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\begin{abstract}
This paper is a continuation of earlier work on generators of simple Lie algebras in arbitrary characteristic \cite{jmb}. We show that, in contrast to classical Lie algebras, simple graded Lie algebras of Cartan type $S,H$ or $K$ never enjoy the ``one-and-a-half generation'' property. The methods rely on a study of centralisers in Cartan type Lie algebras.
\end{abstract}

\section*{Introduction}

In this paper we continue the study of generators of simple Lie algebras initiated in \cite{jmb}. The original motivation for this question comes from group theory. Namely, any finite simple group can be generated by 2 elements; furthermore, one of these elements can be chosen arbitrarily (see for example \cite{Aschbacher,Guralnick}). This last property is referred to as the ``one-and-a-half generation property''. Recently, B. Kunyavski\u\i\ asked whether the analogous property holds for simple Lie algebras (quoted in \cite[Problem 5]{Premet}).

In \cite{jmb}, we provided a partial answer to these questions. We first proved that any simple Lie algebra over an algebraically closed field $\FF$ of characteristic $\neq 2,3$ can be generated by 2 elements. For the 1,5-generation property we have the following information. First, any classical simple Lie algebra (ie. simple complex Lie algebras and their modular analogues) is generated by 1,5 elements. Also, the graded Cartan type Lie algebra $W(m,\nn)$ has the 1,5 generation property if and only if $m = 1$. In the present work we show that the simple graded Cartan type Lie algebras of the remaining types $S,H$ or $K$ are never generated by 1,5 elements.

The methods used in \cite{jmb} to investigate the 1,5 generation property were geometrical in the classical case, and {\it ad hoc} for algebras of Cartan type $W$. In the present paper the approach involves properties of centralisers in filtered Lie algebras. It should be noted that the criterion stated below does not apply for the Lie algebras $W(m,\nn)$.

This paper is organised as follows. We first give a criterion for the 1,5 generation property to fail to be satisfied in a Lie algebra. In the special case of filtered Lie algebras we obtain the following: \\

{\nbf Criterion.} \it
Let $L = L_{\geq -r} \supseteq \ldots \supseteq L_{\geq s} \supseteq (0)$ be a decreasingly filtered Lie algebra, with $r,s \geq 0$. Let $H \subseteq L$ be an ideal of $L$ such that the centraliser $C_L(H) = (0)$. Assume that, for every $h \in H$, the centraliser of $h$ in $L$ intersects the positive part $L_{\geq 1}$ non-trivially. Then $H$ cannot be generated by 1,5 elements. \\
\rm

We proceed with a brief analysis of centralisers in the Lie algebra $W_m = \Der(B_m)$ of derivations of a truncated polynomial ring in $m$ variables. We give a normal form for derivations $D \in W_m$ such that the ring of $D$-constants is $B_m^D = \FF$. Using this normal form we obtain that for such derivations, the centraliser $W_m^D$ has dimension $m$, the lowest possible value.

In Section \ref{section2} we deal with the case of simple graded Lie algebras of Cartan type $S,H$ and $K$. In each case we prove failure of the 1,5-generation property by application of the above criterion on centralisers. \\

Throughout, we will work over an algebraically closed field $\FF$ of characteristic $p > 3$. All maps, derivations, etc.\ under consideration will be $\FF$-linear. We will omit the corresponding subscripts and write for example $\Der$ or $\otimes$ instead of $\Der_\FF$ or $\otimes_\FF$.

\section{Preliminaries}

\subsection{A criterion for one-and-a-half generation}

\subsubsection{} \label{criterion}
Let $L$ be a Lie algebra. Recall that $L$ is {\it generated by 1,5 elements} if, for all $x \in L$ with $x \neq 0$, there exist $y \in L$ such that the generated subalgebra $\Fxy{x,y} = L$. We have the following criterion for the 1,5-generation property to fail: \\

\begin{prop} Let $L$ be a Lie algebra. The following are equivalent:

(i) $L$ is {\bf not} generated by 1,5 elements;

(ii) there exist a dense subset $\Omega \subseteq L$ and a family of subalgebras $(L_i)_{i \in I}$ such that:
\begin{itemize}
\item for all $i \in I$, we have $L_i \neq L$;
\item the union $\bigcup_{i \in I} L_i \supseteq \Omega$;
\item the intersection $\bigcap_{i \in I} L_i \neq (0)$. \\
\end{itemize}
\end{prop}

\proof\ Assume first that $L$ is not generated by 1,5 elements: there exists a nonzero $x \in L$ such that, for all $y \in L$, the subalgebra $L_y = \Fxy{x,y} \neq L$. We have $0 \neq x \in \bigcap_{y \in L} L_y$. Also, $\Omega = L = \bigcup_{y \in L} \{ y \} \subseteq \bigcup_{y \in L} L_y$. Thus property $(ii)$ holds.

Now assume that $(ii)$ is satisfied, and let us show that $(i)$ holds. Let $0 \neq x \in \bigcap_{i \in I} L_i$. Let $U = \{ y \in L\ |\ \Fxy{x,y} = L\}$. This is an open subset of $L$. If $L$ was generated by 1,5 elements, then we would have $U \neq \emptyset$. Since $\Omega$ is dense, there exists an element $y \in \Omega \cap U$. Because $y \in U$, we have $\Fxy{x,y} = L$. On the other hand, since $y \in \Omega$, there exists $i \in I$ such that $y \in L_i$. Since also $x \in L_i$, we have $\Fxy{x,y} \subseteq L_i \neq L$, a contradiction.

\subsubsection{} \label{corollary1,5}
For a Lie algebra $L$, we will use the notation $L^X$ for the centraliser of $X$ in $L$, where $X$ can be an element or a subspace of $L$. \\

\begin{cor}
Let $L = L_{\geq -r} \supseteq \ldots \supseteq L_{\geq s} \supseteq (0)$ be a decreasingly filtered Lie algebra, with $r,s \geq 0$. Let $H \subseteq L$ be an ideal of $L$ satisfying the following conditions:
\begin{itemize}
\item the centraliser $L^H = (0)$;
\item there exists a dense subset $\Omega \subseteq H$ such that, for all $X \in \Omega$, $L^X \cap L_{\geq 1} \neq (0)$.
\end{itemize}
Then $H$ is not generated by 1,5 elements. \\
\end{cor}

{\nbf Remark.} Note that the second condition concerns the centraliser of $X$ in the full algebra $L$, not only in $H$. \\

\proof\ For all $X \in \Omega$, we choose a non-zero element $Y \in L_{\geq 1}$ such that $[X,Y] = 0$. Denote by $H_X = L^Y \cap H$, the space of elements in $H$ which commute to $Y$. Then $X \in H_X$ by construction, so that $\bigcup_{X \in \Omega} H_X \supseteq \Omega$. Since $Y \neq 0$ centralises $H_X$, we must have $H_X \neq H$ according to the first condition. Now, the subalgebra $H$ inherits a decreasing filtration from the one in $L$; let $H_{\geq s'}$ be the non-zero subspace of maximal index. Since $Y \in L_{\geq 1}$, we have $[Y,H_{\geq s'}] \subseteq H_{> s'} = (0)$, whence $\bigcap_{X \in \Omega} H_X \supseteq H_{\geq s'} \neq (0)$. The corollary now follows from the proposition above.

\subsubsection{\bf Remark.} For the Jacobson-Witt algebra $L = H = W_m$ (and, more generally, for the non-restricted $W(m,\nn)$) it can be shown that the second condition of Corollary \ref{corollary1,5} does not hold. Indeed, let $U \subseteq L$ be the defined as:
\begin{equation*}
U = \{ X \in L \ | \ L^X \cap L_{\geq 0} = (0) \}.
\end{equation*}
Equivalently, $X \in U$ if and only if the linear map $(\ad\, X)_{| L_{\geq 0}}: L_{\geq 0} \to L$ is injective. From this it is easy to see that $U$ is open. Let $\cN$ be the regular nilpotent derivation as in \cite[Lemma 3]{Premet_Wn}: using \cite[Lemma 7 (ii)]{Premet_Wn} we can see that $\cN \in U$, so that $U$ is not empty. This precludes the existence of a dense subset $\Omega \subseteq L$ satisfying the second property of the previous corollary.

\subsection{Derivations of truncated polynomial rings}

\subsubsection{} \label{TPrings}
Let $m \geq 1$ be an integer. Let $B_m$ denote the truncated polynomial ring in $m$ variables over $\FF$, that is:
\begin{equation}
B_m = \FF[x_1,\ldots,x_m] / (x_1^p,\ldots,x_m^p).
\end{equation}
The generators of $B_m$ are again denoted by $x_1,\ldots,x_m$, so that the equality $x_i^p = 0$ holds in $B_m$. Recall that $\dim(B_m) = p^m$.

Let $W_m = \Der(B_m)$ be the full derivation algebra. For any $D \in W_m$, we can consider the subring of $D$-constants $B_m^D \subseteq B_m$ to be the space of truncated polynomials annihilated by $D$. We always have $\FF \subseteq B_m^D$.

\subsubsection{}
We introduce some convenient terminology. A derivation $D \in W_m$ is {\it regular nilpotent} if $D^{p^m-1} \neq D^{p^m} = 0$ and {\it regular semi-simple} if $D$ has $p^m$ distinct eigenvalues as a linear map acting on $B_m$. \\

\begin{prop}
Let $D \in W_m$ be a derivation of $B_m$. Assume that $D$ has no non-trivial constants, ie. $B_m^D = \FF$. Then there exists a decomposition $B_m = B' \otimes B''$, with $B' \simeq B_r$ and $B'' \simeq B_{m-r}$, and derivations $\cN \in \Der (B') \simeq W_r$, $\cS \in \Der (B'') \simeq W_{m-r}$ satisfying the following properties:
\begin{enumerate}
\item $\cN \in W_r$ is regular nilpotent and $\cS \in W_{m-r}$ is regular semi-simple;
\item $D = \cN \otimes \id_{B''} + \id_{B'} \otimes \cS \in \Der(B' \otimes B'')$. \\
\end{enumerate}
\end{prop}

\proof\ Let $\chi(t)$ be the characteristic polynomial of $D$ acting in $B_m$ and $\Lambda$ be the set of roots of $\chi(t)$. By \cite[Theorem 1, (ii)]{Premet_Wn}, $\chi(t)$ is a $p$-polynomial, and therefore all its roots have same multiplicity which is a power of $p$, say $p^r$. Furthermore, the set $\Lambda$ is an additive subgroup of $\FF$, of order $p^{m-r}$.

Let $B'$ be the nilspace of $D$, that is the kernel of $D^{p^r}$, and $\cN$ be the derivation of $B'$ induced by $D$. Then $\cN$ is nilpotent and has one-dimensional kernel $\FF$. Hence, there exists a linear basis $\{ z_1,\ldots , z_{p^r}\}$ of $B'$ such that $\cN(z_k) = z_{k-1}$ for all $k > 1$ and $\cN(z_1) = 0$; we may assume that $z_1 = 1$. It is easy to see that the algebra $B'$ has no non-trivial ideal which is stable under $\cN$: a fortiori no non-trivial ideal is stable under all derivations of $B'$. By Harper's theorem \cite{Harper}, $B'$ is isomorphic to some truncated polynomial ring; because $\dim B' = p^r$ we must have $B' \simeq B_r$.

Let $\lambda \in \Lambda$ be an eigenvalue of $D$. Denote by $B^\lambda$ be the corresponding eigenspace and $B^{(\lambda)}$ the characteristic subspace; in particular $\dim(B^{(\lambda)}) = p^r$. Also, let $g_{\lambda}$ be a non-zero eigenvector; for $\lambda = 0$ we can choose $g_0 = 1$. Finally, for all $k \in \{1,\ldots,p^r\}$, set $g_{\lambda,k} = z_k g_{\lambda}$. We have $D(g_{\lambda,k}) = \lambda\, g_{\lambda,k} + g_{\lambda,k-1}$. From this fact we can see that the matrix of the action of $D$ on $B^{(\lambda)}$ is given by a single Jordan block of size $p^r$:
\begin{equation}
\left[ \begin{array}{cccc}
\lambda & 1 & \ldots & 0 \\
0 & \ddots & \ddots  & \\
0 &  & \lambda & 1 \\
0 &  \ldots & 0 & \lambda
\end{array} \right]
\end{equation}
and the elements $g_{\lambda,k}$ for $k \in \{ 1,\ldots,p^r\}$ span the subspace $B^{(\lambda)}$. Furthermore, we have $\dim ( B^{\lambda} ) = 1$.

Let $B'' = \sum_{\lambda \in \Lambda} B^{\lambda}$: clearly, $B''$ is a subalgebra of $B_m$. Since the $g_{\lambda,k} = z_k g_\lambda$, where $\lambda \in \Lambda$ and $k \in \{1,\ldots,p^r \}$, form a linear basis of $B_m$, we can see that the multiplication map induces an isomorphism $B' \otimes B'' \stackrel{\sim}{\to} B_m$. Since $B''$ is generated by eigenvectors of $D$, it is also stable under the derivation $D$. Let $\cS \in \Der(B'')$ be the induced derivation. By construction, $\cS$ is semi-simple, and all eigenspaces of $\cS$ in $B''$ have dimension 1, so that $\cS$ is also regular. Finally, it is easy to check that $D = \cN \otimes \id_{B''} + \id_{B'} \otimes \cS$.

To complete the proof of the proposition it remains to show that $B'' \simeq B_{m-r}$. Let $\gtm \subseteq B_m$ be the maximal ideal of $B_m$. Let $\xi_1,\ldots,\xi_r \in B'$ be a set of generators of $B'$ given by an isomorphism $B_r = \FF[x_1,\ldots,x_r] / (x_1^p,\ldots,x_r^p) \stackrel{\sim}{\to} B'$. Using the fact that $(\xi_1 \cdots \xi_r)^{p-1} \neq 0$, we can see that the linear terms of $\xi_1,\ldots,\xi_r$ are linearly independent modulo $\gtm^2$. Hence, they admit a cobasis $\{ \xi_{r+1},\ldots,\xi_m \}$ for $\gtm$ modulo $\gtm^2$. We obtain a chain of isomorphisms:
\begin{equation}
B_{m-r} \simeq \FF[\xi_{r+1},\ldots,\xi_m] \simeq \frac{B_m}{ \sum_{i = 1}^r \xi_i\, B_m } \simeq \left( \frac{B'}{ \sum_{i = 1}^r \xi_i\, B' } \right) \otimes B'' \simeq \FF \otimes B'' \simeq B''.
\end{equation}
The proposition is proved.

\subsubsection{} \label{CorSmallKernel}
\begin{cor}
Let $D \in W_m$ be such that $B_m^D = \FF$ and $W_m^D$ be the centraliser of $D$ in $W_m$. Then $\dim ( W_m^D ) = m$. \\
\end{cor}

\proof\ We first prove the corollary in two particular cases. First, if $D = \cN$ is regular nilpotent, the equality $\dim ( W_m^D ) = m$ follows from \cite[Lemma 7.2 (ii)]{Premet_Wn}.

Second, assume that $D = \cS$ is regular semi-simple. Then $D$ is contained in some maximal torus $T \subseteq W_m$. Since $D$ is regular, we can see that $T = \sum_{j \geq 0}\FF\, D^{[p^j]}$, and $W_m^D = W_m^T$ the centraliser of $T$. By \cite[Corollary 7.5.17]{Strade}, $W_m^T$ is a Cartan subalgebra of minimal dimension, so that $\dim (W_m^T) = m$ by \cite[Table 3.25]{Benkart}.

Now we turn to the general case. We write $B_m = B' \otimes B''$ and $D = \cN \otimes \id_{B''} + \id_{B'} \otimes \cS$ as in the Proposition. Let $\Delta \in W_m^D$. From the construction in the proof, we have that $B'$ is the nilspace of $D$ and $B''$ the sum of eigenspaces, so that $\Delta$ must stabilise both $B'$ and $B''$. This means that $\Delta$ takes the form $\Delta = \Delta' \otimes \id_{B''} + \id_{B'} \otimes \Delta''$, where $\Delta' \in \Der(B') \simeq W_r$ and $\Delta'' \in \Der(B'') \simeq W_{m-r}$. Furthermore, since $[D,\Delta] = 0$ we also have $[\Delta', \cN] = 0$ and $[\Delta'',\cS] = 0$. It follows readily that:
\begin{equation}
W_m^D = W_r^{\cN} \otimes \id_{B''} +  \id_{B'} \otimes W_{m-r}^{\cS}.
\end{equation}
Since the corollary holds in the regular nipotent and semi-simple cases, we obtain $\dim (W_m^D) = \dim (W_r^{\cN}) + \dim ( W_{m-r}^{\cS} ) = r + (m - r) = m$ as we wanted.

\section{Generators for graded Cartan type algebras}
\label{section2}

\subsection{Divided power algebras and special derivations} $ $\\
\label{SubSectionW}

We recall from \cite[Section 2.1]{Strade} some basic facts and notations about special derivations of divided power algebras.

\subsubsection{}
Let $m \geq 1$ be an integer. For a multi-index $\alpha=(\alpha_1,\ldots,\alpha_m) \in \ZZ^m$, we define the length to be $|\alpha| = \alpha_1+\ldots+\alpha_m$. The factorial and binomial coefficients are given, for $\alpha,\beta \in \NN^m$, by:
\begin{equation}
\alpha! = \alpha_1 ! \cdots \alpha_m ! \quad \mbox{ and } \quad \binom{\alpha}{\beta} = \binom{\alpha_1}{\beta_1} \cdots \binom{\alpha_m}{\beta_m}.
\end{equation}
For all $k \in \{1,\ldots,m\}$, define the $k$-th unit multi-index by $\epsilon_k = (\delta_{1,k},\ldots,\delta_{m,k})$. Furthermore, we set $\un = \epsilon_1 + \ldots + \epsilon_m = (1,\ldots,1)$. The set of multi-indices can be partially ordered in the following fashion: for $\alpha,\beta \in \ZZ^m$, let $\alpha \leq \beta$ if the components $\alpha_i \leq \beta_i$ for all $i \in \{1,\ldots,m \}$. We will also write $\alpha < \beta$ if $\alpha \leq \beta$ and $\alpha \neq \beta$.

\subsubsection{\bf Conventions}
\label{furtherconventions}
In the sequel we will use the following notational conventions. For a multi-index denoted by $\nn \in \NN^m$ we always make the implicit assumption that all components $n_i > 0$. Furthermore, given such a multi-index we also implicitly define $\tau = \tau(\nn) \in \NN^m$ to be $\tau = (p^{n_1}-1,\ldots,p^{n_m}-1)$. In particular, the expression $0 \leq \alpha \leq \tau$ means that the components $0 \leq \alpha_i \leq p^{n_i} -1$ for all $i \in \{ 1,\ldots,m\}$.

\subsubsection{\bf Divided power algebras}
\label{dividedpowers}
Let $m \geq 1$ be an integer. Define the divided power algebra $\cO(m)$ over $\FF$ as follows: as vector space, $\cO(m) = \FF[x^{(\alpha)}\, :\, \alpha \in \NN^m]$ and the multiplication map is given by the formula:
\begin{equation}
(\forall\, \alpha,\beta \in \NN^m)\ :\ x^{(\alpha)} x^{(\beta)} = \binom{\alpha + \beta}{\alpha} x^{(\alpha+\beta)}.
\end{equation}
The algebra $\cO(m)$ has a natural grading over $\ZZ$ such that all elements $x^{(\alpha)}$ are homogeneous, of degree $| \alpha |$. For all muti-indices $\nn \in \NN^m$ , define a linear subspace:
\begin{equation}
\cO(m,\nn) = \operatorname{Span} \left\{x^{(\alpha)}\ |\ 0 \leq \alpha\leq \tau \right\} \subseteq \cO(m).
\end{equation}
Then $\cO(m,\nn)$ is actually a graded subalgebra of $\cO(m)$, of dimension $p^{|\nn|}$.

\subsubsection{} \label{Suggestive}
For divided power algebras, we will use suggestive terminology by analogy with polynomial rings. For example, elements $x^{(\alpha)}$ will be called ``monomials in $x_1,\ldots,x_m$, and a monomial $x^{(\alpha)}$ is ``independent of $x_k$'' if the component $\alpha_k = 0$.

We will also use the following notations. For all $k \in \{ 1,\ldots,m\}$ and $r \geq 0$, set $x_k^{(r)} = x^{(r\, \epsilon_k)}$; when $r = 1$, write $x_k = x_k^{(1)}$. With this notation, we can check that $x^{(\alpha)} = x_1^{(\alpha_1)} \cdots x_m^{(\alpha_m)}$, and $x_k^{(r)} x_k^{(s)} = \binom{r+s}{r} x_k^{(r+s)}$.

\subsubsection{\bf Special derivations}
For all $k \in \{1,\ldots,m\}$, we define the ``partial derivative'' $\d_k \in \Der\, \cO(m,\nn)$ by the rule $\d_k(x^{(\alpha)}) = x^{(\alpha - \epsilon_k)}$. The Lie algebra of special derivations of $\cO(m,\nn)$, denoted by $W(m,\nn)$, is defined to be the sub-$\cO(m,\nn)$-module of $\Der\, \cO(m,\nn)$ generated by these partial derivatives:
\begin{equation} \label{plushaut}
W(m,\nn) = \cO(m,\nn)\, \d_1 \oplus \ldots \oplus  \cO(m,\nn)\, \d_m \subseteq \Der\, \cO(m,\nn).
\end{equation}
The inclusion is proper unless $\nn = \un$. In this case, we have $\cO(m,\un) \simeq B_m$ and $W(m,\un) \simeq W_m$, where $B_m$ and $W_m$ are defined in Section \ref{TPrings}.

\subsubsection{}
Because $\cO(m,\nn)$ is a graded algebra, the Lie algebra $W(m,\nn)$ inherits a natural grading, compatible with the grading on $\cO(m,\nn)$. More precisely, all derivations of the form $x^{(\alpha)} \d_k$ have degree $|\alpha| - 1$. The homogeneous subspace of maximal degree has dimension $m$ and is generated by derivations $x^{(\tau)}\, \d_k$, for $\tau = (p^{n_1}-1,\ldots,p^{n_m}-1)$ and $k \in \{1,\ldots,m\}$. The homogeneous subspace of minimal degree -1 has also dimension $m$ and is generated by the partial derivatives $\d_k$, for $k \in \{1,\ldots,m\}$.

We will frequently use natural notations such as $W(m,\nn)_{\geq k}$ to denote the subspace spanned by homogeneous elements of degree $\geq k$, or $W(m,\nn)_{max}$ to denote the homogeneous subspace of maximal degree. Also, we will use expressions such as ``$D = D_r + \ldots$, where ``$\ldots$'' are higher order terms''. This will mean that $D_r$ is homogeneous of degree $r$, and ``$\ldots$'' is an element of $W(m,\nn)_{> r}$.

\subsubsection{\bf Embeddings of $W$ type algebras.}
\label{Plongements}
Let $m,\nn$ be as above and let $N = |\nn| \in \NN$. By \cite[p. 64]{Strade}, the algebra $\cO(m,\nn)$ is isomorphic to $\cO(N,\un)$ as abstract (ie. non-graded) algebras. Any isomorphism $\sigma:\cO(m,\nn) \stackrel{\sim}{\to} \cO(N,\un)$ gives rise to an isomorphism $\Der\, \cO(m,\nn) \stackrel{\sim}{\to} \Der\, \cO(N,\un) = W(N,\un)$, hence an embedding:
\begin{equation}
\iota: W(m,\nn) \hookrightarrow W(|\nn|,\un).
\end{equation}
Note that, like $\sigma$, this embedding does not respect the natural gradings. By construction, we have the following relations, for all $D \in\, W(m,\nn)$ and $f \in\, \cO(m,\nn)$:
\begin{equation}
\iota(D) = \sigma^{-1} \circ D \circ \sigma
\end{equation}
\begin{equation} \label{iotapreservesmodulestructure}
\iota(f \cdot D) = \sigma(f) \cdot \iota(D).
\end{equation}

\subsection{Lie algebras of type $S$.} $ $\\
\label{SubSectionS}

In this section we focus on simple graded Lie algebras of type $S$. We first recall some constructions; the question of generators is addressed in Theorem \ref{UnEtDemiS}.

\subsubsection{}
Let $m \geq 2$ be an integer and $\nn$ a multi-index. Consider the Lie algebra $W(m,\nn)$ as in \ref{SubSectionW}. Define the {\it divergence of a special derivation} by the rule:
\begin{equation}
\div (\sum_{j = 1}^m f_j\, \d_j) = \sum_{j=1}^m\, \d_j(f_j) \in \cO(m,\nn).
\end{equation}
The divergence map satisfies the following properties, for all $f \in\, \cO(m,\nn)$ and $D,D' \in\, W(m,\nn)$:
\begin{equation} \label{divfD}
\div(f\, D) = f\, \div(D) + D(f),
\end{equation}
\begin{equation} \label{divDD'}
\div\, [D,D'] = D(\div\, D') - D'(\div\, D).
\end{equation}
The Lie algebra $S(m,\nn) \subseteq W(m,\nn)$ is defined as the subspace of divergence-free derivations, ie. $D \in S(m,\nn)$ if and only if $\div (D) = 0$. Then, the derived subalgebra $S(m,\nn)^{(1)}$ is a simple ideal of codimension $m$. Furthermore, both algebras $S(m,\nn)$ and $S(m,\nn)^{(1)}$ are graded subalgebras of $W(m,\nn)$.

For simple Lie algebras of type $S$, it is customary to impose the further restriction $m \geq 3$; for $m = 2$ the Lie algebra $S(2,\nn)^{(1)}$ can actually be identified with a simple Lie algebra of type $H$. However, the results we will prove here do not require such a restriction, so that the case $m = 2$ will also be considered.

\subsubsection{} \label{IotaRespectsDiv}
Recall the embedding $\iota: W(m,\nn) \hookrightarrow W(N, \un)$ described in Section \ref{Plongements}), with $N = \nn$. The following lemma ensures that we can arrange $\iota$ to preserve the divergence maps. \\

\begin{lemma}
There exists an isomorphism $\sigma: \cO(m,\nn) \stackrel{\sim}{\to} \cO(N,\un)$ satisfying the following property. Let  $\iota: W(m,\nn) \hookrightarrow W(N, \un)$ the induced embedding (see \ref{Plongements}). Denote by $\div_{(m,\nn)}$ and $\div_{(N,\un)}$ the divergence maps defined in $W(m,\nn)$ and $W(N,\un)$ respectively. Then, for all $D \in W(m,\nn)$:
\begin{equation} \label{CompatibilityIotaDiv}
\div_{(N,\un)} \big( \iota(D) \big) = \div_{(m,\nn)}(D).
\end{equation}
\end{lemma}

\proof\
We use the usual notations for elements of $\cO(m,\nn)$ and $W(m,\nn)$. For $\cO(N,\un)$ and $W(N,\un)$ we introduce alternative notations. To make descriptions easier, we won't use the standard variables indexed by $\{1,\ldots,N\}$ but rather some suitably doubly indexed variables. We will consider $\cO(N,\un)$ to be generated by variables $\xi_{i,j} = \xi_{i,j}^{(1)}$, where the pairs $(i,j)$ satisfy $i \in \{1,\ldots,m\}$ and $j \in \{ 0,\ldots,n_i - 1\}$. Any element $\Delta \in W(N,\un)$ can be written as:
\begin{equation}
\Delta = \sum_{i,j}\, \varphi_{i,j}\, \delta_{i,j},
\end{equation}
where the coefficients $\varphi_{i,j} \in \cO(N,\un)$ and $\delta_{i,j}$ is the partial derivation defined by $\delta_{i,j}(\xi_{i,j}) = 1$ and $\delta_{i,j}(\xi_{k,l}) = 0$ otherwise. For such a derivation, we have:
\begin{equation}
\div_{(N,\un)}(\Delta) = \sum_{i,j}\, \delta_{i,j}(\varphi_{i,j}).
\end{equation}
The isomorphism $\sigma: \cO(m,\nn) \stackrel{\sim}{\to} \cO(N,\un)$ is uniquely defined by the assignment $x_i^{(p^j)} \mapsto \xi_{i,j}$.

By construction, the embedding $\iota$ preserves multiplication by ``polynomials'', see Equation  (\ref{iotapreservesmodulestructure}). Using the formula  (\ref{divfD}) concerning $\div(fD)$, it is clearly enough for the Lemma to prove the identity (\ref{CompatibilityIotaDiv}) on a generating system of the $\cO(m,\nn)$-module $W(m,\nn)$, for example the partial derivatives $\d_1,\ldots,\d_m$.

Write out the derivation $\iota(\d_k) = \sum_{i,j}\, \psi_{i,j} \, \delta_{i,j}$. In this expression, we have $\psi_{i,j} = \iota(\d_k)(\xi_{i,j}) = \sigma\big( \d_k(x_i^{(p^j)}) \big)$, so that:
\begin{equation}
\psi_{i,j} =
\left\{ \begin{array}{cl}
0 & \mbox{ if }  i \neq k, \\
\sigma(x_k^{(p^j-1)}) & \mbox{ if } i = k.
\end{array} \right.
\end{equation}
Using the decomposition $p^j-1 = (p-1) + (p-1)p + \ldots + (p-1)p^{j-1}$, we can see that
\begin{equation}
x_k^{(p^j-1)} \in \FF\, \left(x_k^{(1)}\right)^{p-1} \cdots \left( x_k^{(p^{j-1})}\right)^{p-1},
\end{equation}
so that $\psi_{k,j} = \sigma(x_k^{(p^j-1)}) \in \FF\, \xi_{k,0}^{p-1} \cdots \xi_{k,j-1}^{p-1}$. It readily follows that $\delta_{k,j} (\psi_{k,j}) = 0$; we finally obtain:
\begin{equation}
\div_{(N,\un)}(\iota(\d_k)) = \sum_{i,j}\, \, \delta_{i,j}(\psi_{i,j}) = 0 = \div_{(m,\nn)}(\d_k),
\end{equation}
as desired. The lemma is proved.

\subsubsection{} \label{UnEtDemiS}
\begin{thm}
Let $m \geq 2$ be an integer and $\nn \in \NN^m$ a multi-index with non-zero entries. The simple Lie algebra $S(m,\nn)^{(1)}$ is not generated by 1,5 elements. \\
\end{thm}

\proof\ First we show that, for all $D \in S(m,\nn)^{(1)}$, the constant ring $\cO(m,\nn)^D \neq \FF$. By Lemma \ref{IotaRespectsDiv}, we have an embedding $S(m,\nn)^{(1)} \hookrightarrow S(N,\un)^{(1)}$ induced by some isomorphism $\cO(m,\nn) \simeq \cO(N,\un)$, so that we may assume that $\nn = \un$. In this case we use the simpler notations $W_m$, $S_m$, $B_m$ instead of $W(m,\un)$, $S(m,\un)$, $\cO(m,\un)$. Let $CS_m = \{ D \in W_m |\ \div(D) \in \FF \}$. We can see that $S_m^{(1)}$ is an ideal of codimension $m+1$ in $CS_m$. Thus, for every $D \in S_m^{(1)}$ the subspace $[D,CS_m] \subseteq CS_m$ has codimension $\geq m+1$; therefore, the centraliser $CS_m^D$ has dimension $\geq m+1$. A fortiori, we have $\dim (W_m^D) > m$: by Corollary \ref{CorSmallKernel}, we obtain $\dim\, B_m^D > 1$.

Now we go back to the general case of $S(m,\nn)^{(1)}$. We will apply Corollary \ref{corollary1,5} with $L = S(m,\nn)$ and $H = S(m,\nn)^{(1)}$. The fact that the centraliser of $H$ in $L$ is trivial follows from the fact that both $H$ and $L$ have trivial centre. Now we turn to the second condition. We endow $S(m,\nn)$ with the filtration defined by its standard grading. Let $\Omega \subseteq S(m,\nn)^{(1)}$ be the set of derivations which have a non-zero homogeneous part of degree $-1$; this is a non-empty Zariski open subset of $S(m,\nn)^{(1)}$. All we need to show is that, for all $D \in \Omega$, there exists a non-zero element $\Delta \in S(m,\nn)_{\geq 1}$ such that $[D,\Delta] = 0$ (we do not require $\Delta$ to belong to the derived subalgebra).

We let $D \in \Omega$ act as a derivation on $\cO(m,\nn)$. Let us check that $\cO(m,\nn)_{\geq 2}^D \neq (0)$, ie. there exist a non-zero element $f \in \cO(m,\nn)$ such that $D(f) = 0$ and $f$ has no homogeneous terms of degree 0 and 1. We showed above that $\cO(m,\nn)^D \neq \FF$, so that there exist $0 \neq f \in \cO(m,\nn)^D$ with no constant term. Write out $f = f_1 + \ldots$, where $f_1$ is homogeneous of degree 1 and ``$\ldots$'' are higher order terms. If $f_1 = 0$, we have $f \in \cO(m,\nn)_{\geq 2}$ and we are done. If $f_1 \neq 0$, we have $f^2 = f_1^2 + \ldots \neq 0$, and we can use $f^2 \in \cO(m,\nn)_{\geq 2}^D$.

Now choose a non-zero element $f \in \cO(m,\nn)_{\geq 2}^D$, and let $\Delta = f\, D \in W(m,\nn)$. Because $D$ has a non-zero term of degree $-1$ and $f \in \cO(m,\nn)_{\geq 2}$, we have $0 \neq \Delta = f\, D \in W(m,\nn)_{\geq 1}$. Using the formula (\ref{divfD}) we obtain $\div(\Delta) = 0$, hence $\Delta \in S(m,\nn)_{\geq 1}$. And finally, we have $[D,\Delta] = [D,f\, D] = 0$. This is what remained to be shown: thus, we can apply Corollary \ref{corollary1,5} and obtain that $S(m,\nn)^{(1)}$ is not generated by 1,5 elements.

\subsection{Lie algebras of type $H$.} $ $\\ \label{SubSectionH}

In this section, we let $m = 2r \geq 2$ be an even integer and $\nn \in \NN^{2r}$ a multi-index with non-zero entries. We keep the general notations for divided power algebra and special derivations introduced in Section \ref{SubSectionW}.

\subsubsection{} \label{introH}
We recall some construction from \cite[Section 4.2]{Strade}. First define a binary operation on $\cO(2r,\nn)$ by setting:
\begin{equation}
\{ f,g \} = \sum_{j=1}^{2r} \sigma(j) \d_j(f)\, \d_{j'}(g),
\end{equation}
where $j' = \left\{ \begin{array}{rcl} j+r & \mbox{if} & 1 \leq j \leq r \\ j-r & \mbox{if} & r < j \leq 2 r \end{array} \right.$ and $\sigma(j) = \left\{ \begin{array}{rcl} +1 & \mbox{if} & 1 \leq j \leq r \\ -1 & \mbox{if} & r < j \leq 2 r \end{array} \right.$.

The bracket $\{.,.\}$ is a Poisson bracket on $\cO(2r,\nn)$, ie. a bi-derivation which is also a Lie bracket. For all $f \in \cO(2r,\nn)$, let $D_H(f)$ be the derivation $D_H(f): g \mapsto \{ f, g \}$. Then $D_H$ is a morphism of Lie algebras $D_H : \cO(2r,\nn) \to W(2r,\nn)$, the kernel of which is $\FF\, 1$. Let $H(2r,\nn)$ denote its image; then the second derived subalgebra $H(2r,\nn)^{(2)}$ is a simple ideal of dimension $p^{|\nn|} - 2$. Note that $H(2r,\nn)^{(2)}$ is a graded subalgebra of $W(2r,\nn)$; for any $0 \neq \alpha \in \NN^{2r}$, the derivation $D_H(x^{(\alpha)})$ is homogeneous, of degree $|\alpha|-2$.

\subsubsection{} \label{UnEtDemiH}
\begin{thm}
The Lie algebra $H(2r,\nn)^{(2)}$ is not generated by 1,5 elements. \\
\end{thm}

\proof\ We will apply Corollary \ref{corollary1,5} to $L = H(2r,\nn)$ and $H = H(2r,\nn)^{(2)}$. It is easy to check that the centraliser of $H$ in $L$ is trivial. Now let $\Omega \subseteq H(2r,\nn)^{(2)}$ be the open set of elements which have a non-zero component of degree $-1$. We want to show that, for all $D \in \Omega$, there exists $0 \neq \Delta \in H(2r,\nn)_\geq 1$ such that $[D,\Delta] = 0$.

We choose an element $f \in \cO(2r,\nn)$ such that $D_H(f) = D$. Since $D_H(1) = 0$, we can assume that $f$ has no constant term; also, since the image $D_H(f)$ has a non-zero component of degree $-1$, $f$ must have a non-zero term of degree 1. Then, the lowest degree term of $f^3$ has degree 3, and clearly $\{ f,f^3 \} = 0$. Let $\Delta = D_H(f^3) \in H(2r,\nn)$. Since $f \in \cO(m,\nn)_{\geq 3}$, we have $\Delta \in H(2r,\nn)_{\geq 1}$; also, we have $[D,\Delta] = D_H(\{ f,f^3\}) = 0$. Finally, the fact that $f^3 \not \in \FF$ ensures that $\Delta \neq 0$: this is what we needed to apply Corollary \ref{corollary1,5}, proving the theorem.

\subsection{Lie algebras of type $K$.} $ $\\

In this section, we let $m = 2r + 1 \geq 3$ be an odd integer and $\nn \in \NN^{2r+1}$ a multi-index with non-zero entries. Again, throughout this section the general notations for divided power algebra and special derivations are the same as in \ref{SubSectionW}.

\subsubsection{}
We recall the constructions from \cite[Section 4.2]{Strade}. As in the case of the Lie algebras of type $H$, we will use the notations $j' = \left\{ \begin{array}{rcl} j+r & \mbox{if} & 1 \leq j \leq r \\ j-r & \mbox{if} & r < j \leq 2 r \end{array} \right.$ and $\sigma(j) = \left\{ \begin{array}{rcl} +1 & \mbox{if} & 1 \leq j \leq r \\ -1 & \mbox{if} & r < j \leq 2 r \end{array} \right.$. (The values of $j'$ and $\sigma(j)$ are not defined for $j = 2r+1$.)

Define a linear map $D_K: \cO(2r+1,\nn) \to W(2r+1,\nn)$ by the formula:
\begin{equation}
D_K(f) = \sum_{j = 1}^{2r} \Bigg( \sigma(j)\, \d_{j}(f) + x_{j'} \d_m(f) \Bigg) \d_{j'}
+ \left( 2f - \sum_{j=1}^{2r}x_j \d_j(f) \right) \d_{2r+1}.
\end{equation}
For $f,g \in \cO(m,\nn)$, we define the {\it contact bracket of $f$ and $g$} by:
\begin{equation} \label{fundamentalK}
\langle f,g \rangle = D_K(f) (g) - 2\, g\, \d_m(f) \in \cO(m,\nn).
\end{equation}
Then $D_K$ is an injective mapping, and for all $f,g \in \cO(2r+1,\nn)$ we have $[D_K(f) , D_K(g)] = D_K(\langle f,g \rangle)$. The Lie algebra $K(2r+1,\nn)$ is defined to be the image of $D_K$; by construction, the contact bracket $\langle.,.\rangle$ endows $\cO(2r+1,\nn)$ with a Lie algebra structure, and the map $D_K:\cO(2r+1) \to K(2r+1,\nn)$ is an isomorphism of Lie algebras. \\

{\nbf Remark.} Unlike the Poisson bracket on $\cO(2r,\nn)$, the contact bracket is not a biderivation of $\cO(2r+1,\nn)$: for example, we have $\langle1,f \rangle = 2 \d_m(f)$ for all $f \in \cO(2r+1,\nn)$, so that $1$ is not central for the contact bracket.

\subsubsection{} \label{Kdegree}
The derived subalgebra $K(2r+1,\nn)^{(1)}$ is a simple Lie algebra. One has $K(2r+1,\nn)^{(1)} = K(2r+1,\nn)$ unless $m + 3 \equiv 0 \mod p$, in which case $K(2r+1,\nn)^{(1)}$ has codimension 1 in $K(2r+1,\nn)$.

The contact algebra admits a grading such that $K(2r+1,\nn) = K_{-2} \oplus K_{-1} \oplus K_0 \oplus \ldots \oplus K_s$. Unlike the other types, there exist non-zero homogeneous elements of degree -2; in particular, the embedding $D_K:K(2r+1,\nn) \hookrightarrow W(2r+1,\nn)$ does not respect the natural degrees. To emphasise the difference, we will call {\it $K$-degree} the degree defined on $K(2r+1,\nn)$.

Let us describe the $K$-degree in $K(2r+1,\nn)$. For all multi-indices $\alpha \in \NN^m$, we let $\| \alpha \| = \alpha_1 + \ldots +\alpha_{2r} + 2 \alpha_{2r+1}$; this amounts to assigning degree 1 to each variable $x_j$ with $j \leq 2r$, and degree 2 to $x_{2r+1}$. Then, the derivation $D_K(x^{(\alpha)})$ is homogeneous, of $K$-degree $\| \alpha \| - 2$. In particular, if we identify $\cO(2r+1,\nn) \equiv K(2r+1,\nn)$ via $D_K$, we have the following equalities:
\begin{equation} \label{sousespacesK}
K_{-2} = \FF \quad ; \quad K_{-1} = \sum_{j = 1}^{2r} \FF\, x_j \quad ; \quad K_0 = \FF\, x_{2r+1} + \sum_{i,j = 1}^{2r}\, \FF\, x_i\, x_j.
\end{equation}

Note that the $K$-degree is not quite additive for the associative product in $\cO(2r+1,\nn)$: actually, we have $K_i\, K_j \subseteq K_{i + j + 2}$ for all $i,j$.

\subsubsection{}
We will often identify the spaces $\cO(2r+1,\nn)$ and $K(2r+1,\nn)$ by means of the map $D_K$. Each element $f \in \cO(2r+1,\nn)$ induces two linear maps on $\cO(2r+1,\nn)$. First, we have $D_K(f) \in \End\, \cO(2r+1,\nn)$, which is actually a special derivation of the divided power algebra $\cO(2r+1,\nn)$. Second, $\ \ad_K(f)$, the adjoint action of $f$ on the Lie algebra $\cO(2r+1,\nn)$ endowed with the contact bracket: for all $f,g \in \cO(2r+1,\nn)$, we have $\ \ad_K(f)(g) = \langle f,g \rangle$. The link betweeen these mappings is given by the following lemma. \\

\begin{lemma}
Let $f \in \cO(2r+1,\nn)$ be arbitrary. Let $\mu_f \in \End\, \cO(2r+1,\nn)$ be the linear map defined by (associative) multiplication by $f$. Then,
\begin{equation}
\ad_K(f) \circ \mu_f = \mu_f \circ D_K(f) \in \End\, \cO(2r+1,\nn).
\end{equation}
\end{lemma}

\proof\ We use the identity (\ref{fundamentalK}): for all $g \in \cO(2r+1,\nn)$, we have:
\begin{eqnarray*}
\langle f,f g \rangle & = & D_K(f)(fg) - 2 \d_m(f)\, fg \\
 & = & f\, D_K(f)(g) + D_K(f)(f)\, g\, -2 \d_m(f)\, f g \\
  & = &f\, D_K(f)(g) + \langle f,f \rangle\, g =  f\, D_K(f)(g),
\end{eqnarray*}
which means exactly that $ \ad_K(f) \circ \mu_f (g) =  \mu_f \circ D_K(f) (g)$.

\subsubsection{} \label{GenericInvariantsK}
\begin{cor}
Assume that $f$ is invertible in $\cO(2r+1,\nn)$, ie. $f$ has a non-zero constant term. Then $D_K(f)$ and $\ad_K(f)$ are conjugate as linear endomorphisms of $\cO(2r+1,\nn)$. \\
\end{cor}

{\nbf Remark.} Since the conjugating map is given by multiplication by $f$, in particular the invariant subspaces are linked by the relation:
\begin{equation}
\cO(2r+1,\nn)^{\ad_K(f)} = f\, \cO(2r+1,\nn)^{D_K(f)}.
\end{equation}

\subsubsection{} \label{centralisateursK}
\begin{lemma}
Let $D \in K(2r+1,\nn)$ and $K(2r+1,\nn)^D$ be its centraliser.
\begin{enumerate}
\item We have $\dim\, K(2r+1,\nn)^D \geq \min\{ 2r+1 , p \}$.
\item If $D$ has a non-zero term of degree $-2$, then $\dim\, K(2r+1,\nn)_{\geq 1}^D \geq 2$. \\
\end{enumerate}
\end{lemma}

\proof\ We first prove \nr 1. Let $\chi_K(t) = \FF[t]$ be the characteristic polynomial of $D$ acting on $K(2r+1,\nn)$ by the adjoint action. Using \ref{GenericInvariantsK} we can see that it is the same as the characteristic polynomial of $D$ acting as a derivation of $\cO(m,\nn)$, so that by \cite[Theorem 1, (ii)]{Premet_Wn}, it is a $p$-polynomial of degree $p^{|\nn|}$. Thus,
\begin{equation}
\chi_K(t) = f(t)^{p^{|\nn|-d}},
\end{equation}
where $f(t)$ is a separable $p$-polynomial of degree $p^d$ for some $d \in \{ 0,\ldots,|\nn|\}$. Let $\ell \in \{ 0,\ldots , |\nn| - d\}$ be such that the minimal annihilating $p$-polynomial of $D$ is $f_\Lambda(t)^{p^\ell}$.

Let $K(2r+1,\nn)^{(D)}$ be the nilspace of $\ad_K(D)$ in $K(2r+1,\nn)$, so that $K(2r+1,\nn)^D \subseteq K(2r+1,\nn)^{(D)}$. We have $\dim\, K(2r+1,\nn)^{(D)} = p^{|\nn| - d}$, and $D$ induces a nilpotent map of index $\leq p^\ell$ on $K(2r+1,\nn)^{(D)}$. This nilpotent map has the following Jordan decomposition:
\begin{equation}
\left[ \begin{array}{ccc}
 J_1 & & \\
 & \ddots & \\
 & & J_k
\end{array} \right],
\end{equation}
and we have the following information:
\begin{itemize}
\item the sizes of the Jordan cells are $p^\ell \geq |J_1| \geq \ldots \geq |J_k|$,
\item the size of the large matrix is $|J_1| + \ldots + |J_k| = p^{|\nn|-d}$,
\item the number of Jordan cells is $k = \dim\, K(2r+1,\nn)^D$.
\end{itemize}

Assume first that $\ell < |\nn| - d$. By the above properties, we have:
\begin{equation}
p^{|\nn|-d} \leq k |J_1| \leq k p^\ell,
\end{equation}
 so that $\dim\, K(2r+1,\nn)^D = k \geq p^{|\nn|-d-\ell} \geq p$.

Now assume that $\ell = |\nn| - d$. Then the minimal annihilating $p$-polynomial of $\ad_K(D)$ has degree $p^{d+\ell} = p^{|\nn|}$, so that the subspace $\mathfrak{D} = \sum_{j \geq 0} \FF\, \ad_K(D)^{p^j} \subseteq \Der \, K(2r+1,\nn)^{(1)}$ has dimension $|\nn|$. By \cite[Theorem 7.1.2]{Strade}, $K(2r+1,\nn) \subseteq \Der\, K(2r+1,\nn)^{(1)}$ is a subspace of codimension $|\nn| - (2r+1)$, so that $\dim\, K(2r+1,\nn) \cap \mathfrak{D} \geq |\nn| - (|\nn|- 2r-1) = 2r+1$. Since $K(2r+1,\nn) \cap \mathfrak{D} \subseteq K(2r+1,\nn)^D$, we obtain $\dim\, K(2r+1,\nn)^D \geq 2r+1$. \\

We turn to the proof of \nr 2. We identify $K(2r+1,\nn)$ with $\cO(2r+1,\nn)$, endowed with the contact bracket: in this identification we write $D = f \in \cO(2r+1,\nn)$. We assume $D$ has a non-zero term of degree -2, so that equivalently $f$ has a non-zero constant term. By Remark \ref{GenericInvariantsK}, we have $K(2r+1,\nn)^{\ad_K(f)} = f\, K(2r+1,\nn)^{D_K(f)}$. We can check that, more generally:
\begin{equation}
K(2r+1,\nn)^{\ad_K(f)}_{\geq d} = f\, K(2r+1,\nn)^{D_K(f)}_{\geq d},
\end{equation}
for all $d \geq -2$. Hence, to prove \nr 2 we only have to find some non-zero $g \in K(2r+1,\nn)^{D_K(f)}_{\geq 1}$. We insist that we are considering the $K$-grading, which starts in degree $-2$, and not the ordinary grading.

By \nr 1, we have $\dim\,  K(2r+1,\nn)^{D_K(f)} = \dim\, K(2r+1,\nn)^{\ad_K(f)} \geq \min\{ 2r+1,p \} \geq 3$. Therefore, $\dim\, K(2r+1,\nn)^{D_K(f)}_{\geq -1} \geq 2$.

Let $g \in K(2r+1,\nn)^{D_K(f)}_{\geq -1}$ be a non-zero element. If $g \in K(2r+1,\nn)_{\geq 1}$ we are done. Otherwise, $g$ has a non-zero homogeneous component of degree $-1$ or $0$. Because $D_K(f)$ is an associative derivation, the subring $\FF[g] \subseteq K(2r+1,\nn)^{D_K(f)}$. Using the description of $K_{-1} + K_0$ derived from (\ref{sousespacesK}) we can check that $\dim\, \FF[g] \cap K(2r+1,\nn)_{\geq 1} \geq 2$, and Property \nr 2 follows.

\subsubsection{} \label{UnEtDemiK}
\begin{thm}
The simple Lie algebra $K(2r+1,\nn)^{(1)}$ is not generated by 1,5 elements. \\
\end{thm}

\proof\ We apply Corollary \ref{corollary1,5} to $L = K(2r+1,\nn)$ and $H = K(2r+1,\nn)^{(1)}$. It is quite straightforward to see that the centraliser of $L^H$ is trivial. Then, let $\Omega \subseteq K(2r+1,\nn)^{(1)}$ be the non-empty open subset of derivations with non-zero term of degree $-2$: by Lemma \ref{centralisateursK}, we obtain that for all $D \in \Omega$, the subspace $K(2r+1,\nn)^D \cap K(2r+1,\nn)_{\geq 1} \neq (0)$, which is what we needed to show. The theorem is proved.

\section*{Acknowledgements}

This research was conducted during the author's stay at the Max-Planck Institute in Bonn in January and February 2008. The paper was subsequently written up while visiting the Collaborative Research Centre 701 at the University of Bielefeld in Spring 2008. He would like to express his gratitude towards both institutions for warm hospitality and support.


TITLE: Generalisation of the Grothendieck construction for presheaves as a lax pullback
QUESTION [6 upvotes]: It is well-known that for any presheaf $F \colon \mathcal{C}^{\mathrm{op}} \to \mathrm{Set}$, the category of elements (obtained by the so-called Grothendieck construction) of $F$ is a comma category $y/\ulcorner F \urcorner$ in the category $\mathrm{CAT}$ of
$$\mathcal{C} \xrightarrow{y} \widehat{\mathcal{C}} \xleftarrow{\ulcorner F \urcorner} 1.$$
Question: does this generalise to presheaves $F \colon \mathcal{C}^{\mathrm{op}} \to \mathrm{Gpd}$ of groupoids (or even categories)?
Mere comma categories yield discrete fibrations, hence won't give the expected answer. So the question is whether some other construction with a similar flavour could do.
Probably, if the construction does generalise, then it will also work for pseudo-functors to $\mathrm{Gpd}$ or $\mathrm{Cat}$, but I'm really interested in strict functors.
Note: I'm half-aware of another universal property of the Grothendieck construction as an oplax colimit. Is it related?

REPLY [1 votes]: Another way of phrasing the isomorphism of categories $\int^{\mathcal{C}}\mathcal{F}\congよ\downarrow[\mathcal{F}]$ is by saying that $\int^{\mathcal{C}}\mathcal{F}$ is the full subcategory of $\mathsf{PSh}(\mathcal{C})_{/\mathcal{F}}$ spanned by the representable presheaves. A quick way to prove this is as follows:

First, note that $よ\downarrow[\mathcal{F}]$ is the category of elements of the functor $\mathsf{Nat}(h_{(-)},\mathcal{F})\colon\mathcal{C}^\mathsf{op}\longrightarrow\mathsf{Sets}$.
Then, the Yoneda lemma gives a natural isomorphism $\mathsf{Nat}(h_{(-)},\mathcal{F})\cong\mathcal{F}$. This is an isomorphism in $\mathsf{PSh}(\mathcal{C})$.
As functors preserve isomorphisms, it follows that $\int^{\mathcal{C}}\colon\mathsf{PSh}(\mathcal{C}^{\mathsf{op}})\longrightarrow\mathsf{Cats}$ sends the isomorphism of presheaves $\mathcal{F}\cong\mathsf{Nat}(h_{(-)},\mathcal{F})$ to an isomorphism of categories $\int^{\mathcal{C}}\mathcal{F}\cong\underbrace{\int^{\mathcal{C}}\mathsf{Nat}(h_{(-)},\mathcal{F})}_{\congよ\downarrow[\mathcal{F}]}$.

To give such a description for the Grothendieck construction, we first pass to bicategories. There, given a pseudopresheaf $\mathcal{F}\colon\mathcal{C}^\mathsf{op}\longrightarrow\mathsf{Cats}_{\mathsf{2}}$ from a bicategory $\mathcal{C}$ to the $2$-category $\mathsf{Cats}_{\mathsf{2}}$ of small categories, functors, and natural transformations, we can repeat the above strategy:

This time, we consider the locally full sub-bicategory of $\mathsf{PseudoPSh}(\mathcal{C})_{/\mathcal{F}}$ spanned by the representable pseudopresheaves. This is the bicategory $よ(\mathcal{C})_{/\mathcal{F}}$ where

and where vertical and horizontal composition of $2$-morphisms is defined as in $\mathsf{PseudoPSh}(\mathcal{C})$ (we also have to specify the associators and unitors, but let's not since this description is already quite long).

Now, note that $よ(\mathcal{C})_{/\mathcal{F}}$ is the bicategory of elements (as defined in arXiv:1212.6283) of the pseudopresheaf $\mathsf{PseudoNat}(\mathsf{h}_{(-)},\mathcal{F})\colon\mathcal{C}^\mathsf{op}\longrightarrow\mathsf{Cats}_{\mathsf{2}}$.
Just as functors preserve isomorphisms, functors of tricategories preserve biequivalences. We can then apply Proposition 3.3.6 of arXiv:1212.6283 to the equivalence $\mathsf{PseudoNat}(\mathsf{h}_(-),\mathcal{F})\cong\mathcal{F}$ provided by the bicategorical Yoneda lemma to obtain a biequivalence $\int^{\mathcal{C}}\mathcal{F}\cong\underbrace{\int^{\mathcal{C}}\mathsf{PseudoNat}(\mathsf{h}_{(-)},\mathcal{F})}_{\cong よ(\mathcal{C})_{/\mathcal{F}}}$.

Finally we connect the above back to the Grothendieck construction. To this end, observe that―just as the category of elements of a presheaf $\mathcal{F}\colon\mathcal{C}^\mathsf{op}\longrightarrow\mathsf{Sets}$ agrees with the Grothendieck construction of $\mathcal{F}_{\mathsf{disc}}\colon\mathcal{C}^\mathsf{op}\longrightarrow\mathsf{Cats}$―the Grothendieck construction of a pseudofunctor $\mathcal{F}\colon\mathcal{C}^\mathsf{op}\longrightarrow\mathsf{Cats}_{\mathsf{2}}$ agrees with the bicategory of elements of $\mathcal{F}\colon\mathcal{C}^{\mathsf{op}}_\mathsf{bi}\longrightarrow\mathsf{Cats}_{\mathsf{2}}$, where $\mathcal{C}^{\mathsf{op}}_\mathsf{bi}$ is the discrete bicategory associated to $\mathcal{C}$ (this is $\mathcal{C}$ but with discrete $\mathsf{Hom}$-categories, i.e. for each $A,B\in\mathrm{Obj}(\mathcal{C})$, we have $\mathsf{Hom}_{\mathcal{C}^{\mathsf{op}}_\mathsf{bi}}(A,B)\overset{\mathrm{def}}{=}\mathrm{Hom}_{\mathcal{C}}(A,B)_{\mathsf{disc}}$).

All this is to say the following: the Grothendieck construction of a pseudofunctor $\mathcal{F}\colon\mathcal{C}^\mathsf{op}\longrightarrow\mathsf{Cats}_{\mathsf{2}}$ is equivalent to the category $よ_\mathsf{disc}(\mathcal{C})_{/\mathcal{F}}$ where

(Sorry for the long answer!)


TITLE: Defining parity on the set of all bijections of ${\mathbb{N}}$
QUESTION [4 upvotes]: Let  $\phi:{\mathbb{N}}\rightarrow {\mathbb{N}}$ be a bijection. Can we extend the notion of parity (of a finite permutation) to $\phi$ ?
In other words, Can we define a group homomorphism $\Lambda $, between the  group of all bijections of $\mathbb{N}$ to the group {1,-1,$\cdot$} such that $\Lambda(\psi)=-1$ if $\psi$ is a "transposition":
There exists $n_{1},n_{2}\in \mathbb{N}$, where $n_{1}\neq n_2$ and $\psi(n_{1})=n_{2} ,\psi(n_{2})=n_{1}$ and ${\psi(x)=x , \forall x \neq n_{1},n_{2} }$.
My guess is that building such a homomorphism requires some version of the axiom of choice / ultra-filters.

REPLY [6 votes]: It would be easier, and equivalent, to consider $\mathbb{Z}.$ But I'll stick to $\mathbb{N}.$
Consider $\sigma=(2\ 3)(4\ 5)(6\ 7)\cdots$ (so $0$ and $1$ are fixed points) and $(0\ 1)\sigma.$ They have the same cycle type but should not have the same parity. Note that fixed points and the particular ordering of $\mathbb{N}$ should be irrelevant.
For a similar example with conjugation consider $\tau=(5\ 7)(9\ 11)(13\ 15)\cdots$ and $\tau'=(1\ 3)(5\ 7)(9\ 11)(13\ 15)\cdots$ which should again not gave the same parity. To turn $\tau$ into $\tau',$ conjugate by this product of two infinite cycles $$(\cdots 13\ 9\ 5\ 1\ 0\ 4\ 8 \cdots) (\cdots 11\ 7\ 3\ 2\ 6\ 20 \cdots).$$


TITLE: Conjugation between commutative subalgebras of a matrix algebra?
QUESTION [10 upvotes]: Let $K$ be an algebraically closed field and $M_n(K)$ the $K$-algebra of all matrices $n\times n$ over $K$. If $L$ and $M$ are two isomorphic commutative subalgebras of $M_n(K)$, it is true that there exists a regular matrix $S\in M_n(K)$ such that $SLS^{-1}=M$. That is, the isomorphism can be chosen to be inner?

REPLY [11 votes]: Two isomorphic subalgebra of $M_n(K)$ do not need to be conjugated. The following example is taken from Exercise 161 of my web site http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf
Set $n=p+q$ with $q>p>0$. Then define  $\mathcal A$ as the subset of $M_n(k)$ made of the matrices with block form
$$\left(\begin{array}{cc} 0_p & 0_{p\times q} \\\\ A & 0_q \end{array}\right).$$
Likewise, ${\cal B}$ is made of the matrices
$$\left(\begin{array}{cc} 0_q & 0_{q\times p} \\\\ B & 0_p \end{array}\right).$$
Both $\cal A$ and $\cal B$ are subalgebras of $M_n(k)$, with dimension $pq$ and the property that $MN=0_n$ for every two elements (of the same algebra). They are obviously isomorphic, because the algebra structure is trivial. But ${\cal A}$ and $\cal B$ are not conjugated in $M_n(k)$. However $\cal B$ is conjugated to ${\cal A}^T$ in $M_n(k)$.


TITLE: Related Rates of Change Ship
QUESTION [2 upvotes]: A ship is 5 km east and 7 km North of a lighthouse. It is moving North at a rate of 12 $kmh^{−1}$ and East at a rate of 16 $kmh^{−1}$. At what rate is its distance from the lighthouse changing?

Looking at how the net resultant velocity vector is $v = (12^2+16^2)^{0.5} = 20$ $kmh^{-1}$, we know that the ship moves exactly 20 km away from the lighthouse each hour. This is what I thought the answer would be, but for some reason, the answer involved the initial position of the ship given ("5 km east and 7 km North"). The correct answer rounded is given as $19.1$ $kmh^{-1}$, which I worked out to be $20 - \frac{(5^2+7^2)^{0.5}}{10}$. Why is this so? Why is the initial position taken into account for the ship's rate of change of distance (aka velocity) from the lighthouse?

REPLY [1 votes]: Let $r$ be the distance from the lighthouse to the ship, so we have
$$r=(x^2+y^2)^{1/2}$$
Where $(x,\,y)$ is the position of the ship, taking $x$ as East direction and $y$ as North direction. Then, the rate of change of the distance is
$$\frac{\mathrm d r}{\mathrm d t}=\frac{x\frac{\mathrm d x}{\mathrm d t}+y\frac{\mathrm d y}{\mathrm d t}}{(x^2+y^2)^{1/2}}$$
Thus, at the given instant,
\begin{align*}
\frac{\mathrm d r}{\mathrm d t}&=\frac{(5 \text{ km})(16\text{ km/h})+(7 \text{ km})(12\text{ km/h})}{((5 \text{ km})^2+(7 \text{ km})^2)^{1/2}}\\[3pt]
&=\frac{164}{\sqrt{74}}\text{ km/h}\\[4pt]
&\approx \boxed{\color{blue}{19.06\text{ km/h}}}
\end{align*}


TITLE: $S$ is of measure zero iff it is of volume zero
QUESTION [2 upvotes]: Can anyone give me a hint how to prove this:

If $S\subseteq \mathbb R^n$ and $S$ is compact then as every open cover has a finite sub-cover , $S$ is of measure zero if and only if $S$ is of volume zero .

REPLY [3 votes]: If $S$ has content (volume) $0$, then for every $\epsilon >0$, there is a finite collection of open rectangles $\{R_1,R_2,\ldots, R_n\}$ covering $S$ such that
$$\sum_{i=1}^{n} vol(R_i) < \epsilon.$$
Thus content $0$ implies measure $0$.
On the other hand, if $S$ has measure $0$, then there is a countable collection of open rectangles covering $S$ with the above property.  Now use the fact that $S$ is compact to conclude $S$ has content $0$.


TITLE: Rational homotopy groups of a projective hypersurface
QUESTION [15 upvotes]: Let $X$ be a smooth projective hypersurface in $\mathbb{P}^n$. Has anyone computed the rational homotopy groups $\pi_i(X)\otimes \mathbb{Q}$ of $X$? I tried Google, but did not find anything.
One could use the formality and Sullivan's minimal model to compute these groups up to some level by hand. And in principle this can be done for all the groups. But is there any nice way of organising these things? For example, give the ranks using some generating functions, or express the generators in a nice way?

REPLY [10 votes]: The paper "On real homotopy properties of complete intersections" by Babenko gives an answer to your question. I will adapt Corollary 1 presented therein to the case of hypersurfaces:
Let $X$ be a degree $d$ hypersurface in $\mathbb{C}\mathbb{P}^{n+1}$, and denote its topological Euler characteristic by $\chi$ (note that $\chi$ depends only on $d$ and $n$). If $\chi = n+1$, then the only rationally non-zero homotopy groups of $X$ are $\pi_2 (X) \otimes \mathbb{Q} = \pi_{2n+1} (X) \otimes \mathbb{Q} \cong \mathbb{Q}$. If $\chi \neq n+1$, then for any $j\geq 1$ we have $$\dim\pi_{j+1}(X)\otimes \mathbb{Q} = \frac{(-1)^j}{j} \sum_{d|j, \ \ d\geq 1} (-1)^d \mu\bigl(\frac{j}{d}\bigr)\sum_{\alpha=1}^{2n-2} \xi_\alpha^{-d},$$
where $\xi_\alpha$ are the roots of $1-(-1)^n(\chi-n-1)(1+z)z^{n-1} + z^{2n-1}$ that are not the root $-1$, and $\mu$ is the Möbius function.
For example, in the case of a $K3$ surface, $n=3$ and $\chi = 24$, and we get $\pi_3(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{252}$, $\pi_4(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{3520}$, $\pi_5(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{57960}$, $\pi_6(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{1020096}$, $\pi_7(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{18664800}$, $\pi_8(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{351204480}$, $\ldots$. This is consistent with Corollary 4.10 in Samik Basu and Somnath Basu's Homotopy groups and periodic geodesics of closed 4-manifolds and the answers to this question What are the higher homotopy groups of a K3 suface?
For convenience, here is a formula for the topological Euler characteristic of a degree $d$ hypersurface in $\mathbb{C}\mathbb{P}^{n+1}$ (simplified thanks to abx's comment); $$\chi = \frac{1}{d}((1-d)^{n+2} + d(n+2) -1).$$


TITLE: Find change of basis matrix
QUESTION [1 upvotes]: I'm asked to find the change of basis matrix from basis $\underline{e}$ to $\underline{f}$ given the following information:
The coordinate relationship is given by:
$$3y_1 = -x_1 + 4x_2 + x_3$$
$$3y_2 = 2x_1 + x_2 + x_3$$
$$3y_3 = 0 + -3x_2 + 0$$
The coordinates $x_i$ belongs to basis $\underline{e}$.
We know that $\underline{e}X_e = \underline{f}AX_e=\underline{f}X_f$ where $X$ is the coordinates in the basis given by the subscript.
$\underline{f}X_f=\underline{f}AX_e$ is exactly what the coordinate relationships say. So to me the transformation matrix is given by:
$$A =\begin{pmatrix}
-1 & 4 & 1\\
2 & 1 & 1\\
0 & -3 & 0
\end{pmatrix}$$
But the answer is the inverse of that.
Can someone explain where my logic fails?

REPLY [2 votes]: I will explain change of basis with a simpler example, perhaps it is easier to see how this inverting comes into being. This may be a bit easier to follow if you accept the philosophy that vectors exist in the space regardless of any bases or coordinate grids. When we choose a basis in the space, the vectors get an algebraic representation given by coordinates. When we're changing basis, we're not changing the vector, but rather the coordinate grid around the vector. At the same time, we do want the transition matrix to act on the coordinates of the vector. It is this that gives rise to the inversion, since algebraically, instead of stretching the grid (which is the geometric interpretation), we're shrinking the vector.
Say we have our space $\Bbb R^3$, and we have two bases $\underline g$ with basis vectors $x_1, x_2, x_3$ and $ \underline h$ with basis vectors $y_1, y_2, y_3$. Since this is a simple example I will just assume that $\underline g$ is using feet and $\underline h$ is using meters, and that their axes agree. That means (with some rounding) that
\begin{align}
y_1 &= 3x_1 + 0x_2 + 0x_3\\
y_2 &= 0x_1 + 3x_2 + 0x_3\tag{*}\\
y_3 &= 0x_1 + 0x_2 + 3x_3
\end{align}
But what is the transition matrix from $\underline g$ to $\underline h$? Say you have a vector $X$, and its representation in the $\underline g$-basis is $X_g = (6, 9, 3)^T$. That means that from the origin we go $6$ feet in the first direction, $9$ feet in the second direction and $3$ feet in the last direction. Converting to meters, this is the same as going $2$ meters in the first direction, $3$ meters in the second direction and $1$ meter in the third direction, which is to say that the representation of $X$ in the $\underline h$-basis is $X_h = (2, 3, 1)^T$.
Writing this in matrix form, if $A$ is the transition matrix from $\underline g$ to $\underline h$, that is to say $X_h = AX_g$ for all vectors $X$, then $A$ needs to divide every entry by $3$, which means that we have
$$
A =
\begin{pmatrix}
\frac13&0&0\\
0&\frac13&0\\
0&0&\frac13
\end{pmatrix}
$$
which is the inverse of the coefficient matrix of $\text{(*)}$ above.


TITLE: How to evaluate $ \lim \limits_{n\to \infty} \sum \limits_ {k=1}^n \frac{k^n}{n^n}$?
QUESTION [28 upvotes]: I can show that the following limit exists but
I am having difficulties to find it. It is
$$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^n}{n^n}$$
Can someone please help me?

REPLY [2 votes]: First, we can rewrite the formula:
$$\frac{\sum_{i=1}^{n}i^n}{n^n} = \sum_{i=1}^{n}\frac{i^n}{n^n} = \sum_{i=0}^{n-1}\frac{(n-i)^n}{n^n} = \sum_{i=0}^{n-1}\left(1-\frac{i}{n} \right)^n$$
We can easily prove the following inequality relations:
$$\frac{1}{\text{e}^i} - \frac{i^2}{n\text{e}^i} \le \frac{1}{\text{e}^i}\left(1-\frac{i}{n} \right)^i\le \left(1-\frac{i}{n} \right)^n \le \frac{1}{\text{e}^i} $$
Therefore, we can get
$$\sum_{i=0}^{n-1}\frac{1}{\text{e}^i}-\frac{1}{n}\sum_{i=0}^{n-1}\frac{i^2}{\text{e}^i}\le\sum_{i=0}^{n-1}\left(1-\frac{i}{n}  \right)^n \le \sum_{i=0}^{n-1}\frac{1}{\text{e}^i}$$
So
$$\lim_{n\to \infty}\sum_{i=0}^{n-1}\frac{1}{\text{e}^i}-\lim_{n\to \infty}\frac{1}{n}\sum_{i=0}^{n-1}\frac{i^2}{\text{e}^i}\le\lim_{n\to \infty}\sum_{i=0}^{n-1}\left(1-\frac{i}{n} \right)^n \le \lim_{n\to \infty}\sum_{i=0}^{n-1}\frac{1}{\text{e}^i}$$
We can easily know that
$$\lim_{n\to \infty}\sum_{i=0}^{n-1}\frac{1}{\text{e}^i} = \frac{\text{e}}{\text{e}-1}$$
$$\lim_{n\to \infty}\frac{1}{n}\sum_{i=0}^{n-1}\frac{i^2}{\text{e}^i} = \left(\lim_{n\to \infty}\frac{1}{n}\right)\cdot \left(\lim_{n\to \infty}\sum_{i=0}^{n-1}\frac{i^2}{\text{e}^i}\right) = 0 \cdot \frac{\text{e}(\text{e}+1)}{(\text{e}-1)^3} = 0$$
Finally, we can get the answer:
$$\lim_{n\to \infty} \frac{\sum_{i=1}^{n}i^n}{n^n} = \lim_{n\to \infty}\sum_{i=0}^{n-1}\left(1-\frac{i}{n}  \right)^n = \frac{\text{e}}{\text{e}-1}$$
For more details, you can watch my video on Bilibili, if you have learned Chinese.
Here is my video link: https://www.bilibili.com/video/BV1yG411g7K1/


TITLE: Understanding continuity in metric space
QUESTION [0 upvotes]: Let $(X,d)$ be a metric space, $x_0 \in X$ and $r>0$ a real number.
The open ball in X of radius r centred on $x_0$ is the set
$B_r (x_0)=\{x \in X : d(x,x_0) <r \}$
The notation if considering more than one metric on X is $B_r ^d (x_0)$
I have a proposition that says f is continuous at $x_0$ if and only if given $\epsilon >0$ there exists a $\delta >0$ such that $f\big(B_{\delta} ^{d_x} (x_0)\big) \subseteq B _{\epsilon} ^{d_y} \big(f(x_0)\big)$
What does this proposition mean?
Also I specifically do not understand what is meant by $f\big(B_{\delta} ^{d_x} (x_0)\big) \subseteq B _{\epsilon} ^{d_y} \big(f(x_0)\big)$

REPLY [1 votes]: The proposition is essentially saying that $f$ is continuous at a point iff for any ball of radius $\epsilon$ centered at $f(x_0)$ in the codomain, there exists a ball of radius $\delta$ centered at $x_0$ in the domain so that that its image is contained in the $\epsilon$ ball in the target.
It's just like the delta epsilon definition of continuity condition but with balls.
As an example consider the identity map from $f:X\to X$. If I take some epsilon ball of radius $\epsilon$ around $f(x_0)$ which in the notation you used would be $B_{\epsilon}(f(x_0))$, there should be some radius $\delta$ around the point $x_0$ that I can consider in the preimage so that all the points $x$ within $\delta$ of $x_0$ i.e. $x\in B_{\delta}(x_0)$ will be mapped to a point within $\epsilon$ of $f(x_0)$ i.e. $d(f(x), f(x_0))<\epsilon$. This is what is meant by $f(B_{\delta}(x_0))\subseteq B_{\epsilon}(f(x_0))$. But since $f$ is the identity map, all we need to do is define a radius less than $\epsilon$ in the domain and we are guaranteed that every point will be mapped to an image within $\epsilon$ of $f(x_0)$ since we are just  mapping a smaller ball into a larger ball. We can see that according the your definition this is continuous at every point.
Hopefully this makes sense :)


TITLE: Graph Theory Paths
QUESTION [0 upvotes]: Prove that every graph $G$ with $|G| < \|G\|$ contains a P$_{4}$. (P$_{4}$ is path with length $3$, $|G|$ is the number of vertices and $\|G\|$ is the number of edges).
Would induction be the only way prove this? Started of with base case $|G|$ greater than or equal to $4$. Then tried induction after that, but got a little lost.

REPLY [1 votes]: It's probably doable by induction, but here's an alternative for the fun of it.
Suppose $G$ has no $P_4$.  We show that $G$ has $|E(G)| \leq |V(G)|$, which proves that statement by contraposition.
For some vertex $v$, denote by $N(v)$ its set of neighbors.
If $v$ has at least 3 neighbors, then all $x \in N(v)$ are of degree one.
Otherwise, take $x,y,z \in N(v)$, and suppose $x$ has a neighbor $x'$.  Then, at least one of $zvxx'$ or $yvxx'$ forms a $P_4$.
So, $v$ and $N(v)$ form a connected component of $G$ with less edges than vertices.
If $v$ has exactly two neighbors $x, y$, then the only possible other neighbor of $x$ is $y$, and vice-versa.
So, $v,x,y$ form a connected component with 3 vertices and at most 3 edges.
If $v$ has one neighbor $v'$, it is either in one of the connected components described above (if $v'$ has degree at least 2), or in a connected component that contains only $v$ and $v'$.
In the end, every connected component has at least as many vertices than edges.  So we must deduce that $|E(G)| \leq |V(G)|$.


text:
\section{A factorization for the weighted GJMS operators}
\label{app:product}

In this appendix we prove the following factorization theorem for the GJMS operators of Riemannian products $(X^{n+1}\times F^m,g_+\oplus g_F)$ of Einstein manifolds $(X^{n+1},g_+)$ and $(F^m,g_F)$ with $\Ric(g_+)=-ng_+$ and $\Ric(g_F)=(m-1)g_F$ when acting on functions which are independent of $F$.  After treating $m$ as a formal variable, this yields Theorem~\ref{thm:weighted_gjms_factorization}.  Our proof is by an adaptation of the proof given by Fefferman and Graham of~\cite[Proposition~7.9]{FeffermanGraham2012} in the case $m=0$, which gives the factorization of the GJMS operators on an Einstein manifold.

\begin{thm}
\label{thm:product}
Let $(X^{n+1},g_+)$ and $(F^m,g_F)$ be two Einstein manifolds satisfying $\Ric(g_+)=-ng_+$ and $\Ric(g_F)=(m-1)g_F$ and let $k\in\bN$.  Denote by $\pi\colon X\times F\to X$ the canonical projection.  Then for any $f\in C^\infty(M)$, it holds that
\begin{equation}
\label{eqn:product}
L_{2k}\left(\pi^\ast f\right) = \left[\prod_{j=1}^k \left(-\Delta - \frac{(n-m+2k-4j+3)(m+n-2k+4j-3)}{4}\right)\right]\left(\pi^\ast f\right)
\end{equation}
where $L_{2k}$ is the $k$-th order GJMS operator of the metric $g_+\oplus h$.
\end{thm}

\begin{proof}

It was shown by Gover and Leitner~\cite{GoverLeitner2009} that the ambient metric of $(X^{n+1}\times F^m,g_+\oplus g_F)$ can be written in the form
\[ \tilde g = 2\rho\,dt^2 + 2t\,dt\,d\rho + t^2g_\rho \]
for
\begin{equation}
\label{eqn:grho}
g_\rho = \left(1-\frac{\rho}{2}\right)^2g_+ + \left(1+\frac{\rho}{2}\right)^2g_F .
\end{equation}
The GJMS operator $P_{2k}$~\cite{GJMS1992} is defined to be the obstruction to formally extending a given function $u_0\in C^\infty(X\times F)$ to a homogeneous function $U$ of degree $w=-\frac{n+m+1}{2}+k$ which is harmonic with respect to the ambient metric.  The extension $U=t^wu$ of $u_0$, where $u=u(x,\rho)$ and $x\in X\times F$, will be harmonic if and only if
\[ 0 = -2\rho u^{\prime\prime} + \left(2(k-1) - \rho\tr_{g_\rho}g_\rho^\prime\right)u^\prime + \Delta_\rho f - \frac{m+n+1-2k}{4}\tr_{g_\rho}g_\rho^\prime u , \]
where ``prime'' denotes differentiation with respect to $\rho$.  In the case $u_0=\pi^\ast f_0$, it follows that $u$ depends only on $X$ and $\rho$; i.e.\ we may write $u=\pi^\ast f$.  Moreover, using the specific form~\eqref{eqn:grho} of the metrics $g_\rho$ in the above display yields
\begin{align*}
0 & = -2\left(1-\frac{\rho}{2}\right)^2\left(1+\frac{\rho}{2}\right)\rho f^{\prime\prime} \\
& \quad + \left(2(k-1)\left(1-\frac{\rho}{2}\right)^2\left(1+\frac{\rho}{2}\right) + (n+1)\rho\left(1-\frac{\rho}{2}\right)\left(1+\frac{\rho}{2}\right) - m\rho\left(1-\frac{\rho}{2}\right)^2\right)f^\prime \\
& \quad + \left(\left(1+\frac{\rho}{2}\right)\Delta + \frac{(n+1)(m+n+1-2k)}{4}\left(1-\frac{\rho}{2}\right)\left(1+\frac{\rho}{2}\right) - \frac{m(m+n+1-2k)}{4}\left(1-\frac{\rho}{2}\right)^2\right)f .
\end{align*}
Differentiating $\ell$ times in $\rho$ and evaluating at $\rho=0$ yields
\begin{align*}
0 & = 2(k-\ell-1)f^{(\ell+1)} + \left(\Delta + \ell(\ell-k+n+1-m) + \frac{(n+1-m)(m+n+1-2k)}{4}\right)f^{(\ell)} \\
& \quad + \frac{\ell}{2}\left(\Delta + (\ell-1)(\ell-k-1+2m) + \frac{m(m+n+1-2k)}{2}\right)f^{(\ell-1)} \\
& \quad + \frac{\ell(\ell-1)}{4}\left((\ell-2)(k-\ell+1-m-n) - \frac{(m+n+1)(m+n+1-2k)}{4}\right)f^{\ell-2} .
\end{align*}
Denote by $(a)_k=a(a+t)\dotsm(a+k-1)=\frac{\Gamma(a+k)}{\Gamma(a)}$.  Multiplying the above equation by $\frac{1}{2}(k-\ell)_\ell$ gives the equivalent formula
\begin{align*}
0 & = (k-\ell-1)_{\ell+1}f^{(\ell+1)} \\
& \quad + \left(\frac{1}{2}\Delta - \frac{\ell(k-\ell)}{2} + \frac{(n+1-m)(m+n+1-2k+4\ell)}{8}\right)(k-\ell)_\ell f^{(\ell)} \\
& \quad + \frac{\ell(k-\ell)}{2}\left(\frac{1}{2}\Delta - \frac{(\ell-1)(k-\ell+1)}{2} + \frac{m(m+n+1-2k+4\ell-4)}{4}\right)(k-\ell+1)_{\ell-1}f^{(\ell-1)} \\
& \quad + \frac{(\ell-1)_2(k-\ell)_2}{8}\left((\ell-2)(k-\ell+2) - \frac{(m+n+1)(m+n+1-2k+4\ell-8)}{4}\right)(k-\ell+2)_{\ell-2}f^{(\ell-2)} .
\end{align*}
In other words, we see that
\begin{equation}
\label{eqn:formula}
(k-\ell)_\ell f^{(\ell)} = q_\ell f
\end{equation}
for $q_\ell$ defined by $q_{-2}=q_{-1}=0$, $q_0=1$, and
\begin{equation}
\label{eqn:q_recursive}
\begin{split}
q_{\ell+1} & = \left(-\frac{1}{2}\Delta + \frac{\ell(k-\ell)}{2} - \frac{(n+1-m)(m+n+1-2k+4\ell)}{8}\right)q_{\ell} \\
& \quad + \frac{\ell(k-\ell)}{2}\left(-\frac{1}{2}\Delta + \frac{(\ell-1)(k-\ell+1)}{2} - \frac{m(m+n+1-2k+4\ell-4)}{4}\right)q_{\ell-1} \\
& \quad + \frac{(\ell-1)_2(k-\ell)_2}{4}\left(-\frac{(\ell-2)(k-\ell+2)}{2} + \frac{(m+n+1)(m+n+1-2k+4\ell-8)}{8}\right)q_{\ell-2} .
\end{split}
\end{equation}
Via a straightforward but tedious induction argument, we find that
\begin{equation}
\label{eqn:q_formula}
\begin{split}
q_\ell & = \sum_{j=0}^\ell\bigg[\binom{\ell}{j}2^{j-\ell}(k-\ell)_{\ell-j}\left(\frac{m+n+1-2k+4j}{2}\right)_{\ell-j} \\
& \qquad \times \prod_{i=1}^{j}\left(-\frac{1}{2}\Delta - \frac{(n-m+2k-4i+3)(m+n-2k+4i-3)}{8}\right)\bigg] .
\end{split}
\end{equation}
On the other hand, \eqref{eqn:formula} implies that the obstruction to finding the harmonic extension $U$ of $u$ is
\[ q_k = \prod_{i=1}^{k}\left(-\frac{1}{2}\Delta - \frac{(n-m+2k-4i+3)(m+n-2k+4i-3)}{8}\right) . \]
Since $2^kq_k$ has leading order term $(-\Delta)^k$, this finishes the proof.
\end{proof}

config: arxiv


TITLE: If $I$ is finitely generated, $R/I$ and $R/J$ are Artinian rings. Then, $R/IJ$ is Artinian
QUESTION [0 upvotes]: Let $R$ be a ring and $I,J$ ideals of $R$. Suppose that $I$ is finitely generated, $R/I$ and $R/J$ are Artinian rings. I'm trying to proof that $R/IJ$ is also an Artinian ring.
My idea: Consider the exact sequence
$$ 0 \to I/IJ \overset{i}{\rightarrow} R/IJ \overset{p}{\rightarrow} R/I \to 0 , $$
where $i$ is the inclusion and $p: R/IJ \to \frac{R/IJ}{I/IJ} \simeq R/I$ the canonical projection. So we have to see that $I/IJ$ and $R/I$ are $R/IJ-$Artinian modules.What implies in $R/IJ$ be Artinian.
I think that $I/IJ$ is Artinian as $R/IJ-$module since $I$ is finitely generated. But I'm not sure about it.
The part that $R/I$ is Artinian as $R/IJ-$module I couldn't see either.
Help?

REPLY [1 votes]: First of all, as the map $R \to R/I$ is surjective, $R/I$ submodules of any $R/I$ module $M$ are the same as $R$ submodules, so $M$ being an Artinian $R/I$ module is equivlant to being an Artinian $R$ module. Similarly for $R/J$ and $R/IJ$
In the exact sequence
$0 \to I/IJ \overset{i}{\rightarrow} R/IJ \overset{p}{\rightarrow} R/I \to 0$
$R/I$ is an Artinian $R$ module, as $R/I$ is an Artinian module over itself.
$I/IJ$ is a finitely generated module over $R/J$ which is an Artinian ring, so it is an Artinian $R/J$ module, thus also an Artinian $R$ module.
So by exactness, we get that $R/IJ$ is an Artinian $R$ module, hence an Artinian $R/IJ$ module, i.e. an Artinian ring.


TITLE: How many of these points can exist on a graph of $f:[0,1]\to [0,1]$?
QUESTION [0 upvotes]: Say that $p$ is a bumpy point if there is a collection of "bumps" limiting to it:

A collection of these bumps and their limit point can lie on any horizontal line (except obviously for the top line $[0,1]\times \{1\}$).
Question: If $f:[0,1]\to [0,1]$ is continuous, then must number of bumpy points on the graph of $f$ be countable?
Easier Question: Does there exist a horizontal line with no bumpy points (other than the top line)?
Note: I do not require that the "bumps" be semi-circles.  They just have to be arcs beginning and ending at the same horizontal line as their limit point, and lie completely above  that horizontal line.

REPLY [1 votes]: Hint: Take the well known continuous function, but no where differentiable