Home » 2011 » April

Monthly Archives: April 2011

Real and Complex projective n-spaces

Sunday, April 17th, 2011 07:47 / Leave a Comment

Real projective n-space

Real projective $n$ space, $\mathbb{R} P^n$ is defined to be the space of all lines through the origin in $\mathbb{R}^{n+1}$ . Each such kind is determined by a non-zero vector in $\mathbb{R}^{n+1}$ , unique up to scalar multiplication, and $\mathbb{R} P^n$ is topologized as the quotient space of $\mathbb{R}^{n+1}-\{0\}$ under the equivalence relation $v \sim \lambda v$ for scalars $\lambda \neq 0$ . We can restrict to vectors of length 1, so $\mathbb{R}P^n$ is also the quotient spaces $\mathbf{S}^n / (v \sim -v)$ , the sphere with antipodal points identified. This is equivalent to saying that $\mathbb{R}P^n$ is the quotient space of a hemisphere $\mathbf{D^n}$ with antipodal points of $\partial D^n$ identified. Since $\partial D^n$ with antipodal points identified is just $\mathbb{R}P^{n-1}$ , we see that $\mathbb{R}P^n$ is obtained from $\mathbb{R}P^{n-1}$ by attaching an $n$ -cell, with the quotient projection $S^{n-1} \to \mathbb{R}P^{n-1}$ as the attaching map. It follows by induction on $n$ that $\mathbb{R}P^n$ has a cell complex structure $e^0 \cup e^1 \cup ... \cup e^n$ with one cell $e^i$ in each dimension $i \leq n$ . Since $\mathbb{R}P^n$ is obtained from $\mathbb{R}P^{n-1}$ by attaching an $n$ -cell, the infinite union $\mathbb{R}P^{\infty}={\bigcup}_n \mathbb{R}P^n$ becomes a Cell Complex with one cell in each dimension. We can view $\mathbb{R}P^{\infty}$ as the space of lines through the origin in $\mathbb{R}^{\infty}={\bigcup}_n \mathbb{R}^n$ .

Complex Projective n-space

Complex projective n-space $\mathbb{C}P^n$ is space of complex lines through the origin in $\mathbb{C}P^{n+1}$ , that is, 1-dimensional vector subspaces of $\mathbb{C}^{n+1}$ . As in the case of Real projective n-space, each line is determined by a non-zero vector in $\mathbb{C}^{n+1}$ , unique up to scalar multiplication, and $\mathbb{C}P^n$ is topologized as the quotient space of $\mathbb{C}^{n+1}-0$ under the equivalence relation $v \sim \lambda v$ for $\lambda = 0$ .

Equivalently, this is the quotient of the unit Sphere $S^{2n+1} \subset \mathbb{C}^{n+1}$ with $v \sim \lambda v$ for $| \lambda |=1$ .

It is also possible to obtain $\mathbb{C}P^n$ as a quotient space of disk $D^{2n}$ under the identifications $v \sim \lambda v$ for $\lambda v \in \partial D^{2n}$ , in the following way:

The vectors in $S^{2n+1} \subset \mathbb{C}^{n+1}$ with last coordinate real and nonnegative are precisely the vectors of the form $(\omega, \sqrt{1-{|\omega|}^2} ) \in \mathbb{C}^n \times \mathbb{C}$ with $|\omega| \leq 1$ . Such vectors form the of the function $\omega \to \sqrt{1-{|\omega|}^2}$ . This is a disk ${D_+}^{2n}$ bounded by the sphere $S^{2n-1} \subset S^{2n+1}$ consisting of vectors $(\omega, 0) \in \mathbb{C}^n \times \mathbb{C}$ with $|\omega|=1$ . Each vector in $S^{2n+1}$ is equivalent under the identifications $v \sim \lambda v$ to a vector in ${D_+}^{2n}$ , and the latter vector is unique if its last coordinate is non-zero. If the last coordinate is zero, we have just the identifications $v \sim \lambda v$ for $v \in S^{2n-1}$ .

From the description of $\mathbb{C}P^n$ as the quotient of ${D_+}^{2n}$ under the identifications $v \sim \lambda v$ for $v \in S^{2n-1}$ , it follows that $\mathbb{C}P^n$ is obtained from $\mathbb{C}P^{n-1}$ by attaching a cell $e^{2n}$ via the quotient map $S^{2n-1} \to \mathbb{C}P^{n-1}$ . So by induction on $n$ , we obtain a cell structure

$\mathbb{C}P^n=e^0 \cup e^2 \cup ...\cup e^{2n}$

with cells only in even dimensions. Similarly, $\mathbb{C}P^{\infty}$ has a cell structure with one cell in each even dimension.

Note: The Sphere $S^n$ has the structure of a cell complex with just two cells, $e^0$ and $e^n$ , the n-cell being attached by the constant map $S^{n-1} \to e^0$ . This is equivalent to regarding $S^n$ as the quotient space $\dfrac{D^n}{\partial D^n}$ .

How to Draw a Cell Complex (or CW Complex)

Let we try to construct a space by following procedure:

Start with a discrete set ${X^0}$ , whose points are regarded as $\mathbf{0}$ -cells.
Inductively, form the $\textbf{n}$ -skeleton $X^n$ from $X^{n-1}$ by attaching $n$ -cells $e^n_{\alpha}$ via maps $\Phi_{\alpha} : S^{n-1} \rightarrow X^{n-1}$ . This means that $X^n$ is the quotient space of the disjoint union $X^{n-1} \, \mathbf{\sqcup_{\alpha}} D^n_{\alpha}$ of $X^{n-1}$ with a collection of n-disks $D^n_\alpha$ under the identifications $x \sim \Phi_{\alpha} (x)$ for $x \in \partial D^n_\alpha$ . Thus as a set

$X^n=X^{n-1} \mathbf{\sqcup_{\alpha}} e^n_\alpha$

where each $e^n_\alpha$ is an open $n$ disk.

One can either stop this inductive process at a finite stage, setting $X=X^n$ for some $n < \infty$ , one can continue indefinitely, setting $X=\bigcup_n X^n$ .

A space $X$ constructed in this way is called a CELL COMPLEX or CW COMPLEX.

Reference:

ALGEBRAIC TOPOLOGY

Allen Hatcher

http://www.math.cornell.edu/~hatcher/#ATI

D’ ALEMBERT’s Test of Convergence of Series

Saturday, April 9th, 2011 11:36 / 1 Comment

Statement

A series $\sum {u_n}$ of positive terms is convergent if from and after some fixed term $\dfrac {u_{n+1}} {u_n} < r < {1}$ , where r is a fixed number. The series is divergent if $\dfrac{u_{n+1}} {u_n} > 1$ from and after some fixed term.

D’ Alembert’s Test is also known as ratio test of convergency of a series.

Definitions for Generally Interested Readers

(Definition 1) An infinite series $\sum {u_n}$ i.e. $\mathbf {u_1+u_2+u_3+....+u_n}$ is said to be convergent if $S_n$ , the sum of its first $n$ terms, tends to a finite limit $S$ as n tends to infinity.
We call $S$ the sum of the series, and write $S=\displaystyle {\lim_{n \to \infty} } S_n$ .
Thus an infinite series $\sum {u_n}$ converges to a sum S, if for any given positive number $\epsilon$ , however small, there exists a positive integer $n_0$ such that
$|S_n-S| < \epsilon$ for all $n \ge n_0$ .
(Definition 2)
If $S_n \to \pm \infty$ as $n \to \infty$ , the series is said to be divergent.
Thus, $\sum {u_n}$ is said to be divergent if for every given positive number $\lambda$ , however large, there exists a positive integer $n_0$ such that $|S_n|>\lambda$ for all $n \ge n_0$ .
(Definition 3)
If $S_n$ does not tends to a finite limit, or to plus or minus infinity, the series is called Oscillatory

Discussions

Let a series be $\mathbf {u_1+u_2+u_3+.......}$ . We assume that the above inequalities are true.

From the first part of the statement:
$\dfrac {u_2}{u_1} < r$ , $\dfrac {u_3}{u_2} < r$ ……… where r <1.
Therefore $\mathbf {{u_1+u_2+u_3+....}= u_1 {(1+\frac{u_2}{u_1}+\frac{u_3}{u_1}+....)}}$
$=\mathbf {u_1{(1+\frac{u_2}{u_1}+\frac{u_3}{u_2} \times \frac{u_2}{u_1}+....)}}$
$< \mathbf {u_1(1+r+r^2+.....)}$
Therefore, $\sum{u_n} < u_1 (1+r+r^2+.....)$
or, $\sum{u_n} < \displaystyle{\lim_{n \to \infty}} \dfrac {u_1 (1-r^n)} {1-r}$
Since r<1, therefore as $n \to \infty , \ r^n \to 0$
therefore $\sum{u_n} < \dfrac{u_1} {1-r}$ =k say, where k is a fixed number.
Therefore $\sum{u_n}$ is convergent.
Since, $\dfrac{u_{n+1}}{u_n} > 1$ then, $\dfrac{u_2}{u_1} > 1$ , $\dfrac{u_3}{u_2} > 1$ …….
Therefore $u_2 > u_1, \ u_3 >u_2>u_1, \ u_4 >u_3 > u_2 >u_1$ and so on.
Therefore $\sum {u_n}=u_1+u_2+u_3+....+u_n$ > $nu_1$ . By taking n sufficiently large, we see that $nu_1$ can be made greater than any fixed quantity.
Hence the series is divergent.

Comments

When $\dfrac {u_{n+1}} {u_n}=1$ , the test fails.

Another form of the test–

The series $\sum {u_n}$ of positive terms is convergent if $\displaystyle {\lim_{n \to \infty}} \dfrac {u_n}{u_{n+1}}$ >1 and divergent if $\displaystyle{\lim_{n \to \infty}} \dfrac {u_n}{u_{n+1}}$ <1.
One should use this form of the test in the practical applications.

A Problem:
Verify whether the infinite series $\dfrac{x}{1.2} + \dfrac {x^2} {2.3} + \dfrac {x^3} {3.4} +....$ is convergent or divergent.

Solution

We have $u_{n+1}= \dfrac {x^{n+1}}{(n+1)(n+2)}$ and $u_n= \dfrac {x^n} {n(n+1)}$
Therefore $\displaystyle {\lim_{n \to \infty}} \dfrac{u_n} {u_{n+1}} = \displaystyle{\lim_{n \to \infty}} (1+\frac{2}{n}) \frac{1}{x} = \frac{1}{x}$
Hence, when 1/x >1 , i.e., x <1, the series is convergent and when x >1 the series is divergent.
When x=1, $u_n=\dfrac{1} {n(n+1)}=\dfrac {1}{n^2} {(1+1/n)}^{-1}$
or, $u_n=\dfrac{1}{n^2}(1-\frac{1}{n}+ \frac {1}{n^2}-.....)$
Take $\dfrac{1}{n^2}=v_n$ Now $\displaystyle {\lim_{n \to \infty}} \dfrac {u_n}{v_n}=1$ , a non-zero finite quantity.
But $\sum {v_n}=\sum {\frac{1}{n^2}}$ is convergent.
Hence, $\sum {u_n}$ is also Convergent.

A General Problem on functions

Saturday, April 2nd, 2011 09:18 / Leave a Comment

Problem

Let $\mathbb{Z}$ denote the set of all integers (as usually it do ).

Consider a function $f : \mathbb{Z} \rightarrow \mathbb{Z}$ with the following properties:
$f (92+x) = f (92-x)$
$f (19 \times 92+x) = f (19 \times 92 -x)$
$f (1992+x)=f (1992-x)$ for all $x \in \mathbb{Z}$ . Is it possible that all positive divisors of $92$ occur as values of $f$ ?
A happy note: $19 \times 92$ is actually 1748 and it is written to retain symmetry in problem.