text:
\section{Groups with cyclic Fitting subgroup} \label{secexpcn}
In this section one wants to investigate groups $G$ such that the Fitting subgroup $F(G)$ is cyclic of order $n$. One wants to know if expansive subgroups with trivial $G$-core exist in such groups.

In this section, one assumes that $G$ is solvable. First note that it is a well known fact that $C_G(F(G)) \leq F(G)$ and therefore $G/F(G)$ injects into $\Out(F(G))$ thus one has the following exact sequence
\begin{figure}[h]
$$\xymatrix@1{1 \ar[r] & C_n \ar[r] &G \ar[r] & S \ar[r] &1} $$
\end{figure}\\
where $S$ is a subgroup of $\Out(C_n) = \Aut(C_n)$. The map $\iota : C_n \rightarrow G$ is the inclusion map. The map $\pi : G \rightarrow S$ sends an element $g$ to the conjugation map $c_g$. Note that this is not enough to ensure that $G$ is Roquette at this stage. One discusses this issue later on, see Theorem \ref{suitex3cond}.

Suppose $n = p_1^{k_1} \dots p_m^{k_m}$ for some primes $p_i$ and integers $k_i$, so $C_n = \prod_{i=1}^m C_{p_i^{k_i}}$. It's a well-known result that $\Aut(C_n) \cong \prod_{i=1}^m \Aut(C_{p_i^{k_i}})$. Recall also that $\Aut(C_{2^k}) \cong C_2 \times C_{2^{k-2}}$ and for an odd prime $p_i$ one has $\Aut(C_{p_i^{k_i}}) \cong C_{p_i-1} \times C_{p_i^{k_i-1}}$. Let $g_{p_i}$ be a generator of $C_{p_i^{k_i}}$ in $C_n$ and define $\alpha_{p_i}$ an element of $\Aut(C_n)$ by $$\alpha_{p_i} : g_{p_i} \mapsto g_{p_i}^{1+p_i^{k_i-1}}$$ and $\alpha_{p_i}(g_{p_j}) = g_{p_j}$ if $j \neq i$. The map $\alpha_{p_i}$ is an element of order $p_i$ if $k_i > 1$, otherwise it is the identity map.

\begin{lem} \label{h1trivial}
Let $p$ be a prime number dividing $n$, then
$$H^1\big(\gen{\alpha_p},C_n\big)  = \left\{
    \begin{array}{ll}
        1 & \text{ if $p$ is odd or $p=2$ and $k>2$,} \\
        C_2 & \text{ if $p=2$ and $k=2$.}
    \end{array}
\right.$$

\end{lem}
\begin{proof}
Decompose $n$ as $n=p^k  \cdot n/p^k$ such that $p$ does not divide $n/p^k$ and let $g$ be a generator of $C_{p^k}$ in $C_n$.

Note that
$$H^1\big(\gen{\alpha_p},C_n\big) \cong H^1\big(\gen{\alpha_p},C_{p^k}\big) \times H^1\big(\gen{\alpha_p},C_{n/p^k}\big)$$ but  $H^1\big(\gen{\alpha_p},C_{n/p^k}\big)$ is trivial because the order of $\gen{\alpha_p}$ and the order of $C_{n/p^k}$ are coprime. Therefore $H^1\big(\gen{\alpha_p},C_n\big)$ is equal to $H^1\big(\gen{\alpha_p},C_{p^k}\big)$. Moreover, recall that in the cyclic case $H^1\big(\gen{\alpha_p},C_{p^k}\big) = \Ker(t) / \im(\alpha_p \cdot \upsilon )$, where $t = \prod_{i=0}^{p-1} \alpha_p^i$, $\upsilon \in \Aut(C_{p^k})$ sends $g$ to $g^{-1}$ and $(\alpha_p \cdot \upsilon)(g) := \alpha_p(g) \upsilon (g)$. Let's start to describe the action of $t$ on $C_{p^k}$.
\begin{eqnarray*}
t(g^j) &=& \prod_{i=0}^{p-1} \alpha_p^i(g)^j = \prod_{i=0}^{p-1} g^{j(1+p^{k-1})^i} = \prod_{i=0}^{p-1}  g^{j+jip^{k-1} + \dots} \\
&=& \prod_{i=0}^{p-1}  g^{j+jip^{k-1}} = g^{pj+jp^{k-1} \frac{p(p-1)}{2} }.
\end{eqnarray*}
The equality between the first and the second line holds because the power of $p$ for the rest of the terms is bigger than $k$. If $p$ is odd the last term is equal to $g^{pj}$. Therefore $t$ sends $g$ to $g^p$, its kernel is $\gen{g^{p^{k-1}}}$. But the image of $\alpha_p \cdot \upsilon$ is also $\gen{g^{p^{k-1}}}$ as $(\alpha_p \cdot \upsilon)(g) =g^{1+p^{k-1}} g^{-1} = g^{p^{k-1}}$.
If $p=2$, the map $t$ sends $g^j$ to $g^{j(p+p^{k-1})}$. The kernel is again $\gen{g^{p^{k-1}}}$ if $k >2$ and $\gen{g^{p^{k-2}}}$ if $k=2$.
As the kernel is a subgroup of the cyclic group $C_{2^k}$ it's easy to check it by hand. The image of $\alpha_p \cdot \upsilon$ is also $\gen{g^{p^{k-1}}}$ as $(\alpha_p \cdot \upsilon)(g) =g^{1+p^{k-1}} g^{-1} = g^{p^{k-1}}$. This leads us to the conclusion.
\end{proof}
\begin{lem} \label{h2trivial}
Let $p$ be a prime number dividing $n$, then
$$H^2\big(\gen{\alpha_p},C_n\big)  = \left\{
    \begin{array}{ll}
        1 & \text{ if $p$ is odd or $p=2$ and $k>2$,} \\
        C_2 & \text{ if $p=2$ and $k=2$.}
    \end{array}
\right.$$

\end{lem}
\begin{proof}
Again, decompose $n$ as $n=p^k  \cdot n/p^k$ such that $p$ does not divide $n/p^k$ and let $g$ be a generator of $C_{p^k}$ in $C_n$. Using the same argument, one has $H^2\big(\gen{\alpha_p},C_n\big) = H^2\big(\gen{\alpha_p},C_{p^k}\big)$. In the cyclic case $H^2\big(\gen{\alpha_p},C_{p^k}\big) = C_{C_{p^k}} (\gen{ \alpha_p}) / \im(t)$, where $t = \prod_i \alpha_p^i$. First, note that $C_{C_{p^k}} (\gen{ \alpha_p}) = \gen{g^p}$. Indeed, it is easy to check that $C_{C_{p^k}} (\gen{ \alpha_p}) \geq \gen{g^p}$ but  $\gen{g^p}$ is a maximal subgroup of  $\gen{g}$ and $g$ is not stabilized by $\alpha_p$ therefore the other inclusion follows. Secondly, using the description of the action of $t$ in Lemma \ref{h1trivial}, one has also that $\im(t) = \gen{g^p}$ if $p$ is odd or $p=2$ and $k>2$ and therefore $H^2\big(\gen{\alpha_p},C_n\big)$ is trivial for these cases. Nevertheless if $p=2$ and $k=2$ then $\im(t) =1$ and so $H^2\big(\gen{\alpha_p},C_n\big) = C_2.$
\end{proof}
\begin{cor} \label{corexistenceD}
Let $p$ be an odd prime or $p = 2$ and $k > 2$. Suppose $\gen{\alpha_p}$ is a subgroup of $S$. Then there exists a subgroup $D$ of $G$ such that $\pi(D) = \gen{\alpha_p}$ and $D\cap C_n =1$.
\end{cor}
\begin{proof}
By Lemma \ref{h2trivial}, there exists only one class of extensions of $\gen{\alpha_p}$ by $C_n$. Therefore the extension $\pi^{-1}(\gen{\alpha_p})$ is the semi-direct product of $C_n$ by a cyclic group of order $p$, which is the subgroup $D$ that we are looking for.
\end{proof}
\begin{lem} \label{lemngcd}
Let $G$ be an extension of $S$ by $C_n$ as above. Let $D$ be a subgroup of $G$ such that $D \cap C_n = 1$, then $N_{C_n}(D) = C_{C_n} (D) = C_{C_n}(\pi(D))$.
\end{lem}
\begin{proof}
For the first equality, let $x$ be an element of $N_{C_n}(D)$. Then, for all $d \in D$ one has $x d x^{-1} \in D.$ But $x d x^{-1} = x \pg{d}{x^{-1}} d$ which belongs to $D$ if, and only if,  $x \pg{d}{x^{-1}}  = 1$ which means that $x =\pg{d}{x}$ . This implies that $x$ is an element of $C_{C_n}(D)$. The other inclusion is trivial.

For the second equality, note that the action of $D$ on $C_n$ is the same as the action of $\pi(D)$ on $C_n$ by definition of the map $\pi$.
\end{proof}
\begin{lem} \label{cproj}
Let $H$ be a subgroup of $S$ and $H_i$ the $i^{\text{th}}$-projection of $H$ on $\Aut(C_n)$. Then
$$C_{C_n} (H) =\prod_{i=1}^m C_{C_{p_i^{k_i}}} (H_i).$$
\end{lem}
\begin{proof}
Recall that $C_n = \prod_{i=1}^m C_{p_i^{k_i}}$. Now this is just a calculation :
\begin{eqnarray*}
C_{C_n} (H)  &=& \{ c = (c_1,\dots, c_m) \in C_n  \tq \pg{h}{c} = c \text{ for all } h \in H \} \\
&=& \{ c = (c_1,\dots, c_m) \in C_n  \tq \pg{h}{c_i} = c_i \text{ for all } i \text{ and for all }h \in H \} \\
&=& \prod_{i=1}^m \{ c_i \in C_{p_i^{k_i}}  \tq \pg{h}{c_i} = c_i \text{ for all } i \text{ and for all }h \in H \} \\
&=& \prod_{i=1}^m \{ c_i \in C_{p_i^{k_i}}  \tq \pg{h_i}{c_i} = c_i \text{ for all } i \text{ and for all }h_i \in H_i \} \\
&=& \prod_{i=1}^m C_{C_{p_i^{k_i}}} (H_i).
\end{eqnarray*}
\end{proof}
\begin{lem} \label{p2}
Let $G$ be the group $C_{2^k} \rtimes C_2$ with $k >2$, where $C_2$ is generated by either $\beta_1 : g \mapsto g^{-1}$ or $\beta_2 : g \mapsto g^{-1+2^{k-1}}$ where $g$ is a generator of $C_{2^k}$. Let $b$ be an element of $C_2$. If the element $g \pg{b}{g^{-1}}$ belongs to $C_{C_{2^k}}(C_2)$ then the only possibility is that $b = 1$.
\end{lem}
\begin{proof}
Note that in both cases $C_{C_{2^k}}(C_2)=  \{ c \in C_{2^k} \tq c^2 = 1\} $. Suppose now that $g \pg{b}{g^{-1}}$ belongs to $C_{C_{2^k}}(C_2)$, where $b$ is an element of $C_2$. So actually, except being the identity, $b$ could only be $\beta_1$ or $\beta_2$ depending in which case we are. One shows that it must imply $b=1$ anyway. Indeed, remark that $g\beta_1(g^{-1}) = g^2$ is not an element of $C_{C_{2^k}}(C_2)$ because $k > 2$. Similary for $g\beta_2(g^{-1}) = g^{2+2^{k-1}}$. Therefore in both cases one must have $b=1$.
\end{proof}
\begin{thm} \label{suitex3cond}
Let $G$ be a group such that there is an exact sequence
\begin{figure}[H]
$$\xymatrix@1{1 \ar[r] & C_n \ar[r] &G \ar[r] & S \ar[r] &1} $$
\end{figure}
where $S$ is a subgroup of $\Aut(C_n)$, the map $\iota : C_n \rightarrow G$ is the inclusion map, the map $\pi : G \rightarrow S$ sends an element $g$ to the conjugation map $c_g$ and the power of the prime $2$ in the decomposition of $n$ is different from $2$. Then

 \begin{enumerate}
\item if $G$ is a Roquette group then $S$ does not contain any subgroup $\gen{ \alpha_p}$ for a prime $p$ dividing $n$.
\item If $S_{p_j}$ does not contain a subgroup $\gen{ \alpha_{p_j}}$ for all prime $p_j$ dividing $n$ then $G$ has no non-trivial expansive subgroup with trivial $G$-core, where $S_{p_j}$ denotes the $j^{\text{th}}$-projection of $S$ on $\Aut(C_n)$.
\item If $G$ has no non-trivial expansive subgroup with trivial $G$-core then $G$ is Roquette.
\end{enumerate}
\end{thm}
\begin{proof}

One proves $1$ by proving the converse. Suppose there exists a prime $p$ such that $\gen{ \alpha_p}$ is a subgroup of $S$. Decompose $n$ as $n=p^{k}  \cdot n/p^{k} $ such that $p$ does not divide $n/p^{k} $ and let $g_p$ be a generator of $C_{p^{k} }$ in $C_n$.
By Corollary \ref{corexistenceD}, let $D$ be a subgroup of $G$ such that $\pi(D) = \gen{ \alpha_p}$ and $D \cap C_n = 1$. By Lemma \ref{lemngcd} and a quick calculation, one has $N_{C_n}(D) = C_{C_n} (\gen{ \alpha_p}) = \gen{(g_p)^p} \times C_{n/p^k}$. Indeed, it is easy to check that $C_{C_n} (\gen{ \alpha_p}) \geq \gen{(g_p)^p} \times C_{n/p^k}$ but  $\gen{(g_p)^p}$ is a maximal subgroup of  $\gen{(g_p)}$ and $g_p$ is not stabilized by $\alpha_p$ therefore the other inclusion follows.  Define the subgroup $E := \gen{(g_p)^{p^{k-1}} } \times D$, which is an elementary abelian $p$-group. One shows that $E$ is a normal subgroup of $G$. Indeed, let $h$ be an element of $G$, then $\pg{h}{E} = \pg{h}{\gen{(g_p)^{p^{k-1}} }} \times \pg{h}{D} = \gen{(g_p)^{p^{k-1}} } \times  \pg{h}{D}$ as $C_n$ is a normal subgroup of $G$. Moreover, by Lemma \ref{h1trivial}, one has $\pg{h}{D} = \pg{c}{D}$ for some $c \in C_{p^k}$, because $\pg{h}{D}$ is a complement of $C_n$ in $C_n \rtimes D$ and the elements of $D$ act trivially on $C_{n/p^k}$. Indeed one has $\pg{h}{D} \cap C_n = \pg{h}{(D \cap \pg{h^{-1}}{C_n})} = \pg{h}{(D \cap C_n)} =1$ and $\pg{h}{D} \subset \pi^{-1} \pi (\pg{h}{D}) = \pi^{-1}(\pg{\pi(h)} {\pi(D)}) =\pi^{-1} (\pi(D)) = C_n \rtimes D$. As $\alpha_p$ acts trivially on $C_{n/p^k}$ so does $D$ and thus one can restrict the conjugation to an element $c \in C_{p^k}$ instead of $C_n$. Now one looks at $\pg{c}{D}$. Let $x$ be an element of $\pg{c}{D}$, then $x = c \pg{d}{c^{-1}} d$ for a $d$ in $D$. Write $c$ as $g_p^i$ and $\pi(d)$ as $ \alpha_p^j$. Recall that the action of $D$ on $C_n$ is the same as $\gen{ \alpha_p}$. Therefore $c\pg{d}{c^{-1}} = c\pg{\alpha_p^j}{c^{-1}} = (g_p)^{-ijp^{k-1}}$ which implies that $\pg{c}{D}$ is included in $\gen{(g_p)^{p^{k-1}} } \times D$ and so $E$ is normal.  Therefore $G$ is not Roquette. This proves\nolinebreak[1] $1$.

For $2$, as before one has $n= p_1^{k_1} \dots p_m^{k_m}$ and moreover to simplify the notation suppose here that  $p_1^{k_1} = 2^k$. Suppose that $S_{p_j}$ does not contain a subgroup $\gen{ \alpha_{p_j}}$ for all prime $p_j$ dividing $n$ and prove $G$ has no non-trivial expansive subgroup with trivial $G$-core.

Let $A$ be a non-trivial subgroup of $G$ with trivial $G$-core. Then $A \cap C_n = 1$ otherwise $A \cap C_n$ would be included in the $G$-core of $A$. So $\pi$ induces an isomorphism between $A$ and a subgroup $H$ of $S$.

The subgroup $H$ is included in $\prod_{i=1}^m H_i$ where $H_i$ is the $i^{\text{th}}$-projection of $H$ on $\Aut(C_n)$. As $A$ is not trivial, so is $H$ and therefore there exists an integer $j$ such that $H_j$ is not trivial. Now one looks at the expansivity condition. Let $e$ be an element in $C_n = \prod_{i=1}^m C_{p_i^{k_i}}$ but not in $C_{C_n}(H)$ and write $e$ as $\prod_{i=1}^m e_{p_i}$ where $e_{p_i} \in C_{p_i^{k_i}}$. More precisely, if $p_j$ is odd, then take $e_{p_i}$ in $C_{C_n}(H)$ and only $e_{p_j}$ not in $C_{C_n}(H)$.  If $p_j = 2$ take the element $e_2$ to be the generator $g$ of $C_{2^k}$ and again $e_{p_i}$ in $C_{C_n}(H)$ if $p_i$ is not equal to $p_j$. Note that if $p_j = 2$, then $k > 2$ as $k=2$ is excluded by assumption and $k=1$ forces $H_2$ to be trivial. As $N_{C_n}(A) =C_{C_n}(H)$ by Lemma \ref{lemngcd}, the element $e$ is not in $N_G(A)$. Let $b$ be an element of $A$. Then $\pg{e}{b} = e \pg{b}{e^{-1}} b$ is an element of $N_G(A)$ if and only if $e \pg{b}{e^{-1}}$ is an element of $N_G(A) \cap C_n = C_{C_n} (H)$, by Lemma \ref{lemngcd}. Our purpose is to show that $e \pg{b}{e^{-1}} = 1$ and therefore $\pg{e}{A} \cap N_G(A) \leq A$. Write $b = \prod_i b_i$ such that $\pi(b_i)$ belongs to $H_i$. This is possible because $\pi$ induces an isomorphism between $A$ and $H$. Then, using Lemma \ref{cproj}, one has that $e \pg{b}{e^{-1}} = \prod_i e_{p_i} \pg{b_i}{e_{p_i}^{-1}} \in C_{C_n} (H) =\prod_i C_{C_{p_i}^{k_i}}(H_i)$, which means that for all $i$ one has $e_{p_i}\pg{b_i}{e_{p_i}^{-1}} \in  C_{C_{p_i}}(H_i)$. But by definition, for all $i$, one has $e_{p_i} \pg{b_i}{e_{p_i}^{-1}}  \in [C_{p_i^{k_i} },H_i]$ and therefore $e_{p_i} \pg{b_i}{e_{p_i}^{-1}}  \in [C_{p^{k_i} },H_i] \cap C_{C_{p_i}^{k_i}}(H_i) $. If $p_i$ is different from $p_j$ then  $\pg{b_i}{e_{p_i}}  = e_{p_i}$, as the element $e_{p_i}$ belongs to $C_{C_n}(H)$. By Theorem $2.3$ of \cite{Gor} on page 177, $[C_{p_j^{k_j} },H_j] \cap C_{C_{p_j}}(H_j)$ is trivial if $p_j$ is different from $2$. The reason is that $H_{j}$ has order prime to $p_j$ because $S_{p_j}$ does not contain a subgroup $\gen{ \alpha_{p_j}}$. For the case $p=2$ one uses Lemma \ref{p2} for $k>2$. Indeed, in this case one has shown that if $g \pg{b_2}{g^{-1}}$ belongs to $C_{C_{2^k}}(C_2)$ then $b_2 = 1$.

 To sum up one has shown that $\pg{b_i}{e_{p_i}}  = e_{p_i}$ for all $i$. This implies that if $\pg{e}{b}$ belongs to $N_G(A)$ then $\pg{e}{b} = e \pg{b}{e^{-1}} b = b$. Finally one concludes that $\pg{e}{A} \cap N_G(A) \leq A$ and therefore $A$ is not expansive. As $A$ was an arbitrary non-trivial subgroup of $G$ with trivial $G$-core, this concludes the proof.

The third fact is a general result, see Corollary 3.7 of \cite{MoT}. One has just to prove the existence of a faithful simple module. Let $L$ be $\Ind_{C_n}^G (\xi)$ where $\xi$ is a primitive $n$th root of unity. This module is irreducible as the conjugate representations of $\xi$ by the action of $G/C_n$ are not isomorphic as it acts by automorphism on $C_n$. Moreover, the condition of primitivity on the root ensures the faithfulness of the induced module so $L$ is the wanted module.
\end{proof}
\begin{rem}{\text{}}

To ensure the non-existence of expansive subgroups with trivial $G$-core in such groups one needs, not only that $S$ does not contain a subgroup $\gen{ \alpha_p}$ for a prime $p$ dividing $n$ but also that $S_{p_j}$ does not contain a subgroup $\gen{ \alpha_{p_j}}$ for the primes $p_j$ dividing $n$. Which means that we do not want a diagonal subgroups of $\prod_{i=1}^m \Aut(C_{p_i^{k_i}})$ in $S$, see Lemma $1.1$ of \cite{The} for the terminology. However note that, depending on $G$, for example if there is no prime $p_i$ dividing $p_j -1$, the two conditions can be equivalent.

\end{rem}


TITLE: Compute the probability that a given box contains more than one sterile seed
QUESTION [0 upvotes]: Can anyone help me to solve this problem?

A farmer estimated that $20\%$ of the pumpkin seeds he produced will not sprout. He sells the seeds in boxes containing $50$ seeds each. Compute the probability that a given box contains more than one sterile seed. Your answer needs to be a number.

I was thinking that if we let $X=$ number of sterile seeds, then we are looking for $P(X>1)=1-P(X \leq 1)$, but after that I do not know what to do.

REPLY [1 votes]: Each seed sprouts with probability $0.8$ and is sterile with probability $0.2$. We have $50$ seeds in the box, and I'm assuming their probabilities of sprouting are independent. We get that
$\Pr(\text{no seed in the box is sterile}) = \Pr(X = 0) = (0.8)^{50}$,
$\Pr(\text{exactly one seed in the box is sterile}) = \Pr(X = 1) = (0.2)\cdot(0.8)^{49}\cdot 50$
Because we have $50$ ways of choosing the sterile seed, see Binomial Distribution.
So in total :
$\Pr(X > 1) = 1 - \left((0.8)^{50} + (0.2)\cdot(0.8)^{49}\cdot 50\right)$


TITLE: $a_1 =π/4$, $a_{n+1}= \cos(a_n)$. Prove that $\lim(a_n) =c$ when $c =\cos(c)$
QUESTION [0 upvotes]: I have been struggling for too long,
Does anyone have an idea how to solve this one ?
$a_1 =π/4\:$ and $a_{n+1}=\cos(a_n)$
Prove that $$\lim_{a_n \to \infty} =c\;\text{ when } \; c = \cos(c)$$

REPLY [0 votes]: If you have
$$a_{n+1}=\cos(a_n)$$ If the limit $L$ exists, it is, by definition, the solution of
$$L=\cos(L)$$ and this is Dottie number.
The value of $a_1$ has nothing to do.
So, you need to prove the convergence of the series.


TITLE: Inconsistency regarding Archimedes' Principle
QUESTION [0 upvotes]: According to the Encyclopedia Britannica, "[...] any body completely or partially submerged in a fluid (gas or liquid) at rest is acted upon by an upward, or buoyant, force the magnitude of which is equal to the weight of the fluid displaced by the body. The volume of displaced fluid is equivalent to the volume of an object fully immersed in a fluid or to that fraction of the volume below the surface for an object partially submerged in a liquid."
The emphasis is mine, and it is that last part that got me thinking about something: What if a body stays afloat on a liquid whose volume is less than that of the object? That seemingly contradicts Archimedes' Principle because the fluid that the body displaces obviously cannot be equivalent to its actual volume; there's not enough liquid.
Take a look at the image* below:

The can has a volume of 350 mL, so it would theoretically displace roughly that much liquid if it were immersed, but this cannot happen because there is only 230 mL of (artificially colored) water. So, how does Archimedes' Principle account for this apparent paradox? Furthermore, why isn't this discussed more often?
(*The source for the picture is an article written in a foreign language, so I won't bother posting it. Interestingly, I could not find a similar setup on websites in English.)

REPLY [4 votes]: "The volume of displaced fluid" doesn't mean the physical volume of fluid that has moved out of the way. (After all, as you say, that doesn't make sense if there's not much fluid.) The phrase actually means the amount of volume that the floating object takes up which is below the fluid's current level.
Since these two definitions coincide most of the time, common sources conflate the two, but the latter is what actually is used, e.g. in a derivation of Archimedes' principle in any physics textbook.


text:
/-
Copyright (c) 2021 OpenAI. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Kunhao Zheng, Stanislas Polu, David Renshaw, OpenAI GPT-f
-/
import mathzoo.imports.miniF2F

open_locale nat rat real big_operators topological_space

axiom imo_1983_p6
  (a b c : ℝ)
  (h₀ : 0 < a ∧ 0 < b ∧ 0 < c)
  (h₁ : c < a + b)
  (h₂ : b < a + c)
  (h₃ : a < b + c) :
  0 ≤ a^2 * b * (a - b) + b^2 * c * (b - c) + c^2 * a * (c - a)

subset_name: curated


TITLE: What is the range space of a homomorphism?
QUESTION [0 upvotes]: I am really confused about the intuitions of the range space of a homomorphism.
My textbook says the range space of a homomorphism $$h: V \to W\ is\ \ \mathscr{R}(h) = \{ h(\overrightarrow{v}) |\overrightarrow{v}\in V \}$$
sometimes denoted h(V). The dimension of the range space is the map's rank.
My understanding: the range space is all members of the V. But isn't V the domain? Then why is the range space referring to the range, which is W in this case? or is it the case that $h(\overrightarrow{v})$ is the image of the domain, so this makes it part of the codomain?
Thank you and apologize if any part of the question is confusing.

REPLY [3 votes]: It's the latter. The range space is the set of elements of $W$ which are of the form $h\left(\vec v\right)$ for some $\vec v\in V.$
For example, consider the map $h:\Bbb R^2\to\Bbb R^3$ given by $h\left([a,b]^T\right)=[a,a,a]^T.$ Geometrically speaking, the range of $h$ is the line through the origin consisting of all points whose coordinates are equal. This is not all of $W$ (which has dimension $3$), nor is it equal to $V$ (which has dimension $2$). The map $h$ has rank $1,$ since its range has dimension $1.$


TITLE: How and why do electric fields occur when we put a voltage difference between two point?
QUESTION [0 upvotes]: When we put a voltage differences between two points on a certain sample, I would like to know how and why the electric field occurs. In addition, I would like to know in what shape does the electric field occur. Does it have the same shape of the electric field which occurs between two point charges + and -?

REPLY [2 votes]: "Why?" questions are always a bit dubious in physics, but let me explain a few words here:
"Voltage" is in itself a difference, and it is (by definition) the difference of electric potential energy. The difference in electric potential energy is by definition the work needed to move a test charge against the electric field. In other words: The electric field is the fundamental thing here, not "Voltage".
Therefore, let's consider the question the other way round: What leads to a difference in potential energy?
Suppose you have an electrically charged object. Suppose you have another charged object, then you can measure a force exerted by the first object on the second. If you move the second object around, you get a force (direction and quantity) at every point in space. Let's call this the "electric field". Instead of any object, let's assume that the object is a pointlike particle, i.e. we take a charged object so very small that we can hardly see it. Then these objects/particles have an electric field which is given by Coulomb's law: A unit charged (i.e. another charged particle with charge $1$) at distance $r$ from the particle will experience the force
$$F(r)=k\frac{q}{r} $$
where $k$ is some constant and $q$ is the charge of the particle. This is the definition of an electric field of a particle and also gives you its "shape". Now by a curious fact of nature, the electric field is additive, which means that if you have more particles, the shape of the field changes by just adding the different forces from each particle at each point. This tells you that the electric field can come in very different shapes, always depending on the distribution of charges. As an example, have a look at the field of four particles here.
Now what about the "voltage"? By another curious fact of nature, if I move a particle from point $A$ to point $B$ in a static electric field, it doesn't matter which way I choose, the work is independent of the way. This is why it makes sense to speak of the work difference between points $A$ and $B$ - which defines "electrostatic potential energy difference" or else voltage.


TITLE: Making proper integral for shaded region
QUESTION [1 upvotes]: I have curve and line, one curve is a parabola intersects at $(-3,0)$, $(-1,0)$, and $(0,1)$ while the line intersects at $(-3,0)$ and $(0,2)$.
Sorry I couldn't upload the picture, but I could describe about the shaded region between these curve and line.

(Condition 1) The area between $(-3,0)$, $(-1,0)$, $(0,1)$, and $(0,2)$ is shaded. The peak of parabola below $x$-axis is not shaded.
(Condition 2) The area between $(0,1)$, $(0,2)$, and $(a,b)$ is shaded. The $a$ must be the value of $x$ when factorized the parabola equation (because $x=-3$ is already here). I don't think the $b$ will be used.

I need to find the area of the shaded region.
Here is my attempt:
I got the equation for the parabola and curve. The parabola is $y=\frac{1}{3}(x+1)(x+3)$ and the line is $y=\frac{2}{3}x+2$, then I defined the integrals are below:
$A_{1}$ is area of Condition 1. This is just integral of the line, which is $$\int_{-3}^{-1}\,{\frac{2}{3}x+2-0}\,dx$$
The 0 is from $y=0$ ($x$-axis).
$A_{2}$ is area of unshaded region between $(-1,0)$, $(0,1)$, and $(0,0)$. The integral is
$$\int_{-1}^{0}\,{\frac{1}{3}(x+1)(x+3)}-0\,dx$$
$A_{3}$ is area of Condition 2. I already got the $a=1$. The integral is subtracting $\mathrm{line-parabola}$, which is
$$\int_{0}^{1}\,{-x^{2}-3x+3}\,dx$$
So the area of shaded region would be
$$A_{1}-A_{2}+A_{3}$$
Is my attempt true?
Thanks!

REPLY [0 votes]: Suppose that you want to find the area of the gray part in the following figure :
$\qquad\qquad\qquad$
Then, your attempt is not true. The area is given by
$$\int_{-3}^{-1}\left(\frac 23x+2\right)dx+\int_{-1}^{1}\left(\left(\frac 23x+2\right)-\left(\frac 13x^2+\frac 43x+1\right)\right)dx$$


TITLE: Does acceleration create gravity since inertial mass is found to be the same as gravitational mass?
QUESTION [1 upvotes]: Does acceleration create gravity since inertial mass is found to be the same as gravitational mass?
If not what would account for acceleration behaving like gravity and vise versa?

REPLY [1 votes]: You are correct, and this is the equivalence principle. As per GR, the inertial and gravitational mass are equivalent.

In the theory of general relativity, the equivalence principle is the equivalence of gravitational and inertial mass, and Albert Einstein's observation that the gravitational "force" as experienced locally while standing on a massive body (such as the Earth) is the same as the pseudo-force experienced by an observer in a non-inertial (accelerated) frame of reference.

https://en.wikipedia.org/wiki/Equivalence_principle
Now you are asking whether acceleration can create gravity. It can create the effects of gravity, like GR time dilation.

Gravitational time dilation is a form of time dilation, an actual difference of elapsed time between two events as measured by observers situated at varying distances from a gravitating mass. The lower the gravitational potential (the closer the clock is from the source of gravitation), the slower time passes, speeding up as the gravitational potential increases (the clock getting away from the source of gravitation). Albert Einstein originally predicted this effect in his theory of relativity and it has since been confirmed by tests of general relativity.[1]

https://en.wikipedia.org/wiki/Gravitational_time_dilation
A very good example is the GR explanation of the twin paradox, where the traveling twin is turning around with the spaceship and at the turn, the acceleration/deceleration is equivalent to a gravitational zone, where the traveling twin's time slows down. Thus, the twin ages slower (during the turn), so when they meet the traveling twin aged less.

The answer is that acceleration is required to resolve the twin paradox and the two observers are not both in inertial frames.
  It is a fundamental principle of general relativity that acceleration and gravity cannot be distinguished.

Twin paradox caused by gravitational difference in space


TITLE: If A and B are positive real numbers and each of the equations: $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real roots, what is the smallest value of A+B
QUESTION [1 upvotes]: Problem
In the equation:$x^2+ax+2b=0$ and $x^2+2bx+a=0$, we have to figure out the sum of a and b by using the following identity:
$P(x)=Q(x)*D(x)+R(x)$
where P(x) is the equations and q(x) is the quotient and d(x) is the divisor and r(x) is the remainder.
My though process
my initial stance was that because these two equations are both equal to zero, then they must equal each other so i set
$x^2+ax+2b=x^2+2bx+a$
which simplifies into:
$ax+2b=2bx+a$
then i knew that that didn't go anywhere so then i set x=0 in the first equation and thought that B was zero but that can't be true
So then i assigned R and S as the roots to the equations where i figured that in general given a simple quadratic equation: $ax^2+bx+c$ with the roots r and s, $r+s=\frac{-b}{a}$ and $rs=\frac{c}{a}$
and going back to the real equation we can set
$r+s=\frac{-a}{1}$ in the equation $x^2+ax+2b=0$ and $rs=\frac{2b}{1}$
and with the same thing to the other equation I got:
$r+s=\frac{-2b}{1}$ and $rs=\frac{a}{1}$
as stated these two equation can be set equal to each other because they are both equal to 0.
This means that $\frac{-a}{1}$=$\frac{-2b}{1}$
and $\frac{2b}{1}$=$\frac{a}{1}$
and thats where i got stuck.

REPLY [1 votes]: HINT:
We need discriminant $\ge0,$
$\implies a^2\ge8b, 4b^2\ge4a\iff b^2\ge a$
$a^4\ge64a\iff a(a-4)(a^2+4a+4^2)\ge0\iff a(a-4)\ge0$ as $a^2+4a+4^2=(a+2)^2+2^2\ge2^2>0$
$\implies$ either $a\le0$ or $a-4\ge0\iff a\ge4$
But $a$ is positive.


TITLE: Under what conditions is $(A+B)(A-B) = (A^2 - B^2)$ for two $n \times n$ matrices $A$ and $B$?
QUESTION [1 upvotes]: so my approach to this problem was
to view $A$ as a matrix of the form $\begin{bmatrix}a_1 & a_2 & a_3 & \dots & a_n\end{bmatrix}$
and $B$ as $\begin{bmatrix}b_1 & b_2 & b_3 & \dots & b_n\end{bmatrix}$
define variable $C = (A+B)$ and $D = (A-B)$
$C = \begin{bmatrix}c_1 & c_2 & c_3 & \dots & c_n\end{bmatrix}$
$D = \begin{bmatrix}d_1 & d_2 & d_3 & \dots & d_n\end{bmatrix}$
$CD = A^2 - B^2$
$Cd_1 + Cd_2 +\dots + Cd_n = (a_1A - b_1B) + (a_2A + b_2B) + (a_3B + b_3B) + \dots + (a_nB + b_nB)$
substitute back for $d$ and $C$
$(a_1-b_1)(A+B) + (a_2 - b_2)(A+B) + \dots = (a_1A - b_1B) + (a_2A + b_2B) + (a_3B + b_3B)+ \dots$
$(a_1A + A_1B -B_1A - b_1B) + (a_2A + A_2B -B_2A - b_2B) + \dots = (a_1A - b_1B) + (a_2A + b_2B) + \dots$
at this point I'm stuck, because I feel like I'm on the wrong track because my answer does not match correctly with the solution. How does one solve this problem?

REPLY [3 votes]: $(A+B)(A-B)=A^2+BA-AB-B^2$ its equal to  $A^2-B^2$ when $AB=BA$ i mean A,B must   commute


TITLE: Is the free R-module on a sheaf of sets still a sheaf?
QUESTION [1 upvotes]: Let $L$ be a sheaf of sets on some site $S$. Let $F$ be the presheaf obtained by composing $L$ with the free R-module functor, i.e. for any object $U$, we define $F(U)$ to be the free $R$-module on the set $L(U)$. Is $F$ a sheaf?

REPLY [6 votes]: Suppose $S$ is a topological space which has two non-empty open subsets $U$ and $V$ with an empty intersection.
Let $L$ be the sheaf of sets that assigns to each open set the singleton set $\lbrace *\rbrace$.
Let $R$ be the integers.  Then the presheaf you describe assigns the integers to every open set, but this is not a sheaf because, for example, the section 2 (over $U$) and the section 3 (over $V$) agreee by default on $U\cap V$ but can't be patched together.


text:
\begin{document}

\title{Twisting operators and centralisers of Lie type groups over local rings}

\author{Zhe Chen}

\address{Department of Mathematics, Shantou University, Shantou, China}

\email{zhechencz@gmail.com}

\begin{abstract}
We extend the classical result asserting that the twisting operator preserves certain Deligne--Lusztig character values for truncated formal power series; along the way we discuss some properties of centralisers. This leads us to the construction of an action of $\mathrm{GL}_n(\mathbb{F}_q[[\pi]]/\pi^r)$ on a Springer fibre intersected by Deligne--Lusztig varieties; we determine the primitivities of the induced cohomological representations for single cycles. The case of $\mathrm{SL}_2$ over finite dual numbers is presented with a criterion on semisimple orbit representations.
\end{abstract}

\maketitle

\tableofcontents

\section{Introduction}\label{section:Intro}

Let $\mathbb{F}_q$ be a finite field with $q$ elements. Consider the finite groups $\mathrm{GL}_n(\mathbb{F}_{q^d})$ and $\mathrm{GL}_n(\mathbb{F}_q)$, where $d\in\mathbb{Z}_{>0}$. A natural question to wonder is: What are the relations between their characters? An answer was given by Shintani \cite{Shintani_twoRemarks} in 1976: He established a bijection from the set of Galois-invariant irreducible characters of $\mathrm{GL}_n(\mathbb{F}_{q^d})$ to the set of irreducible characters of $\mathrm{GL}_n(\mathbb{F}_q)$. This work was later developed into the theory of Shintani descents, which plays an important role in the geometric side of character theory; see \cite{Kawanaka_IrrCharUnitaryGp}, \cite{Lusztig_84_OrangeBook}, \cite{Asai_twisting_op_classicalgroup}, \cite{DM_L-function_Book}, \cite{DM_LusztigFunctorShintaniDescent}, \cite{Shoji_ShintaniAlgGp}, \cite{Shoji_ShintaniSL(n)}, \cite{Deshpande_ShintaniDescent_compositio}.

\vspace{2mm} In the very special case $d=1$, Shintani descent is the natural operation switching the Lang isogeny on the space of class functions
$$f\longmapsto\mathrm{Sh}_G(f)\colon h^{-1}F(h)\mapsto f(F(h) h^{-1}),$$
called the twisting operator (see Definition~\ref{defi:Sh}), whose behaviour can be mysterious for groups admitting disconnected centralisers. In this paper we study this operator and related centraliser properties for connected reductive groups over a (truncated) discrete valuation ring in positive characteristic (like $\mathrm{SL}_n(\mathbb{F}_q[[\pi]]/\pi^r)$).

\vspace{2mm} In Section~\ref{sec:pre} we recall some basic properties of $\mathrm{Sh}_G$. And in Section~\ref{sec:Group over local ring} we make some preparations on reductive groups over formal power series rings and recall the orbit construction.

\vspace{2mm} In Section~\ref{sec:higher DL} we consider two aspects concerning Deligne--Lusztig theory. First, by using a generalised Green function character formula and by generalising a basic property of semisimple centralisers, we extend a result of Asai and Digne--Michel, showing that $\mathrm{Sh}_G$ preserves the higher Deligne--Lusztig character values under a condition on unipotent parts; see Theorem~\ref{thm:main result}.

\vspace{2mm} Second, motivated by a group embedding used in a centraliser discussion (in Proposition~\ref{prop:unipotent centraliser prop}), we observe that $\mathrm{GL}_n(\mathbb{F}_q[[\pi]]/\pi^r)$ can be viewed as a unipotent centraliser; this allows us to construct an action of $\mathrm{GL}_n(\mathbb{F}_q[[\pi]]/\pi^r)$ on the intersection $\mathcal{B}_{u,w}$ of a Springer fibre and a Deligne--Lusztig variety (see Definition~\ref{defi:B_{u,w}}). While both Springer fibres and Deligne--Lusztig varieties are of vital importance in geometric representation theory, it seems these intersections $\mathcal{B}_{u,w}$ have not been considered in literature before. We determine the primitivities of the representations afforded by the cohomology of these varieties for the cycles $w=(1,2,...,z)$; see Theorem~\ref{thm:R_{u,w}}.

\vspace{2mm} In Section~\ref{sec:expl} we study the behaviour of the twisting operator on primitive representations of $\mathrm{SL}_2([[\pi]]/\pi^2)$ in odd characteristic. We find that the semisimple orbit representations are exactly the ones invariant under $\mathrm{Sh}_G$; see Corollary~\ref{coro:semisimple of SL_2 are twist-invariant}.

\vspace{2mm}{\bf Acknowledgement.} The author thanks Alexander Stasinski for several useful comments on an earlier version of this paper, and thanks Yongqi Feng for helpful conversations.

\section{Preliminaries}\label{sec:pre}

In this section we recall some basics of the twisting operator following \cite{DM_L-function_Book}.

\vspace{2mm} Let $G$ be the $\overline{\mathbb{F}}_q$-base change of a connected smooth affine group scheme over ${\mathbb{F}}_q$, and let $F$ be the associated geometric Frobenius endomorphism. Denote by $L\colon x\mapsto x^{-1}F(x)$ the Lang isogeny. The Lang--Steinberg theorem tells us that, for any $g\in G$, there is an $h\in G$ such that $h^{-1}F(h)=g$. Note that, if $g\in G^F$, then $F(h)h^{-1}\in G^F$; via this property one can define a permutation on the conjugacy classes of $G^F$:

\begin{defi}\label{defi:n_F}
Let $g=h^{-1}F(h)\in G^F$ and let $[g]_{G^F}$ be the conjugacy class in $G^F$. Then
$$n_F([g]_{G^F}):=[F(h)h^{-1}]_{G^F}.$$
(Note that this is independent of the choice of $h$.)
\end{defi}

\begin{defi}\label{defi:Sh}
The twisting operator $\mathrm{Sh}_G$ is the linear automorphism on the complex space of class functions of $G^F$ (denoted by $\mathcal{C}(G^F)$) given by composition with $n_F$, that is:
$$\mathrm{Sh}_G(f)=f\circ n_F$$
for $f\in \mathcal{C}(G^F)$.
\end{defi}

\begin{lemm}
The operator $\mathrm{Sh}_G\in \mathrm{End}(\mathcal{C}(G^F))$ is semisimple with eigenvalues being roots of unity. Moreover, $\mathrm{Sh}_G$ preserves the usual inner product on $\mathcal{C}(G^F)$.
\end{lemm}
\begin{proof}
By definition, $\mathrm{Sh}_G$ is a permutation matrix with respect to the basis consisting of characteristic functions on conjugacy classes, from which the first two properties follow. Now the property of being isometric follows from the fact that (see \cite[I.7.2]{DM_L-function_Book}) $n_F$ preserves the cardinals of conjugacy classes.
\end{proof}

A fundamental property of $\mathrm{Sh}_G$ is the following lemma:

\begin{lemm}\label{lemm:a fundamental lemma}
Let $g\in G^F$. Consider the set of $F$-rational elements in the geometric conjugacy class (i.e.\ the $G$-conjugacy class) of $g$ in $G$; partition this set into $G^F$-conjugacy classes. There is a natural bijection between the resulting $G^F$-classes and the $F$-conjugacy classes of the component group $Z_G(g)/Z_G(g)^{\circ}$, such that, under this bijection the action of $n_{F}$ becomes the multiplication by $g$.
\end{lemm}
\begin{proof}
See \cite[IV.1.1]{DM_L-function_Book} or \cite[3.21]{DM1991}.
\end{proof}

From the above it follows immediately that:

\begin{coro}
The order of $\mathrm{Sh}_G$ is the smallest $i$ such that, for any $g\in G^F$, $g^i=1$ in all the $F$-conjugacy classes of $Z_G(g)/Z_G(g)^{\circ}$.
\end{coro}

\begin{lemm}\label{lemm:triviality for commutative groups}
If $G$ is commutative, then $n_F$, and hence $\mathrm{Sh}_G$, are the identity maps.
\end{lemm}
\begin{proof}
This follows from the definition.
\end{proof}

\begin{lemm}\label{lemm:triviality on ss classes}
The permutation $n_{F}$ is the identity on the semisimple classes.
\end{lemm}
\begin{proof}
Let $s\in G^F$ be semisimple, then $s$ is contained in an $F$-stable maximal torus $S$ of $G$ (see e.g.\  \cite[3.16]{DM1991}). The assertion follows from Lemma~\ref{lemm:triviality for commutative groups} by restricting $n_F$ to $S^F$.
\end{proof}

\section{Groups over local rings}\label{sec:Group over local ring}

Let $\mathcal{O}:=\mathbb{F}_q[[\pi]]$, and let $\mathbb{G}$ be a connected reductive group scheme over $\mathcal{O}$. Note that every smooth representation of $\mathbb{G}(\mathcal{O})$ factors through $\mathbb{G}(\mathcal{O}_r)$ for some $r\in \mathbb{Z}_{>0}$, where $\mathcal{O}_r:=\mathbb{F}_q[[\pi]]/\pi^r$. From now on we fix an arbitrary such $r$.

\vspace{2mm} We briefly recall the settings in \cite{Lusztig2004RepsFinRings}: There is a connected algebraic group $G_r$ (Weil restriction), defined over $\mathbb{F}_q$, such that
$$G_r(\overline{\mathbb{F}}_q)\cong \mathbb{G}(\overline{\mathbb{F}}_q[[\pi]]/\pi^r) \ \ \text{and}\ \ G_r^F:=G_r(\overline{\mathbb{F}}_q)^F\cong\mathbb{G}(\mathcal{O}_r),$$
where $F$ is the associated geometric Frobenius endomorphism. For $i\in\mathbb{Z}_{[1,r-1]}$, the reduction map modulo $\pi^i$ induces a surjective morphism between algebraic groups
$$\rho_{r,i}\colon G_r\longrightarrow G_i;$$
we denoted the kernel group (congruence group) by $G^i$. Note that $G_1$ is a connected reductive group over $\overline{\mathbb{F}}_q$ and we have a semi-direct product
$$G_r\cong G_1\ltimes G^1,$$
which is a Levi decomposition. Every closed subgroup scheme of $\mathbb{G}\times_{\mathcal{O}}\overline{\mathbb{F}}_q[[\pi]]/\pi^r$ corresponds to a closed subgroup of $G_r$; for such a subgroup $H\subseteq G_r$, we shall use similar notations $H_i$ and $H^i$. As before, $L\colon g\mapsto g^{-1}F(g)$ denotes the Lang isogeny associated to $F$.  In an algebraic group, we use the conjugation notation $x^y:=y^{-1}xy=:{^{y^{-1}}x}$.

\vspace{2mm} Throughout this paper we would write $G:=G_r$ when there is no confusion.

\vspace{2mm}  An important tool for studying the representations of $G^F=\mathbb{G}(\mathcal{O}_r)$, when $r\geq2$, is the notion of orbits. In the remaining part of this section we assume that $r\geq 2$, and that the Lie algebra $\mathfrak{g}:=\mathrm{Lie}(G_1)$ admits a non-degenerate $G_1$-equivariant bilinear form $\mu\colon \mathfrak{g}\times\mathfrak{g}\rightarrow\overline{\mathbb{F}}_q$ defined over $\mathbb{F}_q$. (This requires $\mathrm{char}(\mathbb{F}_q)$ is not too small unless $\mathbb{G}=\mathrm{GL}_n$; see e.g.\ the discussions in \cite[2]{Let2005book}.)  We fix such a $\mu$ and fix a non-trivial character $\phi\colon \mathbb{F}_q\rightarrow \overline{\mathbb{Q}}_{\ell}^{\times}$. Note that $G^{r-1}$ can be viewed as the additive group of $\mathfrak{g}$; under this viewpoint, the conjugation action of $G$ on $G^{r-1}$ is translated to be the adjoint action of $G_1$ on $\mathfrak{g}$. If $\sigma$ is an irreducible character of $G^F$, then by Clifford's theorem (see \cite[6.2]{Isaacs_CharThy_Book}) we have
$$\sigma|_{(G^{r-1})^F}= e\cdot\left(\sum_{\chi\in\Omega(\sigma)} \chi \right),$$
where $e$ is a positive integer and $\Omega(\sigma)$ is a $G_1^F$-orbit of irreducible characters of the additive group $\mathfrak{g}^F$, so we get a map $\Omega\colon \mathrm{Irr}(G^F)\rightarrow \mathbb{G}(\mathbb{F}_q)\backslash\mathrm{Irr}(\mathfrak{g}^F)$. Taking the dual of $\Omega$ via $\phi(\mu(-,-))$ we get another map:
$$\Omega'\colon \mathrm{Irr}(G^F)\longrightarrow \mathbb{G}(\mathbb{F}_q)\backslash \mathfrak{g}^F.$$
We call $\Omega'(\sigma)$ the orbit of $\sigma$; of particular interest is the case that $\Omega'(\sigma)$ is nilpotent, regular, or semisimple.

\vspace{2mm} Constructing irreducible characters of $G^F$ using orbits attracts many attentions; see e.g.\ the recent works \cite{Stasinski_Stevens_2016_regularRep} and \cite{Krakovski_Onn_Singla_regularchar_2018}, and the references therein. Note that the notion of orbits naturally extends to the whole space $\mathcal{C}(G^F)$: If $f=\oplus_i{\lambda_i}\sigma_i$ is a class function with irreducible components $\sigma_i$, then the formal sum $\Omega'(f):=\sum_i\lambda_i\Omega'(\sigma_i)$ gives a surjective linear map
$$\Omega'\colon \mathcal{C}(G^F)\longrightarrow \mathbb{C}[\mathbb{G}(\mathbb{F}_q)\backslash \mathfrak{g}^F],$$
where $\mathbb{C}[\mathbb{G}(\mathbb{F}_q)\backslash \mathfrak{g}^F]$ denotes the space of formal sums of adjoint orbits.

\begin{prop}\label{prop:Sh preserves orbits}
Let $f$ be a class function of $G_r^F$, then $\Omega'(f)=\Omega'(\mathrm{Sh}_G(f))$.
\end{prop}
\begin{proof}
By construction, $\Omega'(f)$ only depends on the restriction of $f$ to $(G^{r-1})^F$; meanwhile, we have $\mathrm{Sh}_G(f)\mid_{(G^{r-1})^F}=f\mid_{(G^{r-1})^F}$ by Lemma~\ref{lemm:triviality for commutative groups}.
\end{proof}

\section{Deligne--Lusztig constructions}\label{sec:higher DL}

In \cite{Lusztig1979SomeRemarks} and \cite{Lusztig2004RepsFinRings} Lusztig gives the Deligne--Lusztig construction for $\mathbb{G}(\mathcal{O}_r)$. Recently, this work has been studied and generalised along various aspects; see e.g.\ \cite{Sta2009Unramified}, \cite{Sta2011ExtendedDL}, \cite{Chen_2017_InnerProduct}, \cite{Chen_2019_flag_orbit}, \cite{ChanetIvanov_CohomRepParahoricGp}. In this section we study the representations of $\mathbb{G}(\mathcal{O}_r)$ via Deligne--Lusztig theory in two directions, one through the twisting operators and one through the Springer fibres.

\vspace{2mm} For $r=1$, the relations between twisting operators and Deligne--Lusztig theory have been investigated extensively. Here we give a start for general $r\geq 1$. We will work in the setting of \cite{Chen_2017_InnerProduct}, and try to extend the classical result asserting that Deligne--Lusztig character values are invariant under the twisting operator; when $r=1$, this was obtained by Asai \cite{Asai_twisting_op_classicalgroup} for classical groups with connected centre and by Digne--Michel \cite{DM_L-function_Book} for Lusztig inductions in good characteristic. To properly state the result we also need to extend a basic property of semisimple centralisers.

\vspace{2mm} Let $\mathbf{B}=\mathbf{T}\ltimes\mathbf{U}$ be a Borel subgroup of $\mathbb{G}\times_{\mathcal{O}}\overline{\mathbb{F}}_q[[\pi]]/\pi^r$, where $\mathbf{T}$ is a maximal torus and $\mathbf{U}$ is the unipotent radical. Let $B_r, U_r, T_r\subseteq G_r$ be the corresponding algebraic groups; similar to $G=G_r$, we may omit the subscript $r$ and denote them by $B,U,T$. We only concern the case that $T$ is $F$-stable (i.e.\ $FT=T$). Let $\ell$ be a prime not equal to $\mathrm{char}(\mathbb{F}_q)$, and fix a field isomorphism $\mathbb{C}\cong \overline{\mathbb{Q}}_{\ell}$; in particular, we can view $\mathrm{Sh}_G$ as an operator on the $\overline{\mathbb{Q}}_{\ell}$-space of class functions.

\vspace{2mm} For $b\in\mathbb{Z}_{[0,r]}$, write $U^{b,r-b}:=U^b(U^-)^{r-b}$, where $U^-$ denotes the algebraic group corresponding to the opposite of $\mathbf{U}$. Note that $G^F$ (resp.\ $T^F$) acts on $L^{-1}(FU^{b,r-b})$ by left (resp.\ right) translation.
\begin{defi}
For $\theta\in \mathrm{Hom}(T^F,\overline{\mathbb{Q}}_{\ell})$, the corresponding higher Deligne--Lusztig representation of $G^F$ at page $b$ is the virtual representation
$$R_{T,U,b}^{\theta}:=\sum_i(-1)^i H^i_c (L^{-1}(FU^{b,r-b}),\overline{\mathbb{Q}}_{\ell})_{\theta},$$
where $H_c^i(-,\overline{\mathbb{Q}}_{\ell})_{\theta}$ denotes the $\theta$-isotypical part of the compactly supported $i$-th $\ell$-adic cohomology group.
\end{defi}

\begin{remark}
When $b=0$ or $r$, $R_{T,U,b}^{\theta}$ is the representation constructed by Lusztig in \cite{Lusztig2004RepsFinRings}. When $r$ is even and $b=r/2$, $R_{T,U,b}^{\theta}$ is isomorphic to the representation constructed by G\'erardin in \cite{Gerardin1975SeriesDiscretes}. When $B$ is $F$-stable, $R_{T,U,b}^{\theta}\cong \mathrm{Ind}_{T^F(U^{b,r-b})^F}^{G^F}\theta$.
\end{remark}

Let $\mathfrak{R}_{T,U,b}^{\theta}$ be the character of $R_{T,U,b}^{\theta}$.

\begin{thm}\label{thm:main result}
Let $g=su\in G^F$, where $s$ is the semisimple part and $u$ is the unipotent part. Then $u\in Z_{G}(s)^{\circ}$, and if moreover $u\in Z_{Z_{G}(s)^{\circ}}(u)^{\circ}$, then
$$\mathrm{Sh}_G (\mathfrak{R}_{T,U,b}^{\theta})(g)=\mathfrak{R}_{T,U,b}^{\theta} (g).$$
\end{thm}
\begin{proof}
The first part is essentially proved in the arguments in \cite[P2031-2032]{Chen_2018_GreenFunction}; here we include the details for the completeness. Let $c\in G$ be such that $s^c\in T_1\subseteq G_1$, then it is sufficient to show that $u^c\in Z_{G}(s^c)^{\circ}$.

\vspace{2mm} Write $u^c=u_1u_2$ with respect to the decomposition $G=G_1G^1$. By applying the reduction $\rho_{r,1}$ to the identity
$$u_1u_2\cdot s^c=s^c\cdot u_1u_2$$
we get $u_1\in Z_{G_1}(s^c)$, which in turn implies that $u_2\in Z_{G^1}(s^c)$. As $u_1\in Z_{G_1}(s^c)^{\circ}$ (see e.g.\ \cite[2.5]{DM1991}), it is then sufficient to show that $u_2\in Z_{G^1}(s^c)^{\circ}$. Actually we would show that $Z_{G^1}(s^c)=Z_{G^1}(s^c)^{\circ}$.

\vspace{2mm} Write $u_2=tv$ with $t\in T^1$ and $v\in U^1(U^-)^1$; this is the Iwahori decomposition in the sense of \cite[2.2]{Sta2009Unramified}. As $t,s^c\in T$ commute with each other, we see that $v$ also commutes with $s^c$. Now consider the decomposition $v=\prod_{\alpha}v_{\alpha}$ with $v_{\alpha}\in (U_{\alpha})^1$, where $\alpha$ runs over the roots with respect to $T_1$. Then the commutation condition
$$(s^c)^{-1}\cdot v\cdot s^c=v$$
means that, for each $\alpha$, either $v_{\alpha}=1$ or $\alpha(s^c)=1$. Thus $Z_{G^1}(s^c)$ is an affine space, hence connected as desired.

\vspace{2mm} Now we turn to the second part. We shall start with a generalised Green function character formula, and then use a counting method following \cite[IV]{DM_L-function_Book}.

\vspace{2mm} By \cite[3.3]{Chen_2018_GreenFunction} we have:
\begin{equation}\label{formula:induction character formula}
\begin{split}
\mathfrak{R}_{T,U,b}^{\theta} (g)=&\frac{1}{|(Z_{G}(s)^{\circ})^F|} \\
& \cdot  \sum_{   \left\lbrace\substack{   h\in G^F\ \text{s.t.} \\  s\in {^h(T_1)^F}  } \right\rbrace   } \sum_{\tau\in {^h(T^1)^F} }         {\theta}(s^h\cdot \tau^h)\cdot   Q_{{^hT},{^hU}\cap Z_G(s)^{\circ},b}^{Z_{G}(s)^{\circ}}(u,\tau^{-1}),
\end{split}
\end{equation}
where
$$Q_{{^hT},{^hU}\cap Z_G(s)^{\circ},b}^{Z_{G}(s)^{\circ}}(-,-)\colon (\mathcal{U}_{Z_G(s)^{\circ}})^F\times {^h(T^1)^F}\longrightarrow\overline{\mathbb{Q}}_{\ell}$$
is the two-variable Green function in \cite[3.1]{Chen_2018_GreenFunction} (here $\mathcal{U}_{Z_G(s)^{\circ}}$ denotes the unipotent variety of $Z_G(s)^{\circ}$), which is defined by:
\begin{equation*}
(x,y)\longmapsto  \frac{1}{|T^F|}\mathrm{Tr}\left((x,y)  \mid \sum_{i}(-1)^i\cdot H_c^i(Z_G(s)^{\circ}\cap L^{-1}( {^h FU^{r-b,b} }),\overline{\mathbb{Q}}_{\ell})  \right).
\end{equation*}

\vspace{2mm} Since $s$ is contained in an $F$-stable maximal torus (see e.g.\ \cite[3.16]{DM1991}), by Lang--Steinberg theorem and the above first part of the theorem, there are $u'\in Z_{G}(s)^{\circ}$ and semisimple $s'\in Z_{G}(s)^{\circ}$ such that $u'^{-1}F(u')=u$ and $s'^{-1}F(s')=s$. Thus
\begin{equation*}
\begin{split}
n_{F}([g]_{G^F})= & n_{F}([s'^{-1}F(s')u'^{-1}F(u')]_{G^F})=n_{F}([u'^{-1}s'^{-1}F(s')F(u')]_{G^F})\\
= & [F(s')F(u')u'^{-1}s'^{-1}]_{G^F}\\
= & [F(s')s'^{-1}(F(u')u'^{-1})^{s'^{-1}}]_{G^F}\\
= & [s\cdot {^{s'}(F(u')u'^{-1})}]_{G^F}.
\end{split}
\end{equation*}
Note that $s\cdot {^{s'}(F(u')u'^{-1})}$ is a Jordan decomposition, so, evaluating \eqref{formula:induction character formula} with $n_F([g]_{G^F})$ we get
\begin{equation}\label{formula:induction character formula 2}
\begin{split}
& \mathrm{Sh}_G (\mathfrak{R}_{T,U,b}^{\theta})(g)\\
& =\frac{1}{|(Z_{G}(s)^{\circ})^F|} \cdot  \sum_{   \left\lbrace\substack{   h\in G^F\ \text{s.t.} \\  s\in {^h(T_1)^F}  } \right\rbrace   } \sum_{\tau\in {^h(T^1)^F} }         {\theta}(s^h\cdot \tau^h)\cdot   Q_{{^hT},{^hU}\cap Z_G(s)^{\circ},b}^{Z_{G}(s)^{\circ}}( {^{s'}(F(u')u'^{-1})},\tau^{-1}).
\end{split}
\end{equation}

\vspace{2mm} Comparing \eqref{formula:induction character formula} and \eqref{formula:induction character formula 2} it remains to show that: For a given $h\in G^F$ with $s\in {^h(T_1)}$, and $\tau \in {^h(T^1)^F}$, we have
\begin{equation}\label{formula:green}
Q_{{^hT},{^hU}\cap Z_G(s)^{\circ},b}^{Z_{G}(s)^{\circ}}(u,\tau^{-1})=Q_{{^hT},{^hU}\cap Z_G(s)^{\circ},b}^{Z_{G}(s)^{\circ}}({^{s'}(F(u')u'^{-1})},\tau^{-1}).
\end{equation}
Note that, by our assumption on $u$ (that is, $u\in Z_{Z_{G}(s)^{\circ}}(u)^{\circ}$) and Lemma~\ref{lemm:a fundamental lemma}, we have
$$F(u')u'^{-1}={^zu}$$
for some $z\in (Z_G(s)^{\circ})^F$, so ${^{s'}(F(u')u'^{-1})}={^{s''}u}$ where $s'':=s'z$. Hence \eqref{formula:green} can be written as
\begin{equation}\label{formula:green 2}
Q_{{^hT},{^hU}\cap Z_G(s)^{\circ},b}^{Z_{G}(s)^{\circ}}(u,\tau^{-1})=Q_{{^hT},{^hU}\cap Z_G(s)^{\circ},b}^{Z_{G}(s)^{\circ}}({^{s''}u},\tau^{-1}).
\end{equation}
Moreover, by definition $Q_{{^hT},{^hU}\cap Z_G(s)^{\circ},b}^{Z_{G}(s)^{\circ}}(-,-)$ can be viewed as the restriction of a class function on $(Z_G(s)^{\circ})^F\times {^h(T^1)^F}$. In particular, the value $Q_{{^hT},{^hU}\cap Z_G(s)^{\circ},b}^{Z_{G}(s)^{\circ}}({^{s''}u},\tau^{-1})$ is independent of the choice of $s'\in Z_G(s)^{\circ}$, as any two such $s'$'s are different only up to a left translation by some element in $(Z_G(s)^{\circ})^F$. Therefore, as $s\in {^h(T_1)}$, in the below we can assume that $s''\in {^h(T_1)}$. (So we would have ${\tau}=\tau^{s''}$ as both $s'',\tau$ are in the commutative group ${^hT}$.)

\vspace{2mm} By basic properties of Lefschetz numbers (see \cite[1.2]{Lusztig_whiteBk}), showing \eqref{formula:green 2} is equivalent to showing:
\begin{equation}\label{formula:counting points}
\begin{split}
\# \{ x\in Z_G(s)^{\circ}\cap &  L^{-1}( {^h FU^{r-b,b} })  \mid   {^{s''}u}\cdot F^m(x)\cdot \tau^{-1} =x  \} \\
& =\# \{ x\in Z_G(s)^{\circ}\cap L^{-1}( {^h FU^{r-b,b} })  \mid   {u}\cdot F^m(x)\cdot \tau^{-1}=x  \}
\end{split}
\end{equation}
for all $m\in \mathbb{Z}_{>0}$ with $F^m(U^{r-b,b})=U^{r-b,b}$.

\vspace{2mm} Taking the variable change $x\mapsto s''y$, the LHS of \eqref{formula:counting points} becomes
$$\# \{ y\in Z_G(s)^{\circ} \mid   sy^{-1}F(y)\in {^hFU^{r-b,b}} ,  u\cdot s''^{-1}F^m(s'')\cdot F^m(y)\cdot \tau^{-1} =y  \}. $$
(Recall that $s''^{-1}F(s'')=s$ by construction.) Note that
$$s''^{-1}F^m(s'')=s''^{-1}F(s'')\cdot F(s''^{-1})F^2(s'')\cdots F^{m-1}(s''^{-1})F^m(s'')=sF(s)\cdots F^{m-1}(s)=s^m.$$
So the LHS of \eqref{formula:counting points} equals to:
$$\# \{ y\in Z_G(s)^{\circ} \mid   sy^{-1}F(y)\in {^hFU^{r-b,b}} ,  s^mu F^m(y) \tau^{-1} =y  \}. $$
Similarly, via the variable change $x\mapsto ys''$ the RHS of \eqref{formula:counting points} equals to: (Recall that both $s''$ and $\tau$ are in the $F$-stable commutative group ${^hT}$)
\begin{equation*}
\begin{split}
\# \{ y\in Z_G(s)^{\circ} \mid  & s\cdot (y^{-1}F(y))^{F(s'')}\in {^hFU^{r-b,b}} ,  s^mu F^m(y) \tau^{-1} =y  \}\\
& =\# \{ y\in Z_G(s)^{\circ} \mid   s\cdot y^{-1}F(y)\in {^hFU^{r-b,b}} ,  s^mu F^m(y) \tau^{-1} =y  \},
\end{split}
\end{equation*}
where the equality follows from that $F(s'')\in {^hT}$ stabilises ${^hFU^{r-b,b}}$. Thus \eqref{formula:counting points} holds.
\end{proof}

In the statement of the above theorem, note that if $s=1$, then the condition $u\in Z_{Z_{G}(s)^{\circ}}(u)^{\circ}$ becomes $u\in Z_G(u)^{\circ}$, and by Lemma~\ref{lemm:a fundamental lemma} the twisting operator does not affect the character values at such unipotent elements for any character. So we want to ask:

\begin{quest}\label{quest:centraliser quest}
When a unipotent element of $G_r$ lies in the identity component of its centraliser?
\end{quest}

In the case $r=1$, this is well-known (to be always true) if $p$ is a good prime, which is a classical application of the Springer homeomorphism; see \cite[Proposition~5]{McNinch--Sommers_UnipCentGoodChar} and \cite[III.3.15]{Springer--Steinberg_1970_Conj}. For a general $r$ we list some situations in the below.

\begin{prop}\label{prop:unipotent centraliser prop}
Let $g=su\in G_r^F$ be as in Theorem~\ref{thm:main result}.
\begin{itemize}
\item[(1)] If $p$ is good for $G_1$ and $u\in G_1$, then $u\in Z_{G_r}(u)^{\circ}$.

\item[(2)] If $u\in G_r^{[\frac{r+1}{2}]}$, then $u\in Z_{G_r}(u)^{\circ}$. (Here $[-]$ denotes the Gauss floor function.)

\item[(3)] If (i) $g\in G_1$ and $p$ is good for $Z_{G_1}(s)^{\circ}$, or if (ii) $u\in G^{[\frac{r+1}{2}]}$, then the condition ``$u\in Z_{Z_{G}(s)^{\circ}}(u)^{\circ}$'' in Theorem~\ref{thm:main result} holds.

\item[(4)] If $\mathbb{G}=\mathrm{GL}_n$, then $Z_{G}(g)$ is always connected; in particular, $\mathrm{Sh}_G$ is the identity map for $\mathrm{GL}_n$.
\end{itemize}
\end{prop}
\begin{proof}
(1) By \cite[12]{McNinch--Sommers_UnipCentGoodChar} we know $u$ lies in $Z_{G_1}(u)^{\circ}$, which is a closed subgroup of $Z_{G}(u)^{\circ}$.

\vspace{2mm} (2) As $[\frac{r+1}{2}]\geq r/2$, the group $G^{[\frac{r+1}{2}]}$ is abelian, so $u\in G^{[\frac{r+1}{2}]}=Z_{G^{[\frac{r+1}{2}]}}(u)^{\circ}\subseteq Z_{G}(u)^{\circ}$.

\vspace{2mm} (3-(i)) When $g\in G_1$ and $p$ is good for $Z_{G_1}(s)^{\circ}$, we can apply \cite[12]{McNinch--Sommers_UnipCentGoodChar} to $Z_{G_1}(s)^{\circ}$, which asserts that $u$ lies in $Z_{Z_{G_1}(s)^{\circ}}(u)^{\circ}$, a closed subgroup of $Z_{Z_{G}(s)^{\circ}}(u)^{\circ}$.

\vspace{2mm} (3-(ii)) Now assume $u\in G^{[\frac{r+1}{2}]}$. Let $c\in G$ be such that $s^c\in G_1$. Then $(Z_{G}(s^c)^{\circ})^{[\frac{r+1}{2}]}=Z_{G}(s^c)^{\circ}\cap G^{[\frac{r+1}{2}]}$ is a connected unipotent group (see \cite[2.2]{Chen_2018_GreenFunction}), which is abelian as $[\frac{r+1}{2}]\geq r/2$. Therefore (by the first part of Theorem~\ref{thm:main result})
$$u^c\in (Z_{G}(s^c)^{\circ})^{[\frac{r+1}{2}]}=Z_{(Z_{G}(s^c)^{\circ})^{[\frac{r+1}{2}]}}(u^c)^{\circ}\subseteq Z_{Z_{G}(s^c)^{\circ}}(u^c)^{\circ},$$
thus $u\in Z_{Z_{G}(s)^{\circ}}(u)^{\circ}$.

\vspace{2mm} (4) Write an element in $M_n(\overline{\mathbb{F}}_q[[\pi]]/\pi^r)$ as $A_0+A_1\pi+...+A_{r-1}\pi^{r-1}$, then as observed in \cite[5.3]{Chen_2019_flag_orbit}, there is a ring injection from $M_n(\overline{\mathbb{F}}_q[[\pi]]/\pi^r)$ to ${M}_{nr}(\overline{\mathbb{F}}_q)$:
\begin{equation}\label{formula:injection}
A_0+A_1\pi+...+A_{r-1}\pi^{r-1}\longmapsto
\begin{bmatrix}
	A_0     & 0       & 0   & ... & 0   \\
	A_1     & A_0     & 0   & ... & 0   \\
	...     & ...     & ... & ... & ... \\
	A_{r-2} & ...     & A_1 & A_0 & 0   \\
	A_{r-1} & A_{r-2} & ... & A_1 & A_0
\end{bmatrix}.
\end{equation}
Thus, for $g\in\mathrm{GL}_n(\overline{\mathbb{F}}[[\pi]]/\pi^r)$, the condition ``$gx=xg$'' can be viewed as a system of linear equations in the linear space $\overline{\mathbb{F}}_q^{nr}$. So $Z_G(g)$ is an open subset of an affine space, thus connected, and $\mathrm{Sh}_G$ is the identity map by Lemma~\ref{lemm:a fundamental lemma} in this case.
\end{proof}

Now we turn to the other direction. In the remaining part of this section we take $\mathbb{G}$ to be $\mathrm{GL}_n$ and assume that $n,r\geq2$.

\vspace{2mm} The ring injection \eqref{formula:injection} in the above argument restricts to a group embedding from $G_r$ into $\widehat{G_r}:={\mathrm{GL}_{nr}}_{/\overline{\mathbb{F}}_q}$. This realises $G_r$ as a unipotent centraliser:

\begin{lemm}
The image of the above group embedding is the centraliser of the unipotent element
\begin{equation*}
u=\begin{bmatrix}
I_n & 0 &  0 & 0 & ... & 0\\
I_n & I_n & 0 & 0 & ... & 0\\
0 & I_n & I_n & 0 & ... & 0\\
... & ... & ...  & ... & ... & ... \\
0 & ... & 0  & I_n & I_n & 0 \\
0 & 0 & ... & 0 & I_n & I_n\\
\end{bmatrix},
\end{equation*}
whose Jordan normal form is $\mathrm{diag}(J_r,...,J_r)$, where $I_n$ denotes the identity matrix of size $n$ and $J_r$ denotes the unipotent Jordan block of size $r$. In particular, $G_r$ acts on the Springer fibre $\mathcal{B}_u$, the variety of Borel subgroups (of $\widehat{G_r}$) containing $u$, by conjugation.
\end{lemm}
\begin{proof}
This follows from direct computations.
\end{proof}

Now identify identify the flag variety $\mathcal{B}$ of $\widehat{G_r}$ to be the quotient of $\widehat{G_r}$ by the standard upper triangular Borel subgroup of $\widehat{G_r}$, and identify $S_{nr}$ with the Weyl group defined by the diagonal maximal torus inside this Borel. So we can talk about the Borel subgroups at relative positions indexed by $S_{nr}$ and view them as elements in $\mathcal{B}$. This indicates us a natural way to construct a variety out of $\mathcal{B}_u$, on which $G_r^F$, instead of the whole connected group $G_r$, acts:
\begin{defi}\label{defi:B_{u,w}}
Let $\mathcal{B}_{u,w}:=\mathcal{B}_u\cap X_w$, the intersection (in the flag variety $\mathcal{B}$) of $\mathcal{B}_u$ and $X_w$, where $X_w$ denotes the Deligne--Lusztig variety consisting of the Borel subgroups lying on the relative position $w\in S_{nr}$ with their Frobenius image.
\end{defi}
For a fixed $r\in\mathbb{Z}_{>1}$, a representation $R$ of $G_r^F$ is called primitive if $R$ does not factor through the lower level group $G_{r-1}^F$. It would be interesting to figure out for which $w$ the virtual representation
$$R_{u,w}:=\sum_i(-1)^iH_c^i(\mathcal{B}_{u,w},\overline{\mathbb{Q}}_{\ell})$$
of $G_r^F$ is primitive.

\vspace{2mm} Indeed, it can happen that $\mathcal{B}_{u,w}$ is actually empty; even if it is non-empty it can happen that $R_{u,w}$ is non-primitive. We determine these situations for the single cycles $(1,...,z)$:

\begin{thm}\label{thm:R_{u,w}}
Let $w=(1,...,z)\in S_{nr}$ with $z\in[1,nr]$. Then
\begin{itemize}
\item[(i.)] If $z=1$, then $R_{u,w}$ is a primitive permutation representation.
\item[(ii.)] If $1< z<n$, then $R_{u,w}$ is primitive.
\item[(iii.)] If $z=n$, then $R_{u,w}\neq0$ is non-primitive.
\item[(iv.)] If $n< z\leq nr$, then $\mathcal{B}_{u,w}=\emptyset$ (so $R_{u,w}=0$).
\end{itemize}
\end{thm}
\begin{proof}
In the below we would interpret a point in $\mathcal{B}$ as a complete flag; the translation of relative positions in terms of flags can be found in \cite[I.4.3]{AG4} (see also \cite{Webster_RelativePosition}). To make the argument coherent, we would treat (i.) first, then (iv.), (iii.), and finally (ii.).

\vspace{2mm}
From now on $\widehat{G_r}$ is understood as the automorphism group of the $\overline{\mathbb{F}}_q$-space $V$ with the standard basis
\begin{equation}\label{formula:basis}
\{ x_1^{(0)},x_2^{(0)},...,x_n^{(0)},x_1^{(1)},x_2^{(1)},...,x_n^{(r-1)} \}.
\end{equation}
We use $\mathcal{F}\colon V_1\subseteq V_2\subseteq...\subseteq V_{nr}=V$ with $\dim V_i=i$ as a general notation for a complete flag of $V$.

\vspace{2mm} (i.) In this case $\mathcal{B}_{u,w}$ is the finite set of $\mathbb{F}_q$-rational flags of $V$ stabilised by $u$. In particular $R_{u,w}=\overline{\mathbb{Q}}_{\ell}[\mathcal{B}_{u,w}]$ is a permutation representation. Consider the complete flag $\mathcal{F}\colon V_1\subseteq V_2\subseteq...\subseteq V_{nr}=V$ given by
\begin{equation*}
\begin{split}
V_1
& =\langle x_n^{(r-1)} \rangle \subseteq \langle x_n^{(r-1)}, x_{n-1}^{(r-1)} \rangle\subseteq...\subseteq \langle  x_n^{(r-1)}, x_{n-1}^{(r-1)},...,x_2^{(r-1)} \rangle\\
& \subseteq \langle x_n^{(r-1)}, x_{n-1}^{(r-1)},...,x_2^{(r-1)}, x_n^{(r-2)} \rangle \subseteq\langle x_n^{(r-1)}, x_{n-1}^{(r-1)},...,x_2^{(r-1)}, x_n^{(r-2)}, x_{n-1}^{(r-2)} \rangle\\
& \subseteq...\subseteq  \langle x_n^{(r-1)},...,x_2^{(r-1)}, x_n^{(r-2)},...,x_2^{(r-2)}  \rangle   \\
&\subseteq ...\subseteq \langle x_n^{(r-1)},...,x_2^{(r-1)},..., x_n^{(1)},...,x_2^{(1)} \rangle =V_{(n-1)(r-1)}  \\
& \subseteq \langle V_{(n-1)(r-1)}, x_n^{(0)}+x_1^{(r-1)} \rangle=V_{(n-1)(r-1)+1} \\
& \subseteq \langle V_{(n-1)(r-1)+1}, x_1^{(r-1)} \rangle \subseteq \langle V_{(n-1)(r-1)+1}, x_1^{(r-1)}, x_1^{(r-2)} \rangle \\
& \subseteq... \subseteq \langle V_{(n-1)(r-1)+1}, x_1^{(r-1)},...,x_1^{(1)} \rangle=V_{nr-n+1} \subseteq \langle V_{nr-n+1}, x_{n-1}^{(0)} \rangle \\
& \subseteq \langle V_{nr-n+1}, x_{n-1}^{(0)}, x_{n-2}^{(0)} \rangle \subseteq...\subseteq \langle V_{nr-n+1}, x_{n-1}^{(0)},..., x_{1}^{(0)} \rangle=V_{nr}.
\end{split}
\end{equation*}
Note that $\mathcal{F}\in \mathcal{B}_{u,w}$. Also note that, by construction every vector in the piece $V_{(n-1)(r-1)+1}$ has the property that its $n$-th coordinate equals to its $(n(r-1)+1)$-th coordinate. However, this is a property not preserved under the action of $(G^{r-1}_r)^F$. So $\mathcal{F}$ is an element of $\mathcal{B}_{u,w}$ not stabilised by $(G^{r-1}_r)^F$, which implies that $R_{u,w}$ is primitive.

\vspace{2mm}
(iv.) In this case $X_{w}$ is the moduli of complete flags $\mathcal{F}$ whose first $z$ pieces are of the form (see also \cite[2.2]{DL1976})
$$V_1\subseteq V_1+FV_1\subseteq V_1+FV_1+F^2V_1\subseteq \cdots \subseteq ... \subseteq \bigoplus_{j=0}^{z-1}F^jV_1=V_z\cong\overline{\mathbb{F}}_q^{z},$$
where $\dim V_1=1$, and $V_z$ together with all the other pieces are $\mathbb{F}_q$-rational. If this flag is contained in $\mathcal{B}_{u,w}$, then $V_1=\langle v_1 \rangle$ is stabilised by $u$, so we have $uv_1=v_1$ as $u$ is unipotent, which implies that $v_1$ is a vector of the form $(0,...,0,a_1,...,a_n)^T$. Note that the orbit of such a vector under $F$ can only generate a space of dimension at most $n$, smaller than $z$. Thus $\mathcal{B}_{u,w}=\emptyset$.

\vspace{2mm}
(iii.) In this case, similar to the above discussion we have that, any flag $\mathcal{F}$ in $\mathcal{B}_{u,w}$ must be of the following form: (a) The first $n$ pieces are
$$V_1\subseteq V_1+FV_1\subseteq V_1+FV_1+F^2V_1\subseteq \cdots \subseteq ... \subseteq \bigoplus_{j=0}^{n-1}F^jV_1=V_n\cong\overline{\mathbb{F}}_q^{n},$$
where $V_1$ is generated by a vector $v_1$ whose all but the last $n$ coordinates are zero, and (b) $V_n$ and other pieces of $\mathcal{F}$ are $F$-stable. In particular, the $n$-th piece is a fixed space (see the notation of \eqref{formula:basis})
$$V_n=\langle x_1^{(r-1)},x_2^{(r-1)},...,x_n^{(r-1)} \rangle.$$
Now, to form a $V_{n+1}$ (so that the flag lies in $X_{w}$), it suffices to choose an $F$-stable line in the space generated by $\{ x_1^{(0)},x_2^{(0)},...,x_n^{(0)},x_1^{(1)},x_2^{(1)},...,x_n^{(r-2)} \}$; by identifying a line in the vector space as a point in the projective space, this means to choose an $\mathbb{F}_q$-point of $\mathbb{P}^{n(r-1)-1}_{ \overline{\mathbb{F}}_q }$. Continuing this process we see that $\mathcal{B}_{u,w}$ is a non-empty union of irreducible components of $X_w$, and thus isomorphic to a union of components of
$$Y_z\times \prod_{j=1}^{n(r-1)-1} \mathbb{P}^{j}(\mathbb{F}_q),$$
where $Y_z$ is the Coxeter variety, i.e.\ the connected variety of all $v\in \mathbb{P}^{z-1}_{\overline{\mathbb{F}}_q}$ which do not lie on any $\mathbb{F}_q$-rational hyperplane. Note that by construction the subgroup $(G_r^{r-1})^F$ (given by the ``$A_{r-1}$-part'' in the image of the group embedding) does not permute these irreducible components, and in each component the action of $(G_r^{r-1})^F$ is trivial because it fixes every vector of $V_n$. So the subgroup $(G_r^{r-1})^F$ acts on $\mathcal{B}_{u,w}$ trivially, and thus the representations $H_c^i(\mathcal{B}_{u,w},\overline{\mathbb{Q}}_{\ell})$ of $G_r^F$ are all non-primitive. Since the alternating sum of cohomology groups of $Y_z$ is well-known to be non-zero, the vector space $R_{u,w}$ is non-zero, as desired.

\vspace{2mm}
(ii.) Similar to (iii.), in this case $B_{u,w}$ is a disjoint union of copies of $Y_z$, labelled by some $\mathbb{F}_q$-rational partial flags
$$V_z\subseteq V_{z+1}\subseteq...\subseteq V_{nr}$$
with $V_z\subseteq\langle x_1^{(r-1)},...,x_n^{(r-1)} \rangle$, and $(G^{r-1})^F$ fixes the vectors in $V_z$. Thus $R_{u,w}$ is actually a permutation representation of $(G^{r-1})^F$. So, as in the situation of (i.), it suffices to construct a partial flag of the above type on which $(G^{r-1})^F$ acts non-trivially: We can just truncate the flag constructed in (i.), because $z\leq n-1 < (n-1)(r-1)+1$.
\end{proof}

\section{$\mathrm{SL}_2$ over finite dual numbers}\label{sec:expl}

In this section we assume that $\mathbb{G}=\mathrm{SL}_2$, $r=2$, and $p=\mathrm{char}(\mathbb{F}_q)$ odd, and denote by $\varepsilon$ the image of $\pi$ in $\mathcal{O}_r$ (so $\varepsilon^2=0$). Note that $G\cong {\mathrm{SL}_2}_{/\overline{\mathbb{F}}_q}\ltimes \mathrm{Lie}(\mathrm{SL}_2)_{/\overline{\mathbb{F}}_q}$ via the adjoint action.

\vspace{2mm} The only possible degrees of irreducible characters of $G^F$ are (see e.g.\ \cite[Section~3]{Lusztig2004RepsFinRings}): $$1,q,q+1,\frac{q+1}{2},q-1,\frac{q-1}{2},q^2+q,q^2-q,\frac{q^2-1}{2}.$$
Here we only concern the last three ones, as the others are for non-primitive characters (i.e.\ the characters factor through the lower level group $G_1^F=\mathrm{SL}_2(\mathbb{F}_q)$). It is known that, for $\mathrm{SL}_2$, the orbits of primitive characters are always regular and their adjoint centralisers are abelian (see \cite[3.1]{Sta2011ExtendedDL}). In the below we describe how the twisting operator $\mathrm{Sh}_G$ affect these primitive characters; most properties of these characters that we need can be found in \cite{Lusztig2004RepsFinRings} and \cite{Sta2011ExtendedDL}, and we would omit the details when this is the case.

\vspace{2mm} From the work of Lusztig \cite{Lusztig2004RepsFinRings} we know that the degrees $q^2+q$ and $q^2-q$ are those afforded by $\mathfrak{R}_{T,U}^{\theta}$ ($:=\mathfrak{R}_{T,U,r}^{\theta}$) with the $\theta$'s satisfying certain generic condition; the orbits of these characters are regular semisimple. From the work of Stasinski \cite{Sta2011ExtendedDL} we know that most of the characters of degree $\frac{q^2-1}{2}$ are missed by any $\mathfrak{R}_{T,U}^{\theta}$; the orbits of these characters are regular nilpotent.

\vspace{2mm} {Case-I: $\deg=q^2\pm q$.} We want to show that the irreducible characters of these degrees are all invariant under $\mathrm{Sh}_G$. The centraliser of the semisimple part of an element in $G_1={\mathrm{SL}_2}_{/\overline{\mathbb{F}}_q}$ is either a torus or the whole group, so by Theorem~\ref{thm:main result}, Proposition~\ref{prop:unipotent centraliser prop}, and the above discussions, it suffices to show that, if
$A=\begin{bmatrix}
1 & u   \\
0 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
1+a\varepsilon & b\varepsilon   \\
c\varepsilon & 1-a\varepsilon
\end{bmatrix}$
with $u,a,b,c\in\overline{\mathbb{F}}_q$ and $u\neq0$, then $A\in Z_{G}(A)^{\circ}$. It is enough to show that, the subgroup (of $G$) consisting of the elements of the form
$X=\begin{bmatrix}
1 & \upsilon   \\
0 & 1
\end{bmatrix}
\cdot
\begin{bmatrix}
1+x\varepsilon & y\varepsilon   \\
z\varepsilon & 1-x\varepsilon
\end{bmatrix}$ ($\upsilon,x,y,z\in\overline{\mathbb{F}}_q$), that commuting with $A$,
is connected. Indeed, simplifying the equation
$$XA=AX$$
we get
$$z=\frac{c}{u}\upsilon,\ x=\frac{-c}{2u}\upsilon^2+\frac{cu+2a}{2u}\upsilon,$$
so the space of solutions is isomorphic to an affine space, thus connected.

\vspace{2mm} {Case-II: $\deg=\frac{q^2-1}{2}$.} As we mentioned before, these are nilpotent orbit characters. There are two non-trivial nilpotent orbits, with representatives of the form
$o_1=\begin{bmatrix}
0 & \square\in \mathbb{F}_q^{\times} \\
0 & 0
\end{bmatrix}$
and $o_2:=\begin{bmatrix}
0 & \slashed{\square}\in \mathbb{F}_q^{\times} \\
0 & 0
\end{bmatrix}$, respectively, in $\mathfrak{g}^F\cong (G^1)^F$. Note that the centralisers of such elements (in $G^F$) are
$$Z_i:=\left(\{ \pm1 \}\times
\begin{bmatrix}
1 & \mathbb{F}_q \\
0 & 1
\end{bmatrix}\right)\ltimes (G^1)^F\subseteq G^F.$$
Let $\psi_1$ (resp.\ $\psi_2$) be an extension (to $Z_i$) of a fixed irreducible character of $(G^1)^F$ corresponding to $o_1$ (resp.\ $o_2$), then according to Clifford theory (see e.g.\ \cite[6.17]{Isaacs_CharThy_Book}), the nilpotent primitive characters of $\mathrm{SL}_2(\mathbb{F}_q[\varepsilon])$ are precisely
$$R_{\chi,i}:=\mathrm{Ind}_{Z_i}^{G^F}\chi \psi_i\quad (i=1,2),$$
where $\chi$ runs over the irreducible characters of the left factor of $Z_i$. In particular, there are totally $2\times q+2\times q=4q$ such characters, as been indicated in \cite[3.1]{Lusztig2004RepsFinRings}. Now pick an element $X'=\begin{bmatrix}
-1 & x'\varepsilon\neq0 \\
0 & -1
\end{bmatrix}$. Note that the action of $n_F$ multiplies $x'\varepsilon$ in this matrix by a non-square element in $\mathbb{F}_q$. Since $\psi_i$ is non-trivial on exactly one of the orbits of $o_i$, we see that, via the formula of induced characters, the two class functions $\mathrm{Sh}_G(R_{\chi,i})$ and $R_{\chi,i}$ are always different at $X'$.

\vspace{2mm} Summarising Case-I and Case-II we get:

\begin{coro}\label{coro:semisimple of SL_2 are twist-invariant}
Let $R$ be a primitive irreducible character of $\mathrm{SL}_2(\mathbb{F}_q[\varepsilon])$, then $R$ is a semisimple orbit character if and only if $\mathrm{Sh}_G (R)=R$.
\end{coro}

We expect that this criterion can be extended to $\mathrm{SL}_n(\mathbb{F}_q[[\pi]]/\pi^r)$ for any $n,r\geq 2$, in good characteristics.

\bibliographystyle{alpha}
\bibliography{zchenrefs}

\end{document}

config: arxiv


TITLE: Optimization problem using calculus from Khan Academy
QUESTION [1 upvotes]: Consider the function  $f(x)=9−x^2$ for $f(x)≥0$  only.
Let  $T(x)$  equal the area of the shaded isosceles trapezoid that has two vertices on the  x-axis and two vertices on the graph of  $f$, as shown.
What is the maximum value of  $T(x)$?

Question was taken from Khan Academy
Steps I took:
I know that the area of a trapezoid is $\frac { a+b }{ 2 } \cdot h$
In this example, it seems that the height $h$, is $f(x)$ and the width of the top parallel side is $2x$
So I end up with $T(x)=\frac { 2x+b }{ 2 } (9-x^ 2)$
Now I know that I must find the equation of the area of the trapezoid, then find it's derivative, set it to $0$ to find the max and plug that max back into the original area equation. I just don't know how to get past the current step I am at.

REPLY [2 votes]: You have most of the steps. The area can be defined as:
$$A = \frac{2x + b}{2}(9-x^2)$$
Now we must find $b$. Note that we are looking for the maximum area, so choosing the x-intercepts of the functions, as you have indicated in your picture above, will give us the maximum possible area.
Since the zeros of the function are $x = -3, 3$, $b = 3 + 3 = 6$
Now your area is:
$$A = \frac{2x + 6}{2}(9-x^2)$$
Taking the derivative, we get:
$$A' = -3x^2 -6x  +9 = 0$$
Solving for $x$ we get:
$$-3(x+3)(x-1) = 0$$
Therefore, $x = -3, 1$
Testing the points, we find that $x=1$ yields the maximum value:
$$A = \frac{2 + 6}{2}(9-1) = 32$$


text:
\begin{document}

		\author{Christoph Alexander Weitkamp\footremember{ims}{\scriptsize Institute for Mathematical
				Stochastics, University of G\"ottingen,				Goldschmidtstra{\ss}e 7, 37077 G\"ottingen}
			\and
			Katharina Proksch \footremember{UT}{\scriptsize Faculty of Electrical Engineering, Mathematics \& Computer Science, University of Twente,  Hallenweg 19,  7522NH  Enschede}
			\and
			Carla Tameling \footrecall{ims}{}
			\and
			Axel Munk \footrecall{ims} \footnote{\scriptsize Max Planck Institute for Biophysical
				Chemistry, Am Fa{\ss}berg 11, 37077 G\"ottingen}}

\title{Gromov-Wasserstein Distance based Object Matching: Asymptotic Inference}

		\maketitle
\begin{abstract}
	In this paper, we aim to provide a statistical theory for object matching based on the Gromov-Wasserstein distance. To this end, we model general objects as metric measure spaces. Based on this, we propose a simple and efficiently computable asymptotic statistical test for pose invariant object discrimination. This is based on an empirical version of a $\beta$-trimmed lower bound of the Gromov-Wasserstein distance. We derive for $\beta\in[0,1/2)$ distributional limits of this test statistic. To this end, we introduce a novel $U$-type process indexed in $\beta$ and show its weak convergence. Finally, the theory developed is investigated in Monte Carlo simulations and applied to structural protein comparisons.
\end{abstract}

\keywords{Gromov-Wasserstein distance, metric measures spaces, U-processes, distributional limits, protein matching}

\keywordsMSC{Primary: 62E20, 62G20, 65C60 Secondary:  60E05 }

\input{Sections/Dod-Introduction}
\input{Sections/Dod-LimitDistribution}
\input{Sections/Dod-Bootstrap}
\input{Sections/Dod-Simulation}
\appendix

\section{Sketch of the Proofs}\label{appendix:proofs}
	\input{Sections/Dod-Proofs}
\section*{Acknowledgements}
We are grateful to F. M\'emoli, V. Liebscher and M. Wendler for interesting discussions and helpful comments and to C. Br\'echeteau for providing $\mathsf{R}$ code for the application of the DTM-test. Support by the DFG Research Training Group 2088 Project A1 and Cluster of Excellence MBExC 2067 is gratefully acknowledged.
\bibliographystyle{plainnat.bst}
\bibliography{Dod}
\end{document}


TITLE: What is the best way to solve $\lim_{n\to \infty}{(e^{i \theta})^n}$?
QUESTION [1 upvotes]: What is the best way to solve the limit:

$\lim_{n\to \infty}{(e^{i \theta})^n}$

$\theta$ is fixed, but you must have a care for cases $\ \theta > 0 , \ \theta = 0 , \ \theta < 0.$
There is an elegant and clean way to do? Suggestions?

REPLY [1 votes]: The limit does not exist in general.
Take $\theta=\pi$. We have, using $e^{i\theta}=\cos (\theta)+i\sin (\theta)$,
$$
\lim_{n \to \infty}(e^{i\theta})^{2n}=\lim_{n \to \infty}(e^{i\pi})^{2n}=\lim_{n \to \infty}(-1)^{2n}=1
$$ whereas
$$
\lim_{n \to \infty}(e^{i\theta})^{2n+1}=\lim_{n \to \infty}(e^{i\pi})^{2n+1}=\lim_{n \to \infty}(-1)^{2n+1}=-1.
$$
You have a similar case for $\theta=-\pi$.


text:
\subsection{Preliminaries}
\label{sectionPreliminaries}

The second step described in the introduction will use the
Toponogov Comparison Theorem \ref{LowerBoundTCT}. In particular, we use this result to provide bounds on the angles between the
vectors tangent to geodesics emanating from a point $x \in \sigma_M$
to $n$ (that is all but one) of the vertices of $\sigma_M$. These angle bounds are then
used to show that the reduced Gram
matrix associated with these vectors is
non-singular. Here we discuss the observations relating to (reduced)
Gram matrices and bounds on determinants that we will use.

\paragraph{Gram matrices}
\label{sectionGram}

Gram matrices can be applied
to general finite dimensional inner product spaces
($\mathbb{R}^n_G$, if the innerproduct $G=\delta_{ij}$, where $\delta_{ij}$ denotes the Kronecker delta, we shall write $\mathbb{R}^n_{\delta_{ij}} = \mathbb{E}^n$), such as the tangent
spaces ($T_x M$) of Riemannian manifolds with inner product $g(x)$. In this setting we
have
\begin{align} \det ( \langle w_i , w_j \rangle_G ) = \det (G) \det
(w_1 , \ldots , w_n )^2,   \nonumber
\end{align} where $w_1, \ldots, w_n \in \mathbb{R}^n_G$, and $(w_1 , \ldots , w_n )$ denotes the matrix with $w_i$ as columns. One can
already think of the vectors $w_i$ as the tangent vectors $w_i=v_i(x)=\exp_{x}^{-1} (v_i)$ discussed in the introduction. This can be
seen by taking the determinant of
\begin{align} (  \langle w_i, w_j \rangle_G ) =
(w_1, \ldots, w_n )^{t} (G) (w_1, \ldots, w_n ). \nonumber \end{align} This is an expression of the fact that the choise of the metric does not influence linear independence.
Because we will be interested
in the angles (as these feature in the Toponogov Comparison Theorem), we consider instead the Gram matrix associated with the
normalized (we assume that $w_i \neq 0$) vectors $w_i/|w_i|_G$:
\begin{align}
\det ( \cos \theta_{i,j} ) =&  \det \left( \frac{\langle w_i, w_j \rangle_G}{|w_i|_G |w_j|_G } \right) \nonumber\\
  = & \frac{\det((w_1 ,\ldots, w_n ) (G) (w_1, \ldots, w_n ))}{|w_1|_G^2 |w_2|_G^2 \ldots |w_n|_G^2 }, \label{ToConvexHull}
\end{align}
where $\theta_{i,j}$ denotes the angle between $w_i$ and $w_j$. The determinant
of $( \cos \theta_{i,j} )$ is zero if and only if the determinant of
$(w_1, \ldots ,w_n )$ is zero, i.e., if and only if $w_1, \ldots ,w_n$
are linearly dependent. We shall refer to the matrix of cosines of
angles as the
\defn{reduced Gram matrix}.

\paragraph{Bounds on determinants} \label{SectionFriedland}

The following result by Friedland \cite{Friedland}, see also Bhatia
and Friedland \cite{BhatiaFriedland}, Ipsen and Rehman \cite{Ipsen}
and Bhatia \cite{Bhatia} problem I.1.6, will be essential to some
estimates below:
\begin{align}
|det (A+ E ) -\det (A) | \leq n \max \{ \| A \|_p , \| A+ E \| _p \}
^{n-1} \| E \|_p \label{FriedlandBound}
\end{align}
where $A$ and $E$ are $n \times n$-matrices and $\| \cdot \| _p $ is the
$p$-norm, with $1\leq p\leq \infty$, for linear operators: \begin{align} \| A \|_p = \max_{x \in
\mathbb{R}^n } \frac{ | A x|_p }{|x|_p } , \nonumber \end{align}
with $| \cdot |_p $ the $p$-norm on $\mathbb{R}^n$. In our context
$A$ will be the reduced Gram matrix for the Euclidean case and
$E$ the matrix with the small angle deviations from the Euclidean case (or rather the deviations
of their cosines) due to the local geometry, of which each entry is
bounded by some~$\epsilon$.  \newline From \eqref{FriedlandBound} we
see that in this particular context \begin{align} |\det (A+E)| &\geq
|\det A| - n (\max \{ \| A \|_\infty , \| A+ E \| _\infty \} )^{n-1} \| E
\|_\infty \nonumber \\ &\geq |\det(A)| - n
\epsilon ,
\label{formulaFriedland}
\end{align}
where we use that every entry of $A$ and $A+E$ is bounded in absolute value by
$1$ because it they are reduced Gram matrices, a matrices of cosines.

\paragraph{Toponogov comparison theorems and spaces of constant curvature}
\label{subsectionToponogov}

We shall now first give the Toponogov Comparison Theorem and the
definitions which go with it. Here we follow Karcher
\cite{Karcher2}. Then we give some results comparing the cosine rules in small neighbourhoods in spaces of constant curvature to those of Euclidean space. As mentioned in the introduction we assume that we
work in neighbourhoods that lie within the convexity radius unless
mentioned otherwise.

We shall use the notation $\mathbb{H}^n(\curvlowbnd)$ for the space simply connected of dimension $n$ with constant sectional curvature $\curvlowbnd$.
\begin{de}
A \emph{geodesic triangle} $T$ in a Riemannian manifold consists of
three minimizing geodesics connecting three points, sometimes also referred to as vertices. We stress that a geodesic triangle does not include an interior. Assume lower curvature bounds
$\curvlowbnd \leq K$ (or upper bounds $K \leq \curvupbnd$). A triangle with
the same edge lengths as $T$ in $\mathbb{H}^n(\curvlowbnd)$ (or
$\mathbb{H}^n(\curvupbnd)$),
 is called an \emph{Alexandrov triangle} $T_{\curvlowbnd}
$ (or $T_{\curvupbnd}$) associated with T, named after Alexandrov who used these in his study of convex surfaces \cite{Karcher2}. Note that any two choices of $T_{\curvlowbnd}$ (or $T_{\curvupbnd}$) are equivalent due to the constant curvature of the space.
Two edges of a geodesic triangle and
the enclosed angle form a \emph{hinge}; a
\emph{Rauch hinge} in $\mathbb{H}^n(\curvlowbnd)$ (or
$\mathbb{H}^n(\curvupbnd)$) of a given hinge, consists of two geodesics
emanating from a single point with the same lengths and enclosed
angles as the original hinge. The edge closing the Rauch hinge in $\mathbb{H}^n(\curvlowbnd)$ (or
$\mathbb{H}^n(\curvupbnd)$), that is the minimizing geodesic connecting the two endpoint of the geodesics emanating from a single point with the same lengths and enclosed angles as the hinge in a space of arbitrary curvature, will be called the \emph{Rauch edge}. \label{GeodesicT}
\end{de}

\begin{figure}[!htb]
    \centerline{
\mbox{\includegraphics[height=2.2in]{pictures/ToponogovExample}}
    }
\caption{An ellipsoid with a geodesic triangle and the Alexandrov
triangles in the spaces of constant curvatures, in this case both
elliptic spheres.} \label{FigureToponogov1}
\end{figure}

\begin{figure}[!htb]
    \centerline{
\mbox{
\begin{tikzpicture}
  \coordinate [label={below left:$A$}] (A) at (0, 0);
  \coordinate [label={above left:$C$}] (C) at (0.5, 4);
  \coordinate [label={above right:$B$}] (B) at (4, 1.5);
	\draw[thick] (A) to [out=75,in=-80] (C);
 	\draw[thick] (A) to [out=25,in=190] (B);
	\draw[thick] (C) to [out=-50,in=150] (B);
	\draw[thin] (A) ++ (0.271892, 0.126785) arc (25:75:0.3);
	\draw[thin] (C) ++ (0.0520945,-0.295442) arc (-80:-50:0.3);
	\draw[thin] (B) ++ (-0.259808, 0.15) arc (150:190:0.3);
	\node at (0.25,2) {c};
	\node at (2,0.73) {b};
	\node at (2.25,2.75) {a};
  \node at (0.35, 0.35) {$\alpha$};
	\node at (3.4, 1.6) {$\beta$};
	\node at (0.7,3.5) {$\gamma$};
\end{tikzpicture}
}
    }
\caption{Triangle with the standard symbols for angles and lengths}
\label{HingeTriangleAnglePic}
\end{figure}

The Toponogov Comparison Theorem or Triangle Comparison Theorem reads
\begin{thm}[Toponogov Comparison Theorem]
Let $T$ be a
geodesic triangle in $M$ and assume that the sectional curvatures $K$ of $M$ satisfy the bounds
 $ \curvlowbnd \leq K \leq \curvupbnd$.
If $\curvupbnd>0$, assume also
that the triangle perimeter is less then $2 \pi \curvupbnd^{-1/2}$. Then
Alexandrov triangles  $T_{\curvlowbnd}$ and $T_{\curvupbnd}$ exist. Moreover, any angle $\alpha$ of $T$ satisfies
\begin{equation*}
    \alpha_{\curvlowbnd}  \leq \alpha \leq \alpha_{\curvupbnd},
\end{equation*}
where $\alpha_{\curvlowbnd}$ and $\alpha_{\curvupbnd}$ are the corresponding angles in $T_{\curvlowbnd}$ and $T_{\curvupbnd}$ respectively.
The length $c$ of the third edge closing a hinge is bounded in length by the lengths of the Rauch edges, $c_{\curvlowbnd}$ and $c_{\curvupbnd}$, closing the Rauch hinges:
\begin{align}
  c_{\curvlowbnd} \geq c \geq c_{\curvupbnd} . \nonumber
\end{align}
\label{UpperBoundTCT} \label{LowerBoundTCT}
\end{thm}

We also give the cosine rule which is of use in explicit
calculations involving the Toponogov comparison theorem. The cosine
rule for elliptic spaces, that is positive curved spaces of constant curvature, is
given in Section 18.6 of Berger \cite{BergerGeometryII} and Section
12.7 of Coxeter \cite{Coxeter} and reads
\begin{align} \cos \frac{a}{k} = \cos \frac{b}{k} \cos \frac{c}{k} + \sin \frac{b}{k} \sin \frac{c}{k} \cos \alpha,
\label{CosineRuleElliptic}
\end{align} where we assume that the Gaussian (in two dimensions) or sectional curvature $K$ satisfies
$K= 1/k^2$. The cosine
rule for hyperbolic spaces, that is negatively curved spaces of constant curvature, is given in Section 19.3 of Berger
\cite{BergerGeometryII} and Section 12.9 of Coxeter \cite{Coxeter}
to be  \begin{align} \cosh \frac{a}{k} = \cosh \frac{b}{k} \cosh
\frac{c}{k} - \sinh \frac{b}{k} \sinh \frac{c}{k} \cos \alpha,
\label{CosineRuleHyperbolic}
\end{align} where we assume that the Gaussian or sectional curvature is $K=-1/k^2$.

We now prove two lemmas for geodesic triangles in a space of constant curvature. In the following we use the notation
$(c^{\mathbb{E}})^2= a^2 + b^2 - 2 a b \cos \gamma $, to denote the length of the (Rauch) edge closing a hinge with lengths $a$ and $b$ and enclosed angle $\gamma$ in Euclidean space. In general we shall always use the index $\mathbb{E}$ to indicate (comparisons to) Euclidean space.
\begin{lem}
If $M$ is a space of constant curvature and two of the edge-lengths ($a$, $b$) of a hinge in $M$ satisfy
\begin{align}
a, b& \leq d_\textrm{max}/2
& d_\textrm{max}/k & < 1/2, \nonumber
\end{align}
here $d_\textrm{max}$ is some distance bound on the geodesics in the space of constant curvature,
then we have that the length of the (Rauch) edge of the hinge $c$, where
$c^2 = (c^{\mathbb{E}})^2  + E'$, and $|E'| \leq 5 d_\textrm{max}^4 /k^2$ a measure for the deviation from the Euclidean case.  \label{lemSpaceConstantCurvature1}
\end{lem}

\begin{proof}
Taylor's theorem implies that we have that
\begin{align} \sin(y) & = y (1+ \tilde{E}_s(y)) &
\cos (y) &=1-\frac{1}{2} y^2 + E_c(y) \nonumber \\
\sinh(y) & = y (1+ \tilde{E}_{sh}(y)) & \cosh (y) &=1+ \frac{1}{2}
y^2 + E_{ch}(y), \nonumber
\end{align} here $E$ stands for the error. These errors are bounded; if
we assume that $y\leq \phi_m <1$ we have that \begin{align}
|\tilde{E}_s(y)| &\leq \frac{1}{3!} \phi_m^2  \leq \frac{1}{3!} e
\phi_m^2 &
|E_c (y)| & \leq \frac{1}{4!} \phi_m^4 \leq \frac{1}{4!} e \phi_m^4 \nonumber \\
|\tilde{E}_{sh}(y)| & \leq \frac{1}{3!} e \phi_m^2 & |E_{ch} (y)|
&\leq \frac{1}{4!} e \phi_m^4 ,\nonumber \end{align} where $e$ is
Euler's number. It is convenient to use only the weaker bounds found
from the hyperbolic functions as this affords a universal approach. We therefore drop the subscript and write $E$ and $\tilde{E}$.

We define
\begin{align} \phi_1& =a/k & \phi_2&=b/k & \phi_3 &=c/k & \phi_m = d_{\textrm{max}}/k . \nonumber
\end{align}
Using these the cosine rules read
\begin{align} \cos   \phi_3 & = \cos \phi_1  \cos  \phi_2 + \sin \phi_1 \sin  \phi_2 \cos \gamma  & & \textrm{(elliptic)} \nonumber \\
\cosh   \phi_3 & = \cosh \phi_1  \cosh  \phi_2 - \sinh \phi_1 \sinh
\phi_2 \cos \gamma  & & \textrm{(hyperbolic)} .\nonumber
\end{align} We shall now bound $ \phi_3$, assuming $\phi_1 , \phi_2$ and $\gamma$ given.
Because $\phi_1, \phi_2 \leq \phi_m /2<1/4$ we have that
implies that $ \phi_3 \leq  \phi_m$ by the triangle inequality.
We find
\begin{align}\frac{1}{2}  \phi_3^2 + E ( \phi_3) =
\frac{1}{2} \phi_1^2 + \frac{1}{2}  \phi_2^2 + \frac{1}{4} \phi_1^2
 \phi_2^2 + E ( \phi_1) + E ( \phi_2)+ \frac{1}{2} E ( \phi_1)  \phi_2^2 + \frac{1}{2}
E ( \phi_2) \phi_1^2 \nonumber
\\+  E ( \phi_1) E ( \phi_2)- \phi_1  \phi_2 \cos \gamma -  \phi_1  \phi_2 \cos \gamma
(\tilde{E}( \phi_1)+ \tilde{E}( \phi_2)+ \tilde{E}( \phi_1)
\tilde{E}( \phi_2)) \nonumber
\end{align} and thus \begin{align}  \phi_3^2  = &
 \phi_1^2 +   \phi_2^2 - 2 \phi_1  \phi_2 \cos \gamma + \frac{1}{2}
 \phi_1^2  \phi_2^2 + 2 E ( \phi_1) + 2 E ( \phi_2)+  E ( \phi_1)  \phi_2^2 +  E ( \phi_2) \phi_1^2 \nonumber
\\&+  2 E ( \phi_1) E ( \phi_2) - 2  \phi_1  \phi_2 \cos \gamma
(\tilde{E}( \phi_1)+ \tilde{E}( \phi_2)+ \tilde{E}( \phi_1)
\tilde{E}( \phi_2))- E ( \phi_3) \nonumber \\ =& \phi_1^2 + \phi_2^2
- 2 \phi_1  \phi_2 \cos \gamma +E'( \phi_1, \phi_2, \phi_3)
\label{LengthApproxEucl1}
\end{align} with
\begin{align}| E'( \phi_1, \phi_2, \phi_3) | \leq & | \frac{1}{2}
 \phi_1^2  \phi_2^2 + 2 E ( \phi_1) + 2 E ( \phi_2)+  E ( \phi_1)  \phi_2^2 +  E ( \phi_2) \phi_1^2 \nonumber
\\& +  2 E ( \phi_1) E ( \phi_2) - 2  \phi_1  \phi_2 \cos \gamma
(\tilde{E}( \phi_1)+ \tilde{E}( \phi_2)+ \tilde{E}( \phi_1)
\tilde{E}( \phi_2))- E ( \phi_3)| \nonumber \\  \leq &  |
\frac{1}{2} \phi_1^2  \phi_2^2| + 2 |E ( \phi_1)| + 2 |E ( \phi_2)|+
|E ( \phi_1)|  + | E ( \phi_2) | \nonumber
\\& +  | E ( \phi_1)| +
2 | \phi_1  \phi_2| |\tilde{E}( \phi_1)|+ 2 | \phi_1  \phi_2|
|\tilde{E}( \phi_2)|+ 2| \phi_1  \phi_2| | \tilde{E}( \phi_2)|+| E (
\phi_3)| \nonumber
\\ = & | \frac{1}{2} \phi_1^2  \phi_2^2| + 4 |E ( \phi_1)| + 3 |E ( \phi_2)|  \nonumber
\\& +
2 | \phi_1  \phi_2| |\tilde{E}( \phi_1)|+ 2 | \phi_1  \phi_2|
|\tilde{E}( \phi_2)|+ 2| \phi_1  \phi_2| | \tilde{E}( \phi_2)|+| E ( \phi_3)|\nonumber \\
\leq  & \frac{1}{2} \phi_m^4 +\frac{7}{4! } e \phi_m^4 +
\frac{6}{3!} e \phi_m^4+ \frac{1}{4!} e \phi_m^4 \nonumber \\ \leq &
5 \phi_m^4 . \label{BoundErrorLength1}
\end{align}
\end{proof}

We study a geodesic triangle in a space of constant curvature for which we have some estimates on the edge-lengths. To be precise we assume that the edges of the geodesic triangle are themselves the closing (Rauch) edges of some hinges. Moreover, we assume that the conditions of the previous lemma are satisfied so that have the estimates on the deviation of the lengths of these closing edges compared to the value expected in Euclidean space. We provide to have similar bounds on the angles of the geodesic triangle for which we have estimates on the edge-lengths, that is we give bounds on the deviation of the angles of the geodesic triangle compared to the value expected in Euclidean space. To this end we define for some lengths $a^{\mathbb{E}}$, $b^{\mathbb{E}}$ and $c^{\mathbb{E}}$ the angle $\alpha^{\mathbb{E}}$ by
\begin{align} \cos \alpha^{\mathbb{E}} = \frac{(b ^{\mathbb{E}
})^2 + (c^{\mathbb{E}})^2 - (a^{\mathbb{E} })^2 }{ 2
b^{\mathbb{E} } c^{\mathbb{E} }}. \nonumber \end{align}

\begin{lem} \label{lemSpaceConstantCurvature2}
If $M$ is a space of constant sectional curvature and the edge-lengths ($a$, $b$, $c$) of a geodesic triangle in $M$ satisfy
\footnote{We have included the factor $1/k$ to shorten the calculation below}
\begin{align} (a/k)^2 &= (a^{\mathbb{E}}/ k)^2 + E_1, & (b/k)^2 &= (b^{\mathbb{E}}/k)^2 + E_2,& (c/k)^2 &= (c^{\mathbb{E}}/k)^2 + E_3 , \end{align}
with that
$E_1,E_2,E_3 \leq  5 d_\textrm{max}^4 /k^4$,
and
\begin{align} (d_\textrm{max} /k)^{3/2} <a^{\mathbb{E}}/k, b^{\mathbb{E}}/k, c^{\mathbb{E}}/k
< d_\textrm{max} /k < 1/2,
\label{LowerBoundLengths}
\end{align}
then
\begin{align} | \cos \alpha- \cos \alpha^{\mathbb{E}} |  \leq  80
d_\textrm{max}/k. \label{ResultAnglesAppendix} \end{align}
\end{lem}

\begin{proof}
To avoid having to drag along the $1/k$ we shall write $\psi_1= a/k$,  $\psi_1^{\mathbb{E}}= a^{\mathbb{E}}/k$\, et cetera.
Using the previous lemma we see that
$\psi_1$, $\psi_2$, $\psi_3$ satisfy \begin{align} (\psi_1) ^2 =
(\psi_2) ^2+(\psi_3) ^2- \psi_2 \psi_3 \cos \alpha +E_ 4 , \nonumber
\end{align} PErforming a calculation similar to the one in the previous lemma we find that
\begin{align}
| \cos \alpha - \cos \alpha^{\mathbb{E}} | &= \left | \frac{
(\psi^{\mathbb{E}}_2)^2+ (\psi^{\mathbb{E}}_3)^2 -
(\psi^{\mathbb{E}}_1)^2 + E_2+E_3 +E_4 -E_1} {2 \psi^{\mathbb{E}}_2
\psi^{\mathbb{E}}_3 \sqrt{1 + \frac{E_2}{ (\psi^{\mathbb{E}}_2)^2}
}\sqrt{1 + \frac{E_3}{ (\psi^{\mathbb{E}}_3)^2} } } - \frac{
(\psi^{\mathbb{E}}_2)^2+ (\psi^{\mathbb{E}}_3)^2 -
(\psi^{\mathbb{E}}_1)^2 } {2 \psi^{\mathbb{E}}_2 \psi^{\mathbb{E}}_3
 } \right| \nonumber \\ & \leq \left| \frac{
(\psi^{\mathbb{E}}_2)^2+ (\psi^{\mathbb{E}}_3)^2 -
(\psi^{\mathbb{E}}_1)^2 } {2 \psi^{\mathbb{E}}_2 \psi^{\mathbb{E}}_3
 }  \frac{4 E_2}{ (\psi_2^{ \mathbb{E}})^2} \right | + \left| \frac{
(\psi^{\mathbb{E}}_2)^2+ (\psi^{\mathbb{E}}_3)^2 -
(\psi^{\mathbb{E}}_1)^2 } {2 \psi^{\mathbb{E}}_2 \psi^{\mathbb{E}}_3
 }  \frac{4 E_3}{ (\psi_3^{ \mathbb{E}})^2} \right | \nonumber \\ & \phantom{\leq} + \left | \frac{
 E_2+E_3 +E_4 -E_1} {2 \psi^{\mathbb{E}}_2 \psi^{\mathbb{E}}_3 } \right| \left| \left (1 + \frac{2 E_2}{
 (\psi^{\mathbb{E}}_2)^2} \right )
 \left (1 + \frac{E_3}{ (\psi^{\mathbb{E}}_3)^2} \right) \right|  \nonumber \\
& \leq \frac{ 4 E_2}{(\psi_2^{\mathbb{E}})^2} + \frac{ 4
E_3}{(\psi_3^{\mathbb{E}})^2} + 40 \phi_m  \nonumber \\ & \leq  80
\phi_m . \nonumber \end{align}
\end{proof}


TITLE: Given a measure space $ \ ( \Omega, \Sigma, \mu) $ and a sequence $ A_n \in{} \Sigma $ Demonstrate that
QUESTION [1 upvotes]: Guys can someone help me with this demo?
Given a measure space $ \ ( \Omega , \Sigma , \mu ) $ and a sequence $A_n \in{} \Sigma $
Prove that: $ \mu
\ ( \displaystyle\cup{}_{n=1}^{ \infty }A_n ) \leq{} \displaystyle\sum_{n=1}^{\infty }
\mu \ ( A_n ) $
$\mu\left(\bigcup_{k=1}^n A_k\right)\leqslant \sum_{k=1}^n \mu(A_k)$ for all $n \in \mathbb{N}$. Then I can use the fact that $\mu $ is a measure so $\lim_{n\to\infty}\mu\left(\bigcup_{k=1}^n A_k\right )=\mu\left(\bigcup_{k=1}^{\infty }A_k\right)$. it is right? any way to demonstrate it better?

REPLY [3 votes]: Define inductively
$$ B_1 := A_1\quad\text{and}\quad B_n := A_n \setminus \bigcup _{k=1}^{n-1}A_k,\quad n\geqslant 2.$$
It holds that $\bigcup _{k=1}^{\infty} A_k = \bigcup _{k=1}^{\infty} B_k$. The $B_k$ are pairwise disjoint and $B_k\subseteq A_k$ for all $k$.
Consequently,
$$\mu \left ( \bigcup _{k=1}^\infty A_k \right ) = \mu \left ( \bigcup _{k=1}^\infty B_k \right ) = \sum _{k=1}^\infty \mu (B_k) \leqslant \sum _{k=1}^\infty \mu (A_k).$$

To make your approach work note that
$$ C_n :=\bigcup _{k=1}^n A_k \nearrow \bigcup _{k=1}^\infty A_k =: C. $$
Since $\mu$ is continuous from below we have $\mu (C_n) \to \mu (C)$. Hence, from finite subadditivity
$$ \mu \left ( \bigcup _{k=1}^\infty A_k \right ) = \lim _{n\to\infty} \mu \left ( \bigcup _{k=1}^n A_k \right ) \leqslant \lim _{n\to\infty} \sum _{k=1}^n \mu (A_k) = \sum _{k=1}^\infty \mu (A_k). $$