TITLE: The Jeep Problem and Nash's Friends
QUESTION [1 upvotes]: The classical jeep problem is the following. A jeep can carry a maximum load of fuel of 1 gallon, and it travels $l$ miles with $l$ gallons of fuel. The jeep moves along a straight line, and is required to cross a desert $x$ miles wide in the most economic way, that is minimizing the required fuel. Let us say that $x > 0$ is the abscissa of the starting point, and $0$ that of the ending point. At $x$ there is a fuel station, where the jeep can load the fuel, while at any point $y < x$, there is a dumping station where the jeep can dump part of its fuel, in order to load it in a future trip.
The solution in this case was found by Fine, The Jeep Problem, Americ. Math. Monthly, Vol. 54 (1947), 24-31. The minimum required quantity of fuel $f(x)$ is the piecewise linear function having slope $2n+1$ on the interval $\left[ D_n, D_{n+1} \right]$, where $D_0=0$ and
\begin{equation}
D_n = \sum_{k=1}^{n} \frac{1}{2k-1}
\end{equation}
for $n > 1$. From this result Fine easily derives the asymptotic formula for $f(x)$:
\begin{equation}
f(x) = \frac{1}{4} \exp(2x - \gamma) + \mathcal{O}(e^{-2x}),
\end{equation}
where $\gamma$ is Euler's constant.
Now let $N$ be a positive integer and let $g(x,N)$ be the minimum required fuel to cross the desert when the jeep can dump (and subsequently load) the fuel only at the points
\begin{equation}
y_1 = \frac{x}{N}, y_2 = 2 \frac{x}{N}, \dots, y_{N-1}= \frac{N-1}{N} x.
\end{equation}
Suppose to choose for every $x \geq 0$ the positive integer $N(x)$ such that $x^2 / N(x) \rightarrow 0$ as $x \rightarrow \infty$.
Fine states at the end of his paper without proof that $[g(x,N(x)) - f(x)]/f(x)$ is not larger that $1 / 2$ for large $x$, meaning that
\begin{equation}
\limsup_{x \rightarrow \infty} \frac{g(x,N(x))-f(x)}{f(x)} \leq \frac{1}{2}.
\end{equation}
Does anyone know how to prove this result?
Thank you very much for your attention.
PS I discovered the Jeep Problem in the book "A Beautiful Mind" by Sylvia Nasar. In chapter 17, she tells a curious story in which Nash was challenged by one of his friends to give an upper bound for the minimum required quantity of fuel. Nasar writes in the book that "there is no optimal solution to the problem, as it turns out". She is not explicit about what version of the jeep problem Nash and his friends were discussing: it seems from her words that it was the version later analyzed by Rote and Zhang, Optimal Logistics for Expeditions: the Jeep Problem with Complete Refilling. In any case, also for this version there is an optimal solution, and this existence property is true for all the versions I know of the jeep problems (see the references in https://en.wikipedia.org/wiki/Jeep_problem and http://mathworld.wolfram.com/JeepProblem.html ), so I think Nasar simply made a wrong statement (maybe she simply meant that an optimal solution was not known at that time, or simply she quoted a wrong statement by someone).
PSS A very short and elegant proof that $f(x)$ is the minimum required quantity of fuel is found in Gale, The Jeep Once More or Jeeper by the Dozen, Americ. Math. Monthly, 77 (1970), 493-501.

REPLY [0 votes]: Finally, I found the answer to my question.
First of all, we can easily give a recursive solution to the problem as follows.
Let us note that if $P$ is a feasible plan of trips which allows the jeep to arrive at $0$, then we can find another feasible plan $P'$ which arrives at $0$ such that $P'$ is made up a number of trips among $x=y_N$ and $y_{N-1}$, followed by a number of trips among $y_{N-1}$ and $y_{N-2}$, and so on, up to a number of trips among $y_1$ and $y_0=0$.
Indeed, assume that the jeep makes $m$ round trips starting at $x$, followed by a last one-way trip $A_{m+1}$ from $x$ to $y_{N-1}$. The i-th of these round trips is made up of the one-way trip $A_i$ from $x$ to $y_{N-1}$, followed by a round trip $B_i$ starting and ending at $y_{N-1}$, and by a one-way trip $C_i$ from $y_{N-1}$ to $x$. Let $g_i$ be the fuel of the jeep when it starts the trip $A_i$. Suppose we replace the sequence of trips $A_1$, $B_1$, $C_1$, ..., $A_m$, $B_m$, $C_m$, $A_{m+1}$ with the sequence $A_1$, $C_1$, ..., $A_m$, $C_m$, $A_{m+1}$ and deposit the quantity $g_i - 2(x - y_{N-1})$ at $y_{N- 1}$ in $A_i$, $i=1,\dots,m$, and $g_{m+1} - (x - y_{N-1})$ at $y_{N-1}$ in $A_{m+1}$. Then we are now in a position to make the trips $B_1$, $B_2$, ... , $B_m$. When we do this the final configuration is not altered. By induction, we can so get the desired $P'$. We call a plane like $P'$ a standard plan.
Now, if $S$ is a standard plan which realizes the optimal solution, then clearly the plan $S_n$, $n=1,\dots,N-1$ obtained from $S$ deleting the trips starting at $x_m$, for $m > n$, realizes the optimal solution $g(y_n, n)$. So, if $k_n$ is the number of trips in $S$ between $y_{n-1}$ and $y_n$, we have
\begin{equation}
g(y_n,n) - g(y_{n-1},n-1) = (2 k_n - 1) \Delta,
\end{equation}
where $\Delta=x/N$. Moreover, since in the first $k_n - 1$ round trips the maximum fuel that can be deposited at $y_n$ is $1-2 \Delta$, while in the $k_n$-th trip is $1- \Delta$, we must have
\begin{equation}
(k_n - 1)(1 - 2 \Delta) + 1 - \Delta \geq g(y_{n-1},n-1),
\end{equation}
or
\begin{equation}
k_n (1 - 2 \Delta) + 1 + \Delta \geq g(y_{n-1},n-1).
\end{equation}
Now note that, since we want to minimize the consumed fuel, clearly $k_n$ is determined as the least positive integer satisfying the above inequality.
This solves the problem recursively.
Now, note that if $k_n \geq 2$, then we must have
\begin{equation}
(k_n - 1)(1 - 2 \Delta) +  \Delta < g(y_{n-1},n-1),
\end{equation}
from which we get
\begin{equation}
k_n < \frac{g(y_{n-1},n-1) + 1 - 3 \Delta}{1-2 \Delta},
\end{equation}
so that
\begin{equation}
\frac{g(y_n,n) - g(y_{n-1},n-1)}{\Delta} = 2k_n - 1 < \frac{2g(y_{n-1},n-1) + 1 - 4 \Delta}{1-2 \Delta} < \frac{2g(y_{n-1},n-1) + 1}{1-2 \Delta}.
\end{equation}
Since we have
\begin{equation}
g(\Delta \left \lfloor {1/ \Delta} \right \rfloor, \left \lfloor {1/ \Delta} \right \rfloor) = 1,
\end{equation}
we compare $g(y_n,n)$ with the function
\begin{equation}
h(y)= c(N) \exp \left( \frac{2y}{1- 2 \Delta} \right) - \frac{1}{2},
\end{equation}
where
\begin{equation}
c(N) = \frac{3}{2 \exp \left( \frac{ 2 \Delta \left \lfloor {1/ \Delta} \right \rfloor}{1 - 2 \Delta} \right) }.
\end{equation}
The function $h$ is a convex, satisfies the differential equation
\begin{equation}
h'=\frac{2 h + 1}{1-2 \Delta},
\end{equation}
and the initial condition $h(\Delta \left \lfloor {1/ \Delta} \right \rfloor)=1$.
From the above inequality for $g(y_n,n)$ we then get by induction that
\begin{equation}
h(y_n) \geq g(y_n,n),
\end{equation}
for all $n \geq \left \lfloor {1/ \Delta} \right \rfloor$, so that in particular
\begin{equation}
g(x,N(x)) \leq c(N(x)) \exp \left( \frac{2x}{1- 2 \Delta} \right) - \frac{1}{2}.
\end{equation}
Now note that since $x^2 / N(x) \rightarrow 0$ as $x \rightarrow \infty$, we have $\Delta \rightarrow 0$ and $\Delta \left \lfloor {1/ \Delta} \right \rfloor \rightarrow 1$. Moreover we have
\begin{equation}
\lim_{x \rightarrow \infty} \frac{\exp \left( \frac{2x}{1 - 2 \Delta} \right)}{e^{2x}} =
\lim_{x \rightarrow \infty} \exp \left(\frac{4x \Delta}{1 - 2 \Delta} \right) = 1.
\end{equation}
We finally have so
\begin{equation}
\limsup_{x \rightarrow \infty} \frac{g(x,N(x))}{f(x)} \leq
\limsup_{x \rightarrow \infty} \frac{c(N(x)) \exp \left( \frac{2x}{1 - 2 \Delta} \right)}{\frac{1}{4} \exp \left( 2x - \gamma \right) + \mathcal{O}(e^{-2x})} = \frac{6}{e^{2 - \gamma}} < \frac{3}{2}.
\end{equation}
QED


TITLE: Parametric Equations of a surface
QUESTION [0 upvotes]: I need to find the Parametric Equations of the following surface. But I dont know the method. Can someone help me please. $$\frac {x^2}{a^2}+\frac {y^2}{b^2}-\frac {z^2}{c^2}=1$$ $$\frac {x^2}{p}+\frac {y^2}{q}=2z$$ $$\frac {x^2}{a^2}+\frac {y^2}{b^2}-\frac {z^2}{c^2}=0$$

REPLY [0 votes]: In each of these problems, the horizontal cross sections are ellipses. This is to say, when you let $z=v$, you get an equation of the form (after rewriting it)
$$\frac{x^2}{\alpha(v)^2}+\frac{y^2}{\beta(v)^2}=1$$
Then the parametric equation is given by
$$x = \alpha(v)\cos(u), y = \beta(v)\sin(u), z=v$$
For example, for the first problem, you get $\alpha(v) = a\sqrt{1+\frac{v^2}{c^2}}, \beta(v) = b\sqrt{1+\frac{v^2}{c^2}}$.


TITLE: What is the Fourier Transform of the spatial portion of $Ψ(x,t)=A\exp(-b|x-2|)\exp(-iwt)$?
QUESTION [0 upvotes]: What is the Fourier Transform of the spatial portion of $Ψ(x,t)=A\exp(-b|x-2|)\exp(-iwt)$?
I tried applying the regular exponential Fourier transform, but not getting it.
Do you just bring out the $exp(iwt)$? If so, then how do you integrate the $exp(-b|x-2|)exp(-ikx)$ left inside from negative to positive infinity?
Any help will be much appreciated.
Thank you very much!

REPLY [1 votes]: Going straight to the definition of the Fourier transform, we have
\begin{multline}
\mathcal{F}[\Psi](k) = \int_{-\infty}^\infty Ae^{-b|x-2|}e^{iwt}e^{-ikx}dx = Ae^{iwt}\int_{-\infty}^\infty e^{-b|x-2|}e^{-ikx}dx \\= Ae^{iwt}e^{-2ik}\int_{-\infty}^\infty e^{-b|u|}e^{-iku}du = 2Ae^{i(wt-2k)}\int_{0}^\infty e^{-bu}\cos(ku)du = \frac{2Abe^{i(wt-2k)}}{b^2+k^2}
\end{multline}
where the substitution $u = x-2$ was used and the parity of the integrand was used to simplify the last integral.
Also, this wave function is normalizable. The normalization integral is
\begin{multline}
\int_{-\infty}^\infty \Psi^*\Psi dx = \int_{-\infty}^\infty \left[A^*e^{-b|x-2|}e^{-iwt}\right]\left[Ae^{-b|x-2|}e^{iwt}\right] dx \\= |A|^2\int_{-\infty}^\infty e^{-2b|x-2|}dx = |A|^2\int_{-\infty}^\infty e^{-2b|u|}du=\frac{|A|^2}{b}= 1,
\end{multline}
with the same substitution and parity considerations were used here. So $A = \sqrt{b}$.

REPLY [0 votes]: It's linear, so the time portion just comes along for the ride. We can use the table here to note that
$$\mathcal{F}\left\{e^{-a|x|}\right\}=\frac{2a}{a^2+4\pi^2\xi^2}. $$
This is assuming the unitary, ordinary frequency type of FT. You can use a different column if you're using a different convention.
Together with the shifting theorems, that should get you all the way there. Can you continue?


text:
\section{Arthur packets, Galois representations and Hodge classes}
\label{sec:arthur-galois-hodge}

\subsection{Classification}
\label{sec:classification-global}

Let $F$ be a totally real number field and $E$ a totally imaginary quadratic extension of $F$.
Let $\VV$ be an $n$-dimensional hermitian $E$-space and $G = \U(\VV)$ the unitary group of $\VV$.
In this section, we recall the classification of automorphic representations of $G(\A)$, which has been established by Mok \cite{mok} in the quasi-split case, following Arthur's book \cite{arthur}, and has been extended to the general case by Kaletha--Minguez--Shin--White \cite{kmsw}.

More precisely, let $L^2_{\disc}(G)$ be the discrete spectrum of the unitary representation of $G(\A)$ on the Hilbert space $L^2(G(F) \backslash G(\A))$.
Then the decomposition of $L^2_{\disc}(G)$ into near equivalence classes is described as follows.
We say that an irreducible cuspidal automorphic representation $\pi$ of $\GL_m(\A_E)$ is conjugate-self-dual if $\pi^\rho \cong \pi^\vee$, where $\pi^\rho$ and $\pi^\vee$ are the Galois conjugate and the contragredient of $\pi$, respectively.
In this case, exactly one of the Asai $L$-functions $L(s, \pi, \mathrm{As}^+)$ and $L(s, \pi, \mathrm{As}^-)$ has a pole at $s=1$ (see \cite[\S 7]{ggp} for the definition of the Asai representations $\mathrm{As}^\pm$).
For $\epsilon = \pm$, we say that $\pi$ has sign $\epsilon$ if $L(s, \pi, \mathrm{As}^\epsilon)$ has a pole at $s=1$.
We also say that $\pi$ is conjugate-orthogonal (resp.~conjugate-symplectic) if it is conjugate-self-dual with sign $+$ (resp.~$-$).
Consider a formal unordered finite direct sum
\[
 \psi = \bigoplus_i \pi_i \boxtimes \Sym^{d_i-1},
\]
where
\begin{itemize}
 \item $\pi_i$ is an irreducible cuspidal automorphic representation of $\GL_{m_i}(\A_E)$;
 \item $\Sym^{d_i-1}$ is the irreducible representation of $\SL_2(\C)$ of dimension $d_i$.
\end{itemize}
Then $\psi$ is called an elliptic $A$-parameter for $G$ if
\begin{itemize}
 \item $\sum_i m_i d_i = n$;
 \item if $d_i$ is odd, then $\pi_i$ is conjugate-self-dual with sign $(-1)^{n-1}$;
 \item if $d_i$ is even, then $\pi_i$ is conjugate-self-dual with sign $(-1)^n$;
 \item if $(\pi_i, d_i) = (\pi_j, d_j)$, then $i=j$.
\end{itemize}
We attach to $\psi$ an automorphic representation
\[
 \pi_\psi = \bigboxplus_i \left( \pi_i |\det|^{\frac{d_i-1}{2}} \boxplus \pi_i |\det|^{\frac{d_i-3}{2}} \boxplus \cdots \boxplus \pi_i |\det|^{-\frac{d_i-1}{2}} \right)
\]
of $\GL_n(\A_E)$, where $\boxplus$ denotes the isobaric sum.
Then the result of Kaletha--Minguez--Shin--White \cite[Theorem${}^*$ 1.7.1]{kmsw} (which is proved in \cite{kmsw} partially and will be completed in its sequels) says that
\[
 L^2_{\disc}(G) = \bigoplus_\psi L^2_\psi(G),
\]
where $\psi$ runs over elliptic $A$-parameters for $G$ and $L^2_\psi(G)$ is the near equivalence class of irreducible subrepresentations $\pi$ of $L^2_\disc(G)$ such that for almost all places $v$ of $F$, the base change of $\pi_v$ to $\GL_n(E_v)$ is isomorphic to $\pi_{\psi,v}$.

We next describe this decomposition in terms of $L$-groups.
Recall that the $L$-group of $G = \U(\VV)$ is given by
\[
 {}^L G = \GL_n(\C) \rtimes \Gal(\Qbar/F),
\]
where $\Gal(\Qbar/E)$ acts trivially on $\GL_n(\C)$ and the non-trivial element in $\Gal(E/F)$ acts as the automorphism $\theta_n$ defined by
\[
 \theta_n(g) = J_n \cdot {}^t g^{-1} \cdot J_n^{-1}, \qquad
 J_n =
 \begin{pmatrix}
 & & & 1 \\
 & & -1 & \\
 & \varddots & & \\
 (-1)^{n-1} & & &
 \end{pmatrix}.
\]
Put $\tilde{G} = \Res_{E/F} \GL_n$, so that its $L$-group is given by
\[
 {}^L \tilde{G} = \left( \GL_n(\C) \times \GL_n(\C) \right) \rtimes \Gal(\Qbar/F),
\]
where $\Gal(\Qbar/E)$ acts trivially on $\GL_n(\C) \times \GL_n(\C)$ and the non-trivial element in $\Gal(E/F)$ acts as the automorphism $(g_1,g_2) \mapsto (g_2,g_1)$.
Then the base change $L$-homomorphism $\BC:{}^L G \rightarrow {}^L \tilde{G}$ is given by
\[
 \BC(g \rtimes \sigma) = (g, \theta_n(g)) \rtimes \sigma.
\]
Let $\psi$ be an elliptic $A$-parameter for $G$.
Then the local Langlands correspondence induces an (equivalence class of) $A$-parameter $\tilde{\psi}_v : \cL_{F_v} \times \SL_2(\C) \rightarrow {}^L \tilde{G}$ for any place $v$ of $F$, where
\[
 \cL_{F_v} =
 \begin{cases}
  \text{the Weil group of $F_v$} & \text{if $v$ is real;} \\
  \text{the Weil--Deligne group of $F_v$} & \text{if $v$ is finite.}
 \end{cases}
\]
Moreover, by \cite[Theorem 8.1]{ggp}, there exists a unique (equivalence class of) $A$-parameter $\psi_v : \cL_{F_v} \times \SL_2(\C) \rightarrow {}^L G$ such that $\tilde{\psi}_v = {\BC} \circ \psi_v$.
We associate to $\psi_v$ an $L$-parameter $\phi_{\psi_v}: \cL_{F_v} \rightarrow {}^L G$ by
\[
 \phi_{\psi_v}(w) = \psi_v \! \left( w, \mat{|w|_v^{\frac{1}{2}}}{}{}{|w|_v^{-\frac{1}{2}}} \right).
\]
Then $L^2_\psi(G)$ consists of irreducible subrepresentations $\pi$ of $L^2_\disc(G)$ such that the $L$-parameter of $\pi_v$ is $\phi_{\psi_v}$ for almost all $v$.

For our applications, we will consider elliptic $A$-parameters with $n=4$ of the form
\begin{equation}
\label{eq:U(2,2)-A-param}
 \psi = \pi_E \boxtimes \Sym^1,
\end{equation}
where
\begin{itemize}
 \item $\pi$ is an irreducible cuspidal automorphic representation of $\GL_2(\A)$ with central character $\xi_E$;
 \item $\pi_E$ is the base change of $\pi$ to $\GL_2(\A_E)$.
\end{itemize}
We note the following:

\begin{lem}
\begin{enumerate}
\item If $\pi_E$ is cuspidal, then $\pi_E$ is conjugate-orthogonal.
\item If $\pi_E$ is not cuspidal, then $\pi_E = \chi \boxplus \chi^{-1}$ for some conjugate-orthogonal character $\chi$ of $\A_E^\times/E^\times$ such that $\chi^2 \ne 1$.
\end{enumerate}
\end{lem}

\begin{proof}
First assume that $\pi_E$ is not cuspidal.
Then $\pi$ is the automorphic induction of some character $\chi$ of $\A_E^\times/E^\times$, so that $\pi_E = \chi \boxplus \chi^\rho$.
Since $\pi$ is cuspidal, we have $\chi^\rho \ne \chi$.
Also, since the central character of $\pi$ is $\xi_E$, we have $\chi|_{\A^\times} = 1$, i.e.~$\chi$ is conjugate-orthogonal.
Hence the assertion follows.

Next assume that $\pi_E$ is cuspidal.
Put $G' = \GL_2$ and $\tilde{G} = \Res_{E/F} \GL_2$, so that
\[
 {}^L G' = \GL_2(\C) \times \Gal(\Qbar/F), \qquad
 {}^L \tilde{G} = \left( \GL_2(\C) \times \GL_2(\C) \right) \rtimes \Gal(\Qbar/F).
\]
Then the base change $L$-homomorphism $\BC:{}^L G' \rightarrow {}^L \tilde{G}$ is given by
\[
 \BC(g \times \sigma) = (g, g) \rtimes \sigma.
\]
Recall that $\mathrm{As}^+$ is the representation of ${}^L \tilde{G}$ on $\C^2 \otimes \C^2$ defined by
\begin{align*}
 \mathrm{As}^+((g_1, g_2) \rtimes 1)(x \otimes y) & = g_1 x \otimes g_2 y, \\
 \mathrm{As}^+((1, 1) \rtimes \sigma)(x \otimes y) & =
 \begin{cases}
  x \otimes y & \text{if $\sigma \in \Gal(\Qbar/E)$;} \\
  y \otimes x & \text{if $\sigma \notin \Gal(\Qbar/E)$.}
 \end{cases}
\end{align*}
Then we have
\[
 \mathrm{As}^+ \circ \BC \cong \Sym^2 \oplus (\wedge^2 \otimes \xi_E)
\]
as representations of ${}^L G'$.
Since the central character of $\pi$ is $\xi_E$, it follows that
\[
 L(s, \pi_E, \mathrm{As}^+) = L(s, \pi, \Sym^2) \cdot \zeta_F(s).
\]
This implies the assertion.
\end{proof}

\subsection{}
\label{ss:galrep-prep}

To describe the Galois representation on the cohomology of Shimura varieties (see \cite[p.~201]{kot}), we need to introduce some representation of the $L$-group.
Following \cite[\S 5.1]{blasius-rogawski}, we recall its definition.

We first consider the case $F = \Q$ and $G = \GU(n,n)$.
Then
\[
 {}^L G = \hat{G} \rtimes \Gal(\Qbar/\Q), \qquad
 \hat{G} = \GL_{2n}(\C) \times \C^\times,
\]
where $\Gal(\Qbar/E)$ acts trivially on $\hat{G}$ and the non-trivial element in $\Gal(E/\Q)$ acts as the automorphism $\hat{\theta}_{2n}$ defined by
\[
 \hat{\theta}_{2n}(g,\nu) = (\theta_{2n}(g), \nu \cdot \det g).
\]
We define a representation $r$ of $\hat{G}$ on $V = \wedge^n \C^{2n}$ by
\[
 r(g,\nu) = \nu \cdot \wedge^n g.
\]
Since $(\wedge^n \C^{2n})^* \otimes \det \cong \wedge^n \C^{2n}$, there exists a unique automorphism $A$ of $V$ such that
\[
 r(\hat{\theta}_{2n}(g)) \circ A = A \circ r(g), \qquad
 g \in \hat{G},
\]
and such that $A$ fixes the highest weight vector (unique up to scalars) in $V$ with respect to the standard Borel subgroup of $\hat{G}$.
Then we can extend $r$ to ${}^L G$ by setting
\[
 r(1 \rtimes \sigma) =
 \begin{cases}
  \id & \text{if $\sigma \in \Gal(\Qbar/E)$;} \\
  A & \text{if $\sigma \notin \Gal(\Qbar/E)$.}
 \end{cases}
\]

Suppose that $F$ is an arbitrary totally real number field.
Put $G = \GU(\VV)$ with $\dim \VV = 2n$, so that
\[
 {}^LG = \hat{G} \rtimes \Gamma_F, \qquad
 \hat{G} = \GL_{2n}(\C) \times \C^\times, \qquad
 \Gamma_F = \Gal(\Qbar/F),
\]
where the action of $\Gamma_F$ on $\hat{G}$ is defined similarly as above.
Let $G_0 = \Res_{F/\Q} G$.
Then
\[
 {}^LG_0 = \hat{G}_0 \rtimes \Gamma_\Q, \qquad
 \hat{G}_0 = \Ind^{\Gamma_\Q}_{\Gamma_F} \hat{G}
\]
(see \cite[\S 5]{borel-corvallis}).
We may identify $\hat{G}_0$ with $(\hat{G})^d$, where $d = [F:\Q]$ and the action of $\Gamma_\Q$ on $\hat{G}_0$ can be explicated as follows.
Choose a set of representatives $\{ \sigma_1, \dots, \sigma_d \}$ for $\Gamma_\Q/\Gamma_F$ and define an action of $\Gamma_\Q$ on $\{1, \dots, d\}$ so that
\[
 \gamma \sigma_i \Gamma_F = \sigma_{\gamma(i)} \Gamma_F, \qquad
 \gamma \in \Gamma_\Q.
\]
Then $\gamma \in \Gamma_\Q$ acts as the automorphism
\[
 (g_1, \dots, g_d) \mapsto (\gamma_1 \cdot g_{\gamma^{-1}(1)}, \dots, \gamma_d \cdot g_{\gamma^{-1}(d)}),
\]
where
\[
 \gamma_i = \sigma_i^{-1} \gamma \sigma_{\gamma^{-1}(i)} \in \Gamma_F.
\]
We now assume that
\[
 G_v \simeq
 \begin{cases}
  \GU(n,n) & \text{if $v \in \Sigma$;} \\
  \GU(2n) & \text{if $v \in \Sigma_\infty \smallsetminus \Sigma$}
 \end{cases}
\]
for some subset $\Sigma$ of the set of real places $\Sigma_\infty$ of $F$.
Let $F_0$ be the reflex field of the Shimura variety associated to $G_0$.
Namely, $F_0$ is the fixed field of the subgroup
\[
 \{\sigma \in \Gamma_\Q \, | \, \sigma \Sigma = \Sigma \},
\]
where the action of $\Gamma_\Q$ on $\Sigma_\infty$ (regarded as the set of embeddings of $F$ in $\C$) is naturally induced by the inclusion $\Qbar \hookrightarrow \C$.
We identify $\Sigma_\infty$ with $\Gamma_\Q/\Gamma_F$ so that the fixed embedding $F \hookrightarrow \Qbar$ corresponds to the trivial coset $\Gamma_F$.
Then, without loss of generality, we may view $\Sigma$ as a subset $\{\sigma_1 \Gamma_F, \dots, \sigma_s \Gamma_F\}$, where $s = |\Sigma|$.
Let $r$ be the representation of ${}^L G$ on $V = \wedge^n \C^{2n}$ similarly defined as above.
We define a representation $r_0$ of $\hat{G}_0$ on $V_0 = V^{\otimes s}$ by
\[
 r_0(g_1, \dots, g_d) = r(g_1) \otimes \dots \otimes r(g_s).
\]
Then we can extend $r_0$ to $\hat{G}_0 \rtimes \Gamma_{F_0}$ by setting
\[
 r_0(1 \rtimes \gamma)(x_1 \otimes \dots \otimes x_s)
 = r(1 \rtimes \gamma_1) x_{\gamma^{-1}(1)} \otimes \dots \otimes r(1 \rtimes \gamma_s) x_{\gamma^{-1}(s)}.
\]

\subsection{}

We keep the notation of the previous section.
Let $\cL_F$ be the hypothetical Langlands group of $F$ equipped with a surjective homomorphism $\pr: \cL_F \twoheadrightarrow \Gamma_F$.
Let $\psi : \cL_F \times \SL_2(\C) \rightarrow {}^L G$ be an elliptic $A$-parameter.
Then we may decompose $V$ into a direct sum
\[
 V = \bigoplus_i V^i
\]
of representations of $\cL_F$, where
\[
 V^i = \left\{ v \in V \, \left| \, (r \circ \psi) \left( 1, \smat{t}{}{}{t^{-1}} \right) v = t^i v \text{ for all } t \in \C^\times \right. \right\}.
\]
Let $\psi_0: \cL_\Q \times \SL_2(\C) \rightarrow {}^L G_0$ be the elliptic $A$-parameter corresponding to $\psi$.
More explicitly, $\psi_0$ is given by
\[
 \psi_0(\gamma, h) = (\psi'(\gamma_1, h), \dots, \psi'(\gamma_d, h)) \rtimes \pr(\gamma),
\]
where we write $\psi(\sigma, h) = \psi'(\sigma, h) \rtimes \pr(\sigma)$.
As above, we may decompose $V_0$ into a direct sum
\[
 V_0 = \bigoplus_i V^i_0
\]
of representations of $\cL_{F_0}$.
Then we have
\[
 V_0^i = \bigoplus_{i=i_1+\dots+i_s} V^{i_1} \otimes \dots \otimes V^{i_s}.
\]

We now suppose that $\psi = \pi_E \boxtimes \Sym^1$ as in \eqref{eq:U(2,2)-A-param}.
Let $W$ be the $4$-dimensional representation of $\cL_E \times \SL_2(\C)$ induced by $\psi$.
Then we have
\[
 W = W^1 \oplus W^{-1}, \qquad W^1 = W^{-1} = \rho_E,
\]
where $\rho$ is the $2$-dimensional representation of $\cL_F$ associated to $\pi$ and $\rho_E = \rho|_{\cL_E}$.
Hence we see that $V = \wedge^2 W$ decomposes as
\[
 V = V^2 \oplus V^0 \oplus V^{-2},
\]
where
\[
 V^2 = \wedge^2 W^1 = \C, \qquad
 V^0 = W^1 \otimes W^{-1} = (\mathrm{As}^+ \circ \BC)(\rho), \qquad
 V^{-2} = \wedge^2 W^{-1} = \C.
\]
In particular, $V_0^{2s}$ and $V_0^{-2s}$ are both the trivial representation of $\cL_{F_0}$.

\subsection{Local $A$-packets}
\label{ss:local-A-packets}

Let $\psi$ be an elliptic $A$-parameter for $G = \U(\VV)$ and $L^2_\psi(G)$ the near equivalence class associated to $\psi$.
Then the result of Kaletha--Minguez--Shin--White \cite[Theorem${}^*$ 1.7.1]{kmsw} also describes the local-global structure of $L^2_\psi(G)$ with a multiplicity formula.
In particular, if $\pi$ is an irreducible summand of $L^2_\psi(G)$, then for any place $v$ of $F$, the local component $\pi_v$ is an irreducible summand of some representation in $\Pi_{\psi_v}$.
Here $\Pi_{\psi_v}$ is the local $A$-packet associated to $\psi_v$ consisting of certain semisimple representations of $G_v$ of finite length.

Suppose that $v$ is real.
If $G_v$ is quasi-split and $\psi_v$ is ``cohomological'', then it follows from the result of Arancibia--M{\oe}glin--Renard \cite{amr} (the unitary group case had already been treated by Johnson \cite{johnson}) that $\Pi_{\psi_v}$ agrees with the packet constructed by Adams--Johnson \cite{aj}, which we recall below.
From now on, we suppress the subscript $v$ from the notation.

Let $\VV$ be an $n$-dimensional hermitian space over $\C$ of signature $(p,q)$.
Choosing a basis of $\VV$, we may identify the unitary group $G = \U(\VV)$ with
\[
 \U(p,q) = \{ g \in \GL_n(\C) \, | \, {}^t \bar{g} I_{p,q} g = I_{p,q} \},
 \qquad
 I_{p,q} =
 \begin{pmatrix}
  \1_p & \\
  & - \1_q
 \end{pmatrix}.
\]
We define a Cartan involution $\theta$ of $G$ by $\theta(g) = {}^t g^{-1}$.
Let $K \simeq \U(p) \times \U(q)$ be the maximal compact subgroup of $G$ with respect to $\theta$ and $T \simeq \U(1)^n$ the maximal torus of $G$ consisting of diagonal matrices.
Let $B$ be the Borel subgroup of $G_\C \simeq \GL_n(\C)$ (which is not defined over $\R$) consisting of upper triangular matrices.

Let $\fg_0$, $\fk_0$, $\ft_0$ be the Lie algebras of $G$, $K$, $T$, respectively.
We have a Cartan decomposition $\fg_0 = \fk_0 \oplus \fp_0$, where $\fp_0$ is the $(-1)$-eigenspace of $\theta$.
Let $\fg$, $\fk$, $\fp$, $\ft$ be the complexifications of $\fg_0$, $\fk_0$, $\fp_0$, $\ft_0$, respectively.
Let $\fp^{\pm}$ be the $(\pm i)$-eigenspace of the complex structure on $\fp$ defined by
\[
 X \longmapsto J X J^{-1}, \qquad
 J =
 \begin{pmatrix}
  e^{\frac{\pi i}{4}} \1_p & \\
  & e^{-\frac{\pi i}{4}} \1_q
 \end{pmatrix}.
\]
More explicitly, we have $\fg = \fk \oplus \fp^+ \oplus \fp^-$ with
\begin{align*}
 \fk & = \left\{ \left.
 \begin{pmatrix}
  A & 0 \\
  0 & D
 \end{pmatrix}
 \, \right| \, A \in \M_p(\C), \, D \in \M_q(\C) \right\}, \\
 \fp^+ & = \left\{ \left.
 \begin{pmatrix}
  0 & B \\
  0 & 0
 \end{pmatrix}
 \, \right| \, B \in \M_{p,q}(\C) \right\}, \\
 \fp^- & = \left\{ \left.
 \begin{pmatrix}
  0 & 0 \\
  C & 0
 \end{pmatrix}
 \, \right| \, C \in \M_{q,p}(\C) \right\}.
\end{align*}

The packet constructed by Adams--Johnson \cite{aj} consists of certain unitary representations $\pi$ such that $H^*(\fg, K; \pi \otimes F) \ne 0$ for some irreducible finite-dimensional representation $F$ of $G$.
Let $\lambda \in \ft^* \simeq \C^n$ be the highest weight of $F^*$ relative to $B$.
We may write
\[
 \lambda = (\underbrace{\lambda_1, \dots, \lambda_1}_{n_1}, \underbrace{\lambda_2, \dots, \lambda_2}_{n_2}, \dots, \underbrace{\lambda_r, \dots, \lambda_r}_{n_r}) \in \Z^n
\]
with $\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_r$.
Then we consider the $A$-parameter $\psi : \cL_\R \times \SL_2(\C) \rightarrow {}^LG$ whose restriction to $\C^\times \times \SL_2(\C)$ is equal to
\[
 \left( \chi_{\lambda_1 + \rho_1} \boxtimes \Sym^{n_1-1} \right) \oplus \dots \oplus \left( \chi_{\lambda_r + \rho_r} \boxtimes \Sym^{n_r-1} \right),
\]
where
\begin{itemize}
 \item $\rho_i = \frac{1}{2}(- n_1 - \dots - n_{i-1} + n_{i+1} + \dots + n_r)$;
 \item $\chi_\kappa$ is the character of $\C^\times$ defined by $\chi_\kappa(z) = (z/\bar{z})^\kappa$;
 \item $\Sym^{d-1}$ is the irreducible $d$-dimensional representation of $\SL_2(\C)$.
\end{itemize}
This defines a parabolic subgroup $Q$ of $G_\C$ (which is not defined over $\R$) containing $B$ with Levi component $L$ (which is defined over $\R$) such that
\[
 \psi = \xi \circ \psi_L,
\]
where $\xi : {}^L L \rightarrow {}^L G$ is the canonical embedding and $\psi_L :\cL_\R \times \SL_2(\C) \rightarrow {}^L L$ is an $A$-parameter such that the $A$-packet $\Pi_{\psi_L}$ consists of a single $1$-dimensional representation of $L$.
Note that
\[
 L = G \cap \left( \GL_{n_1}(\C) \times \dots \times \GL_{n_r}(\C) \right).
\]
Put
\[
 S = W(L,T) \backslash W(G,T) / W_\R(G,T) \simeq \left( \mathfrak{S}_{n_1} \times \dots \times \mathfrak{S}_{n_r}\right) \backslash \mathfrak{S}_n / \left( \mathfrak{S}_p \times \mathfrak{S}_q \right),
\]
where $W$ and $W_\R$ denote the absolute and relative Weyl groups, respectively.
As a set of representatives for $S$, we can take the set of $w \in \mathfrak{S}_n$ such that
\begin{itemize}
 \item $w^{-1}(i) < w^{-1}(j)$ for $n_1 + \dots + n_{k-1}+1 \le i < j \le n_1 + \dots + n_k$ for $1 \le k \le r$;
 \item $w(i) < w(j)$ for $1 \le i < j \le p$ and for $p+1 \le i < j \le n$.
\end{itemize}
For any $w \in S$, we have a $\theta$-stable parabolic subgroup $Q_w = w^{-1} Q w$ of $G_\C$ with Levi component $L_w = w^{-1} L w$.
Let $\fq_w$ be the Lie algebra of $Q_w$.
Then the Adams--Johnson packet $\Pi^{\mathrm{AJ}}_\psi$ is given by
\[
 \Pi^{\mathrm{AJ}}_\psi = \left\{ A_{\fq_w}(w^{-1} \lambda) \, | \, w \in S \right\}.
\]

We now explicate the $A$-packet $\Pi_\psi$ when $G = \U(2,2)$ and $\psi$ is the localization of a global $A$-parameter as in \eqref{eq:U(2,2)-A-param}.
More precisely, we start with the discrete series representation of $\GL_2(\R)$ of weight $k+1$ with an even integer $k \ge 2$.
Since its base change to $\GL_2(\C)$ is the principal series representation $\Ind(\chi_{\frac{k}{2}} \otimes \chi_{-\frac{k}{2}})$, the associated $A$-parameter $\psi$ is given by
\[
 \psi = \big( \chi_{\frac{k}{2}} \boxtimes \Sym^1 \big) \oplus \big( \chi_{-\frac{k}{2}} \boxtimes \Sym^1 \big),
\]
so that we need to take
\[
 \lambda = \big( \tfrac{k}{2}-1, \tfrac{k}{2}-1, -\tfrac{k}{2}+1, -\tfrac{k}{2}+1 \big).
\]
In this case, we have $S = \{ w_0, w_1, w_2 \}$ with
\[
 w_0 = 1, \qquad
 w_1 = (23), \qquad
 w_2 = (13)(24).
\]
Then
\[
 \Pi_\psi = \Pi^{\mathrm{AJ}}_\psi = \{ A_{\fq_i}(w_i^{-1} \lambda) \, | \, 0 \le i \le 2 \},
\]
where $\fq_i = \fq_{w_i}$.

\begin{prop}
\label{prop:Aq}
We have
\begin{align*}
 \dim H^{p,q}(\fg, K; A_{\fq_0}(w_0^{-1} \lambda) \otimes F) & =
 \begin{cases}
  1 & \text{if $(p,q) = (4,0)$,}  \\
  0 & \text{otherwise,}
 \end{cases} \\
 \dim H^{p,q}(\fg, K; A_{\fq_1}(w_1^{-1} \lambda) \otimes F) & =
 \begin{cases}
  1 & \text{if $(p,q) = (1,1), (3,3)$,}  \\
  2 & \text{if $(p,q) = (2,2)$,}  \\
  0 & \text{otherwise,}
 \end{cases} \\
 \dim H^{p,q}(\fg, K; A_{\fq_2}(w_2^{-1} \lambda) \otimes F) & =
 \begin{cases}
  1 & \text{if $(p,q) = (0,4)$,}  \\
  0 & \text{otherwise,}
 \end{cases}
\end{align*}
where $F$ is the irreducible finite-dimensional representation of $G$ with highest weight $\lambda$.
\end{prop}

\begin{proof}
First, note that $F$ is self-dual, i.e.~$F^*$ has highest weight $\lambda$.
Also, if we write $\fq_i = \fl_i \oplus \fu_i$ with Levi component $\fl_i$ and unipotent radical $\fu_i$, then
\begin{align*}
 \fl_0 & = \left\{
 \begin{pmatrix}
  * & * & 0 & 0 \\
  * & * & 0 & 0 \\
  0 & 0 & * & * \\
  0 & 0 & * & *
 \end{pmatrix}
 \right\}, &
 \fu_0 & = \left\{
 \begin{pmatrix}
  0 & 0 & * & * \\
  0 & 0 & * & * \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0
 \end{pmatrix}
 \right\}, \\
 \fl_1 & = \left\{
 \begin{pmatrix}
  * & 0 & * & 0 \\
  0 & * & 0 & * \\
  * & 0 & * & 0 \\
  0 & * & 0 & *
 \end{pmatrix}
 \right\}, &
 \fu_1 & = \left\{
 \begin{pmatrix}
  0 & * & 0 & * \\
  0 & 0 & 0 & 0 \\
  0 & * & 0 & * \\
  0 & 0 & 0 & 0
 \end{pmatrix}
 \right\}, \\
 \fl_2 & = \left\{
 \begin{pmatrix}
  * & * & 0 & 0 \\
  * & * & 0 & 0 \\
  0 & 0 & * & * \\
  0 & 0 & * & *
 \end{pmatrix}
 \right\}, &
 \fu_2 & = \left\{
 \begin{pmatrix}
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0 \\
  * & * & 0 & 0 \\
  * & * & 0 & 0
 \end{pmatrix}
 \right\}.
\end{align*}
Hence the assertion follows from \cite[Proposition 6.19]{vz}.
\end{proof}

\subsection{The Hodge structure}
\label{sec:the-hodge-structure}

Suppose again that $F$ is a totally real number field and $E$ is a totally imaginary quadratic extension of $F$.
Let $\VV$ be a $4$-dimensional hermitian $E$-space, $G = \GU(\VV)$ the unitary similitude group of $\VV$, and $G' = \U(\VV)$ the unitary group of $\VV$.
We assume that $\disc \VV \notin \N(E^\times)$ but $\disc \VV_v \in \N(E_v^\times)$ for all $v \in \Sigma_\infty$.
Then we may identify $G_v$ with the group
\[
 \{ g \in \GL_4(\C) \, | \, {}^t \bar{g} I_v g = \nu(g) \cdot I_v \},
\]
where
\[
 I_v =
\begin{cases}
 \begin{pmatrix} \1_2 & \\ & -\1_2 \end{pmatrix} & \text{if $v \in \Sigma$;} \\
 \1_4 & \text{if $v \in \Sigma_\infty \smallsetminus \Sigma$}
\end{cases}
\]
for some subset $\Sigma$ of $\Sigma_\infty$.
We further assume that $\Sigma \ne \Sigma_\infty$, so that $G'$ is anisotropic.
For $v \in \Sigma_\infty$, we define a maximal compact subgroup $K_v'$ of $G_v'$ by
\[
 K_v' = \left\{ \left. \begin{pmatrix} a & \\ & d \end{pmatrix} \, \right| \, a, d \in \U(2) \right\}
\]
if $v \in \Sigma$ and $K_v' = G_v'$ if $v \in \Sigma_\infty \smallsetminus \Sigma$.
Put $K_v = F_v^\times \cdot K_v'$, where we regard $F_v^\times \simeq \R^\times$ as a subgroup of $G_v$ via the map $z \mapsto z \1_4$.
Then $K_v$ is a maximal \emph{connected} compact modulo center subgroup of $G_v$.
Put
\begin{align*}
 G_\infty & = \prod_{v \in \Sigma_\infty} G_v, &
 K_\infty & = \prod_{v \in \Sigma_\infty} K_v, \\
 G_\infty' & = \prod_{v \in \Sigma_\infty} G'_v, &
 K_\infty' & = \prod_{v \in \Sigma_\infty} K_v'.
\end{align*}
Let $\fg$ and $\fg'$ be the complexified Lie algebras of $G_\infty$ and $G_\infty'$, respectively.
Let $\Ss = \Res_{\C/\R} \mathbb{G}_m$ and $G_0 = \Res_{F/\Q} G$.
We define a homomorphism $h: \Ss \rightarrow G_{0,\R}$ by $h(z) = (h_v(z))_{v \in \Sigma_\infty}$ with
\[
  h_v(z) =
 \begin{cases}
  \begin{pmatrix} z \1_2 & \\ & \bar{z} \1_2 \end{pmatrix} & \text{if $v \in \Sigma$;} \\
  \1_4 & \text{if $v \in \Sigma_\infty \smallsetminus \Sigma$.}
 \end{cases}
\]
Let $X$ be the $G_0(\R)$-conjugacy class of homomorphisms $\Ss \rightarrow G_{0,\R}$ containing $h$.
Then we have an identification
\[
 X = G_\infty / K_\infty.
\]
For any $v \in \Sigma_\infty$ and any even integer $k \ge 2$, let $(\rho_{v,k}, V_{v,k})$ be the irreducible algebraic representation of $G_v$ such that
\begin{itemize}
 \item $\rho_{v,k}$ has trivial central character;
 \item $\rho_{v,k}' = \rho_{v,k}|_{G_v'}$ is the irreducible finite-dimensional representation of $G_v'$ with highest weight
\[
 \lambda_v = \big( \tfrac{k}{2}-1, \tfrac{k}{2}-1, -\tfrac{k}{2}+1, -\tfrac{k}{2}+1 \big).
\]
\end{itemize}
Let $\uk = (k_v)_{v \in \Sigma_\infty}$ be a tuple of even integers $k_v \ge 2$ and put
\[
 \rho_{\uk} = \bigotimes_{v \in \Sigma_\infty} \rho_{v,k_v}, \qquad
 V_{\uk} = \bigotimes_{v \in \Sigma_\infty} V_{v, k_v}.
\]
For any open compact subgroup $\cK$ of $G(\A_f)$ (where $\A_f$ denotes the ring of finite ad\`eles of $F$), let $\Sh_\cK$ be the Shimura variety associated to $(G_0, X, \cK)$:
\[
 \Sh_\cK = G(F) \backslash X \times G(\A_f) / \cK.
\]
Then $(\rho_{\uk}, V_{\uk})$ gives rise to a local system $\V_{\uk}$ on $\Sh_\cK$.
We have the Hodge decomposition
\[
 H^i(\Sh_\cK, \V_{\uk}) = \bigoplus_{p+q=i} H^{p,q}(\Sh_\cK, \V_{\uk}).
\]

In \S \ref{sec:constr-gl-ex-isom}, we have associated to $\VV$ a quaternion $F$-algebra $B$ and a $3$-dimensional skew-hermitian right $B$-space $\tilde{V}$ such that $\PGU_E(\VV) \simeq \PGU_B(\tilde{V})^0$.
By the above assumption on $\VV$, $B$ is division but $B_v$ is split for all $v \in \Sigma_\infty$.
Let $W$ be the $1$-dimensional hermitian left $B$-space as in \S \ref{ss:theta-setup}.
Then $\GU(W) \simeq B^\times$.
Let $\tau$ be an irreducible unitary automorphic representation of $\GU(W)(\A)^+$ with central character $\xi_E$ such that:
\begin{itemize}
 \item $\tau_v$ is the anti-holomorphic discrete series representation of $\GL_2(\R)^+$ of weight $-k_v-1$ if $v \in \Sigma$;
 \item $\tau_v$ is the holomorphic discrete series representation of $\GL_2(\R)^+$ of weight $k_v+1$ if $v \in \Sigma_\infty \smallsetminus \Sigma$.
\end{itemize}
Let $\Pi = \theta(\tau)$ be the global theta lift of $\tau$ to $\GU(\tilde{V})(\A)$ relative to the standard additive character $\psi$ of $\A/F$ (i.e.~the additive character $\psi$ such that $\psi_v(x) = e^{2 \pi i x}$ for $v \in \Sigma_\infty$).
We assume that $\Pi$ is nonzero.
Then:
\begin{itemize}
 \item $\Pi$ is irreducible by Lemma \ref{lem:irred-theta-similitude};
 \item $\Pi$ has trivial central character by Proposition \ref{prop:spl-hodge}.
\end{itemize}
Hence we may regard $\Pi$ as an irreducible unitary automorphic representation of $G(\A)$ with trivial central character.
Let $S$ be a finite set of places of $F$ including all archimedean places such that:
\begin{itemize}
 \item $G_v$ is unramified over $F_v$;
 \item $\cK_v$ is a hyperspecial maximal compact subgroup of $G_v$;
 \item $\Pi_v$ has a nonzero $\cK_v$-fixed vector.
\end{itemize}
Let $\mathrsfs{H}^S = \mathrsfs{H}(G(\A^S), \cK^S)$ be the Hecke algebra of compactly supported $\cK^S$-biinvariant functions on $G(\A^S)$, where $\A^S = {\prod'_{v \notin S}} F_v$ and $\cK^S = \prod_{v \notin S} \cK_v$.
Then $\mathrsfs{H}^S$ acts on $H^i(\Sh_\cK, \V_{\uk})$.
Put $\Pi^S = \bigotimes'_{v \notin S} \Pi_v$ and
\[
 H^i(\Sh_\cK, \V_{\uk})[\Pi^S] = \{ x \in H^i(\Sh_\cK, \V_{\uk}) \, | \, Tx = \chi(T) x \text{ for all } T \in \mathrsfs{H}^S \},
\]
where $\chi$ is the character of $\mathrsfs{H}^S$ associated to $\Pi^S$.
We define $H^{p,q}(\Sh_\cK, \V_{\uk})[\Pi^S]$ similarly.

\begin{prop}
\label{prop:hodge-only-(d,d)-survives}
We have
\[
 H^{2d}(\Sh_\cK, \V_{\uk})[\Pi^S] = H^{d,d}(\Sh_\cK, \V_{\uk})[\Pi^S],
\]
where $d = |\Sigma|$.
\end{prop}

\begin{proof}
By Matsushima's formula \cite[VII.5.2]{borel-wallach}, we have
\[
 H^{2d}(\Sh_\cK, \V_{\uk}) \simeq \bigoplus_\pi m(\pi) H^{2d}(\fg, K_\infty; \pi_\infty \otimes \rho_{\uk}) \otimes \pi_f^\cK,
\]
where $\pi = \pi_\infty \otimes \pi_f$ runs over equivalence classes of irreducible admissible representations of $(\fg, K_\infty) \times G(\A_f)$, $m(\pi)$ is the multiplicity of $\pi$ in the space of automorphic forms on $G(\A)$, and $\pi_f^\cK$ is the space of $\cK$-fixed vectors in $\pi_f$.
Since this isomorphism is compatible with the Hodge decompositions, it suffices to prove the following: if $m(\pi) > 0$ and $\pi_v \simeq \Pi_v$ for almost all $v$, then
\[
 H^{2d}(\fg, K_\infty; \pi_\infty \otimes \rho_{\uk}) = H^{d,d}(\fg, K_\infty; \pi_\infty \otimes \rho_{\uk}).
\]

Under this assumption, $\pi_v$ has trivial central character for almost all $v$, and hence so is $\pi$ since it is automorphic.
Since $\pi_\infty$ and $\rho_{\uk}$ have trivial central character, we have
\begin{align*}
 H^{2d}(\fg, K_\infty; \pi_\infty \otimes \rho_{\uk}) & = H^{2d}(\fg', K'_\infty; \pi_\infty' \otimes \rho_{\uk}'), \\
 H^{d,d}(\fg, K_\infty; \pi_\infty \otimes \rho_{\uk}) & = H^{d,d}(\fg', K'_\infty; \pi_\infty' \otimes \rho_{\uk}'),
\end{align*}
where $\pi_\infty' = \pi_\infty|_{(\fg', K'_\infty)}$ and $\rho'_{\uk} = \rho_{\uk}|_{G_\infty'}$.
Note that $\pi_\infty'$ and $\rho'_{\uk}$ remain irreducible.

Fix a realization $\mathrsfs{V}$ of $\pi$ in the space of automorphic forms on $G(\A)$.
Let $\mathrsfs{V}|_{G'(\A)}$ be the restriction of $\mathrsfs{V}$ to $G'(\A)$ as functions, so that $\mathrsfs{V}|_{G'(\A)}$ is a nonzero subspace of the space of automorphic forms on $G'(\A)$.
Fix an irreducible component $\pi'$ of $\mathrsfs{V}|_{G'(\A)}$.
Since the natural surjective map $\mathrsfs{V} \rightarrow \mathrsfs{V}|_{G'(\A)}$ is $(\fg', K'_\infty) \times G'(\A_f)$-equivariant, $\pi'_v$ is an irreducible component of $\pi_v|_{G_v'}$ (resp.~$\pi'_v = \pi_v|_{(\fg_v', K_v')}$) if $v$ is finite (resp.~if $v$ is real).

We now compute the $A$-parameter $\psi$ of $\pi'$.
Choose an irreducible unitary cuspidal automorphic representation $\tilde{\tau}$ of $\GL_2(\A)$ so that $\tau$ is an irreducible component of $\tilde{\tau}^B|_{B^\times(\A)^+}$, where $\tilde{\tau}^B$ is the Jacquet--Langlands transfer of $\tilde{\tau}$ to $B^\times(\A)$.
Let $\tau_E$ be the base change of $\tilde{\tau}$ to $\GL_2(\A_E)$.
Note that $\tau_E$ does not depend on the choice of $\tilde{\tau}$.
For almost all $v$, $\tilde{\tau}_v$ is a principal series representation $\Ind(\chi_v \otimes \chi_v^{-1} \xi_{E_v})$ of $\GL_2(F_v)$ for some unramified character $\chi_v$ of $F_v^\times$.
Hence
\[
 \tau_{E,v} = \Ind(\eta_v \otimes \eta_v^{-1})
\]
for almost all $v$, where $\eta_v = \chi_v \circ \N_{E_v/F_v}$.
On the other hand, by Lemmas \ref{l:isom-tori} and \ref{l:theta-unram}, $\pi_v$ is the unique irreducible unramified subquotient of
\[
 \Ind^{G_v}_{B_v}(\eta_v |\cdot|_{E_v}^{\frac{1}{2}} \otimes \eta_v |\cdot|_{E_v}^{-\frac{1}{2}} \otimes \chi_v^{-2})
\]
for almost all $v$, where $B_v$ is the standard Borel subgroup of $G_v$ containing the maximal torus $T_v \simeq (E_v^\times)^2 \times F_v^\times$ as in \S \ref{ss:theta-3}.
Hence $\pi'_v$ is the unique irreducible unramified subquotient of
\[
 \Ind^{G'_v}_{B'_v}(\eta_v |\cdot|_{E_v}^{\frac{1}{2}} \otimes \eta_v |\cdot|_{E_v}^{-\frac{1}{2}})
\]
for almost all $v$, where $B'_v = B_v \cap G_v'$ is the standard Borel subgroup of $G_v'$ containing the maximal torus $T_v \cap G_v' \simeq (E_v^\times)^2$.
Namely, we have
\[
 \psi = \tau_E \boxtimes \Sym^1.
\]

Thus, by the classification of automorphic representations of $G'(\A)$, $\pi_v'$ is an irreducible summand of some representation in the local $A$-packet $\Pi_{\psi_v}$ for all $v$.
In particular, if $v \in \Sigma$, then $\pi_v'$ is one of the representations $A_{\fq_i}(w_i^{-1} \lambda_v)$ as in \S \ref{ss:local-A-packets}, and hence we have
\[
 H^i(\fg_v',K'_v; \pi_v' \otimes \rho'_{v, k_v}) = 0
\]
for $i<2$ and
\[
 H^2(\fg_v',K'_v; \pi_v' \otimes \rho'_{v, k_v}) = H^{1,1}(\fg_v',K'_v; \pi_v' \otimes \rho'_{v, k_v})
\]
by Proposition \ref{prop:Aq}.
From this, we deduce that
\begin{align*}
 H^{2d}(\fg', K'_\infty; \pi_\infty' \otimes \rho_{\uk}')
 & = \bigotimes_{v \in \Sigma}
 H^2(\fg_v', K'_v; \pi_v' \otimes \rho_{v, k_v}')
 \bigotimes_{v \in \Sigma_\infty \smallsetminus \Sigma}
 H^0(\fg_v', K'_v; \pi_v' \otimes \rho_{v, k_v}') \\
 & = \bigotimes_{v \in \Sigma}
 H^{1,1}(\fg_v', K'_v; \pi_v' \otimes \rho_{v, k_v}')
 \bigotimes_{v \in \Sigma_\infty \smallsetminus \Sigma}
 H^0(\fg_v', K'_v; \pi_v' \otimes \rho_{v, k_v}') \\
 & = H^{d,d}(\fg', K'_\infty; \pi_\infty' \otimes \rho_{\uk}').
\end{align*}
This completes the proof.
\end{proof}

\begin{prop}
\label{prop:hodge-is-of-type-(d,d)}
The Hodge structure $H^{2d}(\Sh_\cK, \V_{\uk})[\Pi^S]$ is purely of type $(d,d)$.
\end{prop}

\begin{proof}
From the proof of Prop. \ref{prop:hodge-only-(d,d)-survives}, we need to compute the Hodge structure on
\[
H^{d,d}(\fg', K'_\infty; \pi_\infty' \otimes \rho_{\uk}') \simeq  \bigotimes_{v \in \Sigma} H^{1,1}(\fg_v', K'_v; \pi_v' \otimes \rho_{v, k_v}')
 \bigotimes_{v \in \Sigma_\infty \smallsetminus \Sigma}
 H^0(\fg_v', K'_v; \pi_v' \otimes \rho_{v, k_v}').
 \]
 The term $H^0(\fg_v', K'_v; \pi_v' \otimes \rho_{v, k_v}')$ (for
 $v\in \Sigma_\infty \smallsetminus \Sigma$) is clearly of type $(0,0)$, so we are reduced to showing that
 $H^{1,1}(\fg_v', K'_v; \pi_v' \otimes \rho_{v, k_v}')$ (for $v\in \Sigma$) is of type
 $(1,1)$. The Hodge type can be computed using \cite{zucker}, keeping
 in mind that {\it loc. cit.}  gives the Hodge
 numbers in the standard normalization; they need to be
twisted appropriately to get the Hodge type in the automorphic
(unitary) normalization that we are using.
For ease
 of comparison with \cite{zucker}, we temporarily change notation to
 match that reference. (See also \cite{faltings-bgg}, \cite{harris-motivesvolume}.)

Let $W \cong \mathfrak{S}_4$ and $W_c \cong \mathfrak{S}_2 \times \mathfrak{S}_2$ be the Weyl groups of $\fg = \mathfrak{gl}(4,\C)$ and $\fk = \mathfrak{gl}(2,\C) \oplus \mathfrak{gl}(2,\C)$, respectively.
Let $W^1$ be the set of representatives for $W_c \backslash W$ given by
\[
 \{ w \in W \, | \, w^{-1}(\Delta_c^+) \subset \Delta^+ \},
\]
where $\Delta_c^+$ and $\Delta^+$ are the sets of positive roots in $\fk$ and $\fg$, respectively.
Put
\[
 W^1(p) = \{ w \in W^1 \, | \, \ell(w) = p \},
\]
where $\ell(w)$ is the length of $w$.
Then we can enumerate the elements in $W^1$ as follows:
\[
\begin{array}{c|cccccc}
 p & 0 & 1 & 2 & 2 & 3 & 4 \\ \hline
 w^{-1} & 1 & (23) & (243) & (123) & (1243) & (13)(24)
\end{array}
\]

Fix $v \in \Sigma$ for the rest of the proof. Recall that $k_v$ is a positive even integer and put
\[
 \Lambda = \tfrac{1}{2}(k_v-2,k_v-2,-k_v+2,-k_v+2).
\]
Let $\rho$ be half the sum of positive roots in $\fg$:
\[
 \rho = \tfrac{1}{2}(3,1,-1,-3).
\]

 Let $Z=\{t_\alpha:= (\alpha, \alpha^{-1} ) \in U(1)\times \U(1) \} \subset \U(2)
 \times \U(2)$.
As in \cite{zucker}, \S 1 and \S 4, let $\mu$ and $\lambda$ denote the
highest characters of $Z$ appearing in the adjoint representation
of $\U(2,2)$ and in the representation $\rho'_{v,k_v}$. Then $\mu$ is
just the action on $\fp^+$ and is explicitly given by
\[
\mu (t_\alpha) = \alpha^2.
\]
As for $\lambda$, it agrees with the action of $Z$ on the irreducible
representation of $\U(2) \times \U(2)$ with highest weight $\Lambda$;
since this representation is just $\det^{\frac{k_v}{2}-1} \boxtimes
\det^{-\frac{k_v}{2}+1}$, we see that
\[
\lambda(t_\alpha) = (\alpha^2)^{\frac{k_v}{2}-1}
(\alpha^{-2})^{-\frac{k_v}{2}+1} = \alpha^{2k_v-4}.
\]
Since $\rho'_{v,k_v}$ is self-dual, the lowest character of $Z$
appearing in $\rho'_{v,k_v}$ is simply $\lambda^{-1}$; thus $m=
2k_v-4$, where $m$ is defined as in {\it loc. cit.} equation (4.8).
Here $m$ is the total weight of the Hodge structure on the fiber of
the local system. (To convert to our normalization where the total
weight on the fiber is zero, we must therefore twist the Hodge numbers
below by $(2-k_v,2-k_v)$.)

For completeneess, we consider not just $H^{1,1}$ but all the non-zero
  $H^{p,q}(\fg_v', K'_v; \pi_v' \otimes \rho_{v, k_v}')$, where
  $\pi_v'$ is chosen such that $H^{p,q} \neq 0$.
This space is then the sum of components of multi-degree $(p,q);(r,s)$
where $(r,s)$ with
$r+s=m$ is the bidegree coming from the Hodge structure on the fiber.
By \cite{zucker}, \S
  5,  the $(p,q);(k-p,m+p-k)$ component can only be non-zero if the
action $\tau_Z$ of
$Z$ on the irreducible representation $\tau$ of $\U(2) \times \U(2)$ with
highest weight $w(\Lambda + \rho) - \rho$ is $\lambda - k\mu$
 for some $w \in W^1 (p)$.
Thus we just need to run through the different choices of $(p,q)$ and
$w\in W^1 (p)$.

\begin{itemize}
\item If $(p,q)=(0,4)$ and $w^{-1}=1$, then
\begin{align*}
 w(\Lambda+\rho) & = \tfrac{1}{2}(k_v+1,k_v-1,-k_v+1,-k_v-1), \\
 w(\Lambda+\rho) - \rho & = \tfrac{1}{2}(k_v-2,k_v-2,-k_v+2,-k_v+2),
\end{align*}
so that
\begin{align*}
 \tau & = (\Sym^{0} \otimes {\det}^{\frac{k_v}{2}-1}) \boxtimes (\Sym^{0} \otimes {\det}^{-\frac{k_v}{2}+1}), \\
\tau_Z & : t_\alpha \mapsto (\alpha^2)^{\frac{k_v}{2}-1} (\alpha^{-2})^{-\frac{k_v}{2}+1}=\alpha^{2k_v-4}.
\end{align*}
Then $k=0$, so the Hodge type is
\[
(0,4) + (0,2k_v-4) = (0,2k_v).
\]

\item If $(p,q)=(1,1)$ and $w^{-1}=(23)$, then
\begin{align*}
 w(\Lambda+\rho) & = \tfrac{1}{2}(k_v+1,-k_v+1,k_v-1,-k_v-1), \\
 w(\Lambda+\rho) - \rho & = \tfrac{1}{2}(k_v-2,-k_v,k_v,-k_v+2),
\end{align*}
so that
\begin{align*}
 \tau & = (\Sym^{k_v-1} \otimes {\det}^{-\frac{k_v}{2}}) \boxtimes (\Sym^{k_v-1} \otimes {\det}^{-\frac{k_v}{2}+1}), \\
 \tau_Z & : t_\alpha \mapsto \alpha^{k_v-1} (\alpha^2)^{-\frac{k_v}{2}}(\alpha^{-1})^{k_v-1} (\alpha^{-2})^{-\frac{k_v}{2}+1}=\alpha^{-2}.
\end{align*}
Then $k=k_v-1$, so the Hodge type is
\[
(1,1) + (k_v-1-1, 2k_v-4 +1-(k_v-1) = (k_v-1,k_v-1).
\]

\item If $(p,q)=(2,2)$ and $w^{-1}=(243)$, then
\begin{align*}
 w(\Lambda+\rho) & = \tfrac{1}{2}(k_v+1,-k_v-1,k_v-1,-k_v+1), \\
 w(\Lambda+\rho) - \rho & = \tfrac{1}{2}(k_v-2,-k_v-2,k_v,-k_v+4),
\end{align*}
so that
\begin{align*}
 \tau & = (\Sym^{k_v} \otimes {\det}^{-\frac{k_v}{2}-1}) \boxtimes (\Sym^{k_v-2} \otimes {\det}^{-\frac{k_v}{2}+2}), \\
\tau_Z & : t_\alpha \mapsto \alpha^{k_v} (\alpha^2)^{-\frac{k_v}{2}-1}(\alpha^{-1})^{k_v-2}(\alpha^{-2})^{-\frac{k_v}{2}+2}=\alpha^{-4}.
\end{align*}
Then $k=k_v$, so the Hodge type is
\[
(2,2) + (k_v-2, 2k_v-4 + 2-k_v) = (k_v, k_v).
\]

\item If $(p,q)=(2,2)$ and $w^{-1}=(123)$, then
\begin{align*}
 w(\Lambda+\rho) & = \tfrac{1}{2}(k_v-1,-k_v+1,k_v+1,-k_v-1), \\
 w(\Lambda+\rho) - \rho & = \tfrac{1}{2}(k_v-4,-k_v,k_v+2,-k_v+2),
\end{align*}
so that
\begin{align*}
 \tau & = (\Sym^{k_v-2} \otimes {\det}^{-\frac{k_v}{2}}) \boxtimes (\Sym^{k_v} \otimes {\det}^{-\frac{k_v}{2}+1}), \\
\tau_Z & : t_\alpha \mapsto \alpha^{k_v-2}
(\alpha^2)^{-\frac{k_v}{2}}(\alpha^{-1})^{k_v}(\alpha^{-2})^{-\frac{k_v}{2}+1}=\alpha^{-4}.
\end{align*}
Then $k=k_v$, so the Hodge type is
\[
(2,2) + (k_v-2, 2k_v-4 + 2-k_v) = (k_v, k_v).
\]

\item If $(p,q)=(3,3)$ and $w^{-1}=(1243)$, then
\begin{align*}
 w(\Lambda+\rho) & = \tfrac{1}{2}(k_v-1,-k_v-1,k_v+1,-k_v+1), \\
 w(\Lambda+\rho) - \rho & = \tfrac{1}{2}(k_v-4,-k_v-2,k_v+2,-k_v+4),
\end{align*}
so that
\begin{align*}
 \tau & = (\Sym^{k_v-1} \otimes {\det}^{-\frac{k_v}{2}-1}) \boxtimes (\Sym^{k_v-1} \otimes {\det}^{-\frac{k_v}{2}+2}), \\
\tau_Z & : t_\alpha \mapsto \alpha^{k_v-1}
(\alpha^2)^{-\frac{k_v}{2}-1}(\alpha^{-1})^{k_v-1}(\alpha^{-2})^{-\frac{k_v}{2}+2}=\alpha^{-6}.
\end{align*}
Then $k=k_v+1$, so the Hodge type is
\[
(3,3) + (k_v+1-3, 2k_v-4 + 3-(k_v+1)) = (k_v+1, k_v+1).
\]

\item If $(p,q)=(4,0)$ and $w^{-1}=(13)(24)$, then
\begin{align*}
 w(\Lambda+\rho) & = \tfrac{1}{2}(-k_v+1,-k_v-1,k_v+1,k_v-1), \\
 w(\Lambda+\rho) - \rho & = \tfrac{1}{2}(-k_v-2,-k_v-2,k_v+2,k_v+2),
\end{align*}
so that
\begin{align*}
 \tau & = (\Sym^{0} \otimes {\det}^{-\frac{k_v}{2}-1}) \boxtimes (\Sym^{0} \otimes {\det}^{\frac{k_v}{2}+1}), \\
\tau_Z & : t_\alpha \mapsto
(\alpha^2)^{-\frac{k_v}{2}-1}(\alpha^{-2})^{\frac{k_v}{2}+1}=\alpha^{-2k_v-4}.
\end{align*}
Then $k=2k_v$, so the Hodge type is
\[
(4,0) + (2k_v-4, 2k_v-4 + 4-2k_v) = (2k_v, 0).
\]

\end{itemize}

The relevant case for us is the case $(p,q)=(1,1)$ in which case the overall
Hodge type $(k_v-1,k_v-1)$; twisting it by $(2-k_v, 2-k_v)$, we see
that $H^{1,1}(\fg_v', K'_v; \pi_v' \otimes \rho_{v, k_v}')$ has Hodge type $(1,1)$ as expected.
\end{proof}

Finally, we note that by \S \ref{ss:aut-coh-ls}, we may regard $\Pi$ as an automorphic representation of any of the groups
$\tilde{\mathscr{G}}(\A_F)$, $\mathscr{G}(\A_F)$, $\mathscr{G}_B (\A_F)$, $\tilde{\mathscr{G}}_B (\A_F)$.
Let $S$ and $\cK$ be as in \S \ref{ss:aut-coh-ls} as well. Then we get

\begin{cor}
\label{cor:hodge-is-purely-of-type-(d,d)}
The Hodge structure on $H^{2d}(\Sh_{\tilde{\mathscr{G}}_{B,\cK}}, \V_{\uk})[\Pi^S]$ is purely of type $(d,d)$.
\end{cor}

\subsection{Galois representation}
\label{sec:gal-rep-on-H2}
Finally we state the main result we need on Galois representations, that is motivated by the discussion of
\S \ref{ss:galrep-prep}.
\begin{prop}
\label{prop:galrep-char}
The action of $\Gal(\Qbar/F_\Sigma)$ on
\[
H^{2d} (\Sh_{\tilde{\mathscr{G}},\cK}, \V_{\uk,\ell}) [\Pi^S] (d)
\]
is trivial.
\end{prop}

The proposition above encodes the expected relation between the automorphic form $\Pi$ and the cohomology of the Shimura variety
$\Sh_{\tilde{\mathscr{G}},\cK}$, and is consistent with Cor. \ref{cor:hodge-is-purely-of-type-(d,d)}.
As mentioned in the introduction (Remark \ref{rem:mainremark-intro}, (ii)(b)), the proof of this will appear elsewhere.
See also \S \ref{sec:motivated} for a possible way to bypass the proposition above.

config: arxiv


TITLE: Determining the measure of a central angle in a circle ( subject math GRE exam 9768 Q.23).
QUESTION [0 upvotes]: The question is given in the following picture:

I know that the triangle ABC will be an isosceles triangle but I do not know how to use this knowledge in the solution, could anyone help me please ?

REPLY [2 votes]: HINT
Let the radius of the two circles to be $r$, then
$OA=OB=OC=r=AO=AB=AC$


TITLE: Bundle Morphism which is a diffeomorphism on fibres is a diffeomorphism
QUESTION [1 upvotes]: Let $F_{1}\to E_{1}\overset{\pi_{1}}{\to}\mathcal{M}$ and $F_{2}\to E_{2}\overset{\pi_{2}}{\to}\mathcal{M}$ be two fibre bundles over the same base manifold $\mathcal{M}$. If $H:E_{1}\to E_{2}$ is a bundle morphism such that its restriction $H:E_{1p}\to E_{2p}$ to the fibres $E_{1,2p}:=\pi_{1,2}^{-1}(\{p\})$ is for all $p\in\mathcal{M}$ a diffeomorphism, I have to show that $H$ is a diffeomorphism and hence a bundle isomorphism.
But isn't that obvious? I mean $H$ is by assumption bijective, because $E_{i}$ is the disjoint union of all its fibres and $H$ is a bijective on all of them. Furthermore, $H$ is by definition smooth and since it is a diffeomomorphism for ach fibre, its inverse is smooth restricted to all fibres and hence smooth as a whole, since the total spaces are the disjoint union of all their fibres.

REPLY [3 votes]: Edit: This answer addresses the case in which $H$ is a morphism of topological fiber bundles (i.e. merely continuous). The case where $H$ is required to be smooth is different, but the counterexample shows that the argument in the question is incorrect.
As Thorgott points out in a comment, the claim is false; an extra requirement is needed so that the diffeomorphism of fibers above $p$ "varies smoothly" as we vary $p$.
As a more concrete counterexample, take the bundle $\mathbb{R}^2\to\mathbb{R}$ given by $(x,y)\mapsto x$ and consider the map $\varphi(x,y)=(x,y+f(x))$. This map restricts to a diffeomorphism on each fiber (a translation), even if we chose a function $f$ which is not differentiable.


TITLE: $f :\mathbb R\to \mathbb R$ is a continuous function of period $1$ , then $f$ is uniformly continuous on $\mathbb R$
QUESTION [1 upvotes]: Let $f :\mathbb R\to \mathbb R$ be a continuous function such that $f(x+1)=f(x) , \forall x\in \mathbb R$ i.e. $f$ is of period $1$ , then how to prove that $f$ is uniformly continuous on $\mathbb R$ ?

REPLY [2 votes]: First take $I = [0,2]$. Since $I$ is compact and $f$ is continuous on $I$ we know $f$ is uniformly continuous on $I$. Now let $x,y \in \mathbb{R}$. Let $\epsilon > 0$, since $f$ is uniformly continuous we know $\forall z,w \in I \; \exists \delta' > 0$ where
$$
|z - w| < \delta' \implies |f(x) - f(y)| < \epsilon
$$
Now put $\delta = \min \{ \delta', 1 \}$. For every $x,y \in \mathbb{R}$ so that
$$
| x - y | < \delta
$$
we know that $\exists n \in \mathbb{Z}$ so that $x + n,y+n \in I$. So, finally
$$
\begin{eqnarray}
| x - y | < \delta & \implies & | x + n - y - n | < \delta \\
& \implies & | x + n - y - n | < \delta' \text{ since } \delta \le \delta'\\
& \implies & | f(x+n) - f(y + n) | < \epsilon \\
& \implies & |f(x) - f(y)| < \epsilon
\end{eqnarray}
$$


TITLE: Finding the angular speed of a rolling press?
QUESTION [1 upvotes]: A printing press has a roller 23 inches in diameter. A point on the rollers surface moves at a speed of 90 feet per second. What is the rollers angular speed? Round to the nearest hundredth.
I did c=d*π  which got me c=23π
then I did $90/23π =c/2π$ which after cross multiplying got me 77.24038227. However the answer is 93.91 radians per second.

REPLY [1 votes]: First, the linear speed is feet/sec while the diameter is inches, so you need a factor $12$ that gives you one revolution in $(23 \pi)/(90\cdot 12)$ seconds. Is your angular rate in degrees/second? Another conversion.


TITLE: Diffusion profile for localised release
QUESTION [0 upvotes]: I am trying to solve the 1-D diffusion equation in a tube of finite length that extends from $-L\leq x \leq L$. I have assumed that the radius is much smaller than the length of the tube so that we can neglect variations in the radial direction. Suppose I inject some amount of a substance at $t = 0$ such that the concentration at $x = 0$ is $\text{5000 ppm}$. I am having some trouble formulating the boundary conditions and initial conditions. Here's what I have thought of so far-
$\text{I.C: @ } t=0$ $$C = 5000\delta(x)$$
$\text{B.C: @ } x = 0$ $$\frac{\partial C}{\partial x} = 0 \hspace{10pt} \text{[assuming symmetry about x = 0]}$$
I am not too sure about the 2nd boundary condition. Would it be right to take $\frac{\partial C}{\partial x}$ at $x = L$ if I take an impermeable boundary? The initial condition is also something I am not too sure of since I thought of the Dirac-Delta function has having value $\infty$ at $x = 0$ and $0$ everywhere else. In some places it is defined as being equal to $1$ and $0$ respectively. If the former is true then multiplying it by 5000 doesn't make much sense. Any help is appreciated!

REPLY [1 votes]: The initial condition is correct: the density is infinite, but the integral (the total amount) is finite.
The boundary condition must be imposed at $x=\pm L$: it is the flux that is equal to zero.


TITLE: Does $\sum_{n=1}^{\infty}(-1)^n\frac{n}{\sqrt{n^3 + 6}}$ converge?
QUESTION [2 upvotes]: Test the series for convergence or divergence.
  $$\sum_{n=1}^{\infty}(-1)^n\frac{n}{\sqrt{n^3 + 6}}$$

Because this is an alternating series, I decided to use the alternating series test. This theorem states that if $\lvert a_n \rvert = b_n$ satisfies the conditions:
$b_{n+1} ≤ b_n$ for all $n$
$\lim \limits_{n \to \infty} b_n = 0$
This expression clearly approaches $0$ because $n$ grows asymptotically slower than the denominator of the expression, $\sqrt{n^3 + 6}$, as $n$ approaches $\infty$.
But I think the expression fails to meet the first condition.
While $b_{n+1} ≤ b_n$ is true for all $n>1$, I don't think it is true for $n$. Consider the following values of $b_n$ with values substituted in:
When $n=1$: $$\frac{1}{\sqrt{1^3 + 6}} = \frac{2}{\sqrt{7}} \approx 0.377964473$$
When $n=2$: $$\frac{2}{\sqrt{2^3 + 6}} = \frac{2}{\sqrt{14}} \approx 0.534522483$$
When $n=3$: $$\frac{3}{\sqrt{3^3 + 6}} = \frac{2}{\sqrt{33}} \approx 0.5222329679$$
So, since $0.377964473 < 0.534522483 > 0.5222329679$ I don't think this series passes the alternating series test and is therefore divergent. But I got this question wrong. Apparently, this series converges.
What am I doing wrong? Thanks for your help!

REPLY [3 votes]: Remember exactly what the alternating series test says:

If [some conditions hold] then the series converges.

So, we can never use the alternating series test to conclude that a series diverges; the theorem is silent on the subject of divergence.
The key idea is to find an $N$ such that the absolute value of the terms is monotonically decreasing. That is, if there's some $N$ such that $|a_{n+1}| < |a_n|$ for all $n > N$, then we can use the alternating series test on the sequence
$$\sum_{n = N+1}^\infty a_n,$$
to learn that it converges, and hence the original series must as well (as it's a finite number of terms added to a convergent sequence).
Now your job is to find the $N$ such that the absolute value of terms are monotonically decreasing for $n > N$.


text:
\begin{document}
\begin{frontmatter}

\title{\Large Optimal critical exponent $L^{p}$ inequalities of Hardy type on the sphere via Xiao's method}

\author{Ahmed A. Abdelhakim}
\address{Mathematics Department, Faculty of Science, Assiut University, Assiut 71516, Egypt}
\ead{ahmed.abdelhakim@aun.edu.eg}

\begin{abstract}
First, we correct the proof presented in [Abimbola Abolarinwa, Kamilu Rauf, Songting Yin,
Sharp $L^{p}$ Hardy type and uncertainty
principle inequalities on the sphere, Journal of
Mathematical Inequalities,
13, 4 (2019), 1011 - 1022] and obtain a correct sharp version of an $L^{p}$ Hardy inequality on the sphere
$\mathbb{S}^{n}$ for all $2\leq p<n$. Secondly, we prove sharp critical exponent $L^{n}$ inequalities on the sphere $\mathbb{S}^{n}$ in $\mathbb{R}^{n+1}$, $n\geq 2$. The singularity in this problem is the geodesic distance from an arbitrary point on the sphere.
\end{abstract}

\begin{keyword}
Sobolev Space
\sep $L^{p}$ Hardy inequalities
\sep density agrument
\sep compact manifold
\sep critical exponent
\MSC[2010] 26D10, 35A23, 46E35.
\end{keyword}

\end{frontmatter}
\section{Introduction}
\indent To our best knowledge, the first successful attempt to adapt the ideas introduced in \cite{Kombe}
to obtain inequalities of Hardy type on the $n$-dimensional sphere was that of Xiao's in
\cite{Xiao}. He obtained sharp $L^{2}$
Hardy inequalities on the Euclidean sphere $\mathbb{S}^{n}$, $n\geq 3$. Later, sharp critical case $L^{2}$ results were proved in \cite{ahmed} on the sphere $\mathbb{S}^{2}$ in $\mathbb{R}^{3}$.
Another extension was presented in \cite{xsun}
where subcritical optimal $L^{p}$ inequalities
were proved. The authors in \cite{ahmed,xsun,Xiao} considered the geodesic distance from the pole. In this case, the geodesic distance is precisely the angular variable. Very recently, the author in \cite{Songting}
improved the $L^{2}$ results in \cite{Xiao} by taking the singularity to be the geodesic distance from an arbitrary point on the sphere. This was followed by an attempt to obtain the corresponding sharp $L^{p}$ Hardy inequalities with the general geodesic distance in
\cite{Abimbola1}.
The author has proven in \cite{ahmed2} various sharp $L^{p}$ inequalities of Hardy type on the sphere in both the subcritical and critical exponent cases. The
way we prove sharpness of our inequalities in \cite{ahmed2} takes into account all the constants
involved. \\
\indent Let $u \in C^{\infty}(\mathbb{S}^{n})$, $n\geq 3$. Assume that $d$ denotes the geodesic distance on the sphere from an arbitrary point. It is claimed in (\cite{Abimbola1}, Theorem 1) that the following inequality holds for all $1<p<n$:
\begin{equation}\label{clmlp}
 \left(\frac{n-p}{p}\right)^{p} \int_{\mathbb{S}^{n}}
\frac{|u|^{p}d\sigma_{n}}{|\tan{d}|^{p}}
\leq   \int_{\mathbb{S}^{n}}
|\nabla f|^{p}d\sigma_{n}+
 \left(\frac{n-p}{p}\right)^{p-1}
\int_{\mathbb{S}^{n}}
\frac{|u|^{p}d\sigma_{n}}{\sin^{p-2}{d}}.
\end{equation}
The proof suggested in \cite{Abimbola1} is inaccurate. We point out a missing factor in that proof. Once corrected, the proof no longer implies the inequality (\ref{clmlp}). Interestingly, we obtain
an optimal $L^{p}$ inequality already proved in
\cite{ahmed2} using the divergence theorem, properties of the gradient and Laplacian of the geodesic distance, and H\"{o}lder and Young inequalities. Nevertheless, we believe that the inequality (\ref{clmlp}) probably
holds true.\\
\indent We also show how to adapt Xiao's method to obtain a sharp critical exponent $L^{n}$ Hardy type inequality on ${\mathbb{S}}^{n}$, $n\geq 2$, with the general geodesic distance from an arbitrary point on the sphere.
\section{Preliminaries}
Let $n\geq 2$ and let $\Theta_{n}:=(\theta_j)_{j=1}^{n}\in
[0,\pi]^{n-1}\times[0,2\pi]$.
We can assign to each point on the unit sphere
$\mathbb{S}^{n}$ in $\mathbb{R}^{n+1}$ the spherical coordinates parametrization $\left(x_{m}(\Theta_{n})\right)_{m=1}^{n+1}$, where
\begin{equation*}
x_{m}(\Theta_{n}):=\left\{
                \begin{array}{ll}
                  \cos{\theta_1}, & \hbox{$m=1$;} \vspace{0.1 cm} \\
                  \prod_{j=1}^{m-1}\sin{\theta_j}\cos{\theta_m}, & \hbox{$2\leq m\leq n$;} \vspace{0.15 cm}\\
                  \prod_{j=1}^{n}\sin{\theta_j}, & \hbox{$m=n+1$.}
                \end{array}
              \right.
\end{equation*}
With this representation, the surface gradient $\nabla_{\mathbb{S}^{n}}$ on the sphere $\mathbb{S}^{n}$  is defined by
\begin{equation*}
\nabla_{\mathbb{S}^{n}}  = \frac{\partial }{\partial \theta_1} \widehat{{\theta}_{1}} + \frac{1}{\sin \theta_1} \frac{\partial }{\partial \theta_2} \widehat{{\theta}_{2}}  + \cdots + \frac{1}{\sin \theta_1 \cdots \sin \theta_{n-1}} \frac{\partial }{\partial \theta_{n}} \widehat{{\theta}_{n}},
\end{equation*}
where $\left\{\widehat{{\theta}_{j}}\right\}$
is an orthonormal set of tangential vectors with the vector $\widehat{{\theta}_{j}}$ pointing in the direction of increase of ${\theta}_{j}$. In addition, the Laplace-Beltrami operator $\Delta_{\mathbb{S}^{n}}$ takes the form
\begin{align*}
\nonumber \hspace{-1 cm}\Delta_{\mathbb{S}^{n-1}}  =&\,
 \frac{1}{\sin^{n-2}{\theta_{1}}}
 \frac{\partial }{\partial \theta_1}
 \left(\sin^{n-2}{\theta_{1}} \frac{\partial }{\partial \theta_1} \right)+
\frac{1}{\sin^{2}{\theta_{1}}\sin^{n-3}{\theta_{2}}}
 \frac{\partial }{\partial \theta_2}
 \left(\sin^{n-3}{\theta_{2}} \frac{\partial }{\partial \theta_2} \right)+\\&\nonumber
+\dots+\frac{1}{\sin^{2}{\theta_{1}}\sin^{2}{\theta_{2}}...
 \sin^{2}{\theta_{n-2}}\sin{\theta_{n-1}}}
 \frac{\partial }{\partial \theta_{n-1}}
 \left(\sin{\theta_{n-1}} \frac{\partial }{\partial \theta_{n-1}} \right)+\\&+
  \frac{1}{\sin^{2}{\theta_{1}}\sin^{2}{\theta_{2}}...
 \sin^{2}{\theta_{n-1}}}
 \frac{\partial^2 }{\partial \theta^{2}_{n}}.
\end{align*}
Upon, identifying every point $(x_m(\Theta_{n-1}))_{m=1}^{n+1}\in \mathbb{S}^{n}$ with its parameters $\Theta_{n}$, the geodesic distance $d(\Theta_{n},\Phi_{n})$ from a point $\Phi_{n}
\in \mathbb{S}^{n}$ is defined by
\begin{equation}\label{gdsc}
d(\Theta_{n},\Phi_{n})=\arccos{\left(\sum_{m=1}^{n+1}
x_m(\Theta_{n})x_m(\Phi_{n})\right)}.
\end{equation}
The following properties of the geodesic distance on the sphere are proved in \cite{ahmed2}:
\begin{lem}\label{gradlap}
Suppose $\Phi_{n}$ is a point on the sphere
$\mathbb{S}^{n}$. Assume that $\,d(.,\Phi_{n-1}):
\mathbb{S}^{n}\rightarrow [0,\pi]\,$ is the geodesic distance from $\Phi_{n}$ on $\mathbb{S}^{n}$
defined in (\ref{gdsc}). Then
\begin{eqnarray}
\label{nablagdsc}
\left|\nabla_{\mathbb{S}^{n-1}}d\right|&=&
1,\\
\label{deltagdsc}\Delta_{\mathbb{S}^{n-1}}d
&=&(n-1)
\frac{\cos{d}}{\sin{d}}.
\end{eqnarray}
\end{lem}
We will also need the following basic inequality that can be found in \cite{lindqref}:
\begin{equation}\label{lindq}
|x+y|^{p}\geq |x|^{p}+
p|x|^{p-2}\langle x, y\rangle,\quad x,y\in \mathbb{R}^{n},\;p>1.
\end{equation}
\section{A correction of the proof in \cite{Abimbola1}}
Let $\Phi_{n}\in \mathbb{S}^{n}$, $n\geq 3$, and let $2\leq p<n$.
Let $u\in C^{\infty}(\mathbb{S}^{n})$ and write
\begin{equation}\label{dfn1}
u(\Theta_{n})=
\phi^{\alpha}{(\Theta_{n})}\,{\psi{(\Theta_{n})}},
\end{equation}
where $\phi(\Theta_{n}):=\sin{d(\Theta_{n},\Phi_{n})}$,
$\alpha=-{(n-p)}/{p}$. Clearly $\psi \in C^{\infty}\left(\mathbb{S}^{n}\right)$.
Since the geodesic metric
$d(\Theta_{n},\Phi_{n})=0$ only if $\Theta_{n}=\Phi_{n}$ and $d(\Theta_{n},\Phi_{n})=\pi$ only if
$\Theta_{n},\Phi_{n}$ are antipodal, then
$1/\phi \in C^{\infty}
\left(\mathbb{S}^{n}\setminus\left\{\pm \Phi_{n}\right\}\right)$. Taking the surface gradient of both sides of (\ref{dfn1}), then employing the inequality (\ref{lindq}), it follows that
on $\mathbb{S}^{n}\setminus\left\{\pm \Phi_{n}\right\}$ we have
\begin{align}
\nonumber
\hspace*{-1 cm}\left|\nabla_{\mathbb{S}^{n}} u\right|^{p}&=
\left|\alpha \phi^{\alpha-1}\psi
\nabla_{\mathbb{S}^{n}} \phi+\phi^{\alpha}
\nabla_{\mathbb{S}^{n}} \psi\right|^{p}
\\
&\nonumber \geq |\alpha|^{p}|\phi|^{\alpha p-p}|\psi|^{p}|\nabla_{\mathbb{S}^{n}} \phi|^{p}+\\
\nonumber &\quad+p|\alpha|^{p-2}|\phi|^{(\alpha-1)(p-2)}
|\psi|^{p-2}|\nabla_{\mathbb{S}^{n}} \phi|^{p-2}
\langle \alpha \phi^{\alpha-1}\psi
\nabla_{\mathbb{S}^{n}} \phi,\phi^{\alpha}
\nabla_{\mathbb{S}^{n}} \psi \rangle\\
&\nonumber = |\alpha|^{p}\phi^{\alpha p-p}|\psi|^{p}|\nabla_{\mathbb{S}^{n}} \phi|^{p}+\\
\label{gradf} &\quad
+\alpha|\alpha|^{p-2}
\phi^{\alpha p-p+1}
\left(p|\psi|^{p-2}\psi\right)
|\nabla_{\mathbb{S}^{n}} \phi|^{p-2}
\langle
\nabla_{\mathbb{S}^{n}} \phi,
\nabla_{\mathbb{S}^{n}} \psi \rangle,
\end{align}
since $\phi>0$. So far, our proof is in accordance with that in \cite{Abimbola1}. Since $p>1$, then $|\psi|^{p}$ is differentiable and we have
$\nabla_{\mathbb{S}^{n}}|\psi|^{p}=
p|\psi|^{p-2}\psi \nabla_{\mathbb{S}^{n}}\psi$.
Also, since $1/\phi$ is smooth on
on $\mathbb{S}^{n}\setminus\left\{\pm \Phi_{n}\right\}$ and $\alpha p-p+2=-(n-2)\neq 0$, then we can write $\phi^{\alpha p-p+1}\nabla_{\mathbb{S}^{n}}\phi=
\frac{1}{\alpha p-p+2}\nabla_{\mathbb{S}^{n}}\phi^{\alpha p-p+2}$.
Using this in (\ref{gradf}) implies
\begin{align}
\nonumber
\left|\nabla_{\mathbb{S}^{n}} u\right|^{p}\geq& |\alpha|^{p}\phi^{\alpha p-p}|\psi|^{p}|\nabla_{\mathbb{S}^{n}} \phi|^{p}+\\
&\label{gradf1}+\frac{\alpha|\alpha|^{p-2}}{\alpha p-p+2}|\nabla_{\mathbb{S}^{n}} \phi|^{p-2}
\langle\nabla_{\mathbb{S}^{n}} \phi^{\alpha p-p+2},
\nabla_{\mathbb{S}^{n}} |\psi|^{p} \rangle.
\end{align}
At this point, the factor $|\nabla_{\mathbb{S}^{n}} \phi|^{p-2}$ went unjustifiably missing in \cite{Abimbola1}. Moreover, the vector
$\nabla_{\mathbb{S}^{n}} |\psi|^{p}$ is confused with $\nabla_{\mathbb{S}^{n}} \psi^{p}$.
The next main step in \cite{Abimbola1}
is to write
\begin{equation*}
\langle
\nabla_{\mathbb{S}^{n}} \phi^{\alpha p-p+2},
\nabla_{\mathbb{S}^{n}} |\psi|^{p} \rangle=
\text{div}\left(|\psi|^{p}\nabla_{\mathbb{S}^{n}} \phi^{\alpha p-p+2}\right)-|\psi|^{p}
\Delta_{\mathbb{S}^{n}} \phi^{\alpha p-p+2},
\end{equation*}
then use the divergence theorem that yields
$\int_{\mathbb{S}^{n}}\text{div}\left(|\psi|^{p}
\nabla_{\mathbb{S}^{n}} \phi^{\alpha p-p+2}\right)d\sigma_{n}=0$.
If the divergence theorem is to be used, it should rather be applied
to $|\nabla_{\mathbb{S}^{n}} \phi|^{p-2}$ $\text{div}\left(|\psi|^{p}\nabla_{\mathbb{S}^{n}} \phi^{\alpha p-p+2}\right)$ whose integral does not simply vanish.\\
\indent Let us proceed from (\ref{gradf1}). Substituting for
$\alpha$, $\phi$ and $\psi$, then using (\ref{nablagdsc}), and integrating
both sides over $\mathbb{S}^{n}$, we find
\begin{equation}\label{gradf11}
\int_{\mathbb{S}^{n}}\left|\nabla_{\mathbb{S}^{n}} u\right|^{p}d\sigma_{n}\geq \int_{\mathbb{S}^{n}}\left(\frac{n-p}{p}\right)^{p}
\frac{|u|^{p}}{|\tan{d}|^{p}}d\sigma_{n}+
\frac{1}{n-2}\left(\frac{n-p}{p}\right)^{p-1}
I_{n,p},
\end{equation}
where
\begin{equation*}
I_{n,p}=\int_{\mathbb{S}^{n}}|\cos{d}|^{p-2}
\left\langle
\nabla_{\mathbb{S}^{n}} \frac{1}{\sin^{n-2}{d}},
\nabla_{\mathbb{S}^{n}}\left( |u|^{p}\sin^{n-p}{d} \right)\right\rangle
d\sigma_{n}.
\end{equation*}
Now, observe that, when $p\geq2$, we can make sense of the gradient
\begin{equation*}
\nabla_{\mathbb{S}^{n}}|\cos{d}|^{p-2}\cos{d}=
-(p-1)|\cos{d}|^{p-2}\sin{d}\,
\nabla_{\mathbb{S}^{n}} d.
\end{equation*}
Therefore, using (\ref{deltagdsc}), we may simplify $I_{n,p}$ by integration by parts on the compact manifold $\mathbb{S}^{n}$ to obtain
\begin{eqnarray*}
I_{n,p}&=&(n-2)\int_{\mathbb{S}^{n}}
|u|^{p}\sin^{n-p}{d}\:
\text{div}\left(|\cos{d}|^{p-2}\cos{d}
\frac{\nabla_{\mathbb{S}^{n}}\,d}{\sin^{n-1}{d}}\right)
d\sigma_{n}\\
&=&-(n-2)(p-1)\int_{\mathbb{S}^{n}}
|u|^{p}\sin^{n-p}{d}\frac{|\cos{d}|^{p-2}}
{\sin^{n-2}{d}}
d\sigma_{n}\\
&=&-(n-2)(p-1)\int_{\mathbb{S}^{n}}
\frac{|u|^{p}}{|\tan{d}|^{p-2}}
d\sigma_{n},
\end{eqnarray*}
because, using (\ref{nablagdsc}) and (\ref{deltagdsc}), it turns out that
\begin{equation}\label{bcsnbd}
\left\langle\nabla_{\mathbb{S}^{n}}
\frac{1}{\sin^{n-1}{d}},\nabla_{\mathbb{S}^{n}} d \right\rangle
=-\frac{\Delta_{\mathbb{S}^{n}} d}{\sin^{n-1}{d}},\;
\Theta_{n}\neq\pm \Phi_{n}.
\end{equation}
Plugging the integral $I_{n,p}$ into the inequality (\ref{gradf11}) we deduce the inequality
\begin{align}
\nonumber
\int_{\mathbb{S}^{n}}\left|\nabla_{\mathbb{S}^{n}} u\right|^{p}d\sigma_{n}+&
(p-1)\left(\frac{n-p}{p}\right)^{p-1}\int_{\mathbb{S}^{n}}
\frac{|u|^{p}}{|\tan{d}|^{p-2}}
d\sigma_{n}
\geq \\ &\label{gradf2}\left(\frac{n-p}{p}\right)^{p}\int_{\mathbb{S}^{n}}
\frac{|u|^{p}}{|\tan{d}|^{p}}d\sigma_{n}.
\end{align}
\begin{rem}
The inequality (\ref{gradf2}) is obtained using a different method in \cite{ahmed2}. It is also shown in \cite{ahmed2} that all three coefficients in (\ref{gradf2}) are optimal  for all
$2\leq p<n$, using optimizing sequences
in the Sobolev space $W^{1,p}(\mathbb{S}^{n})$. By density, such functions can be approximated by smooth functions to give optimizing sequences in $C^{\infty}(\mathbb{S}^{n})$.
\end{rem}
\begin{rem}
When $p=2$, the inequality (\ref{gradf2}) reduces to the main inequality proved in (\cite{Songting}, Theorem 1.1).
\end{rem}
\section{Critical $L^{n}(\mathbb{S}^{n})$ Hardy inequality}
We show how to apply Xiao's method \cite{Xiao} to prove an optimal critical exponent Hardy inequality on $\mathbb{S}^{n}$, $n\geq 2$, considering the geodesic distance
$d(.,\Phi_{n})$ defined in (\ref{gdsc}) from an arbitrary point $\Phi_{n}\in \mathbb{S}^{n}$. We will certainly make use of the properties (\ref{nablagdsc})
and (\ref{deltagdsc}).
\begin{thm}
Fix $n\geq 2$ and assume $u \in C^{\infty}(\mathbb{S}^{n})$. Let $\Phi_{n}$
be some point on the sphere $\mathbb{S}^{n}$ and consider the geodesic distance (\ref{gdsc}) from $\Phi_{n}$. Then
\begin{align}
\nonumber
\int_{\mathbb{S}^{n}}\left|\nabla_{\mathbb{S}^{n}} u\right|^{n}d\sigma_{n}+&
(n-1)\left(\frac{n-1}{n}\right)^{n-1}\int_{\mathbb{S}^{n}}
\frac{|u|^{p}}{|\tan{d}|^{n-2}
\left(\log{\frac{e}{\sin{d}}}\right)^{n-1}}
d\sigma_{n}
\geq \\ &\label{gradf3}\left(\frac{n-1}{n}\right)^{n}
\int_{\mathbb{S}^{n}}
\frac{|u|^{n}}{|\tan{d}|^{n}
\left(\log{\frac{e}{\sin{d}}}\right)^{n}}d\sigma_{n}.
\end{align}
\end{thm}
\begin{proof}
Suppose $u\in C^{\infty}(\mathbb{S}^{n})$.
Analogously to (\ref{dfn1}), we can write
\begin{equation*}
u(\Theta_{n})=
\phi^{\alpha}{(\Theta_{n})}\,{\psi{(\Theta_{n})}},
\end{equation*}
where $\psi \in C^{\infty}(\mathbb{S}^{n})$, but we modify the definition of $\alpha$ and that of $\phi$ as follows:
\begin{equation*}
\phi(\Theta_{n}):=
\log{\frac{e}{\sin{d(\Theta_{n},\Phi_{n})}}},\quad
\alpha=\frac{n-1}{n}.
\end{equation*}
We note here that $\phi\in C^{1}\left(\mathbb{S}^{n}\setminus\left\{\pm \Phi_{n}\right\}\right)$ and we have
\begin{equation}\label{grdlogsn}
\nabla_{\mathbb{S}^{n}}
\phi(\Theta_{n})=-\frac{1}{\tan{d(\Theta_{n},\Phi_{n})}}
\nabla_{\mathbb{S}^{n}}{d(\Theta_{n},\Phi_{n})},\quad
\Theta_{n}\neq\pm \Phi_{n}.
\end{equation}
Consequently, in the light of (\ref{nablagdsc}), we see that
\begin{equation}\label{grdlogsn12}
\left|\nabla_{\mathbb{S}^{n}}
\phi(\Theta_{n})\right|=\frac{1}{
\left|\tan{d(\Theta_{n},\Phi_{n})}\right|}.
\end{equation}
We pick up the proof at (\ref{gradf1}) with $p=n$.
Substituting for $\alpha$, $\phi$ and $\psi$, while using (\ref{grdlogsn}) and (\ref{grdlogsn12}), we get
\begin{equation}\label{pc1}
|\alpha|^{n}\phi^{\alpha n-n}|\psi|^{n}|\nabla_{\mathbb{S}^{n}} \phi|^{n}
=\left(\frac{n-1}{n}\right)^{n}
\frac{|u|^{n}}{\left|\tan{d}\right|^{n}
\left(\log{\frac{e}{\sin{d}}}\right)^{n}},\quad
\Theta_{n}\neq \pm \Phi_{n},
\end{equation}
\begin{equation}\label{pc2}
\begin{split}
&\frac{\alpha|\alpha|^{n-2}}{\alpha n-n+2}|\nabla_{\mathbb{S}^{n}} \phi|^{n-2}
\langle\nabla_{\mathbb{S}^{n}} \phi^{\alpha n-n+2},
\nabla_{\mathbb{S}^{n}} |\psi|^{n} \rangle\\
&=
\left(\frac{n-1}{n}\right)^{n-1}
\frac{1}{\left|\tan{d}\right|^{n-2}}
\left\langle\nabla_{\mathbb{S}^{n}} \left(\log{\frac{e}{\sin{d}}}\right),
\nabla_{\mathbb{S}^{n}} |\psi|^{n} \right\rangle\\
&=-
\left(\frac{n-1}{n}\right)^{n-1}
\frac{1}{\left|\tan{d}\right|^{n-2}\tan{d}}
\left\langle\nabla_{\mathbb{S}^{n}} d,
\nabla_{\mathbb{S}^{n}} |\psi|^{n} \right\rangle,\quad
\Theta_{n}\neq \pm \Phi_{n}.
\end{split}
\end{equation}
Taking into account the calculations (\ref{pc1}) and (\ref{pc2}), we integrate both sides of (\ref{gradf1}),  with $p=n$, over $\mathbb{S}^{n}$. It follows that
\begin{equation}\label{gradf301}
\begin{split}
&\int_{\mathbb{S}^{n}}\left|\nabla_{\mathbb{S}^{n}} u\right|^{n}d\sigma_{n}\geq
\left(\frac{n-1}{n}\right)^{n}\int_{\mathbb{S}^{n}}
\frac{|u|^{n}}{|\tan{d}|^{n}
\left(\log{\frac{e}{\sin{d}}}\right)^{n}}d\sigma_{n}-
\left(\frac{n-1}{n}\right)^{n-1}J_{n,p},
\end{split}
\end{equation}
where
\begin{equation*}
J_{n,p}=\int_{\mathbb{S}^{n}}
\frac{\left|\cos{d}\right|^{n-2}\cos{d}}{
\sin^{n-1}{d}}
\left\langle\nabla_{\mathbb{S}^{n}} d,
\nabla_{\mathbb{S}^{n}} |\psi|^{n} \right\rangle
d\sigma_{n}
\end{equation*}
Invoking the divergence theorem, we see that
\begin{equation*}
J_{n,p}=-\int_{\mathbb{S}^{n}}
|\psi|^{n}\text{div}
\left(\frac{\left|\cos{d}\right|^{n-2}\cos{d}}{
\sin^{n-1}{d}}
\nabla_{\mathbb{S}^{n}} d  \right)
d\sigma_{n}
\end{equation*}
Using the identity (\ref{bcsnbd}), we immediately realize
\begin{eqnarray}
\nonumber J_{n,p}&=&-\int_{\mathbb{S}^{n}}
\frac{|u|^{n}}{\left(\log{\frac{e}{
\sin{d(\Theta_{n},\Phi_{n})}}}\right)^{n-1}}
\left\langle \frac{\nabla_{\mathbb{S}^{n}} d}{\sin^{n-1}{d}},
\nabla_{\mathbb{S}^{n}}
{\left|\cos{d}\right|^{n-2}\cos{d}} \right\rangle
d\sigma_{n}\\
&=&\label{fnl}
(n-1)\int_{\mathbb{S}^{n}}
\frac{|u|^{n}}{\left(\log{\frac{e}{
\sin{d(\Theta_{n},\Phi_{n})}}}\right)^{n-1}}
\frac{1}{\left|\tan{d}\right|^{n-2}}
d\sigma_{n}.
\end{eqnarray}
The inequality (\ref{gradf3}) finally
follows from (\ref{gradf301}) and (\ref{fnl}).
\end{proof}
\begin{rem}
The inequality (\ref{gradf3}) is derived in \cite{ahmed2} using a different method.
It is noteworthy that all constants of (\ref{gradf3})
are optimal. This is also proved in \cite{ahmed2}
utilizing optimizing sequences in the Sobolev space
$W^{1,n}(\mathbb{S}^{n})$. The constants are therefore optimal for smooth functions. For, arguing by contradiction, if a constant in (\ref{gradf3}) could be improved for smooth functions, then the improved inequality would also be valid for
$W^{1,n}(\mathbb{S}^{n})$ functions by density.
\end{rem}


TITLE: Is there a solution to this geometry problem?
QUESTION [2 upvotes]: The point $G$ is centroid in the triangle $ABC$.
$m(BGC)=90^o$
$|AB|=8 \;cm$
$|CG|=? \, cm$
I know that $|GD|=|BD|=|DC|$ and $|AG|=|BC|=2|GD|$. I'm stuck after this point. I've tried to find the answer by solving system of equations, but it didn't work out. How do I solve this question?

REPLY [0 votes]: The point $D$ is not uniqely determined. Everything is on a picture:

We have $$4b^2+c^2=16$$
but $$ CG = 2c $$


TITLE: Measure of a set, convergence in measure
QUESTION [1 upvotes]: Let $(f_n)$ be a sequence of measurable real-valued functions which is Cauchy. Select a subsequence $(g_k)$ of $(f_n)$ such that the set $E_k=\{x\in X: |g_{k+1}-g_k(x)|\geq 2^{-k} \}$ is such that $\mu(E_k)<2^{-k}$. Let $F_k=\bigcup_{j=k}^\infty E_j$ so that $\mu(F_k)<2^{-(k-1)}$.
I know that $\mu(E_k)<2^{-(k-1)}$ but i don't know how to prove that $\mu(F_k)<2^{-(k-1)}$

REPLY [3 votes]: $$\mu(F_k)\leq\sum_{j=k}^\infty\mu(E_k)<\frac{2^{-k}}{1-\frac{1}{2}}=2^{-k+1}=2^{-(k-1)}$$


TITLE: Given Quantum Tunnelling, and a High Enough Velocity, Could Two Objects Theoretically Pass Through One Another?
QUESTION [2 upvotes]: I apologize if this has been answered before -- The answer may be out there, but I may just not have the proper terminology to find it.
I was messing around with Universe Sandbox, and noticed that if I flung two objects at one another at the speed of light, they would pass through one another. Obviously, this is a limitation of the game engine and the speed at which it's being simulated, but it did prompt a question for me:
I know that as you approach the speed of light, the length of objects appears shorter. I recall that if you approached earth close to the speed of light, Earth would appear to only be a few miles thick. So, given quantum tunneling, with a velocity close to the speed of light, would it be possible for two macroscopic objects to "quantum tunnel" past one another without any interaction?

REPLY [3 votes]: First, most commonly when people try to apply the tunneling idea to macroscopic objects they provide a rudimentary form of calculation which may be quite wrong. The difficulty is in allowing correctly for things like the Pauli exclusion principle. You can't just treat the wall merely as providing a potential function, because the wall is itself made of matter.
But anyway, even if we treat a simplified model, your idea won't work. It is easiest to see this by treating it in the rest frame of the wall. In this case a fast-moving oncoming entity has lots of energy, which means we don't need to consider tunneling because in any case the oncoming object already has more energy than the height of whatever potential hill the wall can provide. So it will go through the wall not because of some quantum effect, but simply because the wall was not able to exert a force strong enough to reflect it. Basically it will break the wall.


TITLE: Every finite subgroup of symmetric $n\times n$ matrices with nonzero determinant is isomorphic to some $(\mathbb{Z}/2\mathbb{Z})^m$, $0\leq m \leq n$
QUESTION [3 upvotes]: Suppose $G$ is such a subgroup. The products of symmetric matrices are still symmetric matrices, and we have $AB=(AB)^T=B^TA^T=BA$, so any such subgroup is a commuting family of symmetric matrices and can be simultaneously diagonalized by orthogonal matrices. The $(\mathbb{Z}/2\mathbb{Z})^m$ part reminds me of Sylvester law, and I suppose we can scale the length of the columns and rows to make any $A$ congruent to some $diag\{1,1,..,-1,-1,...\}$, (no $0$'s since det $A \neq 0$), and then map $-1$ to $1$ and $1$ to $0$ in the appropriate entry in $(\mathbb{Z}/2\mathbb{Z})^n$. But I don't think we can do the scaling simultaneously for every matrix in $G$.

REPLY [2 votes]: You don't need to do any scaling. You already know that all matrices in $G$ are simultaneously diagonalizable. So you fix one such diagonalization, say it's given by a matrix $R$; the eigenvalues of any $A\in G$ are $\{-1,1\}$ (proof at the end). Then you map $G\to \{-1,1\}^n$ by
$$
A\longmapsto \operatorname{diag}\{R^{-1}AR\}\subset\{-1,1\}^n.
$$
Now you further map, as you say, to $(\mathbb Z/2\mathbb Z)^n$. This way $G$ is isomorphic to a subgroup of $(\mathbb Z/2\mathbb Z)^n$.

Proof that $A\in G$ has eigenvalues in $\{-1,1\}$
As was already said above, we may assume that $A$ is diagonal, and the elements $a_1,\ldots,a_n$ of the diagonal are the eigenvalues of $A$. The $m^{\rm th}$ power of $A$ is still in $G$ and has eigenvalues
$$
a_1^m,\ldots,a_n^m.
$$
As $G$ is finite, these values have to cycle, and so $a_k\in\{-1,1\}$ is the only option, since since symmetric matrices have real eigenvalues and $0$ is disallowed by $A$ being invertible.


TITLE: Does the hyperdeterminant calculate a quantity akin to the volume of a parallelepiped?
QUESTION [10 upvotes]: If $M$ is an $n \times n$ matrix, $|\det(M)|$ is the volume of the $n$-dimensional
parallelepiped spanned by the column vectors of $M$.

(Image from Wikipedia's Determinant article.)

In the 19th-century, Cayley defined the hyperdeterminant of a hypermatrix $H$.
(A hypermatrix
can be viewed as a representation of a
tensor.)
My question is:

Q.
  Does the hyperdeterminant have a geometric interpretation somehow analogous
  to the parallelepiped-volume interpretation of the determinant?

Perhaps an answer resides in the
book by Gelfand, Kapranov, and Zelevinsky entitled Discriminants, Resultants and Multidimensional Determinants (Birkhäuser, Boston, 1994;
MAA link),
which I have yet to
examine.

REPLY [12 votes]: At least, in the classical case treated by Cayley, the $2{\times}2{\times}2$ hyperdeterminant, there is an interpretation in terms of volumes generalizing the classical determinant case.  It goes like this:
First, recall that, when two vector spaces $U$ and $V$ have the same dimension, say, $r$, there is a polynomial map of degree $r$
$$
\mathbf{det}: U\otimes V \to \Lambda^r(U)\otimes\Lambda^r(V)
$$
defined, relative to any basis $u_i$ of $U$ and $v_j$ of $V$, by
$$
\mathbf{det}( a^{ij} u_i\otimes v_j) = \det(a^{ij})
\,(u_1\wedge u_2\wedge\cdots\wedge u_r)\otimes (v_1\wedge v_2\wedge\cdots\wedge v_r)
$$
where $\det$ is the usual determinant of a matrix.  It is easy to see that this definition of $\mathbf{det}$ is independent of the choice of basis $u_i$ of $U$ and $v_\rho$ of $V$.  Note that, since $\Lambda^r(U)$ and $\Lambda^r(V)$ are the top exterior powers of $U$ and $V$, their elements represent 'volume elements' (sometimes called 'mass elements') in $U$ and $V$ respectively, so a tensor $\alpha\in U\otimes V$ gives rise, via $\mathbf{det}(\alpha)$ a way to 'multiply' volume elements in the two vector spaces.
(Of course, in the usual linear algebra interpretation, we consider $\alpha\in U^*\otimes V$ as a linear map from $U$ to $V$, and then
$$
\mathbf{det}(\alpha)\in \Lambda^r(U^*)\otimes \Lambda^r(V)\simeq
\Lambda^r(V)\otimes \Lambda^r(U)^*\simeq\  "\Lambda^r(V)/ \Lambda^r(U)"
$$
can be thought of as a ratio of volume forms on the two vector spaces.  Even more specially, we can take $U=V^*$, and then, because $\Lambda^r(V)\otimes \Lambda^r(V)^*$ is canonically isomorphic to $\mathbb{R}$, $\mathbf{det}(\alpha)$ becomes just a scalar.)
Now, when you have three vector spaces $U$, $V$, and $W$ of the same dimension $r$, you can expand a tensor $\alpha\in U\otimes V\otimes W$ in terms of a basis $u_i$ of $U$, $v_j$ of $V$, and $w_k$ of $W$ as
$$
\alpha = a^{ijk}\,u_i\otimes v_j\otimes w_k
$$
and you can just regard the $w_k$ as 'indeterminates' and define
$$
\mathbf{det}_{UV}(\alpha) = \det(a^{ijk}w_k)\otimes
(u_1\wedge u_2\wedge\cdots\wedge u_r)\otimes(v_1\wedge v_2\wedge\cdots\wedge v_r),
$$
where, now,
$$
\mathbf{det}_{UV}(\alpha) \in S^r(W)\otimes \Lambda^r(U) \otimes \Lambda^r(V).
$$
To go further, you need to look at particular values of $r$.  What Cayley did was consider the fact that, for quadratic forms in $r$ variables, there is a well-defined discriminant mapping $\mathrm{discr}_W:S^2(W)\to S^2(\Lambda^r(W))$ that is a homogeneous polynomial of degree $r$.  It is defined by
$$
\mathrm{discr}_W(q^{ij}w_iw_j) = \det(q^{ij})\,
(w_1\wedge w_2\wedge\cdots\wedge w_r)^2.
$$
In particular, when $r=2$, we can compose to get an element
$$
\mathbf{hdet}_{UVW}(\alpha) = \mathrm{discr}_W(\mathbf{det}_{UV}(\alpha))
\in S^2(\Lambda^2(U))\otimes S^2(\Lambda^2(V)) \otimes S^2(\Lambda^2(W)),
$$
which is a polynomial of degree $4$ in the coefficients of $\alpha$.
This is (the negative of) Cayley's hyperdeterminant in the $2{\times}2{\times}2$
case. Note that it is, indeed, expressed in terms of volume forms on the three vector spaces $U$, $V$, and $W$.  It's just that it is now the product of squares of volume forms.  By the way, it is not hard to show that, if we had, instead, computed $\mathbf{hdet}_{VWU}(\alpha)$, we would have got the same result.
I think that, at least in the $r=2$ case, this is probably the best interpretation of the hyperdeterminant in terms of volumes.
In the case when $(\dim U, \dim V, \dim W) = (2,2,s)$ where $s > 2$, this gives an expression
$$
\mathbf{hdet}_{UVW}(\alpha) = \mathrm{discr}_W(\mathbf{det}_{UV}(\alpha))
\in S^s(\Lambda^2(U))\otimes S^s(\Lambda^2(V)) \otimes S^2(\Lambda^2(W))
$$
of degree $2s$, but, of course, this vanishes identically when $s > 4$.
When you go to higher values of $r$, it's not so clear.  For example, when $r=3$ (again, with all vector spaces of the same dimension $r$), there are the two Aronhold relative invariants:  $Q^4: S^3(W)\to S^4(\Lambda^3W)$ (of degree $4$) and $Q^6:S^3(W)\to S^6(\Lambda^3W)$ (of degree $6$), and so we can define two expressions
$$
\mathbf{Q}^4_{UVW}(\alpha) = Q^4_W(\mathbf{det}_{UV}(\alpha))
\in S^4(\Lambda^3(U))\otimes S^4(\Lambda^3(V)) \otimes S^4(\Lambda^3(W))
$$
of degree $12$ and
$$
\mathbf{Q}^6_{UVW}(\alpha) = Q^6_W(\mathbf{det}_{UV}(\alpha))
\in S^6(\Lambda^3(U))\otimes S^6(\Lambda^3(V)) \otimes S^6(\Lambda^3(W))
$$
of degree $18$.  Thus, these clearly relate powers of volume elements of the underlying vector spaces.  According to the Wikipedia page, though, the hyperdeterminant of $\alpha$ in this case must have degree $36$; presumably it is a polynomial in the above two invariants.


TITLE: How to find the surface area using iterated integrals
QUESTION [2 upvotes]: The prompt is to find the surface area of the surface formed by $z=x^2+y^2$ inside $x^2 + y^2 = 1$

Here's what I did,
$$f(x, y) = x^2 + y^2$$
$$\nabla f(x, y) = <2x, 2y>$$
$$Surface Area = \iint_D \sqrt{(2x)^2 + (2y)^2 + 1} dA$$
But I'm not sure how to find the domain of the region and integral limits

REPLY [1 votes]: You do not have to find anything, the domain $D$ is given: it is the unit disk centered at the origin of the $xy$ plane. By using polar coordinates
$$ \iint_{x^2+y^2\leq 1}\sqrt{4(x^2+y^2)+1}\,dx\,dy = \int_{0}^{2\pi}\int_{0}^{1}\rho\sqrt{4\rho^2+1}\,d\rho\,d\theta=\left[\frac{\pi}{6}(1+4\rho^2)^{3/2}\right]_{0}^{1} $$
it follows that the wanted surface area is $\color{red}{\frac{\pi}{6}\left(5\sqrt{5}-1\right)}$.