TITLE: Triangle geometry - synthetic proof
QUESTION [3 upvotes]: I'm looking for a nice synthetic proof of the following fact.
Consider a non-isosceles triangle, pick a vertex. Assume that the median and the altitude passing through this vertex are isogonal conjugates (i.e. symmetric w.r.t. the bisector of the angle). Prove that the triangle is right-angled in this vertex.

REPLY [1 votes]: Let $\triangle ABC$ have the desired property at $A$. Let $M$, $P$, $N$ be the points where the median, angle bisector, and altitude from $A$ meet side $\overline{BC}$.
Since
$$\angle BAP \cong \angle CAP \qquad \text{and} \qquad \angle MAP \cong \angle NAP$$
we have
$$\angle BAM \cong \angle CAN = C^\prime \qquad \text{and} \qquad
\angle CAM \cong \angle BAN = B^\prime$$
where $B^\prime$ and $C^\prime$ are respective complements of $B$ and $C$.
Invoking the Law of Sines in $\triangle BAM$ and $\triangle CAM$,
$$\underbrace{|\overline{AM}|\;\frac{\sin C^\prime}{\sin B} = |\overline{BM}|}_{\triangle BAM} = \frac{1}{2} |\overline{BC}| = \underbrace{|\overline{CM}| = |\overline{AM}|\; \frac{\sin B^\prime}{\sin C}}_{\triangle CAM}$$
Therefore,
$$\begin{align}
\sin B\sin B^\prime = \sin C \sin C^\prime \quad&\implies\quad \sin B\cos B = \sin C \cos C \\[4pt]
&\implies\quad \sin 2 B = \sin 2 C \\[4pt]
&\implies\quad B = C \quad\text{or}\quad 2B = \pi - 2C \\[4pt]
&\implies\quad B = C \quad\text{or}\quad B + C = \pi/2 \\[4pt]
&\implies\quad B = C \quad\text{or}\quad A = \pi/2
\end{align}$$


TITLE: Equation With Logarithm- How can we solve?
QUESTION [0 upvotes]: How can we solve $e^x-e=\ln x$?

I have tried to subtract exponents, but I can't find the solution. Please help. I know there is a simple method to solve this, but I can't remember it.

REPLY [3 votes]: I don't think there exists a closed form solution for $x$ in terms of elementary functions. However, you can use a numerical method.
I will use the Newton-Raphson method.
The process is as follows:
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \tag{1}$$
Our function to consider is given by:
$$f(x)=e^x-e-\ln(x)$$
Evaluating its derivative gives:
$$f'(x)=e^x-\frac{1}{x}$$
Substituting into $(1)$ gives:
$$x_{n+1}=x_n-\frac{e^{x_n}-e-\ln(x_n)}{e^{x_n}-\frac{1}{x_n}}$$
Applying this iteration by setting appropriate initial values of $x_0$ (An 'educated guess' for the solution), we notice that there exists 2 solutions. As the iterations $n\to \infty$, we see that $x_n\to x$.
By setting $x_0=0.1$, we see the following:
\begin{array}{c|c}n&x_n\\\hline0&0.1\\1&0.177514\\2&0.223620\\3&0.232971\\4&0.233269\\5&0.233270\\6&0.233270\\\end{array}
After many more iterations, we obtain:
$$x \approx 0.2332697479983673$$
There is an obvious solution which is $x=1$. This solution can also be found using the same method. For example, if you set $x_0=1.5$, you will see that the iterations will converge towards $1$.
You can apply this iteration using a spreadsheet or by using more sophisticated software such as MATLAB.


TITLE: Drawing conclusions by NOT using AC.
QUESTION [12 upvotes]: The existence of non-measurable subsets and functions on $\mathbb{R}$ require the use of the axiom of choice. That is, there exist models of ZF in which all subsets of (and hence all functions defined on) $\mathbb{R}$ are measurable. Does this mean that if I can define a subset or function on $\mathbb{R}$ without invoking the axiom of choice, it must be measurable?
Let's say I fully accept AC and am just looking for a quick and dirty way of proving a function is measurable.

REPLY [3 votes]: Solovay showed in the 1960's the consistency of  ZF + axiom of Dependent Choice (for countable sets) + "every set of reals is Lebesgue measurable".  .
This is a formal justification of the idea that any set of reals that can be proved to exist in ZFC without making uncountably many choices ---- i.e., a set that can be constructed in ZF + DC ---- cannot be proved to be non-measurable.
There are also theorems of Shelah, coming out of the analysis of determinacy (AD, Woodin cardinals, etc) to the effect that "any reasonably defined set is Lebesgue measurable".
So it is impossible to produce a Lebesgue non-measurable set without specifically looking for it in the wilds of uncountable AC.
It would be interesting to know what the state of the art is for metatheorems on how broad a range of constructions can be used in ZF+DC, that are guaranteed to stay within the world of provably Lebesgue measurable sets.


TITLE: Matrix reconstruction puzzle
QUESTION [1 upvotes]: Say a reconstruction of matrix $A$ is $A'$  and it's defined as
$ A' = P^TDPA $
where $P$ is an orthonormal matrix, $D$ is a diagonal binary (1 or 0) matrix. In a trivial case, when all diagonal elements are 1, we have a perfect reconstruction ($A'=A$).
Now we constrain the number of 1's in the diagonal entries of $D$ to, say, $n$. How do I find the best $D$ s.t. $Tr(D)=n$ that would minimize $||A-A'||$?
I think I need to inspect the singular vectors of $A$, but I am not sure what to do exactly.

REPLY [1 votes]: $D = argmin_{D\in\text{binary diag}}\quad ||P^\top DPA - A||$
Now let's change basis for A using $P$ (which is orthonormal). Thus $\exists M$ such $A = P^\top MP$. Note that M is deterministic as we know $P$ and $A$.
Making this substitution in the first equation, we get
\begin{align*}
D &= argmin_{D\in\text{binary diag}}\quad ||P^\top DPP^\top MP - P^\top MP||\\
&= argmin_{D\in\text{binary diag}}\quad ||P^\top D MP - P^\top MP||\\
&= argmin_{D\in\text{binary diag}}\quad ||P^\top (D M-M) P||\\
\end{align*}
Now comes the crucial part, since P is orthonormal (equivalent to original basis), this optimization problem is the same as:
\begin{align*}
D &= argmin_{D\in\text{binary diag}}\quad ||D M-M||\\
&= argmin_{D\in\text{binary diag}}\quad ||(D-I) M||\\
\end{align*}
Thus, we choose D such that the 1's correspond to the maximum norm rows of $M$


TITLE: Why is the fibre of each point compact?
QUESTION [2 upvotes]: For a compact covering space, the fibres of the covering map are finite.
I am working on the same question as the one posed in this link, but there was an unanswered question at the end, namely, why is the fibre of EACH point compact. This just so happens to be the question I cannot answer, so I am looking for an answer to this. I didn't want to repost however, so maybe a mod will appropriately be able to deal with this post.
Also, where did the top answerer use the fact that preimages of fibers are compact to complete the problem? I never saw the fact used in his proof.

REPLY [1 votes]: Let $Y\to X$ be the covering map, and let $C$ be a maximal intersection of non-empty closed subsets (that is, if $C'\subset  C$ is closed, then $C=C'$). This implies that for all $x\in C$, and for all open neighbourhoods $V$ of $x$, it holds that $C\subset V$.
Then $f^{-1}(C)$ is closed, and therefore compact (because $Y$ is compact).
Let $\{x_i\}_{i=0}^n$ be an infinite sequence of points with $f(x_i) = x$. It is contained in $f^{-1}(C)$, so it has an accumulation point $y$ with $f(y)=z\in C$. Let $g\colon U\to Y$ be a local inverse with $g(z) = y$ and $g(x)=q$. Then any open neighbourhood of $q$ contains $y$ (because they always contain a $g(V)$ with $V$ a neighbourhood of $x$ and $V$ contains $C$), so $\{x_i\}$ converges to $q$ too.
Therefore $f^{-1}(x)$ has an accumulation point $q$, but we assumed that it was discrete, so we arrive to a contradiction.


TITLE: Find $\mathbf a$ for which the following equation has triple roots
QUESTION [2 upvotes]: Find $\mathbf a \in\mathbb{R}$ for which the following equation has triple roots:
$$ x^4-5x^3+\mathbf ax^2-7x+2=0$$

REPLY [1 votes]: Suppose $b$ is the triple root. Then the polynomial can be factored as $(x-b)^3(x-c)$ and we get
$$
x^4 - (3b+c)x^3 + (3b^2 + 3bc)x^2 - (b^3+3b^2c)x + b^3c
$$
so we need
\begin{cases}
3b+c=5 \\
3b^2+3bc=a\\
b^3+3b^2c=7\\
b^3c=2
\end{cases}
Multiplying the third equation by $b$ and taking into account the fourth equation we get
$$
b^4-7b+6=0
$$
The fourth equation, together with the first, becomes $5b^3-3b^4=2$
$$
3b^4-5b^3+2=0
$$
Obviously $b=1$ is a solution for both. If $b\ne1$ we get
\begin{cases}
b^3+b^2+b-6=0 \\
3b^3-2b^2-2b-2=0
\end{cases}
and, eliminating $b^3$,
$$
5b^2+5b-16=0
$$
Multiplying the top equation by $2$ and summing up,
$$
5b^3-14=0
$$
Since $5b^3+5b^2-16b=0$, we get $5b^2-16b+14=0$, that finally yields $5b-16=-16b+14$, or $b=10/7$, that doesn't satisfy $5b^3=14$.
Therefore $b=1$ and $c=2$, so $a=9$.


TITLE: Why is $\Big\{\frac{1}{n^2}\Big\}_{n \in \mathbb{N}} \bigcup \hspace{0.2cm} \{-1,0 \}$ compact?
QUESTION [0 upvotes]: We have a set on Real numbers:
$$\Big\{\frac{1}{n^2}\Big\}_{n \in \mathbb{N}} \bigcup \hspace{0.2cm}\{-1,0 \}$$
I can't seem to understand why this is a compact set as if we look at the interval I think that there are tiny spaces in between the points of this interval that aren't in it. Isn't it so that a compact set must contain it's points and be somehow continuous?
Does that mean even let's say for example: $A = \{{1,2,3}\}$ that only contains this points is compact? Am I missing something here?

REPLY [3 votes]: First note that points of $\Bbb R$ are compact. Therefore the subset $\{-1\}$ is compact. Now $\lim_{n \to \infty} \frac{1}{n^2} = 0$, therfore $0$ is a limit point of the set $S = \{\frac{1}{n^2}: n \in \Bbb N\}$. Now if $x$ is a limit point of $S$, then there is a subsequence of $\frac{1}{n^2}$ which converges to $x$. But this forces $x = 0$. Therefore $0$ is the only limit point of $S$. Therefore $S\cup \{0$} is closed. Since it is trivially bounded, $S\cup \{0\}$ is compact(Heine Borel Theorem). Therefore $S\cup \{0\}\cup\{-1\}$ is compact.


TITLE: Derivative problem in R^n space
QUESTION [0 upvotes]: I have been trying to solve the following problem:
Let $f,g: \mathbb{R}^n \rightarrow \mathbb{R}$ be differentiable functions and supose $f(x_0)=g(x_0)$, and $f(x)\leq g(x)$ in a certain region with center in $x_0$. Show that the Jacob Matrixes of $f$ and $g$ are equal at $x_0$.

Here's what I have so far:
$\cdot$First I started by defining the function $h(x)=f(x)-g(x)$. Therefore I have that $h(x)\leq 0$ and $h(x_0)=0$
$\cdot$ I write down the definition of the derivative at $x_0$
$$ \lim_{t\to\ 0} \frac{h(x_0 + tv)-h(x_0)}{t} $$
$$ \iff \lim_{t\to\ 0} \frac{h(x_0 + tv)-0}{t} $$
Suposing that $t$ approaches from the right (doesn't make diference), It's easy to conclude that if the limit is positive then $h(x_0+tv) > 0$, which is not possible base on the fact that $h(x)<0$.
What's left to conclude is that $h(x_0+tv)$ can't be negative, but I don't see where to go from here.
Please help me ending this proof or show me a better way to solve this problem.

REPLY [2 votes]: Let $h=g-f$. Then $h \geq 0$ and $h(x_0)=0$ Hence, $h$ has  a local miminimum at $x_0$. This implies that all the partial derivatives of $h$ at $x_0$ are $0$. So $f$ and $g$ have the same partial derivatives at $x_0$.


TITLE: What are the two main ways to prove that a matrix $\Bbb R^{n\times n}$ is definite positive.
QUESTION [0 upvotes]: What are the two main ways to prove that a real $n\times n$ matrix is definite positive?
Is the first way: If a matrix is $n \times n$ symmetric matrix, then the associated quadratic form is positive definite iff the eigenvalues of the matrix are all positive?

REPLY [1 votes]: Quoting from the wikipedia entry in the link jkn gave, your matrix is positive definite if and only if
"Its leading principal minors are all positive. The $k$th leading principal minor of a matrix $M$ is the determinant of its upper-left $k\times k$ sub-matrix. It turns out that a matrix is positive definite if and only if all these determinants are positive. This condition is known as Sylvester's criterion, and provides an efficient test of positive-definiteness of a symmetric real matrix."


TITLE: Path connected subset of $\mathbb R^2$ that is locally connected at none of its points.
QUESTION [0 upvotes]: Let $X$ denote the rational points of the interval $[0,1]\times\{0\}$ of $\mathbb R^2$. Let $T$ denote the union of all line segments joining the point $p=(0,1)$ to points of $X$.

Find a subset of $\mathbb R^2$ that is path connected but is locally connected at none of its points.

My attempts : I got the answer here
I was visualizing the diagram:

im not getting  how  it is  Nowhere locally connected ?
Any hints/solution will be appreciated

REPLY [1 votes]: As Clayton observed, the space pictured in your question is not the space $X$ defined in William Elliot's original answer.
Let us redefine $X$ in a slightly modified form. For $r \in \mathbb{R}$ let $L_r$ denote the line segment from $(0,0)$ to $(1,r)$ if $r \ge 0$ and the line segment from $(1,0)$ to $(0,r)$ if $r \le 0$. Now define
$$X = \bigcup_{q \in \mathbb{Q}} L_q \subset [0,1] \times \mathbb{R} .$$
Note that the original definition was $X = \bigcup_{q \in  [-1,1] \cap \mathbb{Q}} L_q$.
All $L_q$ are path connected subspaces of $X$. Hence those with $q \ge 0$ resp. $q \le 0$ belong to the path component of $(0,0)$ resp. $(0,1)$. But $(0,0), (0,1) \in L_0$ which shows that $X$ is path connected.
Let us check that $X$ is not locally connected at any point $x = (x_1,x_2)\in X$. Define
$$V(x) =
\begin{cases}
   X \backslash \{(0,1)\} & x_1 = 0 \\
   X \backslash \{(0,0)\} & x_1 = 1 \\
   X \backslash \{(0,0), (0,1)\} & 0 < x_1 < 1
\end{cases}
$$
This is an open neighborhood of $x$ in $X$. For each $r \in \mathbb{R}$ the set $([0,1] \times \mathbb{R}) \backslash L_r$ splits into two disjoint nonempty open sets $O_r^\pm$ above and below $L_r$. Hence the sets $V_r^\pm(x) = V(x) \cap O_r^\pm$ are disjoint nonempty open subsets of $V(x)$. Call $r$ admissible for $x$ if $r$ is irrational and $r < 0$ for $x_1 = 0$ resp. $r > 0$ for $x_1 = 1$. For an admissible $r$ we have $L_r \cap V(x) = \emptyset$ and therefore $V(x) = V_r^+(x) \cup V_r^-(x)$.
Now consider any open $U \subset X$ such that $x \in U \subset V(x)$. We shall show that there exists a regular $r$ such that both $U_r^\pm = U \cap V_r^\pm(x)$ are nonempty. This proves that $U$ is not connected.
Since $U$ is open, the set $S(U)  = \{ q \in \mathbb{Q} \mid L_q \cap U \ne \emptyset \}$ is easily seen to be open in $\mathbb{Q}$.
a) $x_1 = 0$. Then $x = (0,q)$ with $q \le 0$. Obviously $q\in S(U)$. Choose $p \in S(U)$ such that $p < q$ and an  irrational $r$ such that $p < r < q$. Then $r$ is admissible for $x$ and both $U_r^\pm$ are nonempty.
b) $x_1 = 1$. Can be treated similarly.
c) $0 < x_1 <1$. Choose $p, q \in S(U)$ such that $p < q$ and an irrational $r$ such that $p < r < q$. Then $r$ is admissible for $x$ and both $U_r^\pm$ are nonempty.


TITLE: Knapsack like problem with nonnegative weight constraint
QUESTION [1 upvotes]: I am dealing with a knapsack-like problem with one difference from the conventional problem: the “weights” can be positive or negative and the constraint is $\sum w_i x_i \ge 0$ instead of $\sum w_i x_i \le W$. The "values" can also be positive or negative.
Can this be transformed to a knapsack problem or is it some other type of combinatorial optimization problem?

REPLY [1 votes]: Assuming $x_i$ is binary, perform a change of variables $\bar{x}_i:=1-x_i$ where $w_i>0$:
\begin{align}
\sum_i w_i x_i
&= \sum_{i:w_i>0} w_i x_i + \sum_{i:w_i<0} w_i x_i \\
&= \sum_{i:w_i>0} w_i(1-\bar{x}_i) + \sum_{i:w_i<0} w_i x_i \\
&= \sum_{i:w_i>0} w_i + \sum_{i:w_i>0} (-w_i)\bar{x}_i + \sum_{i:w_i<0} w_i x_i
\end{align}
So $\sum_i w_i x_i \ge 0$ is equivalent to
$$\sum_{i:w_i>0} w_i\bar{x}_i + \sum_{i:w_i<0} (-w_i) x_i \le \sum_{i:w_i>0} w_i$$


TITLE: Do different prime ideals define different irreducible varieties in general?
QUESTION [0 upvotes]: Over algebraically closed field, we have the dictionaries between irreducible affine varieties and prime ideals. Different prime ideals define different varieties. But when the field is not algebraically closed, e.g. $\mathbf{R}$, we have polynomials $x^2+1$ and $x^2+2$ that define the same empty set.

Are there more examples that different prime ideals define the same (non-empty) variety?
Is is possible for a prime ideal to define a reducible algebraic set?

REPLY [2 votes]: I guess you do not consider points of varieties that are not rational then (from the comments).
In that case, weird things are possible. For example, the answer to $2.$ is positive:
Consider e.g. a finite field $\mathbb{F}$. Then every subset of $\mathbb{A}^n_{\mathbb{F}}$ is closed (it is a finite union of closed points), and the only irreducible subsets are singletons (for the same reason). Thus, and ideal $I$ defines an irreducible subset iff $V(I)$ is a one-point set. So for example,
$$I=(x_1, x_2, \dots, x_i), \;\;i <n$$
is a prime ideal in $\mathbb{F}[x_1, \dots, x_n],$ but $V(I)$ is not irreducible, since it contains more than one point.
For these reasons (which are connected to your previous question) it is preferable to define all your affine varieties in $\mathbb{A}^n_{\overline{\mathbb{F}}}$ instead of $\mathbb{A}^n_{\mathbb{F}}$ (where $\overline{\mathbb{F}}$ denotes the algebraic closure of $\mathbb{F}$) even when you consider polynomials only over $\mathbb{F}$. That is, an affine algebraic set (over $\mathbb{F}$) under this definition is a set of the form
$$V(S)=\{(a_1, a_2, \dots a_n) \in \overline{\mathbb{F}}^n\;|\; \forall f \in S: f(a_1, \dots, a_n)=0\},\;\; S \subseteq \mathbb{F}[x_1, x_2, \dots, x_n].$$
Then the correspondence works similarly as if $\mathbb{F}$ was alg. closed. The downside of this is that some singletons are no longer closed in the ($\mathbb{F}$-)Zariski topology (if you care about such property).


TITLE: Space Completion Theorem for Normed Spaces
QUESTION [3 upvotes]: I went to a functional analysis course this year and my lecturer wrote down this Theorem. Lots of students pointed out it is incorrect, but she insisted it was. I am stating it now and hope someone can give me a good explanation about the confusion.
Here is the statement of the Theorem:
For any normed vector space $(V,\|\cdot\|_V)$, there exists another normed space $(Z,\|\cdot\|_Z)$ such that a map $i:V\rightarrow Z$ such that
(1) $i$ is an isometrical isomorphism
(2) $i(V)$ is dense in $Z$
where $Z$ is unique up to isomorphism.
In this Theorem, nothing said about $Z$ being complete. The students asked if $V$ is $\mathbb{Q}$, what $Z$ should be? $\mathbb{Q}$ fits the definition for $Z$, because $\mathbb{Q}$ is dense in $\mathbb{Q}$, but the lecturer said $\mathbb{Q}$ is not dense. She said something on the line dense means something different in a normed space. Can someone explain to me what this is about? In the Theorem, do we have to have say $Z$ is complete?

REPLY [5 votes]: She was wrong. It’s clear that if $V=\Bbb Q$, then both $Z=\Bbb Q$ and $Z=\Bbb R$ satisfy the stated conclusion. You get uniqueness by requiring $Z$ to be complete. Indeed, that’s the whole point of the theorem: if $V$ is any normed linear space, it has a unique completion (which is a Banach space).
There are actually two important notions of dense subset of $\Bbb R$, but they coincide. One may say that a subset $D$ of $\Bbb R$ is dense with respect to the order $\le$ on $\Bbb R$ if for all $x,y\in\Bbb R$ with $x<y$ there is a $d\in D$ such that $x<d<y$; that’s the order-theoretic notion of dense subset. One may also say that $D$ is dense in $\Bbb R$ if $\operatorname{cl}D=\Bbb R$, where the closure is taken in the usual Euclidean topology of $\Bbb R$; that’s the topological notion of dense subset. But as I said, they coincide, and by either notion $\Bbb Q$ is dense both in itself and in $\Bbb R$.
Added: One normally deals with linear spaces over the complete fields $\Bbb R$ and $\Bbb C$, and I strongly suspect that the theorem was intended to be limited to such spaces. The conclusion should still have specified that $Z$ is complete, however, as uniqueness fails trivially without it: just take any infinite-dimensional normed linear space $V$ that is not complete, and note that her version of the theorem is satisfied both by $V$ and by its completion, which are distinct.


TITLE: Understanding axioms of neighbourhoods in topological spaces
QUESTION [6 upvotes]: Let $X$ be a set of points, along with a certain function $N$ that assigns a subset to every $x \in X.$ Then $X$ together with $N$ define a topological space if $N$ satisfies the following axioms: (quoting below from wikipedia)

If $N$ is a neighbourhood of $x$ (i.e., $N \in N(x)$), then $x \in N$. In other words, each point belongs to every one of its neighbourhoods.
If $N$ is a subset of $X$ and contains a neighbourhood of $x$, then $N$ is a neighbourhood of $x$. I.e., every superset of a neighbourhood of a
  point $x$ in $X$ is again a neighbourhood of $x$.
The intersection of two neighbourhoods of $x$ is a neighbourhood of $x$.
Any neighbourhood $N$ of $x$ contains a neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point of $M$.

Questions:

My first struggle is with the statement of the second axiom, namely that any superset $S\supseteq M$ of a neighbourhood $M$ of $x$ is again a neighbourhood of $x$.

a) Doesn't this mean that everything is a neighbourhood of everything because I can just set $S=X$? But then what good $N$ is at characterizing $X$ in terms of specific subsets of every $x$, if any neighbourhood can be indefinitely enlarged?
b) How to understand the need for such an axiom?

Am I right to interpret the 4th axiom as follows: Given a neighbourhood $N$ of $x,$ it is always possible to extract a smaller neighbourhood $M'\subseteq M$  from $M,$ such that $M'$ is an open subset, where $M$ need not be open itself.
To me it seems that the 2nd and 4th axioms are equivalent statements. What makes them distinct? (or in other words what makes the 4th axiom necessary).

REPLY [8 votes]: The enlargement axiom does not trivialise the theory. It does imply that $X$ is a neighbourhood of every $x \in X$: every $x \in X$ has some $N \in N(x)$ (this is actually part of the axioms/conditions, often forgotten: every $N(x)$ is non-empty) and $N \subseteq X$ so $X \in N(x)$ by axiom 2. And everything inbetween $N$ and $X$ is a neighbourhood too. We cannot go smaller in general, so it's not true that everything becomes a neighbourhood of everything: $X$ is a neighbourhood of every point, but any fixed set $A$ is only a neighbourhood of $x$ if it happens to be the case that $A \in N(x)$, and otherwise it's not.
As to the necessity of the axiom: look at metric spaces, where $A$ is a neighbourhood of $x$ iff there is some $r > 0$ such that $B(x,r) \subseteq A$. The way this is defined already implies that larger sets than $A$ are also neighbourhoods of $x$; the same $r$ will work. So it's not a very surprising axiom. The same can be said for ordered spaces ($A$ is a neighbourhood of $x$iff either there exists $a \in X$ with $x < a$ and $\{z : z < a\} \subseteq A$, or there exists $b \in X$ with $b < x$ and $\{z : b < z \} \subseteq A$ or there exist $a,b \in X$ with $a < x, x < b$ and $\{z: a < z, z < b \} \subseteq A$), where also the enlargement is clearly true. We really are interested in the "small" neighbourhoods of $x$ (for continuity etc.) but in this axiomatisation we want to capture the notion of all possible neighbourhoods, even the not so interesting large ones. We could also axiomatise "basic" neighbourhoods instead, instead of all neighbourhoods, but this is not what we are doing here. In short, the relation of $A$ being a neighbourhood of $x$ is "local" and not affected by enlarging the set.
As to axiom 4, it is true that the "intended" result is that every neighbourhood of $x$ contains an open neighbourhood of $x$, but for that we first need to define what "open" means. We only have a set of sets $\{N(x): x \in X\}$ satisfying the  axioms, no notion of openness. But if you already know the standard axioms for topology in terms of open sets, you want to go back and forth:
If you have a topology $\mathcal{T}$, define for each $x \in X$: $N(x) = \{A \subseteq X: \exists O \in \mathcal{T}: x \in O \subseteq A\}$ and check that this fulfills the axioms (again see that 2 is trivially fulfilled). Call this system $\mathcal{N}(\mathcal{T})$. On the other hand, if we have a system $\mathcal{N}= \{N(x): x \in X \}$ satisfying the axioms, define $\mathcal{T}  = \{O \subseteq X: \forall x \in O: O \in N(x) \}$ and check that this defines a topology (we do not need axiom 4 for this, it turns out). Call this topology $\mathcal{T}(\mathcal{N})$. The axioms 4 is used to show that $\mathcal{N}(\mathcal{T}(\mathcal{N})) = \mathcal{N}$ again. So the neighbourhood system from the topology induced by the neighbourhood system is that same one we started with. It's a fun exercise to check this out in detail. It will enhance your understanding of these axioms. It's also true that $\mathcal{T}(\mathcal{N}(\mathcal{T})) = \mathcal{T}$, which is also fun to prove.


TITLE: VC Dimension generalized to discrete, non-binary, unordered domains?
QUESTION [8 upvotes]: VC dimension is a measure of the complexity of classes of functions $f:X\rightarrow \{0,1\}$ that is closely tied to sample complexity. Fat shattering dimension is a generalization suited to richer ordered domains: i.e. $f:X\rightarrow \mathbb{R}$. Is there a standard generalization of VC-dimension suited to functions with discrete, unordered domains? i.e. $f:X\rightarrow K$ where $K$ is a finite set with no ordering on it.

REPLY [0 votes]: The dimension characterizing multiclass learnability (i.e., the "correct" multiclass analog of VC-dimension) has recently been discovered:
Nataly Brukhim, Daniel Carmon, Irit Dinur, Shay Moran, Amir Yehudayoff:
A Characterization of Multiclass Learnability.
https://arxiv.org/abs/2203.01550


TITLE: Real Period of an Elliptic Curve
QUESTION [4 upvotes]: Trying to work out what the real period of an elliptic curve is as seen in the Birch Swinnerton-Dyer conjecture.
From what I've read, given an elliptic curve E over the rationals, one can associate to it a value $\displaystyle \Omega_{E} = \int_{E(\mathbb{R})}|\omega|$ where $\omega = \dfrac{dx}{2y + a_{1}x + a_{3}}$ as stated in
http://www.math.uci.edu/~asilverb/connectionstalk.pdf
I have also read that this is equal to twice the real period if the elliptic curve has two real components and just equal to the real period otherwise.
What is the real period of an elliptic curve and why is it well defined (that is if I have two isomorphic complex tori, why must they have the same real period)? Also is there a way to compute this integral?

REPLY [3 votes]: The real period can be computed very efficiently using the AGM. See my paper http://dx.doi.org/10.1016/j.jnt.2013.02.002


TITLE: Need a formula / method to get a value between 0 and 1 if a point lies in an area between two rectangles
QUESTION [0 upvotes]: I'm trying to figure out a way to construct a formula or method for the following process:
If Point E is inside the smaller rectangle (red area), the value should be 1.
If Point E is outside the bigger rectangle (blue area), the value should be 0.

If Point E is inside the bigger rectangle, but outside the smaller rectangle (white area), the value should be linearly interpolated between 1 and 0, based on the "difference" of the two rectangles.
The purpose of this is to fade out the edges of a rectangle shaped spotlight in OpenGL.
Any help would be appreciated.
Image link for my problem

REPLY [1 votes]: Since according to your comment the rectangles are described by half their width and height, I assume that they are centered at the origin and aligned with the axes. So the smaller rectangle is $-w_1\le x\le w_1, -h_1\le y\le h_1$ and the larger likewise for $w_2,h_2$. Assume for the moment that $y=0$. Then the smaller rectangle is characterized by $\lvert x\rvert=w_1$ and the larger by $\lvert x\rvert=w_2$. In the latter case you want a result of zero, so start with $f(x,y)=w_2-\lvert x\rvert$. To achieve a value of $1$ at the inner rectangle, change this to $f(x,y)=\frac{w_2-\lvert x\rvert}{w_2-w_1}$. Now you have the linear gradient between these two values, but you still need to clamp things down so you don't exceed the $[0,1]$ range. Use $f(x,y)=\min\left(1,\max\left(0,\frac{w_2-\lvert x\rvert}{w_2-w_1}\right)\right)$ to achieve this clamping. And what about the $y$ coordinate? You could obtain an expression of the same kind for that. Then you can either multiply both expressions, or take the minimum. I'm not sure whether you'd call the former linear, since it's linear in each $x$ and $y$ but since it contains a $xy$ term I'd rather call it quadratic. So I'd go for
$$f(x,y)=\min\left(1,\max\left(0,\frac{w_2-\lvert x\rvert}{w_2-w_1}\right),\max\left(0,\frac{h_2-\lvert y\rvert}{h_2-h_1}\right)\right)$$
Notice that OpenGL has a clamp function, which will take care of that aspect, and a min function to combine the values.


TITLE: prove that if every induced subgraph of G is connected,then G is the complete graph Kn
QUESTION [1 upvotes]: Let G be a graph of order n
prove that if every induced subgraph of G is connected,then G is the complete graph  Kn

REPLY [2 votes]: HINT: If $G$ is not $K_n$, there must be two distinct vertices, $u$ and $v$, of $G$ such that $uv$ is not an edge of $G$.


TITLE: Why FQHE need a lower energy state?
QUESTION [4 upvotes]: There are a lot papers explaining why Laughlin's wavefunction are energetically favorable, but seldom explain why a lower energy state could explain the plateau at $\nu=1/3$. I met at several places claims like: a lower energy state at $\nu=1/3$ will pin the electron density at $\nu=1/3$. But why is that? And what actually it means? when we move $\nu$ from $1/3$ what happens? electrons adjust there distance or new particle been added? And is this a phase transition?
Hope someone familiar this field could give me some help, thanks!

REPLY [1 votes]: The Laughlin state alone doesn't explain the plateau. There is a lot more to the story.
Firstly at filling factor=1/3 the many-body ground state of the interacting electron gas is "approximately" the Laughlin wavefunction. By this I mean that the overlap between the Laughlin state and the numerically found ground state (for any realistic interaction like coulombic) is very large, i.e. their inner product is quite close to 1. Using the plasma analogy one can show that this state corresponds to uniform electron density. (See Girvin's Les Houches notes for details on Plasma analogy.)
Secondly the transport phenomena are decided by charged excitations in the system. For the filling factors 1/3,1/5,1/7,etc. the charged excitations are quasiholes and quasielectrons. While the former has a dip in the density profile at some point Z (say) in the 2D plane, the latter has the opposite thing in its density profile (as opposed to the earlier uniform case). The plasma analogy can again be used to show that these quasiparticles will have fractional e/3 charge in our case. (Atleast for now let us avoid justifying why they are excited states.)
Now lets say we are sitting exactly at 1/3 filling factor and then we add an electron to the system. It will break into 3 quasielectrons which can be separated at no extra energy cost (the idea of fractionalization). Similarly if some more electrons are added they will produce more quasiparticles. Now start thinking in terms of the 'semiclassical percolation picture' that is applied to electrons to explain Integer QHE (Again see Girvin's notes). Instead of electrons we give the same arguments using quasiparticles to explain the plateaus around 1/3 filling factor. The conductivities stop changing when the added quasiparticles are either going into the valleys of the disorder potential or are ending up on shorelines at the 2 well separated edges.
Let me clarify things a bit more. Think of starting with the 1/3 filling factor ground state. Now let us add adiabatically 1 flux quanta through a thin solenoid at the origin of space (See Laughlin's Nobel lecture). He shows that in this process e/3 charge flows towards the origin and gets collected there. Thus we have ended up with an exact ground eigenstate of the original hamiltonian + e/3 charge. So quasiholes are 'charged excitations', not the excited state when sitting at 1/3 filling. In fact the low energy gapped excitations at 1/3 filling are 'neutral collective excitations' (Again see Girvin's notes) and the existence of this gap is necessary for adiabaticity to work fine in the above thought experiment. (In the words of Laughlin the usage of the word quasiparticles here was "unfortunate".) Now if I just move the filling factor a bit in an experiment the new ground state is made up of new "quasiparticles".


text:
\begin{document}

\begin{center}
{\Large \bf SINGULAR POINTS }
\vspace*{2mm}

{\Large \bf OF LIGHTLIKE  HYPERSURFACES}

\vspace*{2mm}

{\Large \bf OF THE DE SITTER SPACE}

\vspace*{3mm}
{\large M. A. Akivis and  V. V. Goldberg}\footnote{This research was partially supported by
the Volkswagen-Stiftung (RiP-program at Oberwolfach).}
\end{center}

\vspace*{5mm}

\begin{abstract}
The authors study singular points  of lightlike hypersurfaces of
the de Sitter space $S^{n+1}_1$ and  the geometry of hypersurfaces
and use them for construction of an invariant normalization and an
invariant affine connection of lightlike hypersurfaces.
\end{abstract}

\setcounter{equation}{0}

\setcounter{section}{-1}

\section{Introduction}

It is well-known that the pseudo-Riemannian manifolds $(M, g)$
of Lorentzian signature play a special role in geometry and
physics, and that they are models of spacetime of
general relativity. At the tangent space $T_x$ of an arbitrary
point $x$ of such a manifold, one can invariantly define a real
isotropic cone $C_x$. From the point of view of physics,
this cone is the light cone: trajectories of light impulses
emanating from the point $x$ are tangent to this cone.

Hypersurfaces  of a Lorentzian manifold $(M, g)$ can be of
three types: spacelike, timelike, and lightlike
(see, for example,  \cite{ON83} or \cite{AG96}).
For definiteness, we will assume that $\dim M =
n + 1$ and $\mbox{{\rm sign}} \; g = (n, 1)$.

The tangent hyperplane to a spacelike hypersurface $U^n$
at any point does not have real
common points with  the light cone $C_x$. This implies that
on $U^n$ a proper Riemannian metric is induced.
The tangent hyperplane to a timelike  hypersurface $U^n$
at any point  intersects  the light cone $C_x$ along
an $(n-1)$-dimensional cone.
This implies that on $U^n$ a pseudo-Riemannian metric
of Lorentzian signature $(n - 1, 1)$ is induced. Finally, the
tangent hyperplane to a lightlike hypersurface $U^n$
at any point is tangent to  the light cones $C_x$. This implies that on $U^n$ a degenerate Riemannian metric
signature $(n - 1, 0)$ is induced.

On spacelike and timelike hypersurfaces of a manifold
of Lorentzian signature, an invariant normalization and
an affine Levi-Civita  connection are induced by
a first-order neighborhood while on lightlike hypersurfaces
one should use differential neighborhoods of higher order
to construct an invariant normalization and
an affine  connection,

From the point of view of physics lightlike hypersurfaces
are of great importance since they are models of different
types of horizons studied in general relativity:
event horizons, Cauchy's horizons, Kruskal's horizons
(see \cite{Ch83} and \cite{MTW73}). This is the reason that
the study of geometric structure of lightlike hypersurfaces
is of interest.

In the current paper we consider lightlike hypersurfaces
in the de Sitter space (a pseudo-Riemannian space
of Lorentzian signature and constant positive curvature),
study their  geometric structure, and prove that there are
singular points and singular submanifolds on them.

The de Sitter space $S_1^{n+1}$ admits a realization on the
exterior of an $n$-dimensional oval hyperquadric $Q^n$ of
a projective space $P^{n+1}$. Thus the de Sitter space
is isometric to a pseudoelliptic space,
$S_1^{n+1} \sim \mbox{{\rm ext}} \; Q^n$.
Since the interior of the
 hyperquadric $Q^n$ is isometric to the hyperbolic geometry
of the Lobachevsky space $H^{n+1}$, $H^{n+1} \sim
\mbox{{\rm int}} \; Q^n$ and the geometry of $Q^n$ itself is
equivalent to that of an  $n$-dimensional conformal
space $C^n$, $C^n \sim Q^n$, the groups of motions
of these three spaces are isomorphic to each other
and are isomorphic to the group ${\bf SO} (n + 2, 1)$
of rotations of a pseudo-Euclidean space $R^{n+2}_1$
of Lorentzian signature. This allows us to apply the apparatus
developed in the book \cite{AG96} for the conformal space $C^n$
to the study of the de Sitter space.

Note also that the geometry of lightlike hypersurfaces
on pseudo-Riemannian manifolds of different signatures
was the subject of  many journal papers and even two
books \cite{DB96} and \cite{Ku96}.
However, the geometry of lightlike
hypersurfaces in the de Sitter space was not studied
in spite of the fact that this geometry has many interesting
geometric features.

In the present paper we study the geometry of the
de Sitter space $S^{n+1}_1$ using its connection with the
geometry of the conformal space. We prove that the geometry of
lightlike hypersurfaces of the space $S^{n+1}_1$ is directly
connected with the geometry of hypersurfaces of the conformal
space $C^n$. The
latter  was studied in detail in the papers of the first
author (see \cite{A52},  \cite{A65})
and also in the book \cite{AG96}. This simplifies the
study of lightlike hypersurfaces of the de Sitter space
$S^{n+1}_1$ and makes possible to apply for their consideration
the apparatus constructed in the conformal theory.

In Section 1 we study the geometry of the
de Sitter space and its connection with the geometry of the
conformal space. Next we study lightlike hypersurfaces
$U^n$ in the space $S^{n+1}_1$, investigate their structure,
and prove that such a hypersurface is tangentially degenerate of
rank $r \leq n - 1$. Its rectilinear or plane
generators form an isotropic fibre bundle on $U^n$.

In Sections 2--5 we investigate lightlike hypersurfaces $U^n$ of
maximal rank, and for their study we use the relationship between
the geometry of such hypersurfaces and the geometry of
hypersurfaces of the conformal space. For a lightlike
hypersurface, we construct the fundamental quadratic forms and
connections determined  by a normalization of a hypersurface by
means of a distribution (the screen distribution) which is
complementary to the isotropic distribution. The screen
distribution plays an important role in the book \cite{DB96}
since it defines a connection on a lightlike hypersurface $U^n$,
and it appears to be important for applications. We prove that
the screen distribution on a lightlike hypersurface  can be
constructed invariantly by means of quantities from a third-order
differential neighborhood, that is, such a distribution is
intrinsically connected with the geometry of a hypersurface.

In Section 5 we study singular points of a  lightlike
hypersurface in the de Sitter space $S^{n+1}_1$, classify them,
and describe the structure of hypersurfaces carrying singular
points of different types. Moreover, we establish the connection
of this classification with that of canal hypersurfaces of the
conformal space.

The principal method of our investigation is the method of
moving frames and exterior differential forms in the form
in which it is presented in the books \cite{AG93} and
\cite{AG96}. All functions considered in the paper are assumed to
be real and
differentiable, and all manifolds are assumed to be smooth
with the possible exception of some isolated singular points and
singular submanifolds.

\section{The de Sitter Space}

\setcounter{equation}{0}

{\bf 1.}
In a projective space $P^{n+1}$ of dimension $n+1$ we consider
an oval hyperquadric $Q^n$. Let $x$ be a point of the space
$P^{n+1}$ with projective coordinates $(x^0, x^1, \ldots ,
x^{n+1})$.  The hyperquadric $Q^n$ is determined by the equations
\begin{equation}\label{eq:1}
(x, x) := g_{\xi\eta} x^\xi x^\eta = 0, \;\;\;\;\;
\xi, \eta = 0, \ldots, n+1,
\end{equation}
whose left-hand side is a quadratic form $(x, x)$
of signature $(n+1, 1)$. The hyperquadric $Q^n$
divides the space $P^{n+1}$ into two parts, external and
internal. Normalize the quadratic form $(x, x)$
in such a way that for the points of the external part
the inequality $(x, x) > 0$ holds. This external domain is
a model of the {\em de Sitter space} $S^{n+1}_1$
(see \cite{Z96}). We will identify the external domain
of $Q^n$ with the space $S^{n+1}_1$. The hyperquadric $Q^n$
is the {\em absolute} of the space $S^{n+1}_1$.

On the hyperquadric $Q^n$ of  the space $P^{n+1}$ the geometry
of a conformal space $C^n$ is realized. The bijective mapping
$C^n \leftrightarrow Q^n$ is called the {\em Darboux mapping},
and the hyperquadric $Q^n$ itself is called the {\em Darboux
hyperquadric}.

Under the Darboux mapping to hyperspheres of the space $C^n$
there correspond cross-sections of the hyperquadric $Q^n$
by hyperplanes $\xi$. But to a hyperplane $\xi$ there corresponds
a point $x$ that is polar-conjugate to $\xi$ with
respect to $Q^n$ and lies outside of $Q^n$, that is,
a point of the space $S^{n+1}_1$. Thus to hyperspheres of the
space $C^n$ there correspond points  of the space $S^{n+1}_1$.

Let $x$ be an arbitrary point of the space $S^{n+1}_1$.
The tangent lines from the point $x$ to the hyperquadric $Q^n$
form a second-order cone $C_x$ with vertex at the point $x$.
This cone is called the {\em isotropic cone}.
For spacetime whose model
is the space $S^{n+1}_1$ this cone is the light cone, and
its generators are lines of propagation of light impulses
whose source coincides with  the point $x$.

The cone $C_x$ separates all straight lines passing through
 the point $x$ into spacelike (not having common points with
the hyperquadric $Q^n$), timelike (intersecting $Q^n$
in two different points), and lightlike (tangent to $Q^n$).
The  lightlike straight lines are generators of the cone $C_x$.

To a spacelike straight line $l \subset S_1^{n+1}$ there
corresponds an elliptic pencil of hyperspheres in the conformal
space $C^n$. All hyperspheres of this pencil pass through a
common $(n-2)$-sphere $S^{n-2}$ (the center of this pencil).
The sphere $S^{n-2}$ is the intersection of the hyperquadric
$Q^n$ and the $(n-1)$-dimensional subspace of the space $P^{n+1}$
which is polar-conjugate to the line $l$ with respect to the
hyperquadric $Q^n$.

To a timelike straight line $l \subset
S_1^{n+1}$ there corresponds a hyperbolic pencil of hyperspheres
in the space $C^n$.
Two arbitrary hyperspheres of this pencil do
not have common points, and the pencil contains two hyperspheres
of zero radius which correspond to the points of intersection of
the straight line $l$ and the hyperquadric $Q^n$.

Finally, to a lightlike straight line $l
\subset S_1^{n+1}$ there corresponds a parabolic pencil of
hyperspheres in the  space $C^n$ consisting of
hyperspheres tangent one to another at a point that is a unique
hypersphere of zero radius belonging to this pencil.

Hyperplanes of the space $S^{n+1}_1$ are also divided
into three types. Spacelike hyperplanes do not have
common points with  the hyperquadric $Q^n$; a timelike
hyperplane intersects   $Q^n$ along a real hypersphere;
and lightlike hyperplanes are tangent to $Q^n$.
Subspaces of any dimension $r, \; 2 \leq r \leq n-1$,
can be also classified in a similar manner.

Let us apply the method of moving frames
to study some questions of differential geometry
of the space $S^{n+1}_1$. With a point $x \in S^{n+1}_1$
we associate a family of projective frames $\{A_0, A_1,
\ldots , A_{n+1}\}$. However, in order to apply formulas derived
in the book \cite{AG96}, we will use the notations used
in this book. Namely, we denote by $A_n$ the vertex
of the moving frame which coincides
 with the point $x$, $A_n = x$; we locate the vertices
$A_0, A_i$, and $A_{n+1}$ at the hyperplane $\xi$ which is polar
conjugate to the point $x$ with respect to
 the hyperquadric $Q^n$, and we assume that the points $A_0$ and
$A_{n+1}$ lie on the hypersphere $S^{n-1} = Q^n \cap \xi$, and
the points $A_i$ are polar-conjugate to the straight line $A_0
A_{n+1}$ with respect to $S^{n-1}$. Since $(x, x) > 0$, we can
normalize the point $A_n$ by the condition $(A_n, A_n) = 1$.
The points  $A_0$ and $A_{n+1}$ are not polar-conjugate  with
respect to  the hyperquadric $Q^n$. Hence we can normalize them
by the condition $(A_0, A_{n+1}) = - 1$. As a result,
the matrix of scalar products of the frame elements
has the form
\begin{equation}\label{eq:2}
\renewcommand{\arraystretch}{1.3}
(A_\xi, A_\eta) = \pmatrix{0 & 0 & 0 & -1 \cr
0 & g_{ij} & 0 & 0 \cr
 0 & 0 & 1 & 0 \cr
-1 & 0 & 0& 0\cr}, \;\;\;\;\; i, j = 1, \ldots, n -1,
\renewcommand{\arraystretch}{1}
\end{equation}
and the quadratic form $(x, x)$ takes the form
\begin{equation}\label{eq:3}
(x, x) = g_{ij} x^i x^j + (x^n)^2 - 2 x^0 x^{n+1}.
\end{equation}
 The quadratic form $g_{ij} x^i x^j $ occurring in
(3) is positive definite.

The equations of infinitesimal displacement of the conformal
frame $\{A_\xi\}, \linebreak \xi = 0, 1, \ldots, n+1$,
we have constructed have the form
\begin{equation}\label{eq:4}
d A_\xi = \omega_\xi^\eta A_\eta, \;\;\;\; \xi, \eta
= 0, 1, \ldots, n+1,
\end{equation}
where by (2), the 1-forms $\omega_\xi^\eta$
  satisfy the following Pfaffian equations:
\begin{equation}\label{eq:5}
\left\{
\renewcommand{\arraystretch}{1.3}
\begin{array}{ll}
\omega_0^{n+1} =  \omega^0_{n+1} = 0, &  \omega_0^0  + \omega_{n+1}^{n+1} = 0, \\
 \omega_i^{n+1} = g_{ij} \omega_0^j, &
\omega_i^0 = g_{ij} \omega_{n+1}^j, \\
\omega_n^{n+1} - \omega_0^n   = 0, & \omega_n^0 - \omega_{n+1}^n = 0, \\
 g_{ij} \omega_n^j + \omega_i^n = 0 , &   \omega_n^n = 0, \\
 dg_{ij} = g_{jk} \omega_i^k + g_{ik} \omega_j^k.
\end{array}
\renewcommand{\arraystretch}{1}
\right.
\end{equation}
These formulas are precisely the formulas derived in the book
\cite{AG96} (see p. 32) for the conformal space $C^n$.

 It follows from (4) that
\begin{equation}\label{eq:6}
dA_n = \omega_n^0 A_0 + \omega_n^i A_i + \omega_n^{n+1} A_{n+1}.
\end{equation}

The differential $dA_n$ belong to the tangent space
$T_x (S_1^{n+1})$, and the 1-forms $\omega_n^0, \omega_n^i$,
and $\omega_n^{n+1}$ form a coframe of this space. The total
number of these forms is $n + 1$, and this number coincides with
the dimension of $T_x (S_1^{n+1})$. The scalar square of the
differential $dA_n$ is the metric quadratic form $\widetilde{g}$ on
the manifold $S_1^{n+1}$. By (2), this quadratic form $\widetilde{g}$
can be written as
$$
\widetilde{g} = (dA_n, dA_n) = g_{ij} \omega_n^i \omega_n^j - 2 \omega_n^0
\omega_n^{n+1}.
$$
Since the first term of this expression is a positive definite
quadratic form, the form $\widetilde{g}$ is of Lorentzian signature
$(n, 1)$. The coefficients of the form $\widetilde{g}$
produce the metric tensor of the space $S^{n+1}_1$ whose matrix is obtained from the
matrix (2) by deleting the $n$th row and the $n$th column.

The quadratic form $\widetilde{g}$  defines
on $S^{n+1}_1$ a pseudo-Riemannian metric of signature $(n, 1)$.
The isotropic cone defined in the space
$T_x (S_1^{n+1})$ by the equation $\widetilde{g} = 0$ coincides with the cone $C_x$
that we defined earlier in the space $S^{n+1}_1$ geometrically.

The 1-forms $\omega_\xi^\eta$ occurring in equations (4)
  satisfy the structure equations of the space $C^n$:
\begin{equation}\label{eq:7}
d \omega_\xi^\eta = \omega_\xi^\zeta \wedge \omega_\zeta^\eta,
\end{equation}
which are obtained by taking exterior derivatives of
equations (4) and which are conditions of complete
integrability of (4). The forms $\omega_\xi^\eta$
are invariant forms of the fundamental group
${\bf PO} (n+2, 1)$ of transformations of the
spaces $H^{n+1}, C^n$, and $S^{n+1}_1$ which is locally
isomorphic to the group ${\bf SO} (n+2, 1)$.

\section{Lightlike Hypersurfaces in the de Sitter Space}

A hypersurface $U^n$ in the de Sitter space $S^{n+1}_1$
is said to be {\em lightlike} if all its tangent hyperplanes are
lightlike, that is, they are tangent to the hyperquadric $Q^n$
which is the absolute of the  space $S^{n+1}_1$.

Denote by $x$ an arbitrary point of the  hypersurface $U^n$,
by $\eta$ the tangent hyperplane to $U^n$ at the point $x,
\eta = T_x (U^n)$, and by $y$ the point of tangency of
the hyperplane $\eta$ with the hyperquadric $Q^n$.
Next, as in Section 1,  denote by $\xi$ the hyperplane
which is polar-conjugate to the point $x$
with respect to the hyperquadric $Q^n$, and  associate
with a point $x$ a family of projective  frames such that
$x = A_n, y = A_0$, the points $A_i, i = 1, \ldots ,
n - 1$, belong to the intersection of the hyperplanes $\xi$ and
$\eta, \; A_i \in \xi \cap \eta$, and the point $A_{n+1}$, as
well as the point $A_0$, is the intersection point of the
quadric $\xi \cap Q^n$ and the
straight line that is polar-conjugate to the $(n-2)$-dimensional subspace spanned by the points $A_i$.
In addition, we normalize the frame vertices
in the same way as this was done in Section 1. Then the matrix of scalar products of the frame elements has the form (2), and the components of infinitesimal displacements of the moving frame
satisfy the Pfaffian equations (5).

Since the hyperplane $\eta$ is tangent to
the  hypersurface $U^n$ at the point $x = A_n$ and does not
contain the point $A_{n+1}$, the differential of
 the point $x = A_n$ has the form
\begin{equation}\label{eq:8}
dA_n =\omega_n^0 A_0 + \omega_n^i A_i,
\end{equation}
 the following equation holds:
\begin{equation}\label{eq:9}
\omega_n^{n+1} = 0,
\end{equation}
and the forms $\omega_n^0$ and $\omega_n^i$ are basis forms of
the  hypersurface $U^n$.
By relations (5), it follows from equation (8) that
\begin{equation}\label{eq:10}
\omega_0^n = 0
\end{equation}
and
\begin{equation}\label{eq:11}
dA_0 =\omega_0^0 A_0 + \omega_0^i A_i.
\end{equation}

Taking  exterior derivative of equation (9), we obtain
\begin{equation}\label{eq:12}
\omega^i_n \wedge \omega_i^{n+1} = 0.
\end{equation}
Since the forms $\omega^i_n$ are linearly independent,
by Cartan's lemma, we find from (12) that
\begin{equation}\label{eq:13}
\omega_i^{n+1} = \nu_{ij} \omega_n^j, \;\;
\nu_{ij} = \nu_{ji}.
\end{equation}
Applying an appropriate formula from (5), we find that
\begin{equation}\label{eq:14}
\omega_0^i = g^{ij}  \omega_j^{n+1} =
g^{ik} \nu_{kj} \omega_n^j,
\end{equation}
where $(g^{ij})$ is the inverse matrix of the matrix $(g_{ij})$.

Now formulas (8) and (11) imply that for
$\omega_n^i = 0$, the point $A_n$ of
the  hypersurface $U^n$ moves along the isotropic straight line
$A_n A_0$, and hence $U^n$ is a ruled hypersurface. In what
follows, we assume that the {\em entire}  straight line
$A_n A_0$ belongs to the  hypersurface $U^n$. Thus the following
theorem holds:

\begin{theorem}
A lightlike hypersurface $U^n$ is the image of the direct
product $M^{n-1} \times l$ of a differentiable manifold
$M^{n-1}$ and a projective line $l$ under the mapping $f:
M^{n-1} \times l \rightarrow P^{n+1}$ into
a projective space $P^{n+1}$: $U^n = f (M^{n-1} \times l)$ which sends
the straight line $l$ to the straight line $A_n A_0 \in P^{n+1}$.
\end{theorem}
Precisely this mapping is the subject of this paper.

In addition, formulas (8) and (11) show that at any point of
a generator of  the  hypersurface $U^n$, its tangent hyperplane
is fixed and coincides with the hyperplane $\eta$. Thus
   $U^n$ is a {\em tangentially degenerate   hypersurface}.

We recall that the {\em rank} of a tangentially degenerate
hypersurface is the number of parameters on which the family of
its tangent hyperplanes depends (see, for example, \cite{AG93},
p. 113). From relations (8) and (11) it follows that the
tangent hyperplane $\eta$ of the hypersurface $U^n$ along
its generator $A_n A_0$ is determined by this generator
and the points $A_i$, $\eta = A_n \wedge A_0 \wedge A_1
\wedge \ldots \wedge A_{n-1}$. The displacement of
this hyperplane is determined by the differentials
(8), (11), and
$$
dA_i = \omega_i^0 A_0 + \omega_i^j A_j + \omega_i^n A_n
+ \omega_i^{n+1} A_{n+1}.
$$
But by (5), $\omega_i^n = - g_{ij} \omega_n^j$, and the forms
$\omega_i^{n+1}$ are expressed according to formulas (13).
From formulas (13) and (14) it follows that the rank of
 a tangentially degenerate   hypersurface $U^n$ is
determined by the rank of the matrix $(\nu_{ij})$ in
terms of which the 1-forms $\omega_i^{n+1}$ and $\omega_0^i$
are expressed. But by (11) and (14) the dimension of
the submanifold $V$ described by the point $A_0$ on the
hyperquadric $Q^n$ is also equal to the rank of the matrix
$(\nu_{ij})$. Thus we have proved
the following result:

\begin{theorem} A lightlike hypersurface of the de Sitter
space $S^{n+1}_1$ is a ruled tangentially degenerate
hypersurface whose rank is equal to the dimension of
the submanifold $V$ described by the point $A_0$ on the
hyperquadric $Q^n$.
\end{theorem}

Denote the rank of the tensor $\nu_{ij}$ and of  the hypersurface
$U^n$ by $r$.  In this and next sections  we will assume that
$r  = n - 1$. The case $r < n -1$ was considered by the authors
in \cite{AG98b}.

For  $r = n - 1$, the hypersurface $U^n$ carries an
$(n-1)$-parameter
family of isotropic rectilinear generators $l = A_n A_0$ along
which the tangent hyperplane $T_x (U^n)$ is fixed. From the point
of view of physics, the isotropic rectilinear generators of
a lightlike hypersurface $U^n$ are trajectories of light
impulses, and the hypersurface $U^n$ itself represents a
{\em light flux} in spacetime.

Since $\mbox{{\rm rank}} \; (\nu_{ij}) = n - 1$,
the submanifold $V$ described by the point $A_0$ on the
hyperquadric $Q^n$ has dimension $n-1$, that is,
$V$ is a hypersurface. We denote it by $V^{n-1}$. The tangent
subspace $T_{A_0} (V^{n-1})$ to $V^{n-1}$ is determined by
the points $A_0, A_1, \ldots , A_{n-1}$. Since $(A_n, A_i)
= 0$, this tangent subspace is polar-conjugate to the
rectilinear generator $A_0 A_n$ of
the lightlike hypersurface $U^n$.

The submanifold $V^{n-1}$ of the
hyperquadric $Q^n$ is the image of a hypersurface of
the conformal space $C^n$ under the Darboux mapping.
 We will denote this hypersurface also by $V^{n-1}$.
In the space $C^n$,
the hypersurface  $V^{n-1}$ is defined by equation (10) which
by (5) is equivalent to equation (9) defining a lightlike
hypersurface $U^n$ in the space $S_1^{n+1}$. To
the rectilinear generator $A_n A_0$ of the hypersurface $U^n$
there corresponds a parabolic pencil of hyperspheres $A_n + sA_0$
tangent to the hypersurface  $V^{n-1}$ (see \cite{AG96}, p. 40).
Thus the following theorem is valid:

\begin{theorem} There exists a one-to-one correspondence
between the set of hypersurfaces of the conformal space $C^n$
and the set of lightlike hypersurfaces of the maximal rank
$r = n-1$ of the de Sitter
space $S^{n+1}_1$. To  pencils of tangent hyperspheres
of the hypersurface  $V^{n-1}$ there correspond isotropic
rectilinear generators of the lightlike hypersurface $U^n$.
\end{theorem}

Note that for lightlike hypersurfaces of the four-dimensional
Minkowski space $M^4$ the result similar to the result of Theorem
2 was obtained  in \cite{K89}.

\section{The Fundamental Forms and Connections on a
Lightlike Hypersurface of the de Sitter Space}

The first fundamental form of a lightlike hypersurface
$U^n$ of the space $S^{n+1}_1$ is a metric quadratic form. It
is defined by the scalar square of the differential $dx$
of a point of this  hypersurface. Since we have $x = A_n$,
by (8) and (2) this scalar square has the form
\begin{equation}\label{eq:15}
(dA_n, dA_n) = g_{ij} \omega_n^i  \omega_n^j = g
\end{equation}
and is a positive semidefinite
differential quadratic form of signature
$(n-1, 0)$. It follows that the system of equations
$ \omega_n^i = 0$ defines on the  hypersurface  $U^n$
a fibration of isotropic lines which, as we showed in Section 2,
coincide with rectilinear generators of this hypersurface.

The second fundamental form of a lightlike hypersurface
$U^n$ determines its deviation from the tangent hyperplane
$\eta$. To find this quadratic form, we compute the part of
the second differential of the point $A_n$ which
does not belong to the tangent hyperplane $\eta = A_0 \wedge A_1
\wedge \ldots \wedge A_n$:
$$
d^2 A_n \equiv \omega_n^i \omega_i^{n+1} A_{n+1} \pmod{\eta}.
$$
This implies that the second fundamental form can be written as
\begin{equation}\label{eq:16}
b = \omega_n^i  \omega_i^{n+1} = \nu_{ij} \omega_n^i  \omega_n^j,
\end{equation}
where we used expression (13) for the form $\omega_i^{n+1}$.
Since we assumed that $\mbox{{\rm rank}} \;
(\nu_{ij}) = n - 1$, the rank of the quadratic form (16)
as well as the rank of the form $g$ is equal to $n - 1$.
The nullspace of this   quadratic form  (see \cite{ON83}, p. 53)
is again determined by  the system of equations
$ \omega_n^i = 0$  and coincides with the isotropic direction
on the hypersurface $U^n$. The reduction of the rank of the
quadratic form $b$ is connected with the tangential degeneracy
of the hypersurface $U^n$. The latter was noted in Theorem 2.

On a hypersurface $V^{n-1}$ of the conformal space $C^n$
that corresponds to a lightlike hypersurface $U^n \subset
S^{n+1}_1$, the quadratic forms (15) and (16) define
the net of curvature lines, that is, an orthogonal
and conjugate net.

To find the connection forms of the hypersurface $U^n$,
we find exterior derivatives of its basis forms
$\omega_n^0$ and $\omega_n^i$:
\begin{equation}\label{eq:17}
\renewcommand{\arraystretch}{1.3}
\left\{
\begin{array}{ll}
d \omega_n^0 =  \omega_n^0 \wedge \omega_0^0
                  + \omega_n^i \wedge \omega_i^0, \\
d \omega_n^i =  \omega_n^0 \wedge \omega_0^i
                 + \omega_n^j \wedge \omega_j^i.
\end{array}
\renewcommand{\arraystretch}{1}
\right.
\end{equation}
This implies that the matrix 1-form
\begin{equation}\label{eq:18}
\omega = \pmatrix{\omega_0^0 & \omega_i^0 \cr
                   \omega_0^i & \omega_j^i \cr}
\end{equation}
defines a torsion-free connection
on the hypersurface $U^n$. To clarify the properties
 of this connection, we find its curvature forms.
Taking exterior derivatives of the forms
(18) and applying equations (5), (7), (9), and (10),
 we obtain
\begin{equation}\label{eq:19}
\renewcommand{\arraystretch}{1.3}
\left\{
\begin{array}{lll}
\Omega_0^0 = d \omega_0^0 - \omega_0^i \wedge \omega_i^0 = 0, \\
\Omega_0^i = d \omega_0^i - \omega_0^0 \wedge \omega_0^i
                   -  \omega_0^j \wedge \omega_j^i = 0, \\
\Omega_i^0 = d \omega_i^0 - \omega_i^0 \wedge \omega_0^0
                   -  \omega_i^j \wedge \omega_j^0
= -g_{ij}  \omega_n^j \wedge \omega_n^0, \\
\Omega_j^i = d \omega_j^i - \omega_j^0 \wedge \omega_0^i
                   -  \omega_j^k \wedge \omega_k^i
                   -  \omega_j^{n+1} \wedge \omega_{n+1}^i
= -g_{jk}  \omega_n^k \wedge \omega_n^i.
\end{array}
\renewcommand{\arraystretch}{1}
\right.
\end{equation}
In these formulas the forms $\omega_j^{n+1}$ and $\omega_0^i$
are expressed in terms of the basis forms $\omega_n^i$, and the
forms $\omega_0^j, \omega_j^i$, and $\omega_i^0$ are
fiber forms. If the principal parameters are fixed, then
these fiber forms are invariant forms of the group $G$
of admissible transformations of frames associated with
a point $x = A_n$ of the hypersurface $U^n$, and the connection
 defined by the form (18) is a $G$-connection.

To assign an affine connection on the hypersurface $U^n$,
it is necessary to make a reduction of
the family of frames in such a way that
the forms $\omega_i^0$ become principal. Denote by
$\delta$ the symbol of differentiation with respect to
the fiber parameters, that is, for a fixed point $x = A_n$
of the hypersurface $U^n$, and by $\pi_\eta^\xi$ the values
of the 1-forms $\omega_\eta^\xi$ for a fixed point $x = A_n$,
that is, $\pi_\eta^\xi = \omega_\eta^\xi (\delta)$. Then
we obtain
$$
\pi_n^0 = 0, \pi_n^i = 0, \pi_i^n = 0, \pi_i^{n+1} = 0.
$$
It follows
\begin{equation}\label{eq:20}
\delta A_i = \pi_i^0 A_0 + \pi_i^j A_j.
\end{equation}

The points $A_0$ and $A_i$ determine the tangent subspace to the
submanifold $V^{n-1}$ described by the point $A_0$ on the
 hyperquadric $Q^n$. If we fix an $(n-2)$-dimensional subspace
$\zeta$ not containing the point $A_0$ in this tangent subspace
and place the points $A_i$ into $\zeta$, then we obtain $\pi_i^0 = 0$.
This means that the forms $\omega_i^0$ become principal, that is,
\begin{equation}\label{eq:21}
\omega_i^0 = \mu_{ij} \omega_n^j + \mu_i \omega_n^0,
\end{equation}
and as a result, an affine connection arises
on the hypersurface $U^n$.

We will call the subspace $\zeta \subset T_{A_0} (V^{n-1})$
the {\em normalizing subspace} of the lightlike
hypersurface $U^n$. We have proved the following result:

\begin{theorem} If in every tangent subspace $T_{A_0} (V^{n-1})$
of  the submanifold $V^{n-1}$ associated with a lightlike
hypersurface $U^n, V^{n-1} \subset Q^n$, a normalizing
 $(n-2)$-dimensional subspace $\zeta$  not containing the point
$A_0$ is assigned, then there arises a torsion-free
affine connection on $U^n$.
\end{theorem}

The last statement follows the first two equations of (19).

The constructed above fibration of normalizing subspaces $\zeta$
defines a distribution $\Delta$ of $(n-1)$-dimensional
elements on a lightlike hypersurface $U^n$. In fact,
the point $x = A_n$ of the hypersurface $U^n$ along with the
subspace $\zeta =  A_1 \wedge \ldots \wedge
A_{n-1}$ define the $(n-1)$-dimensional subspace
which is complementary to the straight line $A_n A_0$ and
lies in the tangent subspace $\eta$ of the hypersurface $U^n$.
Following the book \cite{DB96}, we will call this subspace
the {\em screen}, and the distribution $\Delta$ the
{\em screen distribution}. Since at the point $x$ the screen
is determined by the subspace $A_n A_1 \ldots A_{n-1}$,
the differential equations of the screen distribution
has the form
\begin{equation}\label{eq:22}
\omega_n^0 = 0.
\end{equation}
But by (21)
$$
d\omega_n^0 = \omega_n^i \wedge (\mu_{ij} \omega_n^j
+ \mu_i \omega_n^0).
$$
Hence the screen distribution  is integrable if and only if the tensor $\mu_{ij}$  is symmetric. Thus we arrived
at the following result:

\begin{theorem} The fibration of normalizing subspaces $\zeta$
defines a screen distribution $\Delta$ of $(n-1)$-dimensional
elements on a lightlike hypersurface $U^n$. This  distribution
is integrable if and only if the tensor $\mu_{ij}$
defined by equation $(21)$ is symmetric.
\end{theorem}

Note that the configurations similar to that described in Theorem
5 occurred in the works of the Moscow geometers published in the
1950s. They were called the {\em one-side stratifiable pairs of
ruled surfaces} (see \cite{F56}, \S 30 or \cite{AG93}, p. 187).

\section{An Invariant Normalization of
Lightlike  Hypersurfaces of the de Sitter Space}

In \cite{A52} (see also \cite{AG96}, Ch. 2)
an invariant normalization of a hypersurfaces $V^{n-1}$
of the conformal space $C^n$ was constructed. By Theorem 3, this normalization can be interpreted in terms of the geometry of
the de Sitter space $S_1^{n+1}$.

Taking exterior derivative of equations (10) defining
the hypersurface $V^{n-1}$ in the conformal space $C^n$,
we obtain
$$
 \omega_i^n \wedge  \omega_0^i = 0,
$$
from which by linear independence of the 1-forms
$\omega_0^i$ on $V^{n-1}$ and Cartan's lemma we find that
\begin{equation}\label{eq:23}
\omega_i^n = \lambda_{ij} \omega_0^j, \;\; \lambda_{ij}
= \lambda_{ji}.
\end{equation}
Here and in what follows we retain the notations used
in the study of the geometry of hypersurfaces
of the conformal space $C^n$ in the book \cite{AG96}.

It is not difficult to find relations between the
coefficients $\nu_{ij}$ in formulas (13) and
 $\lambda_{ij}$ in formulas (23). Substituting the values
of the forms $\omega_i^n$ and $\omega_0^j$ from (5) into
(23), we find that
$$
- g_{ij} \omega_n^j = \lambda_{ij} g^{jk} \omega_k^{n+1}.
$$
Solving these equations for $\omega_k^{n+1}$, we obtain
$$
 \omega_i^{n+1} = - g_{ik} \widetilde{\lambda}^{kl} g_{lj}  \omega_n^j,
$$
where $(\widetilde{\lambda}^{kl})$ is the inverse matrix of the
matrix $(\lambda_{ij})$. Comparing these equations with equations (13), we obtain
\begin{equation}\label{eq:24}
\nu_{ij} = - g_{ik} \widetilde{\lambda}^{kl} g_{lj}.
\end{equation}
Of course, in this computation we assumed that the
matrix $(\lambda_{ij})$ is nondegenerate.

Let us clarify the geometric meaning of the vanishing of
$\det (\lambda_{ij})$. To this end, we make an admissible
transformation of the moving frame associated with a point
of a lightlike hypersurface $U^n$ by setting
\begin{equation}\label{eq:25}
\widehat{A}_n = A_n + s A_0.
\end{equation}
The point $\widehat{A}_n$ as the point $A_n$ lies on the
rectilinear generator $A_n A_0$. Differentiating this point and
applying formulas (8) and (11), we obtain
\begin{equation}\label{eq:26}
d\widehat{A}_n = (ds + s\omega_0^0 + \omega_n^0) A_0 +
(\omega_n^i + s \omega_0^i) A_i.
\end{equation}
It follows that in the new frame the forms $\omega_n^i$
become
$$
\widehat{\omega}_n^i = \omega_n^i + s \omega_0^i.
$$
By (5) and (23), it follows that
$$
\widehat{\omega}_n^i = -g^{ik} (\lambda_{kj} - s g_{kj}) \omega_0^j.
$$
This implies that in the new frame the quantities $\lambda_{ij}$
become
\begin{equation}\label{eq:27}
\widehat{\lambda}_{ij} = \lambda_{ij} - s g_{ij}.
\end{equation}
Consider also the matrix $(\widehat{\lambda}^i_j)
= (g^{ik} \widehat{\lambda}_{kj})$. Since $g_{ij}$ is
a nondegenerate tensor, the matrices  $(\widehat{\lambda}^i_j)$
and $(\widehat{\lambda}_{ij})$ have the same rank
$\rho \leq n - 1$.

From equation (26) it follows that
$$
d\widehat{A}_n = (ds + s\omega_0^0 + \omega_n^0) A_0
- \widehat{\lambda}^i_j A_i \omega^j_0.
$$
The differential $d\widehat{A}_n$ is the differential of the
mapping $f: M^{n-1} \times l \rightarrow P^{n+1}$
which was considered in Theorem 1. The linearly independent
forms $\omega_0^i$ are basis forms on
the manifold $M^{n-1}$, and the form
$\widehat{\omega}_n^0 = ds + s\omega_0^0 + \omega_n^0$ containing
a nonhomogeneous parameter $s$ of the projective line $l$ is
a basis form on this line. Thus the matrix
\begin{equation}\label{eq:28}
\pmatrix{1 & 0 \cr
         0 & \widehat{\lambda}^i_j \cr}
\end{equation}
is the Jacobi matrix of this mapping.
Hence the tangent subspace to
the  hypersurface $U^n$ at the point $\widehat{A}_n$
is determined by the points $\widehat{A}_n, A_0$, and
$\widehat{\lambda}^i_j A_i$. At the points,  at which
the rank $\rho$ of the
matrix $(\widehat{\lambda}^i_j)$ is equal to $n-1, \rho = n - 1$,
the tangent subspace to the  hypersurface $U^n$ has dimension $n$, and such points are {\em regular points} of the
hypersurface.
The points,  at which the rank $\rho$ of the
matrix $(\widehat{\lambda}^i_j)$ is reduced,
are {\em singular points} of the  hypersurface $U^n$. The
coordinates
of singular points are defined by the condition
$\det (\widehat{\lambda}^i_j) = 0$ which by (27) is equivalent to
the equation
\begin{equation}\label{eq:29}
\det (\lambda_{ij} - s g_{ij}) = 0,
\end{equation}
the {\em characteristic equation} of the matrix $(\lambda_{ij})$ with respect to the tensor $g_{ij}$. The degree of this equation
is equal to $n - 1$.

In particular, if  $A_n$ is a regular point
of the  hypersurface $U^n$, then the matrix $(\lambda_{ij})$
is nondegenerate, and equation (24) holds. On the other hand,
if  $A_n$ is a singular point
of  $U^n$, then  equation (24) is meaningless.

Since the matrix $(\lambda_{ij})$ is symmetric and
the matrix $(g_{ij})$ defines a positive definite form
of rank $n - 1$, equation (29) has
$n - 1$ real roots if each root is counted as many times
as its multiplicity. Thus on a rectilinear generator $A_n A_0$ of
a lightlike   hypersurface $U^n$ there are $n-1$ real singular points.

By Vieta's theorem, the sum of the roots of equation (29) is
equal to the coefficient in $s^{n-2}$, and this coefficient
is $\lambda_{ij} g^{ij}$. Consider the quantity
\begin{equation}\label{eq:30}
\lambda = \frac{1}{n-1} \lambda_{ij} g^{ij},
\end{equation}
which is the arithmetic mean of the roots of equation (29).
This quantity $\lambda$ allows us to construct  new
quantities
\begin{equation}\label{eq:31}
a_{ij} = \lambda_{ij} - \lambda g_{ij}.
\end{equation}
It is easy to check that the quantities $a_{ij}$
do not depend on the location of the point $A_n$ on the
straight line $A_n A_0$, that is, $a_{ij}$ is invariant with
respect to the transformation of the moving frame defined by
equation (25). Thus the quantities $a_{ij}$ form a tensor
on the  hypersurface $U^n$ defined in its
second-order neighborhood. This tensor satisfies the condition
\begin{equation}\label{eq:32}
a_{ij} g^{ij} = 0,
\end{equation}
that is, it is apolar to the tensor $g_{ij}$.

On the straight line $A_n A_0$ we consider a point
\begin{equation}\label{eq:33}
C = A_n + \lambda A_0.
\end{equation}
It is not difficult to check that this point remains also
fixed when the point $A_n$ moves along the  straight line $A_n
A_0$. Since $\lambda$ is the arithmetic mean of the roots of
equation (29) defining singular points on the  straight line $A_n A_0$, the point $C$ is the {\em harmonic pole}
(see \cite{C50}) of the point $A_0$ with respect to these
singular points. In particular, for $n = 3$, the point $C$ is the
fourth harmonic point to the point $A_0$ with respect to two
singular points of the rectilinear generator $A_3 A_0$ of the lightlike  hypersurface $U^3$ of the de Sitter space $S_1^4$.

In the conformal theory of hypersurfaces, to
the point $C$ there corresponds a hypersphere which is tangent
to the hypersurface at the point $A_0$. This hypersphere
 is called the {\em central tangent hypersphere}
(see \cite{AG96}, pp. 41--42). Since
\begin{equation}\label{eq:34}
(d^2 A_0, C) = a_{ij} \omega_0^i \omega_0^j,
\end{equation}
the cone
$$
a_{ij} \omega_0^i \omega_0^j = 0
$$
with vertex at the point $A_0$ belonging to the tangent
subspace $T_{A_0} (V^{n-1})$ contains the directions along which
the central  hypersphere has a second-order tangency with
the hypersurface $V^{n-1}$ at the point $A_0$. From the apolarity
condition (33) it follows that it is possible to inscribe an orthogonal
$(n-1)$-hedron with vertex at $A_0$ into the cone defined by equation (34).

Now we can construct an invariant normalization of
a lightlike hypersurface $U^n$ of the de Sitter space
$S_1^{n+1}$.
To this end, first we repeat some computations from Ch. 2
of \cite{AG96}.

Taking exterior derivatives of equations (23)
and applying Cartan's lemma, we  obtain the equations
\begin{equation}\label{eq:35}
\nabla \lambda_{ij} +  \lambda_{ij} \omega_0^0 + g_{ij} \omega_n^0 =  \lambda_{ijk} \omega_0^k,
\end{equation}
where
$$
\nabla \lambda_{ij} = d \lambda_{ij} - \lambda_{ik} \omega^k_j
-  \lambda_{kj} \omega^k_i,
$$
and the quantities $ \lambda_{ijk}$ are symmetric
with respect to all three indices.
Equations (35) confirm one more time that the
quantities $ \lambda_{ij} $ do not form a tensor
and depend on a location of the point $A_n$ on the straight line
$A_n A_0$. This dependence is described by a closed form
relation (27). From formulas (35) it follows that the quantity
$\lambda$ defined by equations (30) satisfy the differential
equation
\begin{equation}\label{eq:36}
d \lambda  +  \lambda \omega_0^0 +  \omega_n^0 =  \lambda_k \omega_0^k,
\end{equation}
where
$$
\lambda_k = \frac{1}{n-1} g^{ij} \lambda_{ijk}
$$
(see formulas (2.1.36) and (2.1.37) in the book
\cite{AG96}). The quantities $\lambda_k$ as well as
the quantities $\lambda_{ijk}$ are determined by a
third-order neighborhood of a generator $A_0 A_n$ of
a lightlike hypersurface $U^n \subset S^{n+1}_1$.

The point $C$ lying on the rectilinear generator $A_n A_0$
of the  hypersurface $U^n$ describes a submanifold $W \subset U^n$ when  $A_n A_0$ moves. Let us find the tangent
subspace to $U^n$ at the point $C$. Differentiating equation (33)
and applying formulas (8) and (11), we obtain
$$
dC = (d \lambda + \lambda \omega_0^0 +  \omega_n^0) A_0
+ (\omega_n^i +  \lambda \omega_0^i) A_i.
$$
By (5), (23), (30), and (36), it follows that
 \begin{equation}\label{eq:37}
d C = (\lambda_i A_0  - g^{jk} a_{ki} A_j) \omega_0^i.
   \end{equation}
Define the affinor
 \begin{equation}\label{eq:38}
a^i_j = g^{ik} a_{kj},
 \end{equation}
whose rank coincides with the rank of the tensor $a_{ij}$.
Then equation (37) takes the form
$$
d C = (\lambda_i A_0  -  a_i^j A_j) \omega_0^i.
$$
The points
 \begin{equation}\label{eq:39}
C_i = \lambda_i A_0  -  a_i^j A_j
 \end{equation}
together with the point $C$ define the tangent subspace
to the submanifold $W$ described by the point $C$ on
 the  hypersurface $U^n$.

If the point $C$ is a regular point of
the rectilinear generator $A_n A_0$ of the  hypersurface $U^n$,
then  the rank of the tensor $a_{ij}$ defined by
equations (30) as well as the rank of the affinor $a_j^i$ is
equal to $n-1$. As a result, the points $C_i$ are linearly
independent and together with the point $C$ define
the $(n-1)$-dimensional tangent subspace $T_C (W)$, and the submanifold $W$ itself has dimension $n-1, \dim W = n-1$.

The points $C_i$ also belong to the tangent subspace
$T_{A_0} (V^{n-1})$ and define  the $(n-2)$-dimensional
subspace $\zeta = T_{A_0} (V^{n-1}) \cap T_C (W)$ in it. This
subspace
is a normalizing subspace. Since such a normalizing subspace is
defined in each tangent subspace $T_{A_0} (V^{n-1})$ of the
hypersurface $V^{n-1} \subset Q^n$, there arises the fibration of
these subspaces which by Theorem 4 defines an invariant affine
connection on the lightlike hypersurface $U^n$.
The subspace $T_C (W)$ is determined by a
third-order neighborhood of a generator $A_0 A_n$ of
the  hypersurface $U^n$.

 Thus we proved the following result:

\begin{theorem} If the tensor $a_{ij}$ defined by formula $(40)$
 on a lightlike hypersurface $U^n \subset S^{n+1}_1$
is nondegenerate,
then it is possible to construct the invariant normalization
of $U^n$ by means of the $(n-2)$-dimensional subspaces
$$
\zeta = C_1 \wedge C_2 \wedge \ldots \wedge C_{n-1}.
$$
This normalization induces on $U^n$ an invariant affine
connection intrinsically connected with the geometry of this
hypersurface. The normalization as well as the induced affine
connection are determined by a
third-order neighborhood of a generator $A_0 A_n$ of
the  hypersurface $U^n$.
\end{theorem}

Theorem 5 implies that the invariant normalization
we have constructed defines on $U^n$ an invariant screen
distribution $\Delta$ which is also
intrinsically connected with the geometry of the
hypersurface $U^n$; here $\Delta_x = x \wedge \xi, \; x \in A_n A_0$.

Note that for the hypersurface $V^{n-1}$ of the conformal space $C^n$
a similar  invariant normalization was constructed as far
back as 1953 (see \cite{A52} and also \cite{AG96}, Ch. 2).
In the present paper we gave a new geometric meaning of
this invariant normalization.

\section{Singular Points of Lightlike  Hypersurfaces of
the de Sitter Space}

As we indicated in Section 4, the points
\begin{equation}\label{eq:40}
z =  A_n  + s A_0
\end{equation}
of the rectilinear generator $A_n A_0$ of the lightlike
hypersurface $U^n$ are singular if their nonhomogeneous
coordinate $s$ satisfies the equation
\begin{equation}\label{eq:41}
\det (\lambda_{ij} - s g_{ij}) = 0,
\end{equation}
(In these points the Jacobian of the mapping $f: M^{n-1} \times l \rightarrow P^{n+1}$,
which is equal to the determinant of the matrix (28), vanishes.)
We will investigate in more detail the structure of
a lightlike hypersurface $U^n$ in a neighborhood of its singular
point.

Equation (41) is the characteristic equation of
the matrix $(\lambda_{ij})$ with respect to the tensor
$(g_{ij})$. The degree of this equation is $n-1$, and
since the matrix  $(\lambda_{ij})$  is symmetric and
 the matrix $(g_{ij})$ is also symmetric and
positive definite, then according
to the well-known result of linear algebra, all  roots
of this equation are real, and the matrices $(\lambda_{ij})$
 and  $(g_{ij})$ can be simultaneously reduced
to a diagonal form.

 Denote the roots of the characteristic equation
 by $s_h, h = 1, 2,
\ldots , n - 1$, and denote the corresponding singular points
of the rectilinear generator $A_n A_0$ by
\begin{equation}\label{eq:42}
B_h =  A_n  + s_h A_0.
\end{equation}
These singular points are called {\em foci}
 of the rectilinear generator $A_n A_0$ of a lightlike
hypersurface $U^n$.

It is clear from (42) that the point $A_0$ is not a focus
 of the rectilinear generator $A_n A_0$. This is explained
by the fact that by our assumption $\mbox{{\rm rank}} \;
(\nu_{ij}) = n - 1$, and by (14), on the hyperquadric $Q^n$
the point $A_0$ describes a hypersurface $V^{n-1}$ which is
transversal to the straight lines $A_0 A_n$.

In the conformal theory of hypersurfaces, to
the singular points $B_h$ there correspond the
tangent hyperspheres defining the principal directions at
a point $A_0$ of the hypersurface $V^{n-1}$ of the conformal
space $C^n$ (see \cite{AG96}, p. 56).

We will construct a classification of singular points
of a lightlike hypersurface $U^n$ of the space $S^{n+1}_1$.
We will use some computations that we made while constructing
a classification of canal hypersurfaces in \cite{AG98a}.

Suppose first that $B_1 = A_n + s_1 A_0$ be a singular
point defined by a simple
root $s_1$ of characteristic equation (41),
$s_1 \neq s_h, h = 2, \ldots , n - 1$. For this singular point
we have
\begin{equation}\label{eq:43}
dB_1 = (ds_1 + s_1 \omega_0^0 + \omega_n^0) A_0 -
\widehat{\lambda}_j^i \omega_0^j A_i,
\end{equation}
where
\begin{equation}\label{eq:44}
\widehat{\lambda}_j^i = g^{ik} (\lambda_{kj} - s_1 g_{kj})
\end{equation}
is a degenerate symmetric affinor having a single null
eigenvalue. The matrix of this affinor can be reduced
to a quasidiagonal form
\begin{equation}\label{eq:45}
(\widehat{\lambda}_j^i) = \pmatrix{0 & 0 \cr
             0 & \widehat{\lambda}_q^p \cr},
\end{equation}
where $p, q = 2, \ldots , n - 1$, and $(\widehat{\lambda}_q^p)$
is a nondegenerate symmetric affinor. The matrices
$(g_{ij})$ and $(\lambda_{ij} - s_1 g_{ij})$ are reduced
to the forms
$$
\pmatrix{1 & 0 \cr
         0 & g_{pq}\cr} \;\; \mbox{{\rm and}} \;\;
\pmatrix{0 & 0 \cr
             0 & \widehat{\lambda}_{pq}\cr},
$$
where $(\widehat{\lambda}_{pq}) = (\lambda_{pq} - s_1
g_{pq})$ is a nondegenerate symmetric matrix.

Since the point $B_1$ is defined invariantly on the
generator $A_n A_0$, then it will be fixed if $\omega_0^i = 0$.
Thus it follows from (43) that
\begin{equation}\label{eq:46}
ds_1 + s_1 \omega_0^0 + \omega_n^0 = s_{1i} \omega^i;
\end{equation}
 here and in what follows $\omega^i = \omega_0^i$.
By (45) and (46) relation (43) takes the form
\begin{equation}\label{eq:47}
dB_1 = s_{11} \omega^1 A_0 + (s_{1p} A_0
- \widehat{\lambda}_p^q A_q)
 \omega^p.
\end{equation}
Here the points $C_p = s_{1p} A_0 - \widehat{\lambda}_p^q A_q$
are linearly independent and belong to the tangent subspace
$T_{A_0} (V^{n-1})$.

Consider the submanifold ${\cal F}_1$ described by the singular point
$B_1$ in the space $S^{n+1}_1$. This submanifold is called
the {\em focal manifold} of the hypersurface $U^n$.
Relation (47) shows that two cases are possible:

\begin{description}
\item[1)] $s_{11} \neq 0$. In this case  the submanifold
${\cal F}_1$
is of dimension $n - 1$, and its tangent subspace at the point
$B_1$ is determined by the points $B_1, A_0$, and $C_p$. This
subspace contains the straight line $A_n A_0$, intersects
the hyperquadric $Q^n$, and thus it,  as well as  the submanifold
${\cal F}_1$ itself, is timelike. For $\omega^p = 0$,
the point $B_1$ describes a curve $\gamma$ on the submanifold
${\cal F}_1$
which is tangent to the straight line $B_1 A_0$ coinciding with
the generator $A_n A_0$ of the hypersurface $U^n$.
The curve $\gamma$ is an isotropic curve of the de Sitter
space $S^{n+1}_1$. Thus on
${\cal F}_1$ there arises a fibre bundle of focal lines. The
hypersurface $U^n$ is foliated into an $(n-2)$-parameter
family of torses for which these lines are edges of regressions.
The points $B_1$ are singular points of a kind which is called
a {\em fold}.

If the characteristic equation (41) has distinct roots, then
an isotropic rectilinear generator $l$ of a lightlike hypersurface
$U^n$ carries $n - 1$ distinct foci $B_h, h = 1, \ldots , n-1$.
If for each of these foci the condition of type $s_{11} \neq 0$
holds,
then each of them  describes a focal submanifold ${\cal F}_h$,
carrying conjugate net.  Curves of one family of this net are
tangent
to the straight lines $l$, and this family is isotropic.
On the hypersurface $V^{n-1}$ of the space $C^n = Q^n$ described by
the point $A_0$, to these conjugate nets there correspond the
net of curvature lines.

\item[2)] $s_{11} = 0$. In this case  relation (47) takes
the form
\begin{equation}\label{eq:48}
dB_1 =  (s_{1p} A_0 - \widehat{\lambda}_p^q A_q)  \omega^p,
\end{equation}
and the focal submanifold ${\cal F}_1$
is of dimension $n - 2$. Its tangent subspace at the point
$B_1$ is determined by the points $B_1$ and $C_p$.
An arbitrary point $z$ of this subspace can be written in the
form
$$
z = z^n B_1 + z^p C_p = z^n (A_n + s_1 A_0) + z^p
(s_{1p} A_0 - \widehat{\lambda}_p^qA_q).
$$
Substituting the coordinates of this point into relation (3),
we find that
$$
(z, z) = g_{rs} \widehat{\lambda}_p^r
\widehat{\lambda}_q^s z^p z^q + (z^n)^2 > 0.
$$
It follows that the tangent subspace
$T_{B_1} ({\cal F}_1)$ does not have common points with
the hyperquadric $Q^n$, that is, it is spacelike. Since this
takes place for any point $B_1 \in {\cal F}_1$, the focal
submanifold ${\cal F}_1$ is spacelike.

For $\omega^p = 0$, the point $B_1$ is fixed. The subspace
$T_{B_1} ({\cal F}_1)$ will be fixed too.
On the hyperquadric $Q^n$, the point $A_0$ describes
a curve $q$ which is polar-conjugate to $T_{B_1} ({\cal F}_1)$.
Since $\dim T_{B_1} ({\cal F}_1) = n - 2$, the curve $q$ is
a conic, along which the two-dimensional plane
polar-conjugate to the subspace $T_{B_1} ({\cal F}_1)$  with
respect to the hyperquadric $Q^n$, intersects $Q^n$.
Thus for $\omega^p = 0$, the rectilinear generator $A_n A_0$ of
the hypersurface $U^n$ describes  a two-dimensional
second-order cone  with vertex at the point $B_1$
and the directrix $q$. Hence in the case under consideration
a lightlike hypersurface $U^n$ is foliated into an
$(n-2)$-parameter family of second-order cones whose vertices
describe the $(n-2)$-dimensional focal submanifold
${\cal F}_1$, and
the points $B_1$ are {\em conic} singular points of
the hypersurface $U^n$.

The hypersurface $V^{n-1}$ of the conformal space $C^n$
corresponding to such a lightlike hypersurface $U^n$
is a canal hypersurface which envelops an $(n-2)$-parameter
family of hyperspheres. Such a hypersurface carries a family
of cyclic generators which depends on the same number
of parameters. Such hypersurfaces were investigated in detail
in \cite{AG98a}.
\end{description}

Further let $B_1$ be a  singular point of multiplicity $m$, where
$m \geq 2$, of a rectilinear generator $A_n A_0$
of  a lightlike hypersurface $U^n$ of the space $S^{n+1}_1$
defined by an $m$-multiple root of characteristic equation
(41). We will assume that
\begin{equation}\label{eq:49}
 s_1 = s_2 = \ldots = s_m := s_0, s_0 \neq s_p,
\end{equation}
and also assume that $a, b, c = 1, \ldots, m$
and $p, q, r = m+1, \ldots, n-1$. Then the matrices $(g_{ij})$
and $(\lambda_{ij})$ can be simultaneously
reduced to  quasidiagonal forms
$$
\pmatrix{g_{ab} & 0 \cr
             0 & g_{pq}\cr} \;\; \mbox{{\rm and}} \;\;
\pmatrix{s_0 g_{ab} & 0 \cr
             0 & \lambda_{pq}\cr}.
$$
We also construct the matrix $(\widehat{\lambda}_{ij})
= (\lambda_{ij} - s_0 g_{ij})$. Then
\begin{equation}\label{eq:50}
(\widehat{\lambda}_{ij}) = \pmatrix{0 & 0 \cr
             0 & \widehat{\lambda}_{pq}\cr},
\end{equation}
where $\widehat{\lambda}_{pq} = \lambda_{pq} - s_0 g_{pq}$
is a nondegenerate matrix of order $n - m - 1$.

By relations (50) and formulas (5) and (23) we have
\begin{equation}\label{eq:51}
\omega_a^n - s_0 \omega_a^{n+1} = 0,
\end{equation}
\begin{equation}\label{eq:60}
\omega_p^n - s_0 \omega_p^{n+1} =  \widehat{\lambda}_{pq}
\omega^q.
\end{equation}
Taking exterior derivative of equation (51) and applying relation
(52), we find that
\begin{equation}\label{eq:53}
\widehat{\lambda}_{pq} \omega_a^p \wedge \omega^q
+ g_{ab} \omega^b \wedge (ds_0 + s_0 \omega_0^0
+ \omega_n^0) = 0.
\end{equation}
It follows that the 1-form $ds_0 + s_0 \omega_0^0
+ \omega_n^0$ can be expressed in terms of the basis forms.
We write these expressions in the form
\begin{equation}\label{eq:54}
ds_0 + s_0 \omega_0^0 + \omega_n^0 = s_{0c} \omega^c +
s_{0q} \omega^q.
\end{equation}
Substituting this decomposition into equation (53), we find that
\begin{equation}\label{eq:55}
(\widehat{\lambda}_{pq} \omega_a^p
+ g_{ab} s_{0q} \omega^b) \wedge \omega^q + g_{ab}  s_{0c}
\omega^b \wedge \omega^c  = 0.
\end{equation}
The terms in the left hand side of (55) do not have similar
terms. Hence both terms are equal to 0. Equating to 0
the coefficients of the summands of the second term,
we find that
\begin{equation}\label{eq:56}
 g_{ab}  s_{0c} =  g_{ac}  s_{0b}.
\end{equation}
Contracting this equation with the matrix $(g^{ab})$
which is the inverse matrix of the matrix $(g_{ab})$, we obtain
$$
ms_{0c} = s_{0c}.
$$
Since $m \geq 2$, it follows that
$$
s_{0c} = 0,
$$
and  relation (54) takes the form
\begin{equation}\label{eq:57}
ds_0 + s_0 \omega_0^0 + \omega_n^0 = s_{0p} \omega^p.
\end{equation}

For the  singular point of multiplicity $m$ of the generator
$A_n A_0$ in question the equation (43) can be written in the
form
$$
dB_1 = (ds_0 + s_0 \omega_0^0 + \omega_n^0) A_0 -
\widehat{\lambda}_q^p \omega_0^q A_p.
$$
Substituting decomposition (57) in the last equation,
we find that
\begin{equation}\label{eq:58}
dB_1 = (s_{0p} A_0 - \widehat{\lambda}_p^q A_q)\omega_0^p.
\end{equation}
This relation is similar to equation (48) with the only
difference that in (48) we had $p, q = 2, \ldots, n - 1$, and
in (58) we have $p, q = m+1, \ldots, n - 1$.
Thus the point $B_1$ describes now a spacelike focal
manifold ${\cal F}_1$ of dimension $n-m-1$. For $\omega_0^p = 0$,
the point $B_1$ is fixed, and the point $A_0$ describes an
$m$-dimensional submanifold on the hyperquadric $Q^n$ which
is a cross-section of $Q^n$ by an $(m+1)$-dimensional subspace
that is polar-conjugate to the $(n-m-1)$-dimensional subspace
tangent to the submanifold ${\cal F}_1$.

The point $B_1$ is a  conic singular point of multiplicity $m$ of
a lightlike hypersurface $U^n$, and this hypersurface is foliated
into an $(n-m-1)$-parameter family of $(m+1)$-dimensional
second-order cones circumscribed about
the hyperquadric $Q^n$. The  hypersurface $V^{n-1}$ of the
conformal space $C^n$ that corresponds to such a  hypersurface
$U^n$ is an $m$-canal hypersurface (i.e.,
the envelope of an $(n-m-1)$-parameter family of
hyperspheres), and it carries an $m$-dimensional
spherical generators.

Note also an extreme case when the rectilinear generator
$A_n A_0$ of a lightlike hypersurface $U^n$ carries a single
 singular point of multiplicity $n-1$.
As follows from our consideration
of the cases $m \geq 2$, this singular point is fixed, and
the hypersurface $U^n$ become a second-order hypercone with
vertex at this singular point which is circumscribed about
 the hyperquadric $Q^n$. This hypercone is the isotropic cone
of the space $S^{n+1}_1$. The  hypersurface $V^{n-1}$ of the
conformal space $C^n$ that corresponds to such a  hypersurface
$U^n$ is a hypersphere of the space $C^n$.

The following theorem combines the results of this section:

\begin{theorem}
 A lightlike hypersurface $U^n$ of maximal rank $r=n-1$ of the de
Sitter space $S^{n+1}_1$ possesses $n-1$ real singular points
on each of its rectilinear generators if each of these singular
points  is counted as many times as its multiplicity. The simple
singular points can be of two kinds: a fold and conic. In the
first case the hypersurface $U^n$ is foliated into an $(n-2)$-parameter family of torses, and in the second case it is foliated
into an $(n-2)$-parameter family of second-order cones.
The vertices of these cones
describe the $(n-2)$-dimensional spacelike submanifold
in the space $S^{n+1}_1$. All multiple singular points
of a hypersurface $U^n$ are conic. If a rectilinear
generator of a hypersurface $U^n$ carries a singular point
of multiplicity $m$, $2 \leq m \leq n -1$,
then the hypersurface $U^n$ is foliated into an $(n-m-1)$-parameter family of $(m+1)$-dimensional second-order cones.
The vertices of these cones
describe the $(n-m-1)$-dimensional spacelike submanifold
in the space $S^{n+1}_1$.
The hypersurface $V^{n-1}$ of the conformal space $C^n$
corresponding to a lightlike hypersurface $U^n$ with
 singular points of multiplicity $m$
is a canal hypersurface which envelops an $(n-m-1)$-parameter
family of hyperspheres and has $m$-dimensional spherical
generators.
\end{theorem}

Since lightlike hypersurfaces $U^n$ of
the de Sitter space $S^{n+1}_1$ represent a light flux
(see Section {\bf 2}), its focal submanifolds have the following
physical meaning. If one of them is a lighting submanifold, then
others will be manifolds of concentration of a light flux.
Intensity of concentration depends on multiplicity of a focus
 describing this submanifold.

In the extreme case when an isotropic rectilinear generator
$l = A_n A_0$ of a hypersurface $U^n$ carries one $(n-1)$-multiple
focus, the hypersurfaces $U^n$ degenerates into the light cone
generated by a point source of light. This cone represents a radiating
light flux.

If each isotropic generator $l \subset U^n$ carries two foci
$B_1$ and $B_2$ of multiplicities $m_1$ and $m_2, \,m_1 + m_2 = n - 1, \,
m_1 > 1, \, m_2 > 1$, then these foci describe spacelike submanifolds
${\cal F}_1$ and ${\cal F}_2$ of dimension $n - m_1 - 1$ and
$n - m_2 - 1$, respectively. If one of these submanifolds is a lighting
 submanifold, then on the second one a light flux is concentrated.

config: arxiv


TITLE: Why is it so hard to find out the sign of the cosmological constant?
QUESTION [2 upvotes]: Correct me if I am wrong (it is in fact part of my question :) ) but from the little I know about the cosmological constant, growing existing evidence is still insufficient to tell whether $\Lambda$ is positive, zero or negative.
In case it is true, I would like to know whether there exist any explanations why the growing amount of data keeps failing to give a decisive answer. Seemingly what is known indicates that $\Lambda$, if nonzero, is very very small. Is it true? Is there a good reason for that?
Or, in case it has been already established, could anyone describe the consequences of the current knowledge of $\Lambda$?

REPLY [5 votes]: The big problem with finding the sign of the cosmological constant is that regardless of whether it is positive, negative, or zero, its magnitude seems to be very small.  This means that while we have done a great job of pushing down our error bounds, they still contain both positive, negative, and zero values for the cosmological constant.  As an aside, note that if the cosmological constant is truly zero, then no matter how carefully we measure it we will always have error bounds that contain both positive and negative values, and so we would have to find some other way to conclude that its value is zero.
A comment to the question above asks why the fact that the universe's expansion is accelerating doesn't prove that the cosmological constant has positive sign.  The reason for this is simple: there are different reasons why the universe's expansion could be accelerating.  If the accelerating expansion is due to a cosmological constant, then such a constant would have to be positive.  This is the foundation of the $\Lambda CDM$ model of cosmology, but it isn't the only option.  Dark energy does not have to take the form of a cosmological constant; it could be something else entirely.  If this is the case, the fact that the expansion of the universe is accelerating has no bearing on the sign of the cosmological constant.
Because we know so little about the nature of dark energy, we can't make any clear-cut conclusions about whether or not it is simply a cosmological constant.