text:
\begin{document}

	\title[A bound on the degree of singular vectors for $E(5,10)$]{A bound on the degree of singular vectors for the exceptional Lie superalgebra $E(5,10)$}

	\author{Daniele Brilli}
	\address{Dipartimento di Matematica "Guido Castelnuovo", Universita' di Roma "La Sapienza", Piazzale Aldo Moro 5, Roma, Italia}

	\begin{abstract}
		We use the language of Lie pseudoalgebras to gain information about the representation theory of the simple infinite-dimensional linearly compact Lie superalgebra of exceptional type $E(5,10)$. This technology allows us to prove that the degree of singular vectors in minimal Verma modules is $\leq 14$. A few technical adjustments allow us to refine the bound, proving that the degree must always be $\leq 12$ and it is actually, except for a finite number of cases, $\leq 10$.
	\end{abstract}

	\date{June 22, 2020}

	\maketitle
	\section{Introduction}
	Infinite dimensional linearly compact simple Lie superalgebras over $\C$ were classified by Kac in \cite{K}. Besides Lie algebras in Cartan's list \cite{C}, the complete list consists of ten families and five exceptions, denoted by $E(1,6)$, $E(2,2)$, $E(3,6)$, $E(3,8)$, $E(4,4)$ and $E(5,10)$.\\
	In \cite{KR1} Kac and Rudakov started the study of representations of these exceptional superalgebras following the approach developed by Rudakov for the "non super" case, establishing the language of \emph{generalized Verma modules} and reducing the problem to the description of the so-called \emph{degenerate} modules and the study of \emph{singular vectors}.\\
	They completed the classification of degenerate Verma modules and singular vectors for $E(3,6)$ in \cite{KR1,KR2,KR3}. In the meanwhile they started to investigate the cases of $E(3,8)$ and $E(5,10)$ in \cite{KR4}. For $E(3,8)$ they could apply most of the arguments from \cite{KR2} and found all the degenerate Verma modules; they also described many degenerate $E(5,10)$-modules and conjectured there were no others.\\
	Afterwards Rudakov in \cite{R} related the problem with the study of morphism between Verma modules. He defined a degree of such a morphism and classified all the morphisms of degree $1$ (which correspond to the degenerate modules found in \cite{KR4}) but he also obtained morphisms of degree $2$ and $3$ as products of morphisms of degree $1$ and found morphisms of degree $4$ and $5$. He then conjectured there were no morphisms of higher degree and that his list was exhaustive.\\
	Cantarini and Caselli in \cite{CC} developed some combinatorial aspects of morphisms between Verma modules for $E(5,10)$ that allowed them, in particular, to confirm part of Rudakov's conjecture. Indeed, they showed that the morphisms found by Rudakov up to degree $3$ were the only ones.\\

	We briefly recall the definitions of generalized Verma module and singular vector.\\

	$E(5,10)=\oplus_{i\geq -2} \mfL_i$ is equipped with a $\mathbb{Z}$-grading of depth 2 consistent with the Lie superalgebra structure. In particular, $\mfL_0$ is a Lie algebra isomorphic to $\mathfrak{sl}_5$. If we take an ${sl}_5$-module $V$ and allow $\mfL_+=\oplus_{i>0} \mfL_i$ to act trivially on it, we obtain a module over $\mfL_{\geq 0}=\oplus_{i\geq0} \mfL_i$. We can then consider the induced $\mfL$-module
	\[
	\mfT(V)=U(\mfL)\otimes_{U(\mfL_{\geq 0})} V
	\]
	where $U(\mfL)$ is the universal enveloping algebra of $\mfL$.\\

	$\mfT(V)$ is called \emph{generalized Verma module}. One says it is \emph{minimal} if $V$ is an irreducible $\mathfrak{sl}_5$-module. If $\mfT(V)$ is minimal but not irreducible, it is said to be \emph{degenerate}.\\
	It is possible to define a grading on $\mfT(V)=\oplus_{p\geq 0} \mfT^p(V)$ compatible with that of $\mfL$, meaning that $\mfL_i \cdot \mfT^p(V)\subset \mfT^{p-1} (V)$.\\
	A \emph{singular vector} is an element of $\mfT(V)$ that is killed by $\mfL_+$.\\
	The notions of degeneracy of a minimal Verma module, existence of (non constant) singular vectors and (positive degree) morphisms are all equivalent (see, for instance, \cite[Prop. 3.5]{CC}).\\
	Basically, if one has a positive degree singular vector in a minimal Verma module $\mfT(V)$, the $\mfL$-submodule it generates is a proper submodule, thus $\mfT(V)$ is degenerate. On the other way, a $\mfL$-morphism of positive degree $p$ maps constant vectors, which are automatically singular, to singular vectors of positive degree $p$.\\

	Despite a visible fair amount of understanding of these objects, an explicit bound on the degree of singular vectors (or equivalently morphisms) in the literature is only implicitly conjectured. This article aims to fill this gap and provide such a bound, with the hope that the techniques developed will allow to improve the result and help "attack" these conjectures from "above" as well as from "below".\\

	The idea is the following: the even part of $E(5,10)$ is $S(5)$, the Lie algebra of zero-divergence vector fields in five indeterminates. This algebra is isomorphic to the \emph{annihilation algebra} of a \emph{Lie pseudoalgebra}. Lie pseudoalgebras are Lie algebras in a \emph{pseudotensor category}, hence their name (see \cite{BD,BDK1}). The analogs of Verma modules are called \emph{tensor modules} in this language. A structural correspondence between a Lie pseudoalgebra and its annihilation algebra guarantees a close bond between their representation theories (see \cite[Section 13]{BDK1}).
	(Finite) pseudoalgebras have the upside of having a developed theory (\cite{BDK1,BDK2,BDK3,BDK4,D}) and allow one to talk about singular vectors and their degrees in quite a "manageable" way. For istance, we know that the degree of singular vectors in tensor modules for Lie pseudoalgebras of type S is at most $2$, (Theorem \ref{theorem degrees tensor modules}, see \cite[Section 7]{BDK2} for more details).\\
	The fact that the pseudoalgebraic structure associated with the even part of $E(5,10)$ has such an immediate bound, together with the finiteness of the odd degree, prevent vectors with high enough degree from being singular.  \\

	The paper is organized as follows: in Section \ref{sectionPreliminariesE(5,10)} we recall the basic definitions about $E(5,10)$, generalized Verma modules and singular vectors and set up the notation. In Section \ref{sectionPreliminariesLiepseudoalgebras} we recall the basic notions about Lie pseudoalgebras, annihilation algebras and the correspondence between their representation theory. Section \ref{sectionPrimitiveLiepseudoalgebras} is dedicated to a brief summary of results about representation theory of primitive Lie pseudoalgebras of type $W$ and $S$. In Section \ref{sectionBoundSingularvectors} we build a finite filtration of $S(5)$-submodules on a generalized Verma module and realize their quotients as tensor modules, which allows us to apply the results stated in the previous section and prove our first result. Finally, in Section \ref{sectionBoundRefining} we use a very easy technical lemma to refine the bound.

	\section{E(5,10), Verma modules and notation}\label{sectionPreliminariesE(5,10)}
	We deal with the simple linearly compact Lie superalgebra of exceptionaly type $\mfL =E(5,10)$ and we will use the geometric construction provided in \cite[5.3]{CK}\\
	Let $\mfd = (\C^5)^\ast$ and let $\{ \der_1 ,\dotsc, \der_5\}$ and $\{x_1, \dotsc, x_5\}$ be bases for respectively $\mfd$ and $\mfd^\ast$.\\
	We can realize the even part of $\mfL$ as zero-divergence vector fields in the indeterminates $x_1, \dotsc, x_5$,
	\begin{equation*}
	\mfL_{(0)} = S(5) = \Big\{ D = \nicesum{i=1}^{5} f_i \der_i \, | \, f_i\in C[[x_1, \dotsc, x_5]], \, div(D)=0\Big\},\\
	\end{equation*}
	and the odd part as closed 2-forms in the same indeterminates
	\begin{align*}
	\mfL_{(1)} = d\Omega^1(5) = \Big\{& \omega = \nicesum{i<j=1}^5 f_{ij}\xi_{ij}\, | \, f_{ij}\in \C[[x_1,\dotsc, x_5]], \, \xi_{ij}= \dd x_i\wedge \dd x_j ,\\
	 &\omega = \dd\alpha \text{ for some } \alpha\in \Omega^1(5)\Big\}.
	\end{align*}
	The bracket between even elements is the one of the Lie algebra $S(5)$, while the brackets between even and odds elements are given by the Lie derivative
	\begin{equation*}
	[D,\omega] = \mathcal{L}_D(\omega) = \dd(\iota_D(\omega)),
	\end{equation*}
	where $\iota_D(\omega)$ is the contraction of $\omega$ by $D$.\\
	Finally, the bracket between odd elements is given by wedge product and consequent identification with a zero-divergence vector field:
	\begin{equation*}
	[\omega_1,\,\omega_2] = D \qq{where D is s.t.} \iota_D(v) = \omega_1\wedge\omega_2,
	\end{equation*}
	where $v$ is a fixed volume form. $\dd \omega_1 \wedge \dd\omega_2$ is a closed form and it is easy to check that under this identification it corresponds to a zero-divergence vector field.\\

	In order to give an explicit formula, for $i,j,h,k\in\{1,\dotsc, 5\}$ we set
	\[
	(ijhk)= \begin{cases}
	0 \quad\text{ if } |\{i,j,h,k\}|<4 \\
	l \quad\text{ otherwise, where } \{i,j,h,k, l\} = \{1,\dotsc, 5\};
	\end{cases}
	\]
	we shall adopt the following convention: whenever such an index occurs in an expression and it takes on the value 0, so does the expression.\\

	This way the bracket between two odd elements can be defined as
	\begin{equation}\label{bracketodd}
	[\xi_{ij},\xi_{hk}] = \varepsilon_{(ijhk)}\der_{(ijhk)}
	\end{equation}
	and then extended by $\C[[x_1,\dotsc,x_5]]$-bilinearity, where  $\varepsilon_{(ijhk)}$ is the sign of the permutation $(ijhkl)$ if $(ijhk)=l\neq 0$, $0$ otherwise; in a similar way $\der_{(ijhk)}=\der_l$ if $(ijhk)=l\neq0$, $\der_{(ijhk)}=0$ otherwise.\\

	Setting $deg\, x_i= -deg\, \der_i = 2$ and $deg\, \xi_{hk}=-1$ provides $\mfL=\bigoplus_{i\geq -2}\mfL_{i}$ with a transitive, irreducible $\mathbb{Z}$-grading of depth $2$ consistent with the superalgebra structure, which means that
	\begin{equation}
	[\mfL_n,\mfL_m]\subseteq \mfL_{n+m}
	\end{equation}
	We'll call $\mfL_-=\mfL_{-2}\oplus\mfL_{-1}$, $\mfL_{+}=\bigoplus_{i>0} \mfL_i$ and $\mfL_{\geq 0} = \mfL_{0}\oplus \mfL_{+}$.\\
	We have an isomorphism between $\mfL_0$ and $\sld$ given by
	\begin{equation}
	 x_i\der_j\longmapsto - e^i_j.
	\end{equation}
	Given the fact that $[\mfL_0,\mfL_n]\subseteq \mfL_n$, we can view $\mfL_n$ as an $\sld$-module; in particular, $\mfL_{-2}\cong \mfd$ and $\mfL_{-1} \cong \bigwedge^2 \mfd^\ast =: \mfs$.\\
	It is useful to describe also $\mfL_1$ as a $\sld$-module: it is the highest weight representation in $\mfd^\ast \otimes \bigwedge^2 \mfd^\ast$, (see \cite[Section 4.3]{CK}) and it is generated by the highest weight vector $x_1 \xi_{12}$.\\
	Notice that $\mfL_j = \mfL_1^j$ for $j\geq 1$ and that $\mfL_-\cong \mfd \oplus \mfs$ is a finite dimensional Lie superalgebra whose superbracket is non trivial only when restricted to the odd part, where is given by \eqref{bracketodd}.\\

	This grading extends to the universal enveloping algebra $U(\mfL)$, and in particular to $U(\mfL_-)$. In the latter case, as common practice, the sign of the degree is inverted in order to have a grading over $\mathbb{N}$.\\

	We will use this grading to study generalized Verma modules, which we will introduce here following \cite{R}(see also \cite{KR1,KR2,CC}).\\

	Given a $\sld\cong\mfL_0$-module $V$, we can extend it to a $\mfL_{\geq 0}$-module by letting $\mfL_{+}$ act trivially on it; we can then consider the induced $\mfL$-module
	\[
	\mfT(V)= U(\mfL)\otimes_{U(\mfL_{\geq 0})} V
	\]
	where the action is given by left multiplication.
	\begin{defin}
		Let $V$ be a $\sld$-module. The $\mfL$-module $\mfT(V)$ is called \emph{generalized Verma module}.\\
		If $V$ is a finite-dimensional and irreducible $\sld$-module, we call $\mfT(V)$ \emph{minimal}.\\
		A minimal Verma module is called \emph{non-degenerate} if it is irreducible, \emph{degenerate} otherwise.
	\end{defin}

	\begin{rem}\label{remT(V)}
		Let us notice that, as vector spaces, $\mfT(V)\cong U(\mfL_-)\otimes V$. We will often use this isomorphism omitting the subscript $U(\mfL_{\geq 0})$  on the tensor product.\\
	\end{rem}
	When $V=V(\lambda)$ is an irreducible $\sld$-module of highest weight $\lambda$,  we may use the notation $\mfT(V) = \mfT(\lambda)$. A dominant weight $\lambda$ for $\sld$ will be expressed in terms of a quadruple $[a_1,a_2,a_3,a_4]\in \mathbb{N}^4$ where $\lambda = a_1 \omega_1 + \dotsb + a_4 \omega_4$ and $\omega_1 , \dotsc, \omega_4$ are the fundamental weights of $\sld$.\\
	One should pay special attention because, since we set $\mfd=(\C^5)^\ast$, all the highest weights modules are the duals of the $\mathfrak{sl}_5$ usual ones. For example, in our notation the $\mathfrak{sl}_5$ standard representation, which has highest weight $[1,0,0,0]$, will be $\mfd^\ast$.\\

	The grading of $U(\mfL_-)$ induces one on $\mfT(V)$.\\
	Poincar\'{e}-Birkhoff-Witt Theorem is still true in the superalgebra setting (see \cite[6.1]{M}), so fixed an ordered basis $\{\der_1\dotsc, \der_5, \xi_{12},\dotsc, \xi_{45}\}$ of $\mfL_-$, we can choose as a $PBW$-basis for $U(\mfL_-)$ the monomials $\der^{(I)}\xi^{K}$ where,
	\begin{align}
	\der^{(I)} = \frac{\der_1^{i_1}}{i_1!}\cdots \frac{\der_5^{i_5}}{i_5!}\label{d^I}, \quad &I=(i_1, \dotsc, i_5)\in \mathbb{N}^5\\
	\xi^K =\xi_{12}^{k_{12}}\cdots\xi_{45}^{k_{45}}, \quad &K=(k_{12},\dotsc, k_{45})\in \{ 0,1\}^{10}.
	\end{align}
	The basis elements are, by definition of the grading, homogeneous of degree $p = 2 |I| + |K|$, where $|I|=i_1+ \dotsb + i_5$ and $|K|=k_{12}+\dotsb k_{45}$; they generate the homogeneous subspaces $U^p(\mfL_-)$. We can thus equip $\mfT(V)$ with a grading $\mfT^p(V)= U^p(\mfL_-)\otimes V$.\\
	We should notice that this grading and the grading of $\mfL$ are compatible, by which we mean that
	 \begin{equation}\label{compatiblegradings}
	 \mfL_n \mfT^p(V)\subseteq \mfT^{p-n}(V).
	 \end{equation}
	We will call the elements of $\mfT^p(V)$ \emph{homogeneous vectors of degree $p$}; in particular, we will call the degree $0$ ones \emph{constant}.\\
	For istance, $\mfT^0(V) = \C \otimes V$, $\mfT^1(V) = \mfs \otimes V$, $\mfT^2(V)= \mfd \otimes V + \bigwedge^2(\mfs) \otimes V$, $\mfT^3(V) = \mfd \mfs \otimes V + \bigwedge^3(\mfs)\otimes V$, etc.\\
	If $v\in S^n(\mfd)\bigwedge^m(\mfs)\otimes V$, if we want to keep track of the even and odd degrees, we will say that $v$ has degree $(n|m)$.\\

		Degeneracy of Verma modules can be reformulated in terms of \emph{singular vectors}.
	\begin{defin}
		Let $\mfT(V)$ be a Verma module. $v\in \mfT(V)$ is called a \emph{singular vector} if $\mfL_1v=0$.\\
		The space of singular vectors will be denoted by $sing \, \mfT(V)$.
	\end{defin}
	\begin{ex}
		Any constant vector $v\in \mfT^0(V)$ is singular, since in that case $\mfL_1 v \in \mfT^{-1}(V)=0$.
	\end{ex}
	\begin{rem}
		Take $v\in sing \, \mfT(V)$ and $z\in \mfL_0$. Then, for any $y\in \mfL_1$, $yv=0$ and so
		\[
		y(zv)=[y,z]v + z(yv) = [y,z]v=0
		\]
		where the last identity follows from the fact that $[\mfL_1,\mfL_0]\subseteq \mfL_1$ and the singularity of $v$.\\
		In other terms, $sing\, \mfT(V)$ is a $\mfL_0$-submodule of $\mfT(V)$. In particular, since $\mfT(V)=\bigoplus_p \mfT^p(V)$ as $\mfL_0$-modules, homogeneous components of a singular vector are singular. We will always assume that a singular vector is homogeneous.\\
		The same holds for the weight components of a singular vector. So we will, whenever possible, assume that a singular vector is also a weight vector.
	\end{rem}

	\begin{prop}\label{sing-degenerate}
		A minimal Verma module $\mfT(V)$ is degenerate if and only if it contains non constant singular vectors.
	\end{prop}
	\begin{proof}
		Assume $0\neq v \in sing \mfT^p(V)$ for some $p>0$. Since $\mfL_-$ and $\mfL_0$ can only respectively rise the degree of $v$ or preserve it and since $\mfL_+.v=0$ by assumption, $\mfL v$ is an $\mfL$-submodule of $\mfT(V)$, which is proper because it only contains vectors of degree $\geq p$.\\
		Vice versa, let $W\subset \mfT(V)$ be a non trivial proper $\mfL$-submodule and take $0\neq w \in W$. Since the action of $\mfL_1$ lowers strictly the degree of homogeneous components of $w$, we know that eventually $(\mfL_1)^{n}w=0$ for some finite $n\geq 1$; thus we can assume without loss of generalization that $w$ is singular. Now, if $w$ was constant, by irreducibility of $V$ we would have $\mfL_0 w = \C\otimes V$ and therefore, by iterated action of $\mfL_-$, we would obtain $\mfL w = \mfT(V)$. But $\mfT(V)\supsetneq W \supseteq \mfL w = \mfT(V)$, a contradiction. Hence $w$ is a non constant singular vector.
	\end{proof}
	The proof of the proposition shows vividly how singular vectors "detect" degeneracy of minimal Verma modules.\\

	\begin{ex}\label{exsing}
		Let $V=V([0,0,0,1])\cong \mfd$ and let $ v = \nicesum{i} \xi_{1i} \otimes \der_i\in \mfT(\mfd) $.\\
		A generic element of $\mfL_1$ is of the form $y = x_h \xi_{kl}+x_k \xi_{hl}$ for some $h,k,l\in \{1,\dotsc, 5\}$.\\
		To check if $v$ is singular, we can carry out the computation:
		\begin{align*}
		y \cdot v = &\nicesum{i} x_h[\xi_{kl},\xi_{1i}]\otimes \der_i + x_k[\xi_{hl},\xi_{1i}]\otimes \der_i =\\
		&\nicesum{i} \varepsilon_{(kl1i)} \otimes (x_h\der_{(kl1i)})\der_i + \varepsilon_{(hl1i)} \otimes (x_k\der_{(hl1i)})\der_i=\\
		&\nicesum{i} -\varepsilon_{(kl1i)} \otimes e^h_{(kl1i)}\der_i - \varepsilon_{(hl1i)} \otimes e^k_{(hl1i)}\der_i=\\
		&\nicesum{i} -\varepsilon_{(kl1i)} \otimes \delta^h_i \der_{(kl1i)} - \varepsilon_{(hl1i)} \otimes \delta^k_i \der_{(hl1i)}=\\
		&-\varepsilon_{(kl1h)}\otimes \der_{(kl1h)} - \varepsilon_{(hl1k)}\otimes \der_{(hl1k)}=\\
		&-\varepsilon_{(kl1h)}\otimes \der_{(kl1h)} + \varepsilon_{(kl1h)}\otimes \der_{(kl1h)}=0 .
		\end{align*}
		So $v\in sing \, \mfT^1(V)$.	\footnote{Alternatively, one could have done the following:	notice that $v$ is an highest weight vector, therefore realize that it is sufficient to check that the lowest weight vector of $\mfL_1$, $x_5\xi_{45}$, acts trivially on $v$, which is clearly easier (or at least shorter) (see \cite[Ch.3]{CC}).}
	\end{ex}

	\section{Preliminaries on Lie Pseudoalgebras}\label{sectionPreliminariesLiepseudoalgebras}
	In this section we briefly give the definition of finite Lie pseudoalgebras and of their annihilation algebras and recall their main features following \cite[Section 2]{BDK2}. For a more detailed exposition, see also \cite{BDK1}.\\

	First of all we need a few preliminaries on Hopf algebras.\\
	Let $H$ be a cocommutative Hopf algebra with coproduct $\Delta$, counit $\epsilon$ and antipode $S$.\\
	We will use the Sweedler notation (see \cite{S}), for istance given $h\in H$:
	\begin{align*}
	\Delta(h) &=h_{(1)}\otimes h_{(2)};\\
	(\Delta\otimes id)\Delta(h) &= (id\otimes \Delta)\Delta(h)=h_{(1)}\otimes h_{(2)}\otimes h_{(3)};\\
	(id\otimes S)\Delta(h) &= h_{(1)}\otimes h_{(-2)}, \text{ etc.}
	\end{align*}
	Throughout the paper, $H=U(\mfd)$ will be the universal enveloping algebra of a Lie algebra $\mfd$ of dimension $n$.\\
	In this case the coproduct is given by $\Delta(\der)=\der\otimes 1 + 1\otimes \der$ and the antipode by $S(\der)=-\der$ for $\der\in \mfd$.\\
	Let $\der_1, \dotsc , \der_n$ be a basis of $\mfd$ and take the PBW basis of $H$, $\{\der^{(I)}\}_{I\in \mathbb{N}^{n}}$ as in \eqref{d^I}.\\
	With this choice of basis is easy to show that
	\begin{equation}\label{deltaH}
	\Delta(\der^{(I)})= \nicesum{J+K=I} \der^{(J)}\otimes \der^{(K)}.\
	\end{equation}
	Moreover it lets us define an increasing filtration on $U(\mfd)$
	\begin{equation}
	F^{p} H =span_{\C}\{ \der^{(I)} \; |\; |I|\leq p \}.
	\end{equation}
	Now let $X= H^\ast := Hom(H,\C)$ and let $\{x_I\}$ be a dual basis of $\{ \der^{(I)}\}$, i.e. $\langle x_I, \der^{(J)}\rangle=\delta_I^J$. In particular we indicate the duals of the basis elements of $\mfd\subset H$, $\{\der_i\}$, with $\{x^i\}$, which provides a basis of $\mfd^\ast\subset X$.\\
	$X$ can be viewed as an $H$-bimodule with left and right actions given respectively by
	\begin{align}
	\langle hx, f \rangle &= \langle x, S(h)f\rangle;\label{leftactionHonX}\\
	\langle xh, f \rangle &= \langle x, f S(h)\rangle, \quad \text{for } x\in X,\,h,f\in H.\label{rightactionHonX}
	\end{align}
	Properties of $H$ reflect "dually" on $X$(\cite[Section 2]{BDK1}):
	\begin{itemize}
		\item cocommutativity of $H$ implies commutativity of $X$;\\
		\item from \eqref{deltaH} follows easily that $x_I x_J = x_{I+J}$;\\
		\item using the above, one can identify $X$ with the ring of former power series $\mathcal{O}_n=\C[[x^1, \dotsc, x^n]]$;\\
		\item setting $F_p X=(F^p(H))^{\perp}$ provides a decreasing filtration on $X$.
	\end{itemize}
	Considering the $\{F_p X\}$ as a fundamental system of neighborhoods one can define a topology on $X$ that makes it \emph{linearly compact} (\cite[Chapter 6]{BDK1}) (while considers the discrete topology on $H$). This makes the action of $H$ (and in particular of $\mfd$) on $X$ continuous.\\

	\begin{defin}
		A \emph{Lie ($H$-)pseudoalgebra} $L$ is a left $H$-module endowed with a map, called \emph{pseudobracket}
		\begin{equation*}
		[\cdot \ast \cdot] : L \otimes L \rightarrow (H\otimes H)\otimes_H L
		\end{equation*}
		which has the following properties:
		\begin{description}
			\item[H-bilinearity] $[fa\ast g b] = ((f\otimes g)\otimes_H 1)[a \ast b] \quad\forall a,b\in L,\; g,f\in H;$\\
			\item[Skew-commutativity] $[b\ast a] = - (\sigma\otimes_H 1)[a \ast b]\quad \forall a,b\in L$;\\
			\item[Jacobi] $[a \ast [b\ast c]]- ((\sigma\otimes 1)\otimes_H 1)[b \ast [a\ast c]]=[[a\ast b],\ast c]\quad \forall a,b,c\in L$.
		\end{description}
	where $\sigma:H\otimes H \rightarrow H\otimes$, $f\otimes g \mapsto g\otimes f$ is the permutation of factors and the composition of pseudobrackets in the Jacobi identity are suitable defined in $H^{\otimes 3}\otimes_H L$.\\
	A Lie pseudoalgebra is called \emph{finite} if it is finitely generated as a module over $H$.
	\end{defin}
	The name derives from the fact that this is an algebra in a \emph{pseudotensor category} as introduced in\cite{BD} (see also \cite[Chapter 3]{BDK1}).\\
	\begin{ex}
		Take a Lie algebra $\mathfrak{g}$. One can define the $\emph{current}$ Lie $H$-pseudoalgebra $Cur \, \mathfrak{g} = H\otimes \mathfrak{g}$ as free $H$-module with pseudobracket defined as
		\[
		[(1\otimes a)\ast (1\otimes b)]=(1\otimes 1)\otimes_H [a,b] \quad \text{ for } a,b\in \mathfrak{g}
		\]
		and extended then by $H$-bilinearity.
	\end{ex}
	We will focus on what are called \emph{primitive} Lie pseudoalgebras, which we will define in the next section.\\
	\begin{defin}
		A \emph{representation} of $L$, or \emph{$L$-module}, is a left $H$-module $M$ with an $H$-bilinear map
		\begin{equation*}
		\ast : L \otimes M \rightarrow (H\otimes H)\otimes_H M
		\end{equation*}
		such that
		\begin{equation*}
		[a\ast b]\ast m = a\ast (b\ast m) - ((\sigma\otimes 1)\otimes_H 1)(b\ast(a\ast m)) \quad \forall a,b\in L\, , m\in M.
		\end{equation*}
		An $L$-module is called \emph{finite} if it is finitely generated as an $H$-module.\\
		A subspace $N\subset M$ is an \emph{$L$-submodule} if $L\ast N\subset (H\otimes H)\otimes_H N$.\\
		An $L$-module is \emph{irreducible} if it does not contain any nontrivial proper submodules.
	\end{defin}

	The most important tool in the study of Lie pseudoalgebras (and their representations) is the annihilation algebra.\\

	Define $\mathcal{A}(L) := X\otimes_H L$, where as before $X=H^\ast$ and the right action of $H$ on $X$ is \eqref{rightactionHonX}.\\
	\begin{defin}
	Given $L$ a Lie $H$-pseudoalgebra, its \emph{annihilation algebra} is the Lie algebra $\mfL=\mathcal{A}(L)$ with Lie bracket
	\begin{equation}\label{annihilationbracket}
	[x \otimes_H a, y \otimes_H b]= \sum(x f_i)(y g_i)\otimes_H l_i \quad \text{ where } \quad [a \ast b]= \sum(f_i\otimes g_i)\otimes_H l_i.
	\end{equation}
	\end{defin}
	$H$ acts on $X\otimes_H L$ by \eqref{leftactionHonX} on the first factor. In this way $\mfd\subset U(\mfd)=H$ acts on $\mfL$ by derivations. The semidirect sum $\mfL^e=\mfd\rtimes \mfL$ is called \emph{extended annihilation algebra}.\\

	If $L$ is finite and $L_0$ is a finite-dimensional subspace that generates $L$ as a left $H$-module, we can define a filtration on $\mfL$ induced by the one of $X$,
	\[F_p \mfL= \{ x\otimes_H a \in \mfL \, | \, x\in F_pX \text{ and } a\in L_0\},\]
	which satisfies:
	\begin{equation}\label{l-shift}
	[ F_n \mfL , F_p\mfL ] \subseteq F_{n+p-l} \mfL \; \text{  and  } \; \mfd(F_p \mfL) \subseteq F_{p-1}\mfL,
	\end{equation}
	where $l$ depends only on the choice of $L_0$. We can correct the $l$ shift by setting $\mfL_p = F_{p+l} \mfL$ so that $[\mfL_p,\mfL_n]\subseteq\mfL_{p+n}$. In particular, $\mfL_0$ is a Lie algebra.\\
	We can carry on the filtration to $\mfL^e$ setting $\mfL^e_p = \mfL_p$.\\

	An $\mfL^{e}$-module $V$ is called \emph{conformal} if any $v\in V$ belongs to some
	\[ker_p V:=\{v\in V \, |\, \mfL_p v =0\}.\]

	The next result \cite[Proposition 2.1]{BDK2} we state will be crucial for our purposes.
	\begin{prop}\label{correspondance}
		Any module $V$ over a Lie pseudoalgebra $L$ has a natural structure of a conformal $\mfL^{e}$-module, given by the action of $\mfd$ on $V$ and by
		\begin{equation}\label{(correspondance)}
		(x\otimes_H a) v = \sum\langle x, S(f_i g_{i_{(-1)}})\rangle g_{i_{(2)}}v_i \quad \text{where } a\ast v = \sum (f_i \otimes g_i)\otimes_H v_i
		\end{equation}
		for $a\in L$, $x\in X$, $v\in V$.\\
		Conversely, any conformal $\mfL^{e}$-module $V$ has a natural structure of an $L$-module given by
		\begin{equation}
		a \ast v = \nicesum{I\in \mathbb{N}^n} (S(\der^{(I)})\otimes 1)\otimes_H((x_I\otimes_H a)\cdot v)
		\end{equation}
		Moreover, $V$ is irreducible as a module over $L$ if and only if it is irreducible as a module over $\mfL^e$ (or $\mfL$).
	\end{prop}
\newpage

	\section{Primitive Lie pseudoalgebras of type W and S}\label{sectionPrimitiveLiepseudoalgebras}
	We now define the Lie pseudoalgebra $W(\mfd)$ and its subalgebra $S(\mfd)$ and apply the constructions of the previous section.
	Again, we will follow \cite[]{BDK2}.\\
	\begin{defin}
		The Lie pseudoalgebra $W(\mfd)$ is the free $H$-module $H\otimes \mfd$ with pseudobracket
		\begin{equation}
		[(f\otimes a)\ast (g\otimes b)] = (f \otimes g)\otimes_H(1 \otimes [a,b])-(f \otimes ga)\otimes_H(1\otimes b) + (fb\otimes g)\otimes_H(1\otimes a)
		\end{equation}
	\end{defin}
	Define the $H$-linear map $div: W(\mfd)\rightarrow H$ by $div(\sum h_i \otimes \der_i) = \sum h_i\der_i$. Then
	\[
	S(\mfd) := \{ s\in W(\mfd) \, | \, div(s)=0\}
	\]
	is a subalgebra  of the Lie pseudoalgebra $W(\mfd)$.\\
	In \cite[Proposition 8.1]{BDK1} it is shown that $S(\mfd)$ is generated as an $H$-module by the elements of the form:
	\begin{equation}\label{s_ab}
	s_{ab} = a\otimes b - b\otimes a - 1\otimes [a,b] \quad \text{ for } a,b\in \mfd.
	\end{equation}
	Let $\mathcal{W}=\mathcal{A}(W(\mfd))$ be the annihilation algebra of $W(\mfd)$. Since $W(\mfd)=H\otimes \mfd$,
	\[\mathcal{W}= X\otimes_H (H\otimes \mfd)\equiv X\otimes \mfd.\]
	The Lie bracket of $\mathcal{W}$ is obtained from the pseudobracket of $W(\mfd)$ by \eqref{annihilationbracket}:
	\[
	[x\otimes a, y \otimes b]=xy \otimes[a,b] - x(ya)\otimes b + (xb)y\otimes a \quad \text{for } a,b\in \mfd, \, x,y\in X.
	\]
	The action of $H$ on $\mathcal{W}$ is given by $h(x\otimes a)=hx\otimes a$ and $\mfd$ acts on $\mathcal{W}$ by derivations.\\

	Since $W(\mfd)$ is a free $H$-module, we can choose $L_0=\C\otimes \mfd$ and obtain the induced decreasing filtration on $\mathcal{W}$
	\[
	\mathcal{W}_p=F_p\mathcal{W}=F_p X \otimes_H L_0 \equiv F_p X \otimes \mfd.
	\]
	$\mathcal{W}_{-1}=\mathcal{W}$ and it satisfies \eqref{l-shift} for $l=0$; notice also that $\mathcal{W}/\mathcal{W}_0\cong \C\otimes \mfd$ and $\mathcal{W}_0/\mathcal{W}_1\cong\mfd^*\otimes \mfd$.\\

	$H$ can be endowed with a $W(\mfd)$-pseudoaction given by $(f\otimes a) \ast g= -(f \otimes ga)\otimes_H 1$ where $f,g\in H, \, a\in \mfd$; this induces an action of $\mathcal{W}=X\otimes_H W(\mfd)$ on $X\otimes_H H\equiv X$:
	\begin{equation}\label{actionW(d)onX}
	(x\otimes a)y = -x(ya) \quad \text{ for } a\in \mfd, \, x,y\in X.
	\end{equation}
	Using the fact that $X$ can be identified as $\mathcal{O}_n$ (compatibly with corresponding filtrations and topologies) and that $\mfd$ acts on $X$ by continuous derivations, we can make $\mathcal{W}$ act on $\mathcal{O}_n=\C[[t_1, \ldots t_n]]$ by continuous derivations too. This way we are defining a Lie algebra homomorphism $\varphi:\mathcal{W}\rightarrow W(n)$ where
	\[
	W(n)=Der(\mathcal{O}_n)=\Big\{\nicesum{i=1}^n f_i \frac{\der}{\der t_i} \,|\, f_i\in \C[[t_1, \dotsc, t_n]] \Big\}
	\]
	$W(n)$ has a natural filtration given by
	\[
	F_p W(n)=\Big\{\ \nicesum{i} f_i \frac{\der}{\der t_i} \, | \, f_i\in \C[[t_1, \dotsc, t_n]]_k \, , \, k\leq p \Big\}
	\]
	where $\C[[t_1, \dotsc, t_n]]_k$ is the homogeneous component of degree $k$.\\

	It is proven in \cite{BDK2} that holds the following:
		\begin{enumerate}
			\item $\varphi$ is an isomorphism of Lie algebras;\\
			\item $\varphi(x \otimes a)=\varphi(x)\varphi(a)$ $\forall x\in X, \, a\in \mfd$;\\
			\item $\varphi(1\otimes \der_i)= - \frac{\der}{\der t_i} $ mod $F_0 W(n)$;\\
			\item $\varphi(\mathcal{W}_p)= F_p W(n)$ $\forall p \geq -1$.\\
		\end{enumerate}

	In what follows we will assume that $dim \mfd = n >2$ (which is okay for us since we will apply it for $n=5$).\\
	Let $\mathcal{S }= \mathcal{A}(S(\mfd)) = X\otimes_H S(\mfd)$ be the annihilation algebra of the Lie pseudoalgebra $S(\mfd)$.\\
	The Lie bracket is the one of $\mathcal{W}$, since the canonical injection of $S(\mfd)$ into $W(\mfd)$ induces a Lie algebra homomorphism $\iota: \mathcal{S}\hookrightarrow \mathcal{W}$.\\
	Explicity, if $s=\sum h_i \otimes \der_i \in S(\mfd)\subset W(\mfd) = H\otimes \mfd$,
	\[
	\iota(x\otimes_H s)=\sum xh_i \otimes \der_i \in\mathcal{W}\equiv X\otimes \mfd
	\]
	Choosing $L_0=span_{\C}\{s_{ab} | \, a,b\in \mfd\}$ (where $s_{ab}$ are the ones defined in \eqref{s_ab}) we get a decreasing filtration of $\mathcal{S}$:
	\[
	\mathcal{S}_p=F_{p+1} \mathcal{S}=F_{p+1} X \otimes_H L_0 \quad \text{ for } p\geq-2.
	\]
	$\mathcal{S}_{-2}=\mathcal{S}$ and it satisfies \eqref{l-shift} for $l=1$.\\
	In \cite[Section 8.4]{BDK1} is proven that $\mathcal{S}\cong S(n)$. But we would also like all the filtrations and related topologies defined on these spaces to be compatible.
	In order to do so, we would like to use $\varphi$ defined before for $\mathcal{W}$ which behaves well related to the filtrations.\\
	This can be done but carefully.\\
	First define a map $div:\mathcal{W}\rightarrow X$ as $div(\sum y_i \otimes \der_i)=\sum y_i \der_i$.\\
	It is not difficult to verify that $div([A,B])= A \, div(B) - B\, div(A)$ $\forall A,B\in \mathcal{W}$ (where the action of $\mathcal{W}$ on $X$ is \eqref{actionW(d)onX}), which implies that
	\[
	\overline{\mathcal{S}}=\{ A\in \mathcal{W} \, | \, div(A)=0\}
	\]
	is a Lie subalgebra of $\mathcal{M}$.\\
	In \cite[Section 3.4]{BDK2} is proven first that $\iota:\mathcal{S}\xrightarrow{\sim}\overline{\mathcal{S}}$ in such a way that $\iota(\mathcal{S}_p)=\overline{\mathcal{S}}\cap\mathcal{W}_p$ $\forall p\geq-1$ [Proposition 3.5], then that $\phi$ maps $\overline{\mathcal{S}}$, up to a Lie algebra automorphism $\psi$ induced by a ring automorphism of $\mathcal{O}_n$, to $S(n)\subset W(n)$ [Proposition 3.6].\\
	Finally a Lie algebra isomorphism
	\begin{equation}\label{isoS(n)}
	\phi:\mathcal{S}\xrightarrow{\sim} S(n)\subset W(n)
	\end{equation}
	 such that $\mathcal{S}_p$ maps onto $S(n)\cap F_p W(n)$ is obtained [Corollary 3.3]. In particular we have that $\mathcal{S}_{-2}=\mathcal{S}_{-1}=\mathcal{S}$.\\

	\begin{rem}\label{remark verma modules as S(d) modules}
		Take a generalized Verma module $\mfT(V)=U(\mfL_{-})\otimes V = U(\mfd + \mfs)\otimes V$. It is in particular an $S(5)$-module and, in view of \eqref{isoS(n)}, also an $\mathcal{S}$-module. Furthermore, considering the action of $\mfd$ as left multiplication in $U(\mfd + \mfs)$, we can view $\mfT(V)$ as a $\mathcal{S}^e$-module. Since all these identifications are compatible with the filtrations, \eqref{compatiblegradings} implies that $\mfT(V)$ is a conformal $\mathcal{S}^e$-module. By Proposition \ref{(correspondance)}, it has a natural structure of $S(\mfd)$-module.
	\end{rem}

Finally, we define what are called \emph{tensor modules} for $W(\mfd)$ and $S(\mfd)$ (which are analogs of Verma modules) and extrapolate from \cite{BDK2} what we need.\\

Recall that $\mathcal{W}_0/\mathcal{W}_1\cong \gld$. So if we take a $\gld$-module $V$, we can allow $\mathcal{W}_1$ to act on it trivially and get a $\mathcal{W}_0$-module. After that, we can induce so to get a $\mathcal{W}$-module.\\
But in order to correlate this with the Lie pseudoalgebra $W(\mfd)$, we need to take account of the action of $\mfd$. To do so, we consider the extended annihilation algebra $\mathcal{W}^e$.\\
We call $\mathcal{N}_{\mathcal{W}}$ the normalizer of $\mathcal{W}_p$ in $\mathcal{W}^e$. In \cite{BDK2} is proven that it is independent of $p$ and that $\mathcal{W}^e=\mfd\oplus\mathcal{N}_{\mathcal{W}}$[Proposition 3.3]. Moreover, $\mathcal{W}_1$ acts trivially on any irreducible finite-dimensional conformal $\mathcal{N}$-module (i.e. modules for which every element is killed by some $\mathcal{W}_p$) and $\mathcal{N}_{\mathcal{W}}/\mathcal{W}_1\cong \mfd \oplus \gld$, so that we have a one-to-one correspondence between irreducible finite-dimensional $\mfd\oplus \gld$-modules and irreducible finite-dimensional conformal $\mathcal{N}_{\mathcal{W}}$-modules [Proposition 3.4].\\

In \cite[Section 3.5]{BDK2} totally analogous results are proven for $\mathcal{S}$, whereas $\mathcal{N}_\mathcal{S}/\mathcal{S}_1\cong \mfd \oplus \sld$.\\

Take a finite-dimensional $\mfd \oplus \gld$-module $V$; letting $\mathcal{W}_1$ act as zero on it, we can define an action of $\mathcal{N}_\mathcal{W}$ and then define the $\mathcal{W}^e$-module $T(V)=Ind_{\mathcal{N}_{\mathcal{W}}}^{\mathcal{W}^e} V =\mathcal{W}^e\otimes_{\mathcal{N}_{\mathcal{W}}} V$ which can be identified as an $H$-module with $H \otimes V$ since $\mathcal{W}^e=\mfd \oplus \mathcal{N}_{\mathcal{W}}$.\\
$T(V)$ is called a \emph{tensor module} for $W(\mfd)$.\\
\begin{defin}
	Let $\mathfrak{g}_1$ and $\mathfrak{g}_2$ be Lie algebras and let $V_i$ be $\mathfrak{g}_i$-modules fro $i=1,2$.\\
	We indicate with $V_1\boxtimes V_2$ the $\mathfrak{g}_1\oplus \mathfrak{g}_2$-module $V_1\otimes V_2$ where $\mathfrak{g}_i$ only acts on the $V_i$ factor.
\end{defin}
If $V$ is of the form $\Pi\boxtimes U$, we will also indicate $T(V)=T(\Pi, U)$.\\
We can define on a tensor module $T(V)=H\otimes V$ a filtration as follows:
\begin{equation}
F^p T(V) = F^p H \otimes V \quad\text{for } p\geq -1
\end{equation}
which behaves nicely relatively to the filtration of $\mathcal{W}$:
\begin{lem}\cite[Lemma 6.3]{BDK2}
	For every $p\geq0$ we have:
	\begin{enumerate}
		\item $\mfd \cdot F^p T(V) \subset F^{p+1} T(V)$;
		\item $\mathcal{N}_{\mathcal{W}}\cdot F^p T(V)\subset F^p T(V)$;
		\item $\mathcal{W}_1 \cdot T(V)\subset F^{p-1}T(V)$.
	\end{enumerate}
\end{lem}

For $S(\mfd)$ we do the same construction:\\
take a finite-dimensional $\mfd \oplus \sld$-module $V$, let $\mathcal{S}_1$ act trivially on it so that it has an action of $\mathcal{N}_\mathcal{S}$; then consider the $\mathcal{S}^e$-module $T_{S}(V)=Ind_{\mathcal{N}_{\mathcal{S}}}^{\mathcal{S}^e} V =\mathcal{S}^e\otimes_{\mathcal{N}_{\mathcal{S}}} V$ which again can be identified, as an $H$-module, with $H\otimes V$.\\
In \cite[Theorem 7.3]{BDK2} it is proven that these modules can be obtained as the restriction of tensor modules for $W(\mfd)$, therefore we will call them again \emph{tensor modules} for $S(\mfd)$.\\
If $V$ is of the form $\Pi\boxtimes U$, we will also indicate $T_{S}(V)=T_S(\Pi,U)$.\\
We can define the same filtration we defined in the $W(\mfd)$ case and have the same nice behavior relatively to the filtration of $\mathcal{S}$.\\

It makes sense now to introduce the notion of \emph{singular vectors} for $W(\mfd)$ and $S(\mfd)$.

\begin{defin}
	For a $W(\mfd)$-module $V$, a \emph{singular vector} is an element $v\in V$ such that $\mathcal{W}_1 \cdot v =0$. The space of singular vectors in $V$ in indicated by $sing \, V$.\\
	Analogously one defines singular vectors for $S(\mfd)$-modules.
\end{defin}

\begin{rem}
	By Remark \ref{remark verma modules as S(d) modules} a Verma module has a structure of $S(\mfd)$-module. Again we stretch the fact that all the identifications preserve the filtrations. Therefore, by the last definition, a singular vector for $E(5,10)$ in a Verma module is also  singular for $S(\mfd)$.
\end{rem}

Focusing only on type $S$ now, we summerize all the main results about tensor modules for $S(\mfd)$ and singular vectors in \cite[Section 7]{BDK2}.\\
We will indicate by $\Omega^n$ the $\sld$-module $\bigwedge \mfd^\ast$ with the natural action, where $\Omega^0$ is the trivial module $\C$.
\begin{theo}\label{theorem degrees tensor modules}$ $\newline
	\begin{itemize}
		\item Every irreducible finite $S(\mfd)$-module is a quotient of a tensor module;\\
		\item  Let $\Pi$ (resp. $U$) be an irreducible finite-dimensional module over $\mfd$ (resp. $\sld$). Then the $S(\mfd)$-module $T_S(\Pi,U)$ is irreducible if and only if $U$ is not isomorphic to $\Omega^n$ for any $n\geq 0$;\\
		\item If $V=T_S(\Pi,\Omega^n)$, $n\neq 1$, then $sing \, V \subset F^1 V$;\\
		\item If $V=T_S(\Pi,\Omega^1)$, then $F^1 V \subsetneq sing \, V \subset F^2 V$.\\
	\end{itemize}
\end{theo}

Combining the results above, one gets the following picture:\\
if one looks for irreducible $S(\mfd)$-modules (or, by mean of Proposition \ref{correspondance}, irreducible $\mathcal{S}^e$-modules ) one studies tensor modules. For a tensor module, to be irreducible is equivalent to not contain positive degree singular vectors. Singular vectors can only be found when one induces from irreducible $\sld$-modules $\Omega^n$ and they are found in degree $1$. except for the case $n=1$ where one finds also singular vectors of degree 2.\\

\section{Bound on degree of singular vectors}\label{sectionBoundSingularvectors}
Direct computation of singular vectors is not immediate, especially in high degrees. One thing one can do is trying, as a start, to rule out as many options as possible. This can be done, for example, by looking for restricting conditions on the degree of singular vectors, which is what we are about to do.\\
Recall that by Theorem \ref{theorem degrees tensor modules}, we have a very important result that points in this direction in the "pseudo" setting.\\
Turns out, in order to get a first bound on the degree of singular vectors, it is enough to take into account just the even structure of $\mfL$, exploiting the pseudoalgebra techniques available.\\
To do so, we first define some subspaces of $\mfT(V)$.\\

For $i=0,\dotsc, 10$, let
\begin{equation}
\Gamma_i(V):=\Big\{ v=\sum_{I,K}\der^{(I)}\xi^K \otimes v_{IK}\in \mfT(V) \, | \, |K|\leq i \text{ se } v_{IK}\neq 0 \Big\}.
\end{equation}
In other words, $\Gamma_i(V)$ consists of vectors with "odd degree" at most $i$. We may also describe $\Gamma_i(V)$ as the space generated by the PBW monomials of degree $(n\, |\,j)$ with $j\leq i$.\\
It is easy to check the following properties:
\begin{itemize}
 	\item $\Gamma_i(V)\subseteq \Gamma_{i+1}(V)$;\\
 	\item $\Gamma_0(V)=U(\mfd)\otimes V$;\\
 	\item  $\Gamma_{10}(V)=\mfT(V)$;\\
 	\item  $\Gamma_i(V)$ is a $\mfL_{(0)}\cong S(5)$-submodule of $\mfT(V)$.
 \end{itemize}
Basically, we have built a finite  filtration of $S(5)$-modules of $\mfT(V)$.\\
The last property follows from the fact that the action of an even element in $\mfL$ can only lower the odd degree in $\mfT(V)$. Take for example $y\in \mfL_2, \, v=\der^{(I)}\xi^{K}\otimes v_I^K\in \mfT(V)$.\\
We have
\[
y\cdot v=y\cdot (\der^{(I)}\xi^K\otimes v_{I}^{K})=[y , \der^{(I)}] \xi^K \otimes v_{I}^{K}+ \der^{(I)} [y, \xi^K] \otimes v_{I}^{K}+\der^{(I)}\xi^K \otimes  y\cdot v_{I}^{K} .
\]
Here the third term is $0$ because $\mfL_+$ acts trivially on $V$; the first term consists of elements of degree $(\, |I|-1 \; | \; |K|\, )$; lastly, the second term, once expanded the bracket and sorted everything, can only contribute with elements of degree $(|I| \; | \; |K|-2 )$ or $(|I|+1 \; | \; |K| -4)$. In any case the odd degree cannot increase.\\

These properties allow us to talk about quotients.\\
Let us consider the quotients of $S(5)$-modules $\Gamma_{i}(V)/\Gamma_{i-1}(V)$ for $i=0,\dotsc, 10$ (where we impose $\Gamma_{-1}(V)=0$).\\
As $\sld$-modules, they are isomorphic to $U(\mfd)\otimes (\bigwedge^{i}(\mfs)\otimes V)$ (follows from \cite[Corollary 6.4.5]{M}). The latter look a lot like tensor modules $T(\bigwedge^{i}(\mfs)\otimes V)$ for $S(\mfd)$: if the action of $S(5)$ can be interpreted as the action of the annihilator algebra associated to pseudoaction of $S(\mfd)$, we can put to use Proposition \ref{(correspondance)}.
This is, in fact, possible in the following way.\\

The action of $y\in \mfL_{2j}$ on a class $\overline{u \cdot \xi^K \otimes v}\in \Gamma_{i}(V)/\Gamma_{i-1}(V)$, where $u\in U(\mfd)$, $|K|=i$ and $v\in V$, behaves, depending on $j$, like:\\
\begin{description}
	\item[for $\mathbf{j=-1}$]$y\cdot \overline{u\xi^K \otimes v} =\overline{{(yu)\xi^K \otimes v}}$, since in this case $y\in \mfd\subseteq U(\mfd)$;\\
	\item[ for  \, $\mathbf{j=0}$]$y\cdot \overline{u\xi^K \otimes v} = \overline{[y,u]\xi^K\otimes v + u [y,\xi^K]\otimes v + u \xi^K \otimes (y\cdot v)}$;\\
	\item[ for  \, $\mathbf{j>0}$]$\begin{aligned}[t] y\cdot \overline{u\xi^K \otimes v} =& \overline{[y,u]\xi^K\otimes v + u [y,\xi^K]\otimes v + u \xi^K \otimes (y\cdot v)}\\
																		=&\overline{[y,u]\xi^K\otimes v}
						  \end{aligned}$\\
\end{description}
where the last equality is due to the fact that $y$ lowers the odd degree of at least $1$, sending the second term to $0$ in the quotient, and acts trivially on $V$.\\
Notice that in the case $j=0$, the action of $y\in\mfL_0\cong \sld$ on $\xi^K$ is actually the same as the one on the $\sld$-module $\bigwedge^{i}(\mfs)$ (since the other terms that usually appear in the bracket $[y,\xi^K]$ once sorted are $0$ in the quotient).\\

Summing up, we have that $\mfL_{(0)}=S(5)$ acts on $\Gamma_{i}(V)/\Gamma_{i-1}(V)$
\begin{itemize}
	\item by left multiplication on the $H=U(\mfd)$ factor with the negative degree part;
	\item by the natural action of $\sld$ on $U(\mfd)\otimes(\bigwedge^i(\mfs)\otimes V)$ with the degree $0$ part;
	\item trivially on $\bigwedge^{i}(\mfs)\otimes V$ with the positive degree part.
\end{itemize}

Since this is exactly the action of $\mathcal{S}$ on the tensor module $T_S(\bigwedge^i(\mfs)\otimes V)$ via the isomorphism in \eqref{isoS(n)}, we can state:
\begin{theo}\label{theorem Gamma tensor}
	Let $V$ be a finite-dimensional irreducible $\sld$-module. Then we have an isomorphism of $S(5)\cong X\otimes_H S(\mfd)$-modules
\begin{align}
\varphi:\Gamma_{i}(V)/\Gamma_{i-1}(V) &\xrightarrow{\sim} T_S(\textstyle\bigwedge^{i} (\mfs)\otimes V)

\end{align}
\end{theo}

Take now $v\in sing \, \mfT^p(V)$. It is, in particular, a singular vector for $\mfL_{(0)}=S(5)$ of degree $p$. We indicate the space of such vectors with $sing_{S(5)}\, \mfT(V)$.\\
If we consider $\overline{v}\in\Gamma_{i}(V)/\Gamma_{i-1}(V)$ for a suitable $i=0, \dotsc, 10$, it will still be a singular vector in what we now know is a tensor module. Therefore, by Theorem \ref{theorem degrees tensor modules}, the even degree of $\overline{v}$ must be $\leq 2$. Since the odd degree of a vector cannot be larger than $10$, these ideas, formalized, prevent vectors with a sufficiently high enough degree from being singular.
\begin{theo}\label{theo14}
	Let $\mfT(V)$ be a minimal Verma module and let $v\in sing \, \mfT(V)$.  Then $v$ has degree at most $14$.
\end{theo}
\begin{proof}
We can assume that $v$ is homogeneous of degree $p$.\\
We have, if $p$ is either even or odd:\\

\begin{description}
	\item[p=2n]   $\mfT^p(V)=S^n(\mfd)\otimes V +S^{n -1}(\mfd) \, \bigwedge^2(\mfs)\otimes V+\dotsb + S^{n-5}(\mfd) \bigwedge^{10}(\mfs)\otimes V$;
	\item[p=2n+1]   $\mfT^p(V)=S^{n}(\mfd) \, \mfs \otimes V + S^{n-1}(\mfd)\bigwedge^3(\mfs)\otimes V + \dotsb + S^{n-4}(\mfd) \bigwedge^9(\mfs)\otimes V$.\\
\end{description}

We study the case $p=2n$.\\
Let $0\leq m_0 \leq 10$ be the greatest index such that the term of $v$ in degree $(\frac{p-m_0}{2} | m_0)$ is not $0$ (i.e. the term of $v$ in $S^{(p-m_0)/2}(\mfd)\bigwedge^{m_0}(\mfs)\otimes V$).\\
Therefore $v\in\Gamma_{m_{0}}(V)$ and it is a combination of terms in degrees \[( \textstyle\frac{p-m_0}{2}  |  m_0), \, ( \textstyle\frac{p-m_0}{2} +1  | m_0 -2  ), \dotsc , \, ( \textstyle\frac{p}{2}  |  0  ).\]
We can then consider
\[
\overline{v} \in \Gamma_{m_{0}}(V)/\Gamma_{m_{0}-1}(V)\cong U(\mfd)\otimes(\textstyle\bigwedge^{m_0}(\mfs)\otimes V).
\]
Notice that given $y\in\mfL_{2}$, $y\cdot v\in \mfT^{p-1}(V)$ and the term of degree $(\frac{p-m_0-1}{2} | m_0)$ can be obtained only acting with $y$ on the term of $v$ of degree $(\frac{p-m_0}{2} | m_0)$. Hence, if $v$ is singular for $S(5)$, so must be $\overline{v}$.\\
Recapitulating: if $v\in sing \,\mfT^p(V)$, $\overline{ v }\in U(\mfd)\otimes(\bigwedge^{m_0}(\mfs)\otimes V)=T(\bigwedge^{m_0}(\mfs)\otimes V)$ is a singular vector for $S(5)$ of (even) degree $(p-m_0)/2$ in a tensor module. By Theorem \ref{theorem degrees tensor modules}, the even degree of $\overline{ v }$ must be less than or equal to $2$, which means that $(p-m_0)/2\leq 2$, that is $p\leq m_0 + 4\leq 14$. When $p$ is odd the same argument holds.
\end{proof}

The proof actually tells us more than the statement of the theorem: we can not only estimate the singular vectors' degree, but we can also rule out straightforwardly most of the irreducible $\sld$-modules whose induced modules we expect to possibly contain singular vectors of a certain degree.\\
We recall that in our notation $\Omega^1=\mfd^\ast\cong V([1,0,0,0])$. Similarly $\Omega^2\cong V([0,1,0,0])$, $\Omega^3\cong V([0,0,1,0])$ and $\Omega^4\cong V([0,0,0,1])$.\\

Now apply, for example, the proof's arguments on degree 14:\\
let $v$ be a singular vector of degree $14$ in a minimal Verma module $\mfT(V)$ and consider $\overline{v}\in \Gamma_{10}(V)/\Gamma_{9}(V)\cong U(\mfd)\otimes(\bigwedge^{10}(\mfs)\otimes V)\cong U(\mfd)\otimes V$. If $\overline{ v }\neq 0$, this means that the term of $v$ of degree $(2 | 10)$ is not $0$.\\
We know by Theorem \ref{theorem degrees tensor modules} that we can find singular vectors of (even) degree $2$ in a tensor module $T(V)$ where $V$ is irreducible if and only if $V\cong \Omega^1\cong V([1,0,0,0])$.\\
Therefore, if we assume that $V\ncong \mfd^{\ast}$, $\overline{ v }$ will necessarily be $0$. This implies that $v\in \Gamma_{8}(V)$ and we can consider $\overline{ v }\in U(\mfd)\otimes(\bigwedge^8(\mfs)\otimes V)$, thus obtaining a singular vector in $T(\bigwedge^8(\mfs)\otimes V)$ of (even) degree $(14-8)/2=3$. This cannot happen, so that the only possible solution is $\overline{ v }=0$. Iterating, we discover that $v$ must be $0$.\\
We proved:
\begin{lem}
	If $V\ncong \mfd^\ast\cong V([1,0,0,0])$, $sing \, \mfT^{14}(V)=\lbrace 0 \rbrace$.
\end{lem}
\section{Bound refining}\label{sectionBoundRefining}
	 An extremely simple lemma will be extremely useful:
	 \begin{lem}\label{easylemma}
	 	Let $\mfT(V)$ be a Verma module for $\mfL$. If $v\in sing \, \mfT(V)$ and $\xi\in \mfL_{-1}\cong \mfs$, then $\xi v \in sing_{S(5)} \, \mfT(V)$.
	 \end{lem}
	\begin{proof}
		Take $y\in \mfL_{2}$. Then
		\[
		y \cdot (\xi v) = [y,\xi]\cdot v + \xi (y\cdot v)=0,
		\]
		where the second term is $0$ because $v$ is singular. The same goes for the first term since $[y,\xi]\in \mfL_{1}$.
	\end{proof}
	We apply this new piece of information to the case $V\cong \mfd^*$.\\
	Let $v\in sing \, \mfT^{14}(\mfd^{\ast})$. By the lemma, given any $\xi\in \mfs$, $\xi v$ (which has now degree $15$), is still singular for the action of $S(5)$. Consider the term of degree $(3 | 9)$ and the corresponding $\overline{ \xi v }\in \Gamma_{9}(V)/\Gamma_{8}(V)\cong T(\bigwedge^9(\mfs)\otimes V)$. Like before, it is still singular and has even degree $3$, which implies that $\overline{ \xi v }=0$. We can then consider $\overline{ \xi v }\in \Gamma_{7}(V)/\Gamma_{6}(V)$ and, iterating the argument, obtain that $\xi v$ must be $0$ $\forall \xi \in \mfs$.\\
	We remark that $\mfL_{-2}=[\mfL_{-1},\mfL_{-1}]$ which means that, for any $\der\in \mfd \cong \mfL_{-2}$, we can find $\xi_1,\xi_2\in\mfs\cong\mfL_{-1}$ such that $\der=[\xi_1,\xi_2]$. This in particular implies that $\der v =0$ $\forall \der\in \mfL_{-2}$. Since the action of $\mfL_{-2}$ on a tensor module is simply given by left multiplication, it can only mean that $v=0$.\\
	In conclusion, we showed that even if $V\cong \mfd^\ast$, $\mfT(V)$ cannot have singular vectors of degree 14.\\
	These ideas can be applied systematically to perform a refining of the bound in Theorem \ref{theo14}.

	\begin{theo}\label{raff}
		Let $\mfT(V)$ be a minimal Verma module and let $v\in sing \, \mfT(V)$. Then $v$ has degree $\leq 12$. More precisely:
		\begin{enumerate}
			\item if $V\ncong V([0,0,1,0])$, singular vectors have degree at most $11$;\\
			\item if $V\ncong V(\lambda)$ where $\lambda=[0,0,0,1],\, [0,0,1,0],\,[0,1,0,0], \, [1,0,0,0], \, [0,1,1,0],$ or $[1,0,0,1]$, singular vectors have degree at most $10$;\\
			\item $\begin{aligned}[t]
				if V\ncong V(\mu) \text{ where } \mu = &[0,0,0,0],\,[1,0,0,0],\,[0,0,0,1],\,[0,0,1,0],\, [0,1,0,0], \\
				 									   &[1,1,0,0], \, [0,1,1,0],\, [1,0,0,1], \,[0,0,1,1],\,[1,0,1,0], \\
													   &[0,1,0,1], \, [1,1,0,1],\,[0,2,0,0],\,[2,0,0,0], \, [1,0,2,0], \\
													    &\text{or } [3,0,0,1], \text{ singular vectors have degree at most $9$.}
			\end{aligned}$
		\end{enumerate}
	\end{theo}
	The proof revolves around arguments similar to the previous ones. To that end, because of Theorem \ref{theorem Gamma tensor} and Theorem \ref{theorem degrees tensor modules}, we will need to be able to determine when, given an irreducible $\sld$-module $V$, we can (or rather cannot) find a copy of $V(\omega_i)$ in $\bigwedge^j(\mfs)\otimes V$.\\
	Recall that every irreducible $\sld$-module $V$ has a highest weight vector and that $V$ is uniquely determined by the highest weight. Here, as before, $\omega_1=[1,0,0,0]$, $\omega_2=[0,1,0,0]$, $\omega_3=[0,0,1,0]$, $\omega_4 = [0,0,0,1]$ and $\omega_0 = [0,0,0,0]$.\\
	By Frobenius duality (keeping in mind that these are all finite-dimensional modules), $V(\omega_i)\subseteq \bigwedge^j(\mfs)\otimes V$ if and only if $V(\omega_0)\subseteq \bigwedge^j(\mfs)\otimes V \otimes V(\omega_i)^*$ se e solo se $V^*\subseteq \bigwedge^j(\mfs)\otimes V(\omega_i)^*$ se e solo se $V\subseteq \bigwedge^j(\mfs^*)\otimes V(\omega_i)$.\\
	In the following table we list the highest weights of the irreducible representations that appear in the decomposition of those tensor products. It was obtained using computer software "LiE" ( see \cite{LiE} for further informations).\\

	\begin{tabular}{|c|p{3cm}|p{3cm}|p{3cm}|p{3cm}|}
		\hline
		& $\bigwedge^9(\mfs^*)\otimes V(\omega_i)$ & $\bigwedge^8(\mfs^*)\otimes V(\omega_i)$ & $\bigwedge^7(\mfs^*)\otimes V(\omega_i) $
		& $\bigwedge^6(\mfs^*)\otimes V(\omega_i)$\\
		\hline
		i=0 & [0,1,0,0] & [1,0,1,0] & [0,0,2,0], [2,0,0,1] & [1,0,1,1], [3,0,0,0]\\
		\hline
		i=1 & [0,0,1,0], [1,1,0,0] & [0,1,1,0], [1,0,0,1], [2,0,1,0] & [0,0,1,1], [1,0,2,0], [1,1,0,1], [2,0,0,0], [3,0,0,1] & [0,1,1,1], [1,0,0,2], [1,0,1,0], [2,0,1,1], [2,1,0,0], [4,0,0,0] \\
		\hline
		i=2 &[0,0,0,1], [0,2,0,0], [1,0,1,0] & [0,0,2,0], [0,1,0,1], [1,0,0,0], [1,1,1,0], [2,0,0,1] & [0,0,1,0], [0,1,2,0], [1,0,1,1], [1,1,0,0], [2,1,0,1], [3,0,0,0] & [0,0,2,1], [0,1,0,2], [0,1,1,0], [1,0,0,1], [1,1,1,1], [2,0,0,2], [2,0,1,0], [3,1,0,0]\\
		\hline
		i=3 &[0,0,0,0], [0,1,1,0], [1,0,0,1] &[0,0,1,1], [0,1,0,0], [1,0,2,0], [1,1,0,1], [2,0,0,0] & [0,0,3,0], [0,1,1,1], [1,0,0,2], [1,0,1,0], [2,0,1,1], [2,1,0,0] & [0,0,1,2], [0,0,2,0], [0,1,0,1], [1,0,2,1], [1,1,0,2], [1,1,1,0], [2,0,0,1], [3,0,1,0]\\
		\hline
		i=4 & [0,1,0,1], [1,0,0,0] & [0,0,1,0], [1,0,1,1], [1,1,0,0] & [0,0,2,1], [0,1,1,0], [1,0,0,1], [2,0,0,2], [2,0,1,0] & [0,0,1,1], [1,0,1,2], [1,0,2,0], [1,1,0,1], [2,0,0,0], [3,0,0,1]\\
		\hline
	\end{tabular}
	\begin{proof}[Proof of Theorem \ref{raff}.] We will outline the various steps in a schematic way. The ideas are the same we have already discussed.\\
		\textbf{Degree 13:}\\
		Let $v\in sing \, \mfT^{13}(V)$. We first consider the term of $v$ in degree $(2|9)$; it can be different from zero only if $V$ appears in the decomposition of $\bigwedge^{9}(\mfs^\ast)\otimes V(\omega_1)\cong V([0,0,1,0])\oplus V([1,1,0,0])$. So if $V$ is not isomorphic to one of these two representations, the term in $(2|9)$ of $v$ must be equal to $0$, we look at next term, which is also $0$ because has degree $(3|7)$. Iterating, we deduce that $v=0$.\\
		If $V\cong V([0,0,1,0])$ or $V([1,1,0,0])$, take $\xi\in\mfs$ and consider the term in degree $(2|10)$ of $\xi v$. It can be non zero only if $V$ does not appear in $\bigwedge^{10}(\mfs^\ast)\otimes V(\omega_1) \cong V(\omega_1)$. It follows that it must be $0$ and, iterating, so does $\xi v$  $\forall \xi \in \mfs$ which in turn implies, as we already saw, that $v$ must be $0$. In conclusion, $sing \, \mfT^{13}(V)$ is always $0$.\\
		\textbf{Degree 12:}\\
		Let $v\in sing \, \mfT^{12}(V)$. We consider the term of degree $(1|10)$; it can be non zero only if $V$ appears in $\bigwedge^{10}(\mfs^\ast)\otimes V(\omega_i)\cong\omega_i$ for $i=0,\dotsc, 4$. Therefore if $V$ is not of the form $V(\omega_i)$, we check the term of degree $(2|8)$ which must be $0$ unless a copy of $V$ appears in $\bigwedge^8(\mfs^\ast)\otimes V(\omega_1)= V([0,1,1,0])\oplus V([1,0,0,1]) \oplus V([2,0,1,0])$. In the remaining cases $v=0$.\\
		Now assume $V\cong V([0,1,1,0])$, $V([1,0,0,1])$, $V([2,0,1,0])$ or $V(\omega_i)$ with $i=0,\dotsc, 4$.\\
		Take $\xi\in\mfs$ and consider the term of $\xi v$ of degree $(2|9)$. It cannot be non zero if $V$ does not appear in $\bigwedge^{9}(\mfs^\ast)= V([0,0,1,0])\oplus V([1,1,0,0])$. So if $V\ncong V(\omega_3)$ $\xi v=0$ $\forall \xi \in \mfs$ and again it implies that $v=0$. Therefore the only case in which we cannot rule out the presence of singular vectors of degree $12$ is for $V\cong V(\omega_3)$.\\
		\textbf{Degree 11:}
		Let $v\in sing \, \mfT^{11}(V)$. We consider the term of degree $(1 | 9)$; it can be different from zero when $V$ appears in $\bigwedge^{9}(\mfs^\ast)\otimes V(\omega_i)$ for $i=0,\dotsc, 4$. It happens when $V$ has as highest weight one belonging to the first column of the table.\\
		If we assume $V$ is not isomorphic to any of them, we can move to the next term which has degree $(2|7)$. According to the table, if $V$ is also not isomorphic to $V([0,0,1,1])$, $V([1,0,2,0])$, $V([1,1,0,1])$, $V([2,0,0,0])$ and $V([3,0,0,1])$, $v=0$.\\
		Assume $V$ is isomorphic to one of these representations, a fundamental representation or the trivial one; take $\xi\in\mfs$ and check the term of degree $(1|10)$; if $V$ is not a fundamental representation or the trivial one, we can move to the term of degree $(2|8)$ which will be $0$ unless a copy of $V$ appears in $\bigwedge^8(\mfs^\ast)\otimes V(\omega_1) \cong V([0,1,1,0])\oplus V([1,0,0,1])\oplus V([2,0,1,0])$. Therefore if $V$ is not isomorphic to $V([0,0,0,1])$, $V([0,0,1,0])$, $V([0,1,0,0])$, $V([1,0,0,0])$, $V([0,1,1,0])$ or $V([1,0,0,1])$, $sing \, \mfT^{11}(V)=0$.\\
		\textbf{Degree 10:}\\
		Let $v\in sing \, \mfT^{10}$. The term of $v$ with the greatest odd degree is the one of degree $(0|10)$. In this case we cannot deduce anything, since this term is constant from the point of view of $S(5)$, therefore singular.\\
		We relay again on Lemma \ref{easylemma}: take $\xi\in \mfs$ and consider $\xi v$, which has now degree $11$. This time we can still look at $\overline{ \xi v }\in \Gamma_{9}/(V)\Gamma_{8}(V)\cong T(\bigwedge^9(\mfs)\otimes V))$ which has even degree $1$. Again, we know that it is $0$ unless $V$ appears as an irreducible of $\bigwedge^9(\mfs^\ast)\otimes V(\omega_i)$, $i=0\dotsc, 5$. So if the highest weight of $V$ does not appear in the first column of the table, $\overline{ \xi v }=0$. We can then move to the next term, which has degree $(2|7)$. Here we look at the irreducible modules in $\bigwedge^7(\mfs^\ast)\otimes V(\omega_1)\cong V([0,0,1,1])\oplus V([1,0,2,0])\oplus V([1,1,0,1])\oplus V([2,0,0,0])\oplus V([3,0,0,1])$. So if in addition we ask that $V$ is not isomorphic to these modules, $\xi v=0$ $\forall \xi\in\mfs$ and again we cannot have singular vectors of degree $10$ in $\mfT(V)$.
	\end{proof}


TITLE: Using Cauchy's theorem or Integral formula to evaluate $\oint_C \frac{\sin2z}{6z-\pi} dz,$
QUESTION [3 upvotes]: I have the problem, to use Cauchy's theorem or Cauchy's integral formula to evaluate $\oint_C \dfrac{\sin2z}{6z-\pi} dz$, where $C$ is the circle $\mod (z) = 3$.
I know that the denominator vanishes at z =  $\dfrac{\pi}{6}$, which is within the contour, so the result of the integral is not $0$ (Cauchy's theorem). I have a singularity within the contour, then, and it is not analytic. How then can I use Cauchy's theorem or integral formula for this?

REPLY [4 votes]: No need for the residue theorem, Cauchy's integral formula gives the answer right away:
$$
\oint_C \dfrac{\sin2z}{6z-\pi}\,dz =
\oint_C \dfrac{\frac{\sin2z}{6}}{z-\frac{\pi}{6}}\,dz = 2\pi i\,\frac{\sin \frac{\pi}3}{6} = \frac{\pi i\sqrt{3}}{6}.
$$


TITLE: Matrix Dimension's effect on Positive Definiteness
QUESTION [3 upvotes]: Recently I have been working on a project involving stock returns and portfolio efficiency.  What I thought was a coding error, turned out to be me not understanding matrix math very well.
Long story short, the R function I am using needs a positive definite covariance matrix. I do not have any problems as long as the matrix I am using to calculate the covariance matrix has fewer columns than rows. Once the number of columns equals or exceeds the number of rows, then my covariance matrix is not longer positive definite and my code fails.
A small-scale example. Let matrix A be my original matrix I am using to calculate matrix B, my covariance matrix.  Say A is 3x2. B turns out to be positive definite, and my code works fine. Once I add another column of data to A, making it 3x3, B is no longer positive definite. This same situation occurs no matter the size. 5x2, 8x5, 16x8, 39x38 all produce PD B matrices. But once the number of columns exceeds the number of rows (5x5, 8x11, 39x40), B is no longer PD.
Is this just a matrix thing that I never learned? Is positive definiteness dependent on the matrix dimensions?
Here is the original thread that spurred this question:
https://stackoverflow.com/questions/31007182/portfolio-frontier-not-working-for-large-dataset
If I was unclear please let me know.
Edit:
Here are a few numerical examples.
[-.0108, .0469
.0050, .0193
-.0059, -.0005]

This is an example of my original matrix, A. This leads to a PD B matrix of
[.00006, -.00007
-.00007, .0005]

If A is
[-.0108, .0469, .0190, .0080
.0050, .0193, .0150, .0177
-.0059, -.0005, .0091, -.0023]

Then B will be not PD.

REPLY [1 votes]: Of course, assuming X has more columns than rows, if $\mathbb{E}X=0$ then $\Sigma=X^TX$.  To check positive semi-definiteness, we evaluate
$$y^T\Sigma y=y^TX^TXy=\left\|Xy\right\|_2^2\geq0$$
which must in fact be zero some of the time since $X$ has more columns than rows.
As it turns out mean subtraction actually reduces the rank of your data matrix $X$ by one (proof not forthcoming), so even if $X$ is square, mean subtraction makes it singular and thus it will have non-trivial kernel.  Thus the number of rows of your data matrix X must be strictly greater than the number of columns for the covariance matrix to be strictly positive definite.


TITLE: Fourier Transforms
QUESTION [5 upvotes]: I'm having a terrible time trying to understand Fourier transforms.  I'm very visual so leaving the $X,Y,Z,t$ domain is not working form me :)
I'm trying to figure out the basics at the moment.  Like, taking a Sine wave (they're odd right?) and converting it into its real and imaginary numbers.  I'm pretty sure I got that working, but to make sure, what should the plotted data look like?
Also, how do I find the power spectrum of a transform?  How do I use FT to identify the $n$ most significant frequencies in a signal?  That last question shows how lost I am!
What I have:
I know how to get the real and imaginary numbers from a signal.  I know how to get the phase and the magnitude.  What I need to get is the power spectrum and the most significant frequencies.  Also any dumbed down explanation of what's going on would be very helpful!
Thanks!

REPLY [3 votes]: I'll try and give some intuition from an Electrical engineering perspective. It is useful to think of a Fourier transform as giving you the frequency domain picture of a given signal, i.e., it gives you a picture of both the amplitude and the phase of the different frequency components that make up the signal.
If you are familiar with Fourier series representation of periodic signals, the Fourier transform can be viewed as an extension to non periodic signals. Of course, not all continuous functions have a well defined Fourier transform, but if you look at typical engineering applications, that is not an issue.
When dealing with deterministic signals, the power spectrum is given by the squared magnitude of the Fourier transform. If $f(t)$ is the signal in the (continuous) time domain and $F(\omega)$ is the frequency domain representation, then the Power spectrum is $|F(\omega)|^2$.
When dealing with random signals, the power spectral density is the Fourier transform of the autocorrelation function of the process (Provided the process is wide sense stationary).
The above explanations carry over to the discrete domain as well. So, if you are working with DFTs of signal samples, things are not significantly different.
As for the N most significant frequencies, I think you need to take the DFT of your signal and then look at the frequencies in increasing order of amplitudes.


TITLE: Solving the cubic-exponential Diophantine equation $x^3+3=2^n$
QUESTION [6 upvotes]: The Diophantine equation $x^3+3=2^n$ has the obvious solutions $(-1,1)$,$(1,2)$ and $(5,7)$. I have been wondering if there are any other, but my attempts have been fruitless (I tried factoring it over $\Bbb{Q}(\sqrt[3]{3})$, but I don't know the basic properties of this field, such as what the ring of integers are, the class number etc.). Any help in solving this problem would be greatly appreciated.
Edit: We can actually split this into two, more general equations, namely the elliptic curves
$$
x^3+3=y^2
\quad \text{and} \quad
x^3+3=2y^2
$$
so this opens up another method for solving it.The first one is, in fact, a special case of the infamous Mordell equation.
Edit 2: Looking at this paper by Tzanakis and De Weger :http://www.math.uoc.gr/~tzanakis/Papers/PracticalSolutionThueEq.pdf
I was wondering if we could use the methods explained in Section 3 and extend them to these equations (the methods in the paper, however, require some computational machinery).This could lead to solving a Thue equation, which has only finitely many integral solutions and the general method for solving them is in the paper linked.

REPLY [2 votes]: Two cases:
$1)$ $n=2k$. Then $x^3+3=\left(2^k\right)^2$. But $a^3+3=b^2$ has $2$ integral solutions (http://oeis.org/A081119) $(a,b)=(1,\pm 2)$, so $(x,n)=(1,2)$.
$2)$ $n=2k+1$. Then $(2x)^3+24=\left(2^{k+2}\right)^2$. But $a^3+24=b^2$ has $8$ integral solutions (http://oeis.org/A081119, in particular http://oeis.org/A081119/b081119.txt)(I found them with a program, or you can see these tables)
$$(a,b)=(-2,\pm 4),(1,\pm 5),(10,\pm32),(8158,\pm736844),$$
so $(x,n)=(-1,1),(5,7)$.
Mordell Equations $x^3+k=y^2$ for $k\in\Bbb Z_{\neq 0}$ are fully solved when $|k|<10^4$ (were solved in $1998$; see here), so we can always use this method when the numbers aren't very large.


TITLE: Skolemization problem:∀x∃y∀z∃u (A(x,y,z,u) v B(y,u))
QUESTION [0 upvotes]: Everything in the title but an explanation would be nice beside an answer! I don't really know if I should use constants or functions.

REPLY [2 votes]: Skolemization is performed by replacing every existentially quantified variable $y$ with a term $f(x_i)$ whose function symbol $f$ is new.
If the formula is in prenex normal form, the variables $x_i$ of this term are the variable universally quantified whose quantifiers precede that of $y$.
The function $f$ introduced in this process is called a Skolem function (or Skolem constant if it is of zero arity, i.e. if the existential quantifier to be removed is not preceded by any universal quantifier).
Thus, in order to "skolemize" :

$∀x∃y∀z∃u \ldots$

we need a 1st skolem function $f(x)$ for the $∃y$ existential quantifier (1-arity, because $∃y$ is preceded by $∀x$) and a 2nd one $g(x,z)$ for the $∃u$ existential quantifier (2-arity, because $∃u$ is preceded by $∀x$ and $∀z$).


TITLE: Unique zero solution to a difference equation via Laplace transform
QUESTION [0 upvotes]: We want to prove that
the unique solution to the following difference equation is the null one:
$$
au(x)+b\mathbf{1}_{(0,\frac{1}{2})}(x)u(x+\frac{1}{2})+c\mathbf{1}_{(\frac{1%
}{2},1)}(x)u(x-\frac{1}{2})=0,\text{ }x\in (0,1).
$$
Extending $u$ by zero outside $(0,1)$ and taking the Laplace transform yields
$$
a\int_{0}^{1}e^{-px}u(x)dx+be^{\frac{p}{2}}\int_{\frac{1}{2}%
}^{1}e^{-px}u(x)dx+ce^{-\frac{p}{2}}\int_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,%
\text{ }p\in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
,
$$
that is
$$
\left( a+be^{\frac{p}{2}}\right) \int_{\frac{1}{2}}^{1}e^{-px}u(x)dx+\left(
a+ce^{-\frac{p}{2}}\right) \int_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,\text{ }%
p\in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.
$$
If we let for instance $p=\gamma +4n\pi i,n\in
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $a+ce^{-\frac{\gamma }{2}}=0$ we get
$$
\left( a+be^{\frac{\gamma }{2}}\right) e^{-\gamma }\int_{\frac{1}{2}%
}^{1}e^{-4n\pi ix}u(x)dx=0,
$$
which yields that $u=0$ on $(\frac{1}{2},1)$ if $a+be^{\frac{\gamma }{2}%
}\neq 0$ which is equivalent to $a^2-bc \neq 0$. With the same manner, by choosing this time  $p=\delta +4n\pi
i,n\in
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $a+be^{\frac{\delta }{2}}=0$ we get
$$
\left( a+ce^{-\frac{\delta }{2}}\right) e^{-\delta }\int_{0}^{\frac{1}{2}%
}e^{-4n\pi ix}u(x)dx=0,
$$
so if $a+ce^{-\frac{\delta }{2}}\neq 0$ we get $u=0$ on $(0,\frac{1}{2})$, which is equivalent to $a^2-bc \neq 0$.
I don't know if this kind of reasoning is correct since the Laplace
transform of $u$ is zero on subintervals with different choices of $p.$

REPLY [1 votes]: Your condition can be rewritten as the system of three equations:
$$au(x)+bu(x+1/2)=0\ \forall x\in(0,1/2), \tag{1}$$
$$au(x)+cu(x-1/2)=0\ \forall x\in(1/2,1), \tag{2} $$
$$au(1/2)=0. \tag{3} $$
In turn, (2) can be rewritten as
$$cu(x)+au(x+1/2)=0\ \forall x\in(0,1/2). \tag{2a} $$
So, if the determinant $a^2-bc$ of the system (1)--(2a) of linear equations for $u(x),u(x+1/2)$ is nonzero, then $u(x)=u(x+1/2)=0$ $\forall x\in(0,1/2)$, that is, $u(x)=0$ $\forall x\in(0,1)\setminus\{1/2\}$. Together with (3), this yields $u(x)=0$ $\forall x\in(0,1)$ if $a\ne0$.
The cases when $a^2-bc=0$ or $a=0$ are considered similarly.
In particular, if $a^2-bc=0$ but $a\ne0$, then $a\ne0$ and $b\ne0$, then the system (1)--(3) reduces to the conditions
$$u(x)=-(b/a)u(x+1/2)\ \forall x\in(0,1/2)\tag{1a}$$ and $u(1/2)=0$. So, here one may assign any values to $u$ on $(1/2,1)$, and then use (1a) to determine the values of $u$ on $(1/2,1)$.

The OP requested in a comment that the solution be given in terms of the Laplace transform, say $L$. This can be done as follows.
Of course, to use the Laplace transform, we have to assume that $u$ is integrable on $(0,1)$. Let
$$U(x):=\begin{cases}u(x)&\text{ if }0<x<1/2,\\
0&\text{ if }x>1/2,\end{cases}$$
$$V(x):=\begin{cases}u(x+1/2)&\text{ if }0<x<1/2,\\
0&\text{ if }x>1/2.\end{cases}$$
Then (1) and (2) imply
$$aL(U)+bL(V)=0,$$
$$cL(U)+aL(V)=0.$$
So, if $a^2-bc\ne0$, then $L(U)=L(V)=0$ and hence $U=V=0$ almost everywhere (a.e.), so that $u=0$ a.e., so that $u=0$ if $u$ is continuous.
As shown above, there is no uniqueness if $a^2-bc=0$.


TITLE: Confusion in module structure of indecomposable injectives over quiver path algebra
QUESTION [1 upvotes]: I got very confused while working out the explicit module structure on the indecomposable injective modules $I_i$ over a finite-dimensional quiver path algebra $kQ/I$. I know that the right injective indecomposable $I_i$, where $i$ is a vertex of $Q$, may be described as
$$I_i=\{\text{paths starting at }i \text{ modulo relations}\}^*,$$
where $^*$ denotes the $k$-dual module. But what does it mean? Does this mean that in this module a path $p$ corresponds to a linear map to $k$ denoted $p^*$ which takes $p$ to 1 and other paths to 0? If so, am I right in understanding that multiplication acts by "deletion", i.e. if $p^*\cdot q$ is $s^*$ such that in the quiver $qs=p$ (I am composing paths as maps, i.e. $q$ must start where $s$ begins)?
For example, consider a quiver $Q$ with 2 vertices and one edge $x: 1\to2$. I would say from the above description that in $I_i$ we have $x^*\cdot x=e_1^*$, since $(x^*\cdot x)(e_1)=x^*(x\cdot e_1)=1$?
By the way, I got lost in this because I was trying to understand the criteria for a quiver path algebra to be self-injective, for which I believe it is enough to have a permutation of vertices $\pi$ such that $P_i\cong I_{\pi(i)}$, but I also saw that it is enough to have $\text{soc}(P_i)\cong\text{top}(I_{\pi(i)})$. I would be thankful for any references on this topic!

REPLY [2 votes]: There are several things to unpack here.  Let's review the notions that already appear in your post, and we'll get to a nice description of injective modules at the end of this answer.
First, let's make our conventions explicit.  From your post, I see you are considering right modules, and also that you are composing arrows in a quiver from right to left (that is, if the target of the arrow $x$ is the source of the arrow $y$, then $yx$ is their composition).  Also, to simplify notations, let  $A=kQ/I$.
Before looking at the injective modules, let's describe the indecomposable projective modules.  For each vertex $i$, there is an idempotent $e_i$ in $A$ (corresponding to the path of length zero at $i$); the corresponding indecomposable projective right module is $e_iA$, which is the set of linear combinations of paths of $Q$ ending in $i$ (modulo $I$).  Note that the indecomposable projective left module is $Ae_i$, which is the set of linear combinations of paths of $Q$ starting in $i$ (modulo $I$).
Now, to get injective right modules, we will use the duality functor $(?)^* = \operatorname{Hom}_k(?,k)$  from the category of left $A$-modules to the category of right $A$-modules.  The duality functor sends projective modules to injective modules, and preserves indecomposability.  Thus the indecomposable injective right module corresponding to the vertex $i$ is
$$
  I_i \cong (Ae_i)^* = \operatorname{Hom}_k(Ae_i,k).
$$
The $A$-module structure of $\operatorname{Hom}_k(Ae_i,k)$ is defined as follows: for any $\phi\in \operatorname{Hom}_k(Ae_i,k)$ and any $a\in A$,  $\phi\cdot a$ is the map sending $x$ to $\phi(a\cdot x)$.
In order to obtain a more explicit description of $I_i$, we turn to quiver representations.  There is an equivalence of categories between the category of finite-dimensional right $A$-modules and the category of finite-dimensional representations of the opposite quiver $Q^{op}$ which satisfy the relations in $I$.  This equivalence is fairly straightforward to describe on objects: given a right $A$-module $M$, the corresponding representation is defined by

putting the vector space $Me_i$ at each vertex $i$, and
putting the linear map $Me_j \to Me_i: m \mapsto ma$ at each arrow $a:i\to j$.

Let's apply this to the injective module $I_i$.  For any vertex $j$, we have that
$$
I_ie_j = \operatorname{Hom}_k(Ae_i, k)e_j \cong \operatorname{Hom}_k(e_jAe_i, k).
$$
A basis of $I_ie_j$ is thus given by a dual basis of $e_jAe_i$, the space generated by paths from $i$ to $j$ in $Q$ (modulo $I$).
Finally (and perhaps most importantly in this discussion), for an arrow $a:h\to j$, the corresponding linear map $I_ie_j \to I_ie_h$ is dual to the map $\lambda_a:e_hAe_i \to e_jAe_i$ given by left multiplication by $a$.  In particular, if you fix a basis of $e_hAe_i$ and $e_jAe_i$ and let $[\lambda_a]$ be the matrix of the map $\lambda_a$ in those bases, then the matrix of the dual map $I_ie_j \to I_ie_h$ in the dual bases is the transpose of $[\lambda_a]$.
In particular, if you know a representation $V$ isomorphic to the left projective $A$-module $Ae_i$, then to get a representation isomorphic to the right injective $A$-module $I_i$, all you need to do is take the transpose of all the matrices defining $V$.


TITLE: Assuming existence to show existence - is this a circular argument?
QUESTION [0 upvotes]: The proof to follow is taken from a research paper in risk analysis which is in reviewing. I have been tasked with repeating the results in the paper, under a different risk measure, but I am not sure if the main argument in the paper is correct as it is presented, although I believe the main result should hold.
The context is the following. We are optimizing a risk function over a set of parameters $(a_i)_{i=1}^{m}, (b_i)_{i=1}^{m}$, with particular choices of $(a_i)_{i=1}^{m}, (b_i)_{i=1}^{m}$ denoted as $a^*, b^*$ respectively. The aim is to establish a criterion for optimality of the parameters.
The argument in the paper is as follows:
Begin by assuming that $a^*$ is chosen such that $(1)$ holds, where $(1)$ is a condition on $a^*$. It is then possible to show that there exists some $b^*$ such that a condition $(2)$ on $b^*$ holds. We then show that any $b^*$ satisfying $(2)$ is optimal given that $a^*$ is chosen according to $(1)$.
Assume next that $b^*$ is chosen such that $(2)$ holds. Then, one shows that $a^*$ can only be optimal parameters if $(1)$ holds.
The conclusion is that choosing $b^*$ such that $(2)$ holds is sufficient for obtaining optimal $(a^*, b^*)$.
There are two things that make me uneasy about this argument. Firstly, as we begin by assuming a condition on $a^*$, my instinct is to look next at what happens if $a^*$ does not satisfy $(1)$.
Secondly, it is easy to find $b^*$ such that $(2)$ does not hold. While existence of $b^*$ satisfying $(2)$ given $a^*$ such that $(1)$ holds, it seems to me like we assume existance in the second part of the proof. Then, we seemingly use the assumed existence of $b^*$ such that $(2)$ to show that we must choose $a^*$ such that $(1)$, and hence $b^*$ will exists such that $(2)$. Is this not a circular argument?

REPLY [1 votes]: Saying "any $b^*$ satisfying $(2)$ is optimal given that $a^*$ is chosen according to $(1)$" is a bit ambiguous, as it suggests that there is a closer relationship between $a^*$ and $b^*$ than simply "if an $a^*$ with $(1)$ exists, then a $b^*$ with $(2)$ also exists, but $a^*$ and $b^*$ are linked in no other way"; that is, it suggests that given $a^*$ with $(1)$, there might exist $b^*$ with $(2)$ which is not optimal, but then that $b^*$ is completely "separated" from $a^*$. I can assume that what is most likely meant is that the existence of an $a^*$ satisfying $(1)$ is enough to prove that all $b^*$ satisfying $(2)$ are optimal, but obviously this depends on the actual scenario.
This ambiguity is mirrored in saying "Assume next that $b^*$ is chosen such that $(2)$ holds. Then, one shows that $a^∗$ can only be optimal parameters if $(1)$ holds". Firstly, do you mean that if $(1)$ holds, $a^*$ is optimal, or merely that $a^*$ cannot be optimal if $(1)$ doesn't hold? Secondly, and more importantly, does the entire statement mean that the authors prove that it is enough for a $b^*$ satisfying $(2)$ to exist in order for either all $a^*$ satisfying $(1)$ (even those completely independent from $b^*$) to also be optimal or in order to have as a fact that $a^*$ cannot be optimal if they don't satisfy $(1)$, depending on the answer to the previous question?
In the best case when both issues amount to it there being enough for the corresponding parameters to exist, there remains only to be shown that there exists $a^*$ such that $(1)$ holds, as you noted.


TITLE: Explicit formula of a matrix
QUESTION [0 upvotes]: I have a problem below:
Given a full rank $m\times n$ matrix $H$ with $m<n$. Let $K$ is an $n \times (n-m)$ matrix in which each column is the element of the basis of the subspace $\{x\in \mathbb{R}^n:Hx=0\}.$ Now define
\begin{equation}
P=K(K^TAK)^{-1}K^TA.
\end{equation}
where $A$ is a given non singular symmetric matrix. Can we express $P$ explicitly in term of $H$ and doesn't contain matrix $K$? Thank you in advance. Any help will be appreciated.

REPLY [1 votes]: Try to consider
$$
P=I-Q=I-A^{-1}H^T(HA^{-1}H^T)^{-1}H.
$$
Then
$$
H(P+Q)=H[A^{-1}H^T(HA^{-1}H^T)^{-1}H+K(K^TAK)^{-1}K^TA]=H
$$
and
$$
K^TA(P+Q)=K^TA[A^{-1}H^T(HA^{-1}H^T)^{-1}H+K(K^TAK)^{-1}K^TA]=K^TA.
$$
Hence, if $[H^T,A^TK]$ is not rank-deficient, $P+Q=I$.


text:
/-
Copyright (c) 2022 Joachim Breitner. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Joachim Breitner
-/
import group_theory.order_of_element
import data.finset.noncomm_prod
import data.fintype.card

/-!
# Canonical homomorphism from a finite family of monoids

This file defines the construction of the canonical homomorphism from a family of monoids.

Given a family of morphisms `ϕ i : N i →* M` for each `i : ι` where elements in the
images of different morphisms commute, we obtain a canonical morphism
`monoid_hom.noncomm_pi_coprod : (Π i, N i) →* M` that coincides with `ϕ`

## Main definitions

* `monoid_hom.noncomm_pi_coprod : (Π i, N i) →* M` is the main homomorphism
* `subgroup.noncomm_pi_coprod : (Π i, H i) →* G` is the specialization to `H i : subgroup G`
   and the subgroup embedding.

## Main theorems

* `monoid_hom.noncomm_pi_coprod` coincides with `ϕ i` when restricted to `N i`
* `monoid_hom.noncomm_pi_coprod_mrange`: The range of `monoid_hom.noncomm_pi_coprod` is
  `⨆ (i : ι), (ϕ i).mrange`
* `monoid_hom.noncomm_pi_coprod_range`: The range of `monoid_hom.noncomm_pi_coprod` is
  `⨆ (i : ι), (ϕ i).range`
* `subgroup.noncomm_pi_coprod_range`: The range of `subgroup.noncomm_pi_coprod` is `⨆ (i : ι), H i`.
* `monoid_hom.injective_noncomm_pi_coprod_of_independent`: in the case of groups, `pi_hom.hom` is
   injective if the `ϕ` are injective and the ranges of the `ϕ` are independent.
* `monoid_hom.independent_range_of_coprime_order`: If the `N i` have coprime orders, then the ranges
   of the `ϕ` are independent.
* `subgroup.independent_of_coprime_order`: If commuting normal subgroups `H i` have coprime orders,
   they are independent.

-/

open_locale big_operators

section family_of_monoids

variables {M : Type*} [monoid M]

-- We have a family of monoids
-- The fintype assumption is not always used, but declared here, to keep things in order
variables {ι : Type*} [hdec : decidable_eq ι] [fintype ι]
variables {N : ι → Type*} [∀ i, monoid (N i)]

-- And morphisms ϕ into G
variables (ϕ : Π (i : ι), N i →* M)

-- We assume that the elements of different morphism commute
variables (hcomm : ∀ (i j : ι), i ≠ j → ∀ (x : N i) (y : N j), commute (ϕ i x) (ϕ j y))
include hcomm

-- We use `f` and `g` to denote elements of `Π (i : ι), N i`
variables (f g : Π (i : ι), N i)

namespace monoid_hom

/-- The canonical homomorphism from a family of monoids. -/
@[to_additive "The canonical homomorphism from a family of additive monoids.

See also `linear_map.lsum` for a linear version without the commutativity assumption."]
def noncomm_pi_coprod : (Π (i : ι), N i) →* M :=
{ to_fun := λ f, finset.univ.noncomm_prod (λ i, ϕ i (f i)) $
    by { rintros i - j -, by_cases h : i = j, { subst h }, { exact hcomm _ _ h _ _ } },
  map_one' := by {apply (finset.noncomm_prod_eq_pow_card _ _ _ _ _).trans (one_pow _), simp},
  map_mul' := λ f g,
  begin
    classical,
    convert @finset.noncomm_prod_mul_distrib _ _ _ _ (λ i, ϕ i (f i)) (λ i, ϕ i (g i)) _ _ _,
    { ext i, exact map_mul (ϕ i) (f i) (g i), },
    { rintros i - j - h, exact hcomm _ _ h _ _ },
  end }

variable {hcomm}

include hdec

@[simp, to_additive]
lemma noncomm_pi_coprod_mul_single (i : ι) (y : N i):
  noncomm_pi_coprod ϕ hcomm (pi.mul_single i y) = ϕ i y :=
begin
  change finset.univ.noncomm_prod (λ j, ϕ j (pi.mul_single i y j)) _ = ϕ i y,
  simp only [←finset.insert_erase (finset.mem_univ i)] {single_pass := tt},
  rw finset.noncomm_prod_insert_of_not_mem _ _ _ _ (finset.not_mem_erase i _),
  rw pi.mul_single_eq_same,
  rw finset.noncomm_prod_eq_pow_card,
  { rw one_pow, exact mul_one _  },
  { intros j hj, simp only [finset.mem_erase] at hj, simp [hj], },
end

omit hcomm

/-- The universal property of `noncomm_pi_coprod` -/
@[to_additive "The universal property of `noncomm_pi_coprod`"]
def noncomm_pi_coprod_equiv :
  {ϕ : Π i, N i →* M // pairwise (λ i j, ∀ x y, commute (ϕ i x) (ϕ j y)) }
    ≃ ((Π i, N i) →* M) :=
{ to_fun := λ ϕ, noncomm_pi_coprod ϕ.1 ϕ.2,
  inv_fun := λ f,
  ⟨ λ i, f.comp (monoid_hom.single N i),
    λ i j hij x y, commute.map (pi.mul_single_commute i j hij x y) f ⟩,
  left_inv := λ ϕ, by { ext, simp, },
  right_inv := λ f, pi_ext (λ i x, by simp) }

omit hdec

include hcomm

@[to_additive]
lemma noncomm_pi_coprod_mrange : (noncomm_pi_coprod ϕ hcomm).mrange = ⨆ i : ι, (ϕ i).mrange :=
begin
  classical,
  apply le_antisymm,
  { rintro x ⟨f, rfl⟩,
    refine submonoid.noncomm_prod_mem _ _ _ _ _,
    intros i hi,
    apply submonoid.mem_Sup_of_mem, { use i },
    simp, },
  { refine supr_le _,
    rintro i x ⟨y, rfl⟩,
    refine ⟨pi.mul_single i y, noncomm_pi_coprod_mul_single _ _ _⟩, },
end

end monoid_hom

end family_of_monoids

section family_of_groups

variables {G : Type*} [group G]
variables {ι : Type*} [hdec : decidable_eq ι] [hfin : fintype ι]
variables {H : ι → Type*} [∀ i, group (H i)]
variables (ϕ : Π (i : ι), H i →* G)
variables {hcomm : ∀ (i j : ι), i ≠ j → ∀ (x : H i) (y : H j), commute (ϕ i x) (ϕ j y)}
include hcomm

-- We use `f` and `g` to denote elements of `Π (i : ι), H i`
variables (f g : Π (i : ι), H i)

include hfin

namespace monoid_hom

-- The subgroup version of `noncomm_pi_coprod_mrange`
@[to_additive]
lemma noncomm_pi_coprod_range : (noncomm_pi_coprod ϕ hcomm).range = ⨆ i : ι, (ϕ i).range :=
begin
  classical,
  apply le_antisymm,
  { rintro x ⟨f, rfl⟩,
    refine subgroup.noncomm_prod_mem _ _ _,
    intros i hi,
    apply subgroup.mem_Sup_of_mem, { use i },
    simp, },
  { refine supr_le _,
    rintro i x ⟨y, rfl⟩,
    refine ⟨pi.mul_single i y, noncomm_pi_coprod_mul_single _ _ _⟩, },
end

@[to_additive]
lemma injective_noncomm_pi_coprod_of_independent
  (hind : complete_lattice.independent (λ i, (ϕ i).range))
  (hinj : ∀ i, function.injective (ϕ i)) :
  function.injective (noncomm_pi_coprod ϕ hcomm):=
begin
  classical,
  apply (monoid_hom.ker_eq_bot_iff _).mp,
  apply eq_bot_iff.mpr,
  intros f heq1,
  change finset.univ.noncomm_prod (λ i, ϕ i (f i)) _ = 1 at heq1,
  change f = 1,
  have : ∀ i, i ∈ finset.univ → ϕ i (f i) = 1 :=
    subgroup.eq_one_of_noncomm_prod_eq_one_of_independent _ _ _ _ hind (by simp) heq1,
  ext i,
  apply hinj,
  simp [this i (finset.mem_univ i)],
end

variable (hcomm)

@[to_additive]
lemma independent_range_of_coprime_order [∀ i, fintype (H i)]
  (hcoprime : ∀ i j, i ≠ j → nat.coprime (fintype.card (H i)) (fintype.card (H j))) :
  complete_lattice.independent (λ i, (ϕ i).range) :=
begin
  classical,
  rintros i f ⟨hxi, hxp⟩, dsimp at hxi hxp,
  rw [supr_subtype', ← noncomm_pi_coprod_range] at hxp,
  rotate, { intros _ _ hj, apply hcomm, exact hj ∘ subtype.ext },
  cases hxp with g hgf, cases hxi with g' hg'f,
  have hxi : order_of f ∣ fintype.card (H i),
  { rw ← hg'f, exact (order_of_map_dvd _ _).trans order_of_dvd_card_univ },
  have hxp : order_of f ∣ ∏ j : {j // j ≠ i}, fintype.card (H j),
  { rw [← hgf, ← fintype.card_pi], exact (order_of_map_dvd _ _).trans order_of_dvd_card_univ },
  change f = 1, rw [← pow_one f, ← order_of_dvd_iff_pow_eq_one],
  convert ← nat.dvd_gcd hxp hxi, rw ← nat.coprime_iff_gcd_eq_one,
  apply nat.coprime_prod_left, intros j _, apply hcoprime, exact j.2,
end

end monoid_hom

end family_of_groups

namespace subgroup

-- We have an family of subgroups
variables {G : Type*} [group G]
variables {ι : Type*} [hdec : decidable_eq ι] [hfin : fintype ι] {H : ι → subgroup G}

-- Elements of `Π (i : ι), H i` are called `f` and `g` here
variables (f g : Π (i : ι), H i)

section commuting_subgroups

-- We assume that the elements of different subgroups commute
variables (hcomm : ∀ (i j : ι), i ≠ j → ∀ (x y : G), x ∈ H i → y ∈ H j → commute x y)
include hcomm

@[to_additive]
lemma commute_subtype_of_commute (i j : ι) (hne : i ≠ j) :
  ∀ (x : H i) (y : H j), commute ((H i).subtype x) ((H j).subtype y) :=
by { rintros ⟨x, hx⟩ ⟨y, hy⟩, exact hcomm i j hne x y hx hy }

include hfin

/-- The canonical homomorphism from a family of subgroups where elements from different subgroups
commute -/
@[to_additive "The canonical homomorphism from a family of additive subgroups where elements from
different subgroups commute"]
def noncomm_pi_coprod : (Π (i : ι), H i) →* G :=
  monoid_hom.noncomm_pi_coprod (λ i, (H i).subtype) (commute_subtype_of_commute hcomm)

variable {hcomm}

include hdec

@[simp, to_additive]
lemma noncomm_pi_coprod_mul_single (i : ι) (y : H i) :
  noncomm_pi_coprod hcomm (pi.mul_single i y) = y :=
by apply monoid_hom.noncomm_pi_coprod_mul_single

omit hdec

@[to_additive]
lemma noncomm_pi_coprod_range : (noncomm_pi_coprod hcomm).range = ⨆ i : ι, H i :=
by simp [noncomm_pi_coprod, monoid_hom.noncomm_pi_coprod_range]

@[to_additive]
lemma injective_noncomm_pi_coprod_of_independent (hind : complete_lattice.independent H) :
  function.injective (noncomm_pi_coprod hcomm) :=
begin
  apply monoid_hom.injective_noncomm_pi_coprod_of_independent,
  { simpa using hind },
  { intro i, exact subtype.coe_injective }
end

variable (hcomm)

@[to_additive]
lemma independent_of_coprime_order [∀ i, fintype (H i)]
  (hcoprime : ∀ i j, i ≠ j → nat.coprime (fintype.card (H i)) (fintype.card (H j))) :
  complete_lattice.independent H :=
begin
  simpa using monoid_hom.independent_range_of_coprime_order
    (λ i, (H i).subtype) (commute_subtype_of_commute hcomm) hcoprime,
end

end commuting_subgroups

end subgroup

subset_name: curated


text:
\section{Locally Compact iff Open Neighborhood contains Compact Set}
Tags: Locally Compact Spaces

\begin{theorem}
Let $T = \struct{S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then
:$T$ is [[Definition:Locally Compact Space|locally compact]]
{{iff}}
:$\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is [[Definition:Compact Topological Subspace|compact]]
where $K^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of $K$.
\end{theorem}

\begin{proof}
=== Sufficient Condition ===
Let $T$ be [[Definition:Locally Compact Space|locally compact]].
Let $U \in \tau, x \in U$.
Since $T$ is [[Definition:Locally Compact Space|locally compact]] there exists a [[Definition:Neighborhood Basis|neighborhood basis]] $\mathcal B$ of $x$ consisting of [[Definition:Compact Topological Subspace|compact sets]].
From [[Set is Open iff Neighborhood of all its Points]], $U$ is a [[Definition:Neighborhood (Topology)|neighborhood]] of $x$.
By the definition of a [[Definition:Neighborhood Basis|neighborhood basis]]:
:$\exists K \in \mathcal B : K \subseteq U$.
By the definition of a [[Definition:Neighborhood (Topology)|neighborhood]] of $x$:
:$\exists V \in \tau : x \in V \subseteq K$
By the definition of the [[Definition:Interior (Topology)|interior]] of a [[Definition:Subset|subset]] then:
:$x \in V \subseteq K^\circ$
Thus
:$\exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is [[Definition:Compact Topological Subspace|compact]]
{{qed|lemma}}
=== Necessary Condition ===
Let
:$\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is [[Definition:Compact Topological Subspace|compact]]
Let $\mathcal B$ be the [[Definition:Set|set]] of all [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood (Topology)|neighborhoods]] of $x$.
It is shown that $\mathcal B$ is a [[Definition:Neighborhood Basis|neighborhood basis]] of $x$.
Let $N$ be a [[Definition:Neighborhood (Topology)|neighborhood]] of $x$.
By the definition of a [[Definition:Neighborhood (Topology)|neighborhood]] of $x$:
:$\exists U \in \tau : x \in U \subseteq N$
By assumption:
:$\exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is [[Definition:Compact Topological Subspace|compact]]
By the definition of the [[Definition:Interior (Topology)|interior]] of a [[Definition:Subset|subset]] $K^\circ$ is [[Definition:Open Set/Topology|open]].
By the definition of a [[Definition:Neighborhood (Topology)|neighborhood]] then $K$ is a [[Definition:Neighborhood (Topology)|neighborhood]] of $x$.
Hence $K \in \mathcal B$ and $K \subseteq U \subseteq N$.
As $N$ was arbitrary, then $B$ is a [[Definition:Neighborhood Basis|neighborhood basis]] of $x$ consisting of [[Definition:Compact Topological Subspace|compact sets]].
\end{proof}

config: wiki


TITLE: Open sets do not have measure zero
QUESTION [9 upvotes]: I want to show that if $a\lt b$, then $(a,b)$ is not of measure zero.
My idea was to show that any interval covering $(a,b)$ is such that the sum of the lengths of the intervals is always greater than $b-a$.
So I tried induction. Suppose $n=1$, where $n$ is the number of intervals covering $(a,b)$.
Then the interval consists of a single set of the form, say,  $(c,d)$ such that $(a,b)\subset (c,d)$. If this is the case then obviously, we would have $d-c \gt b-a$, where I have assumed that $c\lt a$ and $b \lt d$.
Now my questions: I am a little bit lost as to how to show it for $n+1$.
Am I even approaching it correctly?
Thanks for your help.

REPLY [16 votes]: First of all, your open set $(a,b)$ contains a closed interval $[c,d]$ with $c \lt d$. By its definition Lebesgue outer measure is monotone, hence it suffices to show that $\mu^{\ast}[c,d] \gt 0$.
I agree with William that induction won't help much. leo's argument is fine, but I prefer a more informal presentation of the argument to see what is going on.
Every cover of $[c,d]$ contains a finite subcover by compactness and dropping superfluous intervals only reduces the total length of the cover, so we may assume that we deal with finitely many intervals to begin with.
So, let $[c,d] \subset I_1 \cup \cdots \cup I_n$. Since $c$ is covered, we must have an interval $(a_1,b_1)$ in our family $\{I_1,\ldots,I_n\}$ with $c \in (a_1,b_1)$, or, equivalently, $a_1 \lt c \lt b_1$. If $b_1 \lt d$ then $b_1$ is covered by another interval $(a_2,b_2)$ in the family $\{I_1,\ldots,I_n\}$, so $a_2 \lt b_1 \lt b_2$. If $b_2 \lt d$ then $b_2$ is covered$\ldots$
Since we started with a finite family of intervals that covers $[c,d]$ we must at some point arrive at an interval $(a_j,b_j)$ with $a_j \lt d \lt b_j$ and we stop there.
In this way we produce a sequence $(a_1,b_1), (a_2,b_2), \ldots, (a_j,b_j)$ of intervals covering $[c,d]$ such that $a_{i+1} \lt b_{i}$ for $i = 1,\ldots,j-1$ and $a_1 \lt c$ as well as $d \lt b_j$. Now we can estimate
$$\begin{align*}
\sum_{i=1}^{n} |I_i| & \geq \sum_{i=1}^{j} (b_i - a_i) =(b_j - a_j) + (b_{j-1} - a_{j-1}) + \cdots + (b_1 - a_1) \\
&=
b_j + \underbrace{(b_{j-1} - a_j)}_{\geq 0} + \underbrace{(b_{j-2} - a_{j-1})}_{\geq 0} + \cdots + \underbrace{(b_1 - a_2)}_{\geq 0} - a_1 \\
&\geq b_j - a_1 \geq  d-c.
\end{align*}$$
This works with an arbitrary finite family of intervals $\{I_1, \ldots,I_n\}$ covering $[c,d]$. Hence this shows that Lebesgue outer measure of $[c,d]$ is at least $\mu^\ast [c,d] \geq d-c$ and since it is clear that it is at most $d-c$, we have $\mu^\ast [c,d] = d-c$.
Finally, we should combine the above with what Carl said, namely that the outer measure of $[a,b]$ is equal to the outer measure of $(a,b)$ and we're done.