text:
\begin{document}

\title[On Triple Veronese Embeddings of $\PP_n$ in the Grassmannians]
{On Triple Veronese Embeddings of $\PP_n$ in the Grassmannians}
\author{Sukmoon Huh}
\address{Korea Institute for Advanced Study \\
Hoegiro 87, Dongdaemun-gu \\
Seoul 130-722, Korea}
\email{sukmoonh@kias.re.kr}

\keywords{Veronese Embedding, Vector Bundle, Projective Space}

\subjclass[msc2000]{Primary: {14D20}; Secondary: {14E25}}

\begin{abstract}
We classify all the embeddings of $\mathbb{P}_n$ in a Grassmannian
$Gr(1,N)$ such that the composition with the Pl\"{u}cker embedding is
given by a linear system of cubics on $\mathbb{P}_n$. As a direct corollary, we prove that every
vector bundle giving such an embedding, splits if $n\geq 3$.
\end{abstract}
\maketitle

\section{Introduction}

Giving a vector bundle $E$ of rank 2 on $\PP_n$ together with an
epimorphism $\Oo_{\PP_n}^{N+1} \rightarrow E$ is equivalent to
giving a morphism from $\PP_n$ to a Grassmannian $Gr(1,N)$. When
$\wedge^2 E=\Oo_{\PP_n}(2)$, Jos\'{e} Carlos Sierra and Luca Ugaglia
\cite{SU} classified all the embeddings of $\PP_n$ in a Grassmannian
$Gr(1,N)$ such that the composition with the Pl\"{u}cker embedding is
given by a linear system of quadrics on $\PP_n$.

In this article, we give similar answers on the vector bundles of
rank 2 giving rise to triple Veronese embeddings of $\PP_n$. The classification is divided into two parts. The first part is about the unstable bundles, and the other part is about the stable bundles. For the second part, we classify all the triple Veronese embeddings of $\PP_2$ and then show that there are no such embeddings in the case of $\PP_n$, $n\geq 3$.

The main statement is as follows:

\begin{theorem}
Let $X\subset Gr(1,N)$ be a triple Veronese embedding of $\PP_n$
given by a vector bundle $E$ of rank 2 on $\PP_n$. Then one of the
following holds:
\begin{enumerate}
\item $E\simeq \Oo_{\PP_n}(a)\oplus \Oo_{\PP_n}(3-a)$, $a=0,1$;
\item $n=2$ and $E$ admits a resolution,
$$0\rightarrow \Oo_{\PP_2}(2) \rightarrow E \rightarrow I_p(1)
\rightarrow 0,$$
where $I_p$ is the ideal sheaf of a point $p\in \PP_2$;
\item $n=2$ and $E\simeq \Omega_{\PP_2}(3)$;
\item $n=2$ and $E$ is a stable vector bundle of rank 2 on $\PP_2$
admitting one of the following resolution;
\begin{enumerate}
\item $0\rightarrow \Oo_{\PP_2}(-1)^{\oplus 3}\rightarrow \Oo_{\PP_2}^{\oplus 5}
\rightarrow E \rightarrow 0;$
\item $0\rightarrow \Omega_{\PP_2}(1)\oplus\Oo_{\PP_2}(-2) \rightarrow
\Oo_{\PP_2}^{\oplus 5} \rightarrow E \rightarrow 0;$
\item $0\rightarrow \Omega_{\PP_2}(1)\oplus \Oo_{\PP_2}(-1)^{\oplus 2}
\rightarrow \Oo_{\PP_2}^{\oplus 6} \rightarrow E \rightarrow 0;$
\item $0\rightarrow \Omega_{\PP_2}(1)^{\oplus 2}\oplus \Oo_{\PP_2}(-1)  \rightarrow
\Oo_{\PP_2}^{\oplus 7} \rightarrow E \rightarrow 0.$
\end{enumerate}
\end{enumerate}
\end{theorem}

As a direct corollary, we can derive a statement that every vector bundle of
rank 2 on $\PP_n$ giving a triple Veronese embedding, splits if
$n\geq 3$.

I am grateful to Jos\'{e} Carlos Sierra for introducing this problem. I thank Prof. Emilia Mezzetti for pointing out the mistake in the list of the theorem (1.1). And I also thank the anonymous referee for pointing out the mistakes in the original version of this paper and a number of corrections and suggestions.

\section{Preliminaries and Examples}
Let $Gr(1,N)$ be a Grassmannian variety which parametrizes all the
projective lines in $\PP_N$.

\begin{definition}
An embedding $\phi : \PP_n \rightarrow Gr(1,N)$ is called a $d$-Veronese
embedding of $\PP_n$ into $Gr(1,N)$ if the composition of $\phi$
with the Pl\"{u}cker embedding of $Gr(1,N)$ is given by a linear
system of degree $d$ on $\PP_n$. When $d=3$, we call it a triple
Veronese embedding.
\end{definition}
Let $n=1$ and $E$ be a vector bundle of rank 2 on $\PP_1$ with
$c_1=d\geq 1$. Due to Grothendieck, $E$ must be a direct sum of two
lines bundles, i.e. $\Oo_{\PP_1} (a) \oplus \Oo_{\PP_1}( d-a )$, $0\leq
a \leq [\frac {d}2]$. $E$ gives a $d$-Veronese embedding of $\PP_1$
into $Gr(1,d+1)$ as the family of lines joining the corresponding
points of two normal rational curves of degree $a$ and $d-a$.

Let $d=1$ and $E$ be a vector bundle of rank 2 on $\PP_n$ with
$c_1=1$ giving a 1-Veronese embedding of $\PP_n$ into a
Grassmannian. For a line $l\subset \PP_n$, we have
$$E|_l\simeq \Oo_l\oplus \Oo_l(1),$$
since $E|_l$ is also globally generated. In particular, $E$ is
uniform and, from \cite{OSS} and \cite{VdV}, $E$ is isomorphic to
$\Oo_{\PP_n}\oplus \Oo_{\PP_n}(1)$, or $\Omega_{\PP_2}(2)$ when
$n=2$. In the first case, $E$ embeds $\PP_n$ into $Gr(1,n+1)$ as the
set of lines passing through a point in $\PP_{n+1}$. In the case
of $E\simeq \Omega_{\PP_2}(2)$, $E$ gives a 1-Veronese
embedding of $\PP_2$ into $Gr(1,2)\simeq \PP_2^*$.

\begin{example}
A vector bundle of rank 2, $E\simeq \Oo_{\PP_n}(a) \oplus
\Oo_{\PP_n}(3-a)$, $a=0,1$, gives a triple Veronese embedding of
$\PP_n$ into $Gr(1,N)$, $N={n+a \choose n}+{n+3-a \choose n}$ as the
family of lines joining the corrresponding points on the two copies
of $\PP_n$, $v_a(\PP_n)$ and $v_{3-a}(\PP_n)$. As a convention, we
assume that $v_0(\PP_n)$ is a point.
\end{example}
\begin{example}
A vector bundle $E$ of rank 2 admitting the following resolution,
$$0\rightarrow \Oo_{\PP_2}(-1)^{\oplus 3}\rightarrow \Oo_{\PP_2}^{\oplus 5}
\rightarrow E \rightarrow 0,$$ gives a triple Veronese embedding of
$\PP_2$ into $Gr(1,4)$. In fact, there exist 3 lines $L_i$, $1\leq i
\leq 3$, in $\PP_4$ with the isomorphisms $\phi_i: \PP_2 \rightarrow
|L_i|$ determined by the resolution, where $|L_i|$ is the projective
space of hyperplanes in $\PP_4$ containing $L_i$. To each point
$x\in \PP_2$, we can associate a line in $\PP_4$ which is the
intersection of all $\phi_i(x)$.  This defines an embedding of
$\PP_2$ into $Gr(1,4)$ (see \cite{DK}).
\end{example}

From now on, we fix $d=3$ and $n\geq 2$.

\section{Classification}

 Let $E$ be a vector bundle of rank 2 on $\PP_n$ with $c_1(E)=3$
and $E$ is globally generated, giving a triple Veronese embedding
$$\phi_{E} : \PP_n \rightarrow Gr (1,N).$$
For a line $l\subset \PP_n$, we have
$$E|_l \simeq \Oo_l(a) \oplus \Oo_l(3-a)$$ with $a=0,1$, since
$E|_l$ is also globally generated. In particular, $h^0(E|_l)=5$.

\begin{proposition}
Let $E$ be an unstable vector bundle of rank 2 on $\PP_n$ with
$c_1(E)=3$, and gives a triple Veronese embedding of $\PP_n$ into
$Gr(1,N)$. Then one of the following holds:
\begin{enumerate}
\item $\Oo_{\PP_n}\oplus \Oo_{\PP_n}(3),$
\item $\Oo_{\PP_n}(1)\oplus \Oo_{\PP_n}(2),$
\item $n=2$ and $E$ is the unique non-trivial extension of the following resolution,
$$0\rightarrow \Oo_{\PP_2}(2) \rightarrow E \rightarrow I_p(1),$$
where $I_p$ is the ideal sheaf of a point $p\in \PP_2$.
\end{enumerate}
\end{proposition}
\begin{proof}
Since $E$ is unstable, we have an exact sequence
\begin{equation}\label{sequn}
0\rightarrow \Oo_{\PP_n}(k) \rightarrow E \rightarrow I_Z(3-k)
\rightarrow 0,
\end{equation}
where $k>1$, $Z$ is a locally complete
intersection of $\PP_n$ with codimension 2 and $I_Z$ is its ideal
sheaf in $\PP_n$.

If $I_Z\simeq \Oo_{\PP_n}$, i.e. the support of $Z$ is empty, then this sequence splits since
$$\Ext^1 (\Oo_{\PP_n}(3-k), \Oo_{\PP_n}(k))\simeq H^1(\Oo_{\PP_n}(2k-3))=0.$$
Since $E$ is globally generated, we have $k=2$ or $3$, and so we get
the vector bundles of the cases (1) and (2)

Now let us assume that the support of $Z$ is not empty.
Let $l\subset \PP_n$ be a line not contained in $Z$. By tensoring with $\Oo_l$, we have
$$0\rightarrow \Tor_1(I_Z(3-k), \Oo_l) \rightarrow \Oo_l(k)
\rightarrow E|_l \rightarrow I_Z (3-k)\otimes \Oo_l \rightarrow 0,$$
and note that $\Tor_1(I_Z(3-k), \Oo_l)=0$, since the restriction of $\Oo_{\PP_n}(k) \rightarrow E$ to $l$ is non-zero and thus injective. Then for some $k'\geq k>1$, we have
$$0\rightarrow \Oo_l(k') \rightarrow E|_l \rightarrow \Oo_l(3-k') \rightarrow 0.$$
Note that $k'>k$ if and only if $I_Z(3-k) \otimes \Oo_l$ is not a
locally free sheaf on $l$, i.e. $l\cap Z \not=\emptyset$.

If $k\geq 3$, then we have an injection from $\Oo_l(k')$ into
$E|_l=\Oo_l(a)\oplus \Oo_l(3-a)$, $a=0$ or 1 for some line $l\subset
\PP_n$ with non-empty intersection with $Z$. But this is not
possible.

Now let us assume that $k=2$ and furthermore $Z$ is not a linear subspace $\PP_{n-2}\subset \PP_n$, i.e. there exists a line $l$ whose intersection with $Z$ is 0-dimensional of length smaller than or equal to $2$.
Then if we tensor the sequence (\ref{sequn}) with $\Oo_l$, then we obtain,
$$0\rightarrow \Oo_l (k'') \rightarrow E|_l \rightarrow \Oo_l(3-k'') \rightarrow 0$$
for $k''\geq 4$, which is not possible because it would embed
$\Oo_l(k'')$ into $\Oo_l(a)\oplus \Oo_l(3-a)$ for $a=0$ or 1. Thus $Z$ is a linear subspace. Now we have
$$c_t(E)=(1+2t)c_t(I_{\PP_{n-2}}(1))=(1+2t)(1-t)^{-1},$$
so $c_3(E)=3$ when $n\geq 3$. It is impossible since $c_3(E)=0$ for vector bundles $E$ of rank 2. Hence, we have $n=2$ and get the exact sequence of the case (3)
$$0\rightarrow \Oo_{\PP_2}(2) \rightarrow E \rightarrow I_p(1) \rightarrow 0.$$
Note that
$$\Ext^1(I_p(1), \Oo_{\PP_2}(2))\simeq H^0(\Oo_p)^* \simeq \CC,$$
and so $E$ is the unique non-trivial extension of the above resolution.
\end{proof}

 Now let us deal with the case when $E$ is stable. Note that the
concepts of stability and semi-stability coincide. The following two
propositions are on the case of the projective plane.

\begin{proposition}\label{prop3.2}
Let $E$ be a stable vector bundle of rank 2 on $\PP_2$ with
$c_1(E)=3$ and $H^0(E(-1))=0$, giving a triple Veronese embedding of
$\PP_2$ into $Gr(1,N)$. Then $E$ admits the following sequence,
$$0\rightarrow \Oo_{\PP_2}(-1)^{\oplus 3} \rightarrow
\Oo_{\PP_2}^{\oplus 5} \rightarrow E \rightarrow 0.$$
\end{proposition}
\begin{proof}
From the exact sequence
$$0\rightarrow E(-1) \rightarrow E \rightarrow E|_l \rightarrow 0$$
for any line $l\subset \PP_2$, we have an injection from $H^0(E)$ to
$H^0(E|_l)$. In particular,
$$h^0(E)\leq h^0(E|_l)=5.$$
Since $E$ is globally generated, we have $h^0(E)\geq 3$.

If $h^0(E)=3$, then we have a sequence
$$0\rightarrow \Oo_{\PP_2}(-3) \rightarrow \Oo_{\PP_2}^{\oplus 3}
\rightarrow E\rightarrow 0,$$ and $E$ defines a map of degree 3 from
$\PP_2$ to $Gr(1,2)\simeq \PP_2^*$, which is clearly not an
embedding.

If $h^0(E)=4$, then we have
$$0\rightarrow F \rightarrow \Oo_{\PP_2}^{\oplus 4} \rightarrow E
\rightarrow 0,$$ where $F$ is a vector bundle of rank 2 on $\PP_2$
with $c_1(F)=-3$. For a line $l\subset \PP_n$, we have a sequence
$$0\rightarrow F|_l \rightarrow \Oo_l^{\oplus 4} \rightarrow E|_l
\rightarrow 0.$$ Note that the map $H^0(E) \rightarrow H^0(E|_l)$
obtained from the above sequence, is injective since $H^0(E(-1))=0$.
Thus we have $H^0(F|_l)=0$ and the only possibility for $F|_l$ is
$\Oo_l(-2)\oplus \Oo_l(-1)$. In particular, $F$ is uniform. By the
result in \cite{VdV}, we have (i) $F\simeq \Omega_{\PP_2}$, or (ii)
$F\simeq \Oo_{\PP_2}(-2)\oplus \Oo_{\PP_2}(-1)$. In the first case,
$c_2(E)$ can be computed to be 6 and $h^2(E)=h^0(E(-6))=0$ since $E$ is stable. Hence we have
$$5=\chi(E)=h^0(E)-h^1(E)=4-h^1(E),$$ which is not
possible. In the case of (ii), we have an embedding of $\PP_2$ into $Gr(1,3)$ with degree 9. From the classification of the congruences in $Gr(1,3)$ with degree up to 9 in \cite{AS}, the Veronese surface of degree 9 cannot be embedded into $Gr(1,3)$. Thus this case is impossible.

If $h^0(E)=5$, we have
$$0\rightarrow F \rightarrow \Oo_{\PP_2}^{\oplus 5} \rightarrow E
\rightarrow 0,$$ where $F$ is a vector bundle of rank 3 on $\PP_2$
with $c_1(F)=-3$. By the same reason as above, we have $H^0(F|_l)=0$
for any line $l\subset \PP_2$. The only possibility for $F|_l$ is
$\Oo_l(-1)^{\oplus 3}$ and in particular $F$ is uniform. By the
theorem(3.2.1) in \cite{OSS}, $F\simeq \Oo_{\PP_2}(-1)^{\oplus 3}$.
\end{proof}
\begin{remark}
\begin{enumerate}
\item When $h^0(E)=4$, we can indeed construct stable vector bundles of rank 2 on $\PP_2$ with $c_1(E)=3$ and $c_2(E)=7$ and a minimal resolution
    $$0\rightarrow \Oo_{\PP_2}(-1) \oplus \Oo_{\PP_2}(-2) \stackrel{A}{\rightarrow} \Oo_{\PP_2}^{\oplus 4} \rightarrow E \rightarrow 0,$$
    where $A=\left(
               \begin{array}{cccc}
                 x & y & z & 0 \\
                 a & b & c & d \\
               \end{array}
             \right)^t
$ with generically chosen $a,b,c,d\in H^0(\PP_2, \Oo_{\PP_2}(2))$.
\item A resolution in the proposition, is called a Steiner resolution. The
stability of a vector bundle $E$ admitting this resolution, can be
easily checked. And it is also well known in \cite{DK} that the
vector bundles admitting a Steiner resolution, form an open Zariski
subset of $M(3,6)$, where $M(c_1,c_2)$ is the moduli space of stable sheaves of rank 2 with the
Chern classes $(c_1, c_2)$ on $\PP_2$.
\end{enumerate}
\end{remark}

 Now let us deal with the case when
$h^0(E(-1))>0$. By the stability of $E$, we have
$$0\rightarrow \Oo_{\PP_2}(1) \rightarrow E \rightarrow I_Z(2)
\rightarrow 0,$$ where $Z$ is a zero dimensional subscheme of
$\PP_2$ with the length $m>0$. Since $I_Z(2)$ is also globally
generated, we have $h^0(I_Z(2))\geq 2$, otherwise we have an
isomorphism between $I_Z(2)$ and $\Oo_{\PP_2}$, which would make $Z$
a conic in $\PP_2$. In particular, $m=|Z|\leq 4$.

 Assume that there exists a line $l\subset \PP_2$ containing three points of $Z$. If we
tensor the above sequence with $\Oo_l$, we have
$$0\rightarrow \Oo_l(k) \rightarrow E|_l \rightarrow
\Oo_l(3-k)\rightarrow 0,$$ where $k\geq 4$. Since
$E|_l=\Oo_l(a)\oplus \Oo_l(3-a)$, $a=0,1$, it is not possible. Thus
no three points are collinear.

Now let us assume that $E(-1)$ is a stable vector bundle of rank 2 on
$\PP_2$ with the Chern classes $c_1=1$, $1\leq m=c_2 \leq 4$ such that $h^0(E(-1))=1$ and the zero locus $Z$ of the non-zero section of $E(-1)$ has the property that no three points of $Z$ are collinear (more precisely, there is no line $l$ such that the length of the scheme theoretic intersection $l\cap Z$ is bigger than 2). Since $h^0(I_Z(1))=0$, so $Z$ is not collinear and $m\geq 3$.

 If $m=3$, then the natural map $H^0(E) \rightarrow H^0(E|_l)$ is surjective
for any line $l\subset \PP_2$ since $H^1(E(-1))=0$. Since $E|_l$ can be $\Oo_l(a)\oplus
\Oo_l(3-a)$ with $a=0,1$, so $E|_l$ is globally generated. In
particular, the natural map $H^0(E|_l) \rightarrow E_p$ is
surjective. Hence the natural composition map $H^0(E) \rightarrow
E_p$ is surjective and so $E$ is globally generated. For any two
points in $\PP_2$, we can consider a line $l$ passing through these
two points and $E|_l$ defines an embedding of $l$ in a Grassmannian.
Hence, $E$ defines a triple Veronese embedding of $\PP_2$.

 If $m=4$, we have $h^0(E)=5$ and $h^0(E(-1))=1$. Note that the
natural restriction map $H^0(E) \rightarrow H^0(E|_l)$ has
1-dimensional kernel. If we tensor the next sequence
$$0\rightarrow F \rightarrow H^0(E)\otimes \Oo_{\PP_2} \rightarrow
E\rightarrow 0,$$ by $\Oo_l$ and take the long exact sequence of
cohomology, we have
$$h^0(F|_l)=h^1(F|_l)=1.$$
The only possibility for $F|_l$ is $\Oo_l\oplus \Oo_l(-1)\oplus
\Oo_l(-2)$ and so we have
$$F\simeq \Omega_{\PP_2}(1)\oplus\Oo_{\PP_2}(-2)~~~ \text{  or   }~~~ S^2\Omega_{\PP_2}(1)$$
since $h^0(F)=0$. Note that
$$c(F)=(1+3t+6t^2)^{-1}=1-3t+3t^2=c(\Omega_{\PP_2}(1)\oplus \Oo_{\PP_2}(-2)),$$
but $c(S^2 \Omega_{\PP_2}(1))=c(\Oo_{\PP_2}(1)^{\oplus 3})^{-1}=1-3t+6t^2$. Thus $F$ is isomorphic to $\Omega_{\PP_2}(1)\oplus \Oo_{\PP_2}(-2)$. If we tensor this resolution with $I_p$ for a point $p\in \PP_2$ and
take the long exact sequence of cohomology, we get
$$h^0(E\otimes I_p)=h^1(F)=h^0(F_p)=3.$$
Hence the natural map $H^0(E)\rightarrow E_p$ is surjective and in
particular, $E$ is globally generated. By the same reason as above,
$E$ also defines a triple Veronese embedding of $\PP_2$.
\begin{remark}
For a stable vector bundle $E\in M(3,6)$, it can be shown that there are four possible cases for the minimal resolution of $E(-1)$:
\begin{enumerate}
\item $0\rightarrow \Oo_{\PP_2}(-3) \rightarrow \Oo_{\PP_2}(-1)^{\oplus 2}\oplus \Oo_{\PP_2} \rightarrow E(-1) \rightarrow 0,$
\item $0\rightarrow \Oo_{\PP_2}(-3)\oplus \Oo_{\PP_2}(-2) \rightarrow \Oo_{\PP_2}(-2)\oplus \Oo_{\PP_2}(-1)^{\oplus 2}\oplus \Oo_{\PP_2} \rightarrow E(-1) \rightarrow 0,$
\item $0\rightarrow \Oo_{\PP_2}(-4) \rightarrow \Oo_{\PP_2}(-3)\oplus \Oo_{\PP_2}^{\oplus 2} \rightarrow E(-1) \rightarrow 0,$
\item $0\rightarrow \Oo_{\PP_2}(-2)^{\oplus 3} \rightarrow \Oo_{\PP_2}(-1)^{\oplus 5} \rightarrow E(-1) \rightarrow 0.$
\end{enumerate}
This gives us a stratification of the 12-dimensional moduli space $M(3,6)\simeq M(-1,4)$. Especially, $h^0(E(-1))=0$ for any $E\in M(3,6)$ of type (4) and these bundles belong to the open stratum, which is isomorphic to the quotient space $\Hom (\Oo_{\PP_2}(-2)^{\oplus 3}, \Oo_{\PP_2}(-1)^{\oplus 5})/\op{GL}_3\times \op{GL}_5$. The bundles $E$ with the resolution (1) describe an locally closed part of codimension 1 in $M(3,6)$, whereas the bundles of type (2) are specialization of bundles of type (1). Types (1) and (2) together describe the part of $M(3,6)$ consisting of bundles $E$ with $h^0(E(-1))=1$. In these cases, $E$ is globally generated and combining the resolution of $E(-1)$ with $0\rightarrow \Oo_{\PP_2} \rightarrow E(-1) \rightarrow I_Z(1)$, we get the resolution
$$0\rightarrow \Oo_{\PP_2}(-3) \rightarrow \Oo_{\PP_2}(-1)^{\oplus 2} \rightarrow I_Z(1) \rightarrow 0,$$
in the case (1), showing that $Z$ is the intersection of two quadrics in general position. In fact, the resolution (1) with the Euler sequence gives us the resolution:
$$0\rightarrow \Omega_{\PP_2}(1)\oplus\Oo_{\PP_2}(-2) \rightarrow
\Oo_{\PP_2}^{\oplus 5} \rightarrow E \rightarrow 0.$$
Similarly, in the case (2), $Z$ is the union of a point and the intersection of a line and a cubic, contradicting our assumption on $Z$, where $I_Z$ is the cokernel of the following matrix:
$$\left(
    \begin{array}{cc}
      x & 0 \\
      p & y \\
      q & z \\
    \end{array}
  \right)
 : \Oo_{\PP_2}(-3)\oplus \Oo_{\PP_2}(-2) \rightarrow \Oo_{\PP_2}(-2) \oplus \Oo_{\PP_2}(-1)^{\oplus 2}.$$
So the case (2) is excluded.
The bundles $E$ of type (3) form a locally closed subset of codimension 5 in $M(3,6)$ and $h^0(E(-1))=2$. But the bundles of this type are not globally generated.
\end{remark}
It is also well-known how the minimal resolution of stable vector bundle $F \in M(-1,c_2)$ look like for small $c_2$. We have $E=F(2)\in M(3,c_2+2)$. If $c_2=1$, then we have the Euler sequence, i.e. $E=\Omega_{\PP_2}(3)$. If $c_2=2$, we have
$$0\rightarrow \Oo_{\PP_2}(-3) \rightarrow \Oo_{\PP_2}(-2)\oplus \Oo_{\PP_2}(-1)^{\oplus 2} \rightarrow F \rightarrow 0.$$
$E$ is globally generated by 7 sections and $c_2(E)=4$. Combining this resolution with the Euler sequence, we obtain the resolution
$$0\rightarrow \Omega_{\PP_2}(1)^{\oplus 2}\oplus \Oo_{\PP_2}(-1)  \rightarrow
\Oo_{\PP_2}^{\oplus 7} \rightarrow E \rightarrow 0.$$
If $c_2=3$, then we have two resolution types: for the generic one, we have
$$0\rightarrow \Oo_{\PP_2}(-3)^{\oplus 2} \rightarrow \Oo_{\PP_2}(-2)^{\oplus 3} \oplus \Oo_{\PP_2}(-1) \rightarrow F \rightarrow 0,$$
and $E$ is globally generated by 6 sections with $c_2(E)=5$. Similarly as above, we obtain
$$ 0\rightarrow \Omega_{\PP_2}(1)\oplus \Oo_{\PP_2}(-1)^{\oplus 2}
\rightarrow \Oo_{\PP_2}^{\oplus 6} \rightarrow E \rightarrow 0.$$
In the special case, we have
$$0\rightarrow \Oo_{\PP_2}(-4) \rightarrow \Oo_{\PP_2}(-3)\oplus \Oo_{\PP_2}(-1)^{\oplus 2} \rightarrow F \rightarrow 0,$$
but $E$ is not globally generated.

Summarizing all the arguments above with the previous proposition, we have the
following statement.

\begin{proposition}
Let $E$ be a stable vector bundle of rank 2 on $\PP_2$ giving a
triple Veronese embedding of $\PP_2$. Then $E$ is an element of
$M(3,c_2)$ with $3\leq c_2 \leq 6$. Conversely, general elements in these moduli spaces, also define triple Veronese embeddings of $\PP_2$.
\end{proposition}

Now let us deal with the case of higher dimensional projective spaces.
At first, let us assume that $n=3$ and take a stable vector bundle $E$ of rank 2 on $\PP_3$ with the Chern classes $(c_1, c_2)$. Then we have the Riemann-Roch formula:
$$\chi (E(m))=\frac{m^3}{3}+(2+\frac{c_1}{2})m^2+(\frac{11}{3}+2c_1+\frac{c_1^2}{2}-c_2)m+1+{{c_1+3}\choose 3}-2c_2-\frac{c_1c_2}{2},$$
which implies the Schwarzenberger condition $c_1c_2\equiv 0 \pmod 2$.

If $c_1(E)=3$ and $H^0(E(-1))\not= 0$, then we have an exact sequence
\begin{equation}\label{seqP3}
0\rightarrow \Oo_{\PP_3} \rightarrow E(-1) \rightarrow I_Z(1) \rightarrow 0,
\end{equation}
where $Z$ is a locally complete intersecting curve in $\PP_3$, since $H^0(E(-2))=0$ due to the stability of $E$.
\begin{lemma}
If $Z$ is smooth, then $E$ does not give a triple Veronese embedding of $\PP_3$ into $Gr(1,N)$.
\end{lemma}
\begin{proof}
By the local fundamental isomorphism, we have
\begin{align*}
\omega_Z=\mathcal{H}om_{\Oo_Z}(\det I_Z/I_Z^2, \Oo_Z(-4))&=\mathcal{H}om_{\Oo_Z}(\det (E(-2))_Z, \Oo_Z(-4))\\
&=\mathcal{H}om_{\Oo_Z}(\Oo_Z(-1), \Oo_Z(-4))\\
&=\Oo_Z(-3).
\end{align*}
Since $c_1(E(-1))=1$ and $d:=\deg (Z)=c_2(E(-1))$, the degree $d$ must be even by the Schwarzenberger condition $c_1c_2 \equiv 0 \pmod 2$. We have
$$\chi(\Oo_Z)=\chi(\Oo_{\PP_3})-\chi(E(-2))+\chi(\Oo_{\PP_3}(-1))=\frac{3}{2}d,$$
where the last equality is from the Riemann-Roch formula. If $Z$ is smooth, then we can let $Z=\cup Z_i$ is a disjoint union of $m$ smooth, connected curves $Z_i$ of genus $g_i\geq 0$ ($i=1, \cdots, m$) and so
$$\frac{3}{2}d = m-\sum_{i=1}^m g_i~,~~~ \text{and}~~~ \sum_{i=1}^m \deg (Z_i)=d.$$
It implies that $m\geq \frac{3}{2}d$ and also $m\leq d$. This is impossible and so $Z$ cannot be smooth.
\end{proof}

\begin{remark}
For example, let $Z=Z_1\cup Z_2$ be the disjoint union of two smooth conics and consider $E$ admitting the sequence
\begin{equation}\label{seqP3-1}
0\rightarrow \Oo_{\PP_3} \rightarrow E \rightarrow I_Z(3) \rightarrow 0.
\end{equation}
From the results in \cite{HS}, we know that $E$ is stable and there exists a line $l$ for which $E|_l\simeq \Oo_l(4)\oplus \Oo_l(-1)$. Indeed, this line $l$ is uniquely determined to be $\Pi_1 \cap \Pi_2$, where $\Pi_i$ is the hyperplane containing $Z_i$. Moreover, we have
$$0\rightarrow \Oo_{\PP_3}(1) \rightarrow E \rightarrow I_{l^2}(2) \rightarrow 0.$$
If we tensor the sequence (\ref{seqP3-1}) with $\Oo_l$, we get
$$0\rightarrow \Oo_l(4) \rightarrow E|_l \rightarrow \Oo_l(-1)$$
and it splits. In particular, the morphism induced by $E$ is not an embedding over $l$.
\end{remark}

\begin{lemma}
If $H^0(E(-1))=0$, then $E$ does not give a triple Veronese embedding of $\PP_3$ into $Gr(1,N)$.
\end{lemma}
\begin{proof}
Clearly, $h^0(E)\geq 3$ since $E$ is globally generated. Thus we have the exact sequence (\ref{seqP3-1}). Using the local fundamental theorem, we obtain that $\omega_Z=\Oo_Z(-1)$. From the Bertini type theorem in the proposition (1.4) of \cite{Hartshorne}, we can assume that $Z$ is a smooth curve. Then $Z=\cup_{i=1}^m Z_i$ is a disjoint union of $m$ smooth and connected curves $Z_i$ of genus $g_i$ ($i=1,\cdots, m)$ and so $Z$ has the degree $d:=\sum_{i=1}^m \deg (Z_i)=c_2(E)$. Moreover, from the Riemann-Roch formula, we have
$$\chi(\Oo_Z)=\chi(\Oo_{\PP_3})-\chi (E(-3))+\chi(\Oo_{\PP_3}(-3))=\frac{d}{2}.$$
If $E$ gives a triple Veronese embedding, then it would induce a triple Veronese embedding of a general hyperplane and so $c_2(E)$ must be at most 6 from the previous results on $\PP_2$. In particular, $d\leq 6$. Since $h^0(I_Z(2))=0$, $Z$ is not contained in a quadric and so $d\geq 3$. Thus we have, by the Schwarzenberger condition, $d=4$ or $6$. First assume that $d=4$. Then we have
$$\sum_{i=1}^m (1-g_i)=2~~~~~\text{and}~~~~~\sum_{i=1}^m \deg(Z_i)=4$$
and the only possibility is $m=2$, $\deg (Z_i)=2$ for $i=1,2$.

But by the previous remark, $E$ does not give an embedding over $l$. Thus $d=6$ and we obtain
$$\sum_{i=1}^m (1-g_i)=3~~ ~~~\text{and}~~ ~~~\sum_{i=1}^m \deg (Z_i)=6.$$
The only possibilities are when $Z$ is the union of
\begin{enumerate}
  \item three disjoint smooth conics, $Z_1, Z_2, Z_3$,
  \item a twisted cubic curve $Z_1$, a smooth conic $Z_2$ and a line $Z_3$,
  \item a plane cubic $Z_1$ and three lines $Z_2,Z_3, Z_4$, or
  \item a rational quartic curve $Z_1$ and two lines $Z_2, Z_3$.
\end{enumerate}
(For the classification of curves in $\PP_3$ with low degrees, see \cite{Hartshorne2}.) But all cases but the first, can be excluded because of the condition that $\omega_Z=\Oo_Z(-1)$. Assume that $Z$ is the union of three disjoint smooth conics. From the previous remark, we can find a line $l\subset \PP_3$ that meets two of the conics of $Z$ at two point for each. By tensoring the sequence (\ref{seqP3-1}) with $\Oo_l$, we obtain $E|_l \simeq \Oo_l(4)\oplus \Oo_l(-1)$. Thus $E$ does not give an embedding.
\end{proof}

\begin{corollary}
Let $E$ be a stable vector bundle of rank 2 on $\PP_n$ with
$c_1(E)=3$ and $n\geq 4$. Then $E$ does not give a triple Veronese
embedding of $\PP_n$ into $Gr(1,N)$.
\end{corollary}
\begin{proof}
Depending on $H^0(E(-1))$, we have the sequence (\ref{seqP3}) or (\ref{seqP3-1}) replacing $\PP_3$ by $\PP_n$. For a general hyperplane $H\simeq \PP_{n-1} \subset \PP_n$, $E|_H$ is also stable. Furthermore, by the Bertini theorem \cite{Hartshorne2}, we can choose $H$ so that $Z\cap H$ is smooth. In other words, we can choose a general 3-plane $H\simeq \PP_3$ for which $E|_H$ is stable and (1) $H^0(E|_H(-1))=0$ or (2) there exists a smooth 2-codimensional $Z'\subset H$ with an extension
$$0\rightarrow \Oo_H \rightarrow E|_H(-1) \rightarrow I_{Z'}(1) \rightarrow 0.$$
From the previous two lemmas, $E|_H$ does not give a triple Veronese embedding.
\end{proof}

The only remaining case to deal with, is when $H^0(E(-1))\not= 0$ where $E$ is a stable vector bundle with the sequence (\ref{seqP3}) and $Z$ is not smooth. By proving that $E$ does not give a triple Veronese embedding, we can prove the following proposition.
\begin{proposition}
Let $E$ be a stable vector bundle of rank 2 on $\PP_n$ with
$c_1(E)=3$ and $n\geq 3$. Then $E$ does not give a triple Veronese
embedding of $\PP_n$ into $Gr(1,N)$.
\end{proposition}
\begin{proof}
Let $d:=\deg (Z)=c_1(E(-1))$. Since $c_1(E)\leq 6$, so $1 \leq d \leq 4$ and so $d=2$ or $4$ by the Schwarzenberger condition. The case when $d=2$ is excluded due to the previous remark. So let us assume that $c_2(E)=d+2=6$. By the local fundamental theorem, we have $\omega_Z=\Oo_Z(-3)$. In particular, the arithmetic genus $p_a$ of $Z$, is $-5$ and so $Z$ cannot be reduced. Indeed, the smallest $p_a$ for a reduced curve $Z\subset \PP_3$ with the degree 4, is $-3$ when $Z$ is the disjoint union of 4 lines.

Now assume that $h^0(I_Z(2))=0$. It implies that $h^0(E)=h^0(\Oo_{\PP_3}(1))=4$. The vector bundle of rank 2 $F:= \ker (\Oo_{\PP_3}^{\oplus 4} \rightarrow E)$ has the Chern classes $c_1(F)=-3$ and $c_2(F)=3$ and it contradicts the Schwarzenberger condition $c_1c_2\equiv 0 \pmod 2$. Thus $h^0(I_Z(2))>0$. Since $Z$ is not reduced, the quadrics containing $Z$ must be double planes in $\PP_3$ or the unions of two planes.

Let us assume that there exists a double plane $2H$ in $\PP_3$ containing $Z$. Then we can assign a triple $\{A,B,C\}$ consisting of a zero-dimensional closed subscheme $A$ and two curves $B$, $C$ in $H$ such that $A\subset B \subset C \subset H$ \cite{HS}. Roughly speaking, $C$ with embedded points $A$ is the intersection of $Z$ with $H$ and $B$ is the residual intersection. If we let $a,b,c$ be the length of $A$ and the degree of $B$, $C$, respectively, then from the proposition (2.1) in \cite{HS}, we have
$$\left\{
    \begin{array}{ll}
      \deg (Z)=4=b+c  \\
      p_a=-5=\frac12 (b-1)(b-2)+\frac12 (c-1)(c-2)+b-a-1
    \end{array}
  \right. $$
So the possibilities are $(a,b,c)=(6,2,2)$ or $(6,1,3)$ since $1\leq b\leq c$. But in either case, $a$ cannot be 6.

Now assume that $Z$ is contained in the union of two planes $H_1\cup H_2$ (see \cite{Hartshorne3}). The only possibility for $Z$ that has not been dealt with yet, is the union of a double line $2(H_1\cap H_2)$ and two lines $L_i\subset H_i$.

Hence $Z=2l+q$, where $q$ is the residual curve of degree 2 whose support does not contain $l$. If we tensor the exact sequence (\ref{seqP3}) with $\Oo_l$, we obtain $E(-2)|_l\simeq I_Z\otimes \Oo_l$. By tensoring the exact sequence,
$$0\rightarrow I_Z \rightarrow I_l \rightarrow \Oo_{l+q}(1) \rightarrow 0,$$
with $\Oo_l$, we obtain
$$0\rightarrow \Oo_l(-3) \rightarrow I_l\otimes \Oo_l \simeq \Oo_l(-1)^{\oplus 2} \rightarrow \Oo_l(1) \rightarrow 0,$$
where $\Oo_l(-3)$ is the image of the map $I_Z\otimes \Oo_l \rightarrow I_l\otimes \Oo_l$ from counting the Chern classes. Since $E(-2)|_l\simeq \Oo_l(a)\oplus \Oo_l(-a-1)$ for some $a\geq 0$, we should have $a=2$ and so $E|_l\simeq \Oo_l(4)\oplus \Oo_l(-1)$, which prevents $E$ from giving a triple Veronese embedding.
\end{proof}

Combining the results so far, we have the main theorem in the
introduction.

A weaker version of Hartshorne's conjecture states that $X\subset
\PP_n$ is a complete intersection if $X$ has codimension 2 and
$n\geq 7$ and due to Serre, this conjecture is equivalent to proving
that every vector bundle of rank 2 on $\PP_n$ splits if $n\geq 7$. Now, from the classification of the triple Veronese embeddings, we have the following statement.
\begin{corollary}
Every vector bundle of rank 2 on $\PP_n$ giving a triple Veronese
embedding, splits if $n\geq 3$.
\end{corollary}

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config: arxiv


TITLE: Baumslag-Solitar Group
QUESTION [1 upvotes]: I am working on an exercise which says Hyperbolic groups do not contain any Baumslag-Solitar group $\operatorname{BS}(m,n)$ as subgroup.
(Hint is $\operatorname{BS}(m,n)$ is not virtually cyclic).
My attempt is I am trying to show that $\operatorname{BS}(m,n)$ contains $\mathbb{Z \oplus Z}$ using the fact that $\operatorname{BS}(m,n)$ is not virtually cyclic where $\operatorname{BS}(m,n)$ denotes the Baumslag-Solitar group.
Any hints are appreciated.
Thanks in advance

REPLY [5 votes]: I'll convert my comments into a sketch proof. I should warn though that this exercise is not easy (the original proof used "biautomaticity", while I'll use a fancy, but standard, theorem). I suspect that there are results in your course notes/book which make the problem easier.
Also, the "hint" given is meaningless. For example, non-abelian free groups are not virtually cyclic, but every hyperbolic group contains a non-abelian free group or is itself virtually cyclic... Possibly the hint meant "centralisers of elements are virtually cyclic", which is in fact the key theorem we'll use.
Firstly, start with the following (standard) result for hyperbolic groups. I've copied the statement from Corollary III.$\Gamma$.3.10 (p462) of Bridson and Haelfliger's book Metric spaces of non-positive curvature (although the proof there is more general then we need, as it is for "semihyperbolic" groups). It is equivalent to the fact that centralisers of elements are virtually cyclic (why?).
Theorem A. Let $H$ be hyperbolic. If $a\in H$ is an element of infinite order then $\langle a\rangle$ has finite index in its centraliser.
Therefore, all we need to do is find for every Baumslag-Solitar group $\operatorname{BS}(m, n)$ an element $g\in \operatorname{BS}(m, n)$ of infinite order such that $\langle g\rangle$ does not have finite index in $C_{\operatorname{BS}(m, n)}(g)$. Lets break this down into a series of facts which need to be proven.

Fact 1. Every non-trivial element of $\operatorname{BS}(m, n)$ has infinite order.

This follows from the fact that $\operatorname{BS}(m, n)$ is an HNN-extension of a torsion-free group (namely $\mathbb{Z}$).

Fact 2. If $\operatorname{BS}(m, n)=\langle a, t\mid t^{-1}a^mt=a^n\rangle$ with $|m|,|n|>1$ or $|m|=|n|=1$ then $C_{\operatorname{BS}(m, n)}(a^n)$ contains a copy of $\mathbb{Z}^2$.

If $|m|,|n|>1$ then $\langle a^n, t^{-1}ata\rangle$ is abelian, and so by Fact 1 is either $\mathbb{Z}$ or $\mathbb{Z}^2$. Use the fact that $\operatorname{BS}(m, n)$ is an HNN-extensions to prove that this subgroup is non-cyclic. The case $|m|=|n|=1$ is similar.

Fact 3. If $\operatorname{BS}(1, n)=\langle a, t\mid t^{-1}at=a^n\rangle$ (so $m=1$) then $C_{\operatorname{BS}(1, n)}(a)$ contains a non-cyclic locally cyclic subgroup.

The hint here is that $\langle t^{p}at^{-p}\mid p\in\mathbb{N}\rangle$ is not finitely generated, but is locally cyclic (every finitely generated subgroup is cyclic). The key observation needed to prove this is that $(t^pat^{-p})^n=t^{p-1}at^{-(p-1)}$).
Facts 2 and 3 mean that $\langle a\rangle$ does not have finite index in $C_{\operatorname{BS}(m, n)}(a)$, while $a$ has infinite order by Fact 1. The result then follows as if $\operatorname{BS}(m, n)$ embeds into a hyperbolic group $H$ then $C_{\operatorname{BS}(m, n)}(a)\leq C_{H}(a)$, contradicting Theorem A.


TITLE: Product of inverse matrix and non-invertible matrix
QUESTION [1 upvotes]: Probably an easy question, but can't find anything about it on the internet and it's not in my book.
I know the product of two inverse matrices is an inverse matrix.
Is it possible to get an inverse matrix if you multiply an inverse matrix with a non invertible matrix?
Same with two non-invertible matrices.

REPLY [1 votes]: In case you haven't seen determinants suppose that $AB$ is invertible. Let $C$ be the inverse of $AB$ then:
\begin{align}
(AB)C = I \\
A(BC) = I
\end{align}
So $A$ is invertible.


TITLE: Cohomology ring of BG
QUESTION [10 upvotes]: Let $G$ be a compact Lie group, let $T$ be a maximal torus, and let $W$ be the Weyl group.  My main question is as follows:

How does one prove that $H^\ast(BG,\mathbb{Q})$ is isomorphic to the $W$-invariant part of $H^\ast(BT,\mathbb{Q}) \cong \mathbb{Q}[[x_1, \ldots, x_n]]$?  This is apparently basic knowledge in algebraic topology, because I keep reading "recall that..." followed by some version of this statement and no references.  But I can't find a proof in any of my textbooks.

I would ideally like a reference which also addresses the following secondary question:

When is the natural map $H^\ast(BG,\mathbb{Z}) \to H^\ast(BT,\mathbb{Z})^W$ an isomorphism, and what can one say about the integral cohomology ring of $BG$ when it is not?  Note the fact that the map above is an isomorphism for $G = U(n)$ is equivalent to the statement that the Chern classes are integral.

Thanks!

REPLY [10 votes]: Q1: Let me first note, that the statement
$$H^\ast(BG,\mathbb{Q}) \cong H^\ast(BT,\mathbb{Q})^W\tag{$\ast$}$$
made in the question requires in addition that $G$ is connected. However, the general case can be reduced to the connected case since
$$H^\ast(BG,\mathbb{Q}) \cong H^\ast(BG_0,\mathbb{Q})^{G/G_0}$$
where $G_0$ is the identity component of $G$.
A text book reference for $(\ast)$ can be found in Hsiang, Cohomology theory of topological transformation groups, Chapter III, §1, Lemma 1.1. The results of the book that are relevant for your question can also be found in the following paper: Richard Gonzales, Localization in equivariant cohomology and GKM theory (cf. Remark 9, Lemma 5). Other approaches and more information can be found in this answer.
Q2: First note that the kernel of the restriction map
$$\rho^\ast: H^\ast(BG;\mathbb{Z}) \to H^\ast(BT;\mathbb{Z})$$
is the tosion subgroup $Tors$ of $H^\ast(BG;\mathbb{Z})$. So, $\rho^\ast$ is injective iff $H^\ast(BG;\mathbb{Z})$ is torsion-free (there is a lot of literature about torsion
of $H^\ast(BG;\mathbb{Z})$ so I won't discuss it here).
A short paper of Feshbach

The image of $H^\ast(BG;\mathbb{Z})$ in $H^\ast(BT;\mathbb{Z})$ for $G$ a compact Lie group with maximal torus $T$. Topology 20(1981) 93-95 doi:10.1016/0040-9383(81)90015-X,

characterizes when the induced map
$$\bar{\rho}^\ast: H^\ast(BG;\mathbb{Z})/Tors \to H^\ast(BT;\mathbb{Z})$$
is an isomorphism:

$\bar{\rho}^\ast$ is an isomorphism iff $H^\ast(BG;\mathbb{Z})/Tors \otimes \mathbb{F}_p$ is an integral domain for each prime $p$.

There is also a counter-example, $\mathrm{Spin}(12)$, that shows that $\bar{\rho}^\ast$ is not always an isomorphism.

REPLY [6 votes]: Notice that there is a sequence of homomorphisms $T \to N \to G$, where $N$ is the maximal torus normaliser (so $W = N/T$).  $W$ acts on $BT$ (because it acts on $T$ by conjugation through group homomorphisms), and there is an equivalence from the classifying space of $N$ to the Borel construction for this action:
$$BN \simeq EW \times_W BT.$$
Consequently, we can compute the cohomology of $BN$ from the Leray-Serre spectral sequence
$$H^\ast(W; H^\ast(BT)) \implies H^\ast(BN).$$
Taking rational cohomology, this spectral sequence is concentrated in group-cohomological degree $0$, since $W$ is a finite group.  Therefore the spectral sequence collapses at $E_2$, which is $H^0(W, H^\ast BT) = H^\ast(BT)^W$.
It therefore suffices to show that the map $BN \to BG$ is an isomorphism in rational cohomology.  If we write $BN$ as $EG / N$, this map is a fibre bundle with fibre $G / N$, so it's enough to show that $G/N$ has the rational homology of a point.
For instance, if $G = SU(2)$, $N = \mathbb{Z} / 2 \ltimes T$, and $T = S^1$.  Then $G/T = \mathbb{C} P^1$, and the action of $\mathbb{Z} / 2$ is antipodal, giving $G / N = \mathbb{R} P^2$, which is indeed rationally a point.  I don't remember the argument in general, but I think this is always true.
Hopefully this indicates how the corresponding integral statement can fail - there can be torsion contributions from the higher group cohomology of $W$, which needs to be exactly cancelled (via a differential in the second spectral sequence above) with a torsion cohomology class from $G/N$.

REPLY [1 votes]: I don't remember where I heard the following proof/sketch:
Using the fibering $G/T\to BT\to BG$ and the fact that the Euler class of $G/T$ is nonzero, we have that $H^\ast(BG)$ embeds into $H^\ast(BT)$ (it composes with the transfer map to be multiplication by the Euler class); and the desired isomorphism comes from the fact that $W$ acts on $H^*(G/T)$ as the regular representation.
This is actually a special case of equivariant cohomology, where we instead use the Borel construction and the fibering $G/T\to M_T\to M_G$.


TITLE: Let $L = \mathbb F_2[X]/\langle X^4 + X + 1 \rangle$ is a field. Show $L^* = L / \{0\} = \langle X \rangle$ is cyclic.
QUESTION [1 upvotes]: Let $L = \mathbb F_2[X]/\langle X^4 + X + 1 \rangle$ is a field. Show $L^* = L / \{0\} = \langle X \rangle$ is cyclic.
I've proven that $X^4 + X + 1$ is irreducible, so $L$ is a field. I also know that $X^5 + X + 1$ is not irreducible. Also I've proven that $|L| = 16$.
Could someone help me out proving that $L^*$ is cyclic ?

REPLY [3 votes]: since the group is of order 15 any element must have order 1,3,5 or 15.
in any field the equation $x^k = 1 $ can have at most k roots, so the number of elements with order less than $15$ is at most $1 + 3 + 5 = 9$
hence there must be an element of order 15
i will add, after the query from OP, that a slightly more sophisticated form of this  argument will show that for any finite field its multiplicative group is cyclic.


text:
\begin{document}
\begin{abstract}
  We give a simple uniqueness criterion (and some derived criteria)
  for holomorphic Abel functions and show that Kneser's real analytic
  Abel function of the exponential is subject to this criterion.
\end{abstract}
\subjclass{Primary 30D05}
\keywords{Abel function; Abel equation; exponential function;
  fractional iterates; holomorphic solution; real analytic}

{This article is a preliminary version of the final article published
in Aequationes Mathematicae with DOI 10.1007/s00010-010-0021-6.
The final publication is available at\\
\url{http://www.springerlink.com/content/u7327835m2850246/}.}

\newpage
\maketitle

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\theoremstyle{definition}
\newtheorem{definition}{Definition}
\newtheorem{criterion}{Criterion}

\newcommand \nS {\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!}
\newcommand \pS {{~}~{~}}

\newcommand {\iLa}[1]{\label{#1}}

\newcommand {\iL}[1]{\iLa{#1}}
\newcommand {\rf}[1]{(\ref{#1})}

\newcommand \sx {\scalebox}

\newcommand \Lc {L^\ast}
\newcommand{\rme}{\text{e}}
\newcommand \rmi {\text{i}}
\newcommand \sexp {{\rm ksexp}}
\newcommand \slog {{\rm kslog}}
\newcommand \ve {\varepsilon}
\newcommand \bN {\mathbb{N}}
\newcommand \bC {\mathbb{C}}
\newcommand \bR {\mathbb{R}}
\newcommand \bZ {\mathbb{Z}}
\newcommand{\rh}{\varrho}
\newcommand{\wo}{\setminus}
\newcommand{\el}{\ell}

\section{Introduction}
There is a lot of discussion about the ``true'' or ``best'' fractional iterates of
the function $\rme^x$ in the (lay-)mathematical community. In 1949
Kneser \cite{Kneser1949} proved the existence of real analytic fractional
iterates. However Szekeres (a pioneer in developing the theory of fractional iteration
\cite{Szekeres1958}) states 1961 in \cite{Szekeres1961}:

{\it ``The solution of Kneser does not really solve the problem of
`best' fractional iterates of $\rme^x$. Quite apart from practical
difficulties involved in the calculation of Kneser's function on the
real axis, there is no indication whatsoever that the function will
grow more regularly to infinity than any other solution. There is
certainly no uniqueness attached to the solution; in fact if $g(x)$ is
a real analytic function with period 1 and $g'(x)+1>0$ (e.g.
$g(x)=\frac{1}{4\pi}\sin( 2\pi x)$ then
$B^\ast(x)=B(x)+g\big(B(x)\big)$

 is also
an analytic Abel function of $\rme^x$ which in general yields a different
solution of the equation.''}

A recent discussion with Prof.\ Jean {\'E}calle supports the impression that
no uniqueness criterion was found up today and that there is even evidence
against the existence of a criterion concerned with the
growth-scale or asymptotic behavior at infinity.

By withdrawing our attention from the purely real analytic behavior
of the Abel function to the behavior in the complex plane we can
succeed in giving a simple uniqueness criterion for Abel functions of
a whole class of real analytic (or
arbitrary holomorphic) functions with two complex fixed points.

We show the usefulness of the criterion by providing an Abel function that satisfies the criterion.
This is the above mentioned by Kneser constructed Abel
function of $\rme^x$ (which can be easily generalized to functions
$b^x$ with $b>\rme^{1/\rme}$).

We have also a suggestion to numerically compute this Abel function
and the corresponding fractional iterates of $\rme^x$ (also of $b^x$ for
$b>\rme^{1/\rme}$ in generalization) by a method developed in
\cite{Kouznetsov2009}. Several other methods to numerically compute
holomorphic fractional iterates of $\rme^x$ or the holomorphic
Abel function have emerged in the past
years (for example one is given in \cite{Walker:abelian}).
A future research goal would be to put them on a thorough
theoretic base (proving convergence and holomorphy) and to verify the
here given uniqueness criterion.

\section{Motivation}
Our original motivation was the investigation of a fourth stage of
operations after the third stage containing power, exponential and
logarithm.

Different terms for such operations were used in the past like:
``generalized exponential'' and ``generalized logarithm'' by Walker
\cite{Walker:generalized}, ``ultra exponential'' and ``infra
logarithm'' by Hooshmand
\cite{Hooshmand2006}, ``super-exponential'' by Bromer
\cite{Bromer1987}, tetration and superlogarithm
\cite{Kouznetsov2009}. In this paper we give them the more succinct
names ``4-exponential'' and ``4-logarithm''.

\begin{definition}[4-exponential]
A {\em 4-exponential} to base $b>0$ is a function $f$ that satisfies
\begin{align}
  f(0)&=1\iL{F01}\\
  f(z+1)&=\exp_b\big(f(z)\big)
  \iL{expb}
\end{align}
for all applicable $z$.
\end{definition}
For any $\tilde{f}$ that only satisfies \eqref{expb} and contains 1 in its
codomain: $\tilde{f}(z_0)=1$, the function $f(z)=\tilde{f}(z+z_0)$ is a
4-exponential.
\begin{definition}[4-logarithm]
A {\em 4-logarithm} to base $b>0$ is a
function $g$ that satisfies the Abel equation \eqref{abel} (see
\cite{Kuczma_etal1990}) (for all applicable $z$) with the following
initial condition:
\begin{align}
  g(1)&=0		\iL{g10}\\
  g(\exp_b(z))&=g(z)+1	\iL{abel}.
\end{align}

\end{definition}
For any $\tilde{g}$ that only satisfies \eqref{abel} and has $1$ in
its domain of definition, the function
$g(z)=\tilde{g}(z)-\tilde{g}(1)$ is a 4-logarithm.
Here we set as usual $\exp_b(z)=b^z=\exp\big(\ln(b)z\big)$.
The inverse of a 4-exponential (if existing) is a 4-logarithm and vice
versa.

On positive integer arguments $z=n$ any 4-exponential $f$ is already determined
to be just the $n$-times application of $\exp_b$ to $1$.
\begin{align}
f(n)=\exp_{b}^{\circ n}(1)=
\underbrace{b^{.^{.^{.^b}}}}_{n \times b}\iL{zexpo}
\end{align}
The question however is how to properly extend the function
real and analytic to non-integer arguments.

The existence of a real analytic strictly
increasing 4-logarithm was proven by Kneser \cite{Kneser1949}. A
non-analytic solution with a uniqueness criterion was given by
Hooshmand in \cite{Hooshmand2006}. A numerical method to compute the real
coefficients of the powerseries development at 0 of a 4-logarithm was
given (though without proof of convergence) by Walker in
\cite{Walker:abelian}. Another numerical method to compute a
real analytic 4-exponential via Cauchy integrals (though also without
convergence proof) was given by Kouznetsov in \cite{Kouznetsov2009}.

A real analytic 4-exponential is expected to have a singularity or
branchpoint at integers $\le -2$ at least on some branch,
because from $f(z+1)=\exp_b(f(z))$ follows
$f(z-1)=\log_b(f(z))$ and by $f(0)=1$ is then $f(-1)=0$
and $f(-2)=\log_b(0)$. To exclude branching we restrict
4-exponentials to
\begin{align}
C_{-2}= \mathbb{C} \backslash  \{x \in \mathbb{R} : x \le -2 \} \iL{C}
\end{align}

It is a conjecture of the authors that holomorphy on the domain
$C_{-2}$ together with $f(z^\ast)=f(z)^\ast$ (complex conjugation) on
$C_{-2}$ implies the uniqueness of the 4-exponential $f$.

From considerations about the uniqueness of
4-logarithms/4-exponentials the following
general uniqueness criterion for Abel functions with two complex
fixed points emerged.

\section{The Uniqueness Criterion}
Before we start we mention some {\em conventions} we use: Usually
curves here are regarded as continuous maps on the open interval
$(-1,1)$. If we however use a curve in a set context then we refer to
the image of the curve, e.g. $\gamma_1\cup \gamma_2 = \gamma_1((-1,1))
\cup \gamma_2((-1,1))$. The disjoint union $C=A\uplus B$ means here
that $C=A\cup B$ and $A\cap B=\emptyset$. The sum $A+z$ of a region
$A\subseteq \bC$ and a number $z\in\bC$ is defined as the region $\{
a+z\colon a\in A\}$. A function being holomorphic
on a non-open set means that there is a neighborhood of each point of
the set where the function is holomorphic. $\log_b(z)=\log(z)/\log(b)$
means the principal branch $-\pi<\Im(\log(z))\le\pi$ of the
logarithm if not stated otherwise. ``Continuable'', ``continuation''
and ``continue'' always refer to {\em analytic} continuation.

\begin{definition}[Abel function, initial
  curve/region]
We call a function $\alpha$, holomorphic on $D$, an {\em Abel function
  of $F$} iff it satisfies the Abel equation
\begin{align}
  \alpha(F(z))&=\alpha(z)+1
  \iL{eq:abel}
\end{align}
for all $z\in D\cap F^{-1}(D)$. $F$ is sometimes called the
{\em base function}.

A curve $\gamma\colon (-1,1)\to\bC$ on which $F$ is
holomorphic is called an {\em initial curve of $F$} iff $\gamma$ and
$F\circ\gamma$ are injective
and disjoint and $\gamma(-1)\neq \gamma(1)$ are two fixed points of
$F$. (To be explicit define $\gamma(\pm 1):=\lim_{t\to\pm 1} \gamma(t)$.)

Under these conditions $\gamma\cup (F\circ \gamma) \cup
\{\gamma(-1),\gamma(1)\}$ is a closed Jordan
curve. We call its inner (bounded) component $C$ joined with $\gamma$
and $F\circ\gamma$ the {\em initial region} of $\gamma$ denoted by
$I_F(\gamma):=\gamma\uplus C\uplus F(\gamma)=\overline{C}\setminus
\{\gamma(-1),\gamma(1)\}$.

\iL{def:initial}
\end{definition}

As a side remark: the following theorem has a slight similarity to the criterion given
in \cite{contreras:2007:remarks}. Both criteria include
injectivity. However the criterion given in
\cite{contreras:2007:remarks} applies to a function with one parabolic
fixed point (moreover a self-map of the unit disk of hyperbolic step zero),
while the criterion here applies to a function with two fixed points
(which can not occur for a holomorphic self-map of some simply
connected region).

\begin{theorem}
Let $\gamma$ be an initial curve of $F$, let $H$ be its initial region and $d\in H$.
There is at most one function $\alpha$ satisfying Criterion
\ref{criterion:A}.
\iL{theorem}
\end{theorem}
\begin{criterion}
  The function $\alpha$ is an on $H$ holomorphic and injective Abel function of
  $F$, $\alpha(d)=0$ and $\bigcup_{k\in\bZ} \left(\alpha(H)+k\right)=\bC$.
\iL{criterion:A}
\end{criterion}
\begin{proof}
Assume there are two such Abel functions $\alpha_1\colon H \leftrightarrow
T_1$ and $\alpha_2\colon H\leftrightarrow T_2$  holomorphic and injective on
$H$. For the rest of this proof we write $\alpha_j$ when referring to $\alpha_1$ as
well as to $\alpha_2$. The inverse function $\alpha_j^{-1}\colon T_j\leftrightarrow H$
satisfies:
\begin{align*}
  \alpha_j^{-1}(z+1)&=F\left(\alpha_j^{-1}(z)\right)
\end{align*}
for all $z$ such that $z,z+1\in T_j$.
So we have two biholomorphic functions
\begin{align*}
  q_1&:=\alpha_2\circ \alpha_1^{-1}\colon T_1\leftrightarrow T_2 &
  q_2:=\alpha_1\circ \alpha_2^{-1}\colon T_2\leftrightarrow T_1
\end{align*}
with the property
\begin{align*}
   q_1(z+1)
  &=\alpha_2\left(\alpha_1^{-1}(z+1)\right)=\alpha_2\left(F\left(\alpha_1^{-1}(z)\right)\right)\\
  &=\alpha_2\left(\alpha_1^{-1}(z)\right)+1 = q_1(z)+1
\end{align*}
for each $z$ with $z,z+1\in T_1$; and generally
\begin{align}
   q_j(z+1)= q_j(z)+1
\iL{eq:delta}
\end{align}
for each $z$ with $z,z+1\in T_j$.

We define $q_{j,k}\colon T_j+k\to\mathbb{C}$ by $q_{j,k}(z+k)=q_j(z)+k$.  By our
property $q_j(z+1)=q_j(z)+1$ the function
$q_{j,k}$ and $q_{j,k+1}$ coincide on the intersection
$(T_j+k)\cap (T_j+k+1)$ which contains the curve $\alpha\circ\gamma + k+1$. In
conclusion $q_j$ can be continued to
the whole complex plane. So lets consider $q_j$ to
be an entire function.

For $z\in T_1$ we have $ q_1(z) = q_2^{-1}(z)$. But the only entire
functions that have an entire inverse are linear functions. By the
values $ q_1(0)=0$ and $ q_1(1)=1$ it can only be the
identity. So $\alpha_2(\alpha_1^{-1}(z))=z$ for $z\in T_1$ and hence $\alpha_2=\alpha_1$
on $H$.
\end{proof}

Now one may argue that the Abel function may depend on the initial
region. This is not the case as shown by the next theorem.

\begin{theorem}
  Let $D$ be a connected set with $d\in D$ and let $F$ be holomorphic
  on $D$, there can be at most one on $D$ holomorphic Abel function
  $\alpha$ of $F$ with the property that $\alpha'(d)\neq 0$ and that $\alpha$
  satisfies Criterion \ref{criterion:A} on some initial region
  $H\subset D$.
\end{theorem}
\begin{proof}
  Assume that there were two such Abel functions $\alpha_1$ and
  $\alpha_2$ such that each $\alpha_j$ is
  holomorphic on $D$ and satisfies Criterion \ref{criterion:A} on the
  initial region $H_j$.
  We follow the proof of Theorem \ref{theorem}, listing only the
  modifications.

  We consider biholomorphic $\alpha_j\colon H_j\leftrightarrow T_j$ and
  holomorphic $q_j\colon T_j\to\bC$ and achieve \eqref{eq:delta}.
  In conclusion each $q_j$ is entire.
  The function $z\mapsto q_1(\alpha_1(z))-\alpha_2(z)$ is holomorphic
  on $D$. It is constantly 0 on $H_1$ and hence on
  $D$. That's why $q_1(0)=q_1(\alpha_1(d))=\alpha_2(d)=0$. Vice versa
  $q_2(0)=0$.

  There is a neighborhood $V$ of $d$ where $\alpha_1$ and $\alpha_2$ is
  injective. Hence $\alpha_j^{-1}$, $q_1=\alpha_2\circ \alpha_1^{-1}$
  and $q_2=\alpha_1\circ \alpha_2^{-1}$ are injective on
  $V'=\alpha_1(V)\cap \alpha_2(V)\ni 0$. But then $q_1=q_2^{-1}$ on
  $V'\cap q_2(V')$.
\end{proof}

Now we want to make the criterion $\bigcup_{k\in\bZ}
\left(\alpha(H)+k\right)=\bC$ a bit more accessible and show that it is a
consequence of $\lim_{t\to\pm 1}\Im(\alpha(\gamma(t)))=\pm\infty$.
Before we start with the actual proof, we need a little
insight into how curves divide the complex plane. We know by the
Jordan curve theorem that each simple closed curve divides the sphere
into two simply connected components. Considering the sphere $\bC\cup
\{\infty\}$ we know that each injective curve $\zeta\colon (-1,1)\to
\bC$
with $\lim_{t\to\pm 1} \zeta(t) = \infty$ divides the complex plane
into two parts; where $\infty$ is the complex infinity and $\lim_{t\to
  \pm 1} \zeta(t) =\infty$ means that for each $r>0$ there is a $t_1$
and $t_0$ such
that $\left|\zeta(t)\right| > r$ for all $t>t_1$ and all $t<t_0$.

A particular subclass of such curves are the injective
curves $\zeta$ with $\lim_{t\to \pm 1}\Im(\zeta(t)) = \pm\infty$. Here
$\infty$ is the
real infinity and $\lim_{t\to \pm 1}\Im(\zeta(t)) = \pm\infty$ means that for
each $u>0$ there are $t_0,\; t_1\in (-1,1)$ such that $\Im(\zeta(t)) >
u$ for all $t>t_1$ and $\Im(\zeta(t))<-u$ for all $t<t_0$.

\begin{definition}[left/right component/ray]
For an injective curve $\zeta\colon (-1,1)\to\bC$ we call a component {\em
  left} (resp.\ {\em right}) if
it contains a left (resp.\ right) ray, where a left (resp.\ right) ray
is a set of the form $\{ z_0\mp x \colon x> 0\}$ for some $z_0\in \bC$.
\end{definition}
The following lemma shows that these properties indeed behave as
expected.
\begin{lemma}[$L$, $R$]
  Let $\zeta\colon (-1,1)\to \bC$ be an injective curve with $\lim_{t\to\pm
    1} \Im(\zeta(t)) = \pm \infty$ and let $P$ and $Q$ be the two
  components the plane is divided into; $\bC =  P\uplus \zeta
  \uplus Q$.

  1) Then either $P$ is left and $Q$ is right or vice versa. We denote the left
  (resp.\ right) component with $L(\zeta)$ (resp.\ $R(\zeta)$).
  Moreover there exist left (resp.\ right) rays contained in the left
  (resp.\ right) component for each prescribed imaginary part $y$.

  2) If $\zeta+d$ is disjoint from $\zeta$ then $\overline{R}(\zeta+d)\subset
  R(\zeta)$ in the case $d>0$ and $\overline{L}(\zeta+d)\subset L(\zeta)$ in the
  case $d<0$. Here the overline means the closure of the component
  (it is equal to the union of the component with $\zeta$).
\iL{lemma}
\end{lemma}
\begin{proof}
  For part 1) of the lemma we show that for each prescribed
  imaginary part $y$ there exists a left ray contained in one component
  and a right ray contained in the other component. We show further
  that the union of all left (resp.\ right) rays such that each is contained in
  either $P$ or $Q$ must be contained in either $P$ or $Q$, which
  ensures the ``either'' in part 1).

  We consider the indices of the intersection of the horizontal line
  $Y=\{x+iy\colon x\in\bR\}$ with the curve $\zeta$,
  $T=\{t\colon \zeta(t)\in Y\}$. Now let $t_0=\inf(T)$ and $t_1=\sup(
  T)$; neither can $t_0=-1$ nor $t_1=1$ because in this case there
  would be a sequence of $t\to\pm 1$ such that $\Im(\zeta(t)) = y$ in
  contradiction to $\Im(\zeta(t))\to \pm \infty$.

  Hence $-1<t_0\le t_1<1$ and $\Im(\zeta(t))>y$ for all
  $t>t_1$ and $\Im(\zeta(t))<y$ for all $t<t_0$. Let
  $x_0=\Re(\zeta(t_0))$ and $x_1=\Re(\zeta(t_1))$ then it is clear that
  $Y_0:=\{x+iy\colon x<x_0\}$ is completely contained in a component as
  well as $Y_1:=\{x+iy\colon x>x_1\}$ is completely contained in a
  component. The compound $Y_0\cup
  \zeta([x_0,x_1])\cup Y_1$ divides the plane into an upper half $H_1$
  which is divided by $\zeta((x_1,1))$ and a lower part $H_0$ which
  is divided by $\zeta((-1,x_0))$.

  If there was a path $\beta\colon [0,1]\to\bC$ that connects
  $Y_0$ with $Y_1$, i.e.\ $\beta(0)\in Y_0$ and $\beta(1)\in Y_1$ then
  we choose $s_0=\beta^{-1}(\sup(\Re(\beta\cap Y_0))+\rmi y)$ and
  $s_1=\beta^{-1}(\inf(\Re(\beta\cap Y_1))+\rmi y)$.
  The restriction of $\beta$ to the
  non-empty interval $(s_0,s_1)$ still connects $Y_0$ to $Y_1$ but
  does neither intersect $Y_0$ nor $Y_1$. As the path is also not
  allowed to intersect $\zeta([t_0,t_1])$ it must either be contained
  in $H_0$ or in $H_1$ but then it would intersect $\zeta((-1,x_0))$
  or $\zeta((x_1,1))$.

  If we have two left rays with the imaginary parts $y_1$ and $y_2$ then
  we can do the above construction of $t_0(y)$ for
  $y=y_1,y_2$. Without restriction let $t_0(y_1)<t_0(y_2)$, then
  $\zeta([t_0(y_1),t_0(y_2)])$ has a minimum of the real part and we
  can connect both left rays by a vertical line with a smaller real
  part.

  Now to part 2: We consider $d>0$. First it is easy to see
  that one point of $\zeta+d$ lies in $R(\zeta)$ and hence the whole
  $\zeta+d$ is contained in $R(\zeta)$. Vice versa $\zeta$ must be
  contained in $L(\zeta+d)$. So $\zeta$ does not intersect
  $R(\zeta+d)$ and we conclude that $R(\zeta+d)$ must be either contained in
  $L(\zeta)$ or in $R(\zeta)$. But $R(\zeta)$ has common points with
  $R(\zeta+d)$ as a right ray $Y$ is contained in $R(\zeta)$ and the
  ray $Y+d$ is contained in $R(\zeta+d)$.
\end{proof}

\begin{criterion}
  Under the preconditions of Theorem \ref{theorem},
  and $\gamma$ being rectifiable, the following criterion implies
  Criterion \ref{criterion:A}:

  The curve $\zeta = \alpha\circ\gamma$ is injective, $\zeta \cap
  (\zeta+1)=\emptyset$ and $\lim_{t\to\pm
    1}\Im(\zeta(t))=\pm\infty$; where $\alpha$ is an on $H$ holomorphic Abel function of $F$ with $\alpha(d)=0$.
\iL{criterion:B}
\end{criterion}
\begin{proof}
The first two conditions of the criterion state that $\alpha$ is
injective on $\gamma\cup (F\circ \gamma)$ because $\alpha\circ F\circ
\gamma=\zeta+1$ by \eqref{eq:abel}.
By a theorem about univalent functions (see \cite{Markushevich1965}
Theorem 4.8) the injectiveness of $\alpha$ on the rectifiable boundary with
finitely many exceptions (e.g.\ $\gamma(-1)$ and $\gamma(1)$) implies
the injectiveness on the enclosed region (inclusive its boundary)
which is $H$.

Now we show that $\bigcup_{k\in\bZ} \left(\alpha(H)+k\right)=\bC$:
The image $\alpha(H)$ is bounded by $\zeta$ and $\zeta+1$ and it must
be simply connected, that's why
\begin{align*}
  \alpha(H) &= \overline{R}(\zeta) \cap \overline{L}(\zeta+1)
\end{align*}
If we now unite consecutive pieces $\alpha(H)+k$, $k\in\bZ$, we get with
$\overline{R}(\zeta-1)\supset \overline{R}(\zeta)$ and
$\overline{L}(\zeta)\subset \overline{L}(\zeta+1)$ from Lemma \ref{lemma} that:
\begin{align*}
  \bigcup_{k=k_0}^{k_1} \left(\alpha(H)+k\right) &=
  \overline{R}(\zeta+k_0)\cap \overline{L}(\zeta+1+k_1)
\end{align*}

By part 1) of Lemma \ref{lemma} we also know that for each
$z=x+i y$ there is a (negative) $k_0\in\bZ$ such that $z\in R(\zeta+k_0)$ (right
ray with starting point $x-k_0+i y$ is contained in $R(\zeta)$)
and a (positive) $k_1\in\bZ$ such that $z\in L(\zeta+k_1)$ (left ray with
starting point $x-k_1+i y$ is contained in $L(\zeta)$). So
$\alpha(H)$ translated by all integers cover the
whole complex plane:
\begin{align}
  \bigcup_{k\in\bZ} \left(\alpha(H)+k\right) &= \bC
\iL{eq:C}
\end{align}
\end{proof}

\begin{criterion}
  Under the preconditions of Theorem \ref{theorem}
  the following criterion implies Criterion \ref{criterion:B}:

  The real function $f(t)=\Im(\alpha(\gamma(t)))$ is strictly
  increasing and $\lim_{t\to\pm 1}f(t) = \pm \infty$; where $\alpha$
  is an on $H$ holomorphic Abel function of $F$ with $\alpha(d)=0$.
\iL{criterion:C}
\end{criterion}
\begin{proof}
  First it is clear that $\zeta:=\alpha\circ\gamma$ is injective as no imaginary
  value can be taken twice.
  Further the correlation $\mu$ given by
  $\Im(\zeta(t))\mapsto \Re(\zeta(t))$, $t\in (-1,1)$, is a welldefined
  function on $\bR$.
  Then $\mu+1>\mu$ is a function not intersecting $\mu$, and hence $\zeta+1$
  does not intersect $\zeta$.
\end{proof}

\section{Application to Kneser's construction}
In this section we apply the uniqueness Criterion \ref{criterion:A} to the
4-logarithm (i.e.\ Abel function of $\exp_b$) $\Psi$ as constructed by
Kneser \cite{Kneser1949}.

\begin{figure}
\begin{center}
\sx{.32}{\begin{picture}(600,460)

\put(20,0){\includegraphics{fig2}}
\put(204,452){\sx{2.5}{$\Im(z)$}}
\put(204,422){\sx{2.5}{$2$}}
\put(204,322){\sx{2.5}{$1$}}
\put(204,222){\sx{2.5}{$0$}}
\put(182,122){\sx{2.5}{$-1$}}
\put(182, 22){\sx{2.5}{$-2$}}
\put(  2,208){\sx{2.5}{$-2$}}
\put(102,208){\sx{2.5}{$-1$}}

\put(317,208){\sx{2.5}{$1$}}
\put(417,208){\sx{2.5}{$2$}}
\put(457,208){\sx{2.5}{$\Re(z)$}}
\put(140,260){\sx{2.5}{$\log_b(H)$}}
\put(280,250){\sx{4}{$H$}}
\put(400,300){\sx{4}{$\exp_b(H)$}}
\put(110,170){\sx{2.5}{$\log_b(\ell)$}}
\put(243,160){\sx{5}{$\ell$}}
\put(320,140){\sx{5}{$b^\ell$}}
\put(520, 40){\sx{6}{$b^{b^\ell}$}}
\put(233,352){\sx{4.4}{$L$}}
\put(233, 82){\sx{4.4}{$\Lc$}}
\end{picture}}
\end{center}
\caption{Contours $\log_b(\ell)$, $\ell$ and $b^\ell$ in the complex $z$-plane,
for base $b=\rme$. \iL{fighen}}
\end{figure}

Related to Kneser's construction we use the choice
\begin{align}
  H=\{z\in \bC\colon \Re(z)\ge \Re(L),\; |z|\le |L|\}\wo \{L,\Lc\}
\iL{H}
\end{align}
as initial region which is depicted in Figure \ref{fighen}
for $b=\rme$ (which is the only base that Kneser considered).
Here $L$ is the fixed point of $\log_b$ in the upper half plane.
The straight line
\begin{align}
  \ell(t)=\Re(L)+\rmi\Im(L)t, \quad -1<t<1,
\iL{ell}
\end{align} between $L$ and its
complex conjugate $\Lc$ is the left boundary of $H$ and $b^{\ell}$ is
the right boundary of $H$.

\begin{lemma}
  If $b>e^{1/e}$ then $\ell$ in \eqref{ell} is an initial curve and
  hence $H$ is an initial region of $\exp_b$.
\end{lemma}
\begin{proof}
  We show that $b^\ell$ is injective and does not intersect $\ell$. By $b^L=L$ we
  know that $b^{\Re(L)}=|b^L|=|L|$ and hence
  \begin{align*}
    b^{\el(t)}=b^{\Re(L)+\rmi\Im(L)t}=|L|\rme^{\rmi\Im(L)\ln(b)t}
  \end{align*}
  which is an arc with radius $|L|$ centered in 0 (shown with a dashed line)
  starting at angle $-\Im(L)\ln(b)$ and ending at angle $\Im(L)\ln(b)$.
  This is true for any non-real conjugated fixed
  point pair of $\exp_b$. For $b>\rme^{1/\rme}$ there are no real fixed points
  of $\exp_b$ and the fixed point pair of $\log_b$, which is the one closest to the real axis, has
  $\pm\Im(L) \in (-\pi/\ln(b),\pi/\ln(b))$. This assures that
  $b^{\ell}$ does not overlap itself, i.e.\ that it is injective.

\end{proof}
Let us --- without proof --- enumerate some counterexamples of initial
curves:
$\log_b\circ \ell$ is initial for $e^{1/e}<b<e^{\pi/2}$ but $\ell$ has
zero or negative real part for $b\ge e^{\pi/2}$ and is hence no more contained in
the default domain of the logarithm. The curve $b^\ell$ is
initial for $b=\rme$ but there are bigger bases where $b^{b^\ell}$ intersects
itself. Each of $\rme^\ell$, $\rme^{\rme^\ell}$,
$\rme^{\rme^{\rme^\ell}}$ are injective but
$\rme^{\rme^{\rme^{\rme^\ell}}}$ is not, so the first two are initial
for $\exp$. The straight line connecting any other conjugated fixed point
pair is not initial because the image under $\exp_b$ is a circle with
radius $|L|$ (winding at least once around 0).

\newcommand{\Ho}{\mathfrak{H}_0}

To get familiar with Kneser's construction we recapitulate the main
steps he does in \cite{Kneser1949} with a slight generalization to bases
$b>\rme^{1/\rme}$. He starts with the K{\oe}nigs function $\chi$ of
$\exp_b$ at the fixed point $L$ and shows that it can be continued to
nearly the whole upper half plane $\mathfrak{H}=\{z\in \bC\colon \Re(z)\ge
0\}\wo\{0,1,b,b^b,b^{b^b},\dots\}$. He also shows that it is injective
on $\mathfrak{H}$ because its inverse can be continued to an entire
function.

Particularly $\chi$ is injective on the ``half'' initial region
\begin{align*}
\Ho&=\{z\in \bC\colon \Re(z)\ge \Re(L),\; \Im(z)\ge 0,\;|z|\le |L|\}\wo
\{L,1\}.
\end{align*}

Note that Kneser initially works with $\Ho$
containing $L$ while he later switches to consider $\Ho$ without $L$
which is the definition we use above. $\Ho$ is related to our initial region $H$ via $\Ho \cup \Ho^\ast  = H \wo \{1\}$.

The K{\oe}nigs function satisfies the Schr\"oder equation
$\chi(b^z)=c\chi(z)$ on $z,b^z\in \mathfrak{H}$ where
$c=\exp_b'(L)=\log(L)$. Next he sets $\psi(z)=\log(\chi(z))$ for a
suitable region of the logarithm which has the cut outside $\chi(

\Ho

)$. It
satisfies $\psi(b^z)=\psi(z)+\log(c)$ on $z\in

\Ho$. Last he conformally maps the union of
$\psi(\mathfrak{H}_0) + k\cdot\log(c)$, $k\in\bZ$, via
the Riemann mapping theorem to the upper halfplane, say with the
conformal map $\rh$.

The resulting function $\Psi = \rh \circ\log\circ \chi$ is at least
defined on $
\Ho

$ and satisfies
$\Psi(b^z)=\Psi(z)+1$ on $
\Ho$. It can even be continued to $z=
1
$. It is real analytic on $\bR\cap
\Ho
$ and can hence analytically continued to the conjugate
region $
\Ho
^\ast$ (and also to the whole real line).

From his construction the following things are
important for us. First:
$\Psi$ is injective and holomorphic on $\Ho\cup\{1\}$ and the
image is contained in the upper halfplane. Hence the conjugate
continuation to $H$ is also injective and holomorphic. Second:
the by integer translated regions $\Psi(\Ho\cup\{1\})$ cover
the whole upper halfplane and hence the by integer translated regions
$\Psi(H)$ cover the whole complex plane.

The application of Theorem \ref{theorem} gives:
\begin{theorem}
  The by Kneser in \cite{Kneser1949} constructed real analytic 4-logarithm
  $\alpha=\Psi$ and each generalization to base $b>\rme^{1/\rme}$
  satisfies Criterion \ref{criterion:A} on $H$ given in \eqref{H}
  (where $d=1$ and $F(z)=b^z$).
\end{theorem}

\section{About Computation of the
  4-Logarithm/4-Exponential}
As Kneser uses the Riemann mapping theorem in his construction,
it is very difficult to approach computationally. Instead we use
here a rather fast numerical method given in \cite{Kouznetsov2009}
utilizing the Cauchy integral formula to compute a 4-exponential which
we denote with $\sexp_b$ and its inverse
with $\slog_b$. This method is originally described for
$b=\rme$ but can be extended to arbitrary bases
$b>\rme^{1/\rme}$. Unfortunately it lacks a proof of convergence (but
if it converges then $\slog_b$ satisfies the Abel equation and is
holomorphic). It
is the conjecture of the authors that this method does converge and
that it satisfies the uniqueness Criterion \ref{criterion:C}.

\begin{figure}
\begin{center}
\sx{3}{\begin{picture}(50,66)

\put(0,0){\includegraphics{fig1b}}
\put(0,0){\includegraphics{fig1c}}
\put(10,64.5){\sx{.34}{$\sexp_{b}(x)$}}
\put(33,65){\sx{.33}{$b=\rme$}}
\put(42,63){\sx{.33}{$b=2$}}

\put(45,20){\sx{.4}{$x$}}
\end{picture}}
\end{center}
\caption{
4-exponential $\sexp_b$ to base  $b=\rme$ (thick solid) and $b=2$
(dashed)

on the real axis.}
\iL{fig01}
\begin{center}
\sx{1.8}{\begin{picture}(140,140)

\put(8,8){\includegraphics{fig3a}}

\put(8,8){\includegraphics{fig3b}}
\put(50,116){\sx{.44}{$\slog(H)$}}
\put(-1,133){\sx{.7}{$\Im(z)$}}
\put(4,108){\sx{.6}{$2$}}
\put(4, 88){\sx{.6}{$1$}}
\put(4, 68){\sx{.6}{$0$}}
\put( 0, 48){\sx{.6}{$-1$}}
\put( 0, 28){\sx{.6}{$-2$}}
\put(23, 2){\sx{.6}{$-2$}}
\put(43, 2){\sx{.6}{$-1$}}
\put(69, 2){\sx{.6}{$0$}}
\put(89, 2){\sx{.6}{$1$}}
\put(109,2){\sx{.6}{$2$}}
\put(120,1){\sx{.7}{$\Re(z)$}}
\end{picture}}
\end{center}
\caption{Contour plot of the function $\sexp_\rme(z)$ showing
  lines of
  constant modulus and of constant phase. The shaded (yellow) region is
  $\slog_\rme(H)$.
Its right boundary is $\slog_\rme\left(\rme^{\ell}\right)$ which
corresponds to $|\sexp_\rme(z)|=|L|$. \iL{figmapa3}
}

\end{figure}
For real values of the argument, this 4-exponential
is plotted in Figure \ref{fig01} for $b=\rme,2$.
The plot of $\slog_\rme$ in the complex plane (depicted in figure
\ref{figmapa3}) hints towards Criterion \ref{criterion:C}.

\subsection{The Fractional Iterates of the Exponential}
The combination of $\sexp$ and $\slog$ allows to define
fractional/continuum iterative powers of the exponential via (see
e.g.\ \cite{Szekeres1958} or \cite{Kuczma_etal1990}):
\begin{align}
\exp_{b}^{\circ c}(z)=\sexp_{b}\big(c+\slog_{b}(z)\big)
\iL{expbcz}
\end{align}
Unlike regular iterates at a hyperbolic fixed point which are analytic at
the fixed point, the above iterates of $\exp_b$ have
a branch point at both complex fixed points. Care must be taken to
determine a principal branch/region/cut of
$\slog_b$ which is then also the principal region of the iterative
power. It should be chosen such that $c+\slog_{b}(z)\in C_{-2}$.
The map of the function $\exp^{\circ 0.5}$ in the complex plane is plotted in reference \cite{Kouznetsov2010},
showing a behavior similar to the iterative square root of the factorial
$\sqrt{~!~}$ computed in \cite{Kouznetsov2010}.

\begin{figure}
\begin{center}
\sx{1.2}

{\begin{picture}(190,190)

\put(6,6){\includegraphics{fig4}}
\put(86,177){\sx{.66}{$c=2$}}
\put(102,169.4){\sx{.66}{$c=1$}}
\put(108,176){\sx{.66}{$c=0.9$}}
\put(134,176){\sx{.66}{$c=0.5$}}
\put(143,169.4){\sx{.66}{$c=0.1$}}
\put(169,170){\sx{.66}{$c=0$}}
\put(170,160){\sx{.66}{$c=-0.1$}}
\put(170,137){\sx{.66}{$c=-0.5$}}
\put(170,121){\sx{.66}{$c=-0.9$}}
\put(170,114){\sx{.66}{$c=-1$}}
\put(171,97){\sx{.66}{$c=-2$}}
\put(-1,171){\sx{.8}{$y=\exp^{\circ c}(x)$}}
\put(1,145){\sx{.8}{3}}
\put(1,125){\sx{.8}{2}}
\put(1,105){\sx{.8}{1}}
\put(1, 85){\sx{.8}{0}}
\put(0, 65){\sx{.8}{-1}}
\put(0, 45){\sx{.8}{-2}}
\put(0, 25){\sx{.8}{-3}}

\put(24, 0){\sx{.8}{-3}}
\put(44, 0){\sx{.8}{-2}}
\put(64, 0){\sx{.8}{-1}}
\put(86, 0){\sx{.8}{0}}
\put(106,0){\sx{.8}{1}}
\put(126,0){\sx{.8}{2}}
\put(146,0){\sx{.8}{3}}
\put(170,0){\sx{.8}{$x$}}
\end{picture}}
\end{center}
\caption{ Function

$y=\exp^{\circ c}(x)$ calculated by equation (\ref{expbcz}) for

$c=0$, $\pm 0.1$, $\pm 0.5$, $\pm 0.9$, $\pm 1$, $\pm 2$ versus $x$. }
\iL{fig14}
\end{figure}

For several {\em real} $c$ we plot $\exp^{\circ c}(x)$ versus
$x$ in Figure \ref{fig14}. For $c=1$ it is indeed the usual exponential,
for $c=0$ it is the identity function, and for $c=-1$ it is the
logarithm.

\bibliographystyle{amsplain}
\bibliography{main}

\end{document}


text:
\section{Causality}\label{S:sec0}

Let ${\mathcal{L}\/}$ be a real or complex linear space and $I$
a totally ordered set
(\,e.g. $I \in \{\mathbb{N}\,, \mathbb{Z}\,, \mathbb{R}\}\,$;
intervals are then naturally defined in $\,I\,$ by means of the order relation.
Of course, we shall think with $\,I\,$ of a discrete or continuous time domain.)
Also, let ${\mathcal{E} \subset \mathcal{L}^{I}\/}$ be a set
of functions such that ${f\in\mathcal{E}\,}$ implies
$\chi_{_{s\leqslant t}}(s) f(s)\in\mathcal{E}\,$,
for every ${t\in I\,}$\,, where $\chi_{_P}(s)$ denotes the
characteristic function of property $\,P\,$\,
(which is 1 if $\,s\,$ shares that property and 0 else).
Sometimes we write simply \quad $[\,f\,]_{_{\leq t}}$ \quad
in the place of \quad $\chi_{_{s\leqslant t}}(s)\;f(s)$.

\begin{ldefinition}\label{D.0.1}
A function
$F:\mathcal{E}\to\mathcal{E}\,$ is called \emph{causal\/},
iff for every ${f\in\mathcal{E}\,}$\\ and ${t \in I\,}$
\vspace{-.25cm}
\begin{equation}\centering\notag
F\,f\,(\,t\,)\;=\;F\;[\;\chi_{_{s\leqslant t}}f\,(\,s\,)\;]\,(\,t\,)\quad .
\end{equation}
Such functions are also called (causal) \emph{propagators}.

\newpage
\begin{remark}
In some respect, causal functions generalize lower triangular matrices
or integral operators such as
\vspace{-.15cm}
\begin{equation}\centering\notag
F\,f\,(\,t\,) \; = \; \int_{-\infty}^{t}K\,(\,t\,-\,s\,)\,f\,(\,s\,)\; ds\;,
\quad
\end{equation}
for instance with a Green's function kernel $K$. \\
Note that in general $\,F:f\mapsto F f\,$ needs not to be linear.

Typically $\,f\,$ represents a state evolving in time
(\,i.e. a \emph{process}\,).
Then causality of $\,F\,$ spells that $\,F\,f\,(\,t\,)\,$
depends on the history of $\,f\,$ only up to present
time $\,t\,$.
\end{remark}
\end{ldefinition}

\hspace{-.53cm}
The proof of the following is easy and left to the reader.
\begin{lproposition}\label{P.0.1}\ \\
For every ${\,t\in I\,}$;
${\,f,\,g\,\in\mathcal{E}\,}$ and causal functions
${F,\,G\,:\,\mathcal{E}\to\mathcal{E}}$
\begin{itemize}
\item[(i)]
$F f\,(\,s\,)\;=\;F\;[\;\chi_{_{r\leqslant t}}(\,r\,)\,f\,(\,r\,)
\;]\,(\,s\,)\,$,
for every $\,s\leqslant\,t$ \;.
\vspace{4pt}
\item[(ii)]
If $\;f\,(s)\;=\;g\,(s)\;$ for \emph{every} $\,s\leqslant\,t\;$, \\
\vspace{-0.45cm}
\begin{equation}\centering\notag
\text{then} \quad F f\,(s)\;=\;F g\,(s)\;\,\text{for every\,}
\,s\leqslant\,t\;.
\end{equation}
\end{itemize}
\begin{itemize}
\item[(iii)]
The product of causal functions
\begin{equation}\centering\notag
\begin{aligned}
FG \, : \, \mathcal{E}\;&\longrightarrow\; \mathcal{E}\\[-2pt]
f\;&\longmapsto\;F\,G\,f\,\;:\,=\;F\,[\,G\,[\,f\,]\,]
\end{aligned}
\end{equation}
is again causal. In fact, if $\mathcal{E}$ is a linear space,
then the causal functions over $\mathcal{E}$ form an algebra.
\end{itemize}
\end{lproposition}


TITLE: Evaluate the limit (if it exists) $\{(\frac{n^3+4n-5}{n^6+2n^2-3},n-7[\frac{n}{7}])\}_{n=1}^{\infty}$
QUESTION [1 upvotes]: $$\left\{\frac{n^3+4n-5}{n^6+2n^2-3},n-7\left[\frac{n}{7}\right]\right\}_{n=1}^{\infty}$$
where $\left[\frac{n}{7}\right]$ is the floor and ceiling function

The sequence converges if each element convarges so:
$$\lim_{n\to \infty} \frac{n^3+4n-5}{n^6+2n^2-3}=\lim_{n\to \infty} \frac{\frac{n^3}{n^6}+\frac{4n}{n^6}+\frac{-5}{n^6}}{\frac{n^6}{n^6}+\frac{2n^2}{n^6}+\frac{-3}{n^6}}=0$$
$$\lim_{n\to \infty}n-7\left[\frac{n}{7}\right]=\infty\cdot \infty=\infty$$
So $$\left\{\left(\frac{n^3+4n-5}{n^6+2n^2-3},n-7\left[\frac{n}{7}\right]\right)\right\}_{n=1}^{\infty}=\{0,\infty\}$$
And therefore there is not limit?

REPLY [2 votes]: The remainder-like function
$$
n-7\left[\frac n7\right]
$$
cycles among the values $\{0,1,2,3,-3,-2,-1\}$. Since one of the components does not converge, the sequence of pairs does not converge.