TITLE: the length of a perpendicular inside of a triangle
QUESTION [1 upvotes]: Let $ABC$ a triangle with its area $S$ and $AB=c$, $BC=a$, $AC=b$.
The median from $A$ intersects the bisecting of the angle $ABC$ in $X$.
If $x=\ $the length of the perpendicular from $X$ on $BC$ than $x$ is:
A: $2S/(a+b+c)$
B: $2S/(2c+a)$
C: $(c+a)/2$
D: $(2c+a)/3$
E: $2S/(2b+a)$
I made a sketch and used the bisecting theorem and get that $AX/XM=2c/a$ and from the median theorem I got the length of $AX$ and $XM$.
With the Pythagoras theorem if I find $QM$ ($Q$ being the projection of $X$ on $BC$) I could find $x$.

REPLY [0 votes]: You are almost there: if $H$ is the perpendicular projection from $A$ on $BC$ and $Q$ is the perpendicular projection from $X$ on $BC$ ($XQ=x$), then:
$$
AH/XQ=AM/XM=1+AX/XM
\quad\text{and}\quad
AH=2S/a.
$$


TITLE: Lebesgue Dominated Convergence Theorem: Clarification
QUESTION [1 upvotes]: Suppose $\{f_n\}$ is a sequence of complex measurable functions such that $$f(x)=\lim\limits_{n\to \infty} f_n(x)$$ exists for every $x\in X$. If there is a function $g\in L(\mu)$ such that $$|f_{n}(x)|\leq g(x)$$ then $f\in L(\mu)$ $$\lim\limits_{n\to \infty} \int_X |f-f_n|d\mu=0$$ and $$\lim\limits_{n\to \infty} \int_X f_nd\mu=\int_X f d\mu$$

My problem lies in understanding what $|f_{n}(x)|\leq g(x)$ means given that it wasn't specified that $\operatorname{Im}(g)=0$. Is it assumed that $g$ is always real valued?

REPLY [2 votes]: Just pretend it said, "If there is a real-valued function $g \in L(\mu)$..."  The author probably thinks it goes without saying that $g$ is real-valued, because otherwise the statement $|f_n(x) | \leq g(x)$ would not make sense.


TITLE: Convergence of quantum cohomology
QUESTION [4 upvotes]: For which manifolds or varieties is quantum cohomology known to converge? Are there any manifolds for which quantum cohomology is known to not converge? I seem to have the impression that quantum cohomology is known to converge for Fano manifolds or maybe toric Fano manifolds, but I don't know if this is actually true.
By "convergence" I mean convergence of, e.g., the genus 0 Gromov-Witten potential.

REPLY [1 votes]: There are many more examples of convergent quantum cohomology generating functions where the proof follows from mirror symmetry. MS for, e.g., the quintic says the QC generating function is the power series expansion of a hypergeometric function, and thus it converges.


TITLE: $C=A+B$ diagonalizable $\implies$ $A$ and $B$ simultaneously diagonalizable?
QUESTION [1 upvotes]: If $A$ and $B$ are simultaneously diagonalizable matrices, then clearly $A+B$ is diagonalizable. But is the converse true ? i.e if $C=A+B$ is diagonalizable, are $A$ and $B$ simultaneously diagonalizable?

REPLY [2 votes]: $A=\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}$ and $B=\begin{pmatrix}
1 & -1 \\
0 & 1
\end{pmatrix}$are not diagonalizable but $A+B$ is diagonal.


TITLE: Show that a set is a right transversal for $S_3$ in $S_4$
QUESTION [3 upvotes]: I am given the set A ={e,(14),(24),(34)}
I'm supposed to show that this set is a right transversal for S3 in S4, meaning that every right coset of S3 contains exactly one element of A.
I'm getting stuck because (14) is not an element in S3...any help would be appreciated.

REPLY [1 votes]: Let $G=S_4,\ H=S_3$ where $H$ is regarded as pemutations of $\{1,2,3\}$ and so is a subset (and a subgroup) of $G$ regarded as permutations of $\{1,2,3,4\}.$
Then none of $(14),(24),(34)$ are in $H,$ so we can form the three possibly distinct right cosets $H(14),\ H(24),\ H(34)$ and none of these are $H.$
If we can show any two of these are in fact distinct, we can finish. Suppose for example that $H(14)=H(24).$ Then multiplying each on the right by $(14)$ and using that $(14)(14)=e,$ we would have $H=H(24)(14)=H(124).$ But that would imply that $(124)$ is in $H$ which it is not.
Other cases are similar. We now have four pairwise disjoint cosets $He,\ H(14),\ H(24),\ H(34)$ each of size 6 so these are the distinct right cosets of $H$ in $G.$ We have already shown that no two of these right cosets can contain two elements of $A=\{e,\ (14),\ (24),\ (34)\}.$


TITLE: LP program: does the decision variable coefficients affect the problem?
QUESTION [0 upvotes]: I just started reading up on linear programming by myself, and am a bit confused by the decision variable coefficients $c_j$, in the objective function $ \sum_j
 c_jx_j$.
Do they matter? I mean, if I replace $4x + 2y$ with $2x + y$, or with $8x + 4y$, i.e. keep them at the same ratio, will it effect the optimum solution, or just the value of the function itself?
I'm not sure, but surely many Simplex method problems would be easier to calculate if we could take advantage of above, if it were true?
Also, if we didn't know the coefficients, how would we graphically solve it? Right now I'm just practicising the intuition with 2 variables, and if I did just call $c_1 = c_1$ and $c_2 = kc_1$, how would I go about solving it graphically, i.e. with the slopes and everything?

REPLY [0 votes]: Provided you don't multiply by a negative value or by zero, two objective functions that are simply multiples of each other will yield the same optimal point (but different values of objective function).


TITLE: How many non-negative integer solutions exist for: $x+y+z=48$ where, $x<y<z$?
QUESTION [2 upvotes]: I want to find the number of non-negative integral solutions to the following:
$x+y+z=48$ where, $x<y<z$.
The answer is apparently 192 and the solution provided is $$\frac{\dbinom{50}{2}-\dbinom{3}{2}\cdot24 - 1}{3!}$$
The method I used was I guessed that the lowest value of $z$ will be 17, I.E. $(15,16,17)$. When $x$ was an odd number, the number of solutions I got were of the form $3K+1$ and when $x$ was even, the number of solutions were of the form $3K-1$. So listing them all down gets you the series: $1,2,4,5,7,8,...,23$
The sum of all the solutions is 192. I did it this way, however this calculation could have been tedious had the number on $R.H.S$ been any larger.
Now after reading the solution, I can understand bits and parts of it. $\binom{50}{2}$ is the total number of solutions , ignoring the $x<y<z$ constraint. The division by 3! I guess is justified by noting that $x$ is the smallest among all the solutions in $\frac{1}{3^{rd}}$ of the cases while in those cases, $\frac{1}{2}$ of them have $y<z$ so we are multiplying by $\frac{1}{6}$. Again we are subtracting 1 because there exists a solution {$16,16,16$} which has to be excluded from the solution set.
My doubts are
$1)$how do you get the "$-\binom{3}{2}\cdot24$" and
$2)$ Is there any different way that one can use to solve this question?
P.S. I have seen similar questions(not exactly the same) on MSE but I am not able to ask my doubts on those posts. So pls point me in the right direction. Thanks!

REPLY [5 votes]: The approach can be summarized as follows.  First, we find the solutions to $x + y + z = 48$ such that $x,y,z$ are all distinct.  Then, we note that each valid $x<y<z$ is represented $3!$ times among this set, since each permutation of $x,y,z$ gives a solution to $x+y+z=48$ with distinct values, hence the division by $3!$.
The "stars and bars" formula tells us that the total number of non-negative solutions to $x+y+z = 48$ is
$$
\binom{48 + 3 - 1}{3 - 1} = \binom{50}2,
$$
as you have stated in the question. Now, we subtract off any solutions where $x,y,z$ are not distinct.
There is one solution where $x = y = z$, namely $x = y = z = 16$.
Now, we want to count the solutions where exactly two are equal.  We can uniquely generate any such solution with the following process:

Select $2$ values that will be equal (for instance, $x$ and $z$).  There are $\binom 32 = 3$ such choices.
Select a common value for the equal pair.  This can be any non-negative integer from $0$ to $24$ inclusive except for $16$ (since choosing $16$ would lead to all three being equal).  There are $24$ valid choices for this value.
Once we have made the above choices, the remaining number is forced, so there are no more choices to be made.

Since the above process uniquely generates all possibilities, we can conclude that there are $\binom 32 \cdot 24$ total solutions to $x+y+z = 48$ in which exactly two values are equal.
With that, we may finally conclude that the desired quanitity is indeed
$$
\frac{\binom{50}2 - \binom 32 \cdot 24 - 1}{3!}.
$$

As for an alternative approach: using the inclusion-exclusion principle would yield the numerator
$$
\begin{align}
[\text{# of triples without repeats}] &=
[\text{total # of triples}]
\\ & \quad - [\text{# of triples with } x=y]
\\ & \quad - [\text{# of triples with } y=z]
\\ & \quad - [\text{# of triples with } x=z]
\\ & \quad + 2\cdot [\text{# of triples with } x=y=z]
\\ & = \binom{50}2 - 3 \cdot 25 + 2,
\end{align}
$$
which is equal to the $\binom{50}2 - \binom 32 \cdot 24 - 1$ we got above.


TITLE: If X is second-countable, then X is Lindelöf space.
QUESTION [1 upvotes]: Suppose that $X$ has countable basis, then every open covering of $X$ contains a countable subcollection covering $X$.
Proof
{$U_i;i\in I$} is open cover for ($X,\tau$) and $B$ = {$V_n;n\in N$} is countable base for topology $\tau$.We have that $X=\cup_i U_i$.For every point $x$ in $X$ let's choose any $U_i$ s.t. $x \in U_i$.Denote that $U_i$ as $U_x$.There exists $V_x \in B$ s.t. $x\in V_x\subset U_x$.Collection of $V_x$ is countable.For every $V_x$ choose $U_x$ s.t. $V_x\subset U_x$.Because $X=\cup_xV_x$ follows that $X=\cup U_x$ and number of $U_x$-s are countable.
I understood proof only last part I can't understand.Why we need that number of $U_x$ to be countable?
We have already open cover for $X$ and found {$V_x$} which is countable subcollection covering $X$.

REPLY [0 votes]: {$U_i;i\in I$} is open cover for ($X,\tau$) and $B$ = {$V_n;n\in N$} is countable base for topology $\tau$.We have that $X=\cup_i U_i$.For every point $x$ in $X$ let's choose any $U_i$ s.t. $x \in U_i$.Denote that $U_i$ as $U_x$.There exists $V_x \in B$ s.t. $x\in V_x\subset U_x$.
Now as these $\{V_{x}\}$ which correspond to the $U_{x}$'s is a subcollection of $B$. Hence is countable as a subset of countable set is countable. Now it's just a matter of indexing. Explicitly we were looking at
$U_{0}=\{U_{x}\in \{U_{i}\}_{i\in I} : x\in V_{x}\subset U_{x}\}$. correspondingly we have $B_{0}=\{V_{x}: V_{x}\subset U_{x}\}$. These sets $B_{0}$ and $U_{0}$ are in bijection and $B_{0}$ being a subset of countable set is countable.
Write out all such $V_{x}$'s as a sequence say $\{V_{1},V_{2},.....\}$. Correspondingly we have $\{U_{1},U_{2},....\}$.
Thus $\{U_{k}\}_{k=1}^{\infty}$ is a countable subcover of $\{U_{i}\}_{i\in I}$


TITLE: Measures induced by integral and by measurable mapping
QUESTION [2 upvotes]: Suppose $(X, \mathcal{F}, u)$ is a measure space.
$f: X \rightarrow \mathbb{R}$ is a measurable mapping. Then $f$ induces a measure $v_f$ on $(X, \mathcal{F})$ as $v_f(A):=\int_A f du$,
$g: X \rightarrow X$ is a measurable mapping. Then $g$ induces  a measure $w_g$ on $(X, \mathcal{F})$ as $w_g(A):=u(g^{-1}(A))$.
My question are:

Are  there  some relations between
$V:=\{ v_f: \forall \text{
    measurable }f: X \rightarrow
    \mathbb{R} \}$ and $W:=\{ w_g:
    \forall \text{ measurable }g: X
    \rightarrow X \}$?
One thing I noticed is that $v_f$
must be absolutely continuous wrt
$u$, while $w_g$ may not be required
as such. Is it true that $V
    \subseteq W$?

Thanks and regards!

REPLY [3 votes]: As stated, (2) is not true. Take $\mu$ be a measure so that $\mu(X) = 1$. Let $f$ be the function $f\equiv 2$. Then $v_f = 2\mu$, and so $v_f(X) = 2$. By the definition of $w_g$, for any self-map $g:X\to X$, $w_g(X) = \mu(g^{-1}(X)) \leq \mu(X) = 1$. Hence $V$ is not a subset of $W$.
The reverse is also not true. Take $\mu$ to be an arbitrary atomless measure. Let $X_0\in \mathcal{F}$ be a set with positive measure, and let $\{x_0\}\in \mathcal{F}$ be a point. Then the map $g:X\to X$ such that $g|_{X\setminus X_0} = Id$, and that $g(y) = x_0$ for any $y\in X_0$ is measurable. But $\mu(\{x_0\}) = 0$, while $w_g(\{x_0\}) = \mu(X_0) > 0$, so $w_g$ is not absolutely continuous w.r.t $\mu$.
In general I don't think there can be any relationships between those two sets you defined: one is the pushforward of $\mu$ under automorphisms of $X$, the other being essentially the class of all absolutely continuous measures w.r.t. $\mu$, it is not clear to me why one may expect the two to be connected, unless perhaps very stringent requirements on what $\mathcal{F}$ and $\mu$ one is allowed to take is made.


text:
\begin{document}

\title{\bfseries\scshape{LIGHTLIKE OSSERMAN SUBMANIFOLDS OF SEMI-RIEMANNIAN MANIFOLDS}}
\author{\bfseries\scshape Cyriaque Atindogbe$^{a}$\thanks{atincyr@imsp-uac.org}, \bfseries\scshape Oscar Lungiambudila$^{a}$\thanks{lungiambudila@yahoo.fr ; lungiaoscar@imsp-uac.org (Corresponding author)} , \bfseries\scshape Jo\"{e}l Tossa$^{a}$\thanks{joel.tossa@imsp-uac.org ; joel.tossa@uac.imsp.bj }\\
$^{a}$Institut de Math\'ematiques et de Sciences Physiques (IMSP)\\Universit\'e d'Abomey-Calavi\\P.B. 613, Porto-Novo, B\'enin}

\date{}
\maketitle \thispagestyle{empty} \setcounter{page}{1}

\begin{abstract}
In this paper, we study
Jacobi operators associated to algebraic curvature maps (tensors)
on lightlike submanifolds $M$. We investigate conditions for an induced Riemann curvature tensor to be an algebraic curvature tensor on $M$. We introduce the notion of lightlike Osserman submanifolds and an example of $2$-degenerate Osserman metric is given. Finally we give some results of symmetry properties on lightlike hypersurfaces from Osserman condition.
\end{abstract}

\noindent {\bf AMS Subject Classification:} 53B25, 53C25, 53C50

\vspace{.08in} \noindent \textbf{Keywords}: Algebraic curvature tensor, Admissible pair of screens, Pseudo-Jacobi operator, Lightlike Osserman submanifold.
\section{Introduction}
The curvature tensor is a central concept in differential geometry.
According to R. Osserman (\cite{Oss}), one could argue that it is a central one.
But the curvature tensor is in general difficult to deal with and the problem which aims to relate
 algebraic properties of the Riemann curvature tensor to the geometry of the manifold is in general
 difficult to be solved. Many authors study the geometric consequences that followed if various
natural operators defined in terms of the curvature tensor are assumed to have constant
eingenvalues on the unit fibre bundle. Osserman has studied the spectral properties of Jacobi
operator in (\cite{Oss}). This operator has been extensively studied in the Riemannian and the
pseudo-Riemannian context. E. Garc\'{\i}a-R\'{\i}o, D. N. Kupeli and R. V\'{a}quez-Lorenzo have studied Osserman condition in Pseudo-Riemannian geometry in (\cite{Ga-Ku-Va}). We refer to (\cite{Ga-Ku-Va}) for an extensive bibliography. In the degenerate geometry, C. Atindogbe and K. L. Duggal have stadied Pseudo-Jacobi operators and introduce the Osserman condition on lightlike hypersurfaces in (\cite{Ati-Dug}). In the present paper we extend this study on $r$-degenerate submanifolds and give some characterization results of symmetry properties on lightlike Osserman hypersurfaces.

Let $(M,g)$ be a semi-Riemannian manifold and $u \in M$. An element $F\in \otimes^{4}T_{u}^{\ast}M$ is said to
be an {\itshape algebraic curvature map (tensor)} on $T_{u}M$ if it satisfies the following symmetries:
\begin{eqnarray}\label{int1}
 F(x,y,z,w)=-F(y,x,z,w)=F(z,w,x,y)\crcr
 F(x,y,z,w)+F(y,z,x,w)+F(z,x,y,w)=0
\end{eqnarray}
In semi-Riemannian geometry, the Riemann curvature tensor $R$ is an algebraic curvature tensor on the tangent space $T_{u}M$ for every point $u \in M$.

For an algebraic curvature map (tensor) $F$ on $T_{u}M$, the associated Jacobi operator $J_{F}(x)$ with respect to $x\in T_{u}M$, is the self-adjoint linear map on $T_{u}M$ characterized by identity
\begin{equation}\label{int2}
 g(J_{F}(x)y,z)=F(y,x,x,z),\quad\forall y,z\in T_{u}M.
\end{equation}
Since $J_{F}(cx)=c^{2}J_{F}(x)$, the natural domains of Jacobi operators
$J_{F}(\cdotp)$ are the unit pseudo-sphere of unit timelike or unit spacelike vectors
$$S_{u}^{\pm}(M) := \left\{x \in T_{u}M : g(x,x) = \pm 1\right\}.$$

The tensor $F\in \otimes^{4}T_{u}^{\ast}M$ is said to be a spacelike
(resp. timelike) Osserman tensor on $T_{u}M$ if $Spec\{J_{F}\}$
is constant on the pseudo-sphere $S_u^{+}(M)$ (resp $S_u^{-}(M)$ ).

In degenerate geometry, it is known that the induced metric on an $r$-degenerate submanifold of a semi-Riemannian manifold has a non-trivial kernel, so the relation (\ref{int2}) is not well defined in the usual way. Also, except a totally geodesic case, in general induced Riemann curvature tensors on a lightlike submanifolds are not algebraic curvature tensors. Therefore, in section 3, we study conditions on a lightlike submanifold to have an induced algebraic Riemann curvature tensor.

In section 4, we have extended pseudo-Jacobi operator on
$r$-degenerate submanifolds, by using non-degenerate metric $\tilde{g}$
associated to the degenerate metric $g$ (see Preliminaries). We introduce and study a class of lightlike Osserman submanifolds. Some results are obtained (Theorem \ref{th4.1}, Proposition \ref{th4.2} and Theorem \ref{th4.3}) and We exhibit an example of $2$-degenerate metric verifying Osserman condition.

It is natural to impose condition on semi-Riemannian manifold that its Riemannian curvature
tensor $R$ be parallel, that is, have vanishing covariant differential, $\nabla R$, where   $\nabla$ is the Levi-Civita connection on semi-Riemannian manifold and $R$ is the corresponding curvature tensor. Such a manifold is said to be locally symmetric. This class of manifolds contains one of manifolds of constant curvature. A semi-Riemannian manifold is called semi-symmetric, if $R\cdot R=0$, which is the integrability condition of $\nabla R=0$. The semi-symmetric manifolds have been classified, in Riemannian case, by Szabo in \cite{Sza1} and \cite{Sza2}. A semi-Riemannian manifold is called Ricci semi-symmetric, if $R\cdot Ric = 0$.

In section 5, we are interested to answer to the following question: ``Are conditions $\nabla R = 0$ and $R\cdot R = 0$ equivalent on lightlike hypersurfaces of semi-Riemannian manifolds?'' These equivalences are not true in general. In virtue of result given by Sahin (\cite{Sahin}, Theorem 4.2), we see that the conditions $\nabla R = 0$ and $R\cdot R = 0$ are equivalent on lightlike hypersurfaces of semi-Euclidean spaces under conditions $Ric(\xi,X)=0$ and $A_{N}\xi$ a vector field non-null. In this paper we give an affirmative answer to this question for lightlike Osserman hypersurfaces of semi-Riemannian manifolds of constant sectional curvature, under condition $A_{N}\xi$ a vector field non-null (Corollary \ref{th5.4}). Also, in the same section, we show that lightlike Osserman hypersurfaces of semi-Riemannian manifolds of constant sectional curvature are Ricci semi-symmetric (Theorem \ref{th5.5}).

\section{Preliminaries}
\subsection{Lightlike submanifolds of semi-Riemannian manifolds}
We follow (\cite{Dug}) for the notations and formulas used in this paper. Let $(\overline{M},\overline{g})$ be an $(m+n)$-dimensional semi-Riemannian manifold of constant
 index $\nu$, $1\leqslant\nu<m+n$ and $M$ be a submanifold of $\overline{M}$ of codimensional $n$. We assume that both $m$ and $n$ are $\geqslant 1$. At a point $u\in M$, we define the orthogonal complement
 $T_{u}M^{\perp}$ of the tangent space $T_{u}M$ by
$$T_{u}M^{\perp}=\{X_{u}\in T_{u}\overline{M} : \overline{g}(X_{u},Y_{u})=0,\ \forall Y_{u}\in T_{u}M\} $$

We put $RadT_{u}M=RadT_{u}M^{\perp}=T_{u}M\cap T_{u}M^{\perp}$. The submanifold $M$ of $\overline{M}$ is
said to be an r-lightlike submanifold (one suppose that the index of $\overline{M}$ is $\nu\geqslant r$), if the mapping
$$RadTM : u \in M\longrightarrow RadT_{u}M$$
defines a smooth distribution on $M$ of rank $r>0$. We call $RadTM$ the radical distribution on
$M$. In the sequel, an $r$-lightlike submanifold will simply be called a lightlike submanifold and
 $g$ is lightlike metric, unless we need to specify $r$.\\
Let $S(TM)$ be a screen distribution which is a semi-Riemannian complematary distribution of $Rad(TM)$ in $TM$, that is,
\begin{equation}\label{pr1}
 TM=RadTM\perp S(TM).
\end{equation}
We consider a screen transversal vector bundle $S(TM^{\perp})$, which is a semi-Riemannian complementary vector bundle of $Rad(TM)$ in $TM^{\perp}$. Since, for any local frame $\{\xi_{i}\}$ of $Rad(TM)$, there exists a local frame $\{N_{i}\}$ of sections with values in the orthogonal complement of $S(TM^{\perp})$ in $S(TM)^{\perp}$ such that $\overline{g}(\xi_{i},N_{j})=\delta_{ij}$ and $\overline{g}(N_{i},N_{j})=0$, it follows that there exists a lightlike transversal vector bundle $ltr(TM)$ locally spanned by $\{N_{i}\}$ (see \cite{Dug}, p144). Let $tr(TM)$ be complementary (but not orthogonal) vector bundle to $TM$ in $T\overline{M}_{|M}$. Then
\begin{equation}\label{pr2}
 tr(TM)=ltr(TM)\perp S(TM^{\perp}),
\end{equation}
\begin{equation}\label{pr3}
T\overline{M}_{|M}=TM\oplus tr(TM)=S(TM)\perp (RadTM\oplus ltr(TM))\perp S(TM^{\perp}).
\end{equation}
Although $S(TM)$ is not unique, it is canonically isomorphic to the factor vector bundle $TM/ RadTM$ (\cite{Kup}).\\
Throughout this paper, we will discuss the dependence (or otherwise) of the results on induced objects and refer to (\cite{Dug}) for their transformation equations. We say that a submanifold $(M,g,S(TM),S(TM^{\perp}))$ of $\overline{M}$ is\\
(1) r-lightlike if $r<min\{m,n\}$;\\
(2) coisotropic if $r=n$, hence, $S(TM^{\perp})=\{0\}$;\\
(3) isotropic if $r=m<n$, hence $S(TM)=\{0\}$;\\
(4) totally lighlike if $r=m=n$, hence $S(TM)=\{0\}=S(TM^{\perp})$.\\
The Gauss and Weingarten equations are
\begin{equation}\label{pr4}
\overline{\nabla}_{X}Y=\nabla_{X}Y+h(X,Y),\quad\forall X,Y\in\Gamma(TM),
\end{equation}
\begin{equation}\label{pr5}
\overline{\nabla}_{X}V=-A_{V}X+\nabla_{X}^{t}V,\quad\forall X\in\Gamma(TM),\ V\in\Gamma(tr(TM)),
\end{equation}
where $\{\nabla_{X}Y,A_{V}X\}$ and $\{h(X,Y),\nabla_{X}^{t}V\}$ belong to $\Gamma(TM)$ and $\Gamma(tr(TM))$, respectively. $\nabla$ and $\nabla^{t}$ are linear connections on $M$ and on the vector bundle $tr(TM)$, respectively. Moreover, we have
\begin{equation}\label{pr6}
\overline{\nabla}_{X}Y=\nabla_{X}Y+h^{l}(X,Y)+h^{s}(X,Y),\quad\forall X,Y\in\Gamma(TM),
\end{equation}
\begin{equation}\label{pr7}
\overline{\nabla}_{X}N=-A_{N}X+\nabla_{X}^{l}N+D^{s}(X,N),\quad\forall X\in\Gamma(TM),\ N\in\Gamma(ltr(TM)),
\end{equation}
\begin{equation}\label{pr8}
\overline{\nabla}_{X}W=-A_{W}X+\nabla_{X}^{s}W+D^{l}(X,W),\quad\forall X\in\Gamma(TM),\ W\in\Gamma(S(TM^{\perp})).
\end{equation}
Denote the projection of $TM$ on $S(TM)$ by $P$. Then, by using (\ref{pr6})-(\ref{pr8}) and taking into account that $\overline{\nabla}$ is a metric connection, we obtain
\begin{equation}\label{pr9}
\overline{g}(h^{s}(X,Y),W)+\overline{g}(Y,D^{l}(X,W))=g(A_{W}X,Y),
\end{equation}
\begin{equation}\label{pr10}
\overline{g}(D^{s}(X,N),W)=\overline{g}(N,A_{W}X)
\end{equation}
From the decomposition (\ref{pr1}) of the tangent bundle of lighlike submanifold, we have
\begin{equation}\label{pr11}
\nabla_{X}Y=\stackrel{*}{\nabla}_{X}PY+\stackrel{*}{h}(X,PY),\quad\forall X,Y\in\Gamma(TM),
\end{equation}
\begin{equation}\label{pr12}
\nabla_{X}\xi=-\stackrel{*}{A}_{\xi}X+\stackrel{*}{\nabla^{t}}_{X}\xi,\quad\forall X\in\Gamma(TM),\ \xi\in\Gamma(RadTM).
\end{equation}
By using the abovee equations, we obtain
\begin{equation}\label{pr13}
\overline{g}(h^{l}(X,PY),\xi)=g(\stackrel{*}{A}_{\xi}X,PY),
\end{equation}
\begin{equation}\label{pr14}
\overline{g}(\stackrel{*}{h}(X,PY),N)=g(A_{N}X,PY),
\end{equation}
\begin{equation}\label{pr15}
\overline{g}(h^{l}(X,\xi),\xi)=0,\qquad\stackrel{*}{A}_{\xi}\xi=0.
\end{equation}
In general, the induced connection $\nabla$ on $M$ is not a metric connection. Since $\overline{\nabla}$ is a metric connection, by using (\ref{pr6}) we get
\begin{equation}\label{pr16}
(\nabla_{X}g)(Y,Z)=\overline{g}(h^{l}(X,Y),Z)+\overline{g}(h^{l}(X,Z),Y),\quad\forall X,Y,Z\in\Gamma(TM).
\end{equation}
However, it is important to note that $\stackrel{*}{\nabla}$ is a metric connection on $S(TM)$.\\
We denote the Riemann curvature tensors of $\overline{M}$ and $M$ by $\overline{R}$ and $R$ respectively. The Gauss equation for $M$ is given by
\begin{eqnarray}\label{pr17}
\overline{R}(X,Y)Z&=&R(X,Y)Z+A_{h^{l}(X,Z)}Y-A_{h^{l}(Y,Z)}X+A_{h^{s}(X,Z)}Y\crcr
&&-A_{h^{s}(Y,Z)}X+(\nabla_{X}h^{l})(Y,Z)-(\nabla_{Y}h^{l})(X,Z)\crcr
&&+D^{l}(X,h^{s}(Y,Z))-D^{l}(Y,h^{s}(X,Z))+(\nabla_{X}h^{s})(Y,Z)\crcr
&&-(\nabla_{Y}h^{s})(X,Z)+D^{s}(X,h^{l}(Y,Z))-D^{s}(Y,h^{l}(X,Z)),
\end{eqnarray}
Therefore,
\begin{eqnarray}\label{pr17.1}
\overline{R}(X,Y,Z,PU)&=&R(X,Y,Z,PU)+\overline{g}(\stackrel{*}{h}(Y,PU),h^{l}(X,Z))-\overline{g}(\stackrel{*}{h}(X,PU),h^{l}(Y,Z)) \crcr
&&+\overline{g}(h^{s}(Y,PU),h^{s}(X,Z))-\overline{g}(h^{s}(X,PU),h^{s}(Y,Z))
\end{eqnarray}
for any $X,Y,Z\in\Gamma(TM)$. Note that for the coisotropic, isotropic and totally lighlike submanifolds, in (\ref{pr17}), we have $h^{s}=0$, $h^{l}=0$ and $h^{l}=h^{s}=0$, respectively.
\subsection{Pseudo-inversion of $r$-degenerate metrics}
In this section we mention (case 1 or 2) to refer to $r$-lightlike submanifolds with $0<r<min\{m,n\}$ or coisotropic submanifolds and also (case 3 or 4) for isotropic submanifolds or totally lightlike submanifolds. We recall from (\cite{Lun}) the following result. Consider on $M$ the local frames $\{\xi_{i}\}$ and $\{N_{i}\}$ of sections of $RadTM$ and $ltr(TM)$ satisfying $\overline{g}(N_{i},\xi_{j})=\delta_{ij}$. Consider on $M$ the 1-forms $\eta_{i},i=1,...,r$ defined by $\eta_{i}(\cdotp)=\overline{g}(N_{i},\cdotp)$. Any vector field $X$ on $M$ is expressed on a coordinate neighbourhood $\mathcal{U}$ as follows,
\begin{equation}\label{pr18}
X=PX+\sum_{i=1}^{r}\eta_{i}(X)\xi_{i}\qquad (case\ 1~or~2)
\end{equation}
\begin{equation}\label{pr19}
X=\sum_{i=1}^{m}\eta_{i}(X)\xi_{i}\qquad\qquad (case\ 3~or~4)
\end{equation}
Now, we define $\flat_{g}$ by
\begin{eqnarray*}
 \flat_{g}:\Gamma(TM)&\longrightarrow& \Gamma(T^{*}M) \crcr
           X & \longmapsto & X^{\flat_{g}}
\end{eqnarray*}
such that for all $Y\in\Gamma(TM)$,
\begin{eqnarray}
&&X^{\flat_{g}}(Y)=g(X,Y)+\sum_{i=1}^{r}\eta_{i}(X)\eta_{i}(Y)\qquad (case\ 1~or~2)\label{pr20} \\
&&X^{\flat_{g}}(Y)=\sum_{i=1}^{m}\eta_{i}(X)\eta_{i}(Y)\qquad\qquad (case\ 3~or~4)\label{pr21}
\end{eqnarray}
The map $\flat_{g}$ is an isomorphism of $\Gamma(TM)$ onto $\Gamma(T^{*}M)$, its inverse is denoted $\sharp_{g}$.  For $X\in\Gamma(TM)$ (resp. $\omega\in\Gamma(T^{*}M)$), $X^{\flat_{g}}$ (resp. $\omega^{\sharp_{g}}$) is called the dual 1-form of $X$ (resp. the dual vector field of $\omega$) with respect to the degenerate metric $g$.\\
We define a $(0,2)$-tensor $\tilde{g}$ by, for any $X,Y\in\Gamma(TM)$,
\begin{equation}\label{pr22}
\tilde{g}(X,Y)= X^{\flat_{g}}(Y)= g(X,Y)+\sum_{i=1}^{r}\eta_{i}(X)\eta_{i}(Y)\qquad (case\ 1~or~2)
\end{equation}
and
\begin{equation}\label{pr23}
\tilde{g}(X,Y)=X^{\flat_{g}}(Y)= \sum_{i=1}^{m}\eta_{i}(X)\eta_{i}(Y)\qquad (case\ 3~or~4)
\end{equation}
Clearly, $\tilde{g}$ defines a non-degenerate metric on $M$. Also, observe that $\tilde{g}$ coincides with $g$ if the latter is non-degenerate. The $(0,2)$-tensor $\tilde{g}^{-1}$, inverse of $\tilde{g}$ is called the pseudo-inverse of $g$. Let consider the local quasi-orthonormal field of frames $\{\xi_{1},...,\xi_{r},X_{r+1},...,X_{m}\}$ and $\{\xi_{1},...,\xi_{r}\}$ on lightlike submanifold $M$ with respect to the decompositions $TM=S(TM)\perp RadTM$ (case 1 or 2 ) and $TM=RadTM$ (case 3 or 4). Using relations (\ref{pr22}) and (\ref{pr23}), we have\\
$$\tilde{g}(\xi_{i},\xi_{j})=\delta_{ij},~1\leqslant i,j\leqslant r\quad and\quad   \tilde{g}(X_{i},X_{j})=g_{ij},~r+1\leqslant i,j\leqslant m,\quad (Case 1~ or~ 2).$$
$$\tilde{g}(\xi_{i},\xi_{j})=\delta_{ij},\quad 1\leqslant i,j\leqslant m,\quad (Case 3~ or~ 4).$$

\section{Algebraic Riemann cuvature tensors}
Contrary to non-degenerate hypersurfaces, the induced Riemann curvature on lightlike submanifold $(M, g, S(T M),S(TM^{\perp}))$ may not have an algebraic curvature tensor. For this, we have the following results.
\begin{proposition}\label{th3.1}
Let $M$ be an isotropic submanifold or a totally lightlike submanifold of a semi-Riemannian manifold $\overline{M}$. Then, the induced Riemann curvature $R$ on $M$ is an algebraic curvature tensor.
\end{proposition}
\textbf{Proof:} Since $TM=RadTM$, we have, $R(X,Y,Z,U)=0,~~\forall X,Y,Z,U\in\Gamma(TM)$.$\qquad\square$

\begin{theorem}\label{th3.2}
Let $M$ be an $r$-lightlike submanifold with $r<min\{m,n\}$ or a coisotropic submanifold of a semi-Riemannian manifold $\overline{M}$, such that the radical distribution $RadTM$ is integrable. If the induced Riemann curvature tensor of $M$ is an algebraic curvature tensor, then at least one of the following holds\\
(1) $h^{l}(X,Y)=0$, for any $X,Y\in\Gamma(TM)$.\\
(2) $A_{N}\xi\in\Gamma(RadTM)$, for any $\xi\in\Gamma(RadTM),~N\in\Gamma(ltr(TM))$\\
(3) $\overline{\nabla}_{\xi}N\in\Gamma(tr(TM))$, for any $\xi\in\Gamma(RadTM),~N\in\Gamma(ltr(TM))$.
\end{theorem}
\textbf{Proof:} Suppose that the radical distribution $RadTM$ is integrable. In virtue of Theorem 2.7 of \cite{Dug}, p162, we have
\begin{equation}\label{eq3.1}
 h^{l}(PX,\xi),\quad\forall\xi\in\Gamma(RadTM),~X\in\Gamma(TM)
\end{equation}
Moreover, suppose that the induced curvature $R$ of $M$ is algebraic. Since $R(X,\xi,Y,Z)=-R(X,\xi,Z,Y)$, for any $\xi\in\Gamma(RadTM)$ and $X,Y,Z\in\Gamma(S(TM))$, from relations (\ref{pr17.1}) and (\ref{eq3.1}), we have
\begin{equation}\label{eq3.2}
\overline{R}(X,\xi,Y,Z)-\overline{g}(\stackrel{*}{h}(\xi,Z),h^{l}(X,Y))=
-\overline{R}(X,\xi,Z,Y)+\overline{g}(\stackrel{*}{h}(\xi,Y),h^{l}(X,Z)).
\end{equation}
Using symmetry of $h^{l}$ and since the Riemann curvature $\overline{R}$ of $\overline{M}$ is algebraic curvature tensor, the relation (\ref{eq3.2}) leads to
\begin{eqnarray*}
\overline{g}(\stackrel{*}{h}(\xi,Z),h^{l}(X,Y))&=&-\overline{g}(\stackrel{*}{h}(\xi,Y),h^{l}(X,Z))=-\overline{g}(\stackrel{*}{h}(\xi,Y),h^{l}(Z,X))\\
&=&\overline{g}(\stackrel{*}{h}(\xi,X),h^{l}(Z,Y))=\overline{g}(\stackrel{*}{h}(\xi,X),h^{l}(Y,Z))\\
&=&-\overline{g}(\stackrel{*}{h}(\xi,Z),h^{l}(Y,X))=-\overline{g}(\stackrel{*}{h}(\xi,Z),h^{l}(X,Y)).
\end{eqnarray*}
Hence,
\begin{equation}\label{eq3.3}
\overline{g}(\stackrel{*}{h}(\xi,Z),h^{l}(X,Y))=0,\quad\forall\xi\in\Gamma(RadTM),~X,Y,Z\in\Gamma(S(TM)).
\end{equation}
Since $\overline{g}$ is a non-degenerate metric, by using (\ref{eq3.1}), we infer from (\ref{eq3.3}) that $\stackrel{*}{h}(\xi,PZ)=0$ or $h^{l}(X,Y)=0$, for any $X,Y,Z\in\Gamma(TM)$. Now, assume that in (\ref{eq3.3}) there exist $X_{0},Y_{0}\in\Gamma(S(TM))$ such that $h^{l}(X_{0},Y_{0})\neq0$. Then $\stackrel{*}{h}(\xi,PZ)=0,~\forall \xi\in\Gamma(RadTM),~Z\in\Gamma(TM)$. This leads to the following
\begin{equation}\label{eq3.4}
\overline{g}(\stackrel{*}{h}(\xi,PZ),N)=g(A_{N}\xi,PZ)=0,~\forall N\in\Gamma(ltr(TM)),~Z\in\Gamma(TM),
\end{equation}
that is $A_{N}\xi\in\Gamma(RadTM),~\forall \xi\in\Gamma(RadTM),~N\in\Gamma(ltr(TM))$. Moreover, if $A_{N}\xi=0$, we have
\begin{equation}\label{eq3.5}
\overline{g}(\stackrel{*}{h}(\xi,PZ),N)=\overline{g}(\overline{\nabla}_{\xi}PZ,N)=
-\overline{g}(\overline{\nabla}_{\xi}N,PZ)=0.
\end{equation}
This lead to $\overline{\nabla}_{\xi}N\in\Gamma(tr(TM))$, for any $\xi\in\Gamma(RadTM),~N\in\Gamma(ltr(TM))$.$\qquad\square$

\begin{corollary}\label{th3.3}
Let $M$ be a coisotropic submanifold of $\overline{M}$, such that the radical distribution $RadTM$ is integrable. If the induced Riemann curvature tensor of $M$ is an algebraic curvature tensor, then at least one of the following holds\\
(1) $M$ is totally geodesic.\\
(2) $A_{N}\xi\in\Gamma(RadTM)$, for any $\xi\in\Gamma(RadTM),~N\in\Gamma(ltr(TM))$\\
(3) $\overline{\nabla}_{\xi}N\in\Gamma(ltr(TM))$, for any $\xi\in\Gamma(RadTM),~N\in\Gamma(ltr(TM))$.
\end{corollary}
If $(M,g)$ is a lightlike hypersurface, the radical distribution $RadTM=TM^{\perp}$ and it is integrable. Also the shape operator $A_{N}$ is $\Gamma(S(TM))$-valued. So, the following holds.
\begin{corollary}\label{th3.4}
Let $(M,g)$ be a lightlike hypersurface of a semi-Riemannian manifold $(\overline{M},\overline{g})$. If the induced Riemann curvature tensor of $M$ is an algebraic curvature tensor, then at least one of the following holds\\
(1) $M$ is totally geodesic.\\
(2) $\overline{\nabla}_{\xi}N\in\Gamma(tr(TM))$, for any $\xi\in\Gamma(RadTM),~N\in\Gamma(tr(TM))$.
\end{corollary}
Let's consider a family of coisotropic submanifolds $M$ such that the local second fundamental forms of the screen distribution $S(TM)$ are related with the local second fundamental forms of $M$ as follows:
\begin{equation}\label{eq3.6}
\stackrel{*}{h}_{i}(X,PY)=\varphi_{i}h_{i}^{l}(X,PY),\quad\forall X,Y\in\Gamma(TM_{|\cal{U}}),
\end{equation}
where each $\varphi_{i}$ is a conformal smooth function on a coordinate neighbourhood $\cal{U}$ in $M$. The following result holds.
\begin{theorem}\label{th3.5} Let $(M,g,S(TM)$ be a coisotropic submanifold of a semi-Riemannian manifold $(\overline{M},\overline{g})$, such that (\ref{eq3.6}) holds and the radical distribution $RadTM$ is integrable. Then, the induced Riemann curvature $R$ of $M$ defines an algebraic curvature tensor if the following holonomy condition is satisfied
$$\overline{R}(X,PY)RadTM\subset RadTM,\quad\forall X,Y\in\Gamma(TM).$$
\end{theorem}
\textbf{Proof:} Consider $M$ such that $RadTM$ is integrable. Since $\stackrel{*}{h}_{i}(X,PY)=\varphi_{i}h_{i}^{l}(X,PY)$, from (\ref{pr17.1}) we have , for any $X,Y,Z,W\in \Gamma(TM)$,
\begin{eqnarray*}
 R(X,Y,Z,PU)&=&\overline{R}(X,Y,Z,PU)+\sum_{i=1}^{r}\varphi_{i}\big\{h_{i}^{l}(X,PU)h_{i}^{l}(Y,Z)-h_{i}^{l}(Y,PU)h_{i}^{l}(X,Z)\big\}\\
            &=& \overline{R}(X,Y,Z,PU)+\mathcal A(X,Y,Z,PU),
\end{eqnarray*}
where $$\mathcal A(X,Y,Z,U)=\sum_{i=1}^{r}\varphi_{i}\big\{h_{i}^{l}(X,U)h_{i}^{l}(Y,Z)-h_{i}^{l}(Y,U)h_{i}^{l}(X,Z)\big\},~\forall X,Y,Z,U\in \Gamma(TM).$$
It is straightforward that $\mathcal A$ verifies the algebraic symmetries of (\ref{int1}). So, $R(X,Y,Z,PU)$ has the required symmetries. On the other hand, for any $X,Y,Z\in \Gamma(TM)$ and $\xi\in RadTM$, we have $R(X,Y,Z,\xi)=-R(Y,X,Z,\xi)=0$. Also, since $h_{i}^{l}(PX,\xi)=0,~\forall X\in\Gamma(TM),\xi\in\Gamma(RadTM)$, we have
$R(Z,\xi,X,Y)=R(Z,\xi,X,PY)=\overline{R}(Z,\xi,X,PY)=-\overline{R}(X,PY,\xi,Z)=0$. This completes the proof.$\qquad\square$\vspace{0.2cm}\\
It is easy to see that a lightlike hypersurface verifying (\ref{eq3.6}) is said to be locally (globally) screen conformal (see \cite{Ati-Dug2}). So, the following holds.
\begin{corollary}\label{th3.6}
Let $(M,g,S(TM))$ be a locally screen conformal lightlike hypersurface of a semi-Riemannian manifold $(\overline{M},\overline{g})$ with ambient holonomy condition
$$\overline{R}(X,PY)RadTM\subset RadTM,\quad\forall X,Y\in\Gamma(TM).$$
Then, the induced Riemann curvature $R$ of $M$ defines an algebraic curvature tensor.
\end{corollary}

\section{Pseudo-Jacobi operators and lightlike Osserman submanifolds}
Let's start by intrinsic interpretation of relation (\ref{int2}) which in pseudo-Riemannian setting characterizes the Jacobi operator $J_{R}(\cdotp)$ associated to an algebraic curvature map (tensor) $R\in \otimes^{4}T_{u}^{\ast}M$, $u\in M$. Indeed, for $x\in S_{u}^{+}(M)$ (or $x\in S_{u}^{-}(M)$), $y$, $w$ in $T_{u}M$, we have,
\begin{equation}\label{eq4.11}
 J_{R}(x)y=R(y,x,x,\bullet)^{\sharp}
\end{equation}
that is
\begin{equation}\label{eq4.12}
 (J_{R}(x)y)^{\flat}(w)=R(y,x,x,w)
\end{equation}
where $\flat$ and $\sharp$ are the usual natural isomorphisms between $T_{u}M$ and its dual $T_{u}^{*}M$, with respect to non-degenerate metric $g$. For degenerate setting, let's consider the associate non-degenerate metric $\tilde{g}$ of $g$ defined by relations (\ref{pr22}) and (\ref{pr23}), and denote by $\flat_{g}$ and $\sharp_{g}$ the natural isomorphisms with respect to the metric $\tilde{g}$. Thus, equivalently the above relations can be written in the form:
\begin{equation}\label{eq4.13}
\tilde{g}(J_{R}(x)y,w)=R(y,x,x,w)
\end{equation}
in which $J_{R}(x)$ is well defined. This leads to the following definition.
\begin{definition}[Pseudo-Jacobi Operator]
Let $(M,g,S(TM),S(TM^{\perp}))$ be a lightlike submanifold of a semi-Riemannian manifold $(\overline{M},\overline{g})$, $u\in M$, $x\in S_{u}^{+}(M)$ (or $x\in S_{u}^{-}(M)$) and $R\in \otimes^{4}T_{u}^{\ast}M$ an algebraic curvature map (tensor) on $T_{u}M$. By {\itshape pseudo-Jacobi operator} associated to $R$ with respect to $x$, we call the self-adjoint linear map $J_{R}(x)$ on $T_{u}M$ defined by
\begin{equation}\label{eq4.14}
 J_{R}(x)y=R(y,x,x,\bullet)^{\sharp_{g}}
\end{equation}
or equivalently
\begin{equation}\label{eq4.15}
 (J_{R}(x)y)^{\flat_{g}}(w)=R(y,x,x,w)
\end{equation}
where $\flat_{g}$ and $\sharp_{g}$ denote the natural isomorphisms between $T_{u}M$ and its dual $T_{u}^{*}M$, with respect to non-degenerate metric $\tilde{g}$ of $g$.
\end{definition}
\textbf{Remark:} For the cases of isotropic submanifold and totally lightlike submanifold, the induced Riemannian curvature tensor $R$ vanish identically on $T_{u}M$ for any $u\in M$. Therefore, the pseudo-Jacobi operator $J_{R}(\cdot)$ associated to $R$ vanish identically on $S_{u}^{\pm}(M)$.\vspace{0.2cm}

It is known by approach developed in (\cite{Dug}) that, the extrinsic geometry of lightlike submanifolds depends on a choice of screen distribution $S(TM)$ and screen transversal vector bundle $S(TM^{\perp})$. Since these screens are not uniquely determined, a well defined concept of Jacobi condition is not possible for an arbitrary lightlike submanifold of a semi-Riemannian manifold. Thus, one must look for a class of pair $\{S(TM),S(TM^{\perp})\}$ of screens for which the induced Riemann curvature and associated pseudo-Jacobi operator have the desired symmetries and properties. In short, we precise the following.
\begin{definition}
A pair $\{S(TM),S(TM^{\perp})\}$ of screen distribution and screen transversal vector bundle on lightlike submanifold $M$ of a semi-Riemannian manifold $\overline{M}$ is said to be \textit{admissible} if its associated induced Riemann curvature $R$ is an algebraic curvature tensor.
\end{definition}
Note that for case of coisotropic submanifold, since $S(TM^{\perp})=\{0\}$, we'll use the concept \textit{admissible screen distribution} $S(TM)$.\\
\textbf{Example }\\
(1) It is obvious that on totally geodesic lightlike submanifold, any pair of screens is admissible.\\
(2) In virtue of Theorem \ref{th3.5}, any coisotropic submanifold $(M,g,S(TM))$ of semi-Riemannian manifold of constant sectional curvature $\overline{M}(c)$ with integrable radical distribution, satisfying relation (\ref{eq3.6}), admits an admissible screen distribution.
\begin{definition}
A lightlike submanifold $(M,g)$ of a semi-Riemannian manifold $(\overline{M},\overline{g})$ of constant index is called timelike (resp. spacelike) Osserman at $u\in M$ if for each admissible pair of screens  $\{S(TM),S(M^{\perp})\}$ and associate induced Riemann curvature $R$, the characteristic polynomial of $J_{R}(x)$ is independent of $x\in S_{u}^{-}(M)$ (resp. $x\in S_{u}^{+}(M)$). Moreover, if this holds at each $u\in M$, then $(M,g)$ is called pointwise Osserman (or Osserman). If this holds independently of the point $u\in M$, $(M,g)$ is called globally Osserman.
\end{definition}
Note that the above definition of Osserman condition extends the definition given in (\cite{Ati-Dug}) for lightlike hypersurface and is independent on the choice of admissible pair of screens. Also, we can show that a lightlike submanifold $(M,g)$ being timelike Osserman at $p\in M$ is equivalent to $(M,g)$ being spacelike Osserman at $p$.\vspace{0.2cm}\\
According to the above remark, we have the following result.
\begin{theorem}
 Let $(M,g)$ be an isotropic submanifold or a totally lightlike submanifold of a semi-Riemannian manifold $(\overline{M},\overline{g})$. Then $(M,g)$ is globally Osserman.
\end{theorem}
In the following example, we exhibit a $2$-degenerate metric which satisfies Osserman condition. We have considered this metric as an induced metric on a coisotropic submanifold.\newpage
\hspace{-0.6cm}\textbf{Basic example}\\
Let $(x,y)=(x_{0},...,x_{p},y_{0},...,y_{p})$ be the usual coordinates on $\mathbb{R}^{2p+2}$. Let $f=f(x_{1},...,x_{p})$ and $h=h(x_{1},...,x_{p})$ be the smooth functions on an open subset $\mathcal{O}\subset\mathbb{R}^{p}$. We define with respect to the natural field of frames $\{\frac{\partial}{\partial x_{0}},...,\frac{\partial}{\partial x_{p}},\frac{\partial}{\partial y_{0}},...,\frac{\partial}{\partial y_{p}}\}$ a $2$-degenerate metric $g_{(f,h)}$ on $M=\mathbb{R}\times\mathcal{O}\times\mathbb{R}^{p+1}$ by
\begin{equation}\label{eqex1.1}
g_{(f,h)}=\sum_{i=1}^{p}(\frac{\partial f}{\partial x_{i}}dx_{0}dx_{i}+\frac{\partial h}{\partial x_{i}}dx_{i}dy_{0})+\sum_{i,j=1}^{p}\{(\frac{\partial f}{\partial x_{i}}\frac{\partial f}{\partial x_{j}}+\frac{\partial h}{\partial x_{i}}\frac{\partial h}{\partial x_{j}})dx_{i}dx_{j}+\delta_{ij}dx_{i}dy_{j}\}.
\end{equation}
The $2$-degenerate manifold $(M,g_{(f,h)})$ arise as a lightlike submanifold in a  $(p+4)$-dimensional semi-Euclidean space $\overline{M}$. Let $\{u_{0},...,u_{p},v_{0},...,v_{p},w_{1},w_{2}\}$ be a basis for a space $\overline{M}$. Define an semi-Euclidean metric $\overline{g}$ of signature $(p+2,p+2)$ on $\overline{M}$ by setting
$$\overline{g}(u_{i},u_{j})=0= \overline{g}(v_{i},v_{j}),\quad
\overline{g}(u_{0},v_{j})=0=\overline{g}(u_{i},v_{0}),\quad0\leqslant i,j\leqslant p.$$
$$\overline{g}(u_{0},w_{1})=1\quad\overline{g}(u_{0},w_{2})=0,\quad
\overline{g}(u_{i},w_{1})=0=\overline{g}(u_{i},w_{2}),\quad 1\leqslant i\leqslant p.$$
$$\overline{g}(v_{0},w_{1})=0\quad\overline{g}(v_{0},w_{2})=1,\quad
\overline{g}(v_{i},w_{1})=0=\overline{g}(v_{i},w_{2}),\quad 1\leqslant i\leqslant p.$$
$$\overline{g}(u_{i},v_{j})=\delta_{ij},\quad \overline{g}(w_{i},w_{j})=\delta_{ij},\quad
 \quad1\leqslant i,j\leqslant p.$$
Let consider the application
\begin{equation}\label{eqex1.2}
F(x,y)=x_{0}~u_{0}+...+x_{p}~u_{p}+y_{0}~v_{0}+...+y_{p}~v_{p}+f~w_{1}+h~w_{2}.
\end{equation}
$F(x,y)$ defines an embedding of $M$ in $\overline{M}$ and $g_{(f,h)}$ is the induced metric on the embedded submaifold $M$.\vspace{0.2cm}\\
FACT 1. By direct calculation using (\ref{eqex1.2}), the tangent space $TM$ is defined by
\begin{eqnarray}
TM&=&Span\{\partial_{0}^{x}=u_{0},~\partial_{1}^{x}=u_{1}+\partial_{1}^{x}f~w_{1}+\partial_{1}^{x}h~w_{2},...,\crcr
&&\qquad\quad\partial_{p}^{x}=u_{p}+\partial_{p}^{x}f~w_{1}+\partial_{p}^{x}h~w_{2},
\partial_{0}^{y}=v_{0},~\partial_{1}^{y}=v_{1},...,\partial_{p}^{y}=v_{p}\},
\end{eqnarray}
where $\partial_{i}^{x}=\frac{\partial}{\partial x_{i}}$ and $\partial_{i}^{y}=\frac{\partial}{\partial y_{i}}$.\\
The radical distribution $RadTM$ of rank $2$ is given by
\begin{equation}\label{eqex1.3}
RadTM=Span
\big\{\xi_{1}=\partial_{0}^{x}-\sum_{i=1}^{p}\partial_{i}^{x}f~\partial_{i}^{y}~,
~\xi_{2}=\partial_{0}^{y}-\sum_{i=1}^{p}\partial_{i}^{x}h~\partial_{i}^{y}\big\}.
\end{equation}
$M$ is a coisotropic submanifold of a semi-Euclidean space $\overline{M}$. The lightlike transversal vector bundle $ltr(TM)$ of $M$ is given by
\begin{equation}\label{eqex1.4}
ltr(TM)=Span
\big\{N_{1}=w_{1}-\frac{1}{2}\xi_{1}~,~N_{2}=w_{2}-\frac{1}{2}\xi_{2}\big\}.
\end{equation}
The corresponding screen distribution $S(TM)$ for the above $ltr(TM)$ is given by
\begin{equation}\label{eqex1.5}
S(TM)=\big\{U_{1},...,U_{p},V_{1},...,V_{p}\big\},
\end{equation}
where $U_{i}=\partial_{i}^{x}-\partial_{i}^{x}f~\partial_{0}^{x}-\partial_{i}^{x}h~\partial_{0}^{y}$ and $V_{i}=\partial_{i}^{y}$.\vspace{0.2cm}\\
FACT 2.  Let's consider on $\overline{M}$ a local quasi-orthogonal frame $\{\xi_{1},\xi_{2},U_{i},V_{i},N_{1},N_{2}\}_{1\leqslant i\leqslant p}$ such that $\{\xi_{1},\xi_{2},U_{i},V_{i}\}_{1\leqslant i\leqslant p}$ is a local frame on $M$ with respect to the decomposition (\ref{pr1}). Using the metric $\overline{g}$, we have
$$\overline{\nabla}_{\xi_{1}}^{\xi_{2}}=\overline{\nabla}_{\xi_{2}}^{\xi_{1}}=0,\quad\overline{\nabla}_{V_{i}}^{\xi_{1}}=\overline{\nabla}_{V_{i}}^{\xi_{2}}=0,~1\leqslant i\leqslant p$$
$$\overline{\nabla}_{U_{i}}^{\xi_{1}}=-\sum_{j=1}^{p}\partial_{i}^{x}\partial_{j}^{x}f~V_{j},\quad
\overline{\nabla}_{U_{i}}^{\xi_{2}}=-\sum_{j=1}^{p}\partial_{i}^{x}\partial_{j}^{x}h~V_{j},~1\leqslant i\leqslant P.$$
Thus, since $h_{i}^{l}(X,Y)=-\overline{g}(\overline{\nabla}_{X}^{\xi_{i}},Y)$, we obtain that the non-vanishing components of $h^{l}$ are
\begin{equation}\label{eqex1.6}
h_{1}^{l}(U_{i},U_{j})=\partial_{i}^{x}\partial_{j}^{x}f,\quad h_{2}^{l}(U_{i},U_{j})=\partial_{i}^{x}\partial_{j}^{x}h,~1\leqslant i,j\leqslant p.
\end{equation}
Also, by straightforward calculation, using the Gauss equation, we obtain that the only non-vanishing components of induced connection $\nabla$ on $M$ are
\begin{eqnarray}\label{eqex1.7}
&& \nabla_{U_{i}}^{U_{j}}=-\frac{1}{2}\partial_{i}^{x}\partial_{j}^{x}f~\xi_{1}
-\frac{1}{2}\partial_{i}^{x}\partial_{j}^{x}h~\xi_{2}
-\sum_{k=1}^{p}\big(\partial_{i}^{x}\partial_{j}^{x}f~\partial_{k}^{x}f
+\partial_{i}^{x}\partial_{j}^{x}h~\partial_{k}^{x}h\big)V_{k},\crcr
&& \nabla_{U_{i}}^{\xi_{1}}=-\sum_{j=1}^{p}\partial_{i}^{x}\partial_{j}^{x}f~V_{j},\quad
\nabla_{U_{i}}^{\xi_{2}}=-\sum_{j=1}^{p}\partial_{i}^{x}\partial_{j}^{x}h~V_{j},\quad1\leqslant i,j\leqslant p.
\end{eqnarray}
FACT 3. By direct calculation, the only non-vanishing components of the induced Riemannian curvature tensor on $M$ are given by
\begin{equation}\label{eqex1.8}
R(U_{i},U_{j})U_{k}=\frac{1}{2}\sum_{l=1}^{p}\big\{ f_{;ik}f_{;jl}-f_{;jk} f_{;il}
+h_{;ik}h_{;jl}-h_{;jk}h_{;il}\big\}V_{l}.
\end{equation}
\begin{equation}\label{eqex1.8}
R(U_{i},U_{j},U_{k},U_{l})=\frac{1}{2}\big\{ f_{;ik}f_{;jl}-f_{;jk} f_{;il}
+h_{;ik}h_{;jl}-h_{;jk}h_{;il}\big\} ,
\end{equation}
where $f_{;ij}=\partial_{i}^{x}\partial_{j}^{x}f$ and $h_{;ij}=\partial_{i}^{x}\partial_{j}^{x}h$.\vspace{0.2cm}\\
FACT 4. In virtue of FACT 3, we can see that for any $X,Y,Z\in\Gamma(TM)$, $R(X,Y)Z\in\Gamma(S(TM))$. Thus for any $X\in S^{\pm}(M)$, $Y,W\in\Gamma(TM)$, we have
$$\tilde{g}(J_{R}(X)Y,W)=R(Y,X,X,W)=g(R(Y,X)X,W)=\tilde{g}(R(Y,X)X,W).$$
Since the metric $\tilde{g}$ is non-degenerate, we infer that, for any $X\in S^{\pm}(M)$
\begin{equation}\label{eqex1.9}
J_{R}(X)Y=R(Y,X)X.
\end{equation}
Let $X=X_{a}\xi_{1}+X_{b}\xi_{2}+X_{1}U_{1}+...+X_{p}U_{p}+X_{p+1}V_{1}+...+X_{2p}V_{p}$ the tangent vector field on $M$, by straightforward calculation, with respect to the local quasi-orthogonal frame $\{\xi_{1},\xi_{2},U_{i},V_{i}\}_{1\leqslant i\leqslant p}$ on $M$, we have
$$J_{R}(X)\xi_{1}=0,\quad J_{R}(X)\xi_{2}=0,\quad J_{R}(X)V_{i}=0,$$
\begin{eqnarray*}
J_{R}(X)U_{i}&=&\sum_{j,k=1}^{p}X_{j}X_{k}R(U_{i},U_{j})U_{k}\\
 &=&\sum_{l=1}^{p}\big\{\frac{1}{2}\sum_{j,k=1}^{p}X_{j}X_{k}\big(
f_{;ik}f_{;jl}-f_{;jk} f_{;il}+h_{;ik}h_{;jl}-h_{;jk}h_{;il}\big)\big\}V_{l}\\
 &=& \sum_{l=1}^{p}\Phi_{li}V_{l},\quad 1\leqslant i \leqslant p,
\end{eqnarray*}
where
$$\Phi_{li}=\frac{1}{2}\sum_{j,k=1}^{p}X_{j}X_{k}\big(
f_{;ik}f_{;jl}-f_{;jk} f_{;il}+h_{;ik}h_{;jl}-h_{;jk}h_{;il}\big),\quad 1\leqslant l,i \leqslant p.$$
Thus, the pseudo-Jacobi operator is given by, for any $X\in S^{\pm}(M)$,
\begin{equation}\label{eqex1.10}
 J_{R}(X)=\left(
\begin{array}{ccccccc}
& & & & & &\\
&O_{p+2,2} & \vdots & O_{p+2,p} & \vdots & O_{p+2,p} & \\

&\cdots & \cdots & \cdots& \cdots&\cdots&\\

&O_{p,2} & \vdots & A_{p,p} & \vdots & O_{p,p} & \\
& & & & & &\\
 \end{array}\right)
\end{equation}
where the submatrix $A_{p,p}=\big(\Phi_{li}\big)_{1\leqslant l,i\leqslant p}$. It follows from the expression of $J_{R}(X)$ that its characteristic polynomial satisfies
\begin{equation}\label{eqex1.11}
P_{\lambda}(J_{R}(X))=det(J_{R}(X)-\lambda I_{2p+2})=\lambda^{2p+2}.
\end{equation}
Thus, all eigenvalues are zero. This proves that the $2$-degenerate submanifold $(M,g_{(f,h)})$ is globally Osserman.
\begin{definition}\cite{Dug-Jin}
 A lightlike submanifold $(M,g)$ of a semi-Riemannian manifold
$(\overline{M},\overline{g})$ is said to be totally umbilical in $\overline{M}$ if there is a smooth transversal vector field $\mathbf H\in \Gamma(tr(TM))$ on $M$, called the \textit{transversal curvature vector field} of $M$, such that for any $X,Y\in\Gamma(TM)$,
\begin{equation}\label{eq4.1}
 h(X,Y)=\mathbf Hg(X,Y).
\end{equation}
\end{definition}
Using (\ref{pr6}), (\ref{pr9}) and (\ref{eq4.1}), it is easy to see that $M$ is \textit{totally umbilical} if and only if on each coordinate neighborhood $\mathcal{U}$, there exist smooth vector fields $H^{l}\in\Gamma(ltr(TM))$ and
$H^{s}\in\Gamma(S(TM^{\perp}))$ such that
\begin{eqnarray}\label{eq4.2}
 &h^{l}(X,Y)=H^{l}g(X,Y),\quad D^{l}(X,W)=0&\crcr
&h^{s}(X,Y)=H^{s}g(X,Y),\quad\forall X,Y\in\Gamma(TM),~~W\in\Gamma(S(TM^{\perp})).&
\end{eqnarray}
Note that for the case of lightlike hypersurface, the relations (\ref{eq4.1}) and (\ref{eq4.2}) are equivalent to
\begin{equation}\label{eq4.2.1}
 B(X,Y)=\rho g(X,Y),\quad\forall X,Y\in\Gamma(TM),
\end{equation}
where $\rho$ is the smooth function on $\mathcal{U}\subset M$ and $B$, the local second fundamental form of $M$.
\begin{definition}
Let $(M,g)$ be an $r$-lightlike submanifold with $r<min\{m,n\}$ or a coisotropic submanifold of a $(m+n)$-dimensional semi-Riemannian manifold $(\overline{M},\overline{g})$. We say that the screen distribution $S(TM)$ is \textit{totally umbilical} if for any section $N$ of $ltr(TM)$ on a coordinate neighbourhood $\mathcal{U}\subset M$, there exists a smooth function $\lambda$ on $\mathcal{U}$ such that
\begin{equation}\label{eq4.3}
\overline{g}(\stackrel{*}{h}(X,PY),N)=\lambda g(X,PY),\quad\forall X,Y\in\Gamma(TM_{|\mathcal{U}}).
\end{equation}
\end{definition}
Note that for the case of $1$-lightlike submanifolds and lightlike hypersurfaces, $A_{N}$ is $\Gamma(S(TM))$-valued. Therefore, using (\ref{pr14}), the relation (\ref{eq4.3}) is equivalent to
\begin{equation}\label{eq4.3.1}
 A_{N}X=\lambda PX,\quad\forall X\in\Gamma(TM_{|\mathcal{U}}).
\end{equation}
We note that by Theorem 2.5 in \cite{Dug}, page 161, $S(TM)$ is integrable. In case $\lambda=0$ on $\mathcal{U}$, we say that $S(TM)$ is totally geodesic.\vspace{0.2cm}\\
In the following, In an ambient space form, we give characterization of a family of admissible pair of screens and we prove that the Ricci tensors associated are symmetric.

\begin{theorem}\label{th4.1}
Let $M$ be an $r$-lightlike submanifold with $r<min\{m,n\}$ or a coisotropic submanifold of a $(m+n)$-dimensional semi-Riemannian manifold of constant sectional curvature $\overline{M}(c)$ that is totally umbilical. Then, the family of admissible pair of screens reduce to the set of totally umbilical screen distributions on $M$. Also, $M$ is pointwise Osserman and for each admissible pair of screens $\{S(TM),S(TM^{\perp})\}$, the associated Ricci tensor is symmetric and $M$ is locally Einstein.
\end{theorem}
\textbf{Proof:} In this proof, we suppose that $RadTM$ is of rank $r\neq1$. For the case of $1$-lightlike submanifold, the proof is similar. For the case of lightlike hypersurface, the proof is given in \cite{Ati-Dug}, Theorem 4.3.\\
Let's consider $M$ a proper totally umbilical, that is, in (\ref{eq4.2}) we have $H^{l}\neq0$ and $H^{s}\neq0$ (in case $r$-lightlike) and $H^{l}\neq0$ (in case coisotropic). Then, using (\ref{pr17}), the induced Riemann curvature tensor takes the form
$$R(X,Y)Z=c\{g(Y,Z)X-g(X,Z)Y\}+A_{h^{l}(Y,Z)}X-A_{h^{l}(X,Z)}Y+A_{h^{s}(Y,Z)}X-A_{h^{s}(X,Z)}Y.$$
Thus, using (\ref{pr9}), (\ref{pr14}) and (\ref{eq4.2}), we obtain, for any $X,Y,Z,U\in\Gamma(TM)$,
\begin{eqnarray}\label{eq4.4}
&R(X,Y,Z,U)=c\{g(Y,Z)g(X,U)-g(X,Z)g(Y,U)\}+g(Y,Z)\overline{g}(\stackrel{*}{h}(X,PU),H^{l})\crcr
&\quad -g(X,Z)\overline{g}(\stackrel{*}{h}(Y,PU),H^{l})+\overline{g}(H^{s},H^{s})\{g(Y,Z)g(X,U) -g(X,Z)g(Y,U)\}.
\end{eqnarray}
Now let's consider an admissible pair of screens $\{S(TM),S(TM^{\perp})\}$ and let $R$ denote the associate induced Riemann curvature tensor. Since
$R(X,Y,Z,U)=R(Z,U,X,Y)$, from (\ref{eq4.4}), we obtain, for any $X,Y,Z,U\in\Gamma(TM)$,
\begin{eqnarray}\label{eq4.5}
&&g(Y,Z)\overline{g}(\stackrel{*}{h}(X,PU),H^{l})-g(X,Z)\overline{g}(\stackrel{*}{h}(Y,PU),H^{l})\crcr
&&\qquad -g(U,X)\overline{g}(\stackrel{*}{h}(Z,PY),H^{l})+g(Z,X)\overline{g}(\stackrel{*}{h}(U,PY),H^{l})=0.
\end{eqnarray}
Since $H^{l}\neq0$, for any $X,U\in\Gamma(TM)$, choose in (\ref{eq4.5}) the vector fields $Y$ and $Z$ such that $g(Y,Z)=1$ and $g(X,Z)=0$. Thus, we get
$$\overline{g}(\stackrel{*}{h}(X,PU),H^{l})=\lambda g(X,PU),\quad\forall X,U\in\Gamma(TM),$$
where $\lambda=\overline{g}(\stackrel{*}{h}(Z,PY),H^{l})$, that is the screen distribution $S(TM)$ is totally umbilical. \\
Conversely, suppose that for any $N\in\Gamma(ltr(TM))$, there exists a smooth function $\lambda$ on $\mathcal{U}\subset M$ such that $\overline{g}(\stackrel{*}{h}(X,PY),N)=\lambda g(X,PY),~~\forall X,Y\in\Gamma(TM)$. Then, using (\ref{pr9}), (\ref{pr14}) and (\ref{eq4.2}) we have, for any $X,Y,Z,U\in\Gamma(TM)$,
\begin{eqnarray}\label{eq4.6}
&R(X,Y,Z,U)=c\{g(Y,Z)g(X,U)-g(X,Z)g(Y,U)\}+\lambda\{g(Y,Z)g(X,PU)\crcr
&\quad -g(X,Z)g(Y,PU)\} +\overline{g}(H^{s},H^{s})\{g(Y,Z)g(X,U) -g(X,Z)g(Y,U)\}.
\end{eqnarray}
Thus, $R$ defines an algebraic curvature tensor, that is $\{S(TM),S(TM^{\perp})\}$ is admissible.\\
Now, let $\{S(TM),S(TM^{\perp})\}$ be an arbitrary admissible pair of screens on $M$. We compute the induced Ricci curvature with respect to $\{S(TM),S(TM^{\perp})\}$. Consider the quasi-orthonormal field of frames $\{E_{1}=\xi_{1},...,E_{r}=\xi_{r},E_{r+1},...,E_{m}\}$ on $TM_{|\mathcal{U}}$. Then, for any $X,Y,Z\in\Gamma(TM)$,
\begin{eqnarray*}
R(X,Y)Z&=&c\{g(Y,Z)X-g(X,Z)Y\}+\lambda\{g(Y,Z)X-g(X,Z)Y\}\\
&&+g(Y,Z)A_{H^{s}}X -g(X,Z)A_{H^{s}}Y.
\end{eqnarray*}
Thus, we have, for any $X,Y\in\Gamma(TM)$,
\begin{eqnarray*}
Ric(X,Y)&=&\sum_{i=r+1}^{m}\varepsilon_{i}g(R(X,E_{i})Y,E_{i})+\sum_{i=1}^{r}g(R(X,\xi_{i})Y,N_{i})\\
&=&(c+\lambda+\overline{g}(H^{s},H^{s}))\{g(X,Y)-(m-r)g(X,Y)\}-r(c+\lambda)g(X,Y)\\
&& -g(X,Y)\sum_{i=1}^{r}\overline{g}(D^{s}(\xi_{i},N_{i}),H^{s})\\
&=&\big\{(1-m+r)(c+\lambda+\overline{g}(H^{s},H^{s}))-r(c+\lambda)
-\sum_{i=1}^{r}\overline{g}(D^{s}(\xi_{i},N_{i}),H^{s})\big\}g(X,Y).
\end{eqnarray*}
Therefore, the induced Ricci curvature is symmetric. Moreover $M$ is locally Einstein.\\
Finally, let's consider $x\in S_{u}^{+}(M)$, $(or~~x\in S_{u}^{-}(M))$, $u\in M$, $y,v\in x^{\perp}$. Then, by using (\ref{eq4.6}), we have
\begin{eqnarray*}
\tilde{g}(J_{R}(x)y,v)&=&R(y,x,x,v)\\
&=&(c+\lambda+\overline{g}(H^{s},H^{s}))g(x,x)g(y,v).
\end{eqnarray*}
Thus, the pseudo-Jacobi operator $J_{R}(x)$ and its characteristic polynomial $P_{t}(J_{R}(x))$ are given by
$$J_{R}(x)y=\pm(c+\lambda+\overline{g}(H^{s},H^{s}))Py,\quad P_{t}(J_{R}(x))=-t\{\pm(c+\lambda+\overline{g}(H^{s},H^{s}))-t\}^{m},$$
for arbitrary given admissible pair of screens. Therefore, $M$ is spacelike(timelike) pointwise Osserman, which completes the proof.
$\quad\square$

\begin{proposition}\label{th4.2}
Let $M$ be an $r$-lightlike submanifold with $r<min\{m,n\}$ or a coisotropic submanifold of a $(m+n)$-dimensional semi-Riemannian manifold $(\overline{M},\overline{g})$, with induced algebraic Riemannian curvature tensor $R$. For any $x\in S_{u}^{+}(M)$, $(x\in S_{u}^{-}(M))$, $u\in M$ we have
$$trace J_{R}(x)=\sum_{i=1}^{r}\eta_{i}(R(x,\xi_{i})x)-Ric(x,x),$$
where $\eta_{i}(\cdotp)=\overline{g}(\cdotp,N_{i})$.
\end{proposition}
\textbf{Proof:} Let's consider the quasi-orthonormal basis  $\{E_{1}=\xi_{1},...,E_{r}=\xi_{r},E_{r+1},...,E_{m}\}$ of $T_{u}M$ such that $S(TM)=Span\{E_{r+1},...,E_{m}\}$. We have
\begin{eqnarray*}
trace J_{R}(x)&=&\sum_{i=1}^{m}\tilde{g}^{ii}\tilde{g}(J_{R}(x)E_{i},E_{i})\\
&=&\sum_{i=r+1}^{m}\tilde{g}^{ii}\tilde{g}(J_{R}(x)E_{i},E_{i})+
    \sum_{i=1}^{r}\tilde{g}(J_{R}(x)\xi_{i},\xi_{i})\\
&=&-\sum_{i=r+1}^{m}\tilde{g}^{ii}\tilde{g}(R(x,E_{i})x,E_{i})\\
&=&-\sum_{i=r+1}^{m}\tilde{g}^{ii}\tilde{g}(R(x,E_{i})x,E_{i})
   -\sum_{i=1}^{r}\tilde{g}(R(x,\xi_{i})x,\xi_{i})+\sum_{i=1}^{r}\tilde{g}(R(x,\xi_{i})x,\xi_{i})\\
&=&\sum_{i=1}^{r}\eta_{i}(R(x,\xi_{i})x)-Ric(x,x).\quad\square
\end{eqnarray*}

It is known (see \cite{Ga-Ku-Va}) that in semi-Riemannian case, being Osserman at a point simplifies the geometry at that point as the manifold is Einstein at that point. Moreover, if the latter is connected and of at least dimension 3, by Schur lemma (\cite{Be}), it is Einstein. For the degenerate case we have the following result.

\begin{theorem}\label{th4.3}
Let $M$ be an $r$-lightlike submanifold with $r<min\{m,n\}$ or a coisotropic submanifold of a $(m+n)$-dimensional semi-Riemannian manifold that is Osserman at $u\in M$. If for a given admissible pair of screens $\{S(TM),S(TM^{\perp})\}$ , the associate induced curvature tensor $R$ verifies,
 $$(1)~~ R(x,\xi_{i})\xi_{j}=0,\quad R(\xi_{i},\xi_{j})x=0,~\forall x\in T_{u}M,~~1\leqslant i,j\leqslant r\qquad\qquad$$
 $$(2)~~|\eta_{i}(R(x,\xi_{i})x)|<\mu\in\mathbb{R},\quad \forall x\in S_{u}^{+}(M)( or~~x\in S_{u}^{-}(M)),~~ 1\leqslant i\leqslant r,$$
then $(M,g,S(TM),S(TM^{\perp}))$ is Einstein at $u\in M$.
\end{theorem}
\textbf{Proof:} Let's consider the quasi-orthonormal basis  $\{E_{1}=\xi_{1},...,E_{r}=\xi_{r},E_{r+1},...,E_{m}\}$ of $T_{u}M$ such that $S(TM)=Span\{E_{r+1},...,E_{m}\}$. Denote by $R'$ and $g'$ the restriction on $S(TM)$ of the induced algebraic curvature tensor $R$ and the induced metric $g$ on $M$, respectively. Since the characteristic polynomial $J_{R}(x)$ is the same for any
$x\in S_{u}^{+}(M)$ $(x\in S_{u}^{-}(M))$, we have $traceJ_{R}(x)$ is bounded on $S_{u}^{+}(M)$, $S_{u}^{-}(M))$. By using Proposition \ref{th4.2}, we have for any $x\in S_{u}^{+}(M)$ $(x\in S_{u}^{-}(M))$,
$$|Ric(x,x)|\leqslant|traceJ_{R}(x)|+\sum_{i=1}^{r}|\eta_{i}(R(x,\xi_{i})x)|$$
It follows that there exist $\alpha\in\mathbb{R}$ such that $|Ric(x,x)|\leqslant\alpha~$ for any $x\in S_{u}^{+}(M)$, $(x\in S_{u}^{-}(M))$.
In particular, we have
$$|Ric'(x,x)|\leqslant\alpha,\quad\forall x\in S^{+}(S(T_{u}M))( x\in  S^{-}(S(T_{u}M) ).$$
Thus, since $(S(T_{u}M),g')$ is non-degenerate, it follows from a well known algebraic result (see \cite{Daj-Nomi}) that there exist $\lambda\in\mathbb{R}$ such that
\begin{equation}\label{eq4.7}
Ric'(x,y)=\lambda g'(x,y),\quad\forall x,y\in S(T_{u}M).
\end{equation}
Also, since $R$ is algebraic, by hypothesis we have, for any $x\in T_{u}M$, $1\leqslant i\leqslant r$,
$$Ric(\xi_{i},x)=\sum_{j=1}^{r}R(\xi_{i},\xi_{j},x,N_{j})+\sum_{j=r+1}^{m}\varepsilon_{j}R(\xi_{i},E_{j},x,E_{j})=\sum_{j=r+1}^{m}\varepsilon_{j}R( E_{j},x,E_{j},\xi_{i})=0,$$
and
$$Ric(x,\xi_{i})=\sum_{j=1}^{r}R(x,\xi_{j},\xi_{i},N_{j})+\sum_{j=r+1}^{m}\varepsilon_{j}R(x,E_{j},\xi_{i},E_{j})=-\sum_{j=r+1}^{m}\varepsilon_{j}R(x, E_{j},E_{j},\xi_{i})=0.$$
Therefore, since $g(\xi_{i},x)=0,~\forall x\in T_{u}M$, the latter and (\ref{eq4.7}) lead to
$$Ric(x,y)=\lambda g(x,y),\quad\forall x,y\in T_{u}M,$$
that is $(M,g,S(TM),S(TM^{\perp}))$ is Einstein at $u\in M$.$\quad\square$

\begin{corollary}\label{th4.4}
Let $(M,g)$ be a lightlike hypersurface of a semi-Riemannian manifold of constant sectional curvature $(\overline{M}(c),\overline{g})$. If $(M,g)$ is Osserman at $u\in M$ then for an admissible screen distribution $S(TM)$, $(M,g,S(TM))$ is Einstein at $u\in M$.
\end{corollary}
\textbf{Proof:} The induced curvature tensor $R$ on $M$ is given by
\begin{equation}\label{eq4.8}
R(X,Y)Z=c\{g(Y,Z)X-g(X,Z)Y\}+B(Y,Z)A_{N}X-B(X,Z)A_{N}Y.
\end{equation}
Since $R(x,\xi)\xi=0=R(\xi,\xi)x,~~\forall x\in T_{u}M$ and $|\eta(R(x,\xi)x)|=|c|,~~\forall x\in S_{u}^{\pm}(M)$, the result is immediate consequence of Theorem \ref{th4.3}.$\quad\square$

\section{Symmetry properties on lightlike hypersurfaces}
Let's consider a lightlike hypersurface $(M,g)$ of an $(m+2)$-dimensional semi-Riemannian manifold of constant sectional curvature $(\overline{M}(c),g)$. By direct calculation, using (\ref{eq4.8}), for a normalizing pair $\{\xi,N\}$, the Ricci tensor on $M$ is given by
\begin{equation}\label{eq5.1}
Ric(X,Y)=mcg(X,Y)+B(X,Y)trA_{N}-B(A_{N}X,Y),\quad\forall X,Y\in\Gamma(TM).
\end{equation}
Now, suppose that $(M,g)$ is pointwise Osserman. In virtue of Corollary \ref{th4.4}, there exist a smooth function $\lambda$ on $\mathcal{U}\subset M$ such that
\begin{equation}\label{eq5.2}
 Ric(X,Y)=\lambda g(X,Y),\quad\forall X,Y\in\Gamma(TM_{|\mathcal{U}}).
\end{equation}
Thus, from relations (\ref{eq5.1}) and (\ref{eq5.2}), we get
\begin{equation}\label{eq5.3}
B(A_{N}\xi,X)=0,\quad\forall X\in\Gamma(TM_{|\mathcal{U}}).
\end{equation}

A lightlike hypersurface $(M,g,S(TM))$ of a semi Riemannian manifold $(\overline{M},\overline{g})$ is said to be locally symmetric \cite{Gun-Sahin-Kil}, if and only if for any $X,Y,Z,T,W\in\Gamma(TM)$ and $N\in\Gamma(tr(TM))$ the following hold
\begin{equation}\label{eq5.4}
 g((\nabla_{W}R)(X,Y)Z,PT)=0 \quad  and \quad \overline{g}((\nabla_{W}R)(X,Y)Z,N)=0.
\end{equation}
That is $(\nabla_{W}R)(X,Y)Z=0$.

\begin{proposition}\label{th5.1}
Let $(M,g)$ be a lightlike hypersurface $(M,g)$ of a semi-Riemannian manifold of constant sectional curvature $(\overline{M}(c),g)$. The lightlike hypersurface $(M,g,S(TM))$ is locally symmetric if and only if it is totally geodesic.
\end{proposition}
\textbf{Proof:} Using relation (\ref{eq4.8}), we obtain, for any $V,X,Y\in\Gamma(TM)$,
\begin{eqnarray}\label{eq5.5}
 \overline{g}((\nabla_{V}R)(X,Y)\xi,N)&=&\overline{g}(R(X,Y)\stackrel{*}{A}_{\xi}V,N)\crcr
&=&c\{B(V,Y)\eta(X)-B(V,X)\eta(Y)\}.
\end{eqnarray}
If $(M,g,S(TM))$ is locally symmetric, by taking $Y=\xi$ into (\ref{eq5.5}), we obtain $B(V,X)=0,~~\forall V,X\in\Gamma(TM)$, that is $M$ is totally geodesic.
Conversely if $M$ is totally geodesic, since $R=\overline{R}_{|TM}$, we obtain  $\nabla_{V}R(X,Y)Z=\overline{\nabla}_{V}\overline{R}(X,Y)Z=0,~~\forall V,X,Y,Z\in\Gamma(TM)$. This complete the proof.$\quad\square$

In what follows, we consider curvature operator on a smooth manifold defined by
\begin{equation}\label{eq5.6}
R(X,Y)=\nabla_{X}\nabla_{Y}-\nabla_{Y}\nabla_{X}-\nabla_{[X,Y]}.
\end{equation}
A lightlike hypersurface $(M,g,S(TM))$ of a semi-Riemannian manifold $(\overline{M},\overline{g})$ is said to be semi-symmetric if the following condition is stisfied (see \cite{Sahin})
\begin{equation}\label{eq5.7}
 (R(V_{1},V_{2})\cdot R)(X,Y,Z,T)=0\quad\forall V_{1},V_{2},X,Y,Z,T\in\Gamma(TM)
\end{equation}
where $R$ is the induced Riemann curvature on $M$.
This is equivalent to
$$-R(R(V_{1},V_{2})X,Y,Z,T)- ... -R(X,Y,Z,R(V_{1},V_{2})T)=0.$$
In general the condition (\ref{eq5.7}) is not equivalent to $(R(V_{1},V_{2})\cdot R)(X,Y)Z=0$ as in the non-degenerate setting. Indeed, by direct calculation we have for any $V_{1},V_{2},X,Y,Z,T\in\Gamma(TM)$,
\begin{eqnarray}\label{eq5.8}
&&(R(V_{1},V_{2})\cdot R)(X,Y,Z,T)=\crcr
&&\qquad\qquad g((R(V_{1},V_{2})\cdot R)(X,Y)Z,T)+(R(V_{1},V_{2}).g)(R(X,Y)Z,T).
\end{eqnarray}

\begin{theorem}\label{th5.2}
Let $(M,g)$ be a (proper) totally umbilical lightlike hypersurface of a semi-Riemannian manifold of constant sectional curvature $(\overline{M}(c),\overline{g})$. Then for an admissible screen distribution $S(TM)$, the lightlike hypersurface $(M,g,S(TM))$ is semi-symmetric.
\end{theorem}
\textbf{Proof:} Suppose that $S(TM)$ is an admissible screen distribution of a totally umbilical lightlike hypersurface $(M,g)$. In virtue of Theorem \ref{th4.1}, the screen distribution $S(TM)$ is totally umbilical. Then, by using relations (\ref{eq4.2.1}) and (\ref{eq4.3.1}), the induced curvature tensor $R$ associated to $S(TM)$ is given by
\begin{equation}\label{eq5.9}
 R(X,Y)Z=c\{g(Y,Z)X-g(X,Z)Y\}+\rho\lambda\{g(Y,Z)PX-g(X,Z)PY\},
\end{equation}
where $\rho$ and $\lambda$ are smooth functions on a coordinate neighbourhood $\mathcal{U}\subset M$.
By straightforward calculation using (\ref{eq5.9}), we obtain
$$(R(V_{1},V_{2})\cdot R)(X,Y,Z,T)=0,\quad\forall V_{1},V_{2},X,Y,Z,T\in\Gamma(TM),$$
that is $(M,g,S(TM))$ is semi-symmetric.$\quad\square$

\begin{theorem}\label{th5.3}
Let $(M,g)$ be a pointwise Osserman lightlike hypersurface of a semi-Riemannian manifold of constant sectional curvature $(\overline{M}(c),\overline{g})$ such that $A_{N}\xi$ is a non-null vector field. Then for an admissible screen distribution $S(TM)$, the lightlike hypersurface $(M,g,S(TM))$ is semi-symmetric if and only if it is totally geodesic.
\end{theorem}
\textbf{Proof:} The induced curvature satisfies relation (\ref{eq4.8}). By straightforward calculation we obtain, for any $V_{1},V_{2},X,Y,Z\in\Gamma(TM)$
\begin{eqnarray}\label{eq5.10}
&&(R(V_{1},V_{2})\cdot R)(\xi,X,Y,Z)=c\big\{B(V_{2},Y)g(V_{1},X)g(A_{N}\xi,Z)\crcr
&&\hspace{2cm}-B(V_{1},Y)g(V_{2},X)g(A_{N}\xi,Z)-B(X,V_{1})g(V_{2},Y)g(A_{N}\xi,Z)\crcr
&&\hspace{2cm}+B(X,V_{2})g(V_{1},Y)g(A_{N}\xi,Z)-B(X,Y)g(A_{N}\xi,V_{1})g(V_{2},Z)\crcr
&&\hspace{2cm}+B(X,Y)g(A_{N}\xi,V_{2})g(V_{1},Z)\big\}-B(V_{2},X)B(A_{N}V_{1},Y)g(A_{N}\xi,Z)\crcr
&&\hspace{2cm}+B(V_{1},X)B(A_{N}V_{2},Y)g(A_{N}\xi,Z)-B(X,A_{N}V_{1})B(V_{2},Y)g(A_{N}\xi,Z)\crcr
&&\hspace{2cm}+B(X,A_{N}V_{2})B(V_{1},Y)g(A_{N}\xi,Z)-B(X,Y)B(V_{2},Z)g(A_{N}\xi,A_{N}V_{1})\crcr
&&\hspace{2cm}+B(X,Y)B(V_{1},Z)g(A_{N}\xi,A_{N}V_{2}).
\end{eqnarray}
So, suppose that $M$ is semi-symmetric, by taking $V_{1}=\xi$ into (\ref{eq5.10}), we get
\begin{eqnarray}
&&B(V_{2},X)B(A_{N}\xi,Y)g(A_{N}\xi,Z)+B(X,A_{N}\xi)B(V_{2},Y)g(A_{N}\xi,Z)\crcr
&&\hspace{4cm}+B(X,Y)B(V_{2},Z)g(A_{N}\xi,A_{N}\xi)=0.
\end{eqnarray}
By assymption and using Corollary \ref{th4.4}, we infer that the relation (\ref{eq5.3}) is satisfied. Thus, we obtain
\begin{equation}\label{eq5.11}
B(X,Y)B(V_{2},Z)g(A_{N}\xi,A_{N}\xi)=0,\quad\forall V_{2},X,Y,Z\in\Gamma(TM).
\end{equation}
Since $A_{N}\xi$ is non-null, by taking $V_{2}=X$ and $Z=Y$ into (\ref{eq5.11}), we infer that $B(X,Y)=0,\forall X,Y\in\Gamma(TM)$, that is $M$ is totally geodesic. Conversely if $M$ is totally geodesic, since $R=\overline{R}_{|TM}$, we obtain  $(R(V_{1},V_{2})\cdot R)(X,Y,Z,T)=(\overline{R}(V_{1},V_{2})\cdot \overline{R})(X,Y,Z,T)=0$, for any  $V_{1},V_{2},X,Y,Z,T\in\Gamma(TM)$. This complete the proof.$\quad\square$\vspace{0.2cm}\\
From Proposition \ref{th5.1} and Theorem \ref{th5.3}, we have the following result.
\begin{corollary}\label{th5.4}
Let $(M,g)$ be a pointwise Osserman lightlike hypersurface of a semi-Riemannian manifold of constant sectional curvature $(\overline{M}(c),\overline{g})$ such that $A_{N}\xi$ is a non-null vector field. Then for an admissible screen distribution $S(TM)$, the lightlike hypersurface $(M,g,S(TM))$ is locally symmetric if and only if it is semi-symmetric.
\end{corollary}

A lightlike submanifold $M$ of a semi-Riemannian manifold $\overline{M}$ is said to be Ricci semi-symmetric if the following condition is satisfied
\begin{equation}\label{eq5.12}
R(V_{1},V_{2}\cdot Ric)(X,Y)=0,~~\forall V_{1},V_{2},X,Y\in\Gamma(TM),
\end{equation}
where $R$ and $Ric$ are induced Riemann curvatrure and Ricci tensor on $M$, respectively. The latter condition is equivalent to
$$-Ric(R(V_{1},V_{2})X,Y)-Ric(X,R(V_{1},V_{2})Y)=0$$

\begin{theorem}\label{th5.5}
Let $(M,g)$ be a pointwise Osserman lightlike hypersurface of a semi-Riemannian manifold of constant sectional curvature $(\overline{M}(c),\overline{g})$. Then for an admissible screen distribution $S(TM)$, the lightlike hypersurface $(M,g,S(TM))$ is Ricci semi-symmetric.
\end{theorem}
\textbf{Proof:} Let $R$ be an induced curvature tensor associated to the admissible screen distribution $S(TM)$.  By virtue of Corollary \ref{th4.4}, there exists a smooth function $\lambda$ on $\mathcal{U}\subset M$ such that $Ric(X,Y)=\lambda g(X,Y),~~\forall X,Y\in\Gamma(TM_{|\mathcal{U}})$. Thus, since $R$ is an algebraic curvature tensor, we have, for any $V_{1},V_{2},X,Y\in\Gamma(TM)$
\begin{eqnarray*}
(R(V_{1},V_{2})\cdot Ric)(X,Y)&=&-Ric(R(V_{1},V_{2})X,Y)-Ric(X,R(V_{1},V_{2})Y)\\
 &=&-\lambda R(V_{1},V_{2},X,Y)-\lambda R(V_{1},V_{2},Y,X)\\
&=&0,
\end{eqnarray*}
that is $(M,g,S(TM))$ is Ricci semi-symmetric.$\quad\square$

config: arxiv


TITLE: Is the localization of such a Noetherian domain at every maximal ideal Artinian?
QUESTION [1 upvotes]: My question here is inspired by this question here. I am not asking how to prove the problem there, but if an alternative approach is possible.
Suppose $A$ is a Noetherian integral domain in which for every maximal ideal $\mathfrak{m}$, the dimension of $\mathfrak{m}/\mathfrak{m}^2$ as a vector space over $A/\mathfrak{m}$ is $1$. Given any maximal ideal $\mathfrak{m}$ of $A$, we know that $A_\mathfrak{m}$ is local and Noetherian with maximal ideal $I = \mathfrak{m}_\mathfrak{m} $.  From here I want to conclude from here that $A_\mathfrak{m}$ is actually an Artinian ring as well. Can I argue along the following lines? I want to prove that there is an $n$ such that  $I^n = I^{n+1}$. This is because if $I$ is like this then I can apply a Nakayama Lemma argument to show that $I$ is nilpotent and then show that $A_\mathfrak{m}$ is Artinian as desired.
However how can I translate the condition on $\mathfrak{m}/\mathfrak{m}^2$ having $A/\mathfrak{m}$ - dimension 1 into something concrete in the localisation?
Perhaps it may well be that $A_\mathfrak{m}$ does not need to be Artinian and I am wrong.
Thanks.

REPLY [5 votes]: A ring is Artinian if and only if it is Noetherian and of dimension zero. The dimension of $A_\mathfrak{m}$ is equal to the height of $\mathfrak{m}$. Because $A$ is Noetherian, $A_\mathfrak{m}$ is as well, so $A_\mathfrak{m}$ is Artinian if and only if $\mathfrak{m}$ has height zero. So you would need $\mathfrak{m}$ to be both maximal and minimal. But in a domain (such as $A$), the only minimal prime is $0$, so this can only happen if $\mathfrak{m}=0$, in which case $A$ is a field...but your $A$ is not a field because $\mathfrak{m}/\mathfrak{m}^2\neq 0$.
In fact, your $A$ is a Dedekind domain. Indeed, the $k(\mathfrak{m}):=A/\mathfrak{m}$-vector spaces $\mathfrak{m}/\mathfrak{m}^2$ and $\mathfrak{m}A_\mathfrak{m}/(\mathfrak{m}A_\mathfrak{m})^2$ are canonically isomorphic, so both are $1$-dimensional by assumption. By Krull's principal ideal theorem, $\dim(A_\mathfrak{m})\leq 1$. I've explained why it's not possible that $\dim(A_\mathfrak{m})=0$, so this inequality is an equality for all $\mathfrak{m}$. Thus each $A_\mathfrak{m}$ is a regular local ring of dimension $1$, i.e., a discrete valuation ring. So $A$ is Dedekind.


TITLE: Prove that $\sin(x^2) = \mathcal{o}(x)$
QUESTION [2 upvotes]: I tried to do this like that:
$$ \sin(x^2) = \mathcal{o}(x) \iff \lim_{x \to 0} \frac{\sin(x^2)}{x} = 0$$
we could get $\sin(x^2)$ from Taylor series.
For $x_0 = 0$,  $T_n = 0$ for every $n$.
So from Peano remainder ($\lim_{x\to 0}r(x)=0$) we have:
$$\lim_{x\to 0}\sin(x^2) = 0$$
which means $$\lim_{x \to 0} \frac{\sin(x^2)}{x} = 0$$
Is this correct?

REPLY [1 votes]: You can use L'Hopital's rule if you want to show $\lim_{x \to 0} \frac{\sin x^2}{x} = 0$, because taking derivatives of numerator and denominator gives $\lim_{x \to 0} \frac{2x \cos x^2}{1} = 0$.


TITLE: Limit of $\frac{(2n-1)!! \sqrt{n}}{n! 2^n}$
QUESTION [0 upvotes]: How to prove that $\lim\limits_{n\to \infty}\frac{(2n-1)!! \sqrt{n}}{n! 2^n}=\frac{1}{\sqrt{\pi}}$ ?
The expression can also be written as
$$ \sqrt{n} \frac{(2n-1)(2n-3)\cdot\cdot\cdot3\cdot1}{2n (2n-2) \cdot\cdot\cdot4\cdot2}=\sqrt{n}\left(1-\frac{1}{2n}\right)\left(1-\frac{1}{2n-2}\right)...\left(1-\frac{1}{2}\right)$$

REPLY [1 votes]: Using the fact that $(2n-1)!!=\frac{(2n)!}{n!2^n}$, it suffices to calculate
$$
\lim_{n\to\infty}\frac{(2n)!\sqrt{n}}{(2^nn!)^2}.
$$
This is now a straightforward application of Stirling's approximation


text:
\begin{document}

\title[Remarks on another proof of Paley--Wiener theorems]{Some remarks on another proof of geometrical Paley--Wiener theorems for the Dunkl transform}

\translator{}

\dedicatory{}

\author{Marcel de Jeu}

\begin{abstract}
We argue that another proof by Trim\`eche of the geometrical form of the
Paley--Wiener theorems for the Dunkl transform is not correct.
\end{abstract}

\date{}

\subjclass[2000]{Primary 33C52; Secondary 33C67}

\keywords{Dunkl transform, Paley--Wiener theorem}

\address{M.F.E.~de~Jeu\\
         Mathematical Institute\\
         Leiden University\\
         P.O. Box 9512\\
         2300 RA Leiden\\
         The Netherlands}

\email{mdejeu@math.leidenuniv.nl}

\urladdr{}

\maketitle

\section{Introduction}
In \cite{trimeche} a proof of the geometrical form of the Paley--Wiener
theorems for the Dunkl transform is presented. It is argued in
\cite{pw_remarks} that this proof is not correct. A new proof has appeared in
\cite{trimeche_2}. In this note we argue that this new proof is not correct.

The material which is presented below has previously been communicated to the
author. It is our opinion that at this moment the geometrical forms of the
Paley--Wiener theorems for the Dunkl transform are still unproven, and that
only partial results have been established \cite{pw_new}.

\section{Arguments}

In \cite{trimeche_2} the geometric form of the Paley-Wiener theorem for the
Dunkl transform is stated for functions as Theorem~5.1. Our arguments concern
the proof of this theorem. In formulating them, we will use the notation and
definitions of \cite{trimeche_2}.

In the proof of Theorem~5.1, the function $\phi$ on $\R^d$ is defined in
equation (5.3) as
\begin{equation}\label{eq:phidefinition}
\phi(x)=\int_{\R^d} f(y)K(iy,x)\frac{\omega_k(y)}{(1+\Vert
y\Vert^2)^p}dy\quad(x\in\R^d).
\end{equation}
Here $f$ is a function of Paley--Wiener type corresponding to a $W$-invariant
compact convex subset $E$ of $\R^d$, as in the statement of Theorem~5.1, and
$p$ is an integer such that $p>\gamma+\frac{d}{2}+1$. Since the constant
$\gamma$ is assumed to be strictly positive in line~-10 of page 4, we see that
that $p\geq 2$.

After a computation involving Riemann sums and contour integration, it is then
concluded on line~9 of page~13 that $\phi$ has support in the set $E$. In
particular, $\phi$ has compact support. Since it has already been observed in
line~10 on page~11 that $\phi$ is smooth, the easy part of the Paley--Wiener
theorem implies that the Dunkl transform of $\phi$ is of Paley--Wiener type. In
particular, the Dunkl transform of $\phi$ has an entire extension to $\C^d$.
The inversion theorem and our equation~\eqref{eq:phidefinition} therefore show
that the function
\begin{equation}\label{eq:hasanextension}
\frac{f(y)}{(1+\Vert y\Vert^2)^p}
\end{equation}
on $\R^d$ has an entire extension to $\C^d$.

Now take $E$ to be an arbitrary closed ball, centered at the origin. Certainly,
if $f$ is the ordinary Fourier transform of a smooth function with support in
$E$, then $f$ is of Paley--Wiener type corresponding to the $W$-invariant
compact convex set $E$, so that the above reasoning applies to $f$.

Combining the previous two paragraphs, we conclude that the function in our
equation~\eqref{eq:hasanextension} has an entire extension from $\R^d$ to
$\C^d$, for all $f$ which are the ordinary Fourier transform of a compactly
supported smooth function. Since $f$ has then itself an entire extension, we
see that the meromorphic function
\[
\frac{f(y)}{(1+(y,y))^p}
\]
on $\C^d$ has a removable singularity along the divisor $\{y\in\C^d\mid
(y,y)=-1\}$, where $(\,.\,,\,.\,)$ is the standard bilinear form on $\C^d$.
Since $p\geq 2>0$, we conclude that each function $f$, which is the entire
extension of the ordinary Fourier transform of a smooth compactly supported
function, vanishes on this divisor.

Now let $f$ be the Fourier transform of a non-zero positive smooth compactly
supported function. Then $f(i,0,\ldots,0)>0$. Since
$((i,0,\ldots,),(i,0,\ldots,))=-1$ we must also have $f(i,0,\ldots,)=0$ by the
reasoning above. This is a contradiction.

\medskip
The above arguments are not based on the details of the computation with
Riemann sums and contour integration, but they are concerned with the
impossibility of the support of $\phi$ being contained in $E$. These arguments
therefore show not only that this computation contains a technical inaccuracy,
but they also show that the technique of this computation can not be corrected,
since the conclusion which the computation is supposed to yield does not hold.


TITLE: How to evaluate this triple integral
QUESTION [0 upvotes]: I'm having some issues to go on with exercise below
$$\iiint_B z \text{ d}V\ $$
where solid $B$ is in the first octant, bounded by the cylinder $y^2+z^2=1$ and planes $y=x$ and $x=0$ .

REPLY [3 votes]: Hint. We have that
$$B=\left\{(x,y,z)\in\mathbb{R}^3\;:\; 0\leq x\leq  y\leq 1,\ 0\leq z\leq \sqrt{1-y^2}\right\}$$
then
$$\iiint_B z \text{ d}V=\int_{y=0}^1\int_{x=0}^y\int_{z=0}^{\sqrt{1-y^2}}z\,dz dx dy.$$
Can you take it from here? The final result should be $1/8$.
Another approach is by cross-sections:
$$\iiint_B z \text{ d}V=\int_{t=0}^1t\cdot \mbox{area}(B\cap \{z=t\}) dt=\int_{t=0}^1t\cdot \frac{1-t^2}{2}\, dt=\frac{1}{8}\left[-(1-t^2)^2\right]_0^1=\frac{1}{8}$$
because $B\cap \{z=t\}$ is a right isosceles triangle with a cathetus of length $\sqrt{1-t^2}$.


TITLE: The rate change of the radius of a coil.
QUESTION [1 upvotes]: Suppose I have a tube of radius $r_0$ that I want to wrap a sheet of length $l$ and thickness $\Delta x$.
Assuming the radius changes only when the paper overlaps the where the previous section overlapped, how does the radius change?
If you are having trouble visualizing this, consider a roll of toilet paper.  How does the radius of the roll change as I remove (or add) a fixed length of paper?

REPLY [0 votes]: The easiest way would be to invert the relation. Since we know that the radius of the roll $r$ increases with respect to the sheet length $l$, we can find $l$ in terms of $r$ first. That's easy because if we look along the axis, the total area taken up by the paper is approximately $π(r^2-{r_0}^2)$, so the total length is about $\frac{π(r^2-{r_0}^2)}{Δx}$. This is a first-order approximation of course, since we assume that $Δx$ is small enough. So the answer would be $r \approx \sqrt{\frac{lΔx}{π}+r_0^2}$


TITLE: If $x\equiv a\pmod{n}$, prove that either $x\equiv a\pmod{2n}$ or $x\equiv a+n\pmod{2n}.$
QUESTION [0 upvotes]: If $x\equiv  a\pmod{n}$, prove that either $x\equiv a\pmod{2n}$ or $x\equiv a+n\pmod{2n}.$

Well I am a bit stuck in this congruence problem. I tried out :-
$n\vert x-a$ so for $2n|x-a$, $x-a$ must be even but I am not able to prove what if $x-a$ is odd, how can I prove that $x-(a+n)$ will be even and divisible by $2n$. Is this the right approach? Though it seems a very vague one, I can't think of any other.

REPLY [3 votes]: Hint: The hypothesis means that $x = a + kn$ for some integer $k$. What can you conclude about $x$ modulo $2n$ when $k$ is even? How about when $k$ is odd?


TITLE: How does one find the axis of rotation for a pure rotation matrix when said matrix is also symmetric?
QUESTION [5 upvotes]: I'm a programmer working on an advanced C++ 3D mathematics library.
Now, things have been going well, in fact, basically everything about the library has been fully implemented, but one final bit of code still alludes me: finding the axis of rotation for a pure rotation matrix when said matrix is also symmetric.
I've got a nice bit of math going when it comes to non-symmetric matrices
Given a non-symmetric 3x3 pure rotation matrix [M]

M = { { a, b, c },
      { d, e, f },
      { g, h, i } }

det(M) = 1

M * T(M) = T(M) * M = I

M =/= T(M)

an eigenvector [u] which sits along the axis of rotation can be found

u = { h - f,
      c - g,
      d - b }

such that its normal is axis of rotation [r] of the matrix

r = u / |u|

but this math breaks the moment you give it a symmetric matrix, as the 'h - f', 'c - g', and 'd - b' portions will all resolve to zero, which obviously isn't the normal vector I want.
Now, I do understand linear algebra, but only a little bit. I've been researching this problem for a few days now, and while there are resources that talk about this, most of them either don't address the problem I'm having, or explain it in a way that my scrub brain simply can't keep up with.
They tell me to do things like 'diagonalize M and solve for u', but I've not a clue what that actually entails doing, let alone in a generalized way, let alone (even more) teaching my C++ library to do it in a generalized way given any symmetric pure rotation matrix.
So ya, that's my plight. Hoping that one of y'all could help bail me out on this one and show me how to solve this problem. :D
Btw, again, this is needed for writing code, so if your answer could be written in a way that addresses that need and also the fact that I'm a linear algebra noobie, that would be awesome.
Thanks in advance!

REPLY [5 votes]: If the matrix is symmetric, then $M^TM=M^2=I$, so it’s a 180-degree rotation about some axis. The method you’re using is known to fail for this angle, as you’ve discovered. (Indeed, among the conditions in the gray block in your question is $M\ne M^T$.) Essentially, it examines the skew-symmetric part of $M$, which is $\sin\theta$ times the “cross-product matrix” of the rotation axis. This unfortunately vanishes when $\theta=\pi$.
This question and this one describe a different method that uses the symmetric part of $M$ instead: Compute $$T = M+M^T-(\operatorname{tr}(M)-1)I.$$ Any nonzero row or column of this matrix is a vector parallel to the rotation axis.

Addendum (2019.11.19): As you noted, “diagonalize $M$ and solve for $u$” isn’t particularly helpful advice: most rotation matrices aren’t diagonalizable over the reals in the first place. The rotation axis is fixed by the rotation, though, so what you can do for any rotation matrix is compute the eigenspace of $1$, i.e., find a null vector of $M-I$. However, the method in your question and the above method are more efficient. Because of roundoff and other computational errors, in practice computing a null vector of a matrix is done by computing its SVD and taking the singular vector that corresponds to the least singular value.


TITLE: Characteristics of a tight string
QUESTION [3 upvotes]: There was a question which I came across in newtons law of motion chapter in the physics book by H.C.Verma which was based upon simple atwood machine (atwood machine consists of two unequal masses connected by string going over a clamped smooth , light pulley).
The question asks us to consider a case where the heavier mass of the atwood machine were stopped for a moment 2 seconds after system is set into motion, and to find time elapsed before string is tight again.
I am willing to do the question by myself but I was unable to understand the condition of tight string. When do we say that the string is tight? Is it at a particular tension or is it when minimum value of tension is just greater than 0?

REPLY [1 votes]: Starting with tightness requirement:
"...Find time elapsed before string is tight again."
Clearly, at some low enough tension a string would still be stretchable to overcome curls in its fibers. The string will be tight/taut provided each mass has a weight in excess of that required to stretch the string to normal length, i.e. no further stretching and below breaking force.
The following sequence of events will happen:
a. both masses begin motion $\rightarrow$ b. 2 seconds elapse $\rightarrow$ c. heavier mass stopped $\rightarrow$ d. string slackens $\rightarrow$ e. lighter mass stops upward motion $\rightarrow$ f. lighter mass drops back down $\rightarrow$ g. lighter mass stops (rope tight/taut)
Assuming the two fore-mentioned are met, and as stated in the question, the pulley is smooth (no friction), then at start of motion, the two masses accelerate at a rate:
$a = F/(m_1+m_2) \tag1$
where the mass weight imbalance is
$F = (m_2-m_1)g\tag2$ At time 2 seconds after start of motion, the velocity of the masses, the heavier one $m_2$ falling, the lighter one $m_1$ rising, is:
$v = u+at \Rightarrow v = 2a \tag3$
substituting Equation $2$ into $1$ and $1$ into $3$:
$v = 2g{{m_2-m_1}\over{m_1+m_2}} \tag4$
As one stops the falling of the heavier mass, then the lighter mass $m_1$ will continue to rise as it decelerates due to gravity. The string loses its tautness. In this process, the final velocity from Equation $4$ above becomes $m_1$'s initial velocity, and its final velocity this time is $0$.
$v = u + at \Rightarrow t=(v-u)/a \Rightarrow t = (0 - (2g{{m_2-m_1}\over {m_1+m_2}}))/g \tag5$
therefore time until lighter mass $m_1$ stops upward motion after the heavier mass $m_2$ is stopped is:
$t = {{2(m_1-m_2)} \over{m_1+m_2}}\tag6$
Finally, the lighter mass will drop to the 'rope tight' position. Now note that since there is no friction in the pulley, and gravity acts symmetrically on the lighter mass as it rises vs as it falls:
(downward velocity of the lighter mass after falling back down to 'rope tight' position) $=$ (upward velocity of lighter mass just after heavier mass was stopped):
$v = u + at$ where v and u are switched for the drop journey of $m_1$ vs its rise journey. The time is therefore the same, and total time is twice the rise time (from Equation 6):
$t_{m_1 rise-and-fall} = 2(t_{m_1 rise}) = {{4(m_1-m_2)} \over{m_1+m_2}}\tag7$


TITLE: Let $[{\bf a} \times {\bf b} \space \; {\bf b} \times {\bf c}\space\; {\bf c} \times {\bf a} ]=k[{\bf a}\space{\bf b}\space{\bf c}]^2$. Find $k$.
QUESTION [2 upvotes]: Let $[{\bf a} \times {\bf b} \space \; {\bf b} \times {\bf c}\space\;  {\bf c} \times {\bf a} ]=k[{\bf a}\space{\bf b}\space{\bf c}]^2$. Find $k$.  Here, $[{\bf u}\ {\bf v} \ {\bf w}]={\bf u}\cdot({\bf v}\times {\bf w})$.

My attempt:
Writing scalar triple product as : $({\bf a} \times {\bf b}) \cdot \big(({\bf b} \times {\bf c}) \times ({\bf c}\times {\bf a})\big)$. Not able to proceed next.

REPLY [1 votes]: As hinted in a comment, we use the identity ${\bf u}\times({\bf v}\times {\bf w})=({\bf u}\cdot {\bf w}){\bf v}-({\bf u}\cdot {\bf v}){\bf w}$ (I proved it here). Let $$\Delta={\bf a}\cdot({\bf b}\times {\bf c})={\bf b}\cdot({\bf c}\times {\bf a})={\bf c}\cdot({\bf a}\times {\bf b}).$$
 First we compute
$$({\bf b}\times {\bf c})\times ({\bf c}\times {\bf a})=\big(({\bf b}\times {\bf c})\cdot {\bf a}\big){\bf c}-\big(({\bf b}\times {\bf c})\cdot {\bf c}){\bf a}=\Delta {\bf c}-0{\bf a}=\Delta {\bf c}.$$
Therefore
$$({\bf a}\times {\bf b})\cdot\big(({\bf b}\times {\bf c})\times ({\bf c}\times {\bf a})\big)=({\bf a}\times {\bf b})\cdot \Delta {\bf c}=\Delta\big(({\bf a}\times {\bf b})\cdot {\bf c}\big)=\Delta^2.$$
Hence $k=1$.


TITLE: Relationship between 'Exit Pupil' and 'Pupil Plane'
QUESTION [1 upvotes]: I am having trouble to understand the relationship between the 'Pupil Plane' from the Fourier Optics Theory and the 'Exit Pupil' from Geometrical Optics. It seems that both are different concepts, but are somehow connected to each other.
My understanding:
Exit Pupil: Image of Aperture Stop (seen from image side)
Pupil Plane: Is positioned at the aperture of the optical system. The 'pupil function' is given by the aperture stop shape at this position. The Fourier transform of this 'pupil function' gives the coherent impulse response of the system.
What I've read and don't understand completely
According to 1:

In an aberration-free coherent imaging system, the light field distribution at the pupil plane (i.e., the back focal plane of the objective lens) is directly proportional to the Fourier transform of the light field at the object plane. Therefore, the spatial coordinates at the object plane and the pupil plane can be expressed as (x, y) and (kx, ky), respectively, with kx and ky as the wave number in the x and y directions. Due to such a Fourier relationship, aberrations of an imaging platform are often characterized at the pupil plane for simplicity [Goodman - Introduction to Fourier Optics].

However, I dont understand why the back focal plane of an optical system should necessarily be a pupil plane. Furthermore, I cannot find the exact quote in [Goodman - Introduction to Fourier Optics].
Acoording to 2:

The pupil function and the pupil planes of a system are not the same thing. The entrance and exit pupils--which together are known as the pupil planes--are the planes in which the images of the system's aperture stop are located. The pupil function, however, is not an image of the aperture stop of the system; it's a 2D Fourier transform of a field.
There is never-the-less a relationship between the pupil function and the plane of the exit pupil. The pupil function represents the relative amplitude and phase of the field on the surface of a so-called reference sphere that intersects the optics axis in the plane of the system's exit pupil.

However, I can't find another source that confirms this statement and I am not sure if this is really correct.
Questions
Question 1: Is my understanding of the defintion of the words 'Exit Pupil' and 'Pupil Plane' correct?
Question 2: What is the relationship between the 'exit pupil' and the 'pupil plane'? Are they at the same position in an optical system? Is there a mathematical expression that states the light field relationship between both planes?

REPLY [1 votes]: Coming at it first from the geometrical side: in any optical system there are several optical spaces separated by focussing surfaces. The first space has the object at one end and the entrance pupil at the other. In each space there are images of these two planes, called image planes and pupil planes (possibly virtual). At the output end we have the exit pupil and the final image. One of the actual pupil planes contains the aperture which limits the ray bundle; this is often but does not have to be the entrance pupil plane. It also is often hard edged, but does not have to be, it may be graded to be a smooth function. So pupil planes are just images, either upstream or downtream, of the actual physical aperture.
The key word in quotation 1 is 'aberration free'. This means that all pupil planes are equivalent geometrically. This is usually rather idealised! The relationship of the light field between image planes (all of them) and pupil planes (all of them) is then a Fourier Transform one. Since a sharp-edged function has an infinite Fourier transform, the distributions in the internal planes may get truncated, leading to further complications.


TITLE: Extension of a bounded linear map
QUESTION [3 upvotes]: Let $K \subset H$ both be Hilbert spaces. I'm trying to prove that for any bounded linear map $T: K \to X$ with $X$ a normed space, there exists an extension $\tilde{T}: H \to X$ such that the norm is preserved ($||T|| = ||\tilde{T}||$). Many of the answers I find on the internet about this have the extra condition that $X$ is Banach, I'm not able to come up with a solution when $X$ is not. What I tried so far is the following:

Extend $T: K -> X$ to $\bar{T}: \bar{K} \to \tilde{X}$ where $\bar{K}$ is the closure of the Hilbert subspace $X$ and $\tilde{X}$ is the completion of the normed space $X$ as we can always associate a Banach space to a normed space. Recalling that $K$ is dense in $\bar{K}$ this gives us an extension via the BLT.

Project $H$ onto the closed convex subset $\bar{K}$; $P_{\bar{K}}: H \to \bar{K}$

Taking the composition of these two maps we get a map $\bar{T} \circ P_\bar{K}: H \to \tilde{X}$.

I think I have to add a map to this to go from $\tilde{X} \to X$ or somehow be able to prove that any complete space gets mapped to a complete space under $T$ (which is not true I think).

Besides proving this I am also trying to find a pair of normed spaces $Y \subset X$, a two-dimensional normed space $Z$ and a bounded linear map $S: Y \to Z$ such that any extension $\tilde{S}: X \to Z$ satisfies $||\tilde{S}|| > ||S||$ indicating that the norm preservation relies on both $K \subset H$ being Hilbert spaces and not just normed spaces. I am also unable to come up with this but I think being able to complete the proof will help with this part too.

REPLY [2 votes]: We start out with the assumption that $K$ is Hilbert, which includes completeness in its definition. Therefore $K$ is already a closed convex subset of $H$ and step 1 is not needed. Just compose projection of $H$ onto $K$ with $T$ to get the desired map.
$$H \stackrel{P_K}{\to} K\stackrel{T}{\to} X$$


TITLE: Why does a voltmeter have to have nearly no current passing through it?
QUESTION [5 upvotes]: I understand that a high resistance of a voltmeter will allow nearly no current to pass through it. However, why is this condition required? Let's consider a simple circuit that consists of a cell, and a resistor. The potential drop across the resistor will be the same as the potential difference across the cell. Now, if we introduce a voltmeter across the resistor, since the voltmeter and the resistor are connected in parallel, the potential drop across the resistor and the voltmeter will be the same, although the current passing through the resistor and the voltmeter will vary depending on their resistances. So, to me, it seems like the potential drop across the resistor (which the voltmeter measures) is independent of the current flowing through the resistor and the voltmeter, and thus independent of the resistance of the voltmeter.
So, why should there be nearly no current flowing through a voltmeter (which is the reason why the voltmeter has to have a very high resistance)?

REPLY [2 votes]: Without Voltmeter
See the circuit given below. Before inserting the voltmeter, the current through the circuit is $ E /(R_1 +R_2)$ where $E$ is the EMF of the battery. So the potential difference across $R_2$ is
$$V_0=E\frac{R_2}{R_1+R_2}$$

Image source
After Voltmeter
Let ther resistance of voltmeter be $R_V$. Then the equivalent resistance of the circuit will be
$$R_{\text{eq}} =R_1 + \frac{R_2 R_V}{R_2+R_V}$$
Therefore the current through the circuit will be
$$I_{\text{total}}= \frac{E}{R_{\text{eq}}}$$
And the current through $R_2$ will be,
$$I_2 = I_{\text{total}} \frac{R_V}{R_2 + R_V}$$
Therefore the potential difference across $R_2$ will be
$$V=I_2 R_2 = I_{\text{total}} \frac{R_V R_2}{R_2 + R_V}= \frac{E}{R_{\text{eq}}}\frac{R_V R_2}{R_2 + R_V} =\frac{E}{ R_1 + \frac{R_2 R_V}{R_2+R_V}}\frac{R_V R_2}{R_2 + R_V}  $$
which simplifies to
$$V= E \frac{R_2 R_V}{R_1 R_2 +R_2 R_V + R_V R_1}$$
Clearly this is different from the original result without the voltmeter($V_0$). Also when $R_V \rightarrow \infty$, $V\rightarrow V_0$ which is expected. At all finite values of $R_V$, $V<V_0$
Intuition
When you connect a voltmeter in parallel, the equivalent resistance of the circuit decreases and the total current increases. But at the same time the current also gets divided across $R_2$ and $R_V$ which results in lower current through $R_2$. The latter effect dominates and a lower potential difference is obtained across $R_2$. So we try to minimize these two effects by increasing $R_V$. This is clearly evident from the mathematics of this scenario.

REPLY [1 votes]: The goal is indeed for the voltmeter to have the smallest possible effect on the rest of circuit. The higher the resistance, the less of an effect. If you imagine the opposite extreme is it more clear: no matter the circuit, if you take the resistance of the voltmeter to zero, the voltage measured will go to zero too (and you'll probably break something).
When I took my AC/DC electronics course, we were forced to use these ancient voltmeters (example). The challenge with these is that the resistance was actually quite low, maybe only 100 Ohms (depending on the setting), so they would dramatically effect the circuit. In order to get the correct voltage (before you attached the voltmeter) you had to (by hand)  work out how the voltmeter affected the circuit and then calculate the correct voltage.


TITLE: (perfect fractional) matching in regular hypergraphs
QUESTION [5 upvotes]: I have been looking in general for results on matching for regular hypergraphs that are not uniform, but I could not find any. Papers I found by internet search seem to focus on uniform hypergraphs (sometimes in addition to being regular). The relative scarcity of results on regular hypergraph matching is somewhat puzzling, since uniform and regular hypergraphs are duals of each other and one would expect that matching may have been studied for both kinds.
I would appreciate if someone can point me to any sources. Of most interest to me are necessary and/or sufficient conditions for existence perfect matchings and closed form expressions or bounds for fractional matchings. I am also curious to know explanations (conceptual, historical etc) for the lack of results of this kind for regular hypergraphs.
Edit: A matching is a collection of pairwise disjoint hyperedges; a perfect matching is a matching that covers all vertices.

REPLY [5 votes]: You are asking many things in the question, and I can only answer part of it:

necessary and/or sufficient conditions for existence perfect matchings

It is unlikely that a nice necessary and sufficient characterization is known, because deciding whether a given 3-regular hypergraph has a perfect matching or not is NP-complete.  Indeed, it is equivalent to the monotone 1-in-3 SAT problem, which is proved to be NP-complete by Schaefer [Sch78].  (In [Sch78], the same problem is called “ONE-IN-THREE SATISFIABILITY” without the word “monotone.”)  To see this equivalence, interpret each hyperedge as a variable and each vertex as a clause.
[Sch78] Thomas J. Schaefer.  The complexity of satisfiability problems.  In Proceedings of the 10th Annual ACM Symposium on Theory of Computing (STOC), pp. 216–226, May 1978.  DOI: 10.1145/800133.804350.