TITLE: How to take cube of $\log$
QUESTION [1 upvotes]: What is the right way to solve $ \displaystyle (\log_2 n)^3 $ .
I want to decompose it in plain $\log$ without exponent.

REPLY [8 votes]: Assuming you mean you need to simplify the function $$f(n) = (\log_2 n)^3$$
There really is no simplification: it is the cube of a function which cannot be further simplified. I.e., any way you might manipulate the function would not result in a more simplified form.
If it is part of an equation you want to solve, please include that in your post. For example, if you have an equation in which $(\log_2 n)^3$ appears, there may be manipulations we can do to solve for $n$. For example, suppose we have the equation
$$(\log_2 n)^3 = 8$$
Then $$(\log_2 n)^3 = 8 \iff \log_2 n = \sqrt[\large 3]{8} = 2 \iff n = 4$$


text:
\section{The MLE with TAR Policies: A Characterization of Efficiency}
\label{sec:var}

In this section, we study the asymptotic variance of the MLE when TAR policies are used to sample and compare the two Markov chains in the experiment.   In Section \ref{ssec:CLT}, we develop a central limit theorem for the MLE estimator when used with TAR policies.  In Section \ref{ssec:optimality}, we use this central limit theorem to give a characterization of policies that are efficient, in the sense that they provide minimum asymptotic variance.

\subsection{A Central Limit Theorem}
\label{ssec:CLT}

A key tool in our analysis is {\em Poisson's equation} from the theory of Markov chains.  Let $P$ be the transition matrix of an irreducible Markov chain on $S$, with corresponding stationary distribution $\pi$; let $\Pi$ be matrix with rows equal to $\pi$, i.e., $\Pi = e \pi$ where $e= (1, \ldots, 1)$.  Further let $r$ be a reward function on $S$; we center $r$ by defining $\tilde{r} = r - e \pi r$.  Recall that Poisson's equation for $\tilde{r}$ under $P$ is:
\begin{equation}
\label{eq:poisson}
(I - P) g = \tilde{r}.
\end{equation}
One solution to the previous equation is given by:
\begin{equation}
\label{eq:PE_solution}
\tilde{g} = (I - P + \Pi)^{-1} r,
\end{equation}
where $(I - P + \Pi)^{-1}$ is the {\em fundamental matrix} associated to $P$.  (In general, the solution to Poisson's equation is not unique; however, the preceding solution is the unique one for which $\pi \tilde{g} = \pi r$.)  The following result is a well-known central limit theorem for finite Markov chains (see, e.g., \cite{asmussen2003applied}, Theorem 7.2).

\begin{proposition}
\label{prop:poisson_CLT}
As $n \to \infty$, the random variable $\frac{1}{\sqrt{n}} \left( r(X_0) + \cdots + r(X_{n-1}) - n \pi r \right)$
converges weakly to a normal random variable with mean zero and variance $\sigma^2(r) = \pi \tilde{g}^2 - \pi (P\tilde{g})^2$, where $\tilde{g}$ is the solution to Poisson's equation in \eqref{eq:PE_solution}.\footnote{Here we use the notation $f^2$ to denote elementwise squaring of the function, i.e., $f^2(x) = f(x)^2$.}
\end{proposition}

Our goal is to obtain a central limit theorem for TAR policies.  Note that there are several complexities in our setting that make this challenging: in general TAR policies allow for adaptive sampling, i.e., the chain chosen by the policy at a given time step can depend on the past history.  In particular, the induced state process may no longer be Markovian as a result.  To further complicate matters, the MLE $\hat{\alpha}_n$ cannot be represented simply as an average sum of rewards over the first $n$ time periods.

Nevertheless, we now present a central limit theorem result analogous to Proposition \ref{prop:poisson_CLT} for MLE estimation with TAR policies.

For $\ell = 1,2$, we define $\tilde{g}(\ell)$ to be the solution to \eqref{eq:PE_solution} for the transition matrix $P(\ell)$ with reward function $r(\ell)$, i.e.:
\[ \tilde{g}(\ell) = (I - P(\ell) + \Pi(\ell))^{-1} r(\ell). \]
In addition, define:
\begin{align}
\sigma^2(\ell,x) &= \Var\bigg(\tilde{g}(\ell,X_1)+R_1\big|X_{0}=x, A_{0}=\ell\bigg)\\
    &=\sum_{y\in S} P(\ell, x, y) [\tilde{g}(\ell,y)-\sum_{z\in S} P(\ell,x,z)\tilde{g}(\ell,z)]^2\\
    &\quad +\sum_{y\in S} P(\ell,x,y)\Var(R_1|X_0=x, X_1=y, A_0=\ell). \label{eq:sigma}
\end{align}

We have the following theorem.

\begin{theorem}\label{thm:CLT}
Suppose that $A$ is a TAR policy, with policy limits $\gamma=(\gamma(\ell,x): \ell=1,2,x\in S)$ that are almost surely  positive.  Let $G=\big(G(\ell,x): x\in S, \ell=1,2\big)$ be a family of independent Gaussian random variables with mean $0$ and unit variance.  Then for the MLE $\hat{\alpha}_n$, there holds
\begin{align}
\label{eq:CLT}
    n^{1/2}(\hat{\alpha}_n-\alpha)\Rightarrow \sum_{x\in S} \frac{\pi(1,x)\sigma(1,x)}{\gamma(1,x)^{1/2}} G(1,x)-\sum_{x\in S}\frac{\pi(2,x)\sigma(2,x)}{\gamma(2,x)^{1/2}}G(2,x)
\end{align}
as $n\rightarrow \infty$, where $G$ is independent of $\gamma$.
\end{theorem}

The full proof of Theorem \ref{thm:CLT} is in Appendix \ref{sec:var_proofs}.  A key idea in the proof is to show, via Poisson's equation, that:
\[     \hat{\alpha}_n(\ell) - \alpha(\ell) = \big(\hat{\pi}_n(\ell)-\pi(\ell)\big)r(\ell)
    =\hat{\pi}_n(\ell)\big(\hat{P}_n(\ell)-P(\ell)\big)\tilde{g}(\ell).
\]
We are then able to apply martingale arguments to analyze the right hand side of the preceding expression, by looking at the difference in the realized state transition, and the expected state transition.  These steps allow us to leverage classical martingale central limit theorem results; see Appendix \ref{sec:var_proofs} for details.

For later reference, the following two results will be useful.  The first applies Theorem \ref{thm:CLT} to show that we can lower bound the scaled asymptotic variance of the MLE.  The second shows that for TAR policies that have constant and positive policy limits, in fact we can exactly obtain the scaled asymptotic variance of the MLE.  We later use these results to show the existence of an optimal TAR policy with constant policy limits.  Proofs of both results are in the Appendix \ref{sec:var_proofs}.

\begin{corollary}
\label{cor:var_lowerbound}
Let $A$ be a TAR policy with almost surely positive policy limits $\gamma$, with associated MLE $\hat{\alpha}_n$.
Then there holds:
\begin{equation}
\label{eq:var_lowerbound}
\liminf_{n \to \infty} n \Var(\hat{\alpha}_n - \alpha) \geq \sum_{\ell=1,2} \sum_{x\in S} \frac{\pi^2(\ell,x)\sigma^2(\ell,x)}{E\{\gamma(\ell,x)\}}.
\end{equation}
\end{corollary}

\begin{corollary}
\label{cor:asymptotic_var_constant_TAR}
Let $A$ be a TAR policy with almost surely constant and positive policy limits $\gamma$.  Let $\hat{\alpha}_n$ be the associated MLE.  Then:
\begin{equation}
    \label{eq:const_policy_limit_var}
 \lim_{n \to \infty} n \Var(\hat{\alpha}_n - \alpha) = \sum_{\ell=1,2} \sum_{x\in S} \frac{\pi^2(\ell,x)\sigma^2(\ell,x)}{\gamma(\ell,x)}.
\end{equation}
\end{corollary}

\subsection{Optimal TAR Policies with the MLE Estimator}
\label{ssec:optimality}

In this section we characterize efficient policies, i.e., those that achieve minimum asymptotic variance within the class of TAR policies with almost surely positive policy limits.  (Note that we know such policies are consistent from Corollary \ref{cor:consistency}, and thus we focus solely on asymptotic variance in considering efficiency.)  We show that the policy limits of any such policy can be characterized via the solution to a particular convex optimization problem.  We require the mild additional assumption in this section that $\sigma(\ell,x) > 0$ for all $\ell,x$; this will hold, e.g., if rewards are random in each state.

We start with the following formal definition of efficiency.
\begin{definition}
\label{def:efficient}
Let $A^*$ be a TAR policy with almost surely positive policy limits, with associated MLE $\hat{\alpha}^*_n$.  We say that $A^*$ is {\em efficient} if it has lower (scaled) asymptotic variance than any other TAR policy with almost surely positive policy limits; i.e., for any such policy $A$ with associated MLE $\hat{\alpha}_n$, there holds:
\begin{equation}
\label{eq:efficient}
 \limsup_{n \to \infty} n \Var(\hat{\alpha}^*_n - \alpha) \leq \liminf_{n \to \infty} n \Var(\hat{\alpha}_n - \alpha).
 \end{equation}
\end{definition}

The following theorem leverages the central limit theorem in Theorem \ref{thm:CLT}, and in particular Corollaries \ref{cor:var_lowerbound} and \ref{cor:asymptotic_var_constant_TAR}, to
give an optimization problem whose solution characterizes the optimal (scaled) asymptotic variance.  The proof of the theorem is in the Appendix.

\begin{theorem}
\label{thm: stationary optimality}
Suppose that for all $\ell,x$, there holds $\sigma(\ell,x) > 0$.  Consider the following (convex) optimization problem:
\begin{align}
    \text{minimize}\ \ \ & \sum_{\ell=1}^{2} \sum_{x\in S} \frac{\pi^2(\ell,x)\sigma^2(\ell,x)}{\kappa(\ell,x)} \label{eq:MLE_opt1}\\
    \text{subject to}\ \ \ & \kappa \in \mathcal{K}. \label{eq:MLE_opt2}
\end{align}
This problem has a unique solution $\kappa^*$, and all entries of $\kappa^*$ are positive.  Any TAR policy $A^*$ that has almost surely constant policy limits $\kappa^*$ is efficient, and the scaled asymptotic variance of the MLE under $A^*$ is given by \eqref{eq:MLE_opt1} evaluated at $\kappa^*$; i.e., if we let $\hat{\alpha}_n^*$ denote the resulting MLE, we have:
\begin{equation}
    \label{eq:Astar_var}
 \lim_{n \to \infty} n \Var(\hat{\alpha}_n^* - \alpha) = \sum_{\ell=1}^{2} \sum_{x\in S} \frac{\pi^2(\ell,x)\sigma^2(\ell,x)}{\kappa^*(\ell,x)}.
\end{equation}

\end{theorem}

config: arxiv


TITLE: The Derivative Operator vs $f'(x)$
QUESTION [1 upvotes]: For the sake of example, suppose $f(x) = \sin(x) $.
Then,
$$f'(x) = \frac{d}{dx}[f(x)] $$
But
$$f'(0) = 1 \neq \frac{d}{dx}[f(0)] = 0. $$
Why is this the case? Usually, when you have two equivalent functions, they are equal for any value of x. So how can the left side not equal the right side in the second line? (I understand the calculus here; I'm simply confused by the notation.)

REPLY [3 votes]: Whereas the expression $f'(x)$ is the derivative of $f(x),$ the expression $f'(0)$ is not the derivative of $f(0).$
Instead, when handling the expression $f'(0),$ the correct order of operation is to differentiate then substitute: \begin{align}f'(0) &= f'(x) \bigg|_{x=0} \\&\neq \frac{\mathrm d}{\mathrm dx}\Big(f(0)\Big).\end{align}
In other words, the $f'$ (let's rename it $g)$ in the expression $f'(0)$ ought to be thought of as a function in its own right.


TITLE: Question about potential function of a vector field
QUESTION [1 upvotes]: Lets say we have conservative vector field $\mathbf{F}(x,y,z)=y\sin{z}\mathbf{i}+x\sin{z}\mathbf{j}+xy\cos{z}\mathbf{k}$
I found it's potential function like this
$$\begin{cases} \frac{\partial{\phi}}{\partial{x}} = y\sin{z} \\ \frac{\partial{\phi}}{\partial{y}} = x\sin{z} \\ \frac{\partial{\phi}}{\partial{z}} = xy\cos{z} \end{cases}\int \to \begin{cases} \phi(x,y,z) = xy\sin{z}+C_1(y,z) \\ \phi(x,y,z) = xy\sin{z}+C_2(x,z) \\ \phi(x,y,z) = xy\sin{z}+C_3(x,y) \end{cases}$$
Stupid question but is the constant of the potential function $C(x,y,z)$ or C. Because in my books solution its C and I think it has to be $C(x,y,z)$.

REPLY [0 votes]: Consider that the function $\phi $ is just one and only function, not three. So in fact, $\ C_1 = C_2 = C_3 $. Since you are demanding it not to be a function of x, nor a function of y or z, then the only choice is that it must be just a number. The $\ C_i $ functions are intermediate steps that follow from the integration with respect to one variable, but get defined when you differentiate wrt to the next one, and compare with the partial derivative you already know (although maybe you have got this just by guessing). Try to differentiate $\phi $ and see that $\frac{\partial{C}}{\partial{x}} = \frac{\partial{C}}{\partial{y}} = \frac{\partial{C}}{\partial{z}} = 0$


TITLE: Prove the set of subsequential limits of ($x_t$) is closed
QUESTION [1 upvotes]: A sequence ($x_t$) in metric space ($X,d$). Let $S$ be the set of subsequential limits of ($x_t$) ($S$ could be empty), prove that $S$ is closed.
I need to prove the problem with one of the characterizations of closed: prove that if ($s_t$) is a sequence of subsequential limits with ($s_t \rightarrow s$), then $s$ is a subsequential limit, i.e. $s\in S$.
Could someone help me with this proof please? Thanks.

REPLY [3 votes]: HINT: This is a ‘follow-your-nose’ proof: try the most straightforward thing, and you find that it works. The most difficult part is keeping the notation straight.
Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $X$, and let $\langle s_k:k\in\Bbb N\rangle$ be a sequence of subsequential limits of $\sigma$ that converges to $s\in X$. For each $k\in\Bbb N$ there is an infinite $N_k\subseteq\Bbb N$ such that the subsequence $\langle x_n:n\in N_k\rangle$ of $\sigma$ converges to $s_k$. (This is one way to write multiple subsequences without getting lost in a lot of indices: $N_k$ is simply the set of indices appearing in the $k$-th subsequence, and we assume that we take them in increasing order, so that we really do get a subsequence of $\sigma$.)
We want to find an infinite $N\subseteq\Bbb N$ such that the subsequence $\langle x_n:n\in N\rangle$ of $\sigma$ converges to $s$. Once we’ve found $N$, we can index it as $N=\{n_k:k\in\Bbb N\}$, where $n_0<n_1<\ldots\;$, and the idea is to choose the integers $n_k$ one at a time so that $\langle x_{n_k}:k\in\Bbb N\rangle$ really does converge to $s$. Specifically, for each $k\in\Bbb N$ we’ll choose $n_k$ so that
$$d(x_{n_k},s)<\frac1{2^k}\;.\tag{1}$$
It’s a recursive construction. Imagine that we’ve already chosen $n_k$ for each $k\le\ell$ in such a way that each satisfies $(1)$, and moreover $n_0<\ldots<n_\ell$; we want to choose $n_{\ell+1}>n_\ell$ such that
$$d(x_{n_{\ell+1}},s)<\frac1{2^{\ell+1}}\;.$$
The sequence $\langle s_k:k\in\Bbb N\rangle$ converges to $s$, so we can find an $m\in\Bbb N$ such that
$$d(s_m,s)<\frac1{2^{\ell+2}}\;.$$
And $\langle x_k:k\in N_m\rangle$ converges to $s_m$ so we can find an $r\in\Bbb N$ such that
$$d(x_k,s_m)<\frac1{2^{\ell+2}}$$
whenever $k\ge r$ and $k\in N_m$.

Can you finish it off from here by specifying how to choose $n_{\ell+1}$, choosing $n_0$, and rounding out the explanation of why the subsequence $\langle x_n:n\in N\rangle$ of $\sigma$ really does converge to $s$?


TITLE: Irreducibility of $x^2 + 1$ in $\Bbb F_p$
QUESTION [1 upvotes]: I can show that in field formed by roots of $x^{4 \cdot 2^n + 1} - x$, the polynomial $x^2 + 1$ factors into linear factors. This gives $p = 4 \cdot 2^n+1\equiv 1 \pmod 4$. But how if $p \equiv 3 \pmod 4$, $x^2 + 1$ is irreducible in $\Bbb F_p$?
So far, I think if $x^2 + 1$ is irreducible in $\Bbb F_p$ then $x(x^p-1)$ should not contain a factor $x^2 + 1$. Any value other than $p=4n+1$ does not give this factor. But why $p \equiv 3 \pmod 4$ in particular? Why not $p \equiv 0 \pmod 4$ or $p \equiv 2 \pmod 4$?
Here it is done in example 1.0.3.

Workout: I was too stupid to realize that $p$ was prime.
  No primes have the form $p = 4n$ or $p = 4n+2 $

REPLY [3 votes]: The polynomial $x^2+1$ factors into linear polynomials if the field has a fourth root of unity.


TITLE: Find all $n$ such that $n \mid x_n$ where $x_n = x_{n-1} + \lfloor n^2 / 4 \rfloor$, $x_0 = 0$
QUESTION [5 upvotes]: $X_n$ is sequence such that $x_n=x_{n-1}+[\frac{n^2}{4}]$ and $x_0=0$. Find all positve integers $n$ for which $x_n$ is divisible by $n$.
[X] means integer part.

REPLY [1 votes]: $$x_{2k}=x_{2k-1}+k^2=x_{2k-2}+2k^2-k=2k^2-k+(2(k-1)^2-(k-1))+x_{2k-4}=\cdots=\sum_{n=1}^{k} 2n^2-\sum_{k=1}^{k} n=\frac{k(k+1)(2k+1)}{3}-\frac{k(k+1)}{2}=\frac{k(k+1)(4k+2-3)}{6}=\frac{k(k+1)(4k-1)}{6}$$
Also
$$a_{2k+1}=a_{2k}+k^2+k\\a_{2k+1}=\frac{k(k+1)(4k-1)}{6}+\frac{6k(k+1)}{6}=\frac{k(k+1)(4k+5)}{6}$$
Now splitting into two cases we can see that
$$2k+1\mid \frac{k(k+1)(4k+5)}{6}$$ Works only for $k=1$ since $\gcd(2k+1,k)=\gcd(2k+1,k+1)=1$ and $\gcd(4k+5,2k+1)\mid  3$ that implies that only possible solution is $2k+1=1$ or $2k+1=3$
$$2k\mid\frac{k(k+1)(4k-1)}{6}$$
Works for $k=12t+7$ and $k=12t+11$,this is because $$\frac{(k+1)(4k-1)}{12}\in\mathbb{N}$$
Putting it all together for $n=3$ and $n=24t+14,n=24t+22$ we have that $n\mid a_n$


TITLE: Noncommuting complex matrices: Existence of a simultaneous eigenvector
QUESTION [3 upvotes]: Let $A$ and $B$ be $n\times n$ matrices with complex entries such that
$AB - BA$ is a linear combination of $A$ and $B$.
I'd like to prove that there exists a non-zero vector $v$ that is an eigenvector of both $A$ and $B$.

REPLY [1 votes]: It is assumed throughout that

The given matrices $A$ and $B$ are linearly independent in the $\mathbb C$-vector space $M_n(\mathbb C)$,
  thus ignoring the unremarkable setup where one matrix is a multiple of the other.

My next guess is that the question doesn't focus upon a vanishing commutator of $A$ and $B$, hence consider
$$0\:\neq\:[A,B]\:=\:AB-BA\:=\:\alpha A+\beta B$$
and at least one of the scalars $\alpha,\beta $ is not zero. Divide by a scalar, say $\alpha\neq0$, then set
$$B'=\frac1\alpha B\;\text{ and }\; A'= A+\beta B'\quad\text{to obtain }\;
\big[\,A',B'\big]\:=\:A'\,.$$
Thus in the sequel we may start w.l.o.g. from
$\,\mathbf{AB-BA=A\qquad\qquad (\star)}$
I decided to spill some more MathJax ink here & now, to give examples and a bit Lie algebra background as well. Who's not willing to go through all this should ...
simply jump: Goto The shared eigenvector !
Examples
how to satisfy the relation above:

If $\,n=2$, then $\,A=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\:$ and
$\;B=\frac 12\begin{pmatrix} -1&0\\ 0&1\end{pmatrix}\;$ do the job.
$\:\begin{pmatrix}1\\ 0\end{pmatrix}$ is a joint eigenvector.
For $n=3$ one may consider
$$A\:=\:\begin{pmatrix}0&a_2&a_3\\ 0&0&0\\ 0&0&0\end{pmatrix}\;,\quad
B\:=\:\begin{pmatrix}b&0&0\\ 0&b+1&0\\ 0&0&b+1\end{pmatrix}$$
where $a_2,a_3,b\in\mathbb C$,and $a_2,a_3$ not both zero.

It is not by accident that $A$ is represented by a nilpotent matrix since this cannot be avoided:
For each $n\in\mathbb N$ one has
$$\big[\,A^n,B\big]\:=\:n\,A^n,$$
the proof by induction using
$\,[AA^{n-1},B] = A[A^{n-1},B] + [A,B]\,A^{n-1}$ (derivation property) is straightforward. Let $\|\cdot\|$ be a submultiplicative matrix norm, then
$$n\|A^n\|\:=\:\|A^nB-BA^n\|\;\leq\;2\|B\| \|A^n\|\quad\forall\,n$$
which implies $A^n=0\,$ at the latest when $\,n>2\|B\|$. Thus $A$ is nilpotent.
The shared eigenvector
for $B$ and $A$ does exist: Pick an eigenvector $w\neq 0$ to an eigenvalue $\mu$ of $B$ such that $\,\mu-1\,$ is no eigenvalue of $B$. Insert $w$ into $(\star)$
$$BAw\:=\:ABw - Aw\:=\:(\mu-1)Aw\quad\Longrightarrow\; Aw\:=\:0\,,$$
thus $w$ is also an eigenvector of $A$, to the eigenvalue zero.
Lie algebra background
$A$ and $B$ span a complex Lie algebra with Lie bracket given by $(\star)$. It is up to (Lie algebra) isomorphism the unique complex Lie algebra, which is 2-dimensional and nonabelian. It is solvable, but not nilpotent.
Lie's theorem guarantees (in any matrix representation) the existence of a shared eigenvector for all elements of a solvable Lie algebra. One implication is that all matrices of a representation may be chosen to be triangular.
Related posts

$AB + BA = A \implies A$ and $B$ have a common eigenvector
Questions about non-abelian 2 dimensional lie algebras
A Problem about Common Eigenvector
Representations of the two dimensional non-abelian Lie algebra @MO
Examples of finite dimensional non simple non abelian Lie algebras @MO


TITLE: Find the values of $a$ for which $f(x)$ is periodic
QUESTION [1 upvotes]: Let a function $f(x)$ be defined as
$$f(x)=x^a+\sin{x}-ax$$
Then find the number of values of $a$ for which $f(x)$ is periodic.

My Attempt:
I was only able to find two values i.e. $a=0, a=1$.
But how can I prove that these are the only solutions.

REPLY [0 votes]: Apparently, $a=0,1$ are two proper values of $a$ satisfying the requirement.
Case 1
Assume that $a>1$ and $f(x)$ has a positive period $T$. Then $f(0)=f(0+nT),(n=1,2,\cdots)$, namely
$$\frac{\sin (nT)}{anT}=-\frac{(nT)^a}{anT}+1=-\frac{(nT)^{a-1}}{a}+1,$$which holds for all $n=1,2,\cdots.$ Hence, let $n \to \infty$ and take the limits of both sides. We obtain $$0=-\infty,$$ which is absurd.
Case 2
Assume that $0<a<1$ and $f(x)$ has a positive period $T'$. Then $f(0)=f(0+nT'),(n=1,2,\cdots)$, namely
$$\frac{\sin (nT')}{anT'}=-\frac{(nT')^a}{anT'}+1=-\frac{(nT')^{a-1}}{a}+1,$$which holds for all $n=1,2,\cdots.$ Hence, let $n \to \infty$ and take the limits of both sides. We obtain $$0=1,$$ which is absurd.
Case 3
Assume that $a<0$ and $f(x)$ has a positive period $T''$. Notice that $f(x)$ has no definition at $x=0.$ Then we should have $f(x_0)=f(x_0+nT''),(x_0 \neq 0,~~~n=1,2,\cdots)$, namely
$$x_0^a+\sin x_0=(x_0+nT)^a+\sin(x_0+nT)-anT\geq (x_0+nT)^a-1-anT$$which holds for all $n=1,2,\cdots.$ Hence, let $n \to \infty$ and take the limits of both sides. We obtain $$x_0^a+\sin x_0 \geq +\infty,$$ which is absurd.
Summing up the three cases, we may know that $a=0,1$ are the all values of $a$ such that $f(x)$ is periodic.


TITLE: Calculate $(f^{-1})'(0)$ of $\int_{\pi}^{x} \cos(\cos x)\,dx$
QUESTION [2 upvotes]: Calculate $(f^{-1})'(0)$ if $$f(x)=\int_{\pi}^{x} \cos(\cos x)\,dx$$

How do I calculate the derivative in $0$ of the inverse function of that integral?
I can't integrate it and I think I don't have too. I thought about using the derivation rule of the inverse function but I'm kind of lost. I read about Bessel function in a similar question but never heard of it and I think its unneeded.
Any help is appreciated!

REPLY [3 votes]: Indeed, you can use the fact that $$(f^{-1})'(0)=\frac{1}{f'(f^{-1}(0))}$$ Notice that $\cos\cos x$ is positive for all $x\in\Bbb R$, since $\cos x\in[-1,1]$ and the nearest zeros of $\cos$ are in $\pm\frac\pi2$.
Therefore, $f$ is a strictly increasing function, which does not leave much choice for what $f^{-1}(0)$ can be...


TITLE: How to solve a recurrence relation with generating functions?
QUESTION [2 upvotes]: I don't really understand how to solve (with generating functions) for the recurrence relation of $$a_n = a_{n-1}+2(n-1)$$ with initial conditions of  $a_1 = 2$ when $n \geq 2$
This is what I was thinking,
\begin{align*}
    g(x) &= a_0x^0+a_1x^1+a_2x^2+...+a_rx^r+...\\
    g(x) &= a_0 + a_1x^1 + \sum^{\infty}_{n=2}a_nx^n\\
    g(x) &= 1 + 2x + \sum^{\infty}_{n=2}{(a_{n-1}+2(n-1))x^n}\\
    g(x) &= 1 + 2x + \sum^{\infty}_{n=2}a_{n-1}x^n + \sum^{\infty}_{n=2}(2n-2)x^n\\
    g(x) &= 1 + 2x + x\sum^{\infty}_{n=1}a_{n-1}x^{n-1} + \sum^{\infty}_{n=2}2n x^n - \sum^{\infty}_{n=2}2x^n\\
    g(x) &= 1 + 2x + x\sum^{\infty}_{m=1}a_{m}x^{m} + \sum^{\infty}_{n=2}2 \binom{n}{1}x^n - \sum^{\infty}_{n=2}2x^n\\
    g(x) &= 1 + 2x + x(g(x) - a_0) + \sum^{\infty}_{n=2}2 \binom{n}{1}x^n - \sum^{\infty}_{n=2}2x^n\\
\end{align*}

REPLY [1 votes]: You certainly do not need generating functions for solving $a_n-a_{n-1}=2(n-1)$: it is enough to sum both sides on $n=1,2,\ldots,N$ to get that $a_n$ depends on $\sum_{k=1}^{n}2(k-1)$. The only issue in your case is that the given values of $a_0$ and $a_1$ do not agree with the given recurrence relation. So, let us assume $a_0=\color{red}{2}$ and $a_{n}=a_{n-1}+2(n-1)$ for any $n\geq 1$. By denoting as $f(x)$ the following generating function
$$ f(x)=\sum_{n\geq 0} a_n x^n = 2+\sum_{n\geq 1}a_n x^n$$
we have
$$ x\cdot f(x) = \sum_{n\geq 0} a_n x^{n+1} = \sum_{n\geq 1} a_{n-1} x^n $$
hence by the recurrence relation
$$ (1-x) f(x) = 2+2\sum_{n\geq 1}(n-1)x^n = 2+\frac{2x^2}{(1-x)^2}$$
where the last identity follows from stars and bars in the form
$$ \frac{1}{(1-x)^{m+1}}=\sum_{n\geq 0}\binom{n+m}{m}x^n.$$
 Since we have
$$ f(x) = \frac{2}{1-x}+\frac{2x^2}{(1-x)^3} $$
stars and bars also gives
$$ a_n = [x^n]f(x) = 2-n+n^2. $$


text:
\begin{document}

\title[Rigidity in de Sitter Space]{A Rigidity Theorem for Spacelike Hypersurfaces in de~Sitter Space}
\author[T. Hasson]{Tristan Hasson}
\address{Tristan Hasson, Department of Mathematical Sciences, University of Durham, Durham DH1 3LE, United Kingdom}
\email{tristan.hasson@durham.ac.uk}
\date{\today}
\subjclass[2010]{53C24}
\begin{abstract}
In this paper we present a rigidity theorem for locally isometric hypersurfaces with a curvature restriction in de~Sitter space. This is an analogue to the case for Riemannian space forms given by Guan and Shen in \cite{GuanShen}.
\end{abstract}
\maketitle

\section{Introduction}
A rigidity theorem for convex regular surfaces in $\mathbb{R}^3$ was proved in 1927 by Cohn-Vossen \cite{CohnVossen}. It states that, given two convex surfaces with the same first fundamental form, these surfaces must differ only by a rigid motion of $\mathbb{R}^3$.

In 2015 Guan and Shen generalised this rigidity theorem to star-shaped hypersurfaces in higher dimensional space forms. Following a similar argument to Guan and Shen, we prove a rigidity result for hypersurfaces in de~Sitter space.

We let $M$ and $\widetilde{M}$ be two spacelike hypersurfaces embedded in de Sitter space (subject to a curvature restriction) for which there exists a local isometry $f:M\rightarrow \widetilde{M}$. We prove that this $f$ must be the restriction of some global isometry of the ambient space. Garding's inequality for hyperbolic polynomials is applied to the second symmetric function of the Weingarten tensor. This together with some integral formulas over the hypersurfaces is used to show that the second fundamental form is preserved under $f$. Hence the first and second fundamental forms are preserved under $f$, which implies $f$ is the restriction of a global isometry.

\section{The Setting}

In this section we introduce the notation and setting for the rigidity theorem. Our ambient space will be de~Sitter space, denoted $(dS^{n+1}, \bar{g})$. We have coordinates $(\rho, \theta)$, where $\theta \in \mathbb{S}^n$. The metric with respect to these coordinates is given by
\begin{equation*}
\bar{g} = ds^2 = -d\rho^2 + \mathrm{cosh}^2(\rho)\sigma,
\end{equation*}
where $\sigma$ denotes the round metric on $\mathbb{S}^n$ with respect to $\theta$. Let $\bar{K}$ denote the sectional curvature of de~Sitter space, and $\bar{\nabla}$ be its Levi-Civita connection. We define the following vector field on de~Sitter space
\begin{equation*}
V=\bar{\phi}(\rho) \frac{\partial}{\partial \rho}.
\end{equation*}
Let $\bar{\phi}(\rho)=\mathrm{cosh}(\rho)$, then define the polar potential function on de~Sitter space as
\begin{equation*}
\bar{\Phi}(\rho) = \int_0^\rho \bar{\phi}(r) dr.
\end{equation*}
Note this means $\bar{\Phi}(\rho)=\mathrm{sinh}(\rho)=\bar{\phi}'(\rho)$.

We will be considering spacelike hypersurfaces in de~Sitter space. These are hypersurfaces whose normal is timelike everywhere, meaning they have positive definite metric everywhere. Spacelike hypersurfaces can be described as graphs over the sphere in de~Sitter space,
\begin{equation*}
\begin{alignedat}{2}
y : \;&\mathbb{S}^n && \to dS^{n+1} \\
&\zeta && \mapsto (y(\zeta), \zeta),
\end{alignedat}
\end{equation*}
with a restriction on the first derivative of $y$ to ensure the normal is timelike. We therefore have canonical coordinates given by $(\zeta)$. In all that follows, we let $(M,g)$ and $(\tilde{M},\tilde{g})$ be locally isometric spacelike hypersurfaces in de~Sitter space, with the isometry denoted
\begin{equation*}
f:M\to\tilde{M}.
\end{equation*}
Let $K$, $\nabla$, $\nu$ and $W(p)$ denote the scalar curvature, Levi-Civita connection, normal and Weingarten map (or shape operator) of $M$ respectively. Similarly define $\tilde{K}$, $\tilde{\nabla}$, $\tilde{\nu}$ and $\tilde{W}(p)$ on $\tilde{M}$. Let $\phi$ and $\tilde{\phi}$ be the restriction of $\bar{\phi}$ to $M$ and $\tilde{M}$ respectively. Similarly let $\Phi$ and $\tilde{\Phi}$ be the restriction of $\bar{\Phi}$.

\section{Hyperbolic Polynomials}

The proof of our rigidity theorem makes vital use of G\r arding's inequality for hyperbolic polynomials. We begin this section by presenting, in our notation, the definitions of G\r arding's theory \cite{Garding}. Results in this brief exposition of G\r arding's theory are given without proof, with references to proofs given. We then go on to apply G\r arding's theory to the case of our rigidity theorem.

\subsection{G\r arding's Theory of Hyperbolic Polynomials}

\begin{definition}
Let $p(x_1, \dots, x_n)$ be a degree $m$ homogeneous polynomial on $\mathbb{R}^n$. If for some $\mathbf{a} \in \mathbb{R}^n$ the univariate polynomial $t \mapsto p(\mathbf{x}+t\mathbf{a})$ has only real solutions for all $\mathbf{x}\in \mathbb{R}^n$, then we say $p$ is hyperbolic with respect to $\mathbf{a}$ (or often simply $p$ is hyperbolic $\mathbf{a}$).
\label{Hyperbolicity}
\end{definition}

\begin{definition}
We define two cones of vectors in $\mathbb{R}^n$:
\begin{equation*}
C(p,\mathbf{a}) \coloneqq \{\mathbf{b}\in \mathbb{R}^n \; | \; \mathrm{the \; roots \; of \;}t \mapsto p(\mathbf{b}+t\mathbf{a}) \mathrm{\; are \; all \; negative}\},
\end{equation*}
\begin{equation*}
S(p, \mathbf{a}) \coloneqq \mathrm{the\ connected\ component\ of\ } \{ \mathbf{x}\in\mathbb{R}^n | p(\mathbf{x})\neq 0\} \mathrm{\ containing\ } \mathbf{a}.
\end{equation*}
\end{definition}

We then have $p$ is hyperbolic with respect to $\mathbf{b}\in C(p,\mathbf{a})=S(p,\mathbf{a})$ (see for example \cite{Renegar}).

\begin{definition}
We define three subspaces of $\mathbb{R}^n$:
\begin{equation*}
E_{C(p,\mathbf{a})}= \{\mathbf{x}\in \mathbb{R}^n \;|\; C(p,\mathbf{a})+\mathbf{x}=C(p,\mathbf{a}) \},
\end{equation*}
\begin{equation*}
L_p = \{ \mathbf{x}\in \mathbb{R}^n \;|\; p(t\mathbf{x}+\mathbf{y})=p(\mathbf{y}),\ \forall\ \mathbf{y},t \},
\end{equation*}
\begin{equation*}
X_{p,\mathbf{a}}= \{ \mathbf{x} \;|\; \mathrm{roots\ of\ } t\mapsto p(t\mathbf{a}+\mathbf{x}) \mathrm{\ are\ all\ zero} \}.
\end{equation*}
\end{definition}

Then we have $E_{C(p,\mathbf{a})}=L_p=X_{p,\mathbf{a}}$ (see for example \cite{Garding}).

\begin{definition}
Let $p(\mathbf{x})$ be a homogeneous polynomial on $\mathbb{R}^n$ of degree $m$. We define the polarized form of $p$ as the unique function $r(\mathbf{x}^1, \dots, \mathbf{x}^m)$, $\mathbf{x}^k \in \mathbb{R}^n$, that is linear in each argument, invariant under permutations of the $\mathbf{x}^k$ and satisfies $r(\mathbf{x}, \dots, \mathbf{x}) = p(\mathbf{x})$.
\label{PolarisedPoly}
\end{definition}

We now give the main result of G\r arding's theory.
\begin{theorem}[G\r arding Inequality for Hyperbolic Polynomials, {\cite[Theorem~5]{Garding}}]
Given $p$ of degree $m>1$, let $p$ be hyperbolic with respect to $\mathbf{a}$ with $p(\mathbf{a})>0$. Let $r$ be the polarised form of $p$, then for $\mathbf{x}^1, \dots, \mathbf{x}^m \in C(p,\mathbf{a})$, the following inequality holds
\begin{equation}
r(\mathbf{x}^1, \dots, \mathbf{x}^m) \geq p(\mathbf{x}^1)^{\frac{1}{m}} \dots p(\mathbf{x}^m)^{\frac{1}{m}}.
\label{GardingEq}
\end{equation}
with equality if and only if $\mathbf{x}^1, \dots , \mathbf{x}^m$ are pairwise proportional modulo $L_p$.
\label{GardingIneq}
\end{theorem}

\subsection{Hyperbolic Polynomials for Rigidity}
\label{HyperbolicRigidity}

This section adapts a result of Guan and Shen applying G\r arding's theory in \cite{GuanShen} to de Sitter space. We will be considering the Weingarten maps (or otherwise shape operators) for all points of a hypersurface, $M$, in de Sitter space, $dS^{n+1}$. These lie in the space of self-adjoint operators (with respect to the induced metric $g$ on $M$) over $T_pM$ for a point $p\in M$, which we will denote $\mathcal{SA}(T_pM)$ (or just $\mathcal{SA}(V)$ for an inner product space $V$).

We define the second symmetric function of a linear map $W$ as
\begin{equation*}
\sigma_2(W)=\sum_{i<j}\kappa_i\kappa_j,
\end{equation*}
where the $\kappa_i$ are the eigenvalues of $W$.
\begin{lemma}
For any given matrix $W$, we have
\begin{equation*}
\sigma_2(W)=\sum_{i<j} w_{ii}w_{jj}-w_{ij}w_{ji}.
\end{equation*}
\label{Sigma2ww}
\end{lemma}

\begin{proof}
First note that the trace of a linear map does not depend on choice of basis. Therefore we have
\begin{equation*}
tr(W)=\sum_i w_{ii} = \sum_i \kappa_{ii}.
\end{equation*}
Then
\begin{equation}
2\sigma_2(W)=2\sum_{i<j}\kappa_i\kappa_j=\sum_{i,j} \kappa_i\kappa_j - \sum_i \kappa_i\kappa_i= tr(W)^2-tr(W^2).
\label{sigma2wproof1}
\end{equation}
Now consider the two traces separately, for components $w_{ij}$ of $W$ under any given basis. We have
\begin{equation}
tr(W)^2=\sum_{i,j} w_{ii} w_{jj} = 2 \sum_{i<j} w_{ii}w_{jj} + \sum_i w_{ii}w_{ii},
\label{sigma2wproof2}
\end{equation}
and
\begin{equation}
tr(W^2)=\sum_{i,j} w_{ij}w_{ji} = 2 \sum_{i<j} w_{ij}w_{ji} + \sum_i w_{ii}w_{ii}.
\label{sigma2wproof3}
\end{equation}
Plugging (\ref{sigma2wproof2}) and (\ref{sigma2wproof3}) into (\ref{sigma2wproof1}) we obtain the result.
\end{proof}

We will now show that $\sigma_2$ is hyperbolic with respect to the identity matrix in $\mathcal{SA}(T_pM)$. We begin by showing the determinant is hyperbolic with respect to the identity in $\mathcal{SA}(T_pM)$ . Since the determinant is of degree $n$, we must show that the univariate polynomial
\begin{equation}
t\mapsto \mathrm{det}(M+tI)
\label{detHyperbolic}
\end{equation}
has $n$ real zeros for all $M\in\mathcal{SA}(T_pM)$. Since we only care if the roots are real, we may as well replace $t$ with $-t$. Then this simply becomes the eigenvalue equation for $M$, but since $M$ is self adjoint we know that all its eigenvalues are real. Therefore (\ref{detHyperbolic}) has $n$ real roots and the determinant is hyperbolic with respect to the identity.

We now state a Lemma of G\r arding \cite{Garding} without proof, followed by a corollary.

\begin{lemma}
Let $p$ be a degree $m$ polynomial on $\mathbb{R}^n$ that is hyperbolic with respect to $\mathbf{a}$. The polynomial
\begin{equation}
q(x_1, \dots, x_n)=\sum_{k=1}^n a_k \frac{\partial}{\partial x_k} p(x_1, \dots, x_n)
\end{equation}
is hyperbolic with respect to $\mathbf{a}$.
\label{GardingLemma1}
\end{lemma}

\begin{corollary}
Given a degree $m>1$ polynomial $p$, hyperbolic with respect to $\mathbf{a}\in \mathbb{R}^n$, the polynomials $\{p_i\}_{i=0}^m$, defined by
\begin{equation*}
p(s\mathbf{a}+\mathbf{x})=\sum_{i=0}^m s^i p_i(\mathbf{x}),
\end{equation*}
are also hyperbolic with respect to $\mathbf{a}$.
\label{SymmetricCor}
\end{corollary}

\begin{proof}
Consider evaluating this polynomial $p(s\mathbf{a}+\mathbf{x})$ at $s=0$. We then have
\begin{equation*}
p(\mathbf{x})=p_0(\mathbf{x})
\end{equation*}
and since $p$ is hyperbolic $\mathbf{a}$, $p_0$ must also be hyperbolic $\mathbf{a}$. Now consider the polynomial
\begin{equation*}
q(s\mathbf{a}+\mathbf{x})=\frac{d}{ds}p(s\mathbf{a}+\mathbf{x})=\sum_{i=1}^m i s^{i-1} p_i(\mathbf{x}).
\end{equation*}
Evaluating this at $s=0$, we obtain
\begin{equation*}
q(\mathbf{x})=p_1(\mathbf{x}).
\end{equation*}
Note that this $q$ is exactly the same $q$ as defined in Lemma $\ref{GardingLemma1}$, so since $p$ is hyperbolic $\mathbf{a}$, we have $q=p_1$ is hyperbolic $\mathbf{a}$. Repeating this process shows that all the $p_i$ are hyperbolic with respect to $\mathbf{a}$.
\end{proof}

Now note that the characteristic polynomial for a matrix satisfies
\begin{equation*}
\begin{split}
(-1)^n \; \mathrm{det}(M-tI)&= \sum_{i=0}^n (-1)^i \; t^{n-i} \; (\mathrm{sum\;of\;all\;combinations\;of\;}i\mathrm{\;eigenvalues}) \\
&= \sum_{i=0}^n (-1)^i \; t^{n-i} \sigma_i(M),
\end{split}
\end{equation*}
where $\sigma_i(M)$ is the $i^{\mathrm{th}}$ symmetric function of the eigenvalues of $M$. Now since the determinant is hyperbolic with respect to the identity, we have $(-1)^n det (M)$ is hyperbolic with respect to $-I$. Then by Corollary \ref{SymmetricCor}, we have that the functions $(-1)^i \; \sigma_i$ are hyperbolic with respect to $-I$. So finally, the symmetric functions, in particular $\sigma_2$, are hyperbolic with respect to the identity.

The following lemma (see \cite{GuanShen}) can easily be derived from the Gauss equation (see for example \cite{DoCarmo}).
\begin{lemma}
Let $K$ and $\bar{K}$ denote the scalar curvatures of $M^n$ and $dS^{n+1}$ respectively. We have yet another expression for $\sigma_2$:
\begin{equation*}
\sigma_2(W)=\frac{n(n-1)}{2}(K-\bar{K}).
\end{equation*}
As a corollary, since the scalar curvature is invariant under local isometries, we have $\sigma_2(W)=\sigma_2(\tilde{W})$.
\label{sigma2KK}
\end{lemma}

From Lemma \ref{sigma2KK} note that if $K>\bar{K}$ for all points on the hypersurface, then $\sigma_2(W)$ is positive for all points on the hypersurface. We now claim that this implies that the Weingarten maps $W(p)$ for every point $p\in M$ all lie in the same hyperbolicity cone. For any point $M \in \mathcal{SA}(T_pM)$, consider the affine line through $M$ in the direction of $I$. Since $\sigma_2$ is hyperbolic with respect to the identity, this affine line must cross the hypersurface $\sigma_2=0$ exactly twice (with multiplicities). Let $t_1< t_2$ be the roots of $t\mapsto \sigma_2(M+tI)$. Since $\sigma_2$ is of degree two, if $t_1\neq t_2$ then $\sigma_2(M+tI)$ changes sign at $t_1$ and at $t_2$, and if $t_1=t_2$ then $\sigma_2(M+tI)$ has the same sign either side of the root. Importantly, note that in either case we have points $\sigma_2(M+tI)$ has the same sign for $t<t_1$ and $t>t_2$, and values $t_1<t<t_2$ have the opposite sign. Now by definition, the points $M+tI$ with $t>t_2$ are in $C(\sigma_2, I)$ and the points $M+tI$ with $t<t_1$ are in $C(\sigma_2, -I)$. Since $C(\sigma_2, I)=S(\sigma_2,I)$, for any affine line $M+tI$, we have $\mathrm{sgn}(\sigma_2(M+tI)) = \mathrm{sgn}(\sigma_2(I)) > 0$ for all points with $t<t_1$ or $t>t_2$, and for any other $t_1<t<t_2$ we have $\mathrm{sgn}(\sigma_2(M+tI))<0$. This means $\sigma_2$ is hyperbolic with respect to any $W\in \mathcal{SA}(T_pM)$ with $\sigma_2(W)>0$. Since we stipulate that $K>\bar{K}$, by Lemma \ref{sigma2KK} $\sigma_2$ is hyperbolic with respect to $W(p)$ for all points $p\in M$. Furthermore, since $\sigma_2(W(p))$ is strictly greater than zero, it always stays in the same connected component of $\{M\in \mathcal{SA}(T_pM) \; | \; \sigma_2(M)\neq 0)\}$. So either $\{W(p) \; | \; p \in M \} \subset S(\sigma_2, I)$ or $\{W(p) \; | \; p \in M \} \subset S(\sigma_2, -I)$.

We will be considering two isometric hypersurfaces $M$ and $\tilde{M}$ and their respective Weingarten tensors $W$ and $\tilde{W}$. We would like Theorem \ref{GardingIneq} to hold for $W(p)$ and $\tilde{W}(f(p))$ for all points $p$ and $f(p)$ identified under the isometry. In order for this to hold, we need that $W(p)$ and $\tilde{W}(f(p))$ both lie in the same hyperbolicity cone, i.e. either $C(\sigma_2, I)$ or $C(\sigma_2, -I)$. Note that $S(\sigma_2, I) = -S(\sigma_2, -I)$.

We now show that by using a global isometry of de Sitter space, we can make sure all Weingarten maps for both hypersurfaces lie in $C(\sigma_2, I)$. Since the hypersurface is spacelike, it must be compact. Therefore, we can find a point $p$ such that the $\rho$ coordinate of $p$ is greater than any other point on the hypersurface. As the hypersurface is described as the graph of a function $f:\mathbb{S}^n\rightarrow \mathbb{R}$, we can calculate the second fundamental form at this point as
\begin{equation*}
B_{ij}=\Gamma^0_{ij}+\frac{\partial^2 f}{\partial \zeta_i \partial \zeta_j},
\end{equation*}
where $\Gamma^0_{ij}$ are Christoffel symbols on de Sitter space running over the $\{\zeta_i\}_{i=1}^n$ coordinates. We have used the fact that $f$ has a global maximum at $p$, meaning that all first derivatives are zero. To find the Weingarten map, we simply contract with the induced inverse metric, $g^{ij}$, of the hypersurface at this point. Calculating the Christoffel symbols, we find
\begin{equation}
\sum_k g^{ik} \Gamma^0_{kj} = \mathrm{cosh}(\rho)\mathrm{sinh}(\rho) \; \delta^i_j,
\label{Gamma0ij}
\end{equation}
where $\delta^i_j$ is the Kronecker delta symbol.

Now consider reflecting this hypersurface in the ``equator'' of de Sitter space, $\rho=0$. This simply amounts to replacing the $\rho$ coordinate of any point $x$ in the hypersurface by $-\rho$. Plugging $-\rho$ in (\ref{Gamma0ij}) we obtain
\begin{equation*}
\mathrm{cosh}(-\rho)\mathrm{sinh}(-\rho) \; \delta^i_j = -\mathrm{cosh}(\rho)\mathrm{sinh}(\rho) \; \delta^i_j.
\end{equation*}
As the $\rho$ coordinate of a point on the hypersurface is given by the function $f$, this reflection is obtained by replacing $f$ with $-f$. So since
\begin{equation*}
\frac{\partial^2 (-f)}{\partial \zeta_i \partial \zeta_j}=-\frac{\partial^2 (f)}{\partial \zeta_i \partial \zeta_j},
\end{equation*}
we have the Weingarten map of the maximal point $p$, under the reflection, is given by
\begin{equation*}
\hat{W}_{ij}=\sum_k g^{ik} \hat{B}_{kj} = -\mathrm{cosh}(\rho)\mathrm{sinh}(\rho) \; \delta^i_j - \sum_k g^{ik} \frac{\partial^2 (f)}{\partial \zeta_k \partial \zeta_j} = -W_{ij}
\label{W=-W}.
\end{equation*}

This shows that by reflecting about the ``equator'' we can move the Weingarten map of the maximum point from $C(\sigma_2, -I)$ to $C(\sigma_2,I)$. So up to global isometry, $W(p), \tilde{W}(f(p)) \in C(\sigma_2, I)$ for all points $p\in M$. Therefore the G\r arding inequality for hyperbolic polynomials holds for all points on the hypersurfaces.

\subsubsection{The Equality Result}
Since $W(p),\tilde{W}(f(p)) \in C(\sigma_2,I)$ for all points $p\in M$, and further $\sigma_2(W)=\sigma_2(\tilde{W})$, we have the inequality
\begin{equation*}
\sigma_{1,1}(W,\tilde{W})\geq \sigma_2(W),
\end{equation*}
where $\sigma_{1,1}(W,\tilde{W})$ denotes the polarised form of $\sigma_2(W)$. As $\sigma_2$ is degree two its polarised form has the following simple form (see \cite{GuanShen}).

\begin{proposition}
The polarised form of $\sigma_2(W)$ as defined in Definition \ref{PolarisedPoly} can be expressed as
\begin{equation*}
\sigma_{1,1}(W,\tilde{W})= \frac{1}{2} \sum_{i,j} \frac{\partial \sigma_2 (W)}{\partial w_{ij}} \tilde{w}_{ij}.
\end{equation*}
\label{Sigma11}
\end{proposition}

Now by Theorem \ref{GardingIneq} we have if
\begin{equation*}
\sigma_{1,1}(W,\tilde{W})= \sigma_2(W)
\end{equation*}
then $W$ and $\tilde{W}$ must be proportional modulo $L_{\sigma_2}$. As such, we would like to know what $L_{\sigma_2}$ actually is, and it turn out that in fact it is trivial.
\begin{proposition}
$L_{\sigma_2}=\mathbf{0}$
\end{proposition}
\begin{proof}
As seen in G\r arding's paper \cite{Garding}, $L_{\sigma_2}$ is the set of $W$ such that the roots of the polynomial
\begin{equation*}
t\mapsto \sigma_2(tI+W)
\end{equation*}
are all zero. Expanding this out, we have
\begin{equation*}
\begin{split}
\sigma_2(tI+W)&=\sum_{i<j} (t+w_{ii})(t+w_{jj})-w_{ij}w_{ji} \\
&=\sum_{i<j} t^2 +(w_{ii}+w_{jj})t+ w_{ii}w_{jj}-w_{ij}w_{ji}.
\end{split}
\end{equation*}
For all the roots of this to be zero, we need
\begin{equation*}
\sum_{i<j} (w_{ii}+w_{jj}) = (n-1) \sum_i w_{ii} =0
\end{equation*}
and
\begin{equation*}
\sum_{i<j} w_{ii}w_{jj} - w_{ij}w_{ji}=0.
\end{equation*}
Put simply, this means we need $\sigma_1(W)=\sigma_2(W)=0$. Now using the definition of $\sigma_2(W)$ in terms of the eigenvalues $\kappa_i$ of $W$, we have
\begin{equation*}
\begin{split}
\sigma_2(W)&=\sum_{i<j} \kappa_i \kappa_j \\
&=\frac{1}{2} \sum_{i\neq j} \kappa_i \kappa_j \\
&=\frac{1}{2} \Big( \sum_{i,j} \kappa_i \kappa_j - \sum_i \kappa_i \kappa_i \Big) \\
&=\frac{1}{2} \Big( \sum_{i} \kappa_i \Big( \sum_j \kappa_j \Big) - \sum_i \kappa_i \kappa_i \Big).
\end{split}
\end{equation*}
Now since $\sigma_1(W)=\sum_i \kappa_i =0$, we have
\begin{equation*}
\sigma_2(W)=-\frac{1}{2} \sum_i \kappa_i \kappa_i.
\end{equation*}
Then clearly the only way for $\sigma_2(W)=0$ is for all the $\kappa_i=0$. Hence $L_{\sigma_2}=\mathbf{0}$.
\end{proof}

Since $\sigma_2(W)=\sigma_2(\tilde{W})$, if $W$ and $\tilde{W}$ are proportional then we must have $W=\tilde{W}$. Hence we have
\begin{equation*}
\sigma_{1,1}(W,\tilde{W})=\sigma_2(W)
\end{equation*}
if and only if $W=\tilde{W}$.

\section{Some Important Integrals}

\subsection{A Few Lemmas}
The following lemmas (see \cite{GuanShen}) give the integrands for the integral equations which follow, these will be vital in the main rigidity proof.

\begin{lemma}
\begin{equation*}
\langle \bar{\nabla}_{e_i} V, e_j \rangle_{dS} + \langle \bar{\nabla}_{e_j} V, e_i \rangle_{dS} = 2 \phi' \bar{g}_{ij}
\end{equation*}
\label{DerivV}
\end{lemma}

\begin{proof}
We will prove this in two stages. First, we show that $(\mathcal{L}_V \bar{g})(e_i,e_j) = \langle \bar{\nabla}_{e_i} V, e_j \rangle_{dS} + \langle \bar{\nabla}_{e_j} V, e_i \rangle_{dS}$, and then that $(\mathcal{L}_V \bar{g})(e_i,e_j) = 2\phi'(\rho)\bar{g}_{ij}$.

So first, since the Lie derivative obeys Leibniz's rule, we have
\begin{equation*}
\mathcal{L}_V (\bar{g}(e_i, e_j)) = (\mathcal{L}_V \bar{g})(e_i,e_j) + \bar{g}(\mathcal{L}_V e_i, e_j) + \bar{g}(e_i, \mathcal{L}_V e_j).
\end{equation*}
Using that the derivative of a function is the directional derivative and that $\mathcal{L}_X Y = [X,Y]$, we have
\begin{equation*}
(\mathcal{L}_V \bar{g})(e_i,e_j) = \langle \mathrm{grad}(\langle e_i, e_j \rangle_{dS}), V \rangle_{dS} - \langle [V, e_i], e_j \rangle_{dS} - \langle e_i, [V, e_j] \rangle_{dS},
\end{equation*}
and taking $V$ into the inner product using compatibiliity with the metric, we obtain
\begin{equation*}
(\mathcal{L}_V \bar{g})(e_i,e_j) = \langle \bar{\nabla}_V e_i, e_j \rangle_{dS} + \langle e_i, \bar{\nabla}_V e_j \rangle_{dS} - \langle [V, e_i], e_j \rangle_{dS} - \langle e_i, [V, e_j] \rangle_{dS}.
\end{equation*}
Now using the symmetry of the connection $\nabla_X Y - \nabla_Y X = [X,Y]$, we have
\begin{equation*}
(\mathcal{L}_V \bar{g})(e_i,e_j) = \langle \bar{\nabla}_{e_i} V, e_j \rangle_{dS} + \langle \bar{\nabla}_{e_j} V, e_i \rangle_{dS}.
\label{VLemma1.5}
\end{equation*}

The second statement $(\mathcal{L}_V \bar{g})(e_i,e_j) = 2\phi'(\rho)\bar{g}_{ij}$, is proved simply by calculation using the Cartan formula for the Lie derivative of a differential form. Hence the lemma is proved.
\end{proof}

\begin{lemma}
Let $\nu$ be the normal to $M$, then we have
\begin{equation*}
\mathrm{Hess}^{M,g}(\Phi)(e_i,e_j) = \phi' g_{ij} + h_{ij} \langle V, \nu \rangle_{dS}.
\end{equation*}
\label{preIntegral}
\end{lemma}

\begin{proof}
We start with the definitions of the Hessian for $\bar{\Phi}$ in de Sitter space and for $\Phi$ in $M$,
\begin{subequations}
\begin{equation}
\mathrm{Hess}^{dS,\bar{g}}(\bar{\Phi})(e_i,e_j) = \langle \mathrm{grad}(\langle \mathrm{grad}(\bar{\Phi}), e_j \rangle_{dS} ), e_i \rangle_{dS} - \langle \mathrm{grad}(\bar{\Phi}), \bar{\nabla}_{e_i} e_j \rangle_{dS},
\label{HessdS}
\end{equation}
\begin{equation}
\mathrm{Hess}^{M,g}(\Phi)(e_i,e_j) = \langle \mathrm{grad}(\langle \mathrm{grad}(\Phi), e_j \rangle_M ), e_i \rangle_M - \langle \mathrm{grad}(\Phi), \nabla_{e_i} e_j \rangle_M.
\label{HessM}
\end{equation}
\end{subequations}
Subtracting (\ref{HessdS}) from (\ref{HessM}), we obtain
\begin{multline*}
\mathrm{Hess}^{M,g}(\Phi)(e_i,e_j) - \mathrm{Hess}^{dS,\bar{g}}(\bar{\Phi})(e_i,e_j) \\
= \langle \mathrm{grad}(\langle \mathrm{grad}(\Phi), e_j \rangle_M ), e_i \rangle_M - \langle \mathrm{grad}(\langle \mathrm{grad}(\bar{\Phi}), e_j \rangle_{dS} ), e_i \rangle_{dS} \\
- \langle \mathrm{grad}(\Phi), \nabla_{e_i} e_j \rangle_M + \langle \mathrm{grad}(\bar{\Phi}), \bar{\nabla}_{e_i} e_j \rangle_{dS}.
\end{multline*}
Now clearly $\langle \mathrm{grad}(\langle \mathrm{grad}(\Phi), e_j \rangle_M ), e_i \rangle_M = \langle \mathrm{grad}(\langle \mathrm{grad}(\bar{\Phi}), e_j \rangle_{dS} ), e_i \rangle_{dS}$, and also $\langle \mathrm{grad}(\Phi), \nabla_{e_i} e_j \rangle_M = \langle \mathrm{grad}(\bar{\Phi}), \nabla_{e_i} e_j \rangle_{dS}$, so we have
\begin{equation*}
\mathrm{Hess}^{M,g}(\Phi)(e_i,e_j) = \mathrm{Hess}^{dS,\bar{g}}(\bar{\Phi})(e_i,e_j) + \langle \mathrm{grad}(\bar{\Phi}), \bar{\nabla}_{e_i} e_j - \nabla_{e_i} e_j \rangle_{dS}.
\end{equation*}
Note that on de Sitter space, $\mathrm{grad}{\bar{\Phi}} = V$, so then from the definition of the Hessian, we obtain
\begin{equation}
\mathrm{Hess}^{M,g}(\Phi)(e_i,e_j) = \langle \bar{\nabla}_{e_i} V, e_j \rangle_{dS} + \langle V, \bar{\nabla}_{e_i} e_j - \nabla_{e_i} e_j \rangle_{dS}.
\label{HessM2}
\end{equation}
Note $\langle V, \bar{\nabla}_{e_i} e_j - \nabla_{e_i} e_j \rangle_{dS} = h_{ij} \langle V, \nu \rangle_{dS}$. Now take this equation and consider swapping $e_i$ and $e_j$ in each term
\begin{equation*}
\mathrm{Hess}^{M,g}(\Phi)(e_j,e_i) = \langle \bar{\nabla}_{e_j} V, e_i \rangle_{dS} + h_{ji} \langle V, \nu \rangle_{dS}.
\end{equation*}
Adding this to (\ref{HessM2}), and using that both the Hessian and second fundamental form are symmetric in $i$ and $j$, we obtain
\begin{equation*}
2 \mathrm{Hess}^{M,g}(\Phi)(e_i,e_j) = \langle \bar{\nabla}_{e_i} V, e_j \rangle_{dS} + \langle \bar{\nabla}_{e_j} V, e_i \rangle_{dS} + 2 h_{ij} \langle V, \nu \rangle_{dS}.
\end{equation*}
Finally by Lemma \ref{DerivV} we obtain the result
\begin{equation*}
\mathrm{Hess}^{M,g}(\Phi)(e_i,e_j) = \phi' g_{ij} + h_{ij} \langle V, \nu \rangle_{dS}.
\end{equation*}
\end{proof}

\subsection{The Integral Equations}
By identifying points of $M$ and $\tilde{M}$ under the isometry $f$, we can treat $\tilde{\phi}$ and $\tilde{\Phi}$ as functions on $M$, i.e. for $x\in M$, $\tilde{\phi}_M: x \mapsto \tilde{\phi}(f(x))$. We will abuse notation and use $\tilde{\phi}$ to denote $\tilde{\phi}_M$ in integrals over $M$. The following theorem is adapted from a lemma in \cite{GuanShen}.

\begin{theorem}
Given an orthonormal frame $\{e_i\}$ on $M$, which can be identified as an orthonormal frame on $\tilde{M}$ under the isometry, the following integral equations hold:
\begin{subequations}
\begin{equation}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \mathrm{Hess}^{M,g}(\Phi)(e_i, e_j) = \int_M (n-1) \tilde{\phi}' \phi' \sigma_1(W) - 2 \tilde{\phi}' \sigma_2(W) \langle V, \nu \rangle,
\label{Integral0a}
\end{equation}
\begin{equation}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(\tilde{W}) \; \tilde{\phi}' \; \mathrm{Hess}^{M,g}(\Phi)(e_i, e_j) = \int_M (n-1) \tilde{\phi}' \phi' \sigma_1(\tilde{W}) - 2 \tilde{\phi}' \sigma_{1,1}(W,\tilde{W}) \langle V, \nu \rangle,
\label{Integral0b}
\end{equation}
\begin{equation}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \phi' \; \mathrm{Hess}^{M,g}(\tilde{\Phi})(e_i, e_j) = \int_M (n-1) \phi' \tilde{\phi}' \sigma_1(W) - 2 \phi' \sigma_{1,1}(\tilde{W}\!,W) \langle \tilde{V}\!, \tilde{\nu} \rangle,
\label{Integral0c}
\end{equation}
\begin{equation}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(\tilde{W}) \; \phi' \; \mathrm{Hess}^{M,g}(\tilde{\Phi})(e_i, e_j) = \int_M (n-1) \phi' \tilde{\phi}' \sigma_1(\tilde{W}) - 2 \phi' \sigma_2(\tilde{W}) \langle \tilde{V}, \tilde{\nu} \rangle.
\label{Integral0d}
\end{equation}
\end{subequations}
\label{IntegralEqn}
\end{theorem}

\begin{proof}
Starting with Lemma \ref{preIntegral} applied to $M$ and $\tilde{M}$, we have
\begin{subequations}
\begin{equation}
\mathrm{Hess}^{M,g}(\Phi)(e_i,e_j) = \phi' g_{ij} + h_{ij} \langle V, \nu \rangle_{dS}
\label{Integral1a}
\end{equation}
\begin{equation}
\mathrm{Hess}^{\tilde{M},\tilde{g}}(\tilde{\Phi})(e_i,e_j) = \tilde{\phi}' \tilde{g}_{ij} + \tilde{h}_{ij} \langle \tilde{V}, \tilde{\nu} \rangle_{dS}
\label{Integral1b}
\end{equation}
\end{subequations}
It is not hard to convince oneself that $\mathrm{Hess}^{\tilde{M},\tilde{g}}(\tilde{\Phi})(e_i,e_j)=\mathrm{Hess}^{M,g}(\tilde{\Phi})(e_i,e_j)$. Now we multiply (\ref{Integral1a}) by $\tilde{\phi}'$ and (\ref{Integral1b}) by $\phi'$. Now take the four combinations of multiplying these two equations by either $\frac{\partial\sigma_2}{\partial w_{ij}}(W)$ or $\frac{\partial\sigma_2}{\partial w_{ij}}(\tilde{W})$, sum over the $i$ and $j$ and integrate over $M$ to obtain
\begin{subequations}
\begin{equation}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \mathrm{Hess}^{M,g}(\Phi)(e_i,e_j) = \int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \Big( \tilde{\phi}' \phi' g_{ij} + \tilde{\phi}' h_{ij} \langle V, \nu \rangle_{dS}\Big),
\label{Integral2a}
\end{equation}
\begin{equation}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(\tilde{W}) \; \tilde{\phi}' \; \mathrm{Hess}^{M,g}(\Phi)(e_i,e_j) = \int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(\tilde{W}) \Big(\tilde{\phi}' \phi' g_{ij} + \tilde{\phi}' h_{ij} \langle V, \nu \rangle_{dS}\Big),
\label{Integral2b}
\end{equation}
\begin{equation}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \phi' \; \mathrm{Hess}^{M,g}(\tilde{\Phi})(e_i,e_j) = \int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \Big(\phi' \tilde{\phi}' \tilde{g}_{ij} + \phi' \tilde{h}_{ij} \langle \tilde{V}, \tilde{\nu} \rangle_{dS}\Big),
\label{Integral2c}
\end{equation}
\begin{equation}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(\tilde{W}) \; \phi' \; \mathrm{Hess}^{M,g}(\tilde{\Phi})(e_i,e_j) = \int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(\tilde{W}) \Big(\phi' \tilde{\phi}' \tilde{g}_{ij} + \phi' \tilde{h}_{ij} \langle \tilde{V}, \tilde{\nu} \rangle_{dS}\Big).
\label{Integral2d}
\end{equation}
\end{subequations}
We now focus on the right hand side of (\ref{Integral2a}). Since $\{e_i\}$ is an orthonormal basis, we have $g_{ij}=\delta_{ij}$, so for the first term we have
\begin{equation}
\sum_{i=j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \tilde{\phi}' \phi'.
\label{Integral2.5}
\end{equation}
The expression for $\frac{\partial \sigma_2}{\partial w_{ij}}(W)$ can easily shown by calculation to be
\begin{equation}
\frac{\partial \sigma_2 (W)}{\partial w_{ij}}=
\begin{cases}
\sum\limits_{k\neq i} w_{kk} & \mathrm{for} \; i=j \\
-w_{ji} & \mathrm{for} \; i\neq j.
\end{cases}
\label{Sigma2Deriv}
\end{equation}
Subsituting this into (\ref{Integral2.5}) gives
\begin{equation*}
\sum_{i} \sum\limits_{k\neq i} w_{kk} \tilde{\phi}' \phi',
\end{equation*}
which is clearly equal to
\begin{equation}
(n-1) \tilde{\phi}' \phi' \sigma_1(W).
\label{Integral3}
\end{equation}
For the second term of the right hand side of (\ref{Integral2a}), note that since $g_{ij}=\delta_{ij}$, we have $h_{ij}=w_{ij}$. Therefore we have
\begin{equation*}
\tilde{\phi}' \langle V, \nu \rangle \sum_{i,j} \frac{\partial \sigma_2(W)}{\partial w_{ij}} w_{ij}.
\end{equation*}
From the expression for $\sigma_{1,1}(W, \tilde{W})$ in Proposition \ref{Sigma11}, this is equal to
\begin{equation*}
2\tilde{\phi}' \langle V, \nu \rangle \sigma_{1,1}(W,W),
\end{equation*}
but as the polarised form of $\sigma_2$, we have $\sigma_{1,1}(W,W)=\sigma_2(W)$. This together with (\ref{Integral3}) gives the right hand side of (\ref{Integral0a}). The remaining three equations are proved along the same lines, hence the theorem is proved.
\end{proof}

\subsection{Symmetry in Tilde}

We have the following proposition from \cite{GuanShen}.

\begin{proposition}
Assume that $W$ is a Codazzi tensor and $\{e_1, \dots, e_n\}$ is an orthonormal frame on $M$, then we have the following identity
\begin{equation*}
\sum_i \Big\langle \mathrm{grad}\Big(\frac{\partial \sigma_2}{w_{ij}}(W)\Big), e_i \Big\rangle=0.
\end{equation*}
\label{gradSigma2}
\end{proposition}

We can now prove our symmetry theorem.

\begin{theorem}
The integral
\begin{equation*}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \mathrm{Hess}^{M,g}(\Phi)(e_i, e_j)
\label{Sym1}
\end{equation*}
is invariant under switching $\tilde{\phi}'$ for $\phi'$ and $\Phi$ for $\tilde{\Phi}$ (henceforth referred to as symmetric in tilde), so that
\begin{equation*}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \phi' \; \mathrm{Hess}^{M,g}(\tilde{\Phi})(e_i, e_j) - \int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \mathrm{Hess}^{M,g}(\Phi)(e_i, e_j) = 0.
\end{equation*}
\label{IntegralSym}
\end{theorem}

\begin{proof}
From the definition of the Hessian, we will start from the integral
\begin{equation*}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \Big( \langle \mathrm{grad}(\langle \mathrm{grad}(\Phi), e_j \rangle ), e_i \rangle - \langle \mathrm{grad}(\Phi), \nabla_{e_i} e_j \rangle \Big).
\end{equation*}
Now taking the $\frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}'$ inside the inner products gives
\begin{equation*}
\int_M \sum_{i,j} \Big\langle \mathrm{grad}(\langle \mathrm{grad}(\Phi), e_j \rangle ), \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; e_i \Big\rangle - \Big\langle \mathrm{grad}(\Phi), \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \nabla_{e_i} e_j \Big\rangle,
\end{equation*}
and integration by parts on the first term gives
\begin{equation*}
\int_M \sum_{i,j} -\langle \mathrm{grad}(\Phi), e_j \rangle \; \mathrm{div}\Big(\frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; e_i \Big) - \Big\langle \mathrm{grad}(\Phi), \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \nabla_{e_i} e_j \Big\rangle.
\end{equation*}
Now by the product rule we have
\begin{multline*}
\int_M \sum_{i,j} -\langle \mathrm{grad}(\Phi), e_j \rangle \; \Big(\frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \mathrm{div}(e_i) - \Big\langle \mathrm{grad}\Big( \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}'\Big), e_i \Big\rangle \Big) \\ - \Big\langle \mathrm{grad}(\Phi), \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \nabla_{e_i} e_j \Big\rangle.
\end{multline*}
Rearranging gives
\begin{multline}
\int_M \sum_{i,j} \langle \mathrm{grad}(\Phi), e_j \rangle \; \Big\langle \mathrm{grad}\Big( \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}'\Big), e_i \Big\rangle \\ - \Big\langle \mathrm{grad}(\Phi), \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \Big( \mathrm{div}(e_i) e_j + \nabla_{e_i} e_j \Big) \Big\rangle.
\label{Sym2}
\end{multline}

We will now examine the two terms in this integral separately. Starting with the first term, since $\mathrm{grad}$ satisfies Leibniz's rule, we have
\begin{equation*}
\int_M \sum_{i,j} -\langle \mathrm{grad}(\Phi), e_j \rangle \; \Big\langle \mathrm{grad}\Big( \frac{\partial \sigma_2}{\partial w_{ij}}(W) \Big) \tilde{\phi}' + \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \mathrm{grad}( \tilde{\phi}'), e_i \Big\rangle.
\end{equation*}
The linearity of the inner product and splitting the integral gives
\begin{multline*}
\int_M \sum_{i,j} -\langle \mathrm{grad}(\Phi), e_j \rangle \; \Big\langle \mathrm{grad}\Big( \frac{\partial \sigma_2}{\partial w_{ij}}(W) \Big) \tilde{\phi}', e_i \Big\rangle \\ + \int_M \sum_{i,j} -\langle \mathrm{grad}(\Phi), e_j \rangle \; \Big\langle \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \mathrm{grad}( \tilde{\phi}'), e_i \Big\rangle.
\end{multline*}
Factoring out $-\langle \mathrm{grad}(\Phi), e_j \rangle$ and $\tilde{\phi}'$ from the sum over $i$ in the first integral gives
\begin{multline}
\int_M \sum_{j} - \Big( \tilde{\phi}' \; \langle \mathrm{grad}(\Phi), e_j \rangle \; \sum_i \Big\langle \mathrm{grad}\Big( \frac{\partial \sigma_2}{\partial w_{ij}}(W) \Big), e_i \Big\rangle \Big) \\ + \int_M \sum_{i,j} -\langle \mathrm{grad}(\Phi), e_j \rangle \; \Big\langle \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \mathrm{grad}( \tilde{\phi}'), e_i \Big\rangle,
\label{Sym3}
\end{multline}
but then by Proposition \ref{gradSigma2} the first integral is zero. After factoring out $\frac{\partial \sigma_2}{\partial w_{ij}}(W)$ in the second integral, since $\tilde{\phi}'=\tilde{\Phi}$ we see that (\ref{Sym3}) is equal to
\begin{equation}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \langle \mathrm{grad}(\Phi), e_j \rangle \langle \mathrm{grad}(\tilde{\Phi}), e_i \rangle.
\label{Sym4}
\end{equation}

Now we will look at the second term of (\ref{Sym2}). First let
\begin{equation*}
X=\sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \Big( \mathrm{div}(e_i) e_j + \nabla_{e_i} e_j \Big),
\end{equation*}
then the second term of (\ref{Sym2}) is equal to
\begin{equation*}
-\int_M \langle \mathrm{grad}(\Phi), \tilde{\phi}' X \rangle.
\end{equation*}
Trivially this is
\begin{equation*}
-\int_M \frac{1}{2}\langle \mathrm{grad}(\Phi), \tilde{\phi}' X \rangle + \frac{1}{2}\langle \mathrm{grad}(\Phi), \tilde{\phi}' X \rangle.
\end{equation*}
Then integration by parts on the second term gives
\begin{equation*}
-\int_M \frac{1}{2}\langle \mathrm{grad}(\Phi), \tilde{\phi}' X \rangle - \frac{1}{2}\Phi \mathrm{div}(\tilde{\phi}' X).
\end{equation*}
By the product rule, this is
\begin{equation*}
-\int_M \frac{1}{2}\langle \mathrm{grad}(\Phi), \tilde{\phi}' X \rangle - \frac{1}{2}\Phi \Big( \tilde{\phi}' \mathrm{div}(X) - \langle \mathrm{grad}(\tilde{\phi}'), X \rangle \Big).
\end{equation*}
Recall we have $\tilde{\phi}'=\tilde{\Phi}$, so after expanding we have
\begin{equation}
-\int_M \frac{1}{2}\langle \mathrm{grad}(\Phi), \tilde{\Phi} X \rangle + \frac{1}{2} \langle \mathrm{grad}(\tilde{\Phi}), \Phi X \rangle- \frac{1}{2}\Phi \tilde{\Phi} \mathrm{div}(X).
\label{Sym5}
\end{equation}

Plugging (\ref{Sym4}) and (\ref{Sym5}) into (\ref{Sym2}), we obtain
\begin{multline}
\int_M \sum_{i,j} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \tilde{\phi}' \; \mathrm{Hess}^{M,g}(\Phi)(e_i, e_j) \\
\begin{aligned}
= \int_M \sum_{i,j} \Big[ & \frac{\partial \sigma_2}{\partial w_{ij}}(W) \langle \mathrm{grad}(\Phi), e_j \rangle \langle \mathrm{grad}(\tilde{\Phi}), e_i \rangle \\
&- \frac{1}{2} \Big\langle \mathrm{grad}(\Phi), \tilde{\Phi} \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \Big( \mathrm{div}(e_i) e_j + \nabla_{e_i} e_j \Big) \Big\rangle \\
&- \frac{1}{2} \Big\langle \mathrm{grad}(\tilde{\Phi}), \Phi \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \Big( \mathrm{div}(e_i) e_j + \nabla_{e_i} e_j \Big) \Big\rangle \\
&+ \frac{1}{2}\Phi \tilde{\Phi} \mathrm{div}\Big( \frac{\partial \sigma_2}{\partial w_{ij}}(W) \; \Big( \mathrm{div}(e_i) e_j + \nabla_{e_i} e_j \Big) \Big) \Big]
\label{Sym6}
\end{aligned}
\end{multline}

Note that since the $\{e_i\}$ are orthonormal, $W=w_{ij}$ is symmetric in $i$ and $j$. Now from the explicit formula for $\frac{\partial \sigma_2}{\partial w_{ij}}(W)$ in (\ref{Sigma2Deriv}), we have that the first term on the right hand side of (\ref{Sym6}) is symmetric in tilde. Clearly the middle two terms together are symmetric in tilde and the last term is as well. Therefore the whole integral is symmetric in tilde.
\end{proof}

\section{The Rigidity Theorem}

Denote by $dS^{n+1, +}$ the region of de Sitter space for which $\rho$ is positive. We can now prove the main rigidity result following the idea for the Riemannian case in \cite{GuanShen}.

\begin{theorem}
Let $M$ and $\tilde{M}$ be two spacelike hypersurfaces in $dS^{n+1,+}$ with $K>\bar{K}$ such that there exists a local isometry $f:M\to \tilde{M}$. Then $f$ is the restriction of some global isometry $F$ of de Sitter space.
\end{theorem}

\begin{proof}
We start with the integral equations in Theorem \ref{IntegralEqn}, subtracting (\ref{Integral0a}) from (\ref{Integral0c}) and (\ref{Integral0b}) from (\ref{Integral0d}), by Theorem \ref{IntegralSym} the right hand sides will cancel and we obtain
\begin{subequations}
\begin{equation}
\int_M \tilde{\phi}' \sigma_2(W) \langle V, \nu \rangle = \int_M \phi' \sigma_{1,1}(\tilde{W},W) \langle \tilde{V}, \tilde{\nu} \rangle,
\label{Rigidity1a}
\end{equation}
\begin{equation}
\int_M \phi' \sigma_2(\tilde{W}) \langle \tilde{V}, \tilde{\nu} \rangle = \int_M \tilde{\phi}' \sigma_{1,1}(W,\tilde{W}) \langle V, \nu \rangle.
\label{Rigidity1b}
\end{equation}
\end{subequations}
Now note that $\sigma_2(W)=\sigma_2(\tilde{W})$ and $\sigma_{1,1}(W,\tilde{W})=\sigma_{1,1}(\tilde{W},W)$, so adding (\ref{Rigidity1a}) and (\ref{Rigidity1b}) and rearranging we obtain
\begin{equation*}
\int_M \Big( \tilde{\phi}' \langle V, \nu \rangle + \phi' \langle \tilde{V}, \tilde{\nu} \rangle \Big) \Big( \sigma_2(W) - \sigma_{1,1}(W,\tilde{W}) \Big) = 0.
\label{Rigidity2}
\end{equation*}
Since $M, \tilde{M}\subset dS^{n+1,+}$, $\phi'$ and $\tilde{\phi}'$ are positive on all of $M$. Furthermore since $M$ and $\tilde{M}$ are spacelike, $\langle V, \nu \rangle$ and $\langle \tilde{V}, \tilde{\nu} \rangle$ are strictly less than zero. So the quantity $\tilde{\phi}' \langle V, \nu \rangle + \phi' \langle \tilde{V}, \tilde{\nu} \rangle$ is strictly less than zero. Now from the inequality in Theorem \ref{GardingIneq} we have that $\sigma_2(W) - \sigma_{1,1}(W,\tilde{W})$ is less than or equal to zero. Therefore the only way the integral in (\ref{Rigidity2}) can be zero is if $\sigma_2(W) - \sigma_{1,1}(W,\tilde{W})$ is identically zero on all of $M$.

Now from the equality result of G\r arding's inequality for hyperbolic polynomials in \S\ref{HyperbolicRigidity}, since $\sigma_2(W) - \sigma_{1,1}(W,\tilde{W})=0$ is zero on all of $M$, we have that $W=\tilde{W}$. Since the first and second fundamental forms are preserved under the map $f$, it must be the restriction of some global isometry $F$, hence the theorem is proved.
\end{proof}

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TITLE: Prove that if W is a subspace of a vector space V
QUESTION [2 upvotes]: Prove that if $W$ is a subspace of a vector space $V$ and $w_1, w_2, ..., w_n$ are in $W$, then $a_1w_1 + a_2w_2 + ... + a_nw_n \in W$ for any scalars $a_1, a_2, ..., a_n$.
My solution is we have $a_iw_i \in W$ for all $i$. And we can get the conclusion that $a_1w_1, a_1w_1 + a_2w_2, a_1w_1 + a_2w_2 + a_3w_3$ are in $W$ inductively.
Any ideas on how to improve this because I feel it is not enough.
Thank you in advance.

REPLY [0 votes]: Since $W$ is a subspace of vector space $V$, then $W$ is closed under addition and scalar multiplication by the definition of subspace.
Now we can inductively prove the statement.
Base:
Let $w_1 \in W$ and let scalar $a_1 \in \digamma$ where $\digamma$ is the field over which $V$ is defined.
We have $a_1w_1 \in W$ because subspace $W$ is closed under scalar multiplication.
Inductive Step:
Let $w_1, w_2, ... , w_n \in W$ and $a_1, a_2, ... , a_n \in \digamma$
Assume $a_1w_1 + a_2w_2 + ... + a_nw_n \in W$ is true.
Let $w_{n+1} \in W$ and let scalar $a_{n+1} \in \digamma$.
We can show that $a_1w_1 + a_2w_2 + ... + a_nw_n + a_{n+1}w_{n+1} \in W$ by closure of addition of subspace $W$.
Therefore, we have established the inductive steps to prove the statement.


TITLE: Direction of Current Flow in Complex Circuit
QUESTION [1 upvotes]: Physics Hypertextbook writes that "The current through the 1 Ω resistor most certainly runs from right to left."
Why is that true?
My approach:
I arbitrarily assigned a counterclockwise direction to the current running through the 1 $\Omega$ resistor. I then applied the junction rule at the node connecting the 2 $\Omega$ Resistor the 20V emf and the wire leading to the 1 $\Omega$ Resistor:
I$_{1\Omega}$ + I$_{2\Omega}$ = I$_{emf}$
where
I$_{2\Omega}$ = 4. If the labeled 3A current loops down and to the right.
Also,
I$_{emf}$ = 3
It can be shown that the current running through the 2 $\Omega$ resistor is 4A:
By the loop rule (for the smaller right-hand circuit): 20V - 2I - 4*3 = 0; I = 4.
If the I$_{emf}$ < 0 (which, by my reasoning, it is), the current actually runs clockwise.

REPLY [2 votes]: The earlier part of the solution (which can be found here) finds that $I_2$ and $I_3$ both have current from right to left.  So the left side must be at lower potential.  The passive path on the top loop must likewise carry current from right to left.
Yes,  $I_2$ is $4A$, but since you haven't calculated the current through the $20V$ source (it isn't $3A$!), you can't use that node to find $I_1$ yet.


TITLE: If $\lim_{n\rightarrow\infty} f(n+1) - f(n) = L$, prove that $\lim_{n\rightarrow\infty} f(n)/n = L$
QUESTION [5 upvotes]: Here's the problem in full. I've stared at it for hours and can't get anywhere, so a hint would be nice.
Suppose that $f:\mathbb N\rightarrow\mathbb R$. If $$\lim_{n\rightarrow\infty}f(n+1)-f(n) = L$$ prove that $\lim_{n\rightarrow\infty}f(n)/n$ exists and equals $L$.
This problem is marked with the word "Cauchy", so I'm guessing Cauchy sequences or something related to them will be necessary, but I can't figure out what.
Edit: It appears as if the Stolz-Cesaro Theorem makes this quite easy. Unfortunately though, we have not discussed this, so I won't be able to use this.

REPLY [1 votes]: Note that $f(n+1)-f(n) = \dfrac{f(n+1)-f(n)}{(n+1)-n}$. Then, by Stolz-Cesàro applied to the sequence $\dfrac{f(n)}{n}$ we have:
$$\lim_{n \to \infty} \dfrac{f(n+1) - f(n)}{n+1 - n} = L \Rightarrow \lim_{n \to \infty} \dfrac{f(n)}{n}= L$$


TITLE: Sobolev embedding for the $L^q$ norm.
QUESTION [7 upvotes]: Suppose $f \in H^1(\mathbb R^2)$, where $H^1$ is the Sobolev space, then how to use this information to bound $\Vert f \Vert_{L^q}$, where $q>2$? It seems like Sobolev embedding, but it's not.

REPLY [3 votes]: This is immediate from the Gagliardo-Nirenberg inequality, or from the Sobolev inequality with fractional exponents, which says in particular that
$$
\|f\|_{L^q} \leq C\|f\|_{H^s} \leq C'\|f\|_{H^1} , \qquad f\in C^1_0(\mathbb{R}^2),
$$
for $2\leq q<\infty$ and $s=1-\frac2q$.


TITLE: Fermat's Little Theorem and Prime Moduli
QUESTION [2 upvotes]: I am given two distinct primes $p$ and $q$, where $$m = p*q$$ Also,
$$ \begin{cases} r\equiv 1\mod p-1\\ r\equiv 1\mod q-1 \end{cases} $$
I have to show that given an integer a, show that $$a^r \equiv a \mod (m)$$
I'm not sure how to get started. I know I can tie this in to Fermat's Little Theorem and I've found something here CRT + Fermat's Little Theorem that was somewhat related to my question but I had a hard time seeing where to go. Just need a little hint to get me in the right direction! Thanks.

REPLY [0 votes]: Since $\begin{cases} r\equiv 1\mod p-1\\ r\equiv 1\mod q-1 \end{cases}$ you can write $r=(p-1)k+1$ and $r=(q-1)l+1$ for some $k,l\in\mathbb{Z}$. Then $a^r=a^{(p-1)k+1}=a\cdot (a^{p-1})^k\equiv a (mod \;p)$ where the last conclusion follows from Fermat's Little Theorem. Similarly you have $a^r\equiv a(mod\; q)$. If you write $a^r=cp+a=dq+a$ you get that $cp=dq$ and since the primes are distinct it must be that $p \; |\;d$ which gives you $a^r=dq+a=pqd_1+a$ which means $a^r\equiv a (mod\; pq)$


TITLE: Solving Lagrange equation.
QUESTION [2 upvotes]: Please help me solving the equation.
I found one of its solutions: $c_1=(xz)/y$
But another one is given as $c_2=(x^3/y)+x$ in the text book. But I cont find. Thank you.

REPLY [2 votes]: Hints:

Rewrite the equation in terms of $\dfrac{dy}{dx}$
Make the substitution $y = v~x$
You end up with a first order equation, use Integrating Factor