{I}_{(l_{1},s),(l_{2},s)} & =\sum_{(i,j):i\leq j}u_{l_{1},i}^{\star}u_{l_{2},i}^{\star}u_{l_{1},j}^{\star}u_{l_{2},j}^{\star}+3\,u_{l_{1},s}^{\star}u_{l_{2},s}^{\star}\sum_{1\leq i\leq d}u_{l_{1},i}^{\star}u_{l_{2},i}^{\star}+5\,u_{l_{1},s}^{\star2}u_{l_{2},s}^{\star2}\\
 & =\frac{1}{2}\langle\bm{u}_{l_{1}}^{\star},\bm{u}_{l_{2}}^{\star}\rangle^{2}+\frac{1}{2}\sum_{1\leq i\leq d}u_{l_{1},i}^{\star2}u_{l_{2},i}^{\star2}+3\,u_{l_{1},s}^{\star}u_{l_{2},s}^{\star}\langle\bm{u}_{l_{1}}^{\star},\bm{u}_{l_{2}}^{\star}\rangle+5\,u_{l_{1},s}^{\star2}u_{l_{2},s}^{\star2},
\end{align*}
and for any $1\leq s_{1}\neq s_{2}\leq d$, $1\leq l_{1},l_{2}\leq d$,
\[
\frac{\sigma_{\min}^{2}}{p}\mathcal{I}_{(l_{1},s_{1}),(l_{2},s_{2})}=u_{l_{1},s_{2}}^{\star}u_{l_{2},s_{1}}^{\star}\langle\bm{u}_{l_{1}}^{\star},\bm{u}_{l_{2}}^{\star}\rangle.
\]
As a consequence, one can express
\begin{align*}
\frac{\sigma_{\min}^{2}}{p}\bm{\mathcal{I}} & =\left[\begin{array}{ccc}
\frac{1}{2}\|\bm{u}_{1}^{\star}\|_{2}^{4}\bm{I}_{d}+\|\bm{u}_{1}^{\star}\|_{2}^{2}\bm{u}_{1}^{\star}\bm{u}_{1}^{\star\top}\\
 & \ddots\\
 &  & \frac{1}{2}\|\bm{u}_{r}^{\star}\|_{2}^{4}\bm{I}_{d}+\|\bm{u}_{r}^{\star}\|_{2}^{2}\bm{u}_{r}^{\star}\bm{u}_{r}^{\star\top}
\end{array}\right]\\
 & \quad+\underbrace{\left[\begin{array}{ccc}
\bm{0} & \cdots & \langle\bm{u}_{r}^{\star},\bm{u}_{1}^{\star}\rangle\bm{u}_{r}^{\star}\bm{u}_{1}^{\star\top}\\
\vdots & \ddots & \vdots\\
\langle\bm{u}_{1}^{\star},\bm{u}_{r}^{\star}\rangle\bm{u}_{1}^{\star}\bm{u}_{r}^{\star\top} & \cdots & \bm{0}
\end{array}\right]}_{=:\,\bm{\Phi}}+\,\bm{\Psi}
\end{align*}
where $\bm{\Phi}$ is a matrix in $\mathbb{R}^{dr\times dr}$ with
all-zero diagonal blocks and $(l_{1,}l_{2})$-th block equal to $\langle\bm{u}_{l_{2}}^{\star},\bm{u}_{l_{1}}^{\star}\rangle\bm{u}_{l_{2}}^{\star}\bm{u}_{l_{1}}^{\star\top}$,
and $\bm{\Psi}$ is a matrix in $\mathbb{R}^{dr\times dr}$ with entries
\begin{align*}
\Psi_{(l,s),(l,s)} & =\frac{1}{2}\sum_{1\leq i\leq d}u_{l,i}^{\star4}+2\,u_{l,s}^{\star2}\|\bm{u}_{l}^{\star}\|_{2}^{2}+5u_{l,s}^{\star4};\quad\text{if}\,\,l_{1}=l_{2},s_{1}=s_{2},\\
\Psi_{(l_{1},s),(l_{2},s)} & =\frac{1}{2}\langle\bm{u}_{l_{1}}^{\star},\bm{u}_{l_{2}}^{\star}\rangle^{2}+\frac{1}{2}\sum_{1\leq i\leq d}u_{l_{1},i}^{\star2}u_{l_{2},i}^{\star2}+2\,u_{l_{1},s}^{\star}u_{l_{2},s}^{\star}\langle\bm{u}_{l_{1}}^{\star},\bm{u}_{l_{2}}^{\star}\rangle+5\,u_{l_{1},s}^{\star2}u_{l_{2},s}^{\star2},\quad\text{if\,\,}l_{1}\neq l_{2},s_{1}=s_{2},\\
\Psi_{(l_{1},s_{1}),(l_{2},s_{2})} & =0,\quad\text{otherwise.}
\end{align*}

In the sequel, we shall control the spectral norm of $\bm{\Phi}$
and $\bm{\Psi}$ separately.
\begin{itemize}
\item For $\bm{\Phi}$, we can bound
\begin{align*}
\|\bm{\Phi}\| & \leq\|\bm{\Phi}\|_{\mathrm{F}}\leq\sqrt{\sum_{1\leq l_{1}\neq l_{2}\leq r}\langle\bm{u}_{l_{1}}^{\star},\bm{u}_{l_{2}}^{\star}\rangle^{2}\big\|\bm{u}_{l_{2}}^{\star}\bm{u}_{l_{1}}^{\star\top}\big\|_{\mathrm{F}}^{2}}\leq\sqrt{\max_{1\leq l_{1}\neq l_{2}\leq r}\langle\bm{u}_{l_{1}}^{\star},\bm{u}_{l_{2}}^{\star}\rangle^{2}\sum_{1\leq l_{1},l_{2}\leq r}\|\bm{u}_{l_{1}}^{\star}\|_{2}^{2}\|\bm{u}_{l_{2}}^{\star}\|_{2}^{2}}\\
 & \leq r\sqrt{\frac{\mu}{d}}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4}=o\Big(\min_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4}\Big),
\end{align*}
where we have used the incoherence condition (\ref{assumption:u-inner-prod})
and the assumptions $r=o\big(\sqrt{d/\mu}\big)$ and $\kappa\asymp1$.
\item As for $\bm{\Psi}$, we note that each block of $\bm{\Psi}$ is a
diagonal matrix. By the incoherence conditions, its entries can be
bounded by
\[
\big|\Psi_{(l,s),(l,s)}\big|\lesssim\|\bm{u}_{l}^{\star}\|_{\infty}^{2}\|\bm{u}_{l}^{\star}\|_{2}^{2}\leq\frac{\mu}{d}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4},
\]
and
\begin{align*}
\big|\Psi_{(l_{1},s),(l_{2},s)}\big| & \lesssim\langle\bm{u}_{l_{1}}^{\star},\bm{u}_{l_{2}}^{\star}\rangle^{2}+\|\bm{u}_{l_{1}}^{\star}\|_{\infty}^{2}\|\bm{u}_{l_{2}}^{\star}\|_{2}^{2}+2\,\|\bm{u}_{l_{1}}^{\star}\|_{\infty}\|\bm{u}_{l_{2}}^{\star}\|_{\infty}\|\bm{u}_{l_{1}}^{\star}\|_{2}\|\bm{u}_{l_{2}}^{\star}\|_{2}\lesssim\frac{\mu}{d}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4},
\end{align*}
thus indicating that
\[
\|\bm{\Psi}\|_{\infty}\lesssim\frac{\mu}{d}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4}.
\]
Given the special structure of $\bm{\Psi}$, one can easily permutation
its columns and rows to arrive at another matrix $\widetilde{\bm{\Psi}}=[\widetilde{\bm{\Psi}}_{i,j}]_{1\leq i,j\leq r}$
such that (1) $\widetilde{\bm{\Psi}}$ is a block diagonal matrix;
(2) $\widetilde{\bm{\Psi}}$ contains $d\times d$ blocks each of
size $r\times r$; (3) each diagonal block $\widetilde{\bm{\Psi}}_{i,i}$
of $\widetilde{\bm{\Psi}}$ has spectral norm at most
\[
\|\widetilde{\bm{\Psi}}_{i,i}\|\leq\|\widetilde{\bm{\Psi}}_{i,i}\|_{\mathrm{F}}\leq r\|\bm{\Psi}\|_{\infty}\lesssim\frac{\mu r}{d}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4}.
\]
Consequently, one can derive
\[
\|\bm{\Psi}\|=\|\widetilde{\bm{\Psi}}\|\leq\max_{1\leq i\leq r}\|\bm{\Psi}_{i,i}\|\lesssim\frac{\mu r}{d}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4}=o\left(\min_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4}\right),
\]
provided that $r=o(d/\mu)$ and $\kappa\asymp1$.
\item Putting the above two estimates together, we conclude that
\begin{align}
\bm{\mathcal{I}} & \preceq\frac{p}{\sigma_{\min}^{2}}\left[\begin{array}{ccc}
\frac{1}{2}\|\bm{u}_{1}^{\star}\|_{2}^{4}\bm{I}_{d}+\|\bm{u}_{1}^{\star}\|_{2}^{2}\bm{u}_{1}^{\star}\bm{u}_{1}^{\star\top}\\
 & \ddots\\
 &  & \frac{1}{2}\|\bm{u}_{r}^{\star}\|_{2}^{4}\bm{I}_{d}+\|\bm{u}_{r}^{\star}\|_{2}^{2}\bm{u}_{r}^{\star}\bm{u}_{r}^{\star\top}
\end{array}\right]+\frac{p}{\sigma_{\min}^{2}}\left(\|\bm{\Phi}\|+\|\bm{\Psi}\|\right)\bm{I}_{dr}\nonumber \\
 & =\big(1+o(1)\big)\frac{p}{\sigma_{\min}^{2}}\left[\begin{array}{ccc}
\frac{1}{2}\|\bm{u}_{1}^{\star}\|_{2}^{4}\bm{I}_{d}+\|\bm{u}_{1}^{\star}\|_{2}^{2}\bm{u}_{1}^{\star}\bm{u}_{1}^{\star\top}\\
 & \ddots\\
 &  & \frac{1}{2}\|\bm{u}_{r}^{\star}\|_{2}^{4}\bm{I}_{d}+\|\bm{u}_{r}^{\star}\|_{2}^{2}\bm{u}_{r}^{\star}\bm{u}_{r}^{\star\top}
\end{array}\right].\label{eq:fisher-eig-UB}
\end{align}
\end{itemize}

\paragraph{Controlling $\|\bm{\mathcal{I}}_{\Omega}-\bm{\mathcal{I}}\|$.}

By construction, $\bm{\mathcal{I}}_{\Omega}-\bm{\mathcal{I}}=\frac{1}{\sigma_{\min}^{2}}\sum_{i\leq j\leq k}(\chi_{i,j,k}-p)\bm{h}_{i,j,k}\bm{h}_{i,j,k}^{\top}$
is a sum of independent zero-mean random matrix in $\mathbb{R}^{dr\times dr}$.
By the incoherence conditions, it is straightforward to bound
\begin{align*}
B & :=\max_{1\leq i\leq j\leq k\leq d}\big\|(\chi_{i,j,k}-p)\bm{h}_{i,j,k}\bm{h}_{i,j,k}^{\top}\big\|\leq\max_{1\leq i\leq j\leq k\leq d}\|\bm{h}_{i,j,k}\|_{2}^{2}\\
 & \,\overset{(\mathrm{i})}{\lesssim}r\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{\infty}^{4}\leq\frac{\mu^{2}r}{d^{2}}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4},
\end{align*}
where (i) arises from the definition of $\bm{h}_{i,j,k}$ in (\ref{eq:h-entry}).
In addition, we also have
\begin{align*}
V & :=\Big\|\sum_{i\leq j\leq k}\mathbb{E}\big[(\chi_{i,j,k}-p)^{2}\big]\|\bm{h}_{i,j,k}\|_{2}^{2}\bm{h}_{i,j,k}\bm{h}_{i,j,k}^{\top}\Big\|\\
 & \,\leq p\max_{1\leq i\leq j\leq k\leq d}\|\bm{h}_{i,j,k}\|_{2}^{2}\Big\|\sum_{1\leq i\leq j\leq k\leq d}\bm{h}_{i,j,k}\bm{h}_{i,j,k}^{\top}\Big\|\\
 & \,\lesssim pr\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{\infty}^{4}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4}\\
 & \,\lesssim\frac{\mu^{2}rp}{d^{2}}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{8}.
\end{align*}
Here, the second inequality arises from (\ref{eq:h-entry}) and (\ref{eq:fisher-eig-UB}),
whereas the third comes from our bound above for $\bm{\mathcal{I}}$.
Invoking the matrix Bernstein inequality, we conclude that with probability
at least $1-O(d^{-10})$,
\begin{align}
\|\bm{\mathcal{I}}_{\Omega}-\bm{\mathcal{I}}\| & \lesssim\frac{B\log d+\sqrt{V\log d}}{\sigma_{\min}^{2}}\lesssim\frac{p}{\sigma_{\min}^{2}}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4}\left\{ \frac{\mu^{2}r\log d}{d^{2}p}+\frac{\mu\sqrt{r\log d}}{d\sqrt{p}}\right\} \nonumber \\
 & =o(1)\frac{p}{\sigma_{\min}^{2}}\min_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4},\label{eq:fisher-dev-UB}
\end{align}
where the last step holds as long as $p\gg\mu^{2}rd^{-2}\log^{2}d$
and $\kappa\asymp1$.

\paragraph{Combining the spectrum of $\bm{\mathcal{I}}_{\Omega}$ and the bound
on $\|\bm{\mathcal{I}}_{\Omega}-\bm{\mathcal{I}}\|$.}

Combining (\ref{eq:fisher-eig-UB}) and (\ref{eq:fisher-dev-UB})
with Weyl's inequality reveals that with probability exceeding $1-O(d^{-10})$,
\[
\bm{\mathcal{I}}_{\Omega}\preceq\big(1+o(1)\big)\frac{p}{\sigma_{\min}^{2}}\left[\begin{array}{ccc}
\frac{1}{2}\|\bm{u}_{1}^{\star}\|_{2}^{4}\bm{I}_{d}+\|\bm{u}_{1}^{\star}\|_{2}^{2}\bm{u}_{1}^{\star}\bm{u}_{1}^{\star\top}\\
 & \ddots\\
 &  & \frac{1}{2}\|\bm{u}_{r}^{\star}\|_{2}^{4}\bm{I}_{d}+\|\bm{u}_{r}^{\star}\|_{2}^{2}\bm{u}_{r}^{\star}\bm{u}_{r}^{\star\top}
\end{array}\right].
\]
By the Woodbury matrix identity, it is straightforward to check
\[
\left(\frac{1}{2}\|\bm{u}_{i}^{\star}\|_{2}^{4}\bm{I}_{d}+\|\bm{u}_{i}^{\star}\|_{2}^{2}\bm{u}_{i}^{\star}\bm{u}_{i}^{\star\top}\right)^{-1}=\frac{2}{\|\bm{u}_{i}^{\star}\|_{2}^{4}}\left(\bm{I}_{d}-\frac{2}{3}\frac{\bm{u}_{i}^{\star}\bm{u}_{i}^{\star\top}}{\|\bm{u}_{i}^{\star}\|_{2}^{2}}\right).
\]
Hence, for any unbiased estimator $\widehat{\bm{U}}$ of $\bm{U}^{\star}$
we have
\begin{equation}
\mathsf{Cov}\big[\mathsf{vec}(\widehat{\bm{U}})\big]\succeq(\bm{\mathcal{I}}_{\Omega})^{-1}\succeq\big(1-o(1)\big)\frac{\sigma_{\min}^{2}}{p}\left[\begin{array}{ccc}
\frac{2}{\|\bm{u}_{1}^{\star}\|_{2}^{4}}\left(\bm{I}_{d}-\frac{2}{3}\frac{\bm{u}_{1}^{\star}\bm{u}_{1}^{\star\top}}{\|\bm{u}_{1}^{\star}\|_{2}^{2}}\right)\\
 & \ddots\\
 &  & \frac{2}{\|\bm{u}_{r}^{\star}\|_{2}^{4}}\left(\bm{I}_{d}-\frac{2}{3}\frac{\bm{u}_{r}^{\star}\bm{u}_{r}^{\star\top}}{\|\bm{u}_{r}^{\star}\|_{2}^{2}}\right)
\end{array}\right].\label{eq:cov-vec-U-LB}
\end{equation}

A few consequences from the above Cram\'er-Rao lower bound are in
order.
\begin{itemize}
\item For each unbiased estimator $\widehat{\bm{u}}_{l}$ of $\bm{u}_{l}^{\star}$,
one necessarily has
\begin{align*}
\mathbb{E}\big[(\widehat{u}_{l,k}-u_{l,k}^{\star})^{2}\big] & \geq\big((\bm{\mathcal{I}}_{\Omega})^{-1})_{(l,k),(l,k)}\geq\big(1-o(1)\big)\frac{2\sigma_{\min}^{2}}{p\|\bm{u}_{l}^{\star}\|_{2}^{4}}\left(1-\frac{2}{3}\frac{u_{l,k}^{\star2}}{\|\bm{u}_{l}^{\star}\|_{2}^{2}}\right)\\
 & \geq\big(1-o(1)\big)\frac{2\sigma_{\min}^{2}}{p\|\bm{u}_{l}^{\star}\|_{2}^{4}}\left(1-\frac{2}{3}\frac{\|\bm{u}_{l}^{\star}\|_{\infty}^{2}}{\|\bm{u}_{l}^{\star}\|_{2}^{2}}\right)\\
 & \overset{(\mathrm{i})}{\geq}\big(1-o(1)\big)\frac{2\sigma_{\min}^{2}}{p\|\bm{u}_{l}^{\star}\|_{2}^{4}}\left(1-\frac{2}{3}\frac{\mu}{d}\right)\\
 & \overset{(\mathrm{ii})}{\geq}\big(1-o(1)\big)\frac{2\sigma_{\min}^{2}}{p\|\bm{u}_{l}^{\star}\|_{2}^{4}}=\big(1-o(1)\big)(\bm{\Sigma}_{k}^{\star})_{l,l},
\end{align*}
where (i) arises from the incoherence assumption (\ref{assumption:incoherence})
and (ii) holds as long as $\mu=o(d)$. This further implies that
\[
\mathbb{E}\big[\|\widehat{\bm{u}}_{l}-\bm{u}_{l}^{\star}\|_{2}^{2}\big]=\sum_{k=1}^{d}\mathbb{E}\big[(\widehat{u}_{l,k}-u_{l,k}^{\star})^{2}\big]\geq\big(1-o(1)\big)\frac{2\sigma_{\min}^{2}d}{p\|\bm{u}_{l}^{\star}\|_{2}^{4}}.
\]
\item Any unbiased estimator $\widehat{T}_{i,j,k}$ of $T_{i,j,k}^{\star}$
necessarily obeys \cite{shao2003mathematical}
\begin{align*}
\mathbb{E}\Big[\big(\widehat{T}_{i,j,k}-T_{i,j,k}^{\star}\big)^{2}\Big] & \geq\left[\frac{\partial T_{i,j,k}^{\star}}{\partial\mathsf{vec}(\bm{U}^{\star})}\right]^{\top}(\bm{\mathcal{I}}_{\Omega})^{-1}\frac{\partial T_{i,j,k}^{\star}}{\partial\mathsf{vec}(\bm{U}^{\star})}\\
 & \overset{(\mathrm{i})}{\geq}\left(1-o(1)\right)\sum_{1\leq s\leq d}\sum_{1\leq l\leq r}\frac{2\sigma_{\min}^{2}}{p\|\bm{u}_{l}^{\star}\|_{2}^{4}}\left(\frac{\partial T_{i,j,k}^{\star}}{\partial u_{l,s}^{\star}}\right)^{2}\\
 & \overset{(\mathrm{ii})}{=}\big(1-o(1)\big)v_{i,j,k}^{\star},
\end{align*}
where (i) uses (\ref{eq:cov-vec-U-LB}), and (ii) follows from (\ref{eq:T-ijk-u-derivative}),
(\ref{def:T-entry-var}) and direct algebraic manipulations.
\item Any unbiased estimator $\widehat{\bm{T}}$ of $\bm{T}^{\star}$ necessarily
satisfies
\begin{align*}
\mathbb{E}\big[\|\widehat{\bm{T}}-\bm{T}^{\star}\|_{\mathrm{F}}^{2}\big] & =\sum_{i,j,k}\mathbb{E}\Big[\big(\widehat{T}_{i,j,k}-T_{i,j,k}^{\star}\big)^{2}\Big]\geq\big(1-o(1)\big)\sum_{i,j,k}v_{i,j,k}^{\star}\\
 & \overset{(\mathrm{i})}{\geq}\big(1-o(1)\big)3\sum_{1\leq i\leq d}\sum_{1\leq j,k\leq d}\sum_{1\leq l\leq r}\frac{2\sigma_{\min}^{2}}{p\|\bm{u}_{l}^{\star}\|_{2}^{4}}u_{l,j}^{\star2}u_{l,k}^{\star2}\\
 & =\big(1-o(1)\big)\frac{6\sigma_{\min}^{2}dr}{p},
\end{align*}
where (i) arises from the definition of $v_{i,j,k}^{\star}$ in (\ref{def:T-entry-var}).
\end{itemize}
\begin{comment}
The next step amounts to studying $h_{i,j,k}(l,s)$. Firstly, it is
straightforward to calculate that, for any $1\leq l\leq r$ and $1\leq s\leq d$,
\begin{align*}
\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}h_{i,j,k}^{2}(l,s) & =\sum_{(i,j):i\leq j,\,i\neq s,\,j\neq s}u_{l,i}^{\star2}u_{l,j}^{\star2}\chi_{i,j,s}+4\,u_{l,s}^{\star2}\sum_{i:i\neq s}u_{l,i}^{\star2}\chi_{i,s,s}+9\,u_{l,s}^{\star4}\chi_{s,s,s}\\
 & =\frac{1}{2}\sum_{(i,j):i\neq j,\,i\neq s,\,j\neq s}u_{l,i}^{\star2}u_{l,j}^{\star2}\chi_{i,j,s}+\sum_{i:i\neq s}u_{l,i}^{\star4}\chi_{i,i,s}+4\,u_{l,s}^{\star2}\sum_{i:i\neq s}u_{l,i}^{\star2}\chi_{i,s,s}+9\,u_{l,s}^{\star4}\chi_{s,s,s}\\
 & \leq\frac{1}{2}\sum_{1\leq i,j\leq d}u_{l,i}^{\star2}u_{l,j}^{\star2}\chi_{i,j,s}+\sum_{1\leq i\leq d}u_{l,i}^{\star4}\chi_{i,i,s}+9\,u_{l,s}^{\star2}\sum_{1\leq i\leq d}u_{l,i}^{\star2}\chi_{i,s,s}.
\end{align*}

\begin{itemize}
\item For the first term, recognizing $\sum_{1\leq i,j\leq d}\mathbb{E}\big[u_{l,i}^{\star2}u_{l,j}^{\star2}\chi_{i,j,s}\big]=p\left\Vert \bm{u}_{l}^{\star}\right\Vert _{2}^{4}$,
we apply the Bernstein inequality shows that with probability at least
$1-O\left(d^{-10}\right)$,
\begin{align*}
\sum_{1\leq i,j\leq d}u_{l,i}^{\star2}u_{l,j}^{\star2}\chi_{i,j,s}-p\left\Vert \bm{u}_{l}^{\star}\right\Vert _{2}^{4} & \lesssim\max_{1\leq i,j\leq d}\big|u_{l,i}^{\star2}u_{l,j}^{\star2}\chi_{i,j,s}\big|\log d+\sqrt{\sum_{1\leq i,j\leq d}\mathrm{Var}\big(u_{l,i}^{\star2}u_{l,j}^{\star2}\chi_{i,j,s}\big)\log d}\\
 & \leq\left\Vert \bm{u}_{l}^{\star}\right\Vert _{\infty}^{4}\log d+\left\Vert \bm{u}_{l}^{\star}\right\Vert _{2}^{2}\left\Vert \bm{u}_{l}^{\star}\right\Vert _{\infty}^{2}\sqrt{p\log d}\\
 & \leq p\left\Vert \bm{u}_{l}^{\star}\right\Vert _{2}^{4}\left\{ \frac{\mu^{2}\log d}{d^{2}p}+\frac{\mu\sqrt{\log d}}{d\sqrt{p}}\right\} ,
\end{align*}
where the last line arises from assumption~(\ref{assumption:u-inf-norm})
and (\ref{eq:defn-mu}). Combined with the sample size condition $p\gg\mu^{2}d^{-3/2}\log d$,
we find
\[
\sum_{1\leq i,j\leq d}u_{l,i}^{\star2}u_{l,j}^{\star2}\chi_{i,j,s}\leq\frac{1}{2}\left(1+o\left(1\right)\right)p\left\Vert \bm{u}_{l}^{\star}\right\Vert _{2}^{4}.
\]
\item As for the remaining two terms, we can invoke the Bernstein inequality
again to bound
\begin{align*}
\sum_{1\leq i\leq d}u_{l,i}^{\star4}\chi_{i,i,s} & \lesssim\sum_{1\leq i\leq d}\mathbb{E}\big[u_{l,i}^{\star4}\chi_{i,i,s}\big]+\max_{1\leq i\leq d}\big|u_{l,i}^{\star4}\chi_{i,i,s}\big|\log d+\sqrt{\sum_{1\leq i\leq d}\mathrm{Var}\big(u_{l,i}^{\star4}\chi_{i,i,s}\big)\log d}\\
 & \leq p\left\Vert \bm{u}_{l}^{\star}\right\Vert _{2}^{2}\left\Vert \bm{u}_{l}^{\star}\right\Vert _{\infty}^{2}+\left\Vert \bm{u}_{l}^{\star}\right\Vert _{\infty}^{4}\log d+\left\Vert \bm{u}_{l}^{\star}\right\Vert _{2}\left\Vert \bm{u}_{l}^{\star}\right\Vert _{\infty}^{3}\sqrt{p\log d}\\
 & \leq p\left\Vert \bm{u}_{l}^{\star}\right\Vert _{2}^{4}\left\{ \frac{\mu}{d}+\frac{\mu^{2}\log d}{d^{2}p}+\frac{\mu^{3/2}\sqrt{\log d}}{d^{3/2}\sqrt{p}}\right\} =o\left(1\right)p\left\Vert \bm{u}_{l}^{\star}\right\Vert _{2}^{4}
\end{align*}
provided $p\gg\mu^{3}d^{-3/2}\log d$ and $\mu\ll\sqrt{d/\log d}$.
Clearly, the upper bound also holds for $u_{l,s}^{\star2}\sum_{1\leq i\leq d}u_{l,i}^{\star2}\chi_{i,s,s}$.
\item As a result, we find that with probability exceeding $1-O\,(d^{-10})$,
\begin{equation}
\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}h_{i,j,k}^{2}(l,s)\leq\frac{1}{2}\left(1+o\left(1\right)\right)p\left\Vert \bm{u}_{l}^{\star}\right\Vert _{2}^{4}.\label{eq:diagonal-Fisher}
\end{equation}
\end{itemize}
Secondly, for any $1\leq l_{1}\neq l_{2}\leq r$ and any $1\leq s\leq d$,
one obtains
\begin{align*}
 & \bigg|\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}h_{i,j,k}(l_{1},s)h_{i,j,k}(l_{2},s)\bigg|\\
 & =\bigg|\sum_{(i,j):i\leq j,\,i\neq s,\,j\neq s}u_{l_{1},i}^{\star}u_{l_{1},j}^{\star}u_{l_{2},i}^{\star}u_{l_{2},j}^{\star}\chi_{i,j,s}+4\,u_{l_{1},s}^{\star}u_{l_{2},s}^{\star}\sum_{i:i\neq s}u_{l_{1},i}^{\star}u_{l_{2},i}^{\star}\chi_{i,s,s}+9\,u_{l_{1},s}^{\star2}u_{l_{2},s}^{\star2}\chi_{s,s,s}\bigg|\\
 & \leq\bigg|\sum_{1\leq i,j\leq d}u_{l_{1},i}^{\star}u_{l_{1},j}^{\star}u_{l_{2},i}^{\star}u_{l_{2},j}^{\star}\chi_{i,j,s}\bigg|+9\,\bigg|u_{l_{1},s}^{\star}u_{l_{2},s}^{\star}\sum_{1\leq i\leq d}u_{l_{1},i}^{\star}u_{l_{2},i}^{\star}\chi_{i,s,s}\bigg|.
\end{align*}

\begin{itemize}
\item For the first term, we use the Bernstein inequality to find that with
probability at least $1-O\left(d^{-10}\right)$,
\begin{align*}
\bigg|\sum_{1\leq i,j\leq d}u_{l_{1},i}^{\star}u_{l_{1},j}^{\star}u_{l_{2},i}^{\star}u_{l_{2},j}^{\star}\chi_{i,j,s}\bigg| & \lesssim\sum_{1\leq i,j\leq d}\mathbb{E}\big[u_{l_{1},i}^{\star}u_{l_{1},j}^{\star}u_{l_{2},i}^{\star}u_{l_{2},j}^{\star}\chi_{i,j,s}\big]+\max_{1\leq i,j\leq d}\big|u_{l_{1},i}^{\star}u_{l_{1},j}^{\star}u_{l_{2},i}^{\star}u_{l_{2},j}^{\star}\big|\log d\\
 & \quad+\sqrt{\sum_{1\leq i,j\leq d}\mathrm{Var}\big(u_{l_{1},i}^{\star}u_{l_{1},j}^{\star}u_{l_{2},i}^{\star}u_{l_{2},j}^{\star}\chi_{i,j,s}\big)\log d}\\
 & \lesssim p\,\langle\bm{u}_{l_{1}}^{\star},\bm{u}_{l_{2}}^{\star}\rangle^{2}+\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{\infty}^{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{\infty}^{2}\log d+\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{2}\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{\infty}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{\infty}\sqrt{p\log d}\\
 & \overset{(\mathrm{i})}{\leq}p\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{2}^{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{2}^{2}\left\{ \frac{\mu}{d}+\frac{\mu^{2}\log d}{d^{2}p}+\frac{\mu\sqrt{\log d}}{d\sqrt{p}}\right\} \overset{(\mathrm{ii})}{=}o\left(1\right)\frac{p}{r}\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{2}^{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{2}^{2},
\end{align*}
where we use assumptions~(\ref{assumption:u-inf-norm})-(\ref{assumption:u-inner-prod})
and (\ref{eq:defn-mu}); (ii) holds as long as $p\gg\mu^{3}rd^{-3/2}\log d$
and $r\ll\sqrt{d/r}.$
\item In a similar manner, we can bound
\begin{align*}
\bigg|u_{l_{1},s}^{\star}u_{l_{2},s}^{\star}\sum_{1\leq i\leq d}u_{l_{1},i}^{\star}u_{l_{2},i}^{\star}\chi_{i,s,s} & \bigg|\lesssim\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{\infty}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{\infty}\left(p\,\langle\bm{u}_{l_{1}}^{\star},\bm{u}_{l_{2}}^{\star}\rangle+\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{\infty}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{\infty}\log d+\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{\infty}\sqrt{p\log d}\right)\\
 & \leq p\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{2}^{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{2}^{2}\left\{ \left(\frac{\mu}{d}\right)^{3/2}+\frac{\mu^{2}\log d}{d^{2}p}+\frac{\mu^{3/2}\sqrt{\log d}}{d^{3/2}\sqrt{p}}\right\} =o\left(1\right)\frac{p}{r}\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{2}^{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{2}^{2}.
\end{align*}
\item Putting these two bounds reveals that with probability at least $1-O\left(d^{-10}\right)$,
\[
\bigg|\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}h_{i,j,k}(l_{1},s)h_{i,j,k}(l_{2},s)\bigg|=o\left(1\right)\frac{p}{r}\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{2}^{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{2}^{2}.
\]
\end{itemize}
In addition, by the symmetric sampling pattern, we can use the triangle
inequality to obtain
\begin{align*}
\bigg|\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}h_{i,j,k}(l_{1},s_{1})h_{i,j,k}(l_{2},s_{2}) & \bigg|\lesssim\bigg|u_{l_{1},s_{2}}^{\star}u_{l_{2},s_{1}}^{\star}\sum_{1\leq k\leq d}u_{l_{1},k}^{\star}u_{l_{2},k}^{\star}\chi_{s_{1},s_{2},k}\bigg|
\end{align*}
for any $1\leq s\neq s_{2}\leq d$ and any $1\leq l_{1},l_{2}\leq r$.
Arguing similarly as in the proof of Lemma~, the following upper
bound holds with probability at least $1-O\left(d^{-20}\right)$,
\begin{align}
\sum_{s_{1}\neq s_{2},1\leq l_{1},l_{2}\leq r}\left|\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}h_{i,j,k}(l_{1},s_{1})h_{i,j,k}(l_{2},s_{2})\right|^{2} & \lesssim\sum_{1\leq l_{1},l_{2}\leq r}\sum_{1\leq s_{1},s_{2}\leq d}u_{l_{1},s_{2}}^{\star2}u_{l_{2},s_{1}}^{\star2}\bigg|\sum_{1\leq k\leq d}u_{l_{1},k}^{\star}u_{l_{2},k}^{\star}\chi_{s_{1},s_{2},k}\bigg|^{2}\nonumber \\
 & \lesssim\asymp o\left(p^{2}\right).\label{eq:FIsher-off-diagonal}
\end{align}

In addition, for any $1\leq s\neq s_{2}\leq d$ and any $1\leq l_{1},l_{2}\leq r$
we can develop the following upper bound
\begin{align*}
\left|\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}h_{i,j,k}(l_{1},s_{1})h_{i,j,k}(l_{2},s_{2})\right| & \lesssim\big|u_{l_{1},s_{1}}^{\star}u_{l_{2},s_{2}}^{\star}\big|\bigg|\sum_{1\leq k\leq d}u_{l_{1},k}^{\star}u_{l_{2},k}^{\star}\chi_{s_{1},s_{2},k}\bigg|\\
 & \leq\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{\infty}^{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{\infty}^{2}\sum_{1\leq k\leq d}\chi_{s_{1},s_{2},k}\\
 & \overset{(\mathrm{i})}{\leq}\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{2}^{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{2}^{2}\frac{\mu^{2}}{d^{2}}\sum_{1\leq k\leq d}\chi_{s_{1},s_{2},k}\\
 & \overset{(\mathrm{ii})}{\leq}\left\Vert \bm{u}_{l_{1}}^{\star}\right\Vert _{2}^{2}\left\Vert \bm{u}_{l_{2}}^{\star}\right\Vert _{2}^{2}\frac{\mu^{2}\left(pd+\log d\right)}{d^{2}}\ind\Big\{\sum_{1\leq k\leq d}\chi_{s_{1},s_{2},k}>0\Big\},
\end{align*}
with probability exceeding $1-O\left(d^{-10}\right)$, where (i) uses
assumption~(\ref{assumption:u-inf-norm}) and (\ref{eq:defn-mu}),
and (ii) arises from the high-probability fact that $\sum_{1\leq k\leq d}\chi_{s_{1},s_{2},k}\lesssim pd+\log d$.
This means that
\begin{align*}
\sum_{s_{1}\neq s_{2},1\leq l_{1},l_{2}\leq r}\left|\sum_{(i,j,k)\in\Omega,\,i\leq j\leq k}h_{i,j,k}(l_{1},s_{1})h_{i,j,k}(l_{2},s_{2})\right|^{2} & \lesssim\left(\frac{pd+\log d}{d^{2}}\right)^{2}\sum_{s_{1}\neq s_{2},1\leq l_{1},l_{2}\leq r}\ind\Big\{\sum_{1\leq k\leq d}\chi_{s_{1},s_{2},k}>0\Big\}\\
 & \lesssim\left(\frac{pd+\log d}{d^{2}}\right)^{2}r^{2}d^{2}p\\
 & \lesssim p^{3}+\frac{p\log^{2}d}{d^{2}}\asymp o\left(p^{2}\right).
\end{align*}

With the above bounds in mind, we are in a position to bound the Fisher
information matrix. For any square matrix $\bm{M}$, denote by $\mathcal{P}_{\mathsf{diag}}(\bm{M})$
a matrix of the same size that extracts out only the diagonal entries
of $\bm{M}$ (so that $\bm{M}-\mathcal{P}_{\mathsf{diag}}(\bm{M})$
represents the off-diagonal components of $\bm{M})$. By combining
(\ref{eq:diagonal-Fisher}) with (\ref{eq:Fisher-Omega}), we see
that
\[
\mathcal{P}_{\mathsf{diag}}\left(\mathcal{I}(\bm{U}^{\star})\right)\preceq\big(1+o(1)\big)\frac{2p}{\sigma_{\min}^{2}}\left[\begin{array}{cccc}
\|\bm{u}_{1}^{\star}\|_{2}^{4}\bm{I}_{d}\\
 & \|\bm{u}_{2}^{\star}\|_{2}^{4}\bm{I}_{d}\\
 &  & \ddots\\
 &  &  & \|\bm{u}_{r}^{\star}\|_{2}^{4}\bm{I}_{d}
\end{array}\right].
\]
Similarly, putting the other estimates together yields
\[
\left\Vert \mathcal{I}(\bm{U}^{\star})-\mathcal{P}_{\mathsf{diag}}\big(\mathcal{I}(\bm{U}^{\star})\big)\right\Vert =o\left(\frac{p}{\sigma_{\min}^{2}}\max_{1\leq l\leq r}\|\bm{u}_{l}^{\star}\|_{2}^{4}\right).
\]
\end{comment}

config: arxiv


TITLE: Can an uncountable union of distinct finite sets be countable or even finite?
QUESTION [0 upvotes]: Does there exist an uncountable set of distinct finite sets such that their union is countable, or even finite?

REPLY [1 votes]: No. Suppose that we have a collection of sets $\{X_i\}_{i \in I}$ for some $I$ an uncountable index set. Recall how we'll define their union:
$$\bigcup_{i \in I} X_i = \{ x \mid \exists i \in I \text{ such that } x \in X_i \}$$
This allows you to very naturally produce an injection $I \to \bigcup X_i$: you can send each $i \in I$ to some $x \in \bigcup X_i$, based on the condition that $x$ is in that very $X_i$. In other words, you're sending each element of the index set, to some element in the union, where that element comes from one of the sets indexed by that element of the index set.
By this, then, $|I| \le | \bigcup X_i |$. So the size of the union must be uncountable if $I$ is.


TITLE: Relation between thermal energy and mass
QUESTION [0 upvotes]: "Suppose you have a glass of water and a beaker of water. They are at the same temperature. The beaker contains twice as much water as the glass. The average kinetic energy of the water molecules is the same in both containers, since they are the same temperature. But there are twice as many water molecules in the beaker as there are in the glass. So the total kinetic energy of the water molecules in the beaker is twice as large as that of the water molecules in the glass. The water in the beaker has twice as much thermal energy as the water in the glass has. If the temperature doesn’t change but the mass of the object increases, the thermal energy in the object increases.  A relatively small sample at a somewhat high temperature such as a cup of tea at its boiling temperature could have less thermal energy than a larger sample such as a pool that's at a lower temperature. If the cup of boiling tea is placed next to the freezing pool, the cup of tea will freeze first because it has less thermal energy than the pool."
So I saw these examples online and I had a doubt whether heat flows from one body to another even if temperature is same just because the larger sample has higher thermal energy?

REPLY [1 votes]: So I saw these examples online and I had a doubt whether heat flows
from one body to another even if temperature is same just because the
larger sample has higher thermal energy?

There's no NET flow of heat between bodies at the same temperature, regardless of their size.
How do we know this? The Laws of Thermodynamics.
How do we know they're true? Hundreds of years of observation without a single violation noted.


text:
\begin{document}

\title{Kostant's problem and parabolic subgroups}

\author{Johan K\aa hrstr\"om}

\maketitle

\abstract{
	Let $\frak g$ be a finite dimensional complex semi-simple
	Lie algebra with Weyl group $W$
	and simple reflections $S$. For $I\subseteq S$ let
	$\frak g_I$ be the corresponding semi-simple subalgebra of $\frak g$.
	Denote by $W_I$ the Weyl group of $\frak g_I$ and let
	$\wcirc$ and $\wIcirc$ be the longest elements of $W$ and $W_I$,
	respectively. In this paper we show that the answer to Kostant's problem,
	i.e. whether the universal enveloping algebra surjects onto the space
	of all ad-finite linear transformations of a given module,
	is the same for the simple highest weight $\frak g_I$-module $L_I(x)$
	of highest weight $x\cdot 0$, $x\in W_I$, as the answer for the
	simple highest weight $\frak g$-module $L(x\wIcirc\wcirc)$
	of highest weight $x\wIcirc\wcirc\cdot 0$.
	We also give a new description of the
	unique quasi-simple quotient of the Verma module $\Delta(e)$ with the
	same annihilator as $L(y)$, $y\in W$.
	}

\section{Introduction}

Let $\frak g=\frak n^-\oplus\frak h\oplus\frak n$ be a
finite dimensional complex semi-simple Lie algebra
with a chosen triangular decomposition, and let $\enva{\frak g}$ be
its universal enveloping algebra. For two $\frak g$-modules $M$ and $N$,
the space $\Hom_\bbC(M, N)$ of linear maps from $M$ to $N$ has
a $\enva{\frak g}$-bimodule structure in the natural way
(see for example~\cite[Kapitel~6]{jantzenEinhullende}), and hence
a $\frak g$-module structure via the adjoint action. The $\frak g$-submodule
of $\Hom_\bbC(M, N)$ consisting of all locally finite elements
is in fact a $\enva{\frak g}$-sub-bimodule, which we denote by
$\calL(M, N)$. As $\enva{\frak g}$ itself is locally finite
under the adjoint action, we have a natural homomorphism of
$\enva{\frak g}$ into $\calL(M, M)$ for every $\frak g$-module $M$, whose
kernel is the annihilator $\Ann M$ of $M$ in $\enva{\frak g}$.
The question raised by Kostant (see for
example~\cite[6.10]{conze}, \cite{josephKostant})
is: for which $\frak g$-modules $M$ is the natural inclusion
\[
	\enva{\frak g}/\Ann M\hookrightarrow\calL(M, M)
\]
a \emph{surjection}.

This is in general a difficult question, and the
answer is not even known for simple highest weight modules.
It is known to have the positive answer for Verma modules (\cite[6.9]{conze} for
simple Verma modules, generalized in \cite[6.4]{josephKostant} for
the general case) and for all quotients
of dominant Verma modules~\cite[6.9]{jantzenEinhullende}.
For semi-simple Lie algebras having roots of different length,
examples of simple highest weight modules
where the answer is negative were found early (see for example~\cite[6.5]{conzeduflo},
\cite[9.5]{josephKostant}). More recently, many examples have also been
found in type $A$ (see~\cite{mazorchukstroppelWedderburn}
and~\cite{kahrstrommazorchuk}).
The answer to Kostant's problem
is a valuable tool for example when determining Goldie rank ratios
(see~\cite{josephGoldieRankOne, josephKostantGoldieRank, josephSumRule}),
and in the study of generalized Verma modules
(see~\cite{milicicsoergel, khomenkomazorchukInduced,
mazorchukstroppelInducedCellModules}).

In this note we investigate how the answer to this question for certain
simple highest weight $\frak g$-modules relates
to the answer for modules of semi-simple subalgebras of $\frak g$.
More precisely,
let $W$ be the Weyl group of $\frak g$, with simple reflections $S$,
determined by the triangular decoposition.
For a subset $I\subseteq S$, let $W_I$ denote the parabolic subgroup
of $W$ generated by $I$, denote by $\frak g_I$ the corresponding
semi-simple subalgebra of $\frak g$, and let $\wcirc$ and $\wIcirc$ denote the longest
elements of $W$ and $W_I$.
For $x\in W$, let $L(x)$ denote the simple
highest weight $\frak g$-modules with highest weight $x\cdot 0$ (see next
section for precise definition),
and similarly, for $x\in W_I$, let $L_I(x)$ denote the simple highest
weight $\frak g_I$-module with highest weight $x\cdot 0$. The main
result of this paper is the following theorem,
which generalizes previous results by
Conze-Berline and Duflo~\cite[2.12 and~6.3]{conzeduflo},
later generalized by Gabber and Joseph~\cite[4.4]{gabberjoseph} (the case when $x=e$),
and Mazorchuk~\cite[Theorem~1]{mazorchukTwistedApproach} (the case when $x$ is a
simple reflection).

\begin{theorem}\label{thm:main}
	Let $x\in W_I$. Then Kostant's problem has the positive answer for
	$L_I(x)$ if and only if Kostant's problem has the positive answer for
	$L(x\wIcirc \wcirc)$.
\end{theorem}

The idea of the proof is as follows. For each $x\in W_I$, there is a unique
quotient $D$ of the dominant Verma module $\Delta_I(e)$ satisfying
$\Ann D = \Ann L_I(x)$. Since Kostant' problem has the positive answer for $D$,
as it is a quotient of a dominant Verma module, we see that
Kostant's problem has the positive answer for $L_I(x)$ if and only if
\begin{equation}\label{eq:intone}
	\calL_I(D, D)\cong\calL_I\bigl(L_I(x), L_I(x)\bigr)
\end{equation}
(where the index $I$ is used to emphasize that objects are defined
with respect to $\frak g_I$ as opposed to $\frak g$).
We show that we can `lift' this situation
by parabolic induction, i.e.
there exists a $\frak g$-module $D'$ for which the answer to Kostant's problem
is positive, and such that
\[
	\calL(D', D')\cong\calL\bigl(L(x\wIcirc\wcirc), L(x\wIcirc\wcirc)\bigr)
\]
holds if and only if~\eqref{eq:intone} holds.

In Section~\ref{sec:D} we give an alternative description of the so-called
\emph{quasi-simple} quotients the dominant Verma module, originally described
in~\cite[Section~5]{josephDixmiersProblem}, which are used as an important
tool in the proof of Theorem~\ref{thm:main}. Finally, in Section~\ref{sec:sl6}
we apply Theorem~\ref{thm:main} to get some new answers to Kostant's problem
for the Lie algebra $\frak{sl}_6$.

\medskip

\noindent{\bf Acknowledgements.}
The author thanks V.~Mazorchuck for fruitful comments, suggestions and discussions.

\section{Notation and preliminaries}

The subset $I$ of $S$ determines a parabolic subalgebra
$\frak p_I$ of $\frak g$, containing $\frak g_I$. The triangular
decomposition of $\frak g$ induces a triangular decomposition
$\frak g_I = \frak n_I^-\oplus\frak h_I\oplus\frak n_I$.
Let $\frak u_I$ be the nilradical of $\frak p_I$, and let $\frak z_I$ be the
orthogonal complement of $\frak h_I$ in $\frak h$ with respect
to the Killing form. We thus have the following decompositions,
\[
\frak h = \frak h_I \oplus \frak z_I,\text{ and }
\frak p_I = \frak g_I\oplus \frak z_I\oplus \frak u_I.
\]

The Weyl group $W$ of $\frak g$ acts on $\frak h^*$ in the natural way $w\lambda$,
but in this setting it is more convenient to consider the so-called `dot action',
given by
\[
	w\cdot \lambda \defeq w(\lambda + \rho) - \rho,
\]
where $\rho$ is the half sum of the positive roots. Similarly we
have both the standard action and dot action of $W_I$ on $\frak h_I^*$.

Let $\catO$ denote the BGG category (see for example~\cite{bgg, humphreys}),
and let $\catO_0$ denote
the principal block of $\catO$, i.e. the full subcategory of $\catO$
consisting of modules that are
annihilated by some power of the maximal ideal of the
center of $\enva{\frak g}$ which annihilates the trivial module.
The simple modules of $\catO_0$ are the simple highest weight modules $L(w)$
of highest weight $w\cdot 0$, where $w$ runs over $W$.
We denote the Verma module with simple head $L(w)$ by $\Delta(w)$,
and the projective cover of $L(w)$ by $P(w)$. Finally, for $w\in W$
we denote
by $\theta_w$ the indecomposable projective functor on $\catO_0$
(see~\cite{bg}) satisfying
\[
	\theta_w\Delta(e) = P(w).
\]
The corresponding objects for $\frak g_I$ are denoted $\catO^I$, $L_I(w)$,
$\calL_I$, etc.

For a subalgebra $\frak a$ of $\frak g$ (here $\frak a$
will be either $\frak h_I$ or $\frak z_I$), a module $M\in\catO$, and
$\lambda\in\frak a^*$, let
\[
	M_\lambda\defeq \bigl\{\,m\in M\,\big\vert\,xm=\lambda(x)m
		\text{ for all $x$ in $\frak a$}\,\bigr\},
\]
and define the support of $M$ with respect to $\frak a$ as
\[
	\Supp_{\frak a} M\defeq
	\bigl\{\,\lambda\in\frak a^*\,\big\vert\,M_\lambda\neq 0\,\bigr\}.
\]

\section{Parabolic induction}

For $\lambda\in\frak z_I^*$, we define the
\emph{induction functor} from $\catO^I$ to $\catO$ by
\[
	\Ind_\lambda M \defeq \enva{\frak g}\otimes_{\enva{\frak p_I}}M^\lambda,
\]
where $M^\lambda$ is the $\frak p_I$-module obtained from $M$
by letting $\frak z_I$ act by $\lambda$, and $\frak u_I$ act by $0$.
We also define the \emph{restriction functor} from $\catO$ to
$\catO^I$ by
\[
	\Res_\lambda M\defeq M_\lambda,
\]
where the action is restricted to $\frak g_I$.

\begin{lemma}\label{lem:indann}
	If $\Ann_{\enva{\frak g_I}} M=\Ann_{\enva{\frak g_I}} N$
	for two $\frak g_I$-modules $M$ and $N$, then
	$\Ann_{\enva{\frak g}}\Ind_\lambda M=\Ann_{\enva{\frak g}}\Ind_\lambda N$
	for any $\lambda\in\frak z_I^*$.
\end{lemma}

\begin{proof}
	We have
	\begin{multline*}
		\Ann_{\enva{\frak p_I}} M^\lambda =
		\bigl(\Ann_{\enva{\frak g_I}} M\bigr)\otimes\enva{\frak z_I}\otimes\enva{\frak u_I}
		+ \enva{\frak g_I}\otimes\ker\lambda\otimes\enva{\frak u_I} \\
		+ \enva{\frak g_I}\otimes\enva{\frak z_I}\otimes\enva{\frak u_I}_{>0},
	\end{multline*}
	where $\enva{\frak u_I}_{>0}$ denotes the elements of $\enva{\frak u_I}$
	of degree at least $1$. Hence
	$\Ann_{\enva{\frak p_I}} M^\lambda = \Ann_{\enva{\frak p_I}} N^\lambda$,
	so the result follows
	from~\cite[Proposition~5.1.7(ii)]{dixmier}.
\end{proof}

Let $\roots_I$ be the simple roots corresponding to $I$.
The fundamental weights of $\frak h_I^*$ dual to $\roots_I$ define a basis $B_I$ of
$\frak z_I^*$, which in turn define a partial order on $\frak z_I^*$
by $\nu\leqslant\lambda$ for $\nu, \lambda\in\frak z_I^*$
if $\lambda-\nu$ is in the non-negative span of $B_I$.
For $\lambda\in\frak z_I^*$ and $M\in\catO$, let
$M_{\not\leqslant\lambda}$ be the submodule of $M$ generated
by all $M_\nu$, $\nu\not\leqslant\lambda$, and define
\[
	M^{\leqslant\lambda} \defeq M/M_{\not\leqslant\lambda}.
\]

Generalising the situation when tensoring Verma modules with finite
dimensional modules, we get the following.

\begin{lemma}\label{lem:tensorfiltration}
	For a finite dimensional $\frak g$-module $V$, $M\in\catO^I$,
	and $\lambda\in\frak z_I^*$,
	the module $V\otimes \Ind_\lambda M$ has a filtration
	\[
		0=M_0\subset M_1\subset\cdots\subset M_k=V\otimes\Ind_\lambda M
	\]
	with
	\[
		M_i/M_{i-1}\cong\Ind_{\lambda+\mu_i}\Bigl(\bigl(\Res_{\mu_i} V\bigr)\otimes M\Bigr),
	\]
	where $\mu_1>\mu_2>\cdots>\mu_k\in\frak z_I^*$ and
	$\Supp_{\frak z_I} V = \{\mu_1, \dots, \mu_k\}$.
\end{lemma}

\begin{proof}
	Let $\mu_1$, $\dots$, $\mu_k\in\frak z_I^*$ be as in the lemma,
	let $B_1$, $\dots$, $B_k$ be bases of $\Res_{\mu_1}V$, $\dots$,
	$\Res_{\mu_k}V$, and
	let $B$ be a basis of $M$. Now define
	\[
		M_i\defeq \sum_{1\leq j\leq i}\enva{\frak g}\Bigl(B_j\otimes
			\bigl(1\otimes_{\enva{\frak p_I}}B\bigr)\Bigr).
	\]
	As in the `standard' case
	(se for instance~\cite[Satz~2.2]{jantzenEinhullende}) we find that
	each $M_i$ is $\enva{\frak u_I^-}$-free over
	\[
		\bigcup_{1\leq j\leq i}B_j\otimes\bigl(1\otimes_{\enva{\frak p_I}} B\bigr).
	\]
	In particular, as $\enva{\frak u_I^-}$-modules we have that
	\[
		M_i/M_{i-1}\cong \enva{\frak u_I^-}\Bigl(B_i\otimes\bigl(1\otimes_{\enva{\frak p_I}} B\bigr)\Bigr).
	\]
	Furthermore, it is straightforward to see that,
	as $\enva{\frak g_I}$-modules,
	\[
		\enva{\frak g_I}\Bigl(B_i\otimes\bigl(1\otimes_{\enva{\frak p_I}}B\bigr)\Bigr)
		\cong \bigl(\Res_{\mu_i}V\bigr)\otimes M,
	\]
	from which the statement follows.
\end{proof}

\begin{corollary}\label{cor:resind}
	For any $\lambda, \mu\in\frak z_I^*$, finite dimensional $\frak g$-module
	$V$, and $M\in\catO^I$, we have
	\[
		\Res_\mu(V\otimes \Ind_\lambda M)^{\leqslant\mu}\cong
		\bigl(\Res_{\mu-\lambda} V\bigr)\otimes M.
	\]
\end{corollary}

\begin{proof}
	If $\mu-\lambda\notin\Supp_{\frak z_I} V$ the result is immediate as both modules
	are zero.
	On the other hand, if $\mu-\lambda\in\Supp_{\frak z_I} V$,
	then by Lemma~\ref{lem:tensorfiltration} the module
	$(V\otimes \Ind_\lambda M)^{\leqslant \mu}$ has a submodule $M'$
	isomorphic to
	\[
		\Ind_\mu\Bigl(\bigl(\Res_{\mu-\lambda}V)\otimes M\Bigr),
	\]
	and
	\[
		\Supp_{\frak z_I}\bigl((V\otimes \Ind_\lambda M)^{\leqslant \mu}/M'\bigr)<\mu,
	\]
	from which the statement follows.
\end{proof}

We now fix $\xi\in\frak z_I^*$ to be the restriction of $\wcirc\cdot 0$
to $\frak z_I$, and let $\catO^\xi$ be the full subcategory of $\catO$
of modules satisfying $\Supp_{\frak z_I} M\leqslant\xi$.
By~\cite[Proposition~11]{mazorchukTwistedApproach},
$\Ind_\xi$ and $\Res_\xi$ induce mutually inverse equivalences
between $\catO^\xi_0$ and $\catO^I_0$, identifying $L_I(x)$ with
$L(x\wIcirc\wcirc)$ and $\Delta_I(x)$ with $\Delta(x\wIcirc\wcirc)$.
Let $\pr_0$ and $\pr^I_0$ denote the projection functors from $\catO$
to $\catO_0$ and $\catO^I$ to $\catO^I_0$, respectively.

\begin{lemma}\label{lem:indrescommute}
	For any $M\in\catO^\xi$ we have
	\[
		\Res_\xi\circ\pr_0(M)\cong\pr^I_0\circ\Res_\xi(M).
	\]
\end{lemma}

\begin{proof}
	Let $\lambda\in\frak h^*$ with $\lambda\vert_{\frak z_I}\leq\xi$.
	If $\lambda\vert_{\frak z_I}<\xi$ then
	\[
		\Res_\xi\circ\pr_0\bigl(L(\lambda)\bigr) =
		\pr^I_0\circ\Res_\xi\bigl(L(\lambda)\bigr) = 0,
	\]
	so assme $\lambda\vert_{\frak z_I}=\xi$.
	We then have that
	\[
		\Res_\xi L(\lambda)\cong L_I\bigl(\lambda\vert_{\frak h_I}\bigr).
	\]
	Furthermore, since $\lambda\vert_{\frak z_I} = (\wcirc\cdot 0)\vert_{\frak z_I}$,
	we have that
	\[
		\pr_0 L(\lambda)\cong
		\begin{cases}
			L(\lambda) & \text{ if $\lambda\in W_I\wcirc\cdot 0$, or equivalently,
					$\lambda\vert_{\frak h_I}\in W_I\cdot 0$,} \\
			0 & \text{otherwise}.
		\end{cases}
	\]
	Hence the statement follows for simple modules since
	\[
		\pr^I_0 L_I\bigl(\lambda\vert_{\frak h_I}\bigr) =
		\begin{cases}
			L_I\bigl(\lambda\vert_{\frak h_I}\bigr)
				& \text{ if $\lambda\vert_{\frak h_I}\in W_I\cdot 0$,} \\
			0 & \text{ otherwise.}
		\end{cases}
	\]
	Now let $M\in\catO^\xi$, and let $M_0\in\catO^\xi_0$ and $M_1\in\catO^\xi$
	be such that
	\[
		M\cong M_0\oplus M_1.
	\]
	By definition, we have
	\begin{equation}\label{eq:projeq1}
		\Res_\xi\circ\pr_0 M \cong \Res_\xi M_0.
	\end{equation}
	Let $L(\lambda)$ be a composition factor of $M_1$.
	If $\lambda\vert_{\frak z_I} < \xi$ then $\Res_\xi L(\lambda)=0$,
	and if $\lambda\vert_{\frak z_I} = \xi$ we must have
	$\lambda\vert_{\frak z_I}\notin W_I\cdot 0$, so
	$\pr^I_0\circ\Res_\xi L(\lambda)=0$. Since both restriction
	and projection are exact it follows that
	\[
		\pr^I_0\circ\Res_\xi M_1=0.
	\]
	On the other hand, since $M_0\in\catO^\xi_0$ we have
	$\Res_\xi M_0\in \catO^I_0$, so
	\[
		\pr^I_0\circ\Res_\xi M_0 \cong \Res_\xi M_0.
	\]
	Since both restriction and projection are additive, it follows that
	\[
		\pr^I_0\circ\Res_\xi M \cong \Res_\xi M_0.
	\]
	Comparing with~\eqref{eq:projeq1} yields the result.
\end{proof}

\section{Proof of Theorem~\ref{thm:main}}

We start by proving the building blocks used in the proof of Theorem~\ref{thm:main}.

\begin{proposition}\label{prop:dimequal}
	For each finite dimensional $\frak g$-module $V$ and $M, N\in\catO_0^I$, we have
	\[
		\Hom_{\frak g}\bigl(V\otimes\Ind_\xi M, \Ind_\xi N\bigr)
		\cong \Hom_{\frak g_I}\bigl(\Res_0V\otimes M, N\bigr).
	\]
\end{proposition}

\begin{proof}
	We have that
	\begin{align*}
		\Hom_{\frak g}\bigl(V\otimes \Ind_\xi M, \Ind_\xi N\bigr)
		&\cong \Hom_{\frak g}\bigl(\pr_0(V\otimes\Ind_\xi M)^{\leqslant\xi},
				\Ind_\xi N\bigr) \\
		&\cong \Hom_{\frak g_I}\bigl(\Res_{\xi}\circ\pr_0(V\otimes\Ind_\xi M)^{\leqslant\xi},
				N\bigr) \\
		&\cong \Hom_{\frak g_I}\bigl(\pr^I_0\circ\Res_{\xi}(V\otimes\Ind_\xi M)^{\leqslant\xi},
				N\bigr) \\
		&\cong \Hom_{\frak g_I}\bigl(\Res_{\xi}(V\otimes\Ind_\xi M)^{\leqslant\xi},
				N\bigr) \\
		&\cong \Hom_{\frak g_I}(\Res_0 V\otimes M, N),
	\end{align*}
	where the first isomorphism follows from the fact that
	$\Ind_\xi N\in\catO_0^\xi$, the second by the adjointness of
	$\Res_\xi$ and $\Ind_\xi$, the third by Lemma~\ref{lem:indrescommute},
	the fourth by the fact that $N\in\catO^I_0$, and the fifth by
	Corollary~\ref{cor:resind}.
\end{proof}

\begin{corollary}\label{cor:dimequal}
	For $M, N\in\catO_0^I$ we have
	\[
		\Hom_{\frak g_I}\bigl(V, \calL_I(M, M)\bigr) \cong
		\Hom_{\frak g_I}\bigl(V, \calL_I(N, N)\bigr)
	\]
	for all finite dimensional $\frak g_I$-modules V if and only if
	\[
		\Hom_{\frak g}\bigl(V', \calL\bigl(\Ind_\xi M, \Ind_\xi M\bigr)\bigr) \cong
			\Hom_{\frak g}\bigl(V', \calL\bigl(\Ind_\xi N, \Ind_\xi N\bigr)\bigr)
	\]
	for all finite dimensional $\frak g$-modules $V'$.
\end{corollary}

\begin{proof}
	For the `only if' part, by Proposition~\ref{prop:dimequal}
	and \cite[6.8~(3)]{jantzenEinhullende} we have
	\begin{align*}
		\Hom_{\frak g}\bigl(V', \calL(\Ind_{\xi}M, \Ind_{\xi}M)\bigr)
		&\cong \Hom_{\frak g}\bigl(V'\otimes \Ind_{\xi}M, \Ind_{\xi}M\bigr) \\
		&\cong \Hom_{\frak g_I}\bigl(\Res_0V'\otimes M, M\bigr) \\
		&\cong \Hom_{\frak g_I}\bigl(\Res_0V'\otimes N, N\bigr) \\
		&\cong \Hom_{\frak g_I}\bigl(\Res_0V', \calL_I(\Ind_{\xi}N, \Ind_{\xi}N)\bigr) \\
		&\cong \Hom_{\frak g}\bigl(V', \calL(\Ind_{\xi}N, \Ind_{\xi}N)\bigr)
	\end{align*}
	for all finite dimensional $\frak g$-modules $V'$.
	Similarly, for the `if' part, we find that
	\[
		\Hom_{\frak g_I}\bigl(\Res_0V', \calL_I(M, M)\bigr)
		\cong\Hom_{\frak g_I}\bigl(\Res_0V', \calL_I(N, N)\bigr)
	\]
	for all finite dimensional $\frak g$-modules $V'$. We need to show
	that this covers all relevant finite dimensional $\frak g_I$-modules.
	We first note that
	\[
		\Hom_{\frak g_I}(V\otimes M, M) \neq 0
	\]
	only if	$V_0 \neq \{0\}$, where $V_0$ denotes the $\frak h_I$-invariant subspace
	of $V$. This follows from the fact that
	\[
		\Supp_{\frak h_I}(V\otimes M)\subseteq \Supp_{\frak h_I}V+\Supp_{\frak h_I}M
	\]
	and, since $M\in\catO_I$,
	\[
		\Supp_{\frak h_I}M\subset\bbZ \roots_I,
	\]
	while, if $V_0=\{0\}$,
	\[
		\Supp_{\frak h_I}V\cap\bbZ\roots_I=\emptyset.
	\]
	On the other hand, extending the highest weight of $V$ from $\frak g_I$
	to $\frak g$ and using the classification of finite dimensional
	$\frak g$-modules (see~\cite[Theorem~7.2.6]{dixmier}
	we have that
	if $V_0\neq\{0\}$ then there is a finite dimensional $\frak g$-module
	$V'$ such that $V$ is a direct summand of $\Res_0 V'$. Now the result
	follows by induction on the dimension of $V$.
\end{proof}

The following crucial observation is due to V.~Mazorchuk.

\begin{proposition}\label{prop:botquot}
	Kostant's problem has the positive answer for any quotient of
	$\Delta(\wIcirc\wcirc)$.
\end{proposition}

\begin{proof}
	Consider a short exact sequence
	\[
		0\rightarrow X\rightarrow \Delta(\wIcirc\wcirc)\rightarrow Y\rightarrow 0.
	\]
	By \cite[Proposition 5]{mazorchukTwistedApproach}, we need to show that
	\[
		\Ext^1_{\catO}\bigl(\Delta(\wIcirc\wcirc), \theta_xX\bigr) = 0
	\]
	for all $x\in W$. Let $C_x$ and $T_x$ denote the completion functor
	and the twisting functor associated with $x\in W$, respectively,
	and let $\calR C_x$ and $\calL T_x$ denote the corresponding right and
	left derived functors.
	They satisfy
	\[
		C_x\Delta(\wcirc)\cong\Delta(x^\inv\wcirc),\text{ and }
		T_x\Delta(x^\inv\wcirc)\cong\Delta(\wcirc),
	\]
	they form mutually inverse equivalences of
	the bounded derived category $\deriv{\catO}$, and they commute
	with projective functors (all this can be found
	in~\cite{andersenstroppel} and~\cite{khomenkomazorchuk}). Hence we have
	\begin{align*}
		\Ext^1_{\catO}\bigl(\Delta(\wIcirc\wcirc), \theta_xX\bigr)
		&\cong \Hom_{\deriv{\catO}}\bigl(\Delta(\wIcirc\wcirc)[-1], \theta_xX\bigr) \\
		&\cong \Hom_{\deriv{\catO}}\bigl(\calR C_{\wIcirc}\Delta(\wcirc)[-1], \theta_xX\bigr) \\
		&\cong \Hom_{\deriv{\catO}}\bigl(\theta_{x^\inv}\Delta(\wcirc)[-1],
			\calL T_{\wIcirc}X\bigr).
	\end{align*}

	To study $\calL T_{\wIcirc}X$,
	we note that $X\in\catO^\xi_0$, and take a projective resolution
	\[
		P^\bullet\epi X:\qquad0\rightarrow P_k\rightarrow\cdots
			\rightarrow P_1\rightarrow P_0\rightarrow X\rightarrow 0,
	\]
	of $X$ in this category.

	In $\deriv{\catO}$ we now have $X\cong P^\bullet$.
	Since $\catO^\xi_0$ is equivalent to
	$\catO^I_0$, since all projective modules in $\catO^I_0$ have Verma flags,
	and since the equivalence maps Verma modules to Verma modules, the
	modules in $P^\bullet$ have Verma flags. Since $T_{\wIcirc}$
	is acyclic on such modules we have $\calL T_{\wIcirc}P^\bullet = T_{\wIcirc}P^\bullet$,
	and hence we have
	\begin{multline*}
		\Hom_{\deriv{\catO}}\bigl(\theta_{x^\inv}\Delta(\wcirc)[-1], \calL T_{\wIcirc}X\bigr) \\
		\cong
		\Hom_{\deriv{\catO}}\bigl(\theta_{x^\inv}\Delta(\wcirc)[-1], T_{\wIcirc}P^\bullet\bigr).
	\end{multline*}
	For $x\in W_I$, let $\tilde P(x\wIcirc\wcirc)$ denote
	the projective cover of the simple $L(x\wIcirc\wcirc)$ in
	$\catO^{\leqslant\lambda}_0$. We have that
	\[
		\Delta(\wIcirc\wcirc) = \tilde P(\wIcirc\wcirc),
	\]
	and, analogous to $\catO_0$, for each $x\in W_I$
	there is a projective functor $\tilde\theta_x$ such that
	\[
		\tilde P(x\wIcirc\wcirc)\cong \tilde\theta_x\Delta(\wIcirc\wcirc).
	\]
	Since twisting functors commute with projective functors
	we have
	\[
		T_{\wIcirc}\tilde P(x\wIcirc\wcirc)
		\cong T_{\wIcirc}\tilde\theta_x\Delta(\wIcirc\wcirc)
		\cong \tilde\theta_x T_{\wIcirc}\Delta(\wIcirc\wcirc)
		\cong \tilde\theta_x \Delta(\wcirc).
	\]
 	Since $\Delta(\wcirc)$ is a tilting module, and projective functors
	take tilting modules to tilting modules, we have that
	$T_{\wIcirc}\tilde P(x\wIcirc\wcirc)$ is a tilting module for all
	$x\in W_I$.
	In particular, $T_{\wIcirc}P^\bullet$ is a complex of tilting modules.
	Similarly, $\theta_{x^\inv}\Delta(\wcirc)$ is a tilting module,
	and hence we have
	\begin{multline*}
		\Hom_{\deriv{\catO}}\bigl(\theta_{x^\inv}\Delta(\wcirc)[-1], T_{\wIcirc}P^\bullet\bigr)\\
		\cong \Hom_{\homot{\catO}}\bigl(\theta_{x^\inv}\Delta(\wcirc)[-1],
				T_{\wIcirc}P^\bullet\bigr),
	\end{multline*}
	by~\cite[Chapter III(2), Lemma~2.1]{happel}, where $\homot{\catO}$ is
	the bounded homotopy category.
	Since $\theta_{x^\inv}\Delta(\wcirc)[-1]$ is concentrated in position $1$,
	and $\calL T_{\wIcirc}P^\bullet$ lies between position $0$ and $-k$, this last
	$\Hom$-space must be zero.
\end{proof}

We can now put the above results together to prove Theorem~\ref{thm:main}.

\begin{proof}[Proof of Theorem~\ref{thm:main}]
	By~\cite[Lemma~3.3]{josephGoldieRankOne},
	there is a (unique)
	quotient $D$ of $\Delta_I(e)$ satisfying
	$\Ann L_I(x)=\Ann D$, and Kostant's problem
	has the positive answer for $D$, since $D$ is a quotient
	of the dominant Verma module (see for example, \cite[6.9]{jantzenModuln}).
	Hence we have
	\begin{equation}\label{eq:lindinjfirst}
		\calL_I(D, D)\cong\enva{\frak g_I}/\Ann D \cong
		\enva{\frak g_I}/\Ann L_I(x)\hookrightarrow\calL_I\bigl(L_I(x), L_I(x)\bigr).
	\end{equation}
	Furthermore, since $L(x\wIcirc\wcirc)\cong \Ind_\xi L_I(x)$ we have
	\[
		\Ann L(x\wIcirc\wcirc) = \Ann\Ind_\xi D
	\]
	by Lemma~\ref{lem:indann}.
	Since $\Ind_\xi D$ is a quotient of
	$\Ind_\xi\Delta_I(\wIcirc)\cong\Delta(\wIcirc\wcirc)$, Kostant's
	problem has the positive answer for $\Ind_\xi D$ by Proposition~\ref{prop:botquot}.
	As above, we have
	\begin{equation}\label{eq:lindinj}
		\calL\bigl(\Ind_\xi D, \Ind_\xi D\bigr)
		\hookrightarrow\calL\bigl(L(x\wIcirc\wcirc), L(x\wIcirc\wcirc)\bigr).
	\end{equation}

	If Kostant's problem has the positive answer for $L(x)$ then the
	injection~\eqref{eq:lindinjfirst} is a bijection, so by
	Corollary~\ref{cor:dimequal} we have
	\[
		\Hom_{\frak g}\Bigl(V, \calL\bigl(\Ind_\xi D, \Ind_\xi D\bigr)\Bigr)
		\cong \Hom_{\frak g}\Bigl(V, \calL\bigl(L(x\wIcirc\wcirc), L(x\wIcirc\wcirc)\bigr)\Bigr)
	\]
	for all finite dimensional $\frak g$-modules $V$. Hence the
	injection~\eqref{eq:lindinj} is a bijection, and Kostant's problem has the
	positive answer for $L(x\wIcirc\wcirc)$. The proof of the converse is
	completely analogous.
\end{proof}

\section{Alternative description of $D$}\label{sec:D}

The module $D$ used in the proof of Theorem~\ref{thm:main} can be described as follows.
If we set $J=\Ann L(x)$, then by \cite[Lemma~3.3]{josephGoldieRankOne},
$J\Delta(e)$ is the unique submodule of $\Delta(e)$ satisfying
\[
	\Ann\bigl(\Delta(e)/J\Delta(e)\bigr) = \Ann L(x).
\]
In particular, $D\defeq \Delta(e)/J\Delta(e)$ is the unique
quotient of $\Delta(e)$ satisfying $\Ann D=\Ann L(x)$.

When beginning this work, the author used a more direct approach
to find the module $D$, inspired by ideas in~\cite{kahrstrommazorchuk}.
Although not necessary for the current exposition, the following result
is interesting in its own right.

\begin{proposition}\label{prop:dquasi}
	Let $x\in W$. The unique quotient $D$ of $\Delta(e)$
	satisfying $\Ann D=\Ann L(x)$ is isomorphic to the image of a
	non-zero homomorphism
	\[
		\Delta(e)\rightarrow\theta_{x}L(x^\inv).
	\]
\end{proposition}

We first note that this image is uniquely defined, since
\begin{align*}
	\dim\Hom_{\frak g}\bigl(\Delta(e), \theta_{x}L(x^\inv)\bigr)
	&= \dim\Hom_{\frak g}\bigl(\theta_{x^\inv}\Delta(e), L(x^\inv)\bigr) \\
	&= \dim\Hom_{\frak g}\bigl(P(x^\inv), L(x^\inv)\bigr) \\
	&= 1.
\end{align*}
To prove Proposition~\ref{prop:dquasi} we need to recall some further theory.

The category $\catO_0$ has a $\bbZ$-graded version
$\catO_0^\bbZ$, in which the modules $L(x)$, $\Delta(x)$ and $P(x)$, for
$x\in W$, all have standard graded lifts (where their heads are
concentrated in degree zero). Furthermore, the projective functors
$\theta_x$, $x\in W$, also have graded lifts, see~\cite{stroppelTranslationFunctors}.
For $M\in\catO_0^\bbZ$
and $i\in\bbZ$,
let $M\langle i\rangle$ denote the graded module defined
by $M\langle i\rangle_j \defeq M_{j-i}$.

The Grothendieck group
of $\catO_0^\bbZ$ is isomorphic to the Hecke algebra $\calH$ of W, i.e. the
free $\bbZ[v, v^\inv]$-module over the basis $\{\,H_x\,\vert\,x\in W\,\}$, where
multiplication is given by $H_xH_y = H_{xy}$ if $\ell(xy)=\ell(x)+\ell(y)$,
and $H_sH_s = H_e + (v^\inv-v)H_s$ for simple reflections $s\in S$.
The Kazhdan-Lusztig basis is a basis of the Hecke algebra,
whose elements we denote by $\underline H_x$, which are self dual
under the duality $H\mapsto\overline{H}$ on $\calH$ given by
$\overline H_x = (H_{x^\inv})^\inv$ and $\overline v = v^\inv$.
We also have the dual Kazhdan-Lusztig basis, whose elements we denote by
$\hat{\underline H}_x$, which is dual to the Kazhdan-Lusztig basis with
respect to the symmetrising trace. We then have
\begin{align*}
	[\Delta(x)] &= H_x, \\
	[P(x)] &= \underline H_x, \\
	[L(x)] &= \hat{\underline H}_x, \\
	[\theta_x\underline{\phantom M}] &=
		\text{ right multiplication by $\underline H_x$, and} \\
	[\underline{\phantom M}\langle i\rangle] &= \text{ multiplication by $v^{-i}.$}
\end{align*}
For a review of this theory, see~\cite{mazorchukstroppelInducedCellModules},
in particular Section~3.

For $x, y\in W$ and $H\in\calH$ let $k^H_{x, y}\in\bbZ[v, v^\inv]$ be such that
\[
	\underline H_xH = \sum_{y\in W}k^H_{x, y}\underline H_y.
\]
The \emph{right preorder} on $W$ is defined by $x\leqslant_R y$ if there exists
an $H\in\calH$ with $k^H_{x, y}\neq 0$. Dually, if
$\hat k^H_{x, y}\in\bbZ[v, v^\inv]$ is such that
\[
	H\underline{\hat H}_x = \sum_{y\in W}\hat k^H_{x, y}\underline{\hat H}_y,
\]
then $x\geqslant_R y$ if and only if there exists a $H\in\calH$
with $\hat k^H_{x, y}\neq 0$ (see~\cite[5.1.16]{lusztigReductive}).
The \emph{left preorder} is defined by $x\leqslant_L y$ if
and only if $x^\inv\leqslant_R y^\inv$.
By~\cite{josephCharacteristicVariety, vogan, kl} we have the important
fact that
\[
	x\leqslant_L y \text{ if and only if } \Ann L(x)\supseteq\Ann L(y).
\]
The
equivalence classes of $\leqslant_R$ and $\leqslant_L$
are called right and left cells, respectively.

For $x, y\in W$, let $h_{x, y}\in\bbZ[v, v^\inv]$ with
\[
	\underline H_y = \sum_{x\in W} h_{x, y} H_x,
\]
and for $x, y, z\in W$, let $k_{x, y, z}\in\bbZ[v, v^\inv]$ with
\[
	\underline H_x\underline H_y =
	\sum_{z\in W}k_{x, y, z}\underline H_z.
\]
Note in particular that $\overline k_{x, y, z}=k_{x, y, z}$.
Now Lusztig's $\luszta$-function on $W$ (see~\cite{lusztigCells})
can be defined as
\[
	\luszta(x) \defeq \max_{y, z\in W}\deg k_{y, z, x}.
\]
It is constant on right cells, and in general
we have (see~\cite[1.3(1)]{lusztigCellsTwo})
\[
	\luszta(x)\leq \mindeg h_{e, x},
\]
where, for $f\in\bbZ[v, v^\inv]$,
$\mindeg f$ is the \emph{minimal degree} of $f$, i.e. the minimal element
$i\in\bbZ$ such that the coefficient of $v^i$ in $f$ is non-zero.
The \emph{Duflo set} $\calD$ (sometimes
called the set of distinguished involutions) is defined
as the set of elements $d\in W$ satisfying
\[
	\luszta(d)=\mindeg h_{e, d}.
\]
By~\cite[Proposition~1.4, Theorem~1.10]{lusztigCellsTwo}, each right
cell contains precisely one Duflo involution. Note that, by the BGG reciprocity,
we have
\[
	[\Delta(e)] = \sum_{x\in W}h_{e, x}[L(x)].
\]
Hence, given a right cell $R$ of $W$, all composition factors on the form
$L(x)$, $x\in R$ of $\Delta(e)$ occur in degree at least $\luszta(x)$,
and there is precisely one such element which occur in degree $\luszta(x)$,
namely the Duflo involution in $R$.

\begin{proof}[Proof of~Proposition~\ref{prop:dquasi}]
	Fix $x\in W$ and denote the image of a non-zero homomorphism from $\Delta(e)$
	to $\theta_xL(x^\inv)$ by $\bar D$. Since $\theta_x$ is exact, applying
	it to
	\[
		P(x^\inv)\epi L(x^\inv)
	\]
	gives
	\begin{equation}\label{eq:pontol}
		\theta_xP(x^\inv)\epi \theta_xL(x^\inv).
	\end{equation}

	Firstly, we have, for some $\hat k_{x^\inv, x, z}\in\bbZ[v, v^\inv]$,
	\[
		\bigl[\theta_xL(x^\inv)\bigr] = \hat{\underline H}_{x^\inv}\underline H_x
		= \sum_{z\in W}\hat k_{x^\inv, x, z}\hat{\underline H}_z
		= \sum_{z\in W}\hat k_{x^\inv, x, z}[L(z)],
	\]
	and $k_{x^\inv, x, z}\neq 0$ implies $z\leqslant_R x^\inv$
	so all composition factors of $\theta_xL(x^\inv)$ are on the form
	$L(y)$, where $y\leqslant_R x^\inv$.
	On the other hand, we have
	\[
		\bigl[\theta_xP(x^\inv)\bigr] = \underline H_{x^\inv}\underline H_x
		= \sum_{z\in W}k_{x^\inv, x, z}\underline H_z
		= \sum_{z\in W}k_{x^\inv, x, z}[P(z)],
	\]
	and $k_{x^\inv, x, z}\neq 0$ implies $z\geqslant_R x^\inv$. Hence the head of
	$\theta_xP(x^\inv)$ has only simple factors on the form $L(y)$, $y\geqslant_R x^\inv$.
	From~\eqref{eq:pontol} it follows that
	$\theta_xL(x^\inv)$ has minimal degree greater than or equal to
	$-\luszta(x^\inv)$, and that the head of $\theta_xL(x^\inv)$ has
	only simple factors on the form $L(y)$, $y\sim_R x^\inv$. Furthermore, since
	$\theta_xL(x^\inv)$ is self-dual, $\theta_xL(x^\inv)$ has maximal degree
	smaller than or equal to $\luszta(x^\inv)$, and all its simple submodules
	are on the form $L(y)$, $y\sim_Rx^\inv$.

	In particular, the maximal degree of $\bar D$ is bounded by $\luszta(x^\inv)$,
	and all simple submodules of $\bar D$ are on the form $L(y)$, $y\sim_Rx^\inv$.
	But the only such submodule occurring on degree $\luszta(x^\inv)$
	or smaller in $\Delta(e)$ is $L(d)$, where $d$ is the unique Duflo involution
	in the same right cell as $x^\inv$,
	occurring precisely once in degree $\luszta(x^\inv)$.
	Hence $\bar D$ has the unique simple submodule $L(d)$, and all other
	simple composition factors are on the form $L(y)$, $y<_Rd$.
	By~\cite[Proposition 6.2 (ii)]{josephKostant} it follows that
	$\Ann\bar D=\Ann L(d)$, and $\Ann L(d)=\Ann L(x)$
	as $d\sim_L x$. Since $D$ is the unique
	quotient of $\Delta(e)$ with this property, we must have $\bar D=D$.
\end{proof}

\section{Kostant's problem for $\mathfrak{sl}_6$}\label{sec:sl6}

In~\cite{kahrstrommazorchuk}, the answer to Kostant's problem was given
for all simple modules in $\catO_0$ for $\mathfrak{sl}_n$, $n\leq 5$,
and partial results were obtained for $\mathfrak{sl}_6$.
In type $A$ the answer to Kostant's problem is a left cell invariant by
\cite[Theorem~60]{mazorchukstroppelInducedCellModules}. Furthermore, since in
type $A$ there is one unique involution in each left cell, it suffices
to solve Kostant's problem for involutions. The Weyl group for $\mathfrak{sl}_6$
is $S_6$, which contains 76 involutions. For 45 of these Kostant's problem
were shown to have the positive answer, for 17 the answer was negative, and for 11
it remained unknown.

We expected that Theorem~\ref{thm:main} would answer many of these 11 unknown cases,
but it actually turned out to answer only two. The involution
$s_1s_2s_1s_5$ is in the same left cell as the element
\[
	s_1s_4\cdot \wIcirc\wcirc,
\]
where $I=\{s_1, s_2, s_3, s_4\}$. By~\cite[Corollary~21]{kahrstrommazorchuk},
Kostant's problem has the positive answer for the $\frak{sl}_5$-module
$L(s_1s_4)$, and hence by Theorem~\ref{thm:main} Kostant's problem has
the positive answer for the $\frak{sl}_6$-module $L(s_1s_2s_1s_5)$.
By symmetry of the Coxeter diagram,
Kostant's problem also has the positive answer for $L(s_1s_4s_5s_4)$.
Hence answer to Kostant's problem is still unknown for the modules
\begin{align*}
&L(s_2s_3s_4s_3s_2), &&L(s_2s_1s_4s_3s_2s_5s_4), &&L(s_1s_3s_2s_4s_3s_2s_1s_5s_4s_3),\\
&L(s_2s_1s_3s_4s_3s_2), &&L(s_1s_2s_3s_2s_4s_3s_2s_1), &&L(s_2s_1s_3s_2s_1s_4s_5s_4s_3s_2),\\
&L(s_2s_4s_3s_2s_5s_4), &&L(s_2s_3s_2s_4s_5s_4s_3s_2), &&L(s_2s_1s_3s_2s_4s_3s_2s_1s_5s_4s_3s_2).
\end{align*}


TITLE: What will be the remainder?
QUESTION [1 upvotes]: I'm stuck with this problem I'm trying to solve from about an hour. Here's the question.
What is the remainder when (3^202)+137 is divided by 101?
There are 4 options -> 36, 45, 56, 11
I want to know the answer of the question with proper and possibly easiest method to solve the problem.
Thanks in advance, waiting for reply. :)

REPLY [2 votes]: HINT $\  101\ $ is prime so a little Fermat $\rm\ \Rightarrow\ \ 3^{101}\ \equiv\ 3\ \ \Rightarrow\ \ 3^{202}\ \equiv\ \ldots\ (mod\ 101)$
Since your comment reveals you are not familiar with modular arithmetic, here is an alternative.
By Fermat's little theorem $101$ divides $\: 3^{101}-3\: $ so it also divides $\rm\ (3^{101}-3)\ (3^{101}+3)\ =\ 3^{202}-9$


TITLE: $\pi$, $\sigma$ - atomic transitions with respect to the magnetic field axis
QUESTION [9 upvotes]: I am confused about the atomic transition with different polarized lights.
  I post the pictures as follows.

There are four cases.
  In case 1, the right-handed circular polarized light propagates along the $\mathbf{B}$ axis. So it induces the $\delta m=+1$ transition(i.e.,$\sigma$+ transition).
In case 2, the left-handed circular polarized light propagates along the $\mathbf{B}$  axis. So it induces the $\delta m=-1$ transition(i.e.,$\sigma$- transition).
In case 3, the light travels normal to the $\mathbf{B}$  axis and it's linear polarized. So it induces the $\delta m=0$ transition(i.e.,$\pi$ transition).
Question a: In case 4, what kind of transition will happen?
Question b: The linear polarized light can be decomposed into $\sigma$+ and $\sigma$- polarized light. If the linear-polarized light travels along the $\mathbf{B}$  axis, there will be $\sigma$+ and $\sigma$- transitions meanwhile. Is it right?

REPLY [8 votes]: a: In case 4, what kind of transition will happen?

In this case, the radiation can excite both $\sigma$ and $\pi$ transitions, although you wouldn't really call them that. More specifically, it can excite transitions which keep $L_z$ constant, and it can also excite transitions which change it by $\Delta L_z=\pm \hbar$.
To see why, simply decompose the circular polarization into a superposition of linear polarizations: one along $z$, and one along (say) $x$. The polarization along $z$ can excite the $\Delta L_z=0$ transitions. The linear polarization along $x$, on the other hand, is itself a superposition of circular polarizations with a $\vec k$ vector along $z$, and each of these excites one $|\Delta L_z|=1$ transition. (Why am I allowed to switch the wave propagation direction? This is all done inside the dipole approximation, which takes the polarization to be uniform, so the atom doesn't actually 'talk' to the propagation direction.)
To do this slightly more rigorously, the transition operator for case 4 is $\hat z+i\hat x$, and it can be decomposed as
$$
\hat z+i\hat x=\hat z+i\left(\frac{\hat x+i\hat y}{2}+\frac{\hat x-i\hat y}{2}\right).
$$
The overall matrix element for the transition is therefore
$$
\left\langle f\middle |\hat z+i\hat x\middle |i\right\rangle
=
\left\langle f\middle |\hat z\middle |i\right\rangle
+i\left\langle f\middle |\frac{\hat x+i\hat y}{2}\middle | i\right\rangle
+i\left\langle f\middle |\frac{\hat x-i\hat y}{2}\middle| i\right\rangle.
$$
Here $\pi$ transitions are those for which $\left\langle f\middle |\hat z\middle |i\right\rangle$ is nonzero, and $\sigma_\pm$ are those for which $\left\langle f\middle |\frac{\hat x\pm i\hat y}{2}\middle | i\right\rangle$ is nonzero. (Don't trust me on the sign conventions, though.)
This then answers your second question pretty directly,

b: The linear polarized light can be decomposed into $\sigma$+ and $\sigma$- polarized light. If the linear-polarized light travels along the $\mathbf{B}$  axis, there will be $\sigma$+ and $\sigma$- transitions meanwhile. Is it right?

From the above simply take
$$
\left\langle f\middle |\hat x\middle |i\right\rangle
=
\left\langle f\middle |\frac{\hat x+i\hat y}{2}\middle | i\right\rangle
+\left\langle f\middle |\frac{\hat x-i\hat y}{2}\middle| i\right\rangle.
$$
and you get your claim.


TITLE: Laurent Series Question: Summation Starts from $-\infty$?
QUESTION [1 upvotes]: Consider $f(z)=\frac{2z-4}{z^2-4z+3}$. Find the Laurent series for $f(z)$ that converges when $|z-1|>2$.

My working:
\begin{align}
f(z)&=\frac{1}{z-3}+\frac{1}{z-1} \\
&=\frac{1}{z-1}\left(\frac{1}{1-\frac{2}{z-1}}\right)+\frac{1}{z-1} \\
&=\sum_{n=0}^{\infty} \left(\frac{2^n}{(z-1)^{n+1}}\right)+\frac{1}{z-1}
\end{align}
Now, the answer provided is, $$\sum_{n=-\infty}^{-1} c_n(z-1)^n, \ \ c_{-1}=2 \ \text{and} \ c_n=2^{-n-1} \ \ n\leq -2.$$
I do not understand in the slightest why this is. What am I missing?

REPLY [1 votes]: $(z-1)^{-k}=\frac{1}{(z-1)^k}$; your answers seem the same up to indexing

REPLY [0 votes]: They are same. But there seem to be typos. It is $c_{-1}$ and $n\leq -2$.


TITLE: How can I use Cauchy integral formula for this integral $g(z)=\int_{C}\frac{s^2+s+1}{s-z}ds$?
QUESTION [2 upvotes]: Let $C$ be the ellipse $9x^2+4y^2=36$ traversed once in the counterclockwise direction. Define the function $g$ by $$g(z)=\int_{C}\frac{s^2+s+1}{s-z}ds.$$
Find $g(4i)$.
Well I know I must find $g(z)$ (that is the integral) before computing $g(4i)$, so I decided to use Cauchy's integral formula $f(z_{0})=\frac{1}{2\pi i}\int_C\frac{f(z)}{z-z_{0}}dz$. This put me into trouble, because I do not how to start. Please i need a hint.
Thanks.

REPLY [1 votes]: $$g(4i)=\int_C \frac{s^2+s+1}{s-4i}ds$$
$\frac{x^2}{4}+\frac{y^2}{9}=1$ is the equation of the vertical ellipse with vertices at $-2, 2, -3i, 3i$. Therefore $4i$ is outside the ellipse and the function $f(s)=\frac{s^2+s+1}{s-4i}$ is analytic within $C$. Then the integral is $0$.


TITLE: Difficult limit involving $n$-th root and $n$-th exponentiation
QUESTION [2 upvotes]: I've encountered this limit on a book without solution and after two days of thinking still I am not able to come up with an answer.

$$\lim_{n\to+\infty}\left(\alpha k^{\frac 1n}-1\right)^n,\: k\in\mathbb R_{>0},\: \alpha\in\mathbb N_{>0}.$$

A full complete and detailed soultion is welcomed. Also hints of course are accepted.
I've tried binomial Formula and to simplify the expression but all has been in vain.

REPLY [2 votes]: First of all, you should prove (if you do not already know) that if $k>0$ then $\lim_{n\to\infty}k^{1/n}=1$. The you have three cases (in the following I assume that $\alpha>0$ is not necessarily an integer):

$0<\alpha<2$: then $|\alpha\,k^{1/n}-1|\to|\alpha-1|<1$ and $\lim_{n\to\infty}\bigl(\alpha\,k^{1/n}-1\bigr)^n=0$.
$\alpha>2$: then $\alpha\,k^{1/n}-1\to\alpha-1>1$ and $\lim_{n\to\infty}\bigl(\alpha\,k^{1/n}-1\bigr)^n=\infty$.
$\alpha=2$: we have a limit of the form $1^\infty$. The case $k=1$ is trival. I will do the case $k>1$ and leave for you the case $0<k<1$. Taking logarithms we see that it is enough to compute
$$
\lim_{n\to\infty}n\ln\bigl(2\,k^{1/n}-1\bigr).
$$
We have
$$\begin{align*}
n\ln\bigl(2\,k^{1/n}-1\bigr)&=\ln\bigl(1+2(k^{1/n}-1)\bigr)\\
&\sim 2\,n\bigl(k^{1/n}-1\bigr)\\
&=2\,n\bigl(e^{\ln k/n}-1\bigr)\\
&\sim 2\ln k
\end{align*}$$
as $n\to\infty$. Thus
$$
\lim_{n\to\infty}\bigl(2\,k^{1/n}-1\bigr)^n=k^2\quad\forall k>1.
$$


TITLE: Closure of real interval in a topology with intervals with rational ends as a bsis.
QUESTION [0 upvotes]: Given a Topolgy over $\mathbb{R}$ with intervals of the form $[p, q), p, q\in\mathbb{Q}$ as a basis, I'm interested in finding the closure of intervals with real ends, such as $\left(-\sqrt{2}, \sqrt{2}\right)$.
I know the closure won't be the interval $\left[-\sqrt{2}, \sqrt{2}\right]$, since $\mathbb{R} - \left[-\sqrt{2}, \sqrt{2}\right]$ cannot be produced out of union of intervals with rational ends. How might I go about finding the closure?

REPLY [2 votes]: The closure is $[-\sqrt{2},\sqrt{2}]$, $\pm \sqrt{2}$ are indeed limit points since we can find points in $(-\sqrt{2},\sqrt{2})$ in any neighbourhood of $\pm \sqrt{2}$, and obviously there can not be any limit point not in $(-\sqrt{2},\sqrt{2})$.Note that since we can take infinite  unions $\mathbb{R}-[-\sqrt{2},\sqrt{2}]$ can be produced out of unions of intervals with rational endpoints.


TITLE: Prove that a group G is finitely generated if and only if there is a surjective homomorphism $F(\{1,..,n \}) \to G$
QUESTION [0 upvotes]: This if from Aluffi's Algebra: Chapter 0. There is an another definition of subgroup generated by a  subset. Here it is:

Let $G$ be a group and $A$ its subset. By universal property of a free group we have a homomorphism $\phi_a:F(A) \to G$ such that $\phi_a(a)=a$. Then its image is the subgroup generated by $A$ in $G$.

REPLY [0 votes]: You have to prove that $\text{Im } \phi_A$, the image of the homomorphism $\phi_A$, is exactly $A$.
One way to prove that is to prove the inclusions
$$A \subseteq \text{Im }\phi_A \text{ and } \text{Im } \phi_A \subseteq A\ .$$
The first inclusion is quite trivial since by hypothesis for every $a \in A \subseteq G$ you have that $a=\phi_A(a) \in \text{Im }\phi_A$, so $A \subseteq \text{Im } \phi_A$.
To prove the other inclusion one simply has to remind that every element of $F(A)$ is (almost) a finite string $(a_1,\dots,a_k)$ of elements of $A$ and that composition in $F(A)$ is given by concatenation.
From this and the fact that $\phi_A$ is an homomorphism it follows that
$$\phi_A((a_1,\dots,a_k))=\phi_A((a_1)\cdots(a_k))=\phi_A((a_1))\cdots\phi_A((a_k))=a_1\dots a_k$$
so every element in the image of $\phi_A$ is a product of elements of $A$, hence it must belong to $A$ itself.
This proves that $\text{Im }\phi_A \subseteq A$, hence the thesis.


TITLE: Fourier series of translated square
QUESTION [1 upvotes]: I can't seem to find the correct Fourier series coefficients ($s_n$) of the following periodic function. I know how to get the Fourier series of the same one that is not vertically translated and has no negative values for any $t$ (Result is $A sinc(n\pi)$), but this one I'm having trouble with. The function is periodic, and during it's period T, is defined with:
$$
  x(t) =
\begin{cases}
A,  & t\in(0,\frac{T}{4}) \\
-A, & t\in(\frac{T}{4},\frac{3T}{4}) \\
A, & t\in(\frac{3T}{4},T)
\end{cases}
$$
How I tried solving this:
$$ s_n = \frac{1}{T}\int_{-T/2}^{T/2}{x(t)e^{-inw_0t}dt} = \frac{1}{T}\Bigg[\int_{-T/2}^{-T/4}{-Ae^{-inw_0t}dt} + \int_{-T/4}^{T/4}{Ae^{-inw_0t}dt} + \int_{T/4}^{T/2}{-Ae^{-inw_0t}dt}\Bigg]$$
Integrating I get:
$$ s_n = \frac{A}{inw_0T}\Bigg[ (e^{inw_0\frac{T}{2}}-e^{-inw_0\frac{T}{2}})-2(e^{inw_0\frac{T}{4}}-e^{-inw_0\frac{T}{4}})\Bigg]$$
Applying Euler's identity and $w_0 = \frac{2\pi}{T}$ and the definition of the $sinc$ function finally I get:
$$ s_n = Asinc(n\pi) - Asinc(\frac{n\pi}{2})$$
That differs from the answer I have in my textbook which says the answer is only $A sinc(\frac{n\pi}{2})$.
It would help if someone could shed some light on where I've done wrong.

REPLY [1 votes]: Since it is said that $x$ is $T$-periodic, you have that
$x(t) = -A$ for $t\in(-T/2,-T/4)$, so you will have to re-evaluate the integrals.
Then, note that
$$\text{sinc}(n\pi) = 0$$


TITLE: Complete expressiveness for language using only the implication connective and propositional constant
QUESTION [1 upvotes]: Show that we can obtain complete expressiveness for a language using only the implication connective if we add to our language the “propositional constant” ⊥ which always has the truth value 0 (false).
I found this problem in Introduction to Logic by Paul Herrick. I'm really stumped on how to prove this.
Any help would be great, thanks in advance!

REPLY [3 votes]: Negation and disjunction are known to be logically complete, so if we can express them in terms of implication and falseness, we are all set.
$$\neg P=P\to\bot$$
$$P\vee Q=(P\to\bot)\to Q$$
and, just to see how to keep building these:
$$P\wedge Q=(P\to(Q\to\bot))\to\bot$$
$$P\equiv Q=((P\to Q)\to((Q\to P)\to\bot))\to\bot$$


TITLE: Is $G$ nowhere-dense in $\mathbb{R}$ iff $\bar{G}^c$ is dense in $\mathbb{R}$?
QUESTION [2 upvotes]: I tried to prove the following theorem whilst having $\mathbb{Z}$ in mind. However, it is by no means a rigorous proof, and because of that, I have the feeling that I may be missing something that cannot be shown by intuitive arguments:
Let $G\subseteq\mathbb{R}$, let $\bar{G}$ stand for the closure of $G$, and let $G^c$ stand for the complement of $G$.

Definitions.

$G$ is dense in $\mathbb{R}$ if and only if $\bar{G}=\mathbb{R}$.
$G$ is nowhere-dense in $\mathbb{R}$ if $\bar{G}$ contains no nonempty open intervals.

Theorem. A set $G$ is nowhere-dense in $\mathbb{R}$ if and only if the complement of $\bar{G}$ is dense in $\mathbb{R}$.
Proof. Let $G$ be nowhere-dense in $\mathbb{R}$. Then $\bar{G}$ contains no nonempty open intervals. This implies that $G$ must be composed of isolated points, and that $\bar{G}^c$ must contain infinitely many nonempty open intervals separated by these isolated points. Therefore, $\bar{\bar{G}^c}=\mathbb{R}$, and $\bar{G}^c$ is dense in $\mathbb{R}$. Conversely, let $\bar{\bar{G}^c}=\mathbb{R}$. It follows that either $\bar{G}^c=\mathbb{R}$ or $\bar{G}^c$ is composed of infinitely many open intervals that are separated by isolated points. If $\bar{G}^c=\mathbb{R}$, then $\bar{G}=G=\emptyset$, which is nowhere-dense in $\mathbb{R}$. If $\bar{G}^c$ is composed of infinitely many open intervals separated by isolated points, then $\bar{G}=G$ is a set composed of isolated points, which is nowhere-dense in $\mathbb{R}$. $\square$
What do you guys think?

REPLY [2 votes]: The statement "This implies that G must be composed of isolated points" is false. To see this, consider the set $\{1,\frac{1}{2},\frac{1}{3},\ldots\}\cup\{0\}$ which is nowhere dense, yet $0$ is not an isolated point.
However, the statement you are trying to prove is correct. Proof: Note that a subset is dense iff it contains points arbitrarily close to any point in the space. Suppose $G$ is nowhere dense in $\mathbb R$, so $\bar{G}$ contains no interval. Then $\bar{G}^c$ intersects every open interval, so for any $r\in\mathbb R$ and $\epsilon>0$ we have $\bar{G}^c\cap(r-\epsilon,r+\epsilon)\neq \emptyset$, thus $\bar{G}^c$ is dense. Suppose $G$ is not nowhere dense in $\mathbb R$, so $\bar{G}$ contains some open interval $I$. Then $I\cap \bar{G}^c=\emptyset$, so $\bar{G}^c$ is not dense.


TITLE: Why is reduction modulo $p$ a group homomorphism on Elliptic Curves?
QUESTION [4 upvotes]: I am reading A. Knapp's book on elliptic curves right now. In Proposition 5.6 the author wants to prove that the reduction map (modulo $p$, where $p$ does not divide the discriminant) of an elliptic curve over $\Bbb Q$ preserves the group structure.
All you need to show is that the chord-tangent construction is preserved, that is, if $E$ is the curve, $r$ is the reduction map and $P \cdot Q$ denotes the third intersection of the line through $P$ and $Q$ with $E$, then
\begin{align*}
r(P) \cdot r(Q) = r(P \cdot Q)
\end{align*}
To prove this, the author uses the following proposition: Given a curve $F$ and a line $L$ in $P_2(\Bbb Q)$ (the two-dimensional prjective space over $\Bbb Q$) and a point $P$, we have $i(P, L, F) \le i(r(P), r(L), r(F))$ where $i$ is the intersection multiplicity.
His proof for the above identity is now as follows: "We apply Proposition 5.5. Since the sum of intersection multiplicities over a line is $\le 3$, the proposition gives $r(P \cdot Q) = r(P) \cdot r(Q)$.
Now here is my problem: What if $P$, $Q$ and $P \cdot Q$ all reduce to the same point modulo $p$? Then the assertion is wrong unless this point is a flex. Clearly, if $L$ is the line through $P$ and $Q$, then its reduction modulo $p$ has intersection multiplicity $1$ with $r(P) = r(Q) = r(P \cdot Q)$. But this does not mean that there is no tangent (over $\Bbb Z_p$) to $r(P)$ so that $r(P) \cdot r(Q) \ne r(P \cdot Q)$, does it?

REPLY [2 votes]: Ok, I found a full and very detailed solution to this problem on A. Knapp's homepage, see here. Apparently, the above problem is indeed a gap in the proof (and a quite big one, as it seems).


TITLE: When is an almost simple group a split extension of its socle?
QUESTION [7 upvotes]: Here an almost simple group is a finite group whose socle (product of all minimal normal subgroups) is a nonabelian simple group. As an extension of its socle, an almost simple group could be split or non-split. For example, there are four groups with socle $L=PSL(2,9)$ other than $L$, i.e. $S_6$, $PGL(2,9)$, $M_{10}$ and $P\Gamma L(2,9)$; the former two are split extensions of $L$ while the latter two are not. Now the question is how to determine all the almost simple groups which is a non-split extension of its socle?

REPLY [4 votes]: See the following paper:
A. Lucchini, F. Menegazzo, M. Morigi.
On the existence of a complement for a finite simple group in its automorphism group. Special issue in honor of Reinhold Baer (1902–1979).
Illinois J. Math. 47 (2003), no. 1-2, 395–418. MR2031330


TITLE: How do I come up with a formula to calculate the possible combinations of this set?
QUESTION [0 upvotes]: Let's say I have the letters a, a, a, b, c, d. I can only arrange them in pairs of 2. If I use up a letter, I cannot use it again. Order doesn't matter.
So the possible sets of pairs I can get are:
(a,a)(a,b)(c,d)
(a,b)(a,c)(a,d)
(a,c)(a,a)(b,d)
(a,d)(a,a)(b,c)
How do I come up with a way to calculate that there would be 4 possiblities? I tried the combination and permutation calculators on the internet but they didn't work because they would either give me 15 probabilities or 30 probababilities.

REPLY [1 votes]: You could consider the cases for whether there is a pair with two $a$'s or not. Then clearly there is only $1$ case where no pair has two $a$'s If a pair does have two $a$'s then another pair must have exactly one and then there are only $3$ ways to arrange the remaining letters. Sometimes there are too many conditions and you must use a combination of techniques (or a more general result than what you currently know).


TITLE: simple question on limits
QUESTION [0 upvotes]: I am getting skeptical about a simple question. It may seem dumb, so I am sorry in advance.
Is the following statement "always" true?
If $\displaystyle{\lim_{i \to \infty}} a_i=0$, then $\displaystyle{\lim_{i \to \infty}} (-1)^ia_i=0$, where $a_i$ are real.
Thanks.

REPLY [2 votes]: HINT
$|(-1)^ia_i|\leq |a_i|\to 0$


TITLE: Multiple combinations from a single set
QUESTION [2 upvotes]: suppose I have a single set of 9 elements {A, B, C, D, E, F, G, H, I} and that I need to create 3 groups of 3 elements each. I know that I can do:
9C3 * 6C3 * 3C3 and that is 1680 different combinations but in those combinations may appear {{A, B, C}, {D, E, F}, {G, H, I}} and {{D, E, F}, {G, H, I}, {A, B, C}}.
Which is the formula to calculate this? I have the notion that I need to divide by something by don't know by what.

REPLY [2 votes]: Your reasoning shows that there are $\binom{9}{3}\binom{6}{3}\binom{3}{3}$ ways to divide our people into three teams of three each, one of the teams to wear blue uniforms,one to wear white, and one to wear red.
Every "nudist camp" division into teams gives rise to $3!$ divisions into uniformed teams. So if $N$ is the number of nudist camp divisions, then: $N=\frac 1{3!}\binom{9}{3}\binom{6}{3}\binom{3}{3}$.
Remark: Alternately,  line up our people in order of age, or student number, or whatever. The first person can choose the rest of her team in $\binom{8}{2}$ ways. The first unchosen person can then choose the rest of her team in $\binom{5}{2}$ ways. So the number of ways is $\binom{8}{2}\binom{5}{2}$.