TITLE: uniform convergence on compact sets in a compactly generated Hausdorff space
QUESTION [1 upvotes]: Let $X$ be a  Hausdorff space that is determined by an upward filtered family $\mathcal{K}=\{K_{\alpha} : \alpha \in I \}$ of compact subsets of $X$, i. e., $X=\cup_{\alpha \in I} K_{\alpha}$ and also :
A subset $F$ is open in $X$ if
and only if $F\cap K_{\alpha}$ is open in $K_{\alpha}$ for all
$\alpha \in I$. (Equivalently,  a subset $F$ is closed in $X$ if
and only if $F\cap K_{\alpha}$ is closed in $K_{\alpha}$ for all
$\alpha \in I$.)
Is the
topology (on $C(X)$) of uniform convergence on compact sets $\{K_{\alpha} :
\alpha \in I \}$  equal to the topology of uniform
convergence on all compact subsets in $X$?
Note that: $C(X)$ is the space of all continuous complex functions on $X$. And also, the
topology (on $C(X)$) of uniform convergence on compact sets $\{K_{\alpha} :
\alpha \in I \}$  equal to the topology induced by the family of seminorms $\{||\cdot||_{ \alpha}\}_{\alpha \in I}$, where $||f||_{
\alpha}:= \sup \{|f(x)| : x \in K_{\alpha} \}$.

REPLY [1 votes]: No.  For instance, let $X=[0,1]$ and let $\mathcal{K}$ be the set of countable closed subsets of $X$ (these determine the topology since a subset of $X$ is closed iff it is sequentially closed).  For each $K\in\mathcal{K}$, choose a function $f_K\in C(X)$ which is $1$ on $K$ but $0$ at some point of $X\setminus K$.  These functions then form a net $(f_K)$ in $C(X)$ which converges uniformly to $1$ on each $K\in\mathcal{K}$ but does not converge to $1$ uniformly on $X$.
More generally, the same construction would work for any compact Hausdorff space $X$ with such a family $\mathcal{K}$ which does not include $X$ itself.


TITLE: basic definition of vector-fields on a manifold
QUESTION [4 upvotes]: Many textbooks introduce vector-fields on a manifold $M$ along the lines of $ X = X_i \frac{\partial}{\partial x_i} $ where -without further ado- $\frac{\partial}{\partial x_i}$ is introduced as a basis vector in Tangent space. Given that the same symbol is widely used as a differential operator in mathematics, should I think about $X$ in terms of an operator rather than a vector?  Or is that notation a clever suggestive trick to help ease calculations - similar to physicists speaking of "ket-" $ | \psi>$ and "bra-" $ < \Phi | $ vectors in a physical Hilbert space that "magically" turn into a scalar-product $ <\Phi|\Psi> $ when "meeting" in a calculation? Similar conceptual difficulties arise for me when in the the definition of 1-forms $\omega = \omega_1 dx $ the object $dx$ is introduced as a "unit" vector when I am used to thinking about $dx$ as an infinitesimal quantity. Cheers!

REPLY [2 votes]: Since no one answered this properly, I'll do it for completeness. Yes, you can regard $X$ as an operator, taking a function $f$ to a function $X(f)$, $p \mapsto X_p(f) = {\rm d}f_p(X_p)$.
Given a coordinate system $\varphi = (x^i)$ for your manifold, we set $$\frac{\partial f}{\partial x^i}(p) \doteq \frac{\partial (f \circ \varphi^{-1})}{\partial x^i}(\varphi(p)),$$where the derivative in the right side is the ordinary partial derivative in Euclidean space. It does not make sense to talk about partial derivatives in an arbitrary manifold without making a choice of coordinate system to transfer the coordinate axes in $\Bbb R^n$ to "coordinate axes" in a manifold. That being said, people denote $\partial/\partial x^i$ by $\partial_i$ to simplify the notation. We define the action of $X = X^i\partial_i$ on a function $f$ by $X(f) = X^i\partial_if$. Indeed, if $(\widetilde{x}^i)$ is another coordinate system in the manifold, and we write $X = \widetilde{X}^i\widetilde{\partial_i}$, then the chain rule gives $$\widetilde{X}^i\widetilde{\partial_i}f = \left(X^j \frac{\partial \widetilde{x}^i}{\partial x^j} \right)\left(\frac{\partial x^k}{\partial \widetilde{x}^i} \partial_kf\right) = X^j \delta^k_j \partial_kf = X^j\partial_jf.$$


text:
\section{Introduction}

This paper is the third part of a trilogy dedicated to the following problem:
given spherically symmetric characteristic
initial data for the Einstein-Maxwell-scalar
field system with a cosmological constant $\Lambda$,
with the data on the outgoing initial null
hypersurface given by a subextremal Reissner-Nordstr\"{o}m black
hole event horizon,
and the remaining data otherwise free,
study the future extendibility of the corresponding
maximal globally hyperbolic development as a  ``suitably regular'' Lorentzian manifold.

We are motivated by the strong cosmic censorship conjecture and the question of determinism in general relativity.
More precisely, the existence
of (non-isometric) extensions of the maximal globally hyperbolic development leads to the breakdown
of global uniqueness for the Einstein equations. If this phenomenon persists for generic initial
conditions then it violates the strong cosmic censorship conjecture.
See the Introduction of Part 1 for a more detailed account of the
mathematical physics context of this work.

In Part~1, we showed the equivalence (under appropriate regularity conditions for the initial data) between the Einstein equations~\eqref{wave_r}$-$\eqref{wave_Omega}
and the system of first order PDE~\eqref{r_u}$-$\eqref{kappa_at_u}. We established existence, uniqueness and identified a breakdown criterion for solutions of this system.
In Part~2, we analyzed the properties of the solution
up to the Cauchy horizon,\footnote{We recall that when it is possible
to isometrically embed the maximal globally hyperbolic development
into a larger spacetime, the boundary of the  maximal globally hyperbolic development is known as the Cauchy horizon.}
 proving, in particular, the stability of the radius function. See Section~\ref{main-one-two} for a summary of our previous results, as well as for the definitions and notation that will be used henceforth.

In this paper we examine the behavior of the renormalized Hawking mass $\varpi$ (see~\eqref{bar_rafaeli}) and the scalar field at the Cauchy horizon. Depending on the control that we have on these quantities, we are able to construct extensions of the metric beyond the Cauchy horizon with different degrees of regularity. The quotient
\[
\Psi:=\frac{k_-}{k_+} > 1,
\]
of the surface gravities (see \eqref{ganhamos}) of the Cauchy and event horizons $r=r_-$ and $r=r_+$ in the reference Reissner-Nordstr\"{o}m black hole
plays an important role in our analysis.

We start by briefly recalling the strategy of Dafermos~\cite{Dafermos1, Dafermos2} to establish mass inflation (that is, blow-up of $\varpi$ at the Cauchy horizon), which naturally generalizes to the case of a non-vanishing cosmological constant. This requires the initial field $\zeta_0$
(see~\eqref{zeta} and~\eqref{RN1}) to satisfy
$$
\zeta_0(u)\geq cu^s\ {\rm for\ some}\ c>0\ {\rm and}\
0<s<\frac\Psi 2-1
$$
(Theorem~\ref{inflation} whose proof is given in Appendix~\ref{prova-massa}).
Since the mass is a scalar invariant involving first derivatives of the metric, its blow up excludes the existence
of spherically symmetric $C^1$ extensions.\footnote{In this paper we will only be concerned with spherically symmetric extensions.} Moreover,
using the techniques of~\cite{DafermosBlack}, one can easily conclude that in this case the Christodoulou-Chru\'sciel inextendibility criterion holds, that is, there is no extension of the metric beyond the Cauchy horizon with Christoffel symbols in $L^2_{\rm loc}$.

The previous approach only allows us to explore a particular subregion of parameter space, corresponding to sufficiently subextremal reference solutions (see the figure below). We proceed by extending the analysis to the full parameter range. First we prove that if the field $\zeta_0$ satisfies the weaker hypothesis
$$
\zeta_0(u)\geq cu^s\ {\rm for\ some}\ c>0\ {\rm and}\ 0<s<\Psi-1,
$$
then either the renormalized mass $\varpi$ or the field $\bigl|\frac\theta\lambda\bigr|$
(see~\eqref{lambda_0} and~\eqref{theta})
blow up at the Cauchy horizon (Theorem~\ref{inflation2}). As a consequence, the Kretschmann curvature scalar also blows up (Remark~\ref{RmkSCC1}).

On the other hand, when the initial field $\zeta_0$ satisfies
$$
|\zeta_0(u)|\leq cu^s\ {\rm for\ some}\ c>0\ {\rm and}\
s>\frac{7\Psi}9-1 > 0,
$$
we show that the mass remains bounded (Theorem~\ref{thmNoMass}). This behavior is in contrast with the standard picture of spherically symmetric gravitational collapse.

The case where no mass inflation occurs is then analyzed in further detail. We construct $C^0$ spherically symmetric extensions of the metric beyond the Cauchy horizon with
the second mixed derivatives of $r$ continuous. There are two natural coordinate choices for the extension, corresponding to either $\lambda=-1$ or $\kappa=1$ (see \eqref{lambda_0} and \eqref{kappa_0})
on the outgoing null ray $u=U$.
Interestingly, these lead to inequivalent $C^2$ structures for the extended manifolds, a fact that is reflected on the behavior of the Christoffel symbols: when the initial data satisfies
$$
c_2u^{s_2}\leq\zeta_0(u)\leq c_1u^{s_1}\ {\rm for\ some}\ \frac{7\Psi}9-1<s_1\leq s_2<\Psi-1,
$$
we prove that one of the Christoffel symbols blows up on $u=U$ at the Cauchy horizon in the $\lambda=-1$ coordinates, but not in the $\kappa=1$ coordinates. Moreover, for both coordinate systems this Christoffel symbol blows up along almost all outgoing null rays, which excludes the existence of $C^{0,1}$ extensions of the metric. Nonetheless, in the $\kappa=1$ coordinates the Christoffel symbols are in $L^2_{{\rm loc}}$
and the field $\phi$ is in $H^1_{{\rm loc}}$ for the whole range of initial data where there is no mass inflation (Corollary~\ref{cc-fails}). Therefore, the Christodoulou-Chru\'sciel inextendibility criterion for strong cosmic censorship does not hold in this setting.

Finally, assuming that
$$
|\zeta_0(u)|\leq cu^s\ {\rm for\ some}\ s>\frac{13\Psi}9-1,
$$
we can bound the field $\frac{\theta}{\lambda}$ at the Cauchy horizon. This allows us to prove
that the solution of the first order system
extends, non-uniquely, to a classical solution beyond the Cauchy horizon (Theorem~\ref{para-la}).
We then show that this solution corresponds to a classical solution of the Einstein
equations extending beyond the Cauchy horizon (Theorem~\ref{fine}).
The metric for this solution is $C^1$ and such that $r\in C^2$ and $\partial_u\partial_v\Omega$
(see~\eqref{metric}) exists and is continuous (Remark~\ref{fail}).
However, we emphasize that the metric does not have to be $C^2$, in spite of the
Kretschmann curvature scalar being bounded.
To the best of our knowledge, these are the first results where the generic existence of extensions as solutions of the Einstein equations is established.

It should be noted that
these results, while valid for all signs of the cosmological constant $\Lambda$, only provide evidence against the strong cosmic censorship conjecture in the case $\Lambda > 0$ (see the discussion in the Introduction of Part~1).

In summary, for a given $\Psi$ and $cu^s\leq\zeta_0(u)\leq Cu^s$, the behavior of the solution at the Cauchy horizon depends on the value of $s$ as described in the following figure.

\begin{center}
\begin{psfrags}
\psfrag{b}{\tiny $\!\frac 97$}
\psfrag{u}{\tiny $\!\frac 92$}
\psfrag{c}{\tiny $2$}
\psfrag{d}{\tiny $3$}
\psfrag{e}{\tiny $4$}
\psfrag{z}{\tiny $\!\!1$}
\psfrag{g}{\tiny $\!\!\!\frac 49$}
\psfrag{h}{\tiny $\!\!1$}
\psfrag{i}{\tiny $\!\!2$}
\psfrag{j}{\tiny $\!\!3$}
\psfrag{k}{\tiny $s=\frac \Psi 2-1$}
\psfrag{m}{\tiny $s=\Psi-1$}
\psfrag{l}{\tiny $s=\frac{7\Psi}9-1$}
\psfrag{n}{\tiny $s=\frac{13\Psi}9-1$}
\psfrag{r}{\tiny $\Psi$}
\psfrag{s}{\tiny $s$}
\psfrag{o}{\tiny $1$}
\psfrag{p}{\tiny $\!\!0$}
\includegraphics[scale=.7]{fig20.eps}
\end{psfrags}
\end{center}
\begin{center}
\begin{psfrags}
\psfrag{e}{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!mass inflation}
\psfrag{u}{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!mass inflation}
\psfrag{b}{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!no mass inflation}
\psfrag{a}{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!no mass inflation}
\psfrag{c}{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!no mass inflation}
\psfrag{d}{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$\theta/\lambda$ bounded}
\psfrag{f}{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$\theta/\lambda$ unbounded}
\psfrag{v}{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! or $\theta/\lambda$ unbounded}
\psfrag{g}{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!smooth extension}
\psfrag{h}{\tiny \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!beyond Cauchy horizon}
\includegraphics[scale=1.2]{fig21.eps}
\end{psfrags}
\end{center}

In Appendix~\ref{appendix-2} we explain how $\Psi$ depends on the physical parameters $r_+$, $r_-$ and $\Lambda$. In particular,
$\Psi$ is a function of $\frac{r_+}{r_-}$ and $\frac{\Lambda r_-^2}3$.

\vspace{3mm}

\noindent {\bf Acknowledgments.} The authors thank M.~Dafermos for bringing
the {\em Epilogue}\/ of~\cite{DafermosBlack} to their attention.

config: arxiv


TITLE: Positive definite matrices?
QUESTION [1 upvotes]: I read that if I add a positive-definite (PD) matrix to a positive-semidefinite (PSD) matrix, then their sum is non-singular (or even PSD). But does that mean that positive-semidefinite matrices are always non-singular? That sounds wrong to me.
Also, is it true that $A^T A$ is always PD or PSD?

REPLY [0 votes]: Recall that an $n \times n$ matrix $A$ is positive-definite if and only if $A = B^T B$ for some $m \times n$ matrix $B$ whose columns are linearly independent. We have therefore that $A^T = (B^T B)^T = B^T (B^T)^T = B^T B = A,$ hence $A$ is symmetric, from which it follows that $A^T A = A^2.$ By the Spectral Theorem, we have that $A$ is orthogonally diagonalizable. Particularly, there exists an invertible matrix $P$ such that $A = PDP^{-1},$ where $D$ is the diagonal matrix whose nonzero entries are the eigenvalues of $A.$ But of course, we have that $$A^2 = (PDP^{-1})^2 = (PDP^{-1})(PDP^{-1}) = PDP^{-1}PDP^{-1} = PDIDP^{-1} = PD^2P^{-1},$$ hence the eigenvalues of $A^2$ are precisely the squares of the eigenvalues of $A.$
Consequently, if $A$ is a positive-definite $n \times n$ matrix, then $A^T A$ is positive-definite. Certainly, we could repeat this entire argument with the relaxed assumption that $A$ is only positive-semidefinite, in which case we would find that $A^T A$ is positive-semidefinite. Of course, a positive-semidefinite matrix could be singular, as it is by definition possible that $0$ is an eigenvalue.


TITLE: Evaluate $\lim_{n\rightarrow\infty} \sqrt{2^n}\int_a^b \cos^n(x+\frac{\pi}{4})\, dx $
QUESTION [0 upvotes]: Evaluate
  $$\lim_{n\rightarrow\infty} \sqrt{2^n}\int_a^b \cos^n \left(x+\frac{\pi}{4}\right)\,dx$$
  knowing that the set of integration $[a,b]$ is included in $(0,\frac{\pi}{2})$.

Now, my attempt was to first notice that the function that we have to integrate is: $\cos^n(x+\frac{\pi}{4})$ and I was thinking that I can use this trigonometric identity :$$\cos(a+b)=\cos(a)\cos(b)+\sin(a)\sin(b)\,,$$ after which I can apply the reduction formula for $\cos^n(x)$. Is this the way to start ?

REPLY [2 votes]: Notice that on $\left[0,\frac{\pi}{2}\right]$ we have
$$1 \geq \sqrt{2}\cos\left(x+\frac{\pi}{4}\right) \geq -1$$
thus by dominated convergence
$$\lim_{n\to\infty} \sqrt{2^n}\int_a^b \cos^n\left(x+\frac{\pi}{4}\right)\:dx = \int_a^b 0 \:dx = 0$$
for any $[a,b] \subset \left[0,\frac{\pi}{2}\right]$


TITLE: Limit as $x\to -\infty$ for function having square root
QUESTION [1 upvotes]: Evaluate
  $$\lim_{x\to  -\infty} \frac{\sqrt{3x^2+2}}{x-2}.$$

My work: when I solved
$$\lim_{x\to -\infty} \frac{\sqrt{3+2/x^2}}{1-2/x}$$
I got answer $\sqrt{3}$
but if we take $|x|$ then I got $-\sqrt{3}$
what should be answer ?
$\sqrt{3}$ or $-\sqrt{3}$ ?

REPLY [3 votes]: $y:=-x.$
Now consider:
$\lim_{y \rightarrow \infty}\dfrac{\sqrt{3y^2 +2}}{-y-2}$


TITLE: Generator of $\mathbb Z/m\mathbb Z\times\mathbb Z/n\mathbb Z$ for coprime $n$ and $m$.
QUESTION [3 upvotes]: For coprime $n$ and $m$ we have that $\mathbb Z/m\mathbb Z\times\mathbb Z/n\mathbb Z\cong\mathbb Z/mn\mathbb Z$. An isomorphism is given by $$a+mn\mathbb Z\mapsto (a+m\mathbb Z,a+n\mathbb Z)$$
Question: Since the cyclic group is generated by (for example) $\bar 1$ and an isomorphism has to map generators to generators to maintain injective, does this mean that $(\bar 1,\bar 1)$ is a generator of the product?

REPLY [1 votes]: If $f\colon G\to G'$ is an isomorphism of cyclic groups, then $f(x)$ is a generator of $G'$ if and only if $x$ is a generator of $G$.

(Easy proof.)
Since, for $m$ and $n$ coprime, the map
$$
a+mn\mathbb{Z}\mapsto(a+m\mathbb{Z},a+n\mathbb{Z})
$$
is (well-defined and) an isomorphism, this implies that
$$
(1+m\mathbb{Z},1+n\mathbb{Z})
$$
is a generator of $\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$.
By the way, every generator of the product is of the form $(a+m\mathbb{Z},a+n\mathbb{Z})$, where $a+mn\mathbb{Z}$ is a generator of $\mathbb{Z}/mn\mathbb{Z}$ (that is, $\gcd(a,mn)=1$).


TITLE: Homotopy equivalence of total space of fibrations
QUESTION [2 upvotes]: I am reading Selick’s book “Introduction to homotopy theory”, and he says in Cor 7.1.5 that for a fibration $p: E \rightarrow B$ with fiber $F$ and $B$ is CW complex, it is equivalent that there exists homotopy retraction $r: E \rightarrow F$ s.t. $rj \simeq 1_F$ where $j:F \rightarrow E$ is inclusion map,  and there exists $r: E \rightarrow F$ s.t. $(r, p): E \rightarrow F \times B$ is homotopy equivalence.
Just before stating this corollary he states that for fibrations $p:X\rightarrow B$ and $q : Y \rightarrow B$ with $B$ : CW-complex and for a map $f: X \rightarrow Y$,  $f$ is homotopy equivalence if and only if $f$ restricts homotopy equivalence on each fiber of $p $ and $q$.
I understand that $(r, p)$ is homotopy equivalence, then there exists homotopy retraction $r'$. But I have no idea why this converse holds. Does anybody prove that?
Thank you.

REPLY [1 votes]: The corollary can only be true if we require the base space $B$ path connected, otherwise there may exist fibers of different homotopy type.
Let $F = p^{-1}(b)$ be the fiber over some $b$. We claim that the following are equivalent:
$(1)$ There exists a homotopy retraction $r : E \to F$.
$(2)$ There exists a map $r : E \to F$ such that $(r,p) : E \to F \times B$ is a fiber homotopy equivalence.
For each map $r : E \to F$ and each $x \in B$ define $r_x = r \mid_{F_x}: F_x = p^{-1}(x) \to F$. What we know is
(a) $r$ is a homotopy retraction if and only if $r_b = r \circ j \simeq 1_F$.
(b) $(r,p)$ is a fiber homotopy equivalence if and only if all $r_x$ are homotopy equivalences.
$(2) \Rightarrow (1)$: By (b) $r_b$ is a homotopy equivalence. Let $\sigma$ be a homotopy inverse for $r_b$. Define $r' = \sigma \circ r$. Then $r'_b = \sigma \circ r_b \simeq 1_F$. i.e. $r'$ is a homotopy retraction.
$(1) \Rightarrow (2)$: By (b) it suffices to show that all $r_x$ are homotopy equivalences. By (a) we know that $r_b$ is one. Let $u : I \to B$ be a path such that $u(0) = b, u(1) = x$. Form the pullback of $(u,p)$. This yields a fibration $p' : E' \to I$ and a map $u' : E' \to E$ such that $u' \circ p = p' \circ u$. The fibers $F'_t = (p')^{-1}(t)$ are mapped by $u'$ homeomorphically onto $F_{u(t)}$. Since $I$ is contractible, there exists a fiber homotopy equivalence $f : F'_0 \times I \to E'$. Define $i_t : F'_0 \to F'_0 \times I, i_t(y) = (y,t)$ and $i'_t : F'_0 \to F'_0 \times \{ t \}, i'_t(y) = (y,t)$. It suffices to show that $F'_0 \stackrel{i'_1}{\rightarrow} F'_0 \times \{ 1 \} \stackrel{f_1}{\rightarrow} F'_1 \stackrel{u'_1}{\rightarrow} F_x \stackrel{r_x}{\rightarrow} F$ is a homotopy equivalence. We have $r_x \circ u'_1 \circ f_1 \circ i'_1 = r \circ u' \circ f \circ i_1 \simeq r \circ u' \circ f \circ i_0 = r_b \circ u'_0 \circ f_0 \circ i'_0$, the latter being a homotopy equivalence.


TITLE: Center of a group with order $p^aq^b$
QUESTION [1 upvotes]: Are there any examples of groups, $G$, such that $|G|=p^aq^b$ (where $p$ and $q$ are distinct primes and $a,b\geq 1$) and $|\mathrm{Z}(G)|=p^a$?
($\mathrm{Z}(G)$ denotes the center of $G$)
I tried seeing if such a group could be constructed by a more general method - by taking the direct product of two groups, $X$ and $Y$, where $|X|=p^a$, $|Y|=q^b$, and $X$ is abelian.
But, of course, $\mathrm{Z}(X\times Y)=\mathrm{Z}(X)\times\mathrm{Z}(Y)$, and since $|Y|$ is a prime power, $Y$ must have a non-trivial center, meaning that $|\mathrm{Z}(X\times Y)|$ will always be some multiple of $p^a$ rather than just $p^a$.
I also looked at some small groups, but I couldn't find any example (because most were Dihedral or Dicyclic groups with trivial/ near-trivial centers).
(I am aware of Burnside theorem, but I can't figure out anything substantive from it in this situation)

REPLY [1 votes]: Unless this is a Sylow/direct products exercise, there's a more elementary argument. If $|Z(G)|=p^a$, then every noncentral element has centralizer of order $p^aq^{b'}$, for some $0< b'<b$; therefore, the class equation yields $p^aq^b=p^a+\sum_iq^{b-b'_i}$: contradiction, because $q$ divides the LHS and $\sum_i$ at the RHS, but not $p^a$.


TITLE: Directional Derivative of a function.
QUESTION [0 upvotes]: I have been trying to do the following question without much success, what seems to be a trivial one, has showed how bad my understanding about multivariable calculus is.
Given $\frac{\partial f}{\partial x}(1,2)=-1$ and $\frac{\partial f}{\partial y}(1,2)=3$
There is a unit vector u=ai+bj such as $\frac{\partial f}{\partial u}(1,2)=4$
According to what I know, I simply had to use the directional derivative formula, which is the partial derivative of a function with respect to a unit vector:
$$\frac{\partial f}{\partial u}(1,2)=\frac{\partial f}{\partial x}(1,2)a+\frac{\partial f}{\partial y}(1,2)b$$
which led to a true(T) statement, but it is false(F).
I really would like to understand what I am missing here.

REPLY [0 votes]: By what you know about the partials of $f$, your equation gives
$$
\frac{\partial f}{\partial u} = -a+3b = 4.
$$
(I am dropping all references to $(1,2)$ out of laziness.)
Solve for $a$ to get $a = 3b-4$.  But wait, as $u = \langle a, b \rangle$ is to be a unit vector you also know that $a^2+b^2=1$.  Plug in $a=3b-4$ and do all the algebra.   What do you see?
When you do this right, you might be surprised by what you get.  It is a great exercise in understanding directional derivatives and gradients to see why you shouldn't be surprised.  At this point you really understand it.


TITLE: The only great book that Bourbaki ever wrote?
QUESTION [18 upvotes]: OK, the title is opinionated and contentious, but I have a definite
question. I know that the title refers to the Bourbaki volume
Groupes et Algèbres de Lie (Chapters 4-6), published in 1968, but

who said that it is the only great book that Bourbaki ever wrote?

The only reference I can find is the 2009 Prize Booklet for the AMS-MAA
Joint Meetings, where no source is given, but I'm sure I've seen the
claim somewhere else.
Edit. I have rolled back the title of this question to almost its original
form, because putting the title in quotes misled some people into thinking
I sought a source for the exact phrase "the only great book that Bourbaki
ever wrote." Rather, I wanted a source (not necessarily unique) for the idea
that Chapters 4-6 of Groupes et Algèbres de Lie  is Bourbaki's one great
book. Gerald's answer and Jim's comment together are exactly what I wanted.

REPLY [4 votes]: In a similar vein, Godement wrote in (1982, p. 6.28; translation):

... the previous lemma, which we have taken from N. Bourbaki, Lie Groups and Lie Algebras, chap. III, (the most unreadable presentation of the theory of Lie groups ever published since Sophus Lie, but fortunately the chapters on semisimple Lie groups and algebras make up for this)

Also Borel (1998):

A good example is provided by Chapters 4, 5, and 6 on reflection groups and root systems.
It started with a draft of about 70 pages on root systems. The author was almost apologetic in presenting to Bourbaki such a technical and special topic, but asserted this would be justified later by many applications. When the next draft, of some 130 pages, was submitted, one member remarked that it was all right, but really Bourbaki was spending too much time on such a minor topic, and others acquiesced. Well, the final outcome is well known: 288 pages, one of the most successful books by Bourbaki. It is a truly collective work, involving very actively about seven of us, none of whom could have written it by himself.


TITLE: Positiveness of the largest Lyapunov exponent
QUESTION [4 upvotes]: Let $\alpha\in \mathbb{R} / \mathbb{Q}$,
\begin{equation}
A(x)=\left(\begin{array}{ll}
\frac{1}{{\lambda}^2}-2 \cos  2\pi x -1& 2\lambda \cos  2\pi x-\frac{1}{{\lambda}} \\
\frac{1}{{\lambda}} & -1
\end{array}\right)
\end{equation}
The Lyapunov exponent (LE) of $(\alpha, A)$ is given by
\begin{equation}
LE(\lambda)=\lim _{n \rightarrow \infty} \frac{1}{n} \int_{\mathbb{R} / \mathbb{Z}} \ln \left\|A_{n}(x)\right\| d x
\end{equation}
where
$$
A_{n}(x)=A(x+(n-1) \alpha) A(x+(n-2) \alpha) \cdots A(x).
$$
How to prove that when $\lambda>0$ is large enough, the Lyapunov exponent is
$LE>0$  ?

REPLY [0 votes]: I think you can use the main result of the paper of Michael Herman "Une méthode pour minorer..." https://link.springer.com/article/10.1007%2FBF02564647 in order to show that LE is positive when $\lambda$ is sufficiently large.


TITLE: Example tripped Kolmogorov and Wiener
QUESTION [1 upvotes]: Assuming the hint is true, I attempt to prove the latter prop:
Assume on the contrary that $\mathscr{L} = \mathscr{R}$.
If $\sigma(Y_0) \subseteq \mathscr{L}$, then $\sigma(Y_0) \subseteq \mathscr{R}$.
Since $\sigma(Y_0)$ and $\mathscr{R}$ are independent, $\sigma(Y_0)$ is independent of itself.
This means $\forall F \in \sigma(Y_0), P(F \cap F) = P(F)P(F) \ \to \ P(F) \in \{0,1\}$.
Choose $F = (Y_0 = 1)$ or $F = (Y_0 = -1)$. We have $P(F) = 1/2 \notin \{0,1\}$
Is that right? I based it off the K 0-1 Law proof here.

REPLY [0 votes]: Yeah, it looks fine to me. –  Shalop Jul 16 at 23:11


TITLE: Prove that there are no positive integer solutions. Please check my proof.
QUESTION [0 upvotes]: Edit: Thank you for the help. It makes much more sense now.
I'm trying to learn proofs on my own, and I would really appreciate any feedback for my proof below. Are there any logical errors? Is there a better way to do this? thanks very much in advance.
Prove that there are no positive integer solutions for $x^2+x+1=y^2$.
My proof:
Suppose there exists a pair of integers $x$ and $y$ such that $x^2+x+1=y^2$.
If $y\in Z$, then $y^2$ is a perfect square. It then follows that $x^2+x+1$ is also a perfect square, so can be written in the form $(x+a)^2$ for some $a \in Z$. Then
$$x^2+x+1=(x+a)^2=x^2+2ax+a^2.$$
Comparing coefficients gives both $a=1$ and $a=\frac{1}{2} \notin Z$. Therefore $x^2+x+1$ is not a perfect square because it cannot be written in this form. Thus $y^2$ is not a perfect square.
Hence $y$ is not an integer and this contradicts the assumption that $x$ and $y$ are integers for $x^2+x+1=y^2$.

REPLY [0 votes]: $$x^2<x^2+x+1 =y^2<(x+1)^2$$
Since $x^2$ and $(x+1)^2$ are two consecutive perfect squares, $x^2+x+1$ is not a perfect square.
There is no positive integer solution.
But for $x=0$, $y=\mp 1$


TITLE: What is the probability that 5 randomly chosen cards in a deck add up to 40 or greater?
QUESTION [4 upvotes]: I have made a probability game, where you have to pull out any 5 cards without looking (from a deck of 52 cards), and if all five cards add up to 40 or more, they player pulling the 5 cards from the deck wins. What is the probability of winning the game?
Face and Ace Card values:
Ace = 1
Jack = 11
Queen = 12
King = 13
The value of the number cards are the ones that are stated on the card: for example a card that has 2 on it has a value of 2.
There is also a grand prize, and it would be if it just added up to 40. How many combinations are there and why?

REPLY [4 votes]: Glen_b's answer is good enough for all practical purposes, but I'll include here the computation of the exact probability (computer-assisted, but I think it could be done by hands with enough time, courage and a few additional tricks). There are 2598960 possibilities of choosing 5 cards among 52. It is fast enough for a computer to test every possibility, but we can be smarter.
When choosing 5 cards among 52, how many cards of different heights can we get? There are 6 possibilities, or partitions (the partition number of 5 minus 1, because we can't get 5 cards of the same height):

(1,1,1,1,1) (all the heights are distinct)
(1,1,1,2) (a pair)
(1,2,2) (double pair)
(1,1,3) (three of a kind)
(2,3) (full house)
(1,4) (four of a kind)

In addition, there are respectively 1024, 384, 144, 64, 24 and 4 different ways of realizing one of these hands when the number of cards at each height are given (e.g. if I tell you that you have one "eigth" and four "King", that corresponds to 4 differents hands, depending on the colour of the "eigth"). These numbers are weights.
Now, for each partition, we compute how many heigths give rise to a sum of at least 40. For instance, for (1,4), we have to find the number of solutions of $x+4y \geq 40$ under the constraints $1 \leq x,y \leq 13$ and $x \neq y$. For the last four partitions, it can be done manually (it is easy for the last two ones), but the computer is very handy for the first two. We find respectively 337, 916, 291, 326, 59 and 64 favourable outcomes. If we multiply by the weights and sum, this yields 760092 favourable outcomes.
Hence, the probability of getting at least 40 is:
$$\frac{761272}{2598960} = \frac{95159}{324870} \simeq 0,293$$
This method may seem complicated, especially given that the problem can be brute-forced, but we have reduced the number of possibilities to check from 2,5 millions to about 5 thousands, not too far from being doable without a computer (and, at least, there are some interesting mathematics).
Edit: the error has been corrected. I just made a typo when I computed $337*1024+916*384+291*144+326*64+59*24+64*4 = 761272$ (my message initially gave this sum at 760092). The reasoning is sound.


TITLE: Unsolved problems in number theory
QUESTION [5 upvotes]: What are the most interesting examples  of unsolved problems in number theory which an 18 year can understand?

REPLY [3 votes]: A problem I particularly like is Broccard's problem.
I investigated this one by myself some time ago and broke the  Berndt&Galway's $2000$ record by a $100x$ factor with an ordinary computer and ordinary computer code. Of course it is not an interesting result because I did not provided better algorithms or theory, just $13$-years better technology.
There are lots of Unsolved problems in number theory and some of them are not paid enough attention.
Try it if you want to!
P.S. And give special thanks to the guy who implemented legendre() in the GMP library :)


TITLE: Prove that the function $e(a,b)=\frac{d(a,b)}{1+d(a,b)} \ where \ a,b\in X$, is a metric
QUESTION [0 upvotes]: I am looking to prove that the following function from the non empty group $X$ is a metric with $d$ already being a metric:
$$e(a,b)=\frac{d(a,b)}{1+d(a,b)} \ \text{where} \ a,b\in X$$
But I am not sure how to proceed with such equations.
UPDATE:
From the comments posted so far, I interpreted the following:

$$d(a,b) \geq 0$$
$$e(a,b) \geq 0 \implies \frac{d(a,b)}{1+d(a,b)} \implies d(a,b) \geq 0 \Rightarrow \text{true}$$
$$d(a,b) = 0 \implies \frac{d(a,b)}{1+d(a,b)} = 0\implies d(a,b) = 0 \implies a=b$$
$$d(a,b)=d(b,a) \implies e(a,b) = e(b,a) \implies \frac{d(a,b)}{1+d(a,b)} = \frac{d(b,a)}{1+d(b,a)}$$
$$d(a,c) \leq d(a,b) + d(c,d) \implies e(a,c) \leq e(a,b) + e(b,c) \implies \frac{d(a,c)}{1+d(a,c)} \leq \frac{d(a,b)}{1+d(a,b)} + \frac{d(b,c)}{1+d(b,c)}$$

Am I following the right path?

REPLY [1 votes]: You are almost there, but there are some issues about the order you've written things in.

Since $d$ is a metric, $d(a,b)\geq 0$ for all $a,b\in X$. It follows that $1+d(a,b)\geq 1$ and $e(a,b)=\frac{d(a,b)}{1+d(a,b)}\geq 0$ for all $a,b\in X$.

2.
$$e(a,b)=0 \iff d(a,b)=0 \iff a=b$$
where the last equivalence comes from the fact that $d$ is a metric.

Since $d$ is a metric $d(a,b)=d(b,a)$ for all $a,b\in X$. It follows that $e(a,b)=e(b,a)$ for all $a,b\in X$.
Lastly, we need to prove that the triangle inequality holds. Take any $a,b,c\in X$. We must show $e(a,b)+e(a,c)\geq e(a,c)$. This is the hard part of this question. There is a nice solution here: If $d$ is a metric, then $d/(1+d)$ is also a metric


TITLE: Partial fraction decomposition of even function
QUESTION [2 upvotes]: I need to do partial fraction decomposition of this function (to solve its integral):
$\frac{t^2}{t^4+4}$
Since $t^4+4=(t^2+2t+2)(t^2-2t+2)$ I would do:
$\frac{t^2}{t^4+4}=A\frac{2t+2}{t^2+2t+2}+B\frac{1}{t^2+2t+2}+C\frac{2t-2}{t^2-2t+2}+D\frac{1}{t^2-2t+2}$
And then continue with calculations.
According to the teacher's notes though, I can instead write it like this because the function is even:
$\frac{t^2}{t^4+4}=A\frac{2t+2}{t^2+2t+2}+B\frac{1}{t^2+2t+2}-A\frac{2t-2}{t^2-2t+2}+B\frac{1}{t^2-2t+2}$
But I can't understand how to get it, knowing that the function is even.
Can you please help me?

REPLY [0 votes]: Since the function is even we have that
$$A\frac{2t+2}{(t+1)^2}+B\frac{1}{(t+1)^2}+C\frac{2t-2}{(t-1)^2}+D\frac{1}{(t-1)^2}$$ is equal to
$$A\frac{-2t+2}{(-t+1)^2}+B\frac{1}{(-t+1)^2}-C\frac{2t+2}{(t+1)^2}+D\frac{1}{(t+1)^2}.$$ Reording terms we get
$$-C\frac{2t+2}{(t+1)^2}+D\frac{1}{(t+1)^2}-A\frac{2t-2}{(-t+1)^2}+B\frac{1}{(-t+1)^2}.$$
Comparing the first and third expressions we get $A=-C, B=D.$

REPLY [0 votes]: Let $t=2u$
$$\dfrac{2u^2}{u^4+1}=\dfrac{u^2-1+u^2+1}{u^4+1}=\dfrac{1-1/u^2}{(u+1/u)^2-2}+\dfrac{1+/u^2}{(u-1/u)^2+2}=?$$


text:
\begin{document}
\maketitle
\begin{abstract}
  In this paper we prove a version of a trace Paley--Wiener theorem
  for tempered representations of  a reductive $p$--adic group. This is applied to complete certain investigation of Shahidi on the proof that
  a Plancherel measure is invariant of a $L$--packet of discrete series.
\end{abstract}

\section{Introduction}

Let $G$ be a reductive $p$--adic group.
Let $Rep(G)$ be the category of smooth admissible complex representations of $G$ of finite length, and let
$R(G)$ be the corresponding Grothendieck group. We write $\Psi(G)$  (resp., $\Psi^u(G)$) for the group of (resp., unitary)
unramified characters of $G$. $\Psi(G)$  has a structure of algebraic variety (a complex tours), and algebra of regular functions
$\mathbb C\left[\Psi(G)\right]$ is generated by evaluations with elements of $G$. The subgroup  $\Psi^u(G)$ is Zariski dense in
 $\Psi(G)$. We say that a complex function is regular on $\Psi^u(G)$ if it is a restriction
of a regular function on $\Psi(G)$. We observe that the restriction map from $\mathbb C\left[\Psi(G)\right]$  into functions on
 $\Psi^u(G)$ is injective since $\Psi^u(G)$ is Zariski dense in
 $\Psi(G)$.

We fix a minimal parabolic subgroup $P_0$, its Levi decomposition  $P_0=M_0U_0$, and, as usual related to these choices,  we fix a set of
standard parabolic subgroups $P=MU$, where $M_0\subset M$, $P=MP_0$. Since the standard parabolic subgroup is determined by the choice of Levi  subgroup,
the normalized parabolic induction $\Ind_P^G(\sigma)$, where $\sigma$ is a smooth representation of $M$, we write as usual $i_{GM}(\sigma)$.

\vskip .2in

 In \cite{bdk}, Bernstein, Deligne, and Kazhdan  proved a trace Paley--Wiener theorem for category $Rep(G)$. We consider
 a full subcategory $Rep_t(G)$ of $Rep(G)$ consisisting of representations having all irreducible subquotients tempered. Let $R_t(G)$ be the corresponding Grothendieck group.
We write
$R^i_t(G)$ for the subgroup generated by $i_{GM}(\sigma)$, where $M$ ranges over standard Levi subgroups of $G$ (including $G$), and
$\sigma$ ranges over a set of square--integrable irreducible representations of $M$. We warn the reader that this notion is
not an analogue of the notion of structly induced modules from (\cite{bdk}, 3.1).
\vskip .2in
We prove the following version of a trace Paley--Wiener theorem:

\begin{Thm}\label{main} Let $f: R_t(G)\longrightarrow \mathbb C$ be the $\mathbb Z$--linear form such that
there exists an open compact subgroup $K\subset G$
which dominates $f$ (i.e., $f$ is non--zero only on those irreducible tempered representations which have a non--trivial space of
$K$--invariant vectors), and, for each standard maximal Levi subgroup $M$ and a
square--integrable modulo center representation  $\sigma$ of $M$,  the the function $\psi\mapsto f(i_{GM}(\psi\sigma))$ is regular on
$\Psi^u(M)$, and for any other proper standard Levi subgroup $N$ we have  $f(i_{GM}(\psi\sigma))=0$ for all $\psi\in \Psi^u(N)$.
Then, there exists $F\in C_c^\infty(G)$ such that $f(\pi)=\text{tr}{(\pi(F))}$, for all $\pi\in R^i_t(G)$.
\end{Thm}

\vskip .2in
An effective version of this theorem is given by Proposition \ref{p}. This constructs correct function needed in the proof of (\cite{Sh}, Proposition 9.3 2))
in the case when $M$ (see notation there is a Levi of a maximal parabolic subgroup. We remark that since Plancherel factors are multiplicative, it is enough to prove
(\cite{Sh}, Proposition 9.3 2)) for maximal Levi subgroup.

\vskip .2in
I would like to than Gordan Savin for turning my attention to this question. A draft of the paper was written while the author visited the Hong Kong University of Science and Technology
in January of 2018. The author would like to thank A. Moy and the  Hong Kong University of Science and Technology for their hospitality.

\section{Proof of Theorem \ref{main}; preliminaries}

This theorem is a proved by reduction to the main result of \cite{bdk}. Let $M$ be a standard Levi subgroup then we write
$\Psi(M)^r$ the group of all unramified characters $\psi$ such that $\psi(G)\subset \mathbb R_{>0}$.

For  a standard Levi subgroup $M$, a tempered
representation $\pi$ of $M$, and  a ``positive character'', with respect to $P$,  $\psi\in \Psi(M)^r$, the module $i_{GM}(\psi\pi)$ is called a
standard module; it has a unique
(Langlands quotient) $L(i_{GM}(\psi\pi))$. The condition of positivity is empty if $M=G$.
By the Langlands classification, every irreducible representation can be expressed in the form  $L(i_{GM}(\psi\pi))$ for unique such
datum $(M, \pi, \psi)$.
The following standard result will be used in the proof:

\begin{Lem}\label{l-1} The standard modules of $G$ form a $\mathbb Z$--basis of $R(G)$.
\end{Lem}
\begin{proof} As in (\cite{clo}, Proposition 1).
\end{proof}

\vskip .2in
In analogy with (\cite{bdk}, 2.1), we make the following definitions.

\vskip .2in
Let $\sigma \in\Irr (M)$ where $M$ is a standard Levi subgroup of $G$.  We define the usual affine  variety attached to $\sigma$
$$
\Irr(M)\supset D(\sigma)=\Psi(M)\sigma=
\Psi(M)/Stab_{\Psi(M)}(\sigma),
$$
where $Stab_{\Psi(M)}(\sigma)$ is a finite group consisiting of all
$\psi\in \Psi(M)$ such that  $\psi\sigma\simeq \sigma$.

\vskip .2in
If $A$ is a maximal split torus in the centre of $M$, then the restriction map $\Psi(M)\rightarrow  \Psi(A)$ is surjective, and the kernel is a finite
group. Therefore, by considering the restriction to $A$ we find that

$$
Stab_{\Psi^u(M)}(\sigma)=Stab_{\Psi(M)}(\sigma).
$$
So, we may consider
$$
D^u(\sigma)\overset{def}{=}\Psi^u(M)/Stab_{\Psi^u(M)}(\sigma)\subset D(\sigma).
$$
It is easy to see that $D^u(\sigma)$ is Zariski dense in $D(\sigma)$.

The action of the Weyl group
$$
W(M)=N_G(M)/M
$$
on $\Psi(M)$ is algebraic. Moreover,  $w\in W(M)$ transforms $Stab_{\Psi(M)}(\sigma)$ onto $Stab_{\Psi(M)}(w(\sigma))$.
So that it maps $D(\sigma)$ (resp., $D^u(\sigma)$) onto $D(w(\sigma))$ (resp., $D^u(w(\sigma))$).

\vskip .2in
Put $D=D(\sigma)$ and $D^u=D^u(\sigma)$. As usual, we consider the group  $W(D)$ of all $w\in W(M)$
such that $w(D)=D$. Explicitly, $W(D)$ consists of all $w\in W(M)$  such that there exists
$\psi_w\in \Psi(G)$ such that
\begin{equation}\label{20000}
w(\sigma)\simeq \psi_w \sigma.
\end{equation}
The character $\psi_w$ is determined uniquely modulo $Stab_{\Psi(M)}(\sigma)$.

The resulting orbit space
$$
D/W(D)
$$
is affine variety with algebra of regular functions given by
$$
\mathbb C\left[D\right]^{W(D)}.
$$

Explicitly, this variety is isomorphic to the orbit space of
$$
\Psi(G)/Stab_{\Psi(M)}(\sigma)
$$
under the action of $W(D)$ given by (see (\ref{20000}))

\begin{equation}\label{20001}
w.\psi Stab_{\Psi(M)}(\sigma)=\psi_w w(\psi) Stab_{\Psi(M)}(\sigma).
\end{equation}
This explicit realization will be important in the next section where we perform explicit
constructions.

One can construct a regular function $D/W(D)$ in the following way:

\begin{Lem}\label{l-2} Let $f\in C_c^\infty(G)$. Then $\psi\longmapsto tr\left(i_{GM}(\psi\sigma)\right)$
belongs to $\mathbb C\left[D\right]^{W(D)}$.
\end{Lem}
\begin{proof} It is standard that this function is regular on $D$. We show that it is $W(D)$--invariant.
Let $w\in W(D)$. By (\cite{bdk}, Lemma 5.3 (iii)), we have
$$
 tr\left(i_{GM}(\psi\sigma)\right)= tr\left(i_{GM}(w(\psi\sigma))\right)
$$
which completes the proof.
\end{proof}

The above explicit description shows that  analogously defined group $W(D^u)$ is a subgroup of
$W(D)$. In fact, those groups are equal if $\sigma$ has unitary central character. In this case
$D^u/W(D)$ is a subset of $D/W(D)$. It is Zariski dense in $D/W(D)$.

\vskip .2in
Now, we begin the preparation for the proof of the theorem.
The following lemma is a fundamental result of Harish--Chandra:

\begin{Lem}\label{l-3} Assume that $M$ and $N$ are standard Levi subgroups of $G$, and  $\sigma$ and $\tau$ are  square--integrable modulo center
representations of $M$ and $N$, respectively.
Then,  $i_{GM}(\sigma)$ and  $i_{GN}(\tau)$ have a common irreducible subrepresentation if and only
if there exists $w\in G$ such that $N=wMw^{-1}$ and $\tau\simeq w(\sigma)$, where $w(\sigma)$ is defined by
$w(\sigma)(n)=\sigma(w^{-1}nw),$ $n\in N$. Moreover, if there exists $w\in G$ such that $N=wMw^{-1}$, then
 $i_{GM}(\sigma)$ and  $i_{GM}(w(\sigma))$ are ismorphic, and in particular equal in  $R_t(G)$.
\end{Lem}
\begin{proof} \cite{w}.
\end{proof}

For irreducible $\pi\in R_t(G)$,
there exists a standard Levi subgroup  $M$ and a square--integrable modulo center $\sigma$ of $M$ such that
$\pi\hookrightarrow i_{GM}(\sigma)$, the pair is $(M, \sigma)$ is unique up to a conjugation by Lemma \ref{l-3}.
We call the equivalence class   $[M, \sigma]$ under conjugation of the pair  $(M, \sigma)$ the $t$--infinitesimal character
of $\pi$. The set of equivalence of such pairs we denote by $\Theta_t(G)$.

\vskip .2in

For a pair $(M, \sigma)$, we define a natural map $\Psi^u(M)\longrightarrow \Theta_t(G)$ given by
$$
\psi\longmapsto [M, \psi \sigma].
$$
The image is called a connected component of $\Theta_t(G)$. We denote it by  $\Theta_t(M, \sigma)$. This map induces a bijection
which enable us to identify
$$
\Theta_t(M, \sigma)=D^u(\sigma)/W(D(\sigma))
$$
which gives an embedding
$$
\Theta_t(M, \sigma)\subset
D(\sigma)/W(D(\sigma))
$$
realizing $\Theta_t(M, \sigma)$ as a Zariski dense subset of affine variety
$D(\sigma)/W(D(\sigma))$.

\vskip .2in
As in (\cite{bdk}, 2.1), we can decompose
$$
R_t(G)=\oplus_{\theta} R_t(G)(\theta),
$$
where $\theta$ ranges over connected components of $\Theta_t(G)$. Here
$$
R_t(G)(\theta)
$$
is generated with all tempered representations which $t$--infinitesimal characters belonging to $\theta$.
\vskip .2in

\vskip .2in
In analogy with \cite{bdk}, we make the following definition. We say that a  $\mathbb Z$--linear form
$f: R_t(G)\longrightarrow \mathbb C$ is {\it good} if
there exists an open compact subgroup $K\subset G$
which dominates $f$ (i.e., $f$ is non--zero only on those irreducible tempered brepresentations which have a non--trivial space of
$K$--invariant vectors), and, for each standard Levi subgroup $M$ and a
square--integrable modulo center representation  $\sigma$ of $M$,  the the function $\psi\mapsto f(i_{GM}(\psi\sigma))$ is regular on
$\Psi^u(G)$ (consequently, using (\cite{bdk}, Lemma 5.3 (iii)), as in the proof of
Lemma \ref{l-2}, it is regular function on $\Theta_t(M, \sigma)$).

 \vskip .2in
The support of a  $\mathbb Z$--linear form $f: R_t(G)\longrightarrow \mathbb C$  is the set of all connected components $\theta$ of
$\Theta_t(G)$ such that $f\neq 0$ on  $R_t(G)(\theta)$.

\begin{Lem}\label{l-4}  Every good functional is supported only on finitely many connected components.
\end{Lem}
\begin{proof} Obvious.
\end{proof}

\vskip .2in
Lemma \ref{l-4} reduces the proof of the theorem to the case where $f$ is supported on only one orbit. We consider this case and even more in the
next section.

\section{Proof of the Theorem; main part}

In this section we fix a standard maximal Levi subgroup $M$, a square--integrable representation $\sigma\in \Irr(M)$, and a regular function
$a\in \mathbb C[\Psi(M)]$ which is invariant under right--translations under $Stab_{\Psi(M)}(\sigma)$ and
$W(D)$--invariant under the action given by  (\ref{20001}) where $D= \Psi(M)\sigma$. This notation is explained in the previous section.
We prove the following proposition:

\vskip .2in
\begin{Prop}\label{p} Under above assumptions, there exists $F\in C_c^\infty(G)$ such that
$$
\text{tr}{(\pi(F))}
=\begin{cases}
a(\psi) \ \ \text{for} \ \ \pi=i_{GM}(\psi\sigma), \ \ \psi\in \Psi(M),\\
0 \ \ \text{for} \ \ \pi=i_{GN}(\psi\tau), \ \ \psi\in \Psi(N),\\
\end{cases}
$$
for any standard proper  Levi subgroup $N$ and a square--integrable representation $\tau$ such that
$\Theta_t(N, \tau)\neq \Theta_t(M, \sigma)$.
\end{Prop}

The proof of Proposition \ref{p} is a generalization of  (\cite{clo}, 4.2, Proposition 1) where the proof of existence of pseudo--coefficients
for semisimple $G$ is given based also on \cite{bdk}.

\vskip .2in

We remark that $\Psi^u(G)$ acts on  $\Psi^u(M)$ in a usual way:
$$
\psi\longmapsto \chi|_M \psi, \ \  \chi \in \Psi^u(G), \ \ \psi\in \Psi^u(M).
$$
For $\psi\in \Psi^u(M)$, the stabilizer
$$
Stab_{\Psi^u(G)}(i_{GM}(\psi\sigma))
$$
is the group of all $\chi \in \Psi^u(G)$ such that
$$
\chi i_{GM}(\psi\sigma)\simeq  i_{GM}(\psi\sigma).
$$
We remind the reader that for all $\chi \in \Psi^u(G)$ we have

$$
\chi i_{GM}(\psi\sigma)\simeq  i_{GM}(\chi|_M \psi\sigma).
$$

\vskip .2in
\begin{Lem}\label{p-2}
 Assume that $\chi \in \Psi^u(G)$  and $\psi\in \Psi^u(M)$.
Then, for each irreducible constituent $\pi$ of $i_{GM}(\psi\sigma)$, the multiplicity of
$\chi\pi$ in $\chi i_{GM}(\psi\sigma)$ is same as that of $\pi$  in $i_{GM}(\psi\sigma)$.
\end{Lem}
\begin{proof} Obvious.
\end{proof}

\vskip .2in
\begin{Lem}\label{p-3}
Assume that for $\chi \in \Psi^u(G)$  and $\psi\in \Psi^u(M)$
there exists an irreducible constituent $\pi$ of $i_{GM}(\psi\sigma)$ such that $\chi\pi$ is an irreducible constituent of $i_{GM}(\psi\sigma)$.
Then,  $\chi\in Stab_{\Psi^u(G)}(i_{GM}(\psi\sigma))$.
\end{Lem}
\begin{proof} First,
$\chi\pi$ is a common constituent of $i_{GM}(\psi\sigma)$ and $i_{GM}(\chi|_M \psi\sigma)$. So, by Lemma \ref{l-3}, there exists $w\in W(M)$
such that
$$
\chi|_M\psi\sigma=w(\psi\sigma).
$$
Then, again  by Lemma \ref{l-3}, we obtain
$$
\chi i_{GM}(\psi\sigma)\simeq  i_{GM}(\psi\sigma).
$$
\end{proof}

\vskip .2in
\begin{Lem}\label{p-4} Let $\psi\in \Psi^u(M)$. Then, we have the following:
\begin{itemize}
\item[(i)] If $\chi\in Stab_{\Psi^u(G)}(i_{GM}(\psi\sigma))$, then
$a(\chi|_M \psi)=a(\psi)$.
\item[(ii)] For each $\eta\in \Psi(G)$ and  $\chi\in Stab_{\Psi^u(G)}(i_{GM}(\psi\sigma))$, we have
$$
a(\chi|_M \eta|_M \psi)=a(\eta|_M \psi).
$$
\end{itemize}
\end{Lem}
\begin{proof}  We prove (i). Since $\chi\in Stab_{\Psi^u(G)}(i_{GM}(\psi\sigma))$, we obtain
$$
 i_{GM}(\chi|_M \psi\sigma) \simeq \chi i_{GM}(\psi\sigma)\simeq  i_{GM}(\psi\sigma).
$$
So,  by Lemma \ref{l-3}, there exists $w\in W(M)$
such that
$$
\chi|_M\psi\sigma=w(\psi\sigma).
$$

This implies
$$
w\in W(D),
$$
and
$$
\chi|_M\psi\sigma=\psi_w w(\psi)\sigma.
$$
Consequently (see (\ref{20001}))
$$
\chi|_M\psi  Stab_{\Psi(M)}(\sigma)= \psi_w w(\psi) Stab_{\Psi(M)}(\sigma)= w.\psi Stab_{\Psi(M)}(\sigma).
$$
This implies $a(\chi|_M \psi)=a(\psi)$. This proves (i).

To prove (ii), we may assume that  $\eta$ is unitary. Then, we obviously have
$$
Stab_{\Psi^u(G)}(i_{GM}(\eta|_M\psi \sigma))=Stab_{\Psi^u(G)}(i_{GM}(\psi \sigma)).
$$
Now, the claim follows from (i).
\end{proof}

\vskip .2in

Now, we define a $\mathbb Z$--linear form $f: R(G)\longrightarrow \mathbb C$ in several steps.

\vskip .2in
\noindent{1.} For each $\Psi^u(G)$--orbit $\cal O$ in $\Psi^u(M)$, we fix a representative $\psi_{\cal O}\in \cal O$ and an irreducible
constituent $\pi_{\cal O}$ in $i_{GM}(\psi_{\cal O}\sigma)$. By Lemma \ref{p-2}, we have
\begin{equation}\label{p-4a}
Stab_{\Psi^u(G)}(\pi_{\cal O})\subset  Stab_{\Psi^u(G)}(i_{GM}(\psi_{\cal O}\sigma)).
\end{equation}
The quotient is finite and if $\chi$ ranges over representatives of
the quotient, then $\chi \pi_{\cal O}$ ranges over the set of all mutually non--equivalent ireducible subrepresentations in
$i_{GM}(\psi_{\cal O} \sigma)$ which are $\Psi^u(G)$--equivalent to $\pi_{\cal O}$. Any of those representations, have the same multiplicity
in  $i_{GM}(\psi_{\cal O} \sigma)$. Let $m_{\cal O}$ be the sum of their multiplicities. We define:

$$
f\left(\chi\pi_{\cal O}\right)=\frac{a(\psi_{\cal O})}{m_{\cal O}}, \ \  \chi\in Stab_{\Psi^u(G)}(i_{GM}(\psi_{\cal O}\sigma)).
$$
Lemma \ref{p-4} shows that this is well--defined.

\vskip .2in
\noindent{2.} For each $\chi \in \Psi^u(G)$, we obviously have
$$
Stab_{\Psi^u(G)}(\chi \pi_{\cal O})=Stab_{\Psi^u(G)}(\pi_{\cal O})
$$
and
$$
Stab_{\Psi^u(G)}(i_{GM}(\chi|_M\psi_{\cal O}\sigma))=Stab_{\Psi^u(G)}(i_{GM}(\psi_{\cal O}\sigma)).
$$
By, Lemma \ref{p-2} and these remarks, the sum of multiplicities of  $\Psi^u(G)$--equivalent representations of $\pi_{\cal O}$
which belong to $i_{GM}(\chi|_M\psi_{\cal O}\sigma)$ is again $m_{\cal O}$. We let
$$
f\left(\chi\pi_{\cal O}\right)=\frac{a(\chi|_M \psi_{\cal O})}{m_{\cal O}}, \ \  \chi \in \Psi^u(G).
$$

\vskip .2in
\noindent{3.} For any other tempered irreducible representation $\pi$ of $G$ we let
$$
f(\pi)=0.
$$

\vskip .2in
\noindent{4.} For any quasi--tempered irreducible representation $\pi$ of $G$, we write $\pi=\chi\pi^u$, where $\chi$ is real unramified character of
$G$ and $\pi^u$ is tempered.  We let
$$
f(\pi)=0
$$
if $\pi^u$ is not in $\Psi^u(G)\pi_{\cal O}$ for any orbit $\cal O$. But if   $\pi^u\in \Psi^u(G)\pi_{\cal O}$ for some $\cal O$, then
we can write  $\pi^u= \psi \pi_{\cal O}$, for some $\psi \in \Psi^u(G)$ uniquely determined modulo  $Stab_{\Psi^u(G)}(\pi_{\cal O})$. We let
$$
f\left(\pi\right)=\frac{a(\chi|_M  \psi|_M  \psi_{\cal O})}{m_{\cal O}}.
$$
Using (\ref{p-4a}) and Lemma \ref{p-4} we see that this is well--defined.

\vskip .2in
\begin{Lem}\label{p-5} $f$ is supported on $\Theta_t(M, \sigma)$.  Moreover,
we have the following:
$$
f(\pi)=\begin{cases}
a(\psi) \ \ \text{for} \ \ \pi=i_{GM}(\psi\sigma), \ \ \psi\in \Psi^u(M),\\
0 \ \ \text{for} \ \ \pi=i_{GN}(\psi\tau), \ \ \psi\in \Psi^u(N),\\
\end{cases}
$$
for any standard Levi $N$ and a square--integrable representation $\tau$ such that
$\Theta_t(N, \tau)\neq \Theta_t(M, \sigma)$.
\end{Lem}

\vskip .2in

\noindent{6.} Finally, we define $f$ on non--tempered Langlands quotients (see Lemma \ref{l-1}).  let  $f$ to be  equal to zero on all standard modules
(induced from proper parabolic subgroups) except at those of the form

$$
i_{GM}(\chi \psi\sigma),
$$
where $\psi\in \Psi^u(M)$, $\chi \in \Psi(M)^r$ ``positive character'', with respect to $P$, where $P$ is standard parabolic subgroup
with Levi $M$. In this case we let
$$
f(i_{GM}(\chi \psi\sigma))=a(\chi \psi).
$$

It is also possible that $\chi \in \Psi(M)^r$ ``positive character'', with respect to $P$. Then, there exists standard maximal parabolic subgroup
$Q$ with standard Levi and $w\in G$ such that $N=wMw^{-1}$ . Now, in $R(G)$, we have
$$
i_{GM}(\chi \psi\sigma)=i_{GN}(w(\chi) w(\psi)w(\sigma)),
$$
and $w(\chi)$ ``positive character'', with respect to $P$. On the standard module $i_{GN}(w(\chi) w(\psi)w(\sigma))$ we let
$$
f(i_{GN}(w(\chi) w(\psi)w(\sigma)))
=a(\chi \psi).
$$
The third case that $\chi \in \Psi(M)^r$ is in neither chamber. Then it is in $\Psi(G)^r$, and
$$
i_{GM}(\chi \psi\sigma)=\chi i_{GM}(\psi\sigma)
$$

\vskip .2in
One can check  the assumptions of (Theorem 1.2, \cite{bdk}) as in \cite{clo}. This completes the proof of Proposition \ref{p}. Having completed the proof of Proposition \ref{p},
Theorem \ref{main} is proved.


TITLE: Finding supremum of a sequence - i can't understand my instructor
QUESTION [0 upvotes]: I have the sequence:
$A = \{ 1 - \frac{1}{n} | n \in \mathbb{N}\}$
I need to find the surpremum with correct mathmatical notation. I got some help from my math instructor, i thought i understood it, but some of the steps i don't understand. I'll make the steps and write when the logic fails for me.
First let's find an upper bound:
$1 - \frac{1}{n} \leq 1$
Check of it is the lowest upper bound:
pick and epsilon $\forall \epsilon > 0 $ or: $1 - \epsilon$
Now comes the step i don't understand. He said:
$\exists n \in \mathbb{N} \space | \space 1 - \epsilon < 1 - \frac{1}{n} \leq 1$
This part i don't understand. Why should $1- \epsilon < 1 - \frac{1}{n}  $? then it is not in the sequence.. Shouldn't we only check it is smaller than 1, and still be inside the sequence?
the last steps he did was:
$\Rightarrow n >  \frac{1}{\epsilon}$
For $n >  \frac{1}{\epsilon}$:
$1 - \epsilon < 1 - \frac{1}{n} \leq 1$
and then this should be the conclusion.. But i'm not sure what we just concluded.
I would very much appreciate your perspective..

REPLY [0 votes]: So we know that $1-\frac1n ≤1$, hence $1$ is an upper bound of A. To show that $1$ is actually the lowest upper bound, we assume that there is a lower one and lead that to a contradiction. If there is a lower upper bound than $1$, it must be lower than $1$ and hence we can write it as $1-\epsilon$ for a (possibly very small) $\epsilon >0$. But then we can choose $n \in \Bbb N$ with $n>\frac1\epsilon$ and we see $$1-\frac1n>1-\frac{1}{\frac1\epsilon}=1-\epsilon.$$
So $1-\epsilon$ can't be an upper bound, because we just found a term of the sequence which is greater.
Hence we conclude that there can't be an upper bound of the form $1-\epsilon$ for $\epsilon >0$ and thus 1 is the lowest upper bound of $A$.


TITLE: What is the value of a scalar raised to a vector?
QUESTION [0 upvotes]: I have been reading about logistic regression and it says that the hypothesis functions is as following:

Theta is a vector that is transposed. If x is a scalar, then their product is a vector. How do you calculate for instance 1^[1 2 3] ?

REPLY [0 votes]: As someone who's studied this before, I can confirm @MattSamuel's explanation is correct. But the original equation's choice to write $\theta\cdot x$ as $\theta^Tx$ involves a slight abuse of notation, as I've discussed before here.


TITLE: Average distance a car makes while looking for a slot in an autopark
QUESTION [1 upvotes]: So I have the following question on my mind but I could not model it mathematically,
Assume we have a car park, which is just composed of a single line of length L, and every slot's width is 1 meter, and also assume that we have X slots in total. Y is the number of free slots inside the car park. Also assume that every slot has equal probability of being free.
Now a new car comes in autopark, what is the average distance it makes given we have Y number of free slots?
For Y=1, the average distance is Length/2, since for every slot probability of being free is 1/X.
Is it possible to obtain a general formula for this problem? Thank you for your help

REPLY [0 votes]: Let's define the distance traveled as $1$ if we park in the first slot, $2$ if we park in the second slot, etc., so if the distance is $d$ then $1 \le d \le L-Y+1$.
The probability that the first slot is occupied is $(L-Y)/L$, the probability that the second slot is also occupied is $(L-Y-1)/(L-1)$, etc., and the probability that slot $d$ is free given that the previous $d-1$ slots are occupied is $Y/(L-d+1)$. So the probability that the first free slot is $d$ is
$$P(d) = \frac{L-Y}{L} \cdot \frac{L-Y-1}{L-1} \cdot \frac{L-Y-2}{L-2} \cdots \frac{L-Y-d+2}{L-d+2} \cdot \frac{Y}{L-d+1}$$
The expected distance travelled is then
$$E(d) = \sum_{d=1}^{L-Y+1} d \; P(d)$$