TITLE: Linear operator to zero power
QUESTION [0 upvotes]: When $T$ is any linear operator acting on a vector space $V$, and $n$ is a natural number, $T^n$ means $T$ applied $n$ times (composition) and that is also a linear operator. That is clear.
When $T$ is a nonzero linear operator acting on a vector space $V$, then $T^0$ is the identity operator $T^0 = I$. But I think that should also be true (true by definition), when $T$ is the zero operator i.e. the operator which sends all vectors to the zero vector.
Why? Because $T^0$ means that we are not applying any operator. So it makes sense to say: OK, all vectors stay unchanged when "applying" $T^0$ even when $T$ is the zero operator. I say "applying" because we're not actually applying anything.
Is that indeed so?
I am asking because this kind of disagrees with what we have for real numbers where $0^0$ is usually left undefined.
EDIT:
What's the context of this question? I was reading a proof for the uniqueness of the Jordan Normal Form. There this expression comes up $2d(\phi^p) - d(\phi^{p-1}) - d(\phi^{p+1})$, where $p$ is a positive integer, and $d$ is the defect of the linear operator in the brackets. The proof is very nice but convoluted and eventually it boils down to proving the uniqueness for a special linear operator which has only $0$ as a characteristic root (as an eigenvalue). So I had some doubts what happens exactly with the expression $\phi^{p-1}$ when $p = 1$, and if we need to put some restrictions on the linear operator $\phi$.

REPLY [4 votes]: We define $T^0=I$ for any linear operator $T:V \to V$ so that the usual laws of exponents hold for composition.  So, you should take this as a definition which is convenient and not anything extraordinarly deep.  Here's why.
For a positive integer $n$ we can define $T^n$ to mean $T \circ T \circ \cdots \circ T$ ($n$ factors), and $T^n$ is again a linear operator on $V$.  You can then check, thanks to associativity of function composition, that we have $T^n \circ T^m = T^{n+m}$ for positive integers $n, m$.  Doesn't that look nice and familiar?
If we extend this definition to include the possibility that $n=0$ by defining $T^0=I$, then the law $T^n \circ T^m = T^{n+m}$ now holds for integers $n,m \geq 0$.
After all, $T^n \circ T^0 = T^n \circ I = T^n$, so we have $T^n \circ T^0 = T^{n+0}$ and all is well.
Even better, if $T$ happens to be invertible we can define $T^{-n}$ for an integer $n \geq 0$ as either $(T^n)^{-1}$ or as $(T^{-1})^n$ as they are both the same, which is a great exercise.  Then the law of exponents
$$
T^n \circ T^m = T^{n+m}
$$
holds for all integers $n, m$.  Not only is this algebraically reminiscient of real number exponents in a nice way, this makes the applications of polynomial relations to linear operators incredibly powerful.
But, note that none of this has anything to do with whether or not $T$ is the zero transformation.  In that case, $T^0=I$ is still the case by definition.  (Note also that this is all about composition of functions and not any sort of real number multiplication/exponentiation, so issues related to $0^0$ in the reals are actually not relevant.)


TITLE: non existence of a continuous function with certain conditions
QUESTION [1 upvotes]: I think there is no a continuous map $f:\mathbb{Q} \rightarrow \mathbb{Q}$ such that
$x,y,z$ are distinct rationals and $f(\{z,y\}) = \{x\}$ and $f^{-1}\{x\} = \{y,z\}$ and $f$ is bijective on $\mathbb{Q} - \{y,z\}$.
My try: If it were one, then take the open basic set $x \in (a,b) = (a,x) \cup \{x\} \cup (x,b)$, then $f^{-1}(a,b)$ = open $\cup \{y,z\} \cup $ open = open $\cup \{y,z\}$ open cann't be open as they are pairewise disjoint. Am I right?

REPLY [0 votes]: To flesh out my hint at a modificaiton of a previous answer, let us suppose for the sake of simplicity at the moment that $x = 0$.  Divide $\mathbb{Q} \setminus \{ 0 \}$ into two sets $S,T$ consisting of those (nonzero) rational numbers whose denominator is even (odd) when put into lowest terms.  Further, let $S_-$ ($S_+$) consist of the positive (negative) elements of $S$, respectively.  And similarly define $T_-$, $T_+$.
Pick an irrational number $\gamma$ between $y$ and $z$ (say $y < \gamma < z$).  Let
$$\begin{align}
U_- &= \mathbb{Q} \cap ( -\infty , y ); & U_+ &= \mathbb{Q} \cap ( y , \gamma ) \\
V_- &= ( \gamma , z ); & V_+ &= \mathbb{Q} \cap ( z , + \infty ).
\end{align}$$
Note that each of the sets $S_-, S_+ , T_- , T_+$ as well as $U_-, U_+, V_-, V_+$ is a densely ordered subset of $\mathbb{Q}$ without maximum or minimum elements, which means that they are all order-isomorphic to $\mathbb{Q}$ (and therefore to each other).  We can then fix order isomorphisms (and hence homeomorphisms):
$$\begin{gather}
f_- : U_- \to S_- \\
f_+ : U_+ \to S_+ \\
g_- : V_- \to T_- \\
g_+ : V_+ \to T_+.
\end{gather}$$ Now define a function $h : \mathbb{Q} \to \mathbb{Q}$ by
$$h(q) = \begin{cases}
f_- ( q ), &\text{if }q \in U_- \\
0,         &\text{if }q = y \\
f_+ ( q ), &\text{if }q \in U_+ \\
g_- ( q ), &\text{if }q \in V_- \\
0,         &\text{if }q = z \\
g_+ ( q ), &\text{if }q \in V_+ \\
\end{cases}$$
It is fairly easy to see that $h$ is continuous, that $h(y) = h(z) = 0$ and that $h$ maps the set $\mathbb{Q} \setminus \{ y,z \}$ bijectively onto $\mathbb{Q} \setminus \{ 0 \}$.
(For arbitrary $x \in \mathbb{Q}$ you can just take $h^\prime (q) = h(q) + x$.)


TITLE: Conditional Probability [10]
QUESTION [0 upvotes]: I have been trying to solve this question for quite some time. The first part is trivial, however, I am having trouble with the second part. The question has to do with conditional probability, but I am not sure with summing up a product of probabilities.

REPLY [1 votes]: The numerator is $P(X>10, X+Y>40)$, where $X$ is the number of birds seen on Monday, and $Y$ is the number seen from Tuesday through Sunday. These are assumed independent. You can write out the numerator by specifying what value $X$ can take:
$$
\begin{align}
P(X>10, X+Y>40)&=\sum_{r=11}^\infty P(X=r, X+Y>40)\\
&=\sum_{r=11}^\infty P(X=r, Y>40-r)\\
&\stackrel{(1)}=\sum_{r=11}^\infty P(X=r)P(Y>40-r)\\
&\stackrel{(2)}=\sum_{r=11}^{40} P(X=r)P(Y>40-r)+\sum_{r=41}^\infty P(X=r)P(Y>40-r)
\end{align}
$$
We use independence in (1). In (2) we split up the sum at the boundary point $r=40$, because if $r>40$ then $40-r$ is negative, so $P(Y>40-r)=1$. This explains the final form that you see.


TITLE: How to show that order of an element is $2$ where the element is $ p-1$ for prime $p$.
QUESTION [2 upvotes]: Question is that we have an element $a$, where $a = p-1$ and $p$ is a prime number, and I need to prove that order of $a$ is $2$.
I know how do we calculate order of an element where we know exact number of elements in a group but I do not understand how do we prove it if we don't know the exact value of $p$.

REPLY [1 votes]: I am going to assume that you are in the group $\mathbb{Z}/p\mathbb{Z}$, and that $p$ is odd.
Now $(p-1)^2 = p^2 - 2p + 1 = 1 \pmod p$, so the order of $p-1$ is indeed $2$.

REPLY [1 votes]: There is a prime number.  We aren't going to be specific, so we are going to call it $p$.  We choose to work in the integers modulo $p$ with multiplication as the operation.  In the integers modulo $p$, there is an element $a$ congruent to $p-1$ modulo $p$.  We ask you to show that the order of $a$ is $2$.
To show that the order of an element, $x$, is $n$, we compute $x^n$ and determine whether that number is the identity of the group.  We also check $x^d$ for all divisors $d$ of $n$ to be certain that the order of $x$ is not actually smaller than $n$.
To determine whether $a$ has order $2$, we must compute
$$  a^2 = (p-1)^2 \pmod{p}  $$
and determine that the result is the identity in the chosen group.  We should still check the divisors of $2$.  (In this problem, this check ensures that we don't accidentally claim that the order of the identity is bigger than $1$ if it happens to be that $a$ is the identity.  Spoiler: this happens in this problem for only one choice of $p$.)

REPLY [1 votes]: I assume
$p \ne 2 \tag 0$
in accord with the comment of J. W. Tanner to the question itself.
In $\Bbb Z_p$, if
$a \equiv p - 1 \mod p, \tag 1$
then
$a^2 = p^2 - 2p + 1 \equiv 1 \mod p; \tag 2$
the order of $a$ is thus $2$ in $\Bbb Z_p$.
Note that (0) and (1) preclude
$a \equiv 1 \mod p. \tag 3$

REPLY [0 votes]: Hint:
$p-1\equiv -1\pmod p,$
so $(p-1)^2\equiv(-1)^2=1\pmod p$


TITLE: How do electrons in a lower energy level have more kinetic energy?
QUESTION [3 upvotes]: From what i've read, electrons in a lower energy level have more kinetic enery and momentum to stop "falling" into the positive nucleus. However, when an electron falls from a higher energy level to a lower one, the loss of energy is emitted as a photon. So how does it gain kinetic energy?
Thanks in advance

REPLY [3 votes]: I think the confusion comes from mixing up two terms: kinetic energy and energy.
Energy is a sum of kinetic energy and potential energy. Energy states, as is well-known, depend on the quantum number $n$ in ascending order: electrons farther away from the nucleus have greater energy. It is why when photons are emitted the electron transits to a closer-to-the-nucleus orbit.
But kinetic energy indeed is higher for the small-$n$ orbits. This is because they have huge negative potential energy, as they are close to the attracting nucleus, which is "compensated" by huge positive kinetic energy.
Btw, I think that the comment by @CountTo10 is incorrect as (s)he also confuses these terms (no offense!). Please correct me if I got something wrong.


TITLE: How would you find the roots to this question?
QUESTION [0 upvotes]: I have a homework problem that I arrived.

With Mathematica, the limit is 0. So by using $\epsilon= 10^{-6}$ (it is -6, not -0, sorry for the cutoff).
$\sin(n^2)/\sqrt{n} <\epsilon =10^{-6}$
So I tried putting that ino Mathematica and no luck so I have a feeling I am approaching this problem the wrong way

REPLY [1 votes]: Hint: $\frac{sin(n^{2})}{\sqrt{n}}\leq\frac{1}{\sqrt{n}}$ and the
latter is monotonic so it suffices to find an $N$ s.t $\frac{1}{\sqrt{N}}<\epsilon$


text:
{\chapter{Unitarizability in the Critical Case
(Corank 1 and 2)} \label{CC-corank12}

From now until \S\ref{basic 0} we fix $\rho\in\Cuspsd$ and $\sigma\in\Cuspcl$.
Let $\alpha=\alpha_{\rho,\sigma}\in\tfrac12 \Z_{\ge0}$ be such that
$$
[\alpha]^{(\rho)}\rtimes\s \quad (=\nu^\alpha\rho\rtimes\s )
$$
is reducible. Recall that we set $\eta=0$ if $\alpha\in\Z$ and $\eta=1$ otherwise so that $\alpha-\frac12\eta\in\Z_{\ge0}$.

For simplicity, we suppress the superscript $\rho$ from the notation. Thus, we write $[x]$ instead of $[x]^{(\rho)}$.

For any $k\ge0$ and $\mathbf{x}=(x_1,\dots,x_k)\in\R^k$, consider the induced representation
$$
\Pi_{\mathbf{x}}:=[x_1]\times\dots \times [x_k]\rtimes\s.
$$
We say that $\mathbf{x}$ is \emph{critical} if $\{[x_1],\dots,[x_k]\}$ forms a segment of
cuspidal representations (possibly with multiplicities) that contains the reducibility point $[\alpha]$.
The goal of this and the following chapters is to study the irreducible subquotients of
$\Pi_{\mathbf{x}}$ and to determine which ones are unitarizable in the critical case for $k\le3$.

Since the composition series of $\Pi_{\mathbf{x}}$ depends only on the orbit of $\mathbf{x}$ under signed permutations,
we may assume without loss of generality that $\mathbf{x}\in \R^k_{++}$, where
$$
\R_{++}^k:=\{\mathbf{x}\in\R^k: 0\leq x_1\leq\dots\leq x_k\},
$$
and $\mathbf{x}$ is critical. Later on, we shall also use the notation
$$
\R_{+}^k:=\{\mathbf{x}\in\R^k: 0\leq x_1,\dots, x_k\}.
$$

\section{Extreme cases} \label{St-oposite}
Suppose that the $x_i$'s are distinct.
We have two extreme cases in this setting. The first is when $x_1=\alpha$.
Then, $\Pi_{x_1,\dots,x_n}$ is a regular (multiplicity one) representation of length $2^n$.
All but two of its irreducible subquotients are non-unitarizable,
the exceptions being the generalized Steinberg representation $\delta([\alpha,\alpha+n-1];\s)$ and its dual.

Now we shall consider the opposite extreme, when
$$
x_n=\alpha \qquad \text{and} \qquad x_1>0.
$$
Once again, one easily sees that $\Pi_{x_1,\dots,x_n}$ is a regular representation of length $2^n$.
We claim that all irreducible subquotients are unitarizable.

It is also easy to describe the Langlands parameters of all the irreducible subquotients.
The tempered parts of Langlands parameters are precisely
$$
\delta([\nu^{\alpha-k+1}\rho],\dots,[ \nu^{\alpha}\rho];\s),\quad k=0,\dots,n.
$$
Then, one gets $\GL$-parts of Langlands parameters from partitions of the remaining exponents
$$
\nu^{\alpha-n+1}\rho, \nu^{\alpha-n+1}\rho,\dots, \nu^{\alpha-k}\rho.
$$

To see that all the irreducible subquotient are unitarizable we argue by induction on $n$. The case $n=0$ is trivial.
Suppose that $n>0$ and take any irreducible subquotient $\pi$ of $\Pi_{(x_1,\dots,x_n)}$.
Clearly, there exists an irreducible subquotient $\pi_-$ of $\Pi_{x_2,\dots,x_{n}}$ such that $\pi\leq [\alpha-n+1]\rtimes\pi_-$.
Observe that $x_2\geq \frac32$, and that by Jantzen decomposition, all the representations
$$
[x]\rtimes\pi_-, \quad 0\leq x<\alpha-1
$$
are irreducible (they are hermitian, since $\rho$ is $F'/F$-selfdual).
Since $\pi_-$ is unitarizable by the inductive hypothesis, we have complementary series, and $\pi$ is at the end of these complementary series.
Therefore, it is unitarizable.

\section{Tempered representations in critical case, corank \texorpdfstring{$\leq 3$}{le3}}

We first describe the irreducible tempered representations that are subquotients of
$$
\Pi_{(x_1,\dots,x_k)},\quad 0\leq x_1\leq \dots\leq x_k, \quad k\leq 3 .
$$

(It is easy to reduce the general case to the critical case.)

First we make a simple observation.

It is a direct consequence of the classification of square-integrable representations in \cite{MR1896238} (cf. \cite{MR2908042},
in particular \S 34) that if $\alpha\in\Z$ and $x_1>0$ or if $\alpha\notin\Z$ and $x_2>\frac12$
then every irreducible tempered subquotients of $\Pi_{\mathbf{x}}$ will be strongly positive (and in particular, square-integrable).

First we shall list all the square-integrable representations.
A direct consequence of \cite{MR1896238} is that if a square-integrable $\pi$ is a subquotient of $\Pi_{\mathbf{x}}$,
then $\mathbf{x}$ is critical.

The classification of square integral representations in \cite{MR1896238} easily implies the following

\begin{proposition} \label{prop: sqrbsc}
The following are the square-integrable representations in the cases $k\le3$.
\begin{enumerate}
\item $k=0:$ $\sigma$ itself.

\item $k=1:$ The representations
$$
\delta([\alpha];\s),\quad \alpha>0.
$$

\item $k=2:$

\begin{enumerate}
\item $\delta_{\spsi}([\alpha-1],[\alpha];\s)$, \quad $\alpha>1$.

\item $\delta([\alpha,\alpha+1];\s)$, \quad $\alpha>0$.

\item $\delta([0,1]_\pm;\s)$, \quad $\alpha=0$.
\end{enumerate}

\item $k=3:$

\begin{enumerate}
\item $\delta_{\spsi}([\alpha-2],[\alpha-1],[\alpha];\s)$, \quad $\alpha>2$.

\item $\delta_{\spsi}([\alpha-1],[\alpha,\alpha+1];\s)$, \quad $\alpha>1$.\footnote{The socle of
$[\alpha-1]\times\delta([\alpha,\alpha+1])\rtimes \s$.}

\item $\delta([\alpha,\alpha+2];\s)$, \quad $\alpha>0$.

\item $\delta([-\frac12,\frac32]_\pm;\s)$, \quad $\alpha=\frac12$.

\item $\delta([0,2]_\pm;\s)$, \quad $\alpha=0$.
\end{enumerate}
\end{enumerate}
\qed
\end{proposition}

Now we shall describe the non-square-integrable tempered subquotients of representations $\Pi_{(x_1,\dots,x_k)}$, $k\leq 3$,
in the critical case.

\begin{proposition} \label{prop: tempbsc}
Suppose that we are in the critical case
(i.e. $\{x_1,\dots,x_k\}$ is a $\Z$-segment in $\Z$ or $\frac12 +\Z$, with possible multiplicities, which contains $\alpha$).
The following are the tempered, but non square-integrable, irreducible subquotients of $\Pi_{\mathbf{x}}$ for $k\le3$.

\begin{enumerate}

\item $k=1:$
$$
\tau_\pm,\quad \text{for} \quad\alpha=0,
$$
where
$$
\rho\rtimes\s=\tau_+\oplus\tau_-.\footnote{We shall denote $\tau_\pm$ also by $\delta([0]_\pm;\s).$}
$$

\item $k=2:$

\begin{enumerate}
\item $\tau([0]_\pm;\delta([1];\s))$,\quad $\alpha=1$, where
\[
[0]\rtimes \delta([1];\s)=\tau([0]_+;\delta([1];\s))\oplus\tau([0]_-;\delta([1];\s)).\footnote{+ sign corresponds to the
representation with $\delta([0,1])\otimes\s$ in its Jacquet module.}
\]

\item $\delta([-\frac12,\frac12]_\pm;\s)$, \quad $\alpha=\frac12$, where
$$
\delta([-\tfrac12,\tfrac12])\rtimes\s=\delta([-\tfrac12,\tfrac12]_+;\s)\oplus
\delta([-\tfrac12,\tfrac12]_-;\s).\footnote{$\delta([-\tfrac12,\tfrac12]_+;\s)$
has $[\tfrac12]\times[\tfrac12]\otimes\s$ in its Jacquet module}
$$

\item $[0] \rtimes\tau_\pm$, \qquad $\alpha=0$, where
$$
\rho\rtimes\s=\tau_+\oplus\tau_-.
$$
\end{enumerate}

\item $k=3:$

\begin{enumerate}
\item $[0]\times[0]\rtimes\tau_\pm$, \qquad $\alpha=0$ \quad $([0]\rtimes\s=\tau_+\oplus\tau_-)$.

\item $[0]\rtimes\tau([0]_\pm;\delta([1];\s))$,\quad $\alpha=1$.

\item $\delta([-1,1]_\pm;\s )$, \qquad $\alpha=1$, where
$$
\delta([-1,1])\rtimes\s=\delta([-1,1] _+;\s )\oplus\delta([-1,1] _-;\s ).\footnote{$ \delta([-1,1] _+;\s )$ is characterised
by the property that it has $\delta([0,1])\times[1]\otimes\s$ in its Jacquet module}
$$

\item $\delta([-1,1]_\pm;\s )$, \qquad $\alpha=0$, where
$$
\delta([-1,1])\rtimes\s =\delta([-1,1] _+;\s )\oplus\delta([-1,1] _-;\s ).\footnote{$ \delta([-1,1] _\pm;\s )$ is characterised
by the property that it has $[1]\times[1]\otimes\tau_\pm$ in its Jacquet module}
$$

\item $[0]\rtimes \tau([0]_\pm;\delta([1];\s))$,\quad $\alpha=1$.

\item $\delta([-\frac12,\frac12])\rtimes \delta([\frac12];\s)$, \qquad $\alpha=\frac12$.

\item $ \tau([0]_\pm;\delta_{\spsi}([1],[2];\s))$, \qquad $\alpha=2$, where
$$
[0]\rtimes\delta_{\spsi}([1],[2];\s)=
\tau([0]_+;\delta_{\spsi}([1],[2];\s))\oplus\tau([0]_-;\delta_{\spsi}([1],[2];\s)).
$$

\item $\tau([0]_\pm;\delta([1,2] ;\s ))$, \qquad $\alpha=1$, where
$$
[0]\rtimes\delta([1,2] ;\s )=\tau([0]_+;\delta([1,2] ;\s ))\oplus
\tau([0]_-;\delta([1,2] ;\s)).\footnote{$\tau([0]_+;\delta_{\spsi}([1],[2];\s))$ is characterised
by the property that it embeds into a representation of the form $[1]\rtimes \lambda$ (see \cite{MR3123571}).}
$$

\item
$[0]\rtimes \delta([0,1]_\pm;\s)$, \qquad $\alpha=0.$
\end{enumerate}
\end{enumerate}
\qed
\end{proposition}

\begin{remark}
For the description of composition series of representations $\Pi_{\mathbf{x}}$, $k\leq3$ in the critical case,
we will use the following non-critical irreducible tempered representation
\begin{enumerate}
\item $k=1 :$

$[0]\rtimes \sigma$, \quad $\alpha \in \{1,2\}$ (for $\Pi_{(0,1,2)}$ as well as for
$\Pi_{(0,1,1)}$ and $\Pi_{(0,0,1)}$ if $\alpha=1$).

\item $k=2 :$

\begin{enumerate}

\item
$[0]\times[0]\rtimes \sigma$, \ \ \ \ \ \quad $\alpha=1$ (for $\Pi_{(0,0,1)}$).

\item $\delta([-\frac12,\frac12])\rtimes \sigma$, \qquad $\alpha=\frac32$ (for $\Pi_{(\frac12,\frac12,\frac32)}$).

\item $[0]\rtimes \delta([2];\sigma), \ \ \ \ \ \quad \alpha=2$ (for $\Pi_{(0,1,2)}$).

\end{enumerate}

\end{enumerate}

\end{remark}

With the above description it is easy to describe $\JH(\Pi_{\mathbf{x}})$ for any given $\mathbf{x}\in\R^3$.
Henceforth, we will do so in the critical case without further explanation.

\section{Composition series in critical case, corank one}
For the rest of this chapter we deal with the case $k\le2$ (and $\mathbf{x}$ critical).

We start with the corank one case.

\subsection{\texorpdfstring{$\mathbf{x}=(\alpha)$}{xalpha} and \texorpdfstring{$\alpha\ge\tfrac12$}{alphage12}}

\begin{proposition}
Suppose that $\alpha\geq\tfrac12$. Then,
\begin{enumerate}
\item In the Grothendieck group we have
$$
\Pi_{(\alpha)}=[\alpha]\rtimes \s=L([\alpha]; \s)+\delta([\alpha]; \s).
$$
\item Both $L([\alpha]; \s)$ and $\delta([\alpha]; \s)$ are unitarizable.
\item $L([\alpha]; \s)^t=\delta([\alpha]; \s)$.
\item We have
$$
\mu^*(\delta([\alpha]; \s))=1\otimes \delta([\alpha]; \s)+[\alpha]\otimes \s,\ \
\mu^*(L([\alpha]; \s))=1\otimes L([\alpha]; \s))+[-\alpha]\otimes \s.
$$
\end{enumerate}
\qed
\end{proposition}

Moreover, by Proposition \ref{JBcomputation}
\footnote{In the sequel we shall conclude Jordan blocks from this proposition
unless otherwise indicated.}
\begin{equation}
\label{eq: Jord-a-sigma}
\Jord_\rho(\delta([\alpha];\s))=\{\eta+1,\eta+3,\dots,2\alpha-3,2\alpha+1 \}.
\end{equation}

\subsection{\texorpdfstring{$\mathbf{x}=(\alpha)$}{xalpha} and \texorpdfstring{$\alpha=0$}{alpha0}}

In this case $[0]\rtimes\s$ is unitarizable and decomposes as
a sum of two irreducible tempered representations
\[
[0]\rtimes\s=\delta ([0]_+;\s)\oplus \delta ([0]_-;\s).
\]
We have
$$
\mu^*(\delta ([0]_\pm;\s))=1\otimes \delta ([0]_\pm;\s)+[0]\otimes \s.
$$
Note that
$$
\delta ([0]_+;\s)^t=\delta ([0]_-;\s).
$$

\section{Composition series in critical case, corank two}

\subsection{\texorpdfstring{$\mathbf{x}=(\alpha,\alpha+1)$}{x01} and \texorpdfstring{$\alpha\ge\tfrac12$}{alphage12}}
\label{Steinberg-place-corank-two}

\begin{proposition} \label{prop-2-a+}
Suppose that $\alpha\geq\tfrac12$. Then,
\begin{enumerate}
\item In the Grothendieck group we have
$$
\Pi_{(\alpha,\alpha+1)}=[\alpha+1] \times[\alpha]\rtimes \s=\pi_1+\pi_2+\pi_3+\pi_4
$$
where
\begin{gather*}
\pi_1= L([\alpha+1] ,[\alpha];\s),\
\pi_2=L([\alpha+1] ;\delta([\alpha];\s))\\
\pi_3=L([\alpha,\alpha+1];\s),\ \pi_4=\delta([\alpha,\alpha+1];\s).
\end{gather*}
\item The representations $\pi_1$ and $\pi_4$ are unitarizable and $\pi_1^t=\pi_4$.
\item The representations $\pi_2$ and $\pi_3$ are not unitarizable and $\pi_2^t=\pi_3$.
\end{enumerate}
\qed
\end{proposition}

We have
\begin{subequations}
\begin{gather}
\mu^*(\pi_1) =1\otimes\pi_1+[-\alpha-1]\otimes L([\alpha];\s)+L([-\alpha-1,-\alpha])\otimes\s.\\
\begin{aligned}
\mu^*(\pi_2)&=1\otimes\pi_2+
[-\alpha-1] \otimes \delta([\alpha];\s) +[\alpha]\otimes[\alpha+1] \rtimes \s
\\&+[\alpha]\times[-\alpha-1] \otimes\s+L([\alpha],[\alpha+1] )\otimes\s,
\end{aligned}\\\label{eq: pi312}
\begin{aligned}
\mu^*(\pi_3)&=1\otimes\pi_3+
[-\alpha]\otimes[\alpha+1] \rtimes\s+[\alpha+1] \otimes L([\alpha];\s)
\\&+\delta([-\alpha-1,-\alpha])\otimes\s+[-\alpha]\times[\alpha+1] \otimes\s,
\end{aligned}\\
\mu^*(\pi_4) =1\otimes\pi_4+[\alpha+1]\otimes\delta([\alpha];\s)+\delta([\alpha,\alpha+1])\otimes\s.
\end{gather}
\end{subequations}

One easily gets that
\begin{equation} \label{eq: jord01}
\Jord_\rho(\pi_4)=\{\eta+1,\eta+3,\dots,2\alpha-3,2\alpha+3 \}.
\end{equation}

Furthermore,
\[
\begin{gathered}
{}[\alpha+1] \rtimes L([\alpha];\s)=\pi_1+\pi_3,\\
[\alpha+1] \rtimes \delta([\alpha];\s)=\pi_2+\pi_4,\\
L([\alpha],[\alpha+1])\rtimes\s=\pi_1+\pi_2,\\
\delta([\alpha,\alpha+1])\rtimes\s=\pi_3+\pi_4.
\end{gathered}
\]

\subsection{\texorpdfstring{$\mathbf{x}=(\alpha,\alpha)$}{x00} and \texorpdfstring{$\alpha\geq 1$}{alphage1}}

\begin{proposition} \label{prop: alphalpha}
Suppose that $\alpha\ge1$. Then,
\begin{enumerate}
\item In the Grothendieck group we have
\[
\Pi_{(\alpha,\alpha)}=\pi_1+\pi_1^t
\]
where
\[
\pi_1=[\alpha] \rtimes \delta([\alpha] ;\s)=L([\alpha] ;\delta([\alpha] ;\s)),\ \
\pi_1^t=[\alpha] \rtimes L([\alpha] ;\s)=L([\alpha] ,[\alpha] ;\s).
\]
\item Neither $\pi_1$ nor $\pi_1^t$ is unitarizable.
\end{enumerate}
\end{proposition}

Clearly,
\[
\Pi_{(\alpha,\alpha)}=[\alpha] \rtimes \delta([\alpha] ;\s)+[\alpha] \rtimes L([\alpha] ;\s).
\]
Since $[\alpha] \rtimes \delta([\alpha] ;\s)$ and $[\alpha] \rtimes L([\alpha] ;\s)$
are irreducible by (\ref{Pr-red-si-item: a-a+1-irr}) of Proposition \ref{Pr-red-si} and \eqref{eq: Jord-a-sigma}, we obtain the first part.

To show that $\pi_1$ is non-unitarizable consider the family of representations
\[
\gamma_s=[s] \rtimes \delta([\alpha] ;\s),\ \ s\in\R.
\]
It is irreducible for $\alpha\le s<\alpha+1$ and $\gamma_{\alpha+1}$ admits a non-unitarizable irreducible quotient by
Proposition \ref{prop-2-a+}. Therefore, $\pi_1=\gamma_\alpha$ is not unitarizable. A similar argument applies to $\pi_1^t$.

\subsection{\texorpdfstring{$\mathbf{x}=(\alpha-1,\alpha)$}{x-10} and \texorpdfstring{$\alpha\geq \tfrac32$}{alphage32}}
\label{subsec-w-prop-2-a-}

\begin{proposition} \label{prop-2-a-}
Assume $\alpha\geq \tfrac32$.
Then,
\begin{enumerate}
\item In the Grothendieck group
\[
\Pi_{(\alpha-1,\alpha)}=\pi_1+\pi_2+\pi_3+\pi_4
\]
where
\begin{gather*}
\pi_1=L([\alpha-1,\alpha];\s),\ \ \pi_2=L([\alpha-1];\delta([\alpha];\s)),\\
\pi_3= L([\alpha-1],[\alpha];\s),\ \ \pi_4=\delta_{\spsi}([\alpha-1],[\alpha];\s).
\end{gather*}
\item $\pi_1,\pi_2,\pi_3,\pi_4$ are unitarizable.
\item $\pi_4^t=\pi_1,\ \ \pi_3^t=\pi_2$.
\end{enumerate}
\end{proposition}

The representation $\Pi_{(\alpha-1,\alpha)}$ is regular and admits the following two decompositions in the Grothendieck group
\[
\Pi_{(\alpha-1,\alpha)}=[\alpha]\times[\alpha-1]\rtimes \s=\Pi_1+\Pi_2=\Pi_3+\Pi_4
\]
where
\begin{gather*}
\Pi_1=[\alpha-1]\rtimes L([\alpha];\s),\ \ \Pi_2=[\alpha-1]\rtimes \delta([\alpha];\s),\\
\Pi_3=\delta([\alpha-1,\alpha])\rtimes\s,\ \ \Pi_4=L([\alpha-1],[\alpha])\rtimes\s.
\end{gather*}
Moreover, we have the decompositions
\begin{equation} \label{eq: dec345}
\Pi_3=\pi_1+\pi_2,\ \ \Pi_4=\pi_3+\pi_4.
\end{equation}
This further implies
\[
\Pi_1=\pi_1+\pi_3,\ \ \Pi_2=\pi_2+\pi_4
\]

Furthermore,
\begin{subequations}
\begin{gather*}
\mu^*(\pi_1)=1\otimes\pi_1+[-\alpha+1]\otimes L([\alpha];\s)
+\delta([-\alpha,-\alpha+1])\otimes\s,\\
\begin{aligned}
\mu^*(\pi_2)&=1\otimes\pi_2
+[\alpha]\otimes[\alpha-1]\rtimes\s+[-\alpha+1]\otimes \delta([\alpha];\s)
\\&+[-\alpha+1]\times[\alpha]\otimes\s +\delta([\alpha-1,\alpha])\otimes\s,
\end{aligned}\\
\begin{aligned}
\mu^*(\pi_3)&=1\otimes\pi_3
+[-\alpha]\otimes [\alpha-1]\rtimes\s +[\alpha-1]\otimes L([\alpha];\s)\\
&+[\alpha-1]\times[-\alpha]\otimes\s+L([-\alpha], [-\alpha+1])\otimes\s,
\end{aligned}\\
\mu^*(\pi_4)=1\otimes\pi_4+[\alpha-1]\otimes \delta([\alpha];\s)+
L([\alpha-1],[\alpha])\otimes\s.
\end{gather*}
\end{subequations}

We also note that
$$
\Jord_\rho(\pi_4)=\{\eta+1,\eta+3,\dots,2\alpha-5,2\alpha-1,2\alpha+1 \},
$$
and the partially defined function $\e$ pertaining to $\pi_4$ takes opposite values at $2\alpha-1$ and $2\alpha+1$
(use the fact that $\pi_4$ is strongly positive; see \cite{MR1896238}).

\subsection{\texorpdfstring{$\mathbf{x}=(0,1)$}{x01} and \texorpdfstring{$\alpha=1$}{alpha1}}

\begin{proposition}
Suppose that $\alpha=1$. Then,
\begin{enumerate}
\item In the Grothendieck group we have
\[
\Pi_{(0,1)}=\pi_1+\pi_2+\pi_3^++\pi_3^-
\]
where
\[
\pi_1=L([1];[0]\rtimes\s),\ \ \pi_2=L([0,1];\s),\ \ \pi_3^\pm=\tau([0]_\pm;\delta([1];\s)).
\]
\item $\pi_1,\pi_2,\pi_3^\pm$ are unitarizable.
\item $\pi_1^t=\pi_3^+$ and $\pi_2^t=\pi_3^-$.
\end{enumerate}
\end{proposition}

Note that
\[
\Pi_{(0,1)}=\Pi_1+\Pi_2
\]
where
\[
\Pi_1=[0]\rtimes L([1];\s),\ \ \Pi_2=[0]\rtimes \delta([1];\s)
\]
and
\[
\Pi_1=\pi_1+\pi_2,\ \ \Pi_2=\pi_3^++\pi_3^-.
\]
Clearly, $\Pi_1$ and $\Pi_2$ are unitarizable.

Moreover,
\begin{equation} \label{eq: 0111}
\delta([0,1])\rtimes\s=\pi_2+\pi_3^+,\ \ L([0],[1])\rtimes\s=\pi_1+\pi_3^-.
\end{equation}

Finally,
\begin{subequations}
\begin{gather}
\begin{aligned}
\mu^*(\pi_1)&=1\otimes \pi_1
+[-1]\otimes[0]\rtimes\s+[0]\otimes L([1];\s)\\
&+2L([-1],[0])\otimes\s +\delta([-1,0])\otimes\s,
\end{aligned}\\
\mu^*(\pi_2)=1\otimes \pi_2+[0]\otimes L([1];\s)+\delta([-1,0])\otimes\s\\
\begin{aligned} \label{01-1-delta+}
\mu^*(\pi_3^+)&=1\otimes\pi_3^++
[1]\otimes[0]\rtimes\s+[0]\otimes\delta([1];\s)
\\&+2\delta([0,1])\otimes\s +L([0],[1])\otimes\s,
\end{aligned}\\\label{eq: 013-}
\mu^*(\pi_3^-)=1\otimes\pi_3^-+[0]\otimes\delta([1];\s)+L([0],[1])\otimes\s.
\end{gather}
\end{subequations}

\subsection{\texorpdfstring{$\mathbf{x}=(\tfrac12,\tfrac12)$}{x1212} and \texorpdfstring{$\alpha=\tfrac12$}{alpha12}}

\begin{proposition} \label{prop: 1212}
Suppose $\alpha=\frac12$. Then,
\begin{enumerate}
\item In the Grothendieck group we have
\[
\Pi_{(\frac12,\frac12)}=\pi_1+\pi_2+\pi_3+\pi_4
\]
where
\begin{gather*}
\pi_1=\delta([-\tfrac12,\tfrac12]_+;\s),\ \ \pi_2=\delta([-\tfrac12,\tfrac12]_-;\s),\\
\pi_3=L([\tfrac12],[\tfrac12];\s),\ \ \pi_4=L([\tfrac12];\delta([\tfrac12];\s)).
\end{gather*}
\item $\pi_1,\pi_2,\pi_3,\pi_4$ are unitarizable.
\item $\pi_1^t=\pi_3$, $\pi_2^t=\pi_4$.
\end{enumerate}
\end{proposition}

We have
\[
\delta([-\tfrac12,\tfrac12])\rtimes\s=\pi_1+\pi_2,\ \
L([-\tfrac12],[\tfrac12])\rtimes\s=\pi_3+\pi_4
\]
and also
\[
[\tfrac12]\rtimes \delta([\tfrac12];\s)=\pi_1+\pi_4,\ \
[\tfrac12]\rtimes L([\tfrac12];\s)=\pi_2+\pi_3.
\]
This implies that $\pi_1^t=\pi_3$, $\pi_2^t=\pi_4$.

Finally,
\begin{subequations}
\begin{gather*}
\begin{aligned}
\mu^*(\pi_1)&=
1\otimes\pi_1+
[\tfrac12]\otimes [\tfrac12]\rtimes \s+
[\tfrac12]\otimes \delta([\tfrac12];\s)\\
&+\delta([-\tfrac12,\tfrac12])\otimes\s+
[\tfrac12]\times[\tfrac12]\otimes\s,
\end{aligned}\\
\mu^*(\pi_2)=
1\otimes\pi_2+
[\tfrac12]\otimes L([\tfrac12];\s)+
\delta([-\tfrac12,\tfrac12]])\otimes\s,\\
\begin{aligned}
\mu^*(\pi_3)&=1\otimes\pi_3+
[-\tfrac12]\otimes [\tfrac12]\rtimes \s
+[-\tfrac12]\otimes L([\tfrac12];\s)
\\&+L([-\tfrac12],[\tfrac12]])\otimes\s+
[-\tfrac12]\times[-\tfrac12]\otimes\s,
\end{aligned}\\
\mu^*(\pi_4)=1\otimes\pi_4+[-\tfrac12]\otimes \delta([\tfrac12];\s)+
L([-\tfrac12],[\tfrac12]])\otimes\s.
\end{gather*}
\end{subequations}

\subsection{\texorpdfstring{$\mathbf{x}=(0,1)$}{x01} and \texorpdfstring{$\alpha=0$}{alpha0}}

\begin{proposition} \label{prop: 001}
Suppose that $\alpha=0$. Then,
\begin{enumerate}
\item
 \label{prop: 001-item: Gg}
 In the Grothendieck group we have
\[
\Pi_{(0,1)}=\pi_1^++\pi_1^-+2\pi_2+\pi_3^++\pi_3^-,
\]
where
$$
\pi_1^\pm=L([1];\delta ([0]_\pm;\s)),\quad
\pi_2= L([0,1];\s),\quad
\pi_3^\pm=\delta([0,1]_\pm;\s).
$$
\item All the irreducible subquotients of $\Pi_{(0,1)}$ are unitarizable.
\item
 \label{prop: 001-item: inv}
We have
\[
(\pi_1^\pm)^t=\pi_3^\mp,\quad \pi_2^t=\pi_2.
\]
\item
\label{prop: 001-item: 01ds}
$
\Jord_\rho(\pi_3^\pm)=\{1,3\}.
$
\end{enumerate}
\end{proposition}

\begin{proof}
By Propositions \ref{prop: sqrbsc} and \ref{prop: tempbsc} it
follow that $\{\pi_1^+,\pi_1^-,\pi_2,\pi_3^+,\pi_3^-\}$ is the Jordan-H\"older series of $\Pi_{(0,1)}$. Note that
\begin{equation} \label{0-2-decomp}
\Pi_{(0,1)}=\Pi_1+\Pi_2=\Pi_3^++\Pi_3^-,
\end{equation}
where
$$
\Pi_1=\delta([0,1])\rtimes\sigma, \ \ \Pi_2=L([0],[1])\rtimes\sigma, \ \ \Pi_3^\pm=[1]\rtimes \delta([0]_\pm;\sigma).
$$
Observe that \eqref{BPLC} implies that $\pi_3^\pm$ and $\pi_2$ are the only possible subquotients of $\Pi_1$. Since $\Pi_1^t=\Pi_2$
and $\Pi_{(0,1)}$ has length $\geq 5$,
we conclude that all three representations are subquotients, which further implies that the Jordan-H\"older series of $\Pi_2$
consists of 3 irreducible representations.
Since each $\pi_3^\pm$ has in its Jacquet module $\delta([0,1])\otimes\sigma$, and the multiplicity of it in the Jacquet module of
$\Pi_{(0,1)}$ is two, we conclude that the multiplicity of each $\pi_3^\pm$ in $\Pi_{(0,1)}$ is one. Therefore,
$
\Pi_1=\pi_3^++\pi_3^-+\pi_2.
$
Both $\pi_1^\pm$ have multiplicity one in $\Pi_{(0,1)}$, which implies
$
\Pi_2=\pi_1^++\pi_1^-+\pi_2.
$
This implies (\ref{prop: 001-item: Gg}) (see \cite[\S5]{MR1212952} for the case of $Sp(4)$).

Multiplicity one of $\pi_1^\pm$ and $\pi_3^\pm$ in $\Pi_{(0,1)}$, together with \eqref{0-2-decomp}, imply $\pi_2^t=\pi_2$.
Each of $\pi_3^\pm$ is a subquotient of $\Pi_3^\pm$ (by definition of $\pi_3^\pm$). Also $\pi_1^\pm$ is a subquotient. This implies
$
\Pi_3^\pm=\pi_1^\pm+\pi_2+\pi_3^\pm.
$
Observe that $(\pi_1^\pm)^t\ne\pi_1^\pm$ and $(\pi_1^\pm)^t\ne\pi_1^\mp$ (since $\Pi_1^t=\Pi_2$), as well as $(\pi_1^\pm)^t\ne\pi_2$
(since $\pi_2^t=\pi_2$). Now $(\Pi_3^\pm)^t=\Pi_3^\mp$ implies $(\pi_1^\pm)^t=\pi_3^\mp$. This completes the proof of (\ref{prop: 001-item: inv}).

All the irreducible subquotients of $\Pi_{(0,1)}$ are unitarizable, since they are subquotients of ends of the complementary series
starting with $\delta([-\frac12,\frac12])\rtimes \sigma$ or $L([-\frac12],[\frac12])\rtimes \sigma$.

Proposition \ref{JBcomputation} implies (\ref{prop: 001-item: 01ds}). The proof is now complete.
\end{proof}

Observe that we have proved above the following equalities (which we shall use later)
\begin{equation}
\label{01-0-half}
\Pi_3^\pm=\pi_1^\pm+\pi_2+\pi_3^\pm,\quad
\Pi_1=\pi_3^++\pi_3^-+\pi_2,\quad
\Pi_2=\pi_1^++\pi_1^-+\pi_2.
\end{equation}
Further, we directly get
\begin{subequations}
\begin{gather}
\label{l1} \mu^*(\pi_3^\pm)= 1\otimes \pi_3^\pm
+[1]\otimes\delta([0]_\pm;\s)
+\delta([0,1])\otimes\s.\\
\label{l2} \mu^*(\pi_1^\pm)=1\otimes \pi_1^\pm+
[-1]\otimes \delta([0]_\pm;\s)+L([-1],[0])\otimes\s.\\
\label{l3} \mu^*(\pi_2)=1\otimes \pi_2+[0]\otimes[1]\rtimes\s+\delta([-1,0])\otimes\s+L([0],[1])\otimes\s.
\end{gather}
\end{subequations}

\subsection{\texorpdfstring{$\mathbf{x}=(0,0)$}{x00} and \texorpdfstring{$\alpha=0$}{alpha=0}}

Here $\Pi_{(0,0)}$ is a unitarizable tempered representation of length two which decomposes as
$$
\Pi_{(0,0)}[0]=[0]\rtimes\delta ([0]_+;\s)\oplus [0]\rtimes\delta ([0]_-;\s).
$$
We have
$$
([0]\rtimes\delta ([0]_+;\s))^t=[0]\rtimes\delta ([0]_-;\s).
$$

\endinput

config: arxiv


TITLE: A functor that “transforms terminal objects in terminal objects”?
QUESTION [1 upvotes]: In Lenstra’s notes on Galois theory for schmes, he defines a Galois category as a category satisfying a variety of properties (page 33) together with a functor $F$.
One of these properties is that

“The functor $F$ transforms terminal objects in terminal objects and ... transforms epimorphisms in epimorphisms.”

The usage of the word “in” is throwing me off. Does this just mean that if $A$ is a terminal object in the category $\mathcal{C}$ that the object $F(A)$ is also terminal?

REPLY [2 votes]: I am pretty certain your guess is right. The terminology is indeed a bit nonstandard (from a category-theoretic point of view). Usually we would say the functor "preserves terminal objects".
The reason I think you are right is because the precise wording in the document is:

The functor $F$ transforms terminal objects in terminal objects and commutes with
  fibred products.

Then this statement is equivalent to saying that $F$ preserves all finite limits (the more common category-theoretic term for "fibred products" is "pullbacks"), something that is common to ask of a functor. Such a functor is also called left exact, on the nLab page you can find more on these left exact functors. Including the claim about the equivalence I just mentioned (Proposition 3.2 on that page, at the time of writing).


text:
\begin{document}

\title[Polarized or int-amplified Endomorphisms]

{Normal projective varieties admitting polarized or int-amplified endomorphisms
}

\author{Sheng Meng and De-Qi Zhang}

\address
{
\textsc{Department of Mathematics} \endgraf
\textsc{National University of Singapore,
Singapore 119076, Republic of Singapore
}}
\email{math1103@outlook.com}
\email{ms@u.nus.edu}
\address
{
\textsc{Department of Mathematics} \endgraf
\textsc{National University of Singapore,
Singapore 119076, Republic of Singapore
}}
\email{matzdq@nus.edu.sg}

\begin{abstract}
Let $X$ be a normal projective variety admitting a polarized or int-amplified endomorphism $f$.
We list up characteristic properties of such an endomorphism and classify
such a variety from the aspects of its singularity,
anti-canonical divisor and Kodaira dimension.
Then we run the equivariant minimal model program with respect to not just the single $f$ but also
the monoid $\SEnd(X)$
of all surjective endomorphisms of $X$,
up to finite-index.
Several applications are given.
We also give both algebraic and geometric characterizations of toric varieties via polarized endomorphisms.
\end{abstract}

\subjclass[2010]{
14E30,

32H50,

08A35,
14M25.

}

\keywords{polarized endomorphism, amplified endomorphism, iteration, equivariant MMP, $Q$-abelian variety, toric variety}

\maketitle

\tableofcontents

\section{Introduction}
In this note, we report our recent results in studying surjective endomorphisms, especially polarized endomorphisms and int-amplified endomorphisms of higher dimensional algebraic varieties in arbitrary characteristic. The main focus is in setting up the equivariant minimal model program (MMP) for such endomorphisms. We will outline the ideas but refer to the original papers for the detailed proofs.
Our approach is more geometric.

For simplicity of the presentation,
{\it we mainly work over an algebraically closed field $k$ of characteristic $0$},
except the last section where the cases of positive characteristic are discussed.

In Section \ref{sec-2}, we introduce various notions, like polarized endomorphisms, (int-) amplified endomorphisms, etc., and state the important properties enjoyed by these endomorphisms.

In Section \ref{sec-3}, we characterize normal projective varieties $X$ admitting amplified endomorphisms from the aspects of their singularities, canonical divisors, and Kodaira dimensions.
These are rough descriptions which will be used later for the theory of the equivariant MMP on $X$.

In Section \ref{sec-4}, we show the equivariance of the MMP for a single polarized or int-amplified endomorphism.

In Section \ref{sec-5}, we show the finiteness of the number of contractible extremal rays for a projective variety $X$ admitting a polarized or int-amplified endomorphism.
This way, we confirm the equivariance of the MMP on $X$ for
a finite-index submonoid of the
monoid $\SEnd(X)$ of all surjective endomorphisms of $X$.

In Section \ref{sec-6}, we give the characterizations of toric varieties in terms of the existence of  almost homogeneity or polarized endomorphisms equipped with an extra ramification condition.

In Section \ref{sec-7}, we list several results which have appeared in the previous sections and also hold true in the case of positive characteristic. Of course, we need to assume some extra natural conditions, like separability of the map, etc.

We refer the readers to the survey paper \cite{Zhsw} which has more number-theoretic flavours and includes many outstanding conjectures.

\par \vskip 1pc
{\bf Acknowledgement.}
The second named author would like to thank the organizing committee for the kind invitation and
warm hospitality during the
International conference : Nevanlinna theory and Complex Geometry in Honor of Le Van Thiem's Centenary,
February - March, 2018, Hanoi, Vietnam.
Both authors would like to thank the referee for the very careful reading and the suggestions to improve and clarify the paper.
The first named author is supported by a Research Assistantship of NUS.
The second named author is supported by an Academic Research Fund of NUS.

\section{Polarized or int-amplified endomorphisms}
\label{sec-2}

Let $X$ be a projective variety.
A Cartier divisor is assumed to be integral, unless otherwise indicated.
Denote by $\Pic(X)$ the group of Cartier divisors modulo linear equivalence and $\Pic^0(X)$ the subgroup of the classes in $\Pic(X)$ which are algebraically equivalent to $0$.
Then
$$\NS(X) := \Pic(X)/\Pic^0(X)$$ is the N\'eron-Severi group.
Denote by
$$\N^1(X):=\NS(X) \otimes_{\Z} \mathbb{R}$$
and
$$\NS_{\mathbb{K}}(X):=\NS(X)\otimes_{\mathbb{Z}} \mathbb{K}$$ for $\mathbb{K} := \mathbb{Q}$, $\mathbb{R}$, or $\C$. So $\NS_{\R}(X) = \N^1(X)$.

Let $n:=\dim(X)$.
We can regard $\N^1(X)$ as the space of numerically equivalent classes of $\R$-Cartier divisors.
Two $\R$-Cartier divisors $D_1$ and $D_2$ are {\it numerically equivalent}, denoted as
$$D_1 \equiv D_2$$ if
their classes $[D_1]$ and $[D_2]$ in $\N^1(X)$ are the same, i.e., if
$$(D_1 - D_2) \cdot C = 0$$ for any curve $C$ on $X$.
Denote by
$$\N_r(X)$$ the space of weakly numerically equivalent classes of $r$-cycles with $\R$-coefficients.
Namely, two $r$-cylces $Z_1$ and $Z_2$ are {\it weakly numerically equivalent},

if
the intersection
$${(Z_1 - Z_2)}\cdot L_1 \cdots L_{n-r} = 0$$ for all Cartier divisors $L_i$ (cf.~\cite[Definition 2.2]{MZ}).
When $X$ is normal, we also call $\N_{n-1}(X)$ the space of weakly numerically equivalent classes of Weil $\R$-divisors.
In this case, $\N^1(X)$ can be regarded as a subspace of $\N_{n-1}(X)$
(cf.~\cite[Lemma 3.2]{Zh-TAMS}).

We recall the definitions of the following cones:
\begin{itemize}
\item $\Amp(X)$ is the cone of ample classes in $\N^1(X)$.
\item $\Nef(X)$ is the cone of nef classes in $\N^1(X)$.
\item $\PE^1(X)$ is the cone of pseudo-effective $\R$-Cartier divisor classes in $\N^1(X)$.
\item $\PE_{n-1}(X)$ is the cone of pseudo-effective Weil $\R$-divisor classes in $\N_{n-1}(X)$.
\end{itemize}
A Weil $\R$-divisor $B$ is said to be {\it big} if $B$ is in the interior part of $\PE_{n-1}(X)$.

Let $f:X\to X$ be a surjective endomorphism.
Then $f$ is a finite morphism of degree $d$.
We may define pullback of cycles for $f$, such that $f^\ast$ induces an automorphism of $\N_r(X)$ and $f_\ast f^\ast=d\,\id$; see \cite[Section 2.3]{Zh-comp}.
Note that the above cones are $f^\ast$-invariant.

We refer to \cite[\S2]{MZ} for more information.

\begin{definition}\label{def-f}
Let $f:X\to X$ be a surjective endomorphism of a projective variety $X$.
\begin{itemize}
\item[(1)]
$f$ is {\it numerically polarized}  if $f^{\ast}L \equiv qL$ for some ample Cartier divisor $L$ and integer $q>1$.
\item[(2)] $f$ is {\it numerically quasi-polarized}  if $f^{\ast}L \equiv qL$ for some big Cartier divisor $L$ and integer $q>1$.
\item[(3)] $f$ is {\it quasi-polarized} if $f^{\ast}L \sim qL$ for some big Cartier divisor $L$ and integer $q>1$.
\item[(4)] $f$ is $q$-{\it polarized}  if $f^{\ast}L \sim qL$ for some ample Cartier divisor $L$ and integer $q>1$.
\item[(5)] $f$ is {\it int-amplified} if $f^*L-L=H$ for some ample Cartier divisors $L$ and $H$.
\item[(6)] $f$ is {\it amplified} if $f^*L-L=H$ for some (not necessarily ample) Cartier divisor $L$ and some ample Cartier divisor $H$.
\end{itemize}
\end{definition}

For convenience, sometimes we simply say $f$ is polarized if it is $q$-polarized for some integer $q>1$.
It is easy to see that ``polarized'' implies ``int-amplified'' and ``int-amplified'' implies ``amplified''.
The notion of an amplified endomorphism $f : X \to X$ was first introduced by Krieger and Reschke (cf.~\cite{Kr}); preceding to this, Fakhruddin \cite[Theorem 5.1]{Fak} has shown that for such $f$, the set of $f$-periodic points is Zarisiki dense in $X$.
There do exist amplified {\it automorphisms} (e.g., automorphisms of positive entropy on abelian surfaces), while the degree of an int-amplified endomorphism is always greater than $1$ (cf.~\cite[Lemma 3.10]{Meng}).

Unlike the polarized property (cf.~\cite[Corollary 3.12]{MZ}),
in general, it is impossible to preserve the amplified automorphism property via a birational equivariant lifting (cf.~\cite[Lemma 4.4]{Kr} and \cite[Theorem 1.2]{Re}).

On one hand, we do not know whether the amplified property can be descended via an equivariant morphism (cf.~\cite[Question 1.10]{Kr}).
On the other hand, int-amplified endomorphisms have all the nice properties enjoyed by the polarized endomorphisms (cf.~\cite[\S 3]{MZ}).

Next we focus on showing the equivalence of the first four polarizations in Definition \ref{def-f}.
The following is a norm criterion for a numerically polarized endomorphism.
\begin{proposition}\label{prop-fx-x}(cf.~\cite[Proposition 2.9]{MZ}, \cite[Proposition 3.1]{CMZ})
Let $\varphi:V\to V$ be an invertible linear map of a positive dimensional real vector space equipped with a norm.
Assume
$$\varphi(C)=C$$ for a convex cone $C\subseteq V$ such that $C$ spans $V$ and its closure $\overline{C}$ contains no line.
Let $q$ be a positive number. Then the conditions (i) and (ii) below are equivalent.
\begin{itemize}
\item[(i)] $\varphi(u)=q u$ for some $u\in C^\circ$ (the interior part of $C$).
\item[(ii)]
There exists a constant $N>0$, such that
$$\frac{||\varphi^i||}{q^i}< N$$ for all $i\in \mathbb{Z}$.
\end{itemize}
Assume further the equivalent conditions (i) and (ii). Then the following are true.
\begin{itemize}
\item[(1)] $\varphi$ is a diagonalizable linear map with all eigenvalues of modulus $q$.
\item[(2)] Suppose $q>1$. Then, for any $v\in V$ such that $\varphi(v)-v\in C$, we have $v\in C$.
\end{itemize}
\end{proposition}

Applying the above criterion to the cones $\Nef(X)$ and $\PE^1(X)$, we now can say the equivalence of ``numerically quasi-polarized'' and ``numerically polarized''; see \cite[Proposition 3.6]{MZ}.
Furthermore, applying \cite[Lemma 2.3]{Na-Zh}, ``numerically polarized'' is equivalent to ``polarized'' by taking $H$ there to be ample.

\begin{theorem}\label{thm-num-lin}(cf.~\cite[Proposition 1.1]{MZ})
Let $f:X\to X$ be a numerically quasi-polarized endomorphism of a projective variety $X$. Then $f$ is polarized.
\end{theorem}

The following are useful criteria for int-amplified endomorphisms.

\begin{proposition}\label{prop-3-equiv}(cf.~\cite[Proposition 3.3]{Meng}) Let $f:X\to X$ be a surjective endomorphism of a projective variety $X$.
Then the following are equivalent.
\begin{itemize}
\item[(1)] The endomorphism $f$ is int-amplified.
\item[(2)] All the eigenvalues of
$$\varphi:=f^*|_{\N^1(X)}$$ are of modulus greater than $1$.
\item[(3)] There exists some big $\R$-Cartier divisor $B$ such that $f^*B-B$ is big.
\item[(4)] If $C$ is a $\varphi$-invariant convex cone in $\N^1(X)$,
then
$$\emptyset\neq(\varphi-\id_{\N^1(X)})^{-1}(C)\subseteq C .$$
\end{itemize}
\end{proposition}

Considering the action $f^*|_{\N_{n-1}(X)}$ and the cone $\PE_{n-1}(X)$, we have similar criteria as follows.

\begin{proposition}\label{prop-4-equiv}(cf.~\cite[Proposition 3.4]{Meng}) Let $f:X\to X$ be a surjective endomorphism of an $n$-dimensional normal projective variety $X$.
Then the following are equivalent.
\begin{itemize}
\item[(1)] The endomorphism $f$ is int-amplified.
\item[(2)] All the eigenvalues of
$$\varphi:=f^*|_{\N_{n-1}(X)}$$ are of modulus greater than $1$.
\item[(3)] There exists some big Weil $\R$-divisor $B$ such that $f^*B-B$ is a big Weil $\R$-divisor.
\item[(4)] If $C$ is a $\varphi$-invariant convex cone in $\N_{n-1}(X)$,
then
$$\emptyset\neq(\varphi-\id_{\N_{n-1}(X)})^{-1}(C)\subseteq C .$$
\end{itemize}
\end{proposition}

By the above criteria, we may easily show that the properties of ``numerically polarized'' and ``int-amplified'' are preserved via every equivariant descending.

\begin{proposition}\label{lem-int-des1}(cf.~\cite[Theorem 3.11]{MZ}, \cite[Lemmas 3.5, 3.6]{Meng}) Let $\pi:X\dasharrow Y$ be a dominant rational map of projective varieties.
Let $f:X\to X$ and $g:Y\to Y$ be two surjective endomorphisms such that
$$g\circ\pi=\pi\circ f .$$
Suppose $f$ is numerically polarized (resp.~int-amplified). Then $g$ is also numerically polarized (resp.~int-amplified).
\end{proposition}

\section{Singularities, anti-canonical divisor and Kodaira dimension}\label{sec-3}

We refer to \cite[Chapters 2 and 5]{KM} for the definitions and the properties of log canonical (lc), Kawamata log terminal (klt), canonical and terminal singularities.

Let $f:X\to X$ be a surjective endomorphism of a normal projective variety $X$.
When $\dim(X)=2$, Wahl \cite[Theorem 2.8]{Wa} showed that
$X$ has at worst lc singularities.
Broustet and H\"oring \cite[Corollary 1.5]{BH} generalized this result to higher dimensions by adding extra assumptions that $f$ is polarized and $X$ is $\Q$-Gorenstein.
We may further weaken the extra assumption ``$f$ is polarized" to the assumption ``$f$ is int-amplified."

\begin{theorem}\label{main-thm-lc}(cf.~\cite[Theorem 1.6]{Meng})
Let $X$ be a $\Q$-Gorenstein normal projective variety over the field $k$ of characteristic $0$ admitting an int-amplified endomorphism.
Then $X$ has at worst lc singularities.
\end{theorem}

When $f$ is polarized and $X$ is smooth, Boucksom, de Fernex and Favre \cite[Theorem C]{BFF} showed that $-K_X$ is pseudo-effective.
Cascini, Meng and Zhang \cite[Theorem 1.1 and Remark 3.2]{CMZ} used a different method to
show further that $-K_X$ is weakly numerically equivalent to an effective Weil $\Q$-divisor without the assumption of $X$ being smooth.
By applying Propositions \ref{prop-3-equiv} and \ref{prop-4-equiv} and the ramification divisor formula, we have the following result for the int-amplified case.

\begin{theorem}\label{main-thm-k}(cf.~\cite[Theorem 1.5]{Meng}) Let $X$ be a normal projective variety admitting an int-amplified endomorphism.
Then $-K_X$ is weakly numerically equivalent to some effective Weil $\Q$-divisor.
If $X$ is further assumed to be $\Q$-Gorenstein, then $-K_X$ is numerically equivalent to some effective $\Q$-Cartier divisor.
\end{theorem}

Let $f:X\to X$ be an amplified endomorphism of a projective variety $X$.
By taking the equivariant Iitaka fibration (cf.~\cite[Theorem A]{NZ09}), we obtain the following constraint on the Kodaira dimension of $X$.

\begin{theorem}\label{lem-amp-kod}(cf.~\cite[Lemma 2.5]{Meng})
Let $f:X\to X$ be an amplified endomorphism of a projective variety $X$.
Then the Kodaira dimension $\kappa(X)\le 0$.
\end{theorem}

\section{Equivariant MMP with respect to an endomorphism}\label{sec-4}

This section generalises and extends results in \cite{Zh-comp} to higher dimensions.

Let
$$(*):X_1\dasharrow X_2\dasharrow\cdots\dasharrow X_r$$
be a finite sequence of dominant rational maps of projective varieties.
Let $f:X_1\to X_1$ be a surjective endomorphism.
We say the sequence $(*)$ is {\it $f$-equivariant} if the following diagram is commutative
$$\xymatrix{
X_1\ar@{-->}[r]\ar[d]^{f_1} &X_2\ar@{-->}[r]\ar[d]^{f_2} &\cdots\ar@{-->}[r] &X_r\ar[d]^{f_r}\\
X_1\ar@{-->}[r] &X_2\ar@{-->}[r] &\cdots\ar@{-->}[r] &X_r\\
}
$$
where $f_1=f$ and all $f_i$ are surjective endomorphisms.

Let $S$ be a set of surjective endomorphisms of $X$.

We say the sequence $(*)$ is {\it $S$-equivariant} if it is $g$-equivariant for any $g\in S$.

We first need the following key lemma for the theory of the equivariant MMP.

\begin{lemma}\label{lem-fin-per}(cf.~\cite[Lemma 8.1]{Meng}, \cite[Lemma 6.1]{MZ})
Let $f:X\to X$  be an int-amplified endomorphism of a projective variety $X$.
Assume $A\subseteq X$ is a closed subvariety with
$$f^{-i}f^i(A) = A$$
for all $i\ge 0$.
Then $A$ is $f^{-1}$-periodic (i.e., $f^{-s}(A)=A$ for some $s>0$).
\end{lemma}

Applying the same argument and proofs of \cite[Lemma 6.2 to Lemma 6.6]{MZ}, we obtain the following theorem.
Note that Proposition \ref{lem-int-des1} and Theorem \ref{thm-num-lin} are needed to show the following $g$ is again polarized or int-amplified.

\begin{theorem}\label{thm-equi-mmp}(cf.~\cite[Theorem 8.2]{Meng}) Let $f:X\to X$ be a polarized (resp.~int-amplified) endomorphism of a $\Q$-factorial lc projective variety $X$.
Let $\pi:X\dasharrow Y$ be a dominant rational map which is
\begin{itemize}
\item[(i)]
either a divisorial contraction, or
\item[(ii)]
a Fano contraction, or
\item[(iii)]
a flipping contraction, or
\item[(iv)]
a flip
\end{itemize}
induced by a $K_X$-negative extremal ray.
Then there exists a polarized (resp.~int-amplified) endomorphism $g:Y\to Y$ such that
$$g\circ\pi=\pi\circ f$$
after replacing $f$ by a positive power.
\end{theorem}

Now assuming the existence of an MMP, we have the following result of equivariant MMP by successively applying Theorem \ref{thm-equi-mmp}.
We will also give a stronger version of the following theorem in the next section (cf. Theorem \ref{main-thm-finite-R}).

\begin{theorem}\label{thm-emmp}
Let $f:X\to X$ be a polarized (or int-amplified) endomorphism of a $\Q$-factorial lc projective variety $X$.
Then any finite sequence of MMP starting from $X$ is $f^s$-equivariant for some $s>0$.
\end{theorem}

The existence of a polarized or int-amplified endomorphism on a variety
exerts strong constraints on the geometry of the variety.
Recall that a normal projective variety $X$ is said to be {\it $Q$-abelian} if there is a finite surjective morphism $\pi:A\to X$ \'etale in codimension $1$ with $A$ being an abelian variety.

\begin{theorem}\label{main-thm-qa}
(cf. \cite[Theorem 3.4]{Na-Zh}, \cite[Theorem 1.21]{GKP}, \cite[Lemma 6.9]{MZ}, \cite[Theorem 1.9]{Meng})
Let $f:X\to X$ be an int-amplified endomorphism of a normal projective variety $X$.
Suppose either $X$ is klt and $K_X$ is pseudo-effective or $X$ is non-uniruled.
Then $X$ is $Q$-abelian.
\end{theorem}

We refer to \cite{KM}, \cite{BCHM} and \cite{Fu15} for details about an MMP.
Together with the previous theorem, one may characterize step by step the equivariant MMP and the action $f^*|_{\N^1(X)}$.
\begin{theorem}\label{main-thm-mmp}(cf.~\cite[Theorem 1.8]{MZ}, \cite[Theorem 1.10]{Meng})
Let $f:X\to X$ be an int-amplified endomorphism of a $\mathbb{Q}$-factorial klt projective variety $X$.
Then, replacing $f$ by a positive power, there exists a finite sequence $(*)$ of $f$-equivariant MMP as at the beginning of the section,
where we set $X_1 := X$, the $X_i \ratmap X_{i+1}$ is a divisorial contraction, flip, or Fano-contraction corresponding to a $K_{X_i}$-negative extremal ray,
and $Y = X_r$ is a $Q$-abelian variety, such that the following assertions hold.
\begin{itemize}
\item[(1)]
If $K_X$ is pseudo-effective, then $X=Y$ and it is $Q$-abelian.
\item[(2)]
If $K_X$ is not pseudo-effective, then for each $i$, $X_i\to Y$ is an equi-dimensional well-defined morphism with every fibre irreducible (and rationally connected if the base field $k$ is further uncountable) and $f_i$ is int-amplified. The $X_{r-1}\to X_r = Y$ is a Fano contraction.
\item[(3)]
$f^*|_{\N^1(X)}$ is diagonalizable over $\mathbb{C}$ if and only if so  is $f_{r}^\ast|_{\N^1(Y)}$.
\item[(4)]
If $f$ is polarized, then $f^*|_{\N^1(X)}$ is a scalar multiplication:
$$f^*|_{\N^1(X)} = q \, \id,$$ if and only if so  is $f_{r}^\ast|_{\N^1(Y)}$.
\end{itemize}
\end{theorem}

\begin{definition}\label{def-q} Let $X$ be a normal projective variety.
\begin{itemize}
\item[(1)]
$q(X):=h^1(X,\mathcal{O}_X)=\dim H^1(X,\mathcal{O}_X)$ (the irregularity).
\item[(2)]
$\tilde{q}(X):=q(\tilde{X})$ with $\tilde{X}$ a smooth projective model of $X$.
\item[(3)]
$q^\natural(X):=\sup\{\tilde{q}(X')\,|\,X'\to X \text{ is finite surjective and \'etale in codimension one}\}$.
\item[(4)]
$\pi_1^{\alg}(X_{\reg})$ is the algebraic fundamental group of the smooth locus $X_{\reg}$ of $X$.
\end{itemize}
\end{definition}

The following result is an application of Theorem \ref{main-thm-mmp}. Note that when $f:X\to X$ is a polarized endomorphism of a projective variety $X$, the action $f^*|_{\N^1_{\C}(X)}$ is always diagonalizable and all the eigenvalues are of the same modulus (cf.~\cite[Proposition 2.9]{MZ}).
However, without the extra assumptions on $X$ as stated, Theorem \ref{main-thm-rc-diag} (2) fails, for a general int-amplified endomorphism;
see Example \ref{exa-fak} given by Najmuddin Fakhruddin.

\begin{theorem}\label{main-thm-rc-diag}(cf.~\cite[Lemma 9.1, Theorem 1.10]{MZ}, \cite[Theorem 1.11]{Meng}) Let $f:X\to X$ be an int-amplified endomorphism of a $\Q$-factorial klt projective variety $X$.
Suppose either $q^\natural(X)=0$ or $\pi_1^{\alg}(X_{\reg})$ is finite (e.g., $X$ is smooth and rationally connected).
Then we have:
\begin{itemize}
\item[(1)] There exists a finite sequence of MMP which ends up with a point.
\item[(2)] There exists some $s>0$, such that $(f^s)^*|_{\N^1(X)}$ is diagonalizable over $\Q$ with all the eigenvalues being positive integers greater than $1$.
\item[(3)] If $f$ is further $q$-polarized, then $(f^s)^*|_{\N^1(X)}=q^s \, \id$.
\end{itemize}
\end{theorem}

\begin{example} {\rm (Fakhruddin)} \label{exa-fak}
\rm{} Let $E$ be an elliptic curve admitting a complex multiplication.
Let
$$S=E\times E .$$
Then $\dim (\N^1(S))=4$.
Let $\sigma:S\to S$ be an automorphism given as:
$$(x,y)\mapsto (x,x+y) .$$
Then $\sigma^*|_{\N^1(S)}$ is not diagonalizable over $\mathbb{C}$.
Let $m_S$ be the multiplication endomorpphism of $S$.
Note that
$$m_S^*|_{\N^1(S)}=m^2\id_{\N^1(S)} .$$
So $f:=\sigma\circ m_S$ is int-amplified for $m > 1$, by Proposition \ref{prop-3-equiv}.
Clearly,
$$f^*|_{\N^1(S)}$$ is not diagonalizable over $\C$.
\end{example}

\section{Equivariant MMP with respect to the monoid $\SEnd(X)$}\label{sec-5}

We use Proposition \ref{prop-finiteclosed} below in proving the results in the rest of this section.
As kindly informed by Professors Dinh and Sibony, this kind of result (with a complete proof) first appeared
in \cite[Section 3.4]{DS03}; \cite[Theorem 3.2]{Dinh09} is a more general form including Proposition \ref{prop-finiteclosed} below, requiring a weaker condition and dealing with also dominant meromorphic self-maps of K\"ahler manifolds;
see comments in \cite[page 615]{DS10} for the history of these results; see also \cite{BD}.

\begin{proposition}\label{prop-finiteclosed}(cf.~\cite[Proposition 3.6]{MZ_PG})
Let $f:X\to X$  be an int-amplified endomorphism of a projective variety $X$.
Then there are only finitely many $f^{-1}$-periodic Zariski closed subsets.
\end{proposition}

Let $X$ be a projective variety and let $C$ be a curve.
Denote by
$$R_C:=\mathbb{R}_{\ge 0}[C]$$
the ray generated by $[C]$ in $\NE(X)$.
Denote by
$$\Sigma_C$$
the union of curves whose classes are in $R_C$.

\begin{definition}\label{def:extrem_ray} Let $X$ be a projective variety.
Let $C$ be a curve such that $R_C$ is an extremal ray in $\NE(X)$.
We say $C$ or $R_C$ is {\it contractible} if there is a surjective morphism
$\pi : X \to Y$ to a projective variety $Y$ such that
the following hold.

\begin{itemize}
\item[(1)] $\pi_*\mathcal{O}_X=\mathcal{O}_Y$.
\item[(2)] Let $C'$ be a curve in $X$.
Then $\pi(C')$ is a point if and only if $[C'] \in R_C$.
\item[(3)] Let $D$ be a $\Q$-Cartier divisor of $X$.
Then $D\cdot C=0$ if and only if $D\equiv \pi^*D_Y$ (numerical equivalence) for some $\Q$-Cartier divisor $D_Y$ of $Y$.
\end{itemize}
\end{definition}

A submonoid $G$ of a monoid $\Gamma$ is said to be of {\it finite-index} in $\Gamma$
if there is a chain
$$G = G_0 \le G_1 \le \cdots \le G_r = \Gamma$$
of submonoids and homomorphisms $\rho_i : G_i \to F_i$ such that $\Ker(\rho_i) = G_{i-1}$ and all $F_i$ are finite
{\it groups}.

Denote by
$$\SEnd(X)$$
the monoid of all surjective endomorphisms of $X$.

\begin{theorem}\label{main-thm-finite-R} (cf.~\cite[Theorem 1.1]{MZ_PG})
Let $X$ be a (not necessarily normal or $\Q$-Gorenstein) projective variety with a polarized (or int-amplified) endomorphism $f$. Then:
\begin{itemize}
\item[(1)]
$X$ has only finitely many (not necessarily $K_X$-negative) contractible extremal rays in the sense of Definition \ref{def:extrem_ray}.
\item[(2)]
Suppose that $X$ is $\Q$-factorial and normal. Then any finite sequence of MMP starting from $X$ is $G$-equivariant for some finite-index submonoid $G$ of $\SEnd(X)$.
\end{itemize}
\end{theorem}

We give the main idea of the proof of Theorem \ref{main-thm-finite-R}.
First, we make use of Lemma \ref{lem-fin-per} to show that the exceptional locus of each contraction is $f^{-1}$-periodic.
By Proposition \ref{prop-finiteclosed}, there are only finitely many $f^{-1}$-periodic closed subvarieties.
Finally, we show that there are only finitely many contractible extremal rays sharing the same exceptional locus.

\begin{theorem}\label{main-thm-GMMP}(cf.~\cite[Theorem 1.2]{MZ_PG})
Let $f:X\to X$ be an int-amplified endomorphism of a $\mathbb{Q}$-factorial klt projective variety $X$.
Then there exist a finite-index submonoid $G$ of $\SEnd(X)$,
a $Q$-abelian variety $Y$, and a $G$-equivariant relative MMP over $Y$
$$X=X_0 \dashrightarrow \cdots \dashrightarrow X_i \dashrightarrow \cdots \dashrightarrow X_r=Y$$
(i.e. $g \in G = G_0$ descends to $g_i \in G_i$ on each $X_i$), such that the following assertions hold.
\begin{itemize}
\item[(1)]
There is a finite quasi-\'etale (i.e., \'etale in
codimension $1$) Galois cover $A \to Y$ from an abelian variety $A$
such that $G_Y := G_r$ lifts to a submonoid $G_A$ of $\SEnd(A)$.
\item[(2)]
If $g$ in $G$ is amplified and
its descending $g_i$ on $X_i$ is int-amplified for some $i$, then $g$ is int-amplified.
The $X_{r-1}\to X_r = Y$ is a Fano contraction.
\item[(3)]
For any subset $H \subseteq G$ and its descending
$$H_Y \subseteq \SEnd(Y)$$
$H$ acts via pullback on $\NS_{\Q}(X)$ or $\NS_{\mathbb{C}}(X)$ as commutative diagonal matrices with respect
to a suitable basis if and only if so does $H_Y$.
\end{itemize}
\end{theorem}

Let
$$\PEnd(X)$$
be the set of {\it all polarized endomorphisms} on $X$, and
let
$$\IEnd(X)$$
be the set of {\it all int-amplified endomorphisms} on $X$.
In our earlier papers, we used $\Pol(X)$ and $\IAmp(X)$
instead of $\PEnd(X)$ and $\IEnd(X)$, respectively.
The new notation here is suggested by the referee.

In general, these sets are not semigroups, i.e., they may not be closed under composition; see \cite[Example 10.4]{Meng}.
When $X$ is rationally connected and smooth,
Theorem \ref{main-thm-rc} below gives the assertion that if $g$ and $h$ are in $\PEnd(X)$ (resp.~$\IEnd(X)$)
then $g^M \circ h^M$ remains in $\PEnd(X)$ (resp.~ $\IEnd(X)$) for some $M>0$ depending only on $X$.
In particular, it answers affirmatively \cite[Question 4.15]{YZ}, ``up to finite-index", when $X$ is rationally connected and smooth.
By \cite[Example 1.7]{MZ_PG}, this extra ``up to finite-index" assumption
is necessary.

\begin{theorem}\label{main-thm-rc} (cf.~\cite[Theorems 1.4, 6.2]{MZ_PG})
Let $X$ be a rationally connected smooth projective variety admitting a polarized (or int-amplified) endomorphism $f$.
Then there are a finite-index submonoid $G \le \SEnd(X)$ and an integer $M > 0$ both depending only on $X$,
such that:
\begin{itemize}
\item[(1)]
$G^*|_{\NS_{\Q}(X)}$ is a commutative diagonal
monoid with respect to a suitable $\Q$-basis $B$ of $\NS_{\Q}(X)$.
Further, for every $g$ in $G$,
the representation matrix $[g^*|_{\NS_{\Q}(X)}]_B$ relative to $B$,
is equal to $$\diag[q_1, q_2, \dots]$$
with integers $q_i \ge 1$.
\item[(2)]
$G \cap \PEnd(X)$ is a subsemigroup of $G$, and consists exactly of those $g$ in $G$
such that
$$[g^*|_{\NS_{\Q}(X)}]_B = \diag[q, \dots, q]$$
for some integer $q \ge 2$.
\item[(3)]
$G \cap \IEnd(X)$ is a subsemigroup of $G$, and consists exactly of those $g$ in $G$
such that
$$[g^*|_{\NS_{\Q}(X)}]_B = \diag[q_1, q_2, \dots]$$
with integers $q_i \ge 2$.
\item[(4)]
We have $h^M\in G$ and that $h^*|_{\NS_{\C}(X)}$ is diagonalizable for every $h \in \SEnd(X)$.
\end{itemize}
\end{theorem}

Let $\Aut(X)$ be the group of all automorphisms of $X$, and $\Aut_0(X)$ its neutral connected component.
By applying Theorems \ref{main-thm-GMMP} and \ref{main-thm-rc}, we have the following result.

\begin{theorem}\label{main-thm-auto}(cf.~\cite[Theorems 1.5, 6.3]{MZ_PG})
Let $X$ be a rationally connected smooth projective variety.
Suppose $X$ admits a polarized (or int-amplified) endomorphism.
Then we have:
\begin{itemize}
\item[(1)]
$\Aut(X)/\Aut_0(X)$ is a finite group. More precisely, $\Aut(X)$ is a linear algebraic group
(with only finitely many connected components).
\item[(2)]
Every amplified endomorphism of $X$ is int-amplified.
\item[(3)]
$X$ has no automorphism of positive entropy (nor amplified automorphism).
\end{itemize}
\end{theorem}

\section{Characterizations of toric varieties}\label{sec-6}

A normal projective variety $X$ is said to be {\it toric} or a {\it toric variety}
if $X$ contains an algebraic torus $T = (k^*)^n$ as an (affine) open dense subset such that
the natural multiplication action of $T$ on itself extends to an action on the whole variety $X$.
In this case, let
$$D:=X\backslash T$$
which is a divisor; the pair $(X,D)$ is said to be a {\it toric pair}.

Let $G$ be a linear algebraic group acting on a normal projective variety $X$.
We say $X$ is {\it $G$-almost homogeneous} if there exists an open dense $G$-orbit in $X$.
Note that $X$ is toric if and only if $X$ is $T$-almost homogeneous for some algebraic torus $T$.
The following result gives a sufficient condition, in terms of a polarized endomorphism,
for an almost homogeneous variety to be toric.
The key ingredient in the proof is the main result in
\cite{BZ}.

\begin{theorem}\label{thm-torus}(cf.~\cite[Theorem 1.1]{MZ-toric})
Let $f:X\to X$ be a polarized endomorphism of a $G$-almost homogeneous normal projective variety $X$ with $G$ being a linear algebraic group.
Assume further the following conditions.
\begin{itemize}

\item[(i)] Let $U$ be the open dense $G$-orbit in $X$ and $D$ the codimension-$1$ part of $X \setminus U$.
The Weil divisor $K_X+D$ is $\Q$-Cartier.
\item[(ii)]
The endomorphism $f$ is $G$-equivariant in the sense: there is a surjective
homomorphism $\varphi: G \to G$ such that
$$f\circ g=\varphi(g)\circ f$$
for all $g$ in $G$.
\end{itemize}
Then $(X, D)$ is a toric pair.
\end{theorem}

Next we are going to state a geometric characterization of a toric pair.
Let $(X,\Delta)$ be a log pair, i.e.,
$X$ is a normal projective variety, $\Delta$ is an effective $\Q$-divisor and the divisor $K_X + \Delta$ is $\Q$-Cartier.
The {\it complexity}
$$c = c(X,\Delta)$$
of the pair $(X,\Delta)$ is defined as
$$c:=\inf\{n+\dim_{\R}(\sum_i \, \R[S_i])-\sum a_i\,|\, \sum a_i S_i\le \Delta, a_i\ge 0, S_i\ge 0\}.$$
Here $\sum_i \, \R[S_i]$ is a subspace, spanned by the Weil divisor classes $[S_i]$, of the real space spanned by all Weil divisor classes on $X$ (modulo algebraic equivalence).
Brown, ${\rm M}^c$Kernan, Svaldi and Zong recently gave a geometric characterization of toric varieties involving the complexity; see \cite[Theorem 1.2]{BMSZ}.
Their result is a special case of a conjecture of Shokurov, which is stated in the relative case (cf.~\cite{Sho}).
A simple version of their result shows that
{\it if $(X,\Delta)$ is a log canonical pair such that $\Delta$ is reduced, $-(K_X + \Delta)$ is nef, and
$c(X,\Delta)$ is non-positive, then $(X,\Delta)$ is a toric pair};
see \cite[Remark 4.4]{MZ-toric}.

Let $X$ be an $n$-dimensional smooth Fano variety of Picard number one
and $D\subset X$ a reduced divisor.
Assume the existence of a non-isomorphic surjective endomorphism $f : X \to X$ such that  $D$ is $f^{-1}$-invariant and $f|_{X\backslash D}$ is \'etale.
Hwang and Nakayama have shown that $X$ is isomorphic to $\mathbb{P}^n$ and $D$ is a simple normal crossing divisor consisting of $n + 1$ hyperplanes;
see \cite[Theorem 2.1]{HN}.
In particular, $(X,D)$ is a toric pair.
Indeed, their argument shows that the complexity $c(X,D)$ is non-positive.

Our Theorem \ref{thm-pitrivial} follows their idea and tries to generalize their result to the singular case. Precisely, let $X$ be a normal projective variety and $D$ a reduced divisor such that the pair
$(X, D)$ is log smooth over an open set $U \subseteq X$,
i.e., $U$ is smooth and $D \cap U$ is of simple normal crossing,
where $X \setminus U$ has codimension $\ge 2$ in $X$.
Let
$$\hat{\Omega}_X^1(\log D) = \iota_*\hat{\Omega}_X^1(\log D)|_U$$
with $\iota: U \to X$ the inclusion, which is a reflexive sheaf and is independent of
the choice of the above $U$.
Our key step is to verify that
$$\hat{\Omega}_X^1(\log D)$$
is free, i.e., isomorphic to $\mathcal{O}_X^{\oplus n}$;
see \cite[Proposition 2.3]{HN} and \cite[Theorem 5.4]{MZ-toric}.

\begin{theorem}\label{thm-pitrivial}(cf.~\cite[Theorem 1.2]{MZ-toric})
Let $X$ be a normal projective variety
which is smooth in codimension $2$, and $D\subset X$ a reduced divisor such that
\begin{itemize}
\item[(i)]
there is a Weil $\Q$-divisor $\Gamma$ such that the pair $(X, \Gamma)$ has only klt singularities;
\item[(ii)]
there is a polarized endomorphism $f:X\to X$ such that $D$ is $f^{-1}$-invariant and $f|_{X\backslash D}$ is quasi-\'etale;
\item[(iii)]
the algebraic fundamental group
$\pi_1^{\alg}(X_{\reg})$ of the smooth locus $X_{\reg}$ of $X$ is trivial
(this holds when $X$ is smooth and rationally connected); and
\item[(iv)]
the irregularity $q(X) := h^1(X, \mathcal{O}_X)$ is zero (this holds when $X$ is rationally connected).
\end{itemize}
Then the complexity $c(X,D)$ is non-positive.
\end{theorem}

An immediate corollary is the following.
\begin{corollary}\label{smooth-cor}(cf.~\cite[Corollary 1.4]{MZ-toric}) Let $X$ be a rationally connected smooth projective variety and $D\subset X$ a reduced divisor.
Suppose $f:X\to X$ is a polarized endomorphism such that $D$ is $f^{-1}$-invariant and $f|_{X\backslash D}$ is quasi-\'etale.
Then $(X,D)$ is a toric pair.
\end{corollary}

We say that a normal projective variety $X$ is of {\it Fano type} if there is a Weil $\Q$-divisor $\Delta$ such that the pair $(X,\Delta)$ has only klt singularities and $-(K_X+\Delta)$ is an ample $\Q$-Cartier divisor.
The assumption below of $X$ being of Fano type is necessary,
since a normal projective toric variety is known to be of Fano type.
The lifting in the following corollary is usually needed too; see \cite[Remark 1.7]{MZ-toric}.

\begin{corollary}\label{toric-lift-cor}(cf.~\cite[Corollary 1.5]{MZ-toric})
Let $f:X\to X$ be a polarized endomorphism of a normal projective variety $X$ of Fano type which is smooth in codimension $2$.
Let $D\subset X$ be an $f^{-1}$-invariant reduced divisor such that $f|_{X\backslash D}$ is quasi-\'etale and $K_X+D$ is $\Q$-Cartier.
Then there exist a quasi-\'etale cover
$$\pi:\widetilde{X}\to X$$
and a polarized endomorphism
$$\widetilde{f}:\widetilde{X}\to \widetilde{X}$$
such that
\begin{itemize}
\item[(1)] the endomorphism $f$ lifts to $\tilde{f}$, i.e.,
$$\pi\circ \widetilde{f}=f\circ \pi ,$$
and
\item[(2)] the pair $(\widetilde{X},\widetilde{D})$ is toric, where $\widetilde{D}=\pi^{-1}(D)$.
\end{itemize}
\end{corollary}

The following well known conjecture is still open (see also Question \ref{que-toric}).
It has been affirmatively solved in dimension $\le 3$.
We refer to \cite{ARV}, \cite{Be}, \cite{CL}, \cite{HM}, \cite{HN}, \cite{Na-Zh} and \cite{PS} for details.

\begin{conjecture}
Let $X$ be a Fano manifold of Picard number one which is different from the
projective space. Then a surjective endomorphism $X\to X$ must be bijective.
\end{conjecture}

Generalizing the above conjecture to arbitrary Picard number, we may ask:

\begin{question}\label{que-toric}
Let $X$ be a rationally connected smooth projective variety of dimension $n \ge 1$
which admits a polarized endomorphism.
Is $X$ (close to) a toric variety?
\end{question}

\section{Cases in positive characteristic}\label{sec-7}

Throughout this section, we always work over the field $k$ of characteristic $p>0$.

\begin{definition} Let $f:X\to X$ be a surjective endomorphism of a projective variety $X$.
We say that  $f$ is {\it separable} if the induced field extension $f^*:k(X)\to k(X)$ is separable.
Denote by $f^{\Gal}: Y\to X$ the Galois closure of $f$.
\end{definition}

Let $f:X\to X$ be a separable surjective endomorphism of a normal projective variety $X$. We have seen that $f$ being ``numerically quasi-polarized'' implies $f$ being ``numerically polarized''; see Proposition \ref{prop-fx-x} which is applied to $\N^1(X)$ and hence is characteristic free.
Furthermore, we have the following result generalizing Theorem \ref{thm-num-lin}.

\begin{theorem}\label{thm-num-lin-p}(cf.~\cite[Theorem 5.1]{CMZ})
Let $f:X\to X$ be a numerically quasi-polarized separable endomorphism of a normal projective variety $X$. Then $f$ is polarized.
\end{theorem}

Propositions \ref{prop-3-equiv} and \ref{lem-int-des1} hold true in the case of positive characteristic since the argument are cone-theoretical.
Let $X$ be an $n$-dimensional normal projective variety.
Then $\N^1(X)$ can be naturally embedded into $\N_{n-1}(X)$ due to the following lemma.
So Proposition \ref{prop-4-equiv} also holds true in the case of positive characteristic.
\begin{lemma}
Let $X$ be a projective variety of dimension $n \ge 2$. Let $M$ a Cartier divisor. Suppose that $H_1\cdots H_{n-1}\cdot M = 0 = H_1\cdots H_{n-2}\cdot M^2$ for some ample Cartier divisors $H_1, \cdots, H_{n-1}$. Then $M\equiv 0$ (numerical equivalence).
\end{lemma}
\begin{proof}
When $n=2$, this lemma follows from the Hodge index theorem.
We then prove by induction on $n$.
Suppose $n>2$.
By taking normalization, we may assume $X$ is normal.
Let $C$ be an irreducible curve on $X$.
Since $n>2$, there exists an integral hypersurface $H\in |mH_1|$ containing $C$ for some $m>0$ (cf.~\cite[Lemma 2.4]{Hu}).
Note that $H_2|_H\cdots H_{n-1}|_H\cdot M|_H = 0 = H_2|_H\cdots H_{n-2}|_H\cdot (M|_H)^2$.
By induction $M\cdot C=M|_H\cdot C=0$.
So the lemma is proved.
\end{proof}

With some additional assumptions and extra work, we can deal with the singularities for the surface case,
extending \cite{Wa} to positive characterisitics.
We refer to \cite[Lemma 4.4]{Ok} for the generalized ramification divisor formula.
In the case of positive characteristic, one requires an additional ``tame'' assumption on $f$ to apply \cite[Proposition 5.20]{KM}; see \cite[Proposition 4.6, Remark 4.7, Example 4.8]{Ok}.

\begin{theorem}\label{thm-w}(cf.~\cite[Theorem 10.2]{CMZ}) Let $f:X\to X$ be a non-isomorphic surjective endomorphism of a normal algebraic surface $X$ such that the degree $\deg f^{\Gal}$ of the Galois closure of $f$ is co-prime to $p$.
Then $X$ is lc.
\end{theorem}

The proof of Theorem \ref{main-thm-k} is based on Propositions \ref{prop-3-equiv} and \ref{prop-4-equiv} and the ramification divisor formula. So we can easily get the following.

\begin{theorem}\label{main-thm-k-p}(cf.~\cite[Theorem 1.1]{CMZ}, \cite[Theorem 1.5]{Meng}) Let $X$ be a normal projective variety admitting a polarized (or int-amplified) separable endomorphism.
Then we have:
\begin{itemize}
\item[(1)]
$-K_X$ is weakly numerically equivalent to some effective Weil $\Q$-divisor.
\item[(2)]
If $X$ is further assumed to be $\Q$-Gorenstein, then $-K_X$ is numerically equivalent to some effective $\Q$-Cartier divisor.
\end{itemize}
\end{theorem}

By a pure algebraic proof, we further have:

\begin{theorem}\label{thm-torsion}(cf.~\cite[Theorem 1.4]{CMZ})
Let $f:X\to X$ be a numerically polarized separable endomorphism of a normal projective variety $X$ with $K_X$ being pseudo-effective and $\Q$-Cartier. Then $f$ is quasi-\'etale and $K_X\sim_{\Q}0$.
\end{theorem}

Following the proof of \cite[Lemma 6.1]{MZ}, \cite[Lemma 6.2]{CMZ} and \cite[Lemma 8.1]{Meng}, we may generalize Lemma \ref{lem-fin-per} as follows.
We refer to \cite[Remark 6.3]{CMZ} to see that the following condition (2) is necessary.
\begin{lemma}(cf.~\cite[Lemmas 3.4 and 3.5]{MZ_PG})
Let $f:X\to X$  be an int-amplified separable endomorphism of a projective variety $X$.
Assume $A\subset X$ is a closed subvariety with
$$f^{-i}f^i(A) = A$$
for all $i\ge 0$.
Assume further either one of the following conditions.
\begin{itemize}
\item[(1)] $A$ is a prime divisor of $X$.
\item[(2)] $p$ and $\deg f$ are co-prime.
\end{itemize}
Then $A$ is $f^{-1}$-periodic.
\end{lemma}

The assumption ``$p$ and $\deg f$ are co-prime'' for the following proposition is also necessary; see \cite[Remark 6.3]{CMZ}.
\begin{proposition}\label{prop-finiteclosed-p}(cf.~\cite[Proposition 3.6]{MZ_PG})
Let $f:X\to X$  be an int-amplified endomorphism of a projective variety $X$.
Suppose $p$ and $\deg f$ are co-prime.
Then there are only finitely many $f^{-1}$-periodic Zariski closed subsets.
\end{proposition}

With the above two results, Theorem \ref{main-thm-finite-R} can also be generalized as follows.

\begin{theorem}\label{main-thm-finite-R-p} (cf.~\cite[Theorem 1.1]{MZ_PG})
Let $X$ be a (not necessarily normal or $\Q$-Gorenstein) projective variety with a polarized (or int-amplified) endomorphism.
Suppose $p$ and $\deg f$ are co-prime.
Then:
\begin{itemize}
\item[(1)]
$X$ has only finitely many (not necessarily $K_X$-negative) contractible extremal rays in the sense of Definition \ref{def:extrem_ray}.
\item[(2)]
Suppose $X$ is $\Q$-factorial normal. Then any finite sequence of MMP starting from $X$ is $G$-equivariant for some finite-index submonoid $G$ of $\SEnd(X)$.
\end{itemize}
\end{theorem}

In the case of positive characteristic, the theory of MMP is still far from being completed and is only known for lc 3-folds with characteristic $p>5$ (cf.~\cite{BW}, \cite{HX}, \cite{Wal} and the references therein).
For the lower dimensional cases, we refer to \cite[Theorem 1.6]{CMZ} and \cite[Theorem 1.8]{CMZ} for versions similar to Theorems \ref{main-thm-mmp}  and \ref{main-thm-rc-diag}, respectively.


TITLE: Symmetry of Uniform Distribution PDF
QUESTION [0 upvotes]: I'm studying probability theory and came across an exercise problem that I couldn't quite understand, even with the solution, and was hoping someone could give me some insight.

Alice, Bob, and Carl arrange to meet for lunch on a certain day. They arrive independently at uniformly distributed times between 1 p.m. and 1:30 p.m. on that day. What is the probability that Carl arrives first?

Apparently the answer is: "By symmetry, the probability that Carl arrives first is $\frac{1}{3}$." But I don't really know how this rather too simple solution came to be.
My initial approach is to divide cases as
$$P(C \lt A \lt B) + P(C \lt B \lt A)$$
and find each probability. However, wouldn't they differ depending on what time Carl arrives, and thus we would have to further condition?

Any feedback is appreciated. Thank you.

REPLY [1 votes]: Without loss of generality, consider a unit cube $ABCDA_1B_1C_1D_1$. Cut the cube along $BB_1D_1D$ and $BCD_1A_1$ to get the quadrilateral pyramid $BAA_1D_1D$, which has the volume:
$$V=\frac13 \cdot S_{AA_1D_1D} \cdot AB=\frac13 \cdot 1^2 \cdot 1=\frac13.$$


TITLE: Multiplicative system in the ring
QUESTION [1 upvotes]: Let $R$ be a commutative ring with unit element. A nonempty subset $S$ of $R$ is called multiplicative system if
$1. \  0\notin S$
$2. \ s_1,s_2\in S$ implies that $s_1s_2\in S$.
Let $n=pq$ where $p,q$ are prime numbers. We know that $\mathbb{Z}_n$ is a ring with addition and multiplication modulo $n$. Consider the $\mathbb{Z}_n^{\times}=\{x: 1\leq x<n, \ \text{gcd}(x,n)=1\}$.
It's easy to check that $\mathbb{Z}_n^{\times}$ is multiplicative system.
What if we consider $\mathbb{Z}_n^{\times}\cup \{p\}$? I have checked that this is not multiplicative system.
Am I right? If yes, can anyone suggest some other example of multiplicative system in $\mathbb{Z}_n$ distinct from $\mathbb{Z}_n^{\times}$?

REPLY [1 votes]: The multiplicative closure of $\{p\}\cup \Bbb{Z}_n^\times$ , namely $\{x: 1\le x< n, \gcd(x,q)=1\}$ is a multiplicative system distinct from $\Bbb{Z}_n^\times$.


TITLE: Derivative definition + question
QUESTION [0 upvotes]: Given the fact that $f$ is differentiable at point $a$ and $f'(a)=5$ calculate the following expression:
$\displaystyle \lim_{h\rightarrow0}\frac{f(a+h)-f(a+3h)}{h} = \,\,\,?$
Is it possible to define the definition of derivative as $\,\,\,\displaystyle \lim_{\Delta\rightarrow0}\frac{f(x)-f(x-\Delta)}{\Delta}\,\,\,$ instead of $\,\,\,\displaystyle \lim_{\Delta\rightarrow0}\frac{f(x+\Delta)-f(x)}{\Delta}\,\,\,$ ?

REPLY [0 votes]: Using Taylor series
$$f(a+n h)=f(a)+h n f'(a)+\frac{1}{2} h^2 n^2 f''(a)+O\left(h^3\right)$$
$$f(a+h)-f(a+3h)=-2 h f'(a)-4 h^2 f''(a)+O\left(h^3\right)$$
$$\frac{f(a+h)-f(a+3h) } h=-2 f'(a)-4 h f''(a)+O\left(h^2\right)$$


TITLE: Linear (in)dependence of distinct nonzero exponential polynomials
QUESTION [0 upvotes]: It is well-known that distinct exponential functions are linearly independent. What is known about the linear (in)dependence of the set of terms of an exponential polynomial? By exponential polynomial I mean a finite sum of terms of the form p_i(x) e^{a_i x} where each p_i is a polynomial. Here the exponential functions should be distinct, so the a_i are distinct; also the polynomials should be nonzero (that is, not identically zero). Under these conditions, is the set of functions p_1(x) e^{a_1 x}, ... , p_n(x) e^{a_n x} linearly independent?

REPLY [0 votes]: The answer is yes. A proof for a more general resulting can be foundation in
http://scholar.google.at/scholar_url?url=https://www.degruyter.com/downloadpdf/j/dema.2008.41.issue-3/dema-2013-0091/dema-2013-0091.xml&hl=en&sa=X&scisig=AAGBfm0VysxlFjgoiWZgO1rxRweNaSSa4Q&nossl=1&oi=scholarr
Edit (2020-03-25): The link above ist broken. Take the following one.
Demonstration mathematica


text:
\section*{Broader Impact}

The content of this research is of broad interest to both researchers and practitioners of computer science and engineering for training deep learning models in machine learning. The method proposed in this paper introduces a new optimization tool that can be adopted for training variety of models such as Convolutional Neural Network (CNN). The proposed optimizer has strong generalizability that include both fast convergence speed and also achieve superior performance compared to the existing off-the-shelf optimizers. The method further introduces a new concept that measures how well the CNN model is trained by probing in different layers of the network and obtain a quality measure for training. This metric can be of broad interest to computer scientists and engineers to develop efficient models that can be tailored on specific applications and dataset.


TITLE: ((a ⇔ b) ⇒ c) ⇔ (a ⇔ (b ⇒ c)) tautology, contradiction, or neither?
QUESTION [1 upvotes]: ((a ⇔ b) ⇒ c) ⇔ (a ⇔ (b ⇒ c)). Prove whether its a tautology, a contradiction, or neither.
My attempt:
((a ⇔ b) ⇒ c) ⇔ (a ⇔ (b ⇒ c))
if I take all F:

((F ⇔ F) ⇒ F) ⇔ (F ⇔ (F ⇒ F))
[(F ⇒ F ∧ F ⇒ F) ⇒ F] ⇔ (F ⇔ T)
(F ∧ F ⇒ F) ⇔ (F ⇒ T ∧ F ⇒ F)
(T ⇒ F) ⇔ (T ∧ F)
F ⇔ F

Then it becomes T
Could someone check if Im going the right direction?
Im kind of stuck here. I know its neither. but I don't how to continue from here.

REPLY [0 votes]: The comments point out that if $a = F$, $c = T$ (and regardless of $b$) the expression is true. It turns out that the expression you wrote has the same truth-value as $a \vee \neg c$.
Mathematica code:
LogicalExpand[Equivalent[Implies[Equivalent[a, b], c], Equivalent[a, Implies[b, c]]]]


TITLE: A Question in Rudin PMA chapter 9
QUESTION [3 upvotes]: If $f$ is a differentiable mapping of a connected open set $E\subset \Bbb{R}^n$ into $\Bbb{R}^m$ and if $f'(x)=0$ for every $x\in E$ prove that $f$ is constant in E.
How can this be done without using convexity? I'm trying to use that $f'(x)=0 \implies f$ is constant in open balls which are convex to show $A=E$ where $A=\{x\in E \mid f(x)=f(x_0\}$ where $x_0$ is fixed. Is there another way to do this with minimal use of connectedness or convexity?

REPLY [4 votes]: First show that $f$ is locally constant using the mean value theorem.
Then choose $x_0 \in E$ and let $F = \{x \in E | f(x) = f(x_0) \}$. Then $F$ is closed because $f$ is continuous, and open because $f$ is locally constant. However, since $E$ is connected, the only subsets of $E$ that are both open and closed are $E$ and $\emptyset$. Hence $F = E$.


TITLE: Does $Add(\kappa,1)^L$ ever collapse cardinals?
QUESTION [9 upvotes]: In general, we know that adding a subset to a regular cardinal $\kappa$ can collapse cardinals.  If, for example, there is $\gamma < \kappa$ with $2^\gamma >\kappa$, then $Add(\kappa,1)$ will collapse $2^\gamma$ to $\kappa$, since every subset of $\gamma$ will appear as a block in the generic.
However, this argument requires that we use the poset defined in $V$.  We can easily construct situations in which  $2^\gamma > \kappa$ but forcing with $Add(\kappa,1)^L$ does not collapse cardinals, by the same techniques that allow us to achieve Easton's Theorem.  If we force over $L$ with $Add(\gamma, \kappa^+) \times Add(\kappa,1)$, for example, then cardinals are preserved.  I am interested to know if this holds in general.

Does $Add(\kappa,1)^L$ ever collapse cardinals?

REPLY [10 votes]: Yes, it follows from a theorem of Mack Stanley (for this special case, in fact from a theorem of Foreman-Magidor-Shelah) that if $0^\sharp$ exists, then forcing with $Add(\kappa, 1)^L$ collapses $\kappa$ into $\omega.$
See Stanley's paper "Forcing disabled" (and Foreman-Magidor-Shelah paper "$0^\sharp$ and some forcing principles", J Symbolic Logic 51 (1986), 39-46). See also Can the Cohen forcing collapse cardinals? and the answer given there.


TITLE: Prime ideals of the ring $\mathbb{Q}[X]/ (X)^3$
QUESTION [0 upvotes]: The following question was part of my abstract algebra quiz and I am unable to solve it.

Let A denote the ring $\mathbb{Q}[X]/ (X)^3$ . Then is there only 1 prime ideal of A?

Ideals of $A$  are $x/\langle x^3 \rangle$ , $x^2 / \langle x^3 \rangle$  and $1$ and $ 1/\langle x^3 \rangle$ .
I proved $x/\langle x^3 \rangle$  prime. THen I tried with $x^2 / \langle x^3 \rangle$. Here any element of $A/(x^2 / \langle x^3 \rangle)$ would be of the form $a_0 + a_1 x$ but I am neither able to prove it prime or not prime.
As other ideals would be of the same form so I am looking for you help to do this part.

REPLY [1 votes]: $A/((X)/(X^3))=(\mathbb{Q}[X]/(X^3))/((X^2)/(X^3))\cong \mathbb{Q}[X]/(X^2)$.
Now $X\neq 0\in \mathbb{Q}[X]/(X^2)$ and $X\cdot X=0\in \mathbb{Q}[X]/(X^2)$,
so $\mathbb{Q}[X]/(X^2)$ is not integral domain.


text:
\section{Introduction and main results}
\label{sec:intro}

\subsection{Some directions in Gromov's vague pseudo-Riemannian conjecture}

Understanding when the isometry group of a pseudo-Riemannian
structure acts non-properly, particularly when the isometry group
$\Iso(M,\g)$ of a compact pseudo-Riemannian manifold $(M,\g)$ is
non-compact, is a paradigmatic case of Gromov's vague conjecture stipulating
that rigid geometric structures with large automorphism groups are
classifiable.
It is also a fundamental question in geometric dynamics that has been
investigated in many works with much progress, especially in the lower
index case, for example the Lorentzian index \cite{AS,DAG,zeghib} or
the authors' results on index two in \cite{BGZ}.
Indeed, let us observe for instance that for any semi\-simple Lie group $G$
with a uniform lattice $\Gamma$, the Killing form endows $M= G/\,\Gamma$
with  a pseudo-Riemannian metric whose isometry group contains $G$, acting
by left-translations, so that $M$ is a homogeneous pseudo-Riemannian
manifold.
However, these are not the only homogeneous pseudo-Riemannian manifolds,
which in general have the form $G/H$ where $H$ is not discrete.
Herein lies the difficulty of the problem.
So, even with an additional homogeneity hypothesis, the pseudo-Riemannian
problem stays intractable.

\subsubsection{Complex framework}

Our approach here in testing Gromov's vague
conjecture is to enrich the pseudo-Riemannian structure by assuming the
existence of a compatible complex structure.
In other words, we consider \emph{pseudo-Hermitian structures}.
More precisely, a \emph{pseudo-Hermitian metric} $\g$ on a complex manifold
$(M, \J)$ is a pseudo-Riemannian metric compatible with $\J$:
$\g(\J\cdot, \J\cdot) = \g(\cdot,\cdot)$.
Here we are interested in the automorphism group of such a structure, which
is the group of holomorphic transformations preserving the pseudo-Hermitian
metric.

To illustrate the effect of the complex structure, let us observe that the
previously introduced pseudo-Riemannian homogeneous examples $G/\,\Gamma$ cannot be
pseudo-Hermitian.
Indeed, the induced complex structure and pseudo-Hermitian product on the
Lie algebra $\frg$ of $G$ are $\Ad(\Gamma)$-invariant and hence
$\Ad(G)$-invariant by the Borel Density Theorem (here we assume $G$ to be
semisimple without compact factors).
For any $x\in\frg$, the adjoint operator $\ad(x)$ is skew with
respect to the pseudo-Hermitian product, and $\ad(\J x) = \J \ad(x)$. Now, the point is that the Lie algebra $\fru(p, q)$ of the unitary group of a
pseudo-Hermitian product of signature $(p, q)$ on $\CC^n$
is totally real in $\Mat(n,\CC)$, that is,
$\fru(p, q) \cap \i \fru(p, q) = \zsp$.

The main result of the present article (see Subsection \ref{results} below)
is the classification of
compact pseudo-Hermitian homogeneous spaces $G/H$ by showing that $G^\circ$
must be compact (compare the discussion following Corollary \ref{mcor:hermitian}).
In the initial example above, $H$ was discrete and we used the
Borel Density Theorem.
All this fails dramatically in the general case, where we
essentially have to deal with ``savage'' groups that are semidirect
products of a compact group by a solvable one.

\subsubsection{Pseudo-K\"ahler case}

The fundamental two-form associated to $\g$ is the differential two-form
defined by $\omega (\cdot,\cdot) = \g(\cdot, \J\cdot)$.
The metric is said to be \emph{pseudo-K\"ahler} if $\omega$ is closed.
Conversely, from the symplectic geometry point of view, pseudo-K\"ahler structures
are special symplectic structures given by a symplectic form $\omega$

calibrated by
a complex structure $\J$,  that is
$\omega(\J\cdot, \J\cdot) = \omega (\cdot,\cdot)$.

Dorfmeister and Guan \cite{DG} proved the compactness of the identity
component of the
holomorphic isometry group of a homogeneous compact pseudo-K\"ahler
manifold, and that the Levi decomposition of this group induces a
holomorphic and metric splitting of the manifold.

The rigidity here comes essentially from the symplectic side.
In fact, Dorfmeister and Guan's theorem is to a large part implied by
Zwart and Boothby's \cite{ZB} results on compact symplectic homogeneous
spaces (compare Appendix \ref{sec:newproof}).

In the pseudo-K\"ahler case,

Gromov's vague conjecture seems to be tractable.
More generally, symplectic actions of finite-dimensional Lie algebras
on compact symplectic manifolds are ``easy'' to handle.
The basic idea

is that the
derived subalgebra consists of Hamiltonian vector fields, from which one
concludes that it is a direct sum of a compact semisimple and an abelian
Lie algebra (see for instance Guan \cite[p.~3362]{guan}, but also
\cite[Section 26]{GS}, \cite[Section 2]{huckleberry}).

\subsection{Statement of the main results}\label{results}

We begin our investigation by considering the more general situation of
\emph{almost} pseudo-Hermitian homogeneous manifolds. Such a manifold
carries a field of complex structures that is not required to be integrable
to a complex structure on the manifold.
Our first main result is:

\begin{mthm}\label{mthm:Levi_compact}
Any almost pseudo-Hermitian homogeneous space $M$ of finite
volume is compact. The maximal semisimple subgroup of the identity component
of the almost complex isometry group of $M$ is compact,
and its solvable radical is nilpotent and at most two-step nilpotent.
\end{mthm}

It is therefore sufficient to henceforth consider compact spaces.

\medskip

In Section \ref{sec:spaces} we find that the only almost
pseudo-Hermitian compact homogeneous manifolds with a transitive action
by a solvable group are complex tori.

\begin{mthm}\label{mthm:solvable}
Let $M$ be a compact almost pseudo-Hermitian homogeneous space with
a transitive effective action by a connected solvable Lie group $G$ of almost complex isometries.
Then $G$ is abelian and compact.
Furthermore, $M$ is a flat complex pseudo-Hermitian torus
$\CC^{p,q}/\,\Gamma$, where $\Gamma$ is a lattice in $\CC^{p,q}$.
\end{mthm}

Upon making the assumption that $M$ is a complex manifold,
we can obtain much stronger statements.
In Section \ref{sec:complex}, we make use of
Tits's \cite{tits} result that
any compact complex homogeneous manifold fibers holomorphically over a
generalized flag manifold with complex parallelizable fibers.
We eventually derive the following structure theorem for holomorphic
isometry groups:

\begin{mthm}\label{mthm:torus_fiber}
Let $M$ be a compact complex homogeneous manifold, and let $G$ be a
closed Lie subgroup of the holomorphic automorphism group of $M$.
Assume that $G$ preserves a
pseudo-Hermitian metric and acts transitively on $M$.
Then the fiber of the Tits fibration of $M$ is a compact complex torus
and the group $G$ is compact.
\end{mthm}

In particular, we obtain that
the identity component of the holomorphic isometry group of a
compact homogeneous pseudo-Hermitian manifold is compact.
It is not clear if this result is still valid if we merely assume an
\emph{almost} pseudo-Hermitian structure on the compact
homogeneous space.

\medskip

Even though we show in Theorem \ref{mthm:torus_fiber} that the identity
component of the holomorphic isometry group is compact, the full
holomorphic isometry group is not necessarily so.
The following example illustrates this.

\begin{example*}
Let $T=\CC^{p,q}/\ZZ^{2p+2q}$ be a complex torus with a pseudo-Hermitian
pro\-duct $\h$ of signature $(p, q)$. The identity component of the
holomorphic isometry group is $T$ itself.
However, the full holomorphic isometry group contains
$\Lambda(\h)= \UU(\h)  \cap\SL(2p+2q,\ZZ)$, where $\UU(\h)$ denotes the unitary group of $\h$.
In the case  of the standard pseudo-Hermitian product

$\h(z,w)= -z_1\bar{w}_1-\ldots-z_p\bar{w}_p + z_{p+1}\bar{w}_{p+1}+\ldots+z_{p+q}\bar{w}_{p+q}$,
this group $\Lambda$ is a lattice in $\SU(\h)$  and thus non-compact (for
$p,q>0$). This fact extends to all pseudo-Hermitian products with rational
coefficients (by the Borel-Harish-Chandra Theorem \cite{BHC}).
\end{example*}

However, in the simply connected case, Theorem \ref{mthm:torus_fiber} extends
to the full holo\-morphic isometry group.

\begin{mcor}\label{mcor:hermitian}
The holomorphic isometry group of a simply connected
compact homogeneous pseudo-Hermitian manifold is compact.
\end{mcor}

A class of examples illustrating Corollary \ref{mcor:hermitian} are
the compact homogeneous pseudo-Hermitian manifolds with non-zero Euler
characteristic.
These are precisely the generalized flag manifolds (cf.~Example
\ref{ex:euler_non0}).

\medskip

\label{geometric.splitting}
Theorem \ref{mthm:torus_fiber} can be interpreted as the identity component
of the holomorphic isometry group of a compact  homogeneous pseudo-Hermitian
manifold being inessential  in the sense that it preserves a positive
definite Hermitian structure. Thus, to classify these objects, start with a
compact complex  manifold homogeneous under the action of  a compact group
$K$, and take any pseudo-Hermitian product invariant under the isotropy of
some point.
By compactness of this isotropy, there always exists an invariant Hermitian product, but existence of higher signature ones depends on the isotropy representation.
The compact group splits, up to finite index, as $G = T \times K$, where
$T$ is abelian and $K$ is semisimple.
In the pseudo-K\"ahler case, this splitting induces a holomorphic and
metric splitting of the manifold as a product of a complex torus and a flag
manifold (a projective homogeneous manifold), see
Borel and Remmert \cite{BR}, Dorfmeister and Guan \cite{DG}.
Without the assumption that the fundamental two-form is closed, we do not
have such a splitting.
This is illustrated by the examples discussed at the end of Section
\ref{sec:complex}.

\medskip

In Appendix \ref{sec:newproof}, we review the results by Zwart and Boothby \cite{ZB}
on compact symplectic homogeneous spaces and by Dorfmeister and Guan \cite{DG} on
compact homogeneous pseudo-K\"ahler manifolds.
We give short proofs based on the techniques developed in this article.
Moreover, in Theorem \ref{thm:metric_symplectic}, we prove a structure
theorem for the intermediate case of a compact symplectic homogeneous space
equipped with an invariant pseudo-Riemannian metric, but without the
assumption of compatibility between the metric and the symplectic form.

\subsection{Additional remarks beyond the homogeneous case}

In order to illustrate how our current homogeneity hypothesis is fairly
reasonable, let us show that any semisimple real Lie group $G$ can act
(non-transitively, but topologically transitive) by preserving a
pseudo-Hermitian structure on a locally homogeneous compact manifold.
Let $\kappa$ be the Killing form on the Lie algebra $\frg$.
On its complexi\-fication $\frg^\CC= \CC \otimes \frg$, consider the
quadratic form $q(x + \i y) = \kappa(x, x) + \kappa(y, y)$, where
$x+ \i y \in \frg^\CC$.
Polarization yields an $\ad(\frg)$-invariant pseudo-Hermitian form.
This allows us to define a left-invariant pseudo-Hermitian metric $\h$ on the group
$G^{\CC}$,
that equals $q$ at the identity element.
By the $\ad(\frg)$-invariance of $q$, $\h$ is invariant under the
right-action of $G$ on $G^\CC$.
If $\Gamma$ is a uniform lattice of $G^\CC$, then $\h$ projects to a
well-defined $G$-invariant pseudo-Hermitian metric on $M= G^\CC /\,\Gamma$.
By Moore's Ergodicity Theorem (see Zimmer \cite[Theorem 2.2.6]{zimmer}),
the $G$-action is ergodic and hence topologically transitive.

These examples are not pseudo-K\"ahler, because, as mentioned above, a Lie
group preserving a symplectic structure on a compact manifold has a product of a
compact group by an abelian group as its derived subgroup.

If $G = \SL(2,\RR)$, then $G^\CC= \SL(2,\CC)$, which

is locally isomorphic to
$\SO(1, 3)$, and so $M$ is the frame bundle of the
hyperbolic three-manifold (or orbi\-fold)
$\HH^3 /\,\Gamma = \SO(3) \backslash \SO(1,3) /\,\Gamma$.
The pseudo-Hermitian metric in this case is of Hermite-Lorentz type, and
in particular has index two when viewed as a pseudo-Riemannian metric.

The whole discussion generalizes to the case where $\kappa$ is replaced
by any $\ad(\frg)$-invariant scalar product and $G$ is any group admitting such a form,
sometimes called a quadratic group (there is a huge family of such groups,
see, for instance Kath and Olbrich \cite{KO1}).
If $G$ is an oscillator group, then, as in the $\SL(2,\RR)$-case,
one gets a Hermite-Lorentz example.
These beautiful and rich examples motivate
the syste\-matic study of isometric actions of Lie groups on
pseudo-Riemannian homogeneous manifolds of index two as initiated by the
authors in \cite{BGZ}.

\subsection*{Notations and conventions}

All Lie algebras $\frg$ under consideration are assumed to be finite-dimensional
and defined over the reals (although most of our algebraic results are valid over any
 field of characteristic zero).

For direct products of Lie algebras $\frg_1$, $\frg_2$ we write
$\frg_1\times\frg_2$, whereas $\frg_1+\frg_2$ or $\frg_1\oplus\frg_2$
refer to sums as vector spaces.

The \emph{solvable radical} $\frr$ of a Lie algebra $\frg$ is the maximal solvable
ideal of $\frg$.  Any semisimple Lie algebra $\frf$ over the real numbers
is a direct product $\frf=\frk\times\frs$, where $\frk$ is a semisimple Lie algebra of \emph{compact type}, meaning its Killing form is definite, and where  $\frs$ is semisimple without factors of compact type.

Any maximal semisimple Lie subalgebra $\frf = \frk \times \frs$ of $\frg$ is called a \emph{Levi subalgebra} of $\frg$. Let us further put
 $\gs=\frs\ltimes\frr$. Then there is  a Levi decomposition
\[
\frg =  \frf \ltimes \frr = (\frk\times\frs)\ltimes\frr = \frk \ltimes \gs .
\]

For a Lie group $G$, we let $G^\circ$ denote the connected component
of the identity.
For a subgroup $H$ of $G$, we write $\Ad_{\frg}(H)$ for the adjoint
representation of $H$ on the Lie algebra $\frg$ of $G$, to distinguish
it from the adjoint representation $\Ad(H)$ on its own Lie algebra
$\frh$. The closed subgroup $H$ is called \emph{uniform} if $G/H$ is compact.

The center of a group $G$, or a Lie algebra $\frg$, is denoted by
$\Zen(G)$, or $\zen(\frg)$, respectively.
Similarly, the centralizer of a subgroup $H$ in $G$
(or a subalgebra $\frh$ in $\frg$) is denoted by $\Zen_G(H)$
(or $\zen_\frg(\frh)$).
The normalizer of $H$ in $G$ is denoted by $\Nor_G(H)$, that of $\frh$ in $\frg$
by $\nor_{\frg}(\frh)$.

\subsection*{Acknowledgements}

Wolfgang Globke was partially supported by the Australian Research Council grant {DE150101647} and the Austrian Science Fund FWF grant I 3248.