TITLE: Fractional change in volume from scale-factor
QUESTION [0 upvotes]: I was given the following question which I am unable to get a seemingly correct answer from:

A body expands linearly by a factor $\alpha$ due to an increase in temperature. Because of the expansion, a point at $\vec{r}$ is displaced to the point $\vec{r}+\vec{h}$. Calculate $\nabla \cdot \vec{h}$. By what fraction does the volume increase?

From the question we can deduce that:
$$\vec{r}+\vec{h}=\alpha\vec{r}$$
We can take the divergence of both sides and use the linearity of $\nabla$ to write:
$$\nabla \cdot (\vec{r}+\vec{h})=\alpha \nabla \cdot \vec{r}\implies \nabla \cdot \vec{r} + \nabla \cdot \vec{h} = \alpha(\nabla \cdot \vec{r})$$
We can then rearrange and use $\nabla \cdot \vec{r}=3$ to get:
$$\nabla \cdot \vec{h}=3(\alpha - 1)$$
However, I am unsure how to use this to calculate the fractional volume increase. I tried:
$$\frac{V_{\text{new}}}{V_{\text{old}}}=\frac{\iiint_{V}\nabla\cdot\vec{h}\:\mathrm{d}V}{\iiint_{V}\:\mathrm{d}V}=3(\alpha - 1)$$
However, this is clearly wrong as we would expect the answer to be $\frac{V_{\text{new}}}{V_{\text{old}}} \in \mathcal{O}(\alpha^{3})$, so I'd appreciate any hints to get me back on the right track.
Thanks in advance!

REPLY [0 votes]: We know that the fractional increase in volume is:
$$\frac{a^{3}V_{\text{old}}-V_{\text{old}}}{V_{\text{old}}}=a^{3}-1$$
We can write:
$$a^{3}-1=(1+(a-1))^3-1$$
Using the binomial expansion, $(1+x)^{n}=1+nx+\mathcal{O}(x^{2})$ we have:
$$a^{3}-1=1+3(a-1)+\mathcal{O}(a-1)^{2}-1=3(a-1)+\mathcal{O}(a-1)^{2}$$
If we have $a \approx 1$, then we have:
$$a^{3}-1\approx 3(a-1)$$
As required.


TITLE: Rate of change in Vector Calc
QUESTION [0 upvotes]: Looking for some help with this vector calc review question:
In what direction at the point $(2,0)$ does the function $ f(x,y) = xy $ have rate of change $-1$? Anad are there directions in which the rate is $ -2 $
Not quite sure how to approach this problem. Do I need to take the partial derivatives of $ f(x,y) $ which would be $ fx = y $ and $ fy = x $ and then plug my values in at the point $ (2,0) $  just looking for some help

REPLY [0 votes]: Note that it asks "in what direction" hinting at a directional derivative. For that we need the gradient of $f$ which is
$$\nabla f=<y,x>$$
Then we are looking for a unit vector such that
$$\begin{align}
\nabla f(2,0)\cdot\hat u &= <0,2>\cdot\hat u \\
&= -1
\end{align}$$
Since $\hat u$ is a unit vector it is of the form
$$\hat u=\frac{<x,y>}{\sqrt{x^2+y^2}}$$
Plugging this into the above equation we find that the components of $\hat u$ must satisfy
$$y=-\frac{1}{2}\sqrt{x^2+y^2}\qquad (1)$$
Here $x$ can be thought of as a free variable, so we solve for $y$ in terms of it. Doing this we get
$$y=\frac{x}{\sqrt{3}}\qquad x<0$$
Or we could say that $y=-|x|/\sqrt{3}$, this is because $y$ must be negative. Plugging in this definition of $y$ into $\hat u$ we find that
$$\begin{align}
\hat u &=\frac{<x,-|x|/\sqrt{3}>}{\sqrt{x^2/3+x^2}} \\
&=\frac{<x,-|x|/\sqrt{3}>}{2|x|(1/\sqrt{3})} \\
&=\left<\frac{\sqrt{3}}{2}\cdot\frac{x}{|x|},-\frac{1}{2}\right> \\
&= \frac{1}{2}\left<\pm\sqrt{3},-1\right>
\end{align}$$
Are the directions in which the rate of change of $f$ at $(2,0)$ are $-1$. A similar process will show that the rate of change is $-2$ in the direction
$$\hat v=<0,-1>$$


text:
\begin{document}

\maketitle

It is well-known that an abelian variety  is (absolutely) simple
or is isogenous to a self-product of an (absolutely) simple
abelian variety if and only if the center of its endomorphism
algebra is a field. In this paper we prove that the center is a
field if the field of definition of points of prime order $\ell$
is ``big enough".

The paper is organized as follows. In \S \ref{endo} we discuss
Galois properties of points of  order $\ell$ on an abelian variety
$X$ that imply that its endomorphism algebra $\End^0(X)$ is a
central simple algebra over the field of rational numbers. In \S
\ref{tate} we prove that similar Galois properties for two abelian
varieties $X$ and $Y$ combined with the linear disjointness of the
corresponding fields of definitions of points of  order $\ell$
imply that $X$ and $Y$ are non-isogenous (and even $\Hom(X,Y)=0$).
In \S \ref{jac} we give applications to endomorphism algebras of
hyperelliptic jacobians.
 In \S \ref{mult} we prove that if $X$ admits multiplications by a
 number field $E$ and the dimension of the centralizer of $E$ in
 $\End^0(X)$ is ``as large as possible" then $X$ is an abelian
 variety of CM-type isogenous to a self-product of an absolutely
 simple abelian variety.

 Throughout the paper we will freely use the following
 observation
 \cite[p. 174]{MumfordAV}: if an abelian variety $X$
 is isogenous to a self-product $Z^d$ of an abelian variety $Z$
 then a choice of an isogeny between  $X$ and $Z^d$ defines an isomorphism
 between $\End^0(X)$ and the  algebra
 $\M_d(\End^0(Z))$ of $d\times d$ matrices over $\End^0(Z)$. Since
 the center of $\End^0(Z)$ coincides with the center of
 $\M_d(\End^0(Z))$,
 we get an isomorphism between the center of $\End^0(X)$ and  the center of $\End^0(Z)$
 (that does not depend on the choice of an isogeny). Also
 $\dim(X)=d\cdot\dim(Z)$; in particular, both $d$ and $\dim(Z)$ divide
 $\dim(X)$.

\section{Endomorphism algebras of abelian varieties}
\label{endo} Throughout this paper $K$ is a field. We write $K_a$
for its algebraic closure and $\Gal(K)$ for the absolute Galois
group $\Gal(K_a/K)$. We write $\ell$ for a prime different from
$\fchar(K)$. If $X$ is an abelian variety of positive dimension
 over $K_a$ then we write $\End(X)$ for the ring of all its
$K_a$-endomorphisms and $\End^0(X)$ for the corresponding
$\Q$-algebra $\End(X)\otimes\Q$. If $Y$ is (may be, another)
abelian variety over $K_a$ then we write $\Hom(X,Y)$ for the group
of all $K_a$-homomorphisms from $X$ to $Y$. It is well-known that
 $\Hom(X,Y)=0$ if and only if $\Hom(Y,X)=0$.

 If $n$ is a positive
integer that is not divisible by $\fchar(K)$ then we write $X_n$
for the kernel of multiplication by $n$ in $X(K_a)$. It is
well-known \cite{MumfordAV} that  $X_n$ is a free $\Z/n\Z$-module
of rank $2\dim(X)$. In particular, if $n=\ell$ is a prime then
$X_{\ell}$ is an $\F_{\ell}$-vector space of dimension $2\dim(X)$.

 If $X$ is defined over $K$ then $X_n$ is a Galois
submodule in $X(K_a)$. It is known that all points of $X_n$ are
defined over a finite separable extension of $K$. We write
$\bar{\rho}_{n,X,K}:\Gal(K)\to \Aut_{\Z/n\Z}(X_n)$ for the
corresponding homomorphism defining the structure of the Galois
module on $X_n$,
$$\tilde{G}_{n,X,K}\subset
\Aut_{\Z/n\Z}(X_{n})$$ for its image $\bar{\rho}_{n,X,K}(\Gal(K))$
and $K(X_n)$ for the field of definition of all points of $X_n$.
Clearly, $K(X_n)$ is a finite Galois extension of $K$ with Galois
group $\Gal(K(X_n)/K)=\tilde{G}_{n,X,K}$. If $n=\ell$  then we get
a natural faithful linear representation
$$\tilde{G}_{\ell,X,K}\subset \Aut_{\F_{\ell}}(X_{\ell})$$
of $\tilde{G}_{\ell,X,K}$ in the $\F_{\ell}$-vector space
$X_{\ell}$.

\begin{rem}
\label{ell2}
 If $n=\ell^2$ then there is the  natural surjective
homomorphism
$$\tau_{\ell,X}:\tilde{G}_{\ell^2,X,K}\twoheadrightarrow \tilde{G}_{\ell,X,K}$$
corresponding to the field inclusion $K(X_{\ell})\subset
K(X_{\ell^2})$; clearly, its kernel is a finite $\ell$-group.
Every prime dividing $\#(\tilde{G}_{\ell^2,X,K})$ either divides
$\#(\tilde{G}_{\ell,X,K})$ or is equal to $\ell$. If $A$ is a
subgroup in  $\tilde{G}_{\ell^2,X,K}$ of index $N$ then its image
$\tau_{\ell,X}(A)$ in $\tilde{G}_{\ell,X,K}$ is isomorphic to
$A/A\bigcap\ker(\tau_{\ell,X})$. It follows easily that the index
of  $\tau_{\ell,X}(A)$ in $\tilde{G}_{\ell,X,K}$ equals $N/\ell^j$
where $\ell^j$ is  the index of $A\bigcap\ker(\tau_{\ell,X})$ in
$\ker(\tau_{\ell,X})$. In particular, $j$ is a nonnegative
integer.
\end{rem}

We write $\End_K(X)$ for the ring of all $K$-endomorphisms of $X$.
We have
$$\Z=\Z\cdot 1_X \subset \End_K(X) \subset \End(X)$$ where $1_X$
is the identity automorphism of $X$. Since $X$ is defined over
$K$, one may associate with every $u \in \End(X)$ and $\sigma \in
\Gal(K)$ an endomorphism $^{\sigma}u\ \in \End(X)$ such that
$^{\sigma}u (x)=\sigma u(\sigma^{-1}x)$ for $x \in X(K_a)$ and we
get the  group homomorphism
$$\kappa_{X}: \Gal(K) \to \Aut(\End(X));
\quad\kappa_{X}(\sigma)(u)=\ ^{\sigma}u \quad \forall \sigma \in
\Gal(K),u \in \End(X).$$ It is well-known that $\End_K(X)$
coincides with the subring of $\Gal(K)$-invariants in $\End(X)$,
i.e., $\End_K(X)=\{u\in \End(X)\mid\  ^{\sigma}u\ =u \quad \forall
\sigma \in \Gal(K)\}$. It is also well-known that $\End(X)$
(viewed as a group with respect to addition) is a free commutative
group of finite rank and $\End_K(X)$ is its {\sl pure} subgroup,
i.e., the quotient $\End(X)/\End_K(X)$ is also  a free commutative
group of finite rank.
 All  endomorphisms of $X$ are defined over a finite separable
 extension of $K$.
More precisely \cite{Silverberg}, if $n\ge 3$ is a positive
integer not divisible by $\fchar(K)$ then all the endomorphisms of
$X$ are defined over $K(X_n)$; in particular,
$$\Gal(K(X_n)) \subset \ker(\kappa_{X})\subset \Gal(K).$$
 This implies that if
$\Gamma_K: =\kappa_{X}(\Gal(K)) \subset \Aut(\End(X))$ then there
exists a surjective homomorphism $\kappa_{X,n}:\tilde{G}_{n,X}
\twoheadrightarrow \Gamma_K$ such that the composition
$$\Gal(K)\twoheadrightarrow \Gal(K(X_{n})/K)=
\tilde{G}_{n,X}\stackrel{\kappa_{X,n}}{\twoheadrightarrow}
\Gamma_K$$ coincides with $\kappa_X$ and
$$\End_K(X)=\End(X)^{\Gamma_K}.$$
Clearly,  $\End(X)$ leaves invariant the subgroup $X_{\ell}\subset
X(K_a)$. It is well-known that $u\in \End(X)$ kills $X_{\ell}$
(i.e. $u(X_{\ell})=0$) if and only if $u \in \ell\cdot\End(X)$.
This gives us a natural embedding
$$\End_K(X)\otimes\Z/\ell\Z\subset \End(X)\otimes\Z/\ell\Z\
\hookrightarrow \End_{\F_{\ell}}(X_{\ell});$$  the image of
$\End_K(X)\otimes\Z/\ell\Z$ lies in the centralizer of the Galois
group, i.e., we get an embedding
$$\End_K(X)\otimes\Z/\ell\Z
\hookrightarrow
\End_{\Gal(K)}(X_{\ell})=\End_{\tilde{G}_{\ell,X,K}}(X_{\ell}).$$
The next easy assertion seems to be well-known (compare with Prop.
3 and its proof on pp. 107--108 in \cite{Mori2}) but quite useful.

\begin{lem}
\label{overK}
 If $\End_{\tilde{G}_{\ell,X,K}}(X_{\ell})=\F_{\ell}$
then $\End_K(X)=\Z$.
\end{lem}

\begin{proof}
It follows that the $\F_{\ell}$-dimension of
$\End_K(X)\otimes\Z/\ell\Z$ does not exceed $1$. This means that
the rank of the free commutative group $\End_K(X)$ does not exceed
$1$ and therefore is $1$. Since $\Z\cdot 1_X \subset \End_K(X)$,
it follows easily that $\End_K(X)=\Z\cdot 1_X=\Z$.
\end{proof}

\begin{lem}
\label{overK2}
 If $\End_{\tilde{G}_{\ell,X,K}}(X_{\ell})$ is a field
then $\End_K(X)$ has no zero divisors, i.e, $\End_K(X)\otimes\Q$
is a division algebra over $\Q$.
\end{lem}

\begin{proof}
It follows that $\End_K(X)\otimes\Z/\ell\Z$ is also a field and
therefore has no zero divisors. Suppose that $u,v$ are non-zero
elements of $\End_K(X)$ with $uv=0$. Dividing (if possible) $u$
and $v$ by suitable powers of $\ell$ in $\End_K(X)$, we may assume
that both $u$ and $v$ do not lie in $\ell \End_K(X)$ and induce
non-zero elements in $\End_K(X)\otimes\Z/\ell\Z$ with zero
product. Contradiction.
\end{proof}

 Let us put $\End^0(X):=\End(X)\otimes\Q$. Then
 $\End^0(X)$ is a semisimple
finite-dimensional $\Q$-algebra \cite[\S 21]{MumfordAV}.  Clearly,
the natural map $\Aut(\End(X)) \to \Aut(\End^0(X))$ is an
embedding. This allows us to view $\kappa_{X}$ as a homomorphism
$$\kappa_{X}: \Gal(K) \to \Aut(\End(X))\subset \Aut(\End^0(X)),$$
whose image coincides with $\Gamma_K\subset \Aut(\End(X))\subset
\Aut(\End^0(X))$;
  the subalgebra $\End^0(X)^{\Gamma_K}$  of
 $\Gamma_K$-invariants coincides with $\End_K(X)\otimes\Q$.

\begin{rem}
\label{split}
\begin{itemize}
\item[(i)]
 Let us split
the semisimple $\Q$-algebra $\End^0(X)$ into a finite direct
product $\End^0(X)= \prod_{s\in \I} D_s$
 of simple $\Q$-algebras $D_s$. (Here $\I$ is identified with the
 set of minimal two-sided ideals in $\End^0(X)$.)
Let $e_s$ be the identity element of $D_s$. One may view $e_s$ as
an idempotent in $\End^0(X)$. Clearly,
$$1_X=\sum_{s\in \I} e_s\in \End^0(X), \quad e_s e_t=0 \ \forall s\ne t.$$
There exists a positive integer $N$ such that all $N \cdot e_s$
lie in $\End(X)$. We write $X_s$ for the image $X_s:=(Ne_s) (X)$;
it is an abelian subvariety in $X$ of positive dimension.  The sum
map
$$\pi_X:\prod_s X_s \to X, \quad (x_s) \mapsto \sum_s x_s$$
is an isogeny. It is also clear that the intersection $D_s\bigcap
\End(X)$ leaves $X_s \subset X$ invariant. This gives us a natural
identification $D_s \cong \End^0(X_s)$.  One may easily check that
each $X_s$ is isogenous to a self-product of  (absolutely) simple
abelian variety and if $s\ne t$ then $\Hom(X_s,X_t)=0$.
\item[(ii)]
We write $C_s$ for the center of $D_s$. Then $C_s$ coincides with
the center of $\End^0(X_s)$ and is therefore either a totally real
number field of degree dividing $\dim(X_s)$ or a CM-field of
degree dividing $2\dim(X_s)$ \cite[p. 202]{MumfordAV}; the center
$C$ of $\End^0(X)$ coincides with $\prod_{s\in \I}C_s=\oplus_{s\in
S}C_s$.
\item[(iii)]
All the sets $$\{e_s\mid s\in \I\}\subset \oplus_{s\in \I}\Q\cdot
e_s\subset \oplus_{s\in \I}C_s=C$$ are stable under the Galois
action $\Gal(K)
\stackrel{\kappa_X}{\longrightarrow}\Aut(\End^0(X))$. In
particular, there is a continuous homomorphism from $\Gal(K)$ to
the group $\Perm(\I)$ of permutations of $\I$ such that its kernel
contains $\ker(\kappa_X)$ and
$$e_{\sigma(s)}=\kappa_X(\sigma)(e_s)=\ ^{\sigma}e_s ,
\ ^{\sigma}(C_s)\ =C_{\sigma(s)}, \ ^{\sigma}(D_s)=D_{\sigma(s)}
 \quad \forall \sigma\in \Gal(K), s\in \I.$$
It follows that $X_{\sigma(s)}=Ne_{\sigma(s)}(X)=\sigma (Ne_s
(X))=\sigma(X_s)$; in particular, abelian subvarieties $X_s$ and
$X_{\sigma(s)}$ have the same dimension and $u\mapsto\ ^{\sigma}u$
gives rise to an isomorphism of $\Q$-algebras
$\End^0(X_{\sigma(s)})\cong \End^0(X_s)$.
\item[(iv)]
If $J$ is a non-empty Galois-invariant subset in $\J$ then the sum
$\sum_{s\in J}Ne_s$ is Galois-invariant and therefore lies in
$\End_K(X)$. If $J'$ is another Galois-invariant subset of $\I$
that does not meet $J$ then $\sum_{s\in J}Ne_s$ also lies in
$\End_K(X)$ and $\sum_{s\in J}Ne_s \sum_{s\in J'}Ne_s=0$. Assume
that $\End_K(X)$ has no zero divisors. It follows that $\I$ must
consist of one Galois orbit; in particular, all $X_s$ have the
same dimension equal to $\dim(X)/\#(\I)$. In addition, if $t\in
\I$, $\Gal(K)_t$ is the stabilizer of $t$ in $\Gal(K)$ and $F_t$
is the subfield of $\Gal(K)_t$-invariants in the separable closure
of $K$ then it follows easily that $\Gal(K)_t$ is an open subgroup
of index $\#(\I)$ in $\Gal(K)$, the field extension $F_t/K$ is
separable  of degree $\#(\I)$ and $\prod_{s\in S}X_s$ is
isomorphic over $K_a$ to the Weil restriction $\Res_{F_t/K}(X_t)$.
This implies that $X$ is isogenous over $K_a$ to
$\Res_{F_t/K}(X_t)$.
\end{itemize}
\end{rem}

\begin{thm}
\label{center1} Suppose  that $\ell$ is a prime, $K$ is a field of
characteristic $\ne \ell$. Suppose that $X$ is an abelian variety
of positive dimension $g$ defined over  $K$.  Assume that
$\tilde{G}_{\ell,X,K}$ contains a  subgroup $\mathcal{G}$ such
 $\End_{\mathcal{G}}(X_{\ell})$ is a field.

Then one of the following  conditions holds:

\begin{enumerate}
\item[(a)]
The center of  $\End^0(X)$ is a field. In other
 words, $\End^0(X)$ is a simple $\Q$-algebra.
\item[(b)]
\begin{itemize}
\item[(i)]
The prime $\ell$ is odd;
\item[(ii)]
there exist a positive integer $r>1$ dividing $g$, a field $F$
with
$$K\subset K(X_{\ell})^{\mathcal{G}}=:L\subset F\subset K(X_{\ell}), \quad [F:L]=r$$
 and a $\frac{g}{r}$-dimensional abelian variety
$Y$ over $F$ such that $\End^0(Y)$ is a simple $\Q$-algebra, the
$\Q$-algebra $\End^0(X)$ is isomorphic to the direct sum of $r$
copies of $\End^0(Y)$ and the Weil restriction $\Res_{F/L}(Y)$ is
isogenous over $K_a$ to $X$. In particular, $X$ is isogenous over
$K_a$ to a product of $\frac{g}{r}$-dimensional abelian varieties.
In addition, $\mathcal{G}$ contains a subgroup of index $r$;
\end{itemize}
 \item[(c)]
\begin{itemize}
\item[(i)]
The prime $\ell=2$;
\item[(ii)]
there exist a positive integer $r>1$ dividing $g$,  fields $L$ and
$F$ with
$$K\subset K(X_{4})^{\mathcal{G}}\subset L\subset F\subset K(X_{4}), \quad [F:L]=r$$ and a
$\frac{g}{r}$-dimensional abelian variety $Y$ over $F$ such that
$\End^0(Y)$ is a simple $\Q$-algebra,  the $\Q$-algebra
$\End^0(X)$ is isomorphic to the direct sum of $r$ copies of
$\End^0(Y)$ and the Weil restriction $\Res_{F/L}(Y)$ is isogenous
over $K_a$ to $X$. In particular, $X$ is isogenous over $K_a$ to a
product of $\frac{g}{r}$-dimensional abelian varieties.In
addition, there exists a nonnegative integer $j$ such that $2^j$
divides $r$ and $\mathcal{G}$ contains a subgroup of index
$\frac{r}{2^j}>1$.
\end{itemize}
\end{enumerate}
\end{thm}

\begin{proof}
We will use notations of Remark \ref{split}.  Let us put $n=\ell$
if $\ell$ is odd and $n=4$ if $\ell=2$. Replacing $K$ by
$K(X_{\ell})^{{\mathcal{G}}}$, we may and will assume that
$$\tilde{G}_{\ell,X,K}={\mathcal{G}}.$$
If $\ell$ is odd then let us put $L=K$ and
$H:=\Gal(K(X_{\ell})/K)={\mathcal{G}}=\Gal(L(X_{\ell})/L)$.

If $\ell=2$ then we choose a subgroup $\HH \subset
\tilde{G}_{4,X,K}$ of smallest possible order such that
$\tau_{2,X}(\HH)=\tilde{G}_{2,X,K}={\mathcal{G}}$ and put
$L:=K(X_{4})^{\HH}\subset K(X_{4})$. It follows easily that
$L(X_{4})=K(X_{4})$ and
 $\Gal(L(X_{2})/L)=\Gal(K(X_{2})/K)$, i.e.,
$$\HH=\tilde{G}_{4,X,L},\quad
\tilde{G}_{2,X,L}={\mathcal{G}}.$$ The minimality property of
$\HH$ combined with Remark \ref{ell2} implies that if $H\subset
\tilde{G}_{4,X,L}$ is a subgroup of index $r>1$ then
$\tau_{2,X}(H)$ has index $\frac{r}{2^j}>1$ in $\tilde{G}_{2,X,L}$
for some nonnegative index $j$.

In light of Lemma \ref{overK2}, $\End_L(X)$ has no zero divisors.
It follows from  Remark \ref{split}(iv) that $\Gal(L)$ acts on
$\I$ transitively. Let us put $r=\#(\I)$. If $r=1$ then $\I$ is a
singleton and $\I=\{s\}, X=X_s, \End^0(X)=D_s,C=C_s$. This means
that  assertion (a) of Theorem \ref{center1} holds true.

Further we assume that $r>1$. Let  us choose $t\in \I$ and put
$Y:=X_t$. If $F:=F_t$ is the subfield of $\Gal(L)_t$-invariants in
the separable closure of $K$ then it follows from  Remark
\ref{split}(iv) that $F_t/L$ is a separable degree $r$ extension,
$Y$ is defined over $F$ and $X$ is isogenous over $L_a=K_a$ to
$\Res_{F/L}(Y)$.

Recall (Remark \ref{split}(iii)) that $\ker(\kappa_X)$ acts
trivially on $\I$. It follows that  $\Gal(L(X_n))$ acts trivially
on $\I$. This implies that $\Gal(L(X_n))$ lies in $\Gal(L)_t$.
Recall that $\Gal(L)_t$ is an open subgroup of index $r$ in
$\Gal(L)$ and $\Gal(L(X_n))$ is a normal open subgroup in
$\Gal(L)$. It follows that $H:=\Gal(L)_t/\Gal(L(X_n))$ is a
subgroup of index $r$ in
$$\Gal(L)/\Gal(L(X_n))=\Gal(L(X_n)/L)=\tilde{G}_{n,X,L}.$$
If $\ell$ is odd then $n=\ell$ and
$\tilde{G}_{n,X,L}=\tilde{G}_{\ell,X,L}=\GG$  contains a subgroup
of index $r>1$. It follows from Remark \ref{split} that assertion
(b) of Theorem \ref{center1} holds true.

If $\ell=2$ then $n=4$ and $\tilde{G}_{n,X,L}=\tilde{G}_{4,X,L}$
contains a subgroup $H$ of index $r>1$. But in this case we know
(see the very beginning of this proof) that
$\tilde{G}_{2,X,L}=\GG$ and $\tau_{2,X}(H)$ has index
$\frac{r}{2^j}>1$ in $\tilde{G}_{2,X,L}$ for some nonnegative
integer $j$. It follows from Remark \ref{split} that  assertion
(c) of Theorem \ref{center1} holds true.
\end{proof}

Before stating our next result, recall that a {\sl perfect} finite
group $\GG$ with center $\ZZ$ is called {\sl quasi-simple} if the
quotient $\GG/\ZZ$ is a simple nonabelian group. Let $H$ be a
non-central normal subgroup in quasi-simple ${\mathcal{G}}$. Then
the image of $H$ in simple ${\mathcal{G}}/\ZZ$ is a non-trivial
normal subgroup and therefore coincides with ${\mathcal{G}}/\ZZ$.
This means that ${\mathcal{G}}=\ZZ H$. Since ${\mathcal{G}}$ is
perfect, ${\mathcal{G}}=[{\mathcal{G}},{\mathcal{G}}]=[H,H]\subset
H$. It follows that ${\mathcal{G}}=H$. In other words,  every
proper normal subgroup in a quasi-simple group is central.

\begin{thm}
\label{center2} Suppose  that $\ell$ is a prime, $K$ is a field of
characteristic different from $\ell$. Suppose that $X$ is an
abelian variety of positive dimension $g$ defined over  $K$.   Let
us assume that $\tilde{G}_{\ell,X,K}$ contains a subgroup
$\mathcal{G}$ that enjoys the following properties:

\begin{enumerate}
\item[(i)]
$\End_{\mathcal{G}}(X_{\ell})=\F_{\ell}$;
\item[(ii)]
The group $\mathcal{G}$ does not contain a subgroup of index $2$.
\item[(iii)]
The only normal subgroup in $\mathcal{G}$ of index dividing $g$ is
$\mathcal{G}$ itself.
\end{enumerate}

 Then one of the following two conditions (a) and (b) holds:
\begin{enumerate}
\item[(a)] There exists a positive integer $r>2$ such that:
\begin{itemize}
\item[(a0)]
$r$ divides $g$ and $X$ is isogenous over $K_a$ to a product of
$\frac{g}{r}$-dimensional abelian varieties;
\item[(a1)]
If $\ell$ is odd then
 $\mathcal{G}$  contains a subgroup of index $r$;
\item[(a2)]
If $\ell=2$ then there exists a nonnegative integer $j$ such that
 $\mathcal{G}$  contains a subgroup of index $\frac{r}{2^j}>1$.
\end{itemize}

\item[(b)]
\begin{itemize}
\item[(b1)]
The center of  $\End^0(X)$ coincides with $\Q$. In other
 words, $\End^0(X)$ is a matrix algebra either over $\Q$ or over a
 quaternion $\Q$-algebra.
\item[(b2)]
 If $\mathcal{G}$ is perfect and $\End^0(X)$ is a matrix algebra  over
 a quaternion $\Q$-algebra $\H$ then $\H$ is unramified at every
 prime not dividing  $\#(\mathcal{G})$.
\item[(b3)]
Let $\ZZ$ be the center of $\mathcal{G}$.  Suppose that
$\mathcal{G}$ is quasi-simple, i.e. it is perfect and  the
quotient ${\mathcal{G}}/\ZZ$ is a simple group. If $\End^0(X) \ne
\Q$ then there exist a perfect finite (multiplicative) subgroup
$\Pi\subset\End^0(X)^{*}$ and a surjective homomorphism $\Pi
\twoheadrightarrow {\mathcal{G}}/\ZZ$.
 \end{itemize}
 \end{enumerate}
\end{thm}

\begin{proof}
Let $C$ be the center of  $\End^0(X)$. Assume that  $C$  is not a
field. Applying Theorem \ref{center1}, we conclude that the
condition (a) holds.

 Assume now that  $C$   is  a field.
 We need to prove (b).
Let us define $n$ and $L$ as in the beginning of the proof of
Theorem \ref{center1}. We have
$$\GG=\tilde{G}_{\ell,X,L},\quad
\End_{\tilde{G}_{\ell,X,L}}(X_{\ell})=\F_{\ell}.$$ In addition, if
$\ell=2$ and  $H\subset \tilde{G}_{4,X,L}$ is a subgroup of index
$r>1$ then $\tau_{2,X}(H)$ has index $\frac{r}{2^j}>1$ in
$\tilde{G}_{2,X,L}=\GG$ for some nonnegative integer $j$. This
implies that the only normal subgroup in
$\tilde{G}_{n,X,L}=\tilde{G}_{4,X,L}$ of index dividing $g$ is
$\tilde{G}_{n,X,L}$ itself. It is also clear that
$\tilde{G}_{n,X,L}$ does not contain a subgroup of index $2$. It
follows from Remark \ref{ell2} that if $\GG$ is perfect then
$\tilde{G}_{4,X,L}$ is also perfect and every prime dividing
$\#(\tilde{G}_{4,X,L})$ must divide $\#(\GG)$, because (thanks to
a celebrated theorem of Feit-Thompson) $\#(\GG)$ must be even. (If
$\ell$ is odd then $n=\ell$ and $\tilde{G}_{n,X,L}=\GG$.)

It follows from Lemma \ref{overK} that $\End_L(X)=\Z$ and
therefore $\End_L(X)\otimes\Q=\Q$. Recall that
$\End_L(X)\otimes\Q=\End^0(X)^{\Gal(L)}$ and
$\kappa_X:\Gal(L)\to\Aut(\End^0(X))$ kills $\Gal(L(X_n))$. This
gives rise to the homomorphism
$$\kappa_{X,n}:\tilde{G}_{n,X,L}=\Gal(L(X_n)/L)=\Gal(L)/\Gal(L(X_n))\to\Aut(\End^0(X))$$
with
$\kappa_{X,n}(\tilde{G}_{n,X,L})=\kappa_X(\Gal(L))\subset\Aut(\End^0(X))$
and $\End^0(X)^{\tilde{G}_{n,X,L}}=\Q$. Clearly, the action of
$\tilde{G}_{n,X,L}$ on $\End^0(X)$ leaves invariant the center $C$
and therefore defines a homomorphism $\tilde{G}_{n,X,L}\to
\Aut(C)$ with $C^{\tilde{G}_{n,X,L}}=\Q$. It follows that $C/\Q$
is a Galois extension and the corresponding map
$$\tilde{G}_{n,X,L}\to \Aut(C)=\Gal(C/\Q)$$
is surjective.
 Recall  that $C$ is either a totally real
number field of degree dividing $g$ or a purely imaginary
quadratic extension of a totally real number field $C^{+}$ where
$[C^{+}:\Q]$ divides $g$ . In the case of totally real $C$ let us
put $C^{+}:=C$. Clearly, in both cases $C^{+}$ is the largest
totally real subfield of $C$ and therefore the action of
$\tilde{G}_{n,X,L}$ leaves $C^{+}$ stable, i.e. $C^{+}/\Q$ is also
a Galois extension.  Let us put $r:=[C^{+}:\Q]$. It is known
\cite[ p. 202]{MumfordAV}
 that $r$ divides $g$.  Clearly, the Galois
group $\Gal(C^{+}/\Q)$ has order $r$ and we have a surjective
homomorphism (composition)
$$\tilde{G}_{n,X,L}\twoheadrightarrow
\Gal(C/\Q)\twoheadrightarrow\Gal(C^{+}/\Q)$$ of
$\tilde{G}_{n,X,L}$ onto order $r$ group $\Gal(C^{+}/\Q)$.
Clearly, its kernel is a normal subgroup of index $r$ in
$\tilde{G}_{n,X,L}$. This contradicts our assumption if $r>1$.
Hence $r=1$, i.e. $C^{+}=\Q$.
  It follows that either $C=\Q$ or $C$ is an
imaginary quadratic field and $\Gal(C/\Q)$ is a group of order
$2$. In the latter case we get the surjective homomorphism from
$\tilde{G}_{n,X,L}$ onto $\Gal(C/\Q)$, whose kernel is a subgroup
of order 2 in $\tilde{G}_{n,X,L}$, which does not exist. This
proves that $C=\Q$. It follows from Albert's classification
\cite[p. 202]{MumfordAV} that $\End^0(X)$ is either a matrix
algebra $\Q$ or a matrix algebra $\M_d(\H)$ where $\H$ is a
quaternion $\Q$-algebra. This proves  assertion (b1) of Theorem
\ref{center2}.

Assume, in addition, that $\GG$ is perfect. Then, as we have
already seen, $\tilde{G}_{n,X,L}$ is also perfect. This implies
that $\Gamma:=\kappa_{X,n}(\tilde{G}_{n,X,L})$ is a finite perfect
subgroup of $\Aut(\End^0(X))$ and every prime dividing
$\#(\Gamma)$ must divide $\#(\tilde{G}_{n,X,L})$ and therefore
divides $\#(\GG)$. Clearly,
$$\Q=\End^0(X)^{\Gamma} \eqno(1).$$
Assume that $\End^0(X)\ne\Q$. Then $\Gamma\ne \{1\}$. Since
$\End^0(X)$ is a  central simple $\Q$-algebra, all its
automorphisms are inner, i.e.,
$\Aut(\End^0(X))=\End^0(X)^*/\Q^{*}$. Let $\Delta
\twoheadrightarrow \Gamma$ be the universal central extension of
$\Gamma$. It is well-known \cite[Ch. 2, \S 9]{Suzuki} that
$\Delta$ is a finite perfect group and the set of prime divisors
of $\#(\Delta)$ coincides with the set of prime divisors of
$\#(\Gamma)$ . The universality property implies that the
inclusion map $\Gamma\subset \End^0(X)^{*}/\Q^{*}$ lifts
(uniquely) to a homomorphism $\pi:\Delta \to \End^0(X)^{*}$. The
equality (1) means that the centralizer of $\pi(\Delta)$ in
$\End^0(X)$ coincides with $\Q$ and therefore $\ker(\pi)$ does not
coincide with $\Delta$. It follows that the image $\Gamma_0$ of
$\ker(\pi)$ in $\Gamma$ does not coincide with the whole $\Gamma$.
It also follows that if $\Q[\Delta]$ is the group $\Q$-algebra of
$\Delta$ then $\pi$ induces the $\Q$-algebra homomorphism $\pi:
\Q[\Delta] \to \End^0(X)$ such that the centralizer of the image
$\pi(\Q[\Delta])$ in $\End^0(X)$ coincides with $\Q$.

I claim that $\pi(\Q[\Delta])=\End^0(X)$ and therefore $\End^0(X)$
is isomorphic to a direct summand of $\Q[\Delta]$. This claim
follows easily from the next lemma that will be proven later in
this section.

\begin{lem}
\label{sur} Let $E$ be a field of characteristic zero, $T$ a
semisimple finite-dimensional $E$-algebra, $S$ a
finite-dimensional central simple $E$-algebra, $\beta: T \to S$ an
$E$-algebra homomorphism that sends $1$ to $1$. Suppose that the
centralizer of the image $\beta(T)$ in $S$ coincides with the
center $E$. Then $\beta$ is surjective, i. e. $\beta(T)=S$.
\end{lem}

In order to prove  (b2), let us assume that $\End^0(X)=\M_d(\H)$
where $\H$ is a quaternion $\Q$-algebra. Then
 $\M_d(\H)$ is isomorphic to a direct summand of $\Q[\Delta]$. On
the other hand, it is well-known that if $q$ is a prime not
dividing  $\#(\Delta)$ then
$\Q_q[\Delta]=\Q[\Delta]\otimes_{\Q}\Q_q$ is a direct sum of
matrix algebras over (commutative) fields. It follows that
$\M_d(\H)\otimes_{\Q}\Q_q$  also splits. This proves the assertion
(b2).

In order to prove  (b3), let us assume that ${\mathcal{G}}$ is a
quasi-simple finite group with center $\ZZ$. Let us put
$\Pi:=\pi(\Delta)\subset \End^0(X)^{*}$. We are going to construct
a surjective homomorphism $\Pi \twoheadrightarrow
{\mathcal{G}}/\ZZ$. In order to do that, it suffices to construct
a surjective homomorphism $\Gamma \twoheadrightarrow
{\mathcal{G}}/\ZZ$. Recall that there are surjective homomorphisms
$$\tau:\tilde{G}_{n,X,L} \twoheadrightarrow
\tilde{G}_{\ell,X,L}={\mathcal{G}},\quad
\kappa_{X,n}:\tilde{G}_{n,X,L} \twoheadrightarrow\Gamma.$$ (If
$\ell$ is odd then $\tau$ is the identity map; if $\ell=2$ then
$\tau=\tau_{2,X}$.) Let $H_0$ be the kernel of
$\kappa_{X,n}:\tilde{G}_{n,X,L} \twoheadrightarrow\Gamma$.
Clearly,
$$\tilde{G}_{n,X,L}/H_0\cong \Gamma \eqno(2).$$ Since $\Gamma\ne \{1\}$, we have $H_0\ne
\tilde{G}_{n,X,L}$. It follows  that $\tau(H_0)\ne {\mathcal{G}}$.
The surjectivity of $\tau:\tilde{G}_{n,X,L} \twoheadrightarrow
{\mathcal{G}}$ implies that $\tau(H_0)$ is normal in
${\mathcal{G}}$ and therefore lies in the center $\ZZ$. This gives
us the surjective homomorphisms
$$\tilde{G}_{n,X,L}/H_0 \twoheadrightarrow
\tau(\tilde{G}_{n,X,L})/\tau(H_0)={\mathcal{G}}/\tau(H_0)\twoheadrightarrow
{\mathcal{G}}/\ZZ,$$ whose composition is a surjective
homomorphism $\tilde{G}_{n,X,L}/H_0\twoheadrightarrow
{\mathcal{G}}/\ZZ$. Using (2), we get the desired surjective
homomorphism $\Gamma\twoheadrightarrow {\mathcal{G}}/\ZZ$.
\end{proof}

\begin{proof}[Proof of Lemma \ref{sur}]
Replacing $E$ by its algebraic closure $E_a$ and tensoring $T$ and
$S$ by $E_a$, we may and will assume that $E$ is algebraically
closed. Then $S=\M_n(E)$ for some positive integer $n$. Clearly,
$\beta(T)$ is a direct sum of say, $b$ matrix algebras over $E$
and the center of $\beta(T)$ is isomorphic to a direct sum of $b$
copies of $E$. In particular, if $b>1$ then the centralizer of
$\beta(T)$ in $S$ contains the $b$-dimensional center of
$\beta(T)$ which gives us the contradiction. So, $b=1$ and
$\beta(T)\cong \M_k(E)$ for some positive integer $k$. Clearly, $k
\le n$; if the equality holds then we are done. Assume that $k<n$:
we need to get a contradiction. So, we have
$$1 \in E \subset \beta(T)\cong \M_k(E)\hookrightarrow \M_n(E)=S.$$
This provides $E^n$ with a structure of faithful $\beta(T)$-module
in such a way that $E^n$ does not contain a non-zero submodule
with trivial (zero) action of $\beta(T)$. Since $\beta(T)\cong
\M_k(E)$, the $\beta(T)$-module $E^n$ splits into a direct sum of
say, $e$  copies of a simple faithful $\beta(T)$-module $W$ with
$\dim_E(W)=k$. Clearly, $e=n/k>1$. It follows easily that the
centralizer of $\beta(T)$ in $S=\M_n(E)$ coincides with
$$\End_{\beta(T)}(W^e)=\M_e(\End_{\beta(T)}(W))=\M_e(E)$$
and has $E$-dimension $e^2>1$. Contradiction.
\end{proof}

\begin{cor}
\label{odd} Suppose  that $\ell$ is a prime, $K$ is a field of
characteristic different from $\ell$. Suppose that $X$ is an
abelian variety of positive dimension $g$ defined over  $K$.   Let
us assume that $\tilde{G}_{\ell,X,K}$ contains a perfect subgroup
$\mathcal{G}$ that enjoys the following properties:
\begin{enumerate}
\item[(a)]
$\End_{\mathcal{G}}(X_{\ell})=\F_{\ell}$;
\item[(b)]
The only  subgroup of index dividing $g$ in $\mathcal{G}$ is
$\mathcal{G}$ itself.
\end{enumerate}
If $g$ is odd then either $\End^0(X)$ is a matrix algebra over
$\Q$ or $p=\fchar(K)>0$ and $\End^0(X)$ is a matrix algebra
$\M_d(\H_p)$
 over a quaternion $\Q$-algebra $\H_p$  that is
ramified exactly at $p$ and $\infty$ and $d>1$. In particular, if
$\fchar(K)$ does not divide $\#(\mathcal{G})$ then $\End^0(X)$ is
a matrix algebra over $\Q$.
\end{cor}
\begin{proof}[Proof of Corollary \ref{odd}]
Let us assume that $\End^0(X)$ is {\sl not} isomorphic to a matrix
algebra over $\Q$. Then $\End^0(X)$ is (isomorphic to) a matrix
algebra $\M_d(\H)$ over
 a quaternion $\Q$-algebra $\H$. This means that there exists an
 absolutely simple abelian variety $Y$ over $K_a$ such that $X$ is
 isogenous to $Y^d$ and $\End^0(Y)=\H$. Clearly, $\dim(Y)$ is odd.
It follows from Albert's classification \cite[p. 202]{MumfordAV}
that $p:=\fchar(K_a)=\fchar(K)> 0$.  By Lemma 4.3 of \cite{Oort},
if there exists a prime $q\ne p$ such that $\H$ is unramified at
$q$ then $4=\dim_{\Q}\H$ divides $2\dim(Y)$. Since $\dim(Y)$ is
odd, $2\dim(Y)$ is not divisible by $4$ and therefore $\H$ is
unramified at all primes different from $p$. It follows from the
theorem of Hasse-Brauer-Noether  that $\H \cong \H_p$.

Now, assume that $d=1$, i.e. $\End^0(X)=\H_p$. We know that
$\End^0(X)^{*}=\H_p^*$ contains a nontrivial finite perfect group
$\Pi$. But this contradicts to the following elementary statement,
whose proof will be given later in this section.

\begin{lem}
\label{Hp}
 Every finite subgroup in $\H_p^*$ is
solvable.
\end{lem}

Hence $\End^0(X)\ne \H_p$, i.e. $d>1$.

 Assume now that $p$ does {\sl not} divide
$\#(\mathcal{G})$. It follows from Theorem \ref{center2} that $\H$
is unramified at $p$. This implies that $\H$ can be ramified only
at $\infty$ which could not be the case. The obtained
contradiction proves that $\End^0(X)$ is a matrix algebra over
$\Q$.
\end{proof}

\begin{proof}[Proof of Lemma \ref{Hp}]
If $p\ne 2$ then $\H_p^*\subset (\H_p\otimes_{\Q}\Q_2)^{*}\cong
\GL(2,\Q_2)$ and if $p=2$ then $\H_2^*\subset
(\H_2\otimes_{\Q}\Q_3)^{*}\cong \GL(2,\Q_3)$. Since every finite
subgroup in $\GL(2,\Q_2)$ (resp. $\GL(2,\Q_3)$) is conjugate to a
finite subgroup in $\GL(2,\Z_2)$ (resp. $\GL(2,\Z_3)$), it
suffices to check that every finite subgroup in $\GL(2,\Z_2)$ and
$\GL(2,\Z_3)$ is solvable.

Recall that both $\GL(2,\F_2)$ and $\GL(2,\F_3)$ are solvable and
use the Minkowski-Serre lemma (\cite[pp. 124--125]{SerreL}; see
also \cite{SZ}).  This lemma asserts, in particular, that if $q$
is an odd prime  then the kernel of the reduction map
$\GL(n,\Z_q)\to\GL(n,\F_q)$ does not contain nontrivial elements
of finite order and that all periodic elements in the kernel of
the reduction map $\GL(n,\Z_2)\to\GL(n,\F_2)$ have order $1$ or
$2$.

 Indeed, every
finite subgroup $\Pi\subset \GL(2,\Z_3)$ maps injectively in
$\GL(2,\F_3)$ and therefore is solvable. If $\Pi\subset
\GL(2,\Z_2)$ is a finite subgroup then the kernel of the reduction
map $\Pi\to \GL(2,\F_2)$ consists of elements of order $1$ or $2$
and therefore is an elementary commutative $2$-group. Since the
image of the reduction map is solvable, we conclude that $\Pi$ is
solvable.
\end{proof}

\begin{cor}
\label{int} Suppose  that $\ell$ is a prime, $K$ is a field of
characteristic different from $\ell$. Suppose that $X$ is an
abelian variety of dimension $g$ defined over  $K$.  Let us put
$g'=\max(2,g)$. Let us assume that $\tilde{G}_{\ell,X,K}$ contains
a perfect subgroup $\mathcal{G}$ that enjoys the following
properties:
\begin{enumerate}
\item[(a)]
$\End_{\mathcal{G}}(X_{\ell})=\F_{\ell}$;
\item[(b)]
The only  subgroup of index dividing $g$ in $\mathcal{G}$ is
$\mathcal{G}$ itself.
\item[(c)] If  $\ZZ$ is the center of $\mathcal{G}$
 then ${\mathcal{G}}/\ZZ$ is a simple nonabelian group.
\end{enumerate}
Suppose that $\End^0(X)\cong \M_d(\Q)$ with $d>1$. Then there
exist a perfect finite  subgroup $\Pi\subset \GL(d,\Z)$ and a
surjective homomorphism $\Pi \twoheadrightarrow
{\mathcal{G}}/\ZZ$.
\end{cor}
\begin{proof}[Proof of Corollary \ref{int}]
Clearly, $\End^0(X)^{*}=\GL(n,\Q)$. One has only to recall that
every finite subgroup in $\GL(n,\Q)$ is conjugate to a finite
subgroup in $\GL(n,\Z)$ \cite[p. 124]{SerreL} and apply Theorem
\ref{center2}(iii).
\end{proof}

\section{Homomorphisms of abelian varieties}

\label{tate}
\begin{thm}
 \label{hom}
 Let $\ell$ be a prime, $K$ a field of characteristic different from
 $\ell$, $X$ and $Y$   abelian varieties of positive dimension defined over $K$.
 Suppose that the following conditions hold:

\begin{enumerate}
\item[(i)]
The extensions $K(X_{\ell})$ and $K(Y_{\ell})$  are linearly
disjoint over $K$.
\item[(ii)]
 $\End_{\tilde{G}_{\ell,X,K}}(X_{\ell})=\F_{\ell}$.
\item[(iii)]
The centralizer of $\tilde{G}_{\ell,Y,K}$ in
$\End_{\F_{\ell}}(Y_{\ell})$ is a field.
\end{enumerate}

Then either $\Hom(X,Y)=0,  \Hom(Y,X)=0$ or $\fchar(K)>0$ and both
abelian varieties $X$ and $Y$ are supersingular.
\end{thm}

\begin{rem} Theorem \ref{hom} was proven in \cite{ZarhinSh}  under  an
addititional assumption that the Galois modules $X_{\ell}$ and
$Y_{\ell}$ are simple. \end{rem}

In order to prove Theorem \ref{hom},  we need first to discuss the
notion of Tate module. Recall \cite{MumfordAV,SerreAb,ZP} that
this is a $\Z_{\ell}$-module $T_{\ell}(X)$
 defined as
the projective limit of Galois modules $X_{\ell^m}$. It is
well-known that $T_{\ell}(X)$ is a free $\Z_{\ell}$-module of rank
$2\dim(X)$ provided with the continuous action
$$\rho_{\ell,X}:\Gal(K) \to \Aut_{\Z_{\ell}}(T_{\ell}(X)).$$
There is  the natural isomorphism of
Galois modules
$$X_{\ell}=T_{\ell}(X)/\ell T_{\ell}(X) \eqno(3),$$
so one may view $\tilde{\rho}_{\ell,X}$ as the reduction of
$\rho_{\ell,X}$ modulo $\ell$.
 Let us put
$$V_{\ell}(X)=T_{\ell}(X)\otimes_{\Z_{\ell}}\Q_{\ell};$$
it is a $2\dim(X)$-dimensional $\Q_{\ell}$-vector space. The group
$T_{\ell}(X)$ is naturally identified with the $\Z_{\ell}$-lattice
in $V_{\ell}(X)$
 and the inclusion
$\Aut_{\Z_{\ell}}(T_{\ell}(X))\subset
\Aut_{\Q_{\ell}}(V_{\ell}(X))$ allows us to view $V_{\ell}(X)$ as
representation of $\Gal(K)$ over $\Q_{\ell}$. Let $Y$ be (may be,
another) abelian variety of positive dimension defined over $K$.
Recall \cite[\S 19]{MumfordAV}  that $\Hom(X,Y)$ is a free
commutative group of finite rank. Since $X$ and $Y$
 are defined over $K$, one may associate with every $u \in \Hom(X,Y)$
and $\sigma \in \Gal(K)$ an endomorphism $^{\sigma}u\ \in
\Hom(X,Y)$ such that
$$^{\sigma}u (x)=\sigma u(\sigma^{-1}x)\quad \forall x \in X(K_a)$$
and we get the group homomorphism
$$\kappa_{X,Y}: \Gal(K) \to \Aut(\Hom(X,Y));
\quad\kappa_{X,Y}(\sigma)(u)=\ ^{\sigma}u \quad \forall \sigma \in
\Gal(K),u \in \Hom(X,Y),$$ which provides the finite-dimensional
$\Q_{\ell}$-vector space $\Hom(X,Y)\otimes\Q_{\ell}$ with the
natural structure of Galois module.

There is a natural structure of  Galois module on the
$\Q_{\ell}$-vector space
$\Hom_{\Q_{\ell}}(V_{\ell}(X),V_{\ell}(Y))$  induced by the Galois
actions on  $V_{\ell}(X)$ and  $V_{\ell}(Y)$. On the other hand,
there is a natural  embedding of Galois modules \cite[\S
19]{MumfordAV},
$$\Hom(X,Y)\otimes\Q_{\ell}\subset\Hom_{\Q_{\ell}}(V_{\ell}(X),V_{\ell}(Y)),$$
whose image must be a $\Gal(K)$-invariant $\Q_{\ell}$-vector
subspace. It is also clear that
$\Hom_{\Z_{\ell}}(T_{\ell}(X),T_{\ell}(Y))$ is a Galois-invariant
$\Z_{\ell}$-lattice in
$\Hom_{\Q_{\ell}}(V_{\ell}(X),V_{\ell}(Y))$. The equality (3)
gives rise to a natural isomorphism of Galois modules
$$\Hom_{\Z_{\ell}}(T_{\ell}(X),T_{\ell}(Y))\otimes_{\Z_{\ell}}\Z_{\ell}/\ell
\Z_{\ell}=\Hom_{\F_{\ell}}(X_{\ell},Y_{\ell}) \eqno(4).$$

\begin{proof}[Proof of Theorem \ref{hom}]
Let $K(X_{\ell},Y_{\ell})$ be the compositum of the fields
 $K(X_{\ell})$ and $K(Y_{\ell})$. The linear disjointness of
$K(X_{\ell})$ and $K(Y_{\ell})$ means that
$$\Gal(K(X_{\ell},Y_{\ell})/K)= \Gal(K(Y_{\ell})/K)\times
\Gal(K(X_{\ell})/K).$$ Let
$X_{\ell}^*=\Hom_{\F_{\ell}}(X_{\ell},\F_{\ell})$ be the dual of
$X_{\ell}$ and $\bar{\rho}_{n,X,K}^{*}:\Gal(K)
\to\Aut(X_{\ell}^{*})$ the dual of $\bar{\rho}_{n,X,K}$. One may
easily check that
$\ker(\bar{\rho}_{n,X,K}^{*})=\ker(\bar{\rho}_{n,X,K})$ and
therefore we have an isomorphism of the images
$$\tilde{G}^{*}_{\ell,X,K}:=\bar{\rho}_{n,X,K}^{*}(\Gal(K))\cong
\bar{\rho}_{n,X,K}(\Gal(K)))=\tilde{G}_{\ell,X,K}.$$ One may also
easily check that the centralizer of $\Gal(K)$ in
$\End_{\F_{\ell}}(X_{\ell}^{*})$ still coincides with $\F_{\ell}$.
It follows that if $A_1$ is the $\F_{\ell}$-subalgebra in
$\End_{\F_{\ell}}(X_{\ell}^*)$ generated by
$\tilde{G}^{*}_{\ell,X,K}$ then its centralizer in
$\End_{\F_{\ell}}(X_{\ell}^*)$ coincides with $\F_{\ell}$.  Let us
consider the Galois module
$W_1=\Hom_{\F_{\ell}}(X_{\ell},Y_{\ell})=X_{\ell}^{*}\otimes_{\F_{\ell}}
Y_{\ell}$ and denote by $\tau$ the homomorphism
 $\Gal(K)\to\Aut(W_1)$ that defines the Galois module structure on
 $W_1$.
One may easily check that $\tau$ factors through
 $\Gal(K(X_{\ell},Y_{\ell})/K)$ and the image of $\tau$ coincides
 with the image of
$$\tilde{G}^{*}_{\ell,X,K}\times \tilde{G}_{\ell,X,Y}\subset
\Aut(X_{\ell}^{*})\times
\Aut(Y_{\ell})\to\Aut(X_{\ell}^{*}\otimes_{\F_{\ell}}
Y_{\ell})=\Aut(W_1).$$

Let $A_2$ be the $\F_{\ell}$-subalgebra in
$\End_{\F_{\ell}}(Y_{\ell})$ generated by $\tilde{G}_{\ell,Y,K}$.
Recall that the centralizer of $\Gal(K)$ in
$\End_{\F_{\ell}}(Y_{\ell})$ is a field, say $\F$. Clearly,  the
centralizer of $A_2$ in $\End_{\F_{\ell}}(Y_{\ell})$ coincides
with $\F$. One may easily check that the subalgebra of
$\End_{\F_{\ell}}(W_1)$ generated by the image of $\Gal(K)$
coincides with
$$A_1\otimes_{\F_{\ell}}A_2\subset
\End_{\F_{\ell}}(X_{\ell}^*)\otimes_{\F_{\ell}}\End_{\F_{\ell}}(Y_{\ell})=\End_F(X_{\ell}^{*}\otimes_{\F_{\ell}}
Y_{\ell})=\End_{\F_{\ell}}(W_1).$$ It follows from Lemma (10.37)
on p. 252 of \cite{CR} that the centralizer of
$A_1\otimes_{\F_{\ell}}A_2$ in
$\End_F(X_{\ell}^{*}\otimes_{\F_{\ell}} Y_{\ell})$ coincides with
$\F_{\ell}\otimes_{\F_{\ell}}\F=\F$. This implies that the
centralizer of $\Gal(K)$ in
$\End_F(X_{\ell}^{*}\otimes_{\F_{\ell}}
Y_{\ell})=\End_{\F_{\ell}}(W_1)$ is the field $\F$.

Let us consider the $\Q_{\ell}$-vector space
$V_1=\Hom_{\Q_{\ell}}(V_{\ell}(X),V_{\ell}(Y))$ and the free
$\Z_{\ell}$-module $T_1=\Hom_{\Z_{\ell}}(T_{\ell}(X),T_{\ell}(Y))$
provided with the natural structure of Galois modules. Clearly,
$T_1$ is a Galois-stable $\Z_{\ell}$-lattice in $V_1$. By (4),
there is a natural isomorphism of Galois modules $W_1=T_1/\ell
T_1$. Let us denote by $D_1$  the centralizer of $\Gal(K)$ in
$\End_{\Q_{\ell}}(V_1)$. Clearly, $D_1$ is a finite-dimensional
$\Q_{\ell}$-algebra. Therefore in order to prove that $D_1$ is a
division algebra, it suffices to check that $D_1$  has no zero
divisors.

Suppose that $D_1$ has zero divisors, i.e. there are non-zero
$u,v\in D_1$ with $uv=0$. We have $u,v \subset D_1 \subset
\End_{\Q_{\ell}}(V_1)$. Multiplying $u$ and $v$ by proper powers
of $\ell$, we may and will assume that $u (T_1)\subset T_1,
v(T_1)\subset T_1$ but $u(T_1)$ is {\sl not} contained in $\ell
T_1$ and $v(T_1)$ is {\sl not} contained in $\ell T_1$. This means
that $u$ and $v$ induce {\sl non-zero} endomorphisms
$\bar{u},\bar{v}\in \End(W_1)$ that commute with $\Gal(K)$ and
$\bar{u}\bar{v}=0$. Since both $\bar{u}$ and $\bar{v}$ are
non-zero elements of the field $\F$, we get a contradiction that
proves that $D_1$  has no zero divisors and therefore is a
division algebra.

{\sl End of the proof of Theorem} \ref{hom}. We may and will
assume that $K$ is finitely generated over its prime subfield
(replacing $K$ by its suitable subfield). Then the conjecture of
Tate \cite{Tate} (proven by the author in characteristic $>2$
\cite{ZarhinI,ZarhinT}, Faltings in characteristic zero
\cite{Faltings1,Faltings2} and Mori in characteristic $2$
\cite{MB}) asserts that the natural representation of $\Gal(K)$ in
$V_{\ell}(Z)$ is completely reducible for any abelian variety $Z$
over $K$. In particular,   the natural representations of
$\Gal(K)$ in $V_{\ell}(X)$ and $V_{\ell}(Y)$ are completely
reducible. It follows easily that the dual Galois representation
in $\Hom_{\Q_{\ell}}(V_{\ell}(X),\Q_{\ell})$ is also completely
reducible.  Since $\Q_{\ell}$ has characteristic zero, it follows
from a theorem of Chevalley \cite[p. 88]{Chevalley} that the
Galois representation in the tensor product
$\Hom_{\Q_{\ell}}(V_{\ell}(X),\Q_{\ell})\otimes_{\Q_{\ell}}V_{\ell}(Y)=
\Hom_{\Q_{\ell}}(V_{\ell}(X),V_{\ell}(Y))=:V_1$ is completely
reducible. The complete reducibility implies easily that $V_1$ is
an irreducible Galois representation, because the centralizer is a
division algebra. Recall that $\Hom(X,Y)\otimes\Q_{\ell}$ is a
Galois-invariant subspace in
$\Hom_{\Q_{\ell}}(V_{\ell}(X),V_{\ell}(Y))=V_1$. The
irreducibility of $V_1$ implies that either
$\Hom(X,Y)\otimes\Q_{\ell}=0$ or $\Hom(X,Y)\otimes\Q_{\ell}=V_1$.

 If $\Hom(X,Y)\otimes\Q_{\ell}=0$ then $\Hom(X,Y)=0$ and therefore
$\Hom(Y,X)=0$.

If $\Hom(X,Y)\otimes\Q_{\ell}=V_1$ then the rank of the free
commutative group $\Hom(X,Y)$ coincides with the dimension of the
$\Q_{\ell}$-vector space $V_1$. Clearly, $V_1$ has dimension
$4\dim(X)\dim(Y)$. It is proven in Proposition 3.3 of
\cite{ZarhinSh} that if $A$ and $B$ are abelian varieties over an
algebraically closed field ${\mathcal{K}}$ and the rank of
$\Hom(A,B)$ equals $4\dim(A)\dim(B)$ then
$\fchar({\mathcal{K}})>0$ and both $A$ and $B$ are supersingular
abelian varieties. Applying this result to $X$ and $Y$, we
conclude that $\fchar(K)=\fchar(K_a)>0$ and both $X$ and $Y$ are
supersingular abelian varieties.
\end{proof}

\section{Hyperelliptic jacobians}
\label{jac} In this section we deal with the case of $\ell=2$.
 Suppose that $\fchar(K)\ne 2$. Let $f(x)\in K[x]$ be a polynomial
of degree $n\ge 3$ without multiple roots. Let $\RR_f\subset K_a $
be the set of roots of $f$. Clearly, $\RR_f$ consists of $n$
elements. Let $K(\RR_f)\subset K_a$ be the splitting field of $f$.
Clearly, $K(\RR_f)/K$ is a Galois extension and we write $\Gal(f)$
for its Galois group $\Gal(K(\RR_f)/K)$. By definition,
$\Gal(K(\RR_f)/K)$ permutes elements of  $\RR_f$; further we
identify $\Gal(f)$ with the corresponding subgroup of
$\Perm(\RR_f)$ where $\Perm(\RR_f)$ is the group of permutations
of $\RR_f$.

We write $\F_2^{\RR_f}$ for the $n$-dimensional $\F_2$-vector
space of maps $h:\RR_f \to \F_2$. The space $\F_2^{\RR_f}$ is
provided with a natural action of $\Perm(\RR_f)$ defined as
follows. Each $s \in \Perm(\RR_f)$ sends a map
 $h:\RR_f\to \F_2$ to  $sh:\alpha \mapsto h(s^{-1}(\alpha))$. The permutation module $\F_2^{\RR_f}$
contains the $\Perm(\RR_f)$-stable hyperplane
$$(\F_2^{\RR_f})^0=
\{h:\RR_f\to \F_2\mid\sum_{\alpha\in \RR_f}h(\alpha)=0\}$$ and the
$\Perm(\RR_f)$-invariant line $\F_2 \cdot 1_{\RR_f}$ where
$1_{\RR_f}$ is the constant function $1$. Clearly,
$(\F_2^{\RR_f})^0$ contains $\F_2 \cdot 1_{\RR_f}$ if and only if
$n$ is even.

 If  $n$ is even then
let us define the $\Gal(f)$-module
$Q_{\RR_f}:=(\F_2^{\RR_f})^0/(\F_2 \cdot 1_{\RR_f})$. If $n$ is
odd then let us put $Q_{\RR_f}:=(\F_2^{\RR_f})^0$. If $n \ne 4$
the natural representation of $\Gal(f)$ is faithful, because in
this case the natural homomorphism
$\Perm(\RR_f)\to\Aut_{\F_2}(Q_{\RR_f})$ is injective.

\begin{rem}
\label{klemm}
 It is known \cite[Satz 4]{Klemm},  that
$\End_{\Gal(f)}(Q_{\RR_f})=\F_2$ if either $n$ is odd and
$\Gal(f)$ acts doubly transitively on $\RR_f$ or $n$ is even and
$\Gal(f)$ acts  $3$-transitively on $\RR_f$.
\end{rem}

The canonical surjection $\Gal(K)\twoheadrightarrow
\Gal(K(\RR_f)/K)=\Gal(f)$ provides $Q_{\RR_f}$ with a natural
structure of $\Gal(K)$-module. Let $C_f$ be the hyperelliptic
curve $y^2=f(x)$ and $J(C_F)$ its jacobian. It is well-known that
$J(C_F)$ is a $\left[\frac{n-1}{2}\right]$-dimensional abelian
variety defined over $K$. It is also well-known that the
$\Gal(K)$-modules $J(C_f)_2$ and $Q_{\RR_f}$ are isomorphic (see
for instance \cite{Poonen,SPoonen,ZarhinTexel}). It follows that
if $n\ne 4$ then
$$\Gal(f)=\tilde{G}_{2,J(C_f)}.$$
It follows from Remark \ref{klemm} that if either $n$ is odd and
$\Gal(f)$ acts doubly transitively on $\RR_f$ or $n$ is even and
$\Gal(f)$ acts  $3$-transitively on $\RR_f$ then
$$\End_{\tilde{G}_{2,J(C_f)}}(J(C_f)_2))=\F_2.$$
It is also clear that $K(J(C_f)_2))\subset K(\RR_f)$. (The
equality holds if  $n\ne 4$.)

The next assertion follows immediately from Theorem \ref{center2},
Corollaries \ref{odd} and \ref{int} (applied to
$X=J(C_f),\ell=2,\GG=\Gal(f)$).

\begin{thm}
\label{double}
 Let $K$ be a field of characteristic different from
$2$, let $n\ge 5$ be an integer, $g=\left[\frac{n-1}{2}\right]$
and $f(x)\in K[x]$ a polynomial of degree $n$. Suppose that either
$n$ is odd and $\Gal(f)$ acts doubly transitively on $\RR_f$ or
$n$ is even and $\Gal(f)$ acts  $3$-transitively on $\RR_f$.
Assume also that $\Gal(f)$ is a simple nonabelian group that does
not contain a subgroup of index dividing $g$ except $\Gal(f)$
itself. If $g$ is odd then $\End^0(J(C_f))$ enjoys one of the
following properties:
\begin{itemize}
\item[(i)]
$\End^0(J(C_f))$ is isomorphic to the matrix algebra $\M_d(\Q)$
where $d$ divides $g$. If $d>1$ there exist a finite perfect group
$\Pi\subset\GL(d,\Z)$ and a surjective homomorphism
$\Pi\twoheadrightarrow\Gal(f)$ such that every prime dividing
$\#(\Pi)$ also divides $\#(\Gal(f))$.
\item[(ii)]
$p:=\fchar(K)$ is a prime dividing $\#(\Gal(f))$ and
$\End^0(J(C_f))$ is isomorphic to the matrix algebra $\M_d(\H_p)$
where $d>1$ divides $g$.
\end{itemize}
\end{thm}

\begin{ex}
\label{a5} Suppose that $n=5$ and $\Gal(f)$ is the alternating
group $\A_5$ acting doubly transitively on $\RR_f$. Clearly, $g=2$
and $\Gal(f)$ is a simple nonabelian group without subgroups of
index $2$.  Applying Theorem \ref{double}, we conclude that
$\End^0(J(C_f))$ is either $\Q$ or $\M_2(\Q)$ or $\M_2(\H)$ where
$\H$ is a quaternion $\Q$-algebra unramified outside
$\{\infty,2,3,5\}$; in addition $\H\cong \H_p$ if
$p:=\fchar(K)>0$. Suppose that $\End(J(C_f))\ne \Z$ and therefore
$\End^0(J(C_f))\ne \Q$. If $\End^0(J(C_f))=\M_2(\Q)$ then
$\GL(2,\Q)=\M_2(\Q)^{*}$ contains a finite group, whose order
divides $5$, which is not the case. This implies that
$\End^0(J(C_f))=\M_2(\H)$. This means that $J(C_f)$ is
supersingular and therefore  $p:=\fchar(K)>0$. This implies that
$p=3$ or $p=5$.

We conclude that either $\End(J(C_f))= \Z$ or $\fchar(K)\in
\{3,5\}$ and $J(C_f)$ is a supersingular abelian varietiy. In
fact, it is known \cite{ZarhinBSMF} that if $\fchar(K)=5$ then
$\End(J(C_f))= \Z$. On the other hand, one may find a
supersingular $J(C_f)$ in characteristic $3$ \cite{ZarhinBSMF}.
\end{ex}

Example \ref{a5} is a special case of the following general result
proven by the author \cite{ZarhinMRL,ZarhinMMJ,ZarhinBSMF}. {\sl
Suppose that $n\ge 5$ and $\Gal(f)$ is the alternating group
$\A_n$ acting on $\RR_f$. If $\fchar(K)=3$ we assume additionally
that $n\ge 7$. Then $\End(J(C_f))= \Z$}.

We refer the reader to
~\cite{Mori1,Mori2,Katz1,Katz2,Masser,KS,ZarhinMRL,ZarhinMRL2,ZarhinMMJ,ZarhinP,ZarhinPAMS,ZarhinSt}
for a discussion of other known results about, and examples of,
hyperelliptic jacobians without complex multiplication.

\begin{cor}
\label{l27} Suppose that $n=7$ and
$\Gal(f)=\SL_3(\F_2)\cong\PSL_2(\F_7)$ acts doubly transitively on
$\RR_f$. Then $\End^0(J(C_f))=\Q$ and therefore $\End(J(C_f))=\Z$.
\end{cor}
\begin{proof}
We have $g=\dim(J(C_f))=3$. Since $\PSL_2(\F_7)$ is a simple
nonabelian group it does not contain a subgroup of index $3$. So,
we may apply Theorem \ref{double}. We obtain that if
$\End^0(J(C_f))\ne\Q$ then either $\End^0(J(C_f))=\M_3(\Q)$ and
there exist a finite perfect group $\Pi\subset\GL(3,\Z)$ and a
surjective homomorphism
$\Pi\twoheadrightarrow\Gal(f)=\PSL_2(\F_7)$ or
$\End^0(J(C_f))=\M_3(\H_p)$ where $p=\fchar(K)$ is either $3$ or
$7$. The  case of $\End^0(J(C_f))=\M_3(\H_p)$ means that $J(C_f)$
is supersingular,  which is not true \cite[Th. 3.1]{ZarhinBSMF}.
Hence $\End^0(J(C_f))=\M_3(\Q)$ and $\GL(3,\Z)$ contains a finite
group, whose order is divisible by $7$. It follows that
$\GL(3,\Z)$ contains an element of order  $7$, which is not true.
The obtained contradiction proves that $\End^0(J(C_f))=\Q$ and
therefore $\End(J(C_f))=\Z$.
\end{proof}

\begin{cor}
\label{l211} Suppose that $n=11$ and $\Gal(f)=\PSL_2(\F_{11})$
acts doubly transitively on $\RR_f$. Then $\End^0(J(C_f))=\Q$ and
therefore $\End(J(C_f))=\Z$.
\end{cor}
\begin{proof}
We have $g=\dim(J(C_f))=5$. It is known \cite{Atlas} that
$\PSL_2(\F_{11})$ is a simple nonabelian subgroup  not containing
a subgroup of index $5$. So, we may apply Theorem \ref{double}. We
obtain that if $\End^0(J(C_f))\ne\Q$ then either
$\End^0(J(C_f))=\M_5(\Q)$ and there exist a finite perfect group
$\Pi\subset\GL(5,\Z)$ and a surjective homomorphism
$\Pi\twoheadrightarrow\Gal(f)=\PSL_2(\F_{11})$ or
$\End^0(J(C_f))=\M_5(\H_p)$ where $p=\fchar(K)$ is either $3$ or
$5$ or $11$.

Assume that  $\End^0(J(C_f))=\M_5(\Q)$. Then $\GL(5,\Z)$ contains
a finite group,  whose order is divisible by $11$. It follows that
$\GL(5,\Z)$ contains an element of order $11$, which is not true.
Hence $\End^0(J(C_f))\ne\M_5(\Q)$.

Assume that $\End^0(J(C_f))=\M_5(\H_p)$ where $p$ is either $3$ or
$5$ or $11$. This implies that  $J(C_f)$ is a supersingular
abelian variety.

Notice that every homomorphism from simple $\PSL_{2}(\F_{11})$ to
$\GL(4,\F_2)$ is trivial, because $11$ divides
$\#(\PSL_{2}(\F_{11}))$  but $\#(\GL(4,\F_2))$ is {\sl not}
divisible by $11$. Since $4=g-1$, it follows from Theorem 3.3 of
\cite{ZarhinBSMF} (applied to $g=5,
X=J(C_f),G=\Gal(f)=\PSL_{2}(\F_{11})$) that there exists a central
extension $\pi_1:G_1 \to \PSL_{2}(\F_{11})$ such that $G_1$ is
perfect,  $\ker(\pi_1)$ is a cyclic group of order $1$ or $2$ and
$\M_5(\H_p)$ is a direct summand of the group $\Q$-algebra
$\Q[G_1]$. It follows easily that  $G_1=\PSL_{2}(\F_{11})$ or
$\SL_{2}(\F_{11})$. It is known \cite{J,F} that
$\Q[\PSL_{2}(\F_{11})]$ is a direct sum of matrix algebras over
fields. Hence $G_1=\SL_{2}(\F_{11})$ and the direct summand
$\M_5(\H_p)$ corresponds to a faithful ordinary irreducible
character $\chi$ of $\SL_{2}(\F_{11})$ with degree $10$ and
$\Q(\chi)=\Q$. This implies that in notations of \cite[\S 38]{DA},
 $\chi=\theta_j$ where $j$ is an odd integer such that $1\le j
\le \frac{11-1}{2}=5$ and either $6j$ is divisible by $11+1=12$ or
$4j$ is divisible by $12$ (\cite{F}, Th. 6.2 on p. 285). This
implies that $j=3$ and $\chi=\theta_3$. However, the direct
summand attached to $\theta_3$ is ramified at $2$ (\cite[the case
(c) on p. 4]{J};  \cite[theorem  6.1(iii) on p. 284]{F}). Since $p
\ne 2$, we get a contradiction which
  proves that $J(C_f)$ is not
supersingular. This implies that $\End^0(J(C_f))=\Q$ and therefore
$\End(J(C_f))=\Z$.
\end{proof}

\begin{cor}
\label{m12} Suppose that $n=12$ and $\Gal(f)$ is the Mathieu group
$\M_{12}$ acting  $3$-transitively on $\RR_f$. Then
$\End(J(C_f))=\Z$.
\end{cor}
\begin{proof}
Let $\alpha$ be a root of $f(x)$ and $K_1=K(\alpha)$. Clearly, the
stabilizer of $\alpha$ in $\Gal(f)=\M_{12}$ is $\PSL_2(\F_{11})$
acting doubly transitively on the roots of
$f_1(x)=\frac{f(x)}{x-\alpha}\in K_1[x]$. Let us put
$h(x)=f_1(x+\alpha)\in K_1[x],  h(x)=x^{11} h(1/x)\in K_1[x]$.
Clearly, $\deg(h_1)=11$ and  $\Gal(h_1)=\PSL_2(\F_{11})$ acts
doubly transitively on the roots of $h_1$. By Corollary
\ref{l211}, $\End(J(C_{h_1}))=\Z$. On the other hand, the standard
substitution $x_1=1/(x-\alpha),  y_1=y/(x-\alpha)^6$ establishes a
birational isomorphism between $C_f$ and $C_{h_1}:y_1^2=h_1(x_1)$.
This implies that $J(C_f)\cong J(C_{h_1})$ and therefore
$\End(J(C_f))=\Z$.
\end{proof}

In characteristic zero the assertions of Corollaries \ref{l27},
\ref{l211} and \ref{m12} were earlier proven in
\cite{ZarhinBSMF,ZarhinTexel}.

\begin{cor}
\label{mat} Suppose that $\deg(f)=n$ where $n=22,23$ or $24$ and
$\Gal(f)$ is the corresponding (at least) $3$-transitive Mathieu
group $\MM_n\subset\Perm(\RR_f)\cong \Sn$. Then $\End(J(C_f))=\Z$.
\end{cor}
\begin{proof}
First, assume that $n=23$ or $24$. We have $g=\dim(J(C_f))=11$. It
is known that both $\MM_{23}$ and $\MM_{24}$ do not contain a
subgroup of index $11$ \cite{Atlas}. So, we may apply Theorem
\ref{double} and obtain that if $\End(J(C_f)\ne\Z$ then
$\End^0(J(C_f))\ne\Q$ and one of the following conditions holds:

\begin{itemize}
\item[(i)]
$\End^0(J(C_f))=\M_{11}(\Q)$ and there exist a finite perfect
group $\Pi\subset\GL(11,\Z)$ and a surjective homomorphism
$\Pi\twoheadrightarrow\Gal(f)=\MM_n$;
\item[(ii)]
$p=\fchar(K)\in\{3,5,7,11,23\}$ and
$\End^0(J(C_f))=\M_{11}(\H_p)$.
\end{itemize}

Assume that the condition (i) holds. Then
$\End^0(J(C_f))=\M_{11}(\Q)$ and $\GL(11,\Z)$ contains a finite
group, whose  order is divisible by $23$. It follows that
$\GL(11,\Z)$ contains an element of order  $23$, which is not
true. The obtained contradiction proves that the condition (i) is
not fulfilled.

Hence  the condition (ii) holds. Then
$p=\fchar(K)\in\{3,5,7,11,23\}$ and  there exist a finite perfect
subgroup $\Pi\subset \End^0(J(C_f))^{*}=\GL(11,\H_p)$ and a
surjective homomorphism $\pi:\Pi\twoheadrightarrow\MM_n$.
Replacing $\Pi$ by a suitable subgroup, we may and will assume
that no proper subgroup of $\Pi$ maps onto $\MM_{n}$. By tensoring
$\H_p$ to the field of complex numbers (over $\Q$), we obtain an
embedding
$$\Pi \subset \GL(11,\H_p)\subset\GL(22,\C).$$
In particular, the (perfect) group $\Pi$ admits a non-trivial
projective  $22$-dimensional representation over $\C$. Recall that
$\MM_{n}$ has Schur's multiplier $1$ (since $n=23$ or $24$)
\cite{Atlas} and therefore all its projective representations are
(obtained from) linear representations. Also, all nontrivial
linear representations of $\MM_{24}$ have dimension $\ge 23$,
because the smallest dimension of a nontrivial linear
representation of $\MM_{24}$ is $23$. It follows from results of
Feit--Tits \cite{FT} that $\Pi$ cannot have  a non-trivial
projective representation of dimension $<23$. This implies that $n
\ne 24$, i.e. $n=23$.

Recall that $22$ is the smallest possible dimension of a
nontrivial representation of $\MM_{23}$ in characteristic zero,
because its every irreducible representation in characteristic
zero has dimension $\ge 22$ \cite{Atlas}. It follows from a
theorem of Feit--Tits (\cite{FT}, pp. 1 and \S 4; see also
\cite{KL}) that the projective representation
$$\Pi \to \GL(11,\H_p)/\Q^{*}\subset\GL(22,\C)/\C^{*}$$
factors through $\ker(\pi)$. This means that $\ker(\pi)$ lies in
$\Q^{*}$ and therefore $\Pi$ is a central extension of $\MM_{23}$.
Now the perfectness of $\Pi$ implies that $\pi$ is an isomorphism,
i.e. $\Pi\cong\MM_{23}$.

Let us consider the natural homomorphism $\Q[\MM_{23}]\cong
\Q[\Pi] \to \M_{11}(\H_p)$ induced by the inclusion $\Pi\subset
\M_{11}(\H_p)^{*}$. It is surjective, because otherwise one may
construct a (complex) nontrivial representation of $\MM_{23}$ of
dimension $<22$. This implies that $\M_{11}(\H_p)$ is isomorphic
to a direct summand of $\Q[\MM_{23}]$. But this is not true, since
Schur indices of all irreducible representations of $\MM_{23}$ are
equal to $1$ \cite[\S 7]{F} and therefore $\Q[\MM_{23}]$ splits
into a direct sum of matrix algebras over fields. The obtained
contradiction proves that the condition (ii) is not fulfilled. So,
$\End(J(C_f))=\Z$.

Now let $n=22$.  Then $g=10$. It is known that $\MM_{22}$ is a
simple nonabelian group
 not containing a subgroup of index $10$ \cite{Atlas}. Let us assume that
 $\End^0(J(C_f))\ne\Q$.
 Applying Theorem
\ref{center2}, we conclude that there exists a positive integer
$d$ dividing $10$ such that either $d>1$ and
$\End^0(J(C_f))=\M_d(\Q)$ or $\End^0(J(C_f))=\M_d(\H)$ where $\H$
is a quaternion $\Q$-algebra unramified outside $\infty$ and the
prime divisors of $\#(\MM_{22})$. In addition, there exist a
finite perfect subgroup $\Pi\subset \End^0(J(C_f))^{*}$ and a
surjective homomorphism $\pi:\Pi\twoheadrightarrow\MM_{22}$.
Replacing $\Pi$ by a suitable subgroup, we may and will assume
(without losing the perfectness) that no proper subgroup of $\Pi$
maps onto $\MM_{n}$.

 By Lemma 3.13 on pp. 200--201 of \cite{ZarhinP}, every
homomorphism from $\Pi$ to $\PSL(10,\R)$ is trivial. The
perfectness of $\Pi$ implies that every homomorphism from $\Pi$ to
$\PGL(10,\R)$ is trivial. Since $\M_d(\Q)^*=\GL(d,\Q)\subset
\GL(10,\R)$, we conclude that $\End^0(J(C_f))\ne \M_d(\Q)$ and
therefore $\End^0(J(C_f))=\M_d(\H)$.

If $d=10$ then $p:=\fchar(K)>0$ and  $J(C_f)$ is a supersingular
abelian variety.

 Assume that $d \ne 10$, i.e. $d=1,2$ or $5$. If $H$ is unramified at $\infty$
 then there exists an embedding $\H \hookrightarrow \M_2(\R)$.
 This gives us the embeddings
$$\Pi\subset \M_d(\H)^{*}\hookrightarrow \M_{2d}(\R)^{*}=\GL(2d,\R)\subset\GL(10,\R)$$
and therefore there is a nontrivial homomorphism from $\Pi$ to
$\PGL(10,\R)$. The obtained contradiction proves that $\H$ is
ramified at $\infty$.

There exists an embedding $\H\hookrightarrow \M_4(\Q)\subset
\M_4(\R)$. This implies that if $d=1$ or $2$ then there are
embeddings
$$\Pi\subset \M_d(\H)^{*}\hookrightarrow \M_{4d}(\R)^{*}=\GL(4d,\R)\subset\GL(10,\R)$$
and therefore there is a nontrivial homomorphism from $\Pi$ to
$\PGL(10,\R)$. The obtained contradiction proves that $d=5$. This
means that there exists an abelian surface $Y$ over $K_a$ such
that $J(C_f)$ is isogenous to $Y^5$ and $\End^0(Y)=\H$. However,
there do not exist abelian surfaces, whose endomorphism algebra is
a definite quaternion algebra over $\Q$. This result is well-known
in characteristic zero (see, for instance \cite{OZ}); the positive
characteristic  case was done by Oort \cite[Lemma 4.5 on p.
490]{Oort}. Hence $d\ne 5$. This implies that $d=10$ and $J(C_f)$
is a supersingular abelian variety.

 Since
$\MM_{22}$ is a simple group and $11\mid\#(\MM_{22})$, every
homomorphism from $\MM_{22}$ to $\GL(9,\F_2)$ is trivial, because
$\#(\GL(9,\F_2))$ is not divisible by $11$. Since $9=g-1$, it
follows from Theorem 3.3 of \cite{ZarhinBSMF} (applied to $g=10,
X=J(C_f),G=\Gal(f)=\MM_{22}$) that there exists a central
extension $\pi_1:G_1 \to \MM_{22}$ such that $G_1$ is perfect,
$\ker(\pi_1)$ is a cyclic group of order $1$ or $2$ and there
exists a faithful $20$-dimensional absolutely irreducible
representation of $G_1$ in characteristic zero. However, such a
central extension with $20$-dimensional irreducible representation
does not exist \cite{Atlas}.
\end{proof}

Combining Corollary \ref{mat} with previous author's results
\cite{ZarhinTexel,ZarhinMMJ} concerning small Mathieu groups, we
obtain the following statement.

\begin{thm}
\label{m} Suppose that $n\in\{11,12,22,23,24\}$ and $\Gal(f)$ is
the corresponding  Mathieu group $\MM_n\subset\Perm(\RR_f)\cong
\Sn$. Then $\End(J(C_f))=\Z$.
\end{thm}

In characteristic zero  the assertion of Theorem \ref{m} was
earlier proven in \cite{ZarhinTexel,ZarhinP}.

\begin{thm}
Suppose that $n=15$ and $\Gal(f)$ is the alternating group $\A_7$
acting doubly transitively on $\RR_f$. Then either
$\End(J(C_f))=\Z$ or $J(C_f)$ is isogenous over $K_a$ to a product
of elliptic curves.
\end{thm}

\begin{proof}
We have $g=7$. Unfortunately, $\A_7$ has a subgroup of index $7$.
However, $\A_7$ is simple nonabelian and does not have a normal
subgroup of index $7$. Applying Theorem \ref{center2} to
$X=J(C_f),g=7, \ell=2,\GG=\Gal(f)=\A_7$,
 we obtain that either $J(C_f)$ is isogenous to a product of
 elliptic curves (case (a))
  or $\End^0(J(C_f))$ is a central simple
 $\Q$-algebra (case (b)). If $\End^0(J(C_f))$ is a matrix algebra over $\Q$
 then either $\End^0(J(C_f))=\Q$ (i.e., $\End(J(C_f))=\Z$) or $\End^0(J(C_f))=\M_7(\Q)$
 (i.e., $J(C_f)$ is isogenous to the $7$th power of an elliptic curve
 without complex multiplication).

 If the central simple
 $\Q$-algebra $\End^0(J(C_f))$ is not a matrix algebra over $\Q$ then there
 exists a quaternion $\Q$-algebra $\H$ such that either
 $\End^0(J(C_f))=\H$ or $\End^0(J(C_f))=\M_7(\H)$. If
 $\End^0(J(C_f))=\M_7(\H)$ then $J(C_f)$ is a supersingular abelian
 variety and therefore is isogenous to a product of elliptic
 curves.

 Let us assume that $\End^0(J(C_f))=\H$. We need to arrive to a
 contradiction. Since $7=\dim(J(C_f))$ is odd,
 $p=\fchar(K)>0$. The same arguments as in the proof of Corollary
 \ref{odd} tell us that $\H=\H_p$. By Theorem \ref{center2}(b3), there
 exist a perfect finite group $\Pi\subset\End^0(J(C_f))^*=\H_p^{*}$ and a
 surjective homomorphism $\Pi\twoheadrightarrow\A_7$. But Lemma \ref{Hp} asserts
 that every finite subgroup in $\H_p^{*}$ is solvable. The obtained
 contradiction proves that $\End^0(J(C_f))\ne\H_p$.
\end{proof}

\begin{thm}
\label{l2q} Suppose that $n=q+1$ where $q\ge 5$ is a prime  power
that is congruent to $\pm 3$ modulo $8$. Suppose that
$\Gal(f)=\PSL_2(\F_q)$ acts doubly transitively on $\RR_f$ (where
$\RR_f$ is identified with the projective line $\P^1(\F_q)$). Then
$\End^0(J(C_f))$ is a simple $\Q$-algebra, i.e. $J(C_f)$ is either
absolutely simple or isogenous to a power of an absolutely simple
abelian variety.
\end{thm}

\begin{proof}
Since $n=q+1$ is even, $g=\frac{q-1}{2}$. It is known
\cite{Mortimer} that the $\Gal(f)=\PSL_2(\F_q)$-module $Q_{\RR_f}$
is simple and the centralizer of $\PSL_2(\F_q)$ in
$\End_{\F_2}(Q_{\RR_f})$ is the field $\F_4$. On the other hand,
$\PSL_2(\F_q)$ is a simple nonabelian group: we need to inspect
its subgroups. The following statement will be proven later in
this section.

\begin{lem}
\label{sub} Let $q\ge 5$ be a  power of an odd prime. Then
$\PSL_2(\F_q)$ does not contain a subgroup of index dividing
$\frac{q-1}{2}$ except $\PSL_2(\F_q)$ itself.
\end{lem}

 Recall that
$\tilde{G}_{2,J(C_f)}=\Gal(f)=\PSL_2(\F_q)$. Now Theorem \ref{l2q}
follows readily from Theorem \ref{center1} combined with Lemma
\ref{sub}.
\end{proof}

\begin{proof}[Proof of Lemma \ref{sub}]
Since $\PSL_2(\F_q)$ is a simple nonabelian subgroup, it does not
contain a subgroup of index $\le 4$ except $\PSL_2(\F_q)$ itself.
This implies that in the course of the proof we may assume that
$\frac{q-1}{2}\ge 5$, i.e., $q\ge 11$.

 Recall that $\#(\PSL_2(\F_q))=(q+1)q(q-1)/2$. Let $H \ne
\PSL_2(\F_q)$ be a subgroup in $\PSL_2(\F_q)$. The list of
subgroups in $\PSL_2(\F_q)$ given in \cite[theorem 6.25 on p.
412]{Suzuki} tells us that $\#(H)$ divides either $q\pm 1$ or
$q(q-1)/2$ or $60$ or $(b+1)b(b-1)$ where $b<q$ is a positive
integer such that $q$ is an integral power of $b$. This implies
that if the index of $H$ is a divisor of $\frac{q-1}{2}$ then
either

\begin{enumerate}
\item
$(q+1)q$ divides  $60$

or
\item
$\frac{(q+1)q(q-1)}{2} \le \frac{q-1}{2}
(\sqrt{q}+1)\sqrt{q}(\sqrt{q}-1)=\frac{q-1}{2}(q-1)\sqrt{q}$.
\end{enumerate}

In the case (1) we have $q=5$  which contradicts our assumption
that $q\ge 11$. So, the case (2) holds. Clearly, $(q+1)\sqrt{q}
\le (q-1)$ which is obviously not true.
\end{proof}

\begin{thm}
\label{homjac} Let $K$ be a field of characteristic different from
$2$. Suppose that $f(x)$ and $h(x)$ are polynomials in $K[x]$
enjoying the following properties:
\begin{itemize}
\item[(i)]
$\deg(f)\ge 3$ and the Galois group $\Gal(f)$ acts doubly
transitively on the set $\RR_f$ of roots of $f$. If $\deg(f)$ is
even then this action is $3$-transitive;
\item[(ii)]
 $\deg(h)\ge 3$ and the Galois group $\Gal(h)$ acts doubly
transitively on the set $\RR_h$ of roots of $h$. If $\deg(h)$ is
even then this action is $3$-transitive;
\item[(iii)]
The splitting fields $K(\RR_f)$ of $f$ and $K(\RR_h)$ of $h$ are
linearly disjoint over $K$.
\end{itemize}
Let $J(C_f)$ be the jacobian of the hyperelliptic curve
$C_f:y^2=f(x)$ and $J(C_h)$ be the jacobian of the hyperelliptic
curve $C_h:y^2=h(x)$. Then either $\Hom(J(C_f),J(C_h))=0,
\Hom(J(C_h),J(C_f))=0$ or $\fchar(K)>0$ and both $J(C_f)$ and
$J(C_h)$ are supersingular abelian varieties.
\end{thm}

\begin{proof}
Let us put $X=J(C_f), Y=J(C_h)$. The transitivity properties imply
that $\End_{\tilde{G}_{2,X}}(X_2)=\F_2$ and
$\End_{\tilde{G}_{2,Y}}(Y_2)=\F_2$. The linear disjointness of
$K(\RR_f)$ and $K(\RR_h)$ implies that the fields
$K(X_2)=K((J(C_f)_2)\subset K(\RR_f)$ and
$K(Y_2)=K((J(C_h)_2)\subset K(\RR_h)$ are also linearly disjoint
over $K$. Now the assertion follows readily from Theorem \ref{hom}
with $\ell=2$.
\end{proof}

\section{Abelian varieties with multiplications}
\label{mult}

Let $E$ be a number field. Let $(X, i)$ be a pair consisting of an
abelian variety $X$ of positive dimension over $K_a$ and an
embedding $i:E \hookrightarrow  \End^0(X)$. Here $1\in E$ must go
to $1_X$. It is well known \cite{Ribet2} that the degree $[E:\Q]$
divides $2\dim(X)$, i.e.
$$d=d_X:=\frac{2\dim(X)}{[E:\Q]}$$
is a positive integer. Let us denote by $\End^0(X,i)$ the
centralizer of $i(E)$ in $\End^0(X)$. The image $i(E)$ lies in the
center of the finite-dimensional $\Q$-algebra $\End^0(X,i)$. It
follows that $\End^0(X,i)$ carries a natural structure of
finite-dimensional $E$-algebra. If $Y$ is (possibly) another
abelian variety over $K_a$ and $j:E \hookrightarrow \End^0(Y)$ is
an embedding that sends $1$ to  $1_Y$ then we write
$$\Hom^0((X,i),(Y,j))=\{u \in \Hom^0(X,Y)\mid ui(c)=j(c)u \quad
\forall c\in E\}.$$ Clearly, $\End^0(X,i)=\Hom^0((X,i),(X,i))$. If
$m$ is a positive integer then we write $i^{(m)}$ for the
composition $E\hookrightarrow \End^0(X)\subset \End^0(X^m)$ of $i$
and the diagonal inclusion $\End^0(X)\subset
\End^0(X^m)=\M_m(\End^0(X))$. We have
$$\End^0(X^m,i^{(m)})=\M_m(\End^0(X,i))\subset\M_m(\End^0(X))=\End^0(X^m).$$

\begin{rem}
\label{ss} The $E$-algebra $\End^0(X,i)$ is semisimple. Indeed, in
notations of Remark \ref{split} $\End^0(X)= \prod_{s\in\I} D_s$
where all  $D_s=\End^0(X_s)$ are simple $\Q$-algebras. If
$\pr_s:\End^0(X) \twoheadrightarrow D_s$ is the corresponding
projection map and
  $D_{s,E}$ is the centralizer of $\pr_s i(E)$ in $D_s$ then one may easily check that
$\End^0(X,i)=\prod_{s\in\I} D_{s,E}$. Clearly, $\pr_s i(E)\cong E$
is a simple $\Q$-algebra. It follows from Theorem 4.3.2 on p. 104
of \cite{Herstein} that $D_{s,E}$ is also a {\sl simple}
$\Q$-algebra. This implies  that $D_{s,E}$ is a {\sl simple}
$E$-algebra and therefore $\End^0(X,i)$ is a semisimple
$E$-algebra. We write $i_s$ for the composition $ \pr_s i: E
\hookrightarrow \End^0(X)\twoheadrightarrow D_{s} \cong
\End^0(X_s)$. Clearly, $D_{s,E}=\End^0(X_s,i_s)$ and
$$\End^0(X,i)=\prod_{s\in\I} \End^0(X_s,i_s) \eqno(5) .$$
It follows that $\End^0(X,i)$ is a simple $E$-algebra if and only
if $\End^0(X)$ is a simple $\Q$-algebra, i.e., $X$ is isogenous to
a self-product of (absolutely) simple abelian variety.
\end{rem}

\begin{thm}
\label{maxE}
\begin{itemize}
\item[(i)]
$\dim_E(\End^0((X,i)) \le \frac{4\cdot\dim(X)^2}{[E:\Q]^2}$;
\item[(ii)]
Suppose that
 $\dim_E(\End^0((X,i)) =
\frac{4\cdot\dim(X)^2}{[E:\Q]^2}$.
 Then:
\begin{itemize}
\item[(a)]
 $X$ is isogenous to a self-product of an
(absolutely) simple abelian variety. Also $\End^0((X,i)$ is a
central simple $E$-algebra, i.e., $E$ coincides with the center of
$\End^0((X,i)$. In addition, $X$ is an abelian variety of CM-type.
\item[(b)]
There exist an abelian variety $Z$, a positive integer $m$, an
isogeny $\psi: Z^m \to X$ and an embedding $k: E \hookrightarrow
\End^0(Z)$ that sends $1$ to $1_Z$ such that:

\begin{enumerate}
\item
$\End^0(Z,k)$ is a central division algebra over $E$ of dimension
$\left(\frac{2\dim(Z)}{[E:\Q]}\right)^2$ and $\psi \in
\Hom^0((Z^r,k^{(m)}),(X,i))$.
\item
If $\fchar(K_a)=0$ then $E$ contains a CM subfield and
$2\dim(Z)=[E:\Q]$. In particular, $[E:\Q]$ is even.
\item
If $E$ does not contain a CM-field (e.g., $E$ is a totally real
number field) then $\fchar(K_a)>0$ and $X$ is a supersingular
abelian variety.
\end{enumerate}

\end{itemize}
\end{itemize}
\end{thm}

\begin{proof}
Recall that $d=2\dim(X)/[E:\Q]$. First, assume that $X$ is
isogenous to a self-product of an absolutely simple abelian
variety, i.e., $\End^0(X,i)$ is a simple $E$-algebra. We need to
prove that
$$N:=\dim_E(\End^0(X,i))\le d^2.$$
Let $C$ be the center of $\End^0(X)$.
 Let $E'$ be the center of
$\End^0(X,i)$. Clearly,
$$C \subset E'\subset \End^0(X,i)\subset\End^0(X).$$

Let us put $e=[E':E]$. Then $\End^0(X,i)$ is a {\sl central}
simple $E'$-algebra of dimension $N/e$. Then there exists a
central division $E'$-algebra $D$ such that $\End^0(X,i)$ is
isomorphic to the matrix algebra $\M_m(D)$ of size $m$ for some
positive integer $m$. Dimension arguments imply that
$$m^2\dim_{E'}(D)=\frac{N}{e},\quad \dim_{E'}(D)=\frac{N}{e m^2}.$$
Since $\dim_{E'}(D)$ is a square,
$$\frac{N}{e}=N_1^2, \quad N=e N_1^2, \quad \dim_{E'}(D)=\left(\frac{N_1}{m}\right)^2$$
for some positive integer $N_1$.
 Clearly,  $m$ divides
$N_1$.

 Clearly, $D$ contains a (maximal) field extension $L/E'$ of
degree $\frac{N_1}{m}$ and $\End^0(X,i)\cong \M_m(D)$ contains
every field extension $T/L$ of degree $m$. This implies that
$$\End^0(X) \supset\End^0(X,i)\supset T$$
and the number field $T$ has degree $[T:\Q]=[E':\Q]\cdot
\frac{N_1}{m} \cdot m=[E:\Q]e N_1$. But $[T:\Q]$ must divide
$2\dim(X)$ (see \cite[proposition 2 on p. 36]{Shimura}); if the
equality holds then $X$ is an abelian variety of CM-type. This
implies that $e N_1$ divides $d=\frac{2\dim(X)}{[E:\Q]}$. It
follows that $(eN_1)^2$ divides $d^2$; if the equality holds then
$[T:\Q]=2\dim(X)$ and therefore $X$ is an abelian variety of
CM-type. But $(eN_1)^2=e^2 N_1^2=e (e
N_1^2)=eN=e\cdot\dim_E(\End^0(X,i))$. This implies that
$\dim_E(\End^0(X,i)) \le \frac{d^2}{e} \le d^2$, which proves (i).

Assume now that
 $\dim_E(\End^0(X,i))=d^2$.  Then $e=1$ and
$$(eN_1)^2=r^2, N_1=d,\ [T:\Q]=[E:\Q]e N_1=[E:\Q] d=2\dim(X);$$
in particular, $X$ is an abelian variety of CM-type. In addition,
since $e=1$, we have $E'=E$, i.e. $\End^0(X,i)$ is a {\sl central}
simple $E$-algebra. We also have $C\subset E$ and
$$\dim_{E}(D)=\dim_{E'}(D)=\left(\frac{N_1}{m}\right)^2=\left(\frac{d}{m}\right)^2.$$
Since $E$ is the center of $D$, it is also  the center of the
matrix algebra $\M_m(D)$. Clearly, there exist an abelian variety
$Z$ over $K_a$, an embedding $j:D \hookrightarrow \End^0(Z)$ and
an isogeny $\psi:Z^m \to X$ such that the induced isomorphism
$$\psi_{*}: \End^0(Z^m) \cong \End^0(X),\ u\mapsto \psi u \psi^{-1}$$ maps
$j(\M_m(D)):=\M_m(j(D)) \subset \M_m(\End^0(Z))=\End^0(Z^m)$ onto
$\End^0(X,i)$. Since $E$ is the center of $\M_m(D)$ and $i(E)$ is
the center of $\End^0(X,i)$, the isomorphism $\psi_{*}$ maps $j(E)
\subset j(\M_m(D))=\M_m(j(D))\subset \End^0(Z^m)$
 onto $i(E) \subset \End^0(X)$. In other words,
 $\psi_{*}j(E)=i(E)$.
 It follows that there exists an
 automorphism $\sigma$ of the
 field $E$ such that
 $i=\psi_{*}j\sigma$ on $E$. This implies that if we put
 $k:=j\sigma:E \hookrightarrow \End^0(Z)$
 then $\psi \in\Hom((Z^m,k^{(m)}), (X,\psi))$.

 Clearly, $k(E)=j(E)$ and therefore $j(D)\subset \End^0(Z,k)$.
 Since $\M_m(\End^0(Z,k))\cong \End^0(X,i)\cong \M_m(D)$, the
 dimension arguments imply that $j(D)= \End^0(Z,k)$ and therefore
 $\End^0(Z,k)\cong D$ is a division algebra. We have
 $$\dim(Z)=\frac{\dim(X)}{m}, \quad
 \dim_E(D)=\left(\frac{d}{m}\right)^2=\left(\frac{2\dim(X)}{[E:\Q]m}\right)^2=\left(\frac{2\dim(Z)}{[E:\Q]}\right)^2.$$

Let $B$ be an absolutely {\sl simple} abelian variety over $K_a$
such that $X$ is isogenous to a self-product $B^r$ of $B$ where
the positive integer $r=\frac{\dim(X)}{\dim(B)}$. Then $\End^0(B)$
is a central division algebra over $C$; we define a positive
integer $g_0$ by $\dim_C(\End^0(B))=g_0^2$. Since $\End^0(X)$
contains a field of degree $2\dim(X)$, it follows from
Propositions 3 and 4 on pp. 36--37 in \cite{Shimura} (applied to
$A=X,K=C,g=g_0,m=\dim(B),f=[C:\Q]$) that $2\dim(B)=[C:\Q]\cdot
g_0$. Let $T_0$ be a maximal subfield in the $g_0^2$-dimensional
central division algebra $\End^0(B)$. Well-known properties of
maximal subfields of division algebras imply that $T_0$ contains
the center $C$ and $[T_0:C]=g_0$. It follows that $[T_0:\Q]=[C:\Q]
[T_0:C]=[C:\Q] \cdot g_0=2\dim(B)$ and therefore $\End^0(B)$
contains a field of degree $2\dim(B)$. This implies that $B$ is an
absolutely simple abelian variety of CM-type; in terminology of
\cite{Oort2}, $B$ is an absolutely simple abelian variety with
{\sl sufficiently many complex multiplications}.

 Assume now that $\fchar(K_a)=0$. We need to check that $2\dim(Z)=[E:\Q]$ and $E$ contains a
 CM-field.
 Indeed, since $D$ is a division algebra, it follows from Albert's classification
 \cite{MumfordAV,Oort} that $\dim_{\Q}(D)$
divides $2\dim(Z)=\frac{2\dim(X)}{m}=[E:\Q]\frac{d}{m}$. On the
other hand,
$\dim_{\Q}(D)=[E:\Q]\dim_{E}(D)=[E:\Q]\left(\frac{d}{m}\right)^2$.
Since  $m$ divides $d$, we conclude that $\frac{d}{m}=1$, i.e.,
$\dim_{E}(D)=1,  D=E,  2\dim(Z)=[E:\Q]$. In other words,
$\End^0(Z)$ contains the field $E$ of degree $2\dim(Z)$. It
follows from Theorem 1 on p. 40 in \cite{Shimura} (applied to
$F=E$) that $E$ contains a CM-field.

Now let us drop the assumption about $\fchar(K_a)$ and assume
instead that $E$  does {\sl not} contain a CM subfield. It follows
that
 $\fchar(K)>0$.  Since $C$ lies in $E$, it is totally real.
Since $B$ is an absolutely simple abelian variety with {\sl
sufficiently many complex multiplications}  it is isogenous to an
absolutely simple abelian variety $W$ defined over a finite field
\cite{Oort2} and $\End^0(B)\cong \End^0(W)$. In particular, the
center of $\End^0(W)$ is isomorphic to $C$ and therefore is a
totally real number field.  It follows from the Honda--Tate theory
\cite{Tate2} that $W$ is a supersingular elliptic curve and
therefore  $B$ is also a supersingular elliptic curve. Since $X$
is isogenous to $B^r$, it is  a supersingular abelian variety.

Now let us consider the case of arbitrary $X$. Applying the
already proven case of  Theorem \ref{maxE}(i) to each $X_s$, we
conclude that
$$\dim_E(\End^0(X_s,i)) \le
\left(\frac{2\dim(X_s)}{[E:\Q]}\right)^2.$$  Applying (5), we
conclude that
\begin{gather*}
\dim_E(\End^0(X,i)) =\sum_{s\in\I}\dim_E(\End^0(X_s,i_s))\le\\
\sum_{s\in\I}\left(\frac{2\dim(X_s)}{[E:\Q]}\right)^2\le
\frac{(2\sum_{s\in\I}
\dim(X_s))^2}{[E:\Q]^2}=\frac{(2\dim(X))^2}{[E:\Q]^2}.
\end{gather*}
 It follows
that if the equality
$\dim_E(\End^0(X,i))=\frac{(2\dim(X))^2}{[E:\Q]^2}$ holds then the
set $\I$ of indices $s$ is a singleton, i.e. $X=X_s$ is isogenous
to a self-product of  an absolutely simple abelian variety.

\end{proof}

config: arxiv


TITLE: Why the trace of the Canonical Stress Tensor (Generalization of the Hamiltonian) implies that the photons have mass?
QUESTION [0 upvotes]: In Chapter 12.10 B) of “Classical Eletrodynamics” of Jackson, it is said that

The two final criticisms of $T^{\alpha \beta}$ are that it involves the potentials explicitly, and so is not gauge invariant, and its trace ( $T^{\alpha}_{\alpha}$) is not zero, as required for zero mass photons.

Could someone explain me why the fact that the trace of the tensor have to be zero in order that the photons have zero mass?

REPLY [3 votes]: Set $\mu_0=c=1$. Without a current, a massive photon's Lagrangian density can be taken as$$\mathcal{L}=-\frac14F_{ab}F^{ab}-\frac12m^2A_aA^a,$$from which one can show (proof is an exercise)$$T^{ab}=-F^{ac}F^b_{\;\;c}+\frac14\eta^{ab}F_{cd}F^{cd}+m^2(\tfrac12\eta^{ab}A_cA^c-A^aA^b),$$so in $4$-dimensional spacetime$$T^a_{\;\;a}=m^2A_aA^a.$$