TITLE: Joint Probability of Random Variables
QUESTION [0 upvotes]: Suppose I took measurements $\{X_i\}$, which are all independent and they follow a normal distribution $X_i\sim N(\mu,\sigma)$.  I am asked for the joint probability of all of the measurements.  Based on joint probability, I expect:
$$Prob(x_i=X_i)=\prod Prob(x_i)$$
$$=\prod\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$
Where I use $\prod$ as a product over all $i$.  Am I correct in saying this, or am I missing something important?

REPLY [0 votes]: Yes you are correct, but small correction.
$$ P(X_i= x_i) = \prod P(x_i)= \prod \frac{1}{\sigma \sqrt{2\pi}} e^{-(1/2\sigma^2)(x_i-\mu)^2}
 = \left(\frac{1}{\sigma \sqrt{2\pi}}\right)^n e^{-(1/2\sigma^2)\sum (x_i-\mu)^2} $$


TITLE: Switching a parametric form of a plane to a non-parametric form
QUESTION [0 upvotes]: So I have a parameterized representation of a plane:
$$ (1,0,2) + t(-1,0,1) + u(0,1,3) : t ϵ R, u ϵ R $$
I want to swicth it to the equation $ax + by + cz = d$ in order to find the intersection line with a plane given by the equation $x + 2y + 3z = 6$
I know how to do it for a line in $R^2$ but I guess it is not the same for plane.
can anybody give a hint?

REPLY [0 votes]: You have $(x,y,z)=(1,0,2)+t(-1,0,1)+u(0,1,3)$
which yelds a system of equations:
\begin{equation}
x=1-t\\
y=u\\
z=2+t+3u
\end{equation}
You have to solve the system for $u$ and $t$ using one line at a time. In the end you will have your cartesian equation.


text:
\begin{document}
\maketitle

\begin{abstract} We construct small covers and quasitoric
manifolds over $n$-dimensional simple polytopes which allow proper
colorings of facets with $n$ colors. We calculate Stiefel-Whitney
classes of these manifolds as obstructions to immersions and
embeddings into Euclidean spaces. The largest dimension required
for embedding is achieved in the case $n$ is a power of two.

\hspace{0 mm} \textbf{2010 Mathematics Subject Classification}:
    57N35,     52B12.

\hspace{0 mm} \textbf{Keywords}: immersions, quasitoric manifolds,
simple polytopes, colorings, Stiefel-Whitney classes.

\end{abstract}

\renewcommand{\thefootnote}{}
\footnotetext{This work was supported by Ministry of Education,
Science and Technological developments of Republic of Serbia
(Grants 174020 and 174034) and by Ministry of Science and
Technology of Republic of Serbska, Bosnia and Herzegovina (Grant
19/6-020/961-120/14).}

\section{Introduction}

\subsection{Colorings of simple polytopes}

An $n$-dimensional convex polytope is \textit{simple} if the
number of facets which are meeting at each vertex is equal to $n$.
The \textit{proper coloring} into $k$ colors of a simple polytope
$P^n$ is a map $$h: \{F_1, \dots, F_m\}\rightarrow
\{1,\ldots,k\}$$ of its set of facets such that $h (F_i)\neq h
(F_j)$ for each two intersecting facets. The \textit{chromatic
number} $\chi (P^n)$ of a simple polytope $P^{n}$ is the least $k$
for which there exists a proper coloring of $P^n$ into $k$ colors.
The chromatic numbers of the simplex $\Delta^n$, the cube $I^n$
and the permutahedron $\Pi^n$ (Figure \ref{bojenja}) are
$$\chi (\Delta^n)=n+1, \ \ \ \chi (I^n)=n, \ \ \ \chi
(\Pi^n)=n.$$

\begin{figure}[h!h!]
\centerline{\includegraphics[width=\textwidth]{bojenja.pdf}}
\caption{The colorings of the cube and the permutahedron}
\label{bojenja}
\end{figure}

Obviously, $\chi (P^n)\geq n$ for any simple polytope $P^n$. The
chromatic number of a polygon is clearly equal to $2$ or $3$,
depending on the parity of the number of its faces. By famous Four
Color Theorem chromatic numbers of $3$-dimensional simple
polytopes are equal to $3$ or $4$. In general case, for $n\geq 4$
it does not hold $\chi (P^n)\leq n+1$. Moreover, there are simple
polytopes such that their chromatic numbers are equal exactly to
numbers of their facets. The polars of the cyclic polytopes with
$m$ vertices $C^n (m)$ are examples of such type (see
\cite[Example~0.6]{5}).

We consider the class of $n$-dimensional simple polytopes with the
chromatic number equal to $n$. Denote this class by $\mathcal{C}$.
The class $\mathcal{C}$ is closed with respect to products (see
\cite[Construction~1.12]{2}) and connected sums (see
\cite[Construction~1.13]{2}). Also from any given simple polytope
$P^n$ by truncation over all its faces we obtain a simple polytope
$Q^n$ which belongs to the class $\mathcal{C}$. The complete
description of the class $\mathcal{C}$ is obtained in \cite{9}. A
simple $n$-polytope $P^n$ admits a proper coloring in $n$ colors
if and only if every its $2$-face has an even number of edges (see
\cite[Theorem 16]{9}).

\subsection{Small covers and quasitoric manifolds}

Quasitoric manifolds and they real analogues called small covers
are introduced by Davis and Januszkiewicz in \cite{3}. Their
geometric and algebraic topological properties are closely related
to combinatorics of simple polytopes. The following definitions
and notions are extensively elaborated in \cite{2}. Denote by
$$G_d=\left\{\begin{array}{rl}
S^{0}, & d=1,\\
S^{1},  & d=2,\end{array} \right. \ R_d=\left\{\begin{array}{rl}
\mathbb{Z}_2, & d=1,\\
\mathbb{Z},  & d=2,\end{array}\right. \
\mathbb{K}_d=\left\{\begin{array}{rl}
\mathbb{R}, & d=1,\\
\mathbb{C},  & d=2,\end{array} \right.$$ where $S^{0}=\{-1,+1\}$
and $S^{1}=\{z\mid |z|=1\}$ are multiplicative subgroups of real
and complex numbers, $\mathbb{Z}$ is the ring of integers and
$\mathbb{Z}_2=\{0,1\}$ is the ring of integers modulo $2$.

The group $G_d^{n}$ acts standardly on $\mathbb{K}_d^{n}$ by
$(t_1,\ldots,t_n)\cdot(x_1,\ldots,x_n)=(t_1x_1,\ldots,t_nx_n)$.
The action of $G_d^{n}$ on a smooth $dn$-dimensional manifold
$M^{dn}$ is locally standard if for any point of $M^{dn}$ there is
a $G_d^{n}$-invariant neighborhood which is weakly equivariantly
diffeomorphic to some open $G_d^{n}$-invariant subset of
$\mathbb{K}_d^{n}$ with the standard action of $G_d^{n}$. Recall
that two $G_d^{n}$-manifolds are weakly equivariantly
diffeomorphic if there is an automorphism $\omega:G_d^n\rightarrow
G_d^n$ and a diffeomorphism $f:M_1^{dn}\rightarrow M_2^{dn}$ such
that $f(g\cdot x)=\omega(g)\cdot f(x)$ for any $g\in G_d^n$ and
$x\in M_1^{dn}$. A smooth $dn$-dimensional $G_d^{n}$-manifold
$M^{dn}$ with a locally standard action of the group $G_d^{n}$
such that the orbit space $M^{dn}/G_d^{n}$ is diffeomorphic, as a
manifold with corners, to a simple $n$-dimensional polytope
$P^{n}$ is called {\it small cover} over $P^{n}$ if $d=1$,
correspondingly {\it quasitoric manifolds} over $P^{n}$ if $d=2$.
Let $\pi:M^{dn}\rightarrow P^{n}$ be the projection map and
$\{F_1,\ldots,F_m\}$ be the set of facets of the polytope $P^{n}$.
The inverse images $M^{d(n-1)}_j=\pi^{-1}(F_j)$ are
$G_d^{n}$-invariant submanifolds of codimension $d$ called {\it
characteristic submanifolds}. To each characteristic submanifold
$M_j^{d(n-1)}$ corresponds the isotropy subgroup
$G(F_j)=G_{\lambda_j}$ of the rank one, where
$G_{\lambda_j}=\{(1,\ldots,1),((-1)^{\lambda_{1j}},\ldots,(-1)^{\lambda_{nj}})\}$
for some
$\lambda_j=(\lambda_{1j},\ldots,\lambda_{nj})\in\mathbb{Z}_2^{n}\setminus\{0\}$
if $d=1$ and $G_{\lambda_j}=\{(e^{2\pi
i\lambda_{1j}t},\ldots,e^{2\pi i\lambda_{nj}t})\mid
t\in\mathbb{R}\}$ for some primitive vector
$\lambda_j=(\lambda_{1j},\ldots,\lambda_{nj})\in\mathbb{Z}^{n}$
defined up to the sign, if $d=2$. In this way, the action of the
group $G_d^{n}$ on $M^{dn}$ defines the {\it characteristic map}
$$l_d:\{F_1, \dots, F_m\}\rightarrow R^n_d,$$ which to a facet
$F_j$ of the polytope $P^{n}$ assigns a primitive vector
$\lambda_j=(\lambda_{1j},\ldots,\lambda_{nj})\in R^n_d$. Denote by
$\Lambda= (\lambda_1, \dots, \lambda_m)$ the matrix formed by
these vectors. Then $\det \Lambda_{{(V)}}=\pm 1$ for any vertex
$V\in P^{n}$, where $\Lambda_{(V)}$ is the submatrix of the matrix
$\Lambda$ formed by vectors corresponding to facets which are
intersecting in that vertex. A pair $(P^{n}, \Lambda)$ satisfying
this condition on submatrices is called a {\it characteristic
pair}. The manifold $M^{dn}$ is reconstructible from the
characteristic pair $(P^{n}, \Lambda)$ up to weak equivariant
diffeomorphism in the following way (see \cite{3} and
\cite[Construction~6.18]{2}). Let $F_{i_1}\cap\ldots\cap F_{i_k}$
be the minimal face containing a point $q\in P^{n}$. To the point
$q$ we associate the subgroup
$G(q)=G_{\lambda_{i_1}}\times\cdots\times G_{\lambda_{i_k}}$.
Define $M(P^{n}, \Lambda)=G_d^{n}\times P^{n}/\sim$, where
$(t_1,p)\sim(t_2,q)$ if and only if $p=q$ and $t_1t_2^{-1}\in
G(q)$.

Though not all simple polytopes allow characteristic maps (see
\cite[Example~1.15]{3}), to considered polytopes
$P^{n}\in\mathcal{C}$ in a simple way can be assigned a
characteristic matrix $\Lambda$ if each color $i\in\{1,\ldots,n\}$
is identified with the vector $e_i$ of the standard basis in
$R_d^{n}$.

The cohomology ring of a small cover and a quasitoric manifold
$M^{dn}, d=1,2$ is described in the following way. Let
$v_j=D[M_j^{d(n-1)}], j=1,\ldots,m$ be cohomology classes which
are Poincare duals to fundamental classes of characteristic
manifolds. The characteristic matrix $\Lambda=(\lambda_1, \dots,
\lambda_m)$ defines following linear forms
$$\theta_i:=\sum_{j=1}^m \lambda_{i j} v_j, i=1,\ldots,n,$$ where
$\lambda_j=(\lambda_{1 j}, \dots, \lambda_{n j})^t \in R_d^n,
j=1,\ldots,m$. Let $\mathcal{J}$ be the ideal in $R_d[v_1, \dots,
v_m]$ generated by elements $\theta_1,\ldots,\theta_n$ and
$\mathcal{I}$ be the Stanley-Reisner ideal of the polytope
$P^{n}$, which is generated by monomials $v_{i_1}\cdots v_{i_k}$
whenever $F_{i_1}\cap\ldots F_{i_k}=\emptyset$ in $P^{n},
i_1<\ldots<i_k$. Then for the cohomology ring the following
isomorphism holds (see \cite{3})

\begin{equation} \label{djf} H^\ast (M^{dn},R_d)\simeq R_d[v_1, \dots,
v_m]/(\mathcal{I}+\mathcal{J}).\end{equation} The same formula
$(\ref{djf})$ holds with $\mathbb{Z}_2$ coefficients if $d=2$. The
total Stiefel-Whitney class is determined by the following
Davis-Januszkiewicz's formula

\begin{equation}
\label{djfswc} w (M^{dn})= \prod_{i=1}^m (1+v_i)\in H^\ast
(M^{dn}; \mathbb{Z}_2),\end{equation} where in the case $d=2$ all
classes $v_i$ are regarded as $\mathbb{Z}_2$-restrictions of
corresponding integral classes.

\subsection{Immersions and Embeddings}

Immersions and embeddings of manifolds are a classical topic in
algebraic topology and differential topology. We consider
immersions and embeddings in the smooth category. For a manifold
$M^{n}$ define the numbers $imm (M^n)$ and $em(M^n)$ as the
smallest dimensions of Euclidean spaces in which this manifold is
immersed and embedded, correspondingly. In accordance with
Whitney's theorem for each smooth manifold $M^{n}$ is satisfied
$imm(M^n)\leq 2n-1$ and $em(M^n)\leq 2n$. On the other hand, the
dual Stiefel-Whitney classes $\overline{w}_i$ serve as
obstructions to immersions and embeddings of manifolds in
Euclidean spaces. Recall that dual Stiefel-Whitney classes
$\overline{w}_i(M^{n})$ are characteristic classes of the stable
normal bundle of a manifold $M^{n}$.

\begin{theorem}[see \cite{4}]\label{imeem} For a smooth manifold $M^{n}$ let $k:=\max \{i\mid
\w_i(M^n)\neq 0\}$. Then $\imm (M^n)\geq n+k \, \mbox{\,and\,}\,
\Em (M^n)\geq n+k+1.$
\end{theorem}

The Stiefel-Whitney classes are also obstructions to so called
totally skew embeddings introduced by Ghomi and Tabachnikov in
\cite{6}. A manifold $M^{n}$ is totally skew embedded in the
Euclidean space $\mathbb{R}^{N}$ if any two tangent lines in
different points of $M^{n}$ are skew lines in $\mathbb{R}^{N}$. It
is proved in the paper \cite{6} that the number $N(M^{n})$ defined
as the smallest dimension of Euclidean spaces for which there is a
totally skew embedding of $M^{n}$, satisfies
\[2n+2\leq N(M^n)\leq 4n+1.\] The better lower bound is obtained
in \cite{7}.
\begin{theorem}[see \cite{7}, Proposition~1, Corollary~4]\label{skewteo}
If $$k=\max \{i\mid \overline{w}_i(M^n)\neq 0\}$$ then $N(M^n)\geq
2n+2k+1.$
\end{theorem}

The immersions and embeddings of quasitoric manifolds over cubes
are studied in \cite{1}.

In this paper we prove the following results.

\begin{theorem}\label{main} Let $n$ be a power of two and $P^n$ be a simple convex
$n$-dimensional polytope which allows a proper coloring in $n$
colors.
\begin{itemize}
\item[(1)] There is a small cover $M^{n}$ over the polytope
$P^{n}$ which satisfies following identities $imm(M^{n})=2n-1$ and
$em(M^{n})=2n$.

\item[(2)] There is a quasitoric manifold $M^{2n}$ over the
polytope $P^{n}$ which satisfies following inequalities
$imm(M^{2n})\geq 4n-2$ and $em(M^{2n})\geq 4n-1$. Moreover, for
$n\geq 3$ in both relation the equality holds.
\end{itemize}
\end{theorem}

We give an explicit construction of manifolds $M^{n}$ and $M^{2n}$
from Theorem \ref{main}. By using combinatorial properties of the
simple polytope $P^n$ and Davis-Januszkiewicz formula (\ref{djf})
and (\ref{djfswc}), we describe the cohomology rings $H^*(M^{dn},
\mathbb{Z}_2), d=1,2$ and calculate Stiefel-Whitney classes
$w_k(M^{dn}), d=1,2$. When $n$ is a power of two, we prove that
classes $\w_{d(n-1)}( M^{dn}), d=1,2$ are nontrivial, the claim
that implies Theorem \ref{main}. Also from this claim and Theorem
\ref{skewteo} for totally skew embeddings of constructed manifolds
immediately follows

\begin{corollary} If $n$ is a power of two then
$$ N(M^{n})\geq 4n-1, \ \ \ N(M^{2n})\geq 8n-3.$$
\end{corollary}

The obtained small covers $M^{n}$ are a new class of manifolds for
which the numbers $N(M^{n})$ are equal to $4n-1, 4n,$ or $4n+1$.
So far, only known examples with this property were real
projective spaces (see \cite{7}).

Cohen \cite{8} in 1985 resolved positively the famous {\em
Immersion Conjecture}, by showing that each compact smooth
$n$-manifold for $n>1$ can be immersed in
$\mathbb{R}^{2n-\alpha(n)}$, where $\alpha(n)$ is the number of
1's in the binary expansion of $n$. The products of real
projective spaces are examples of manifolds that achieve the upper
bounds. We construct new examples of this type in the class of
small covers.

\begin{theorem}\label{exist}
For every positive integer $n$ there is a small cover $M^{n}$ and
a quasitoric manifold $M^{2n}$ over some simple $n$-dimensional
polytope which allows a proper coloring in $n$ colors such that
\begin{eqnarray*}
imm(M^{n})=2n-\alpha(n), \ \ \ imm(M^{2n})\geq 4n-2\alpha(n) \\
em(M^{n})=2n-\alpha(n)+1, \ \ \ em(M^{2n})\geq 4n-2\alpha(n)+1, \\
N(M^{n})\geq 4n-2\alpha(n)+1, \ \ \ N(M^{n})\geq 8n-4\alpha(n)+1.
\end{eqnarray*}
\end{theorem}

We conjecture that the statement of Theorem \ref{exist} is
satisfied for any simple $n$-dimensional polytope which is
properly colored by $n$ colors.

\begin{conjecture}\label{conj1}
Let $P^n$ be a simple convex $n$-dimensional polytope which allows
a proper coloring in $n$ colors. Then there exist a small cover
$M^{n}$ and a quasitoric manifold $M^{2n}$ over the polytope $P^n$
which satisfy

\begin{eqnarray*}
imm(M^{n})= 2n-\alpha(n),\;\;\;\;em(M^{n})= 2n-\alpha(n)+1,\\
imm(M^{2n})\geq  4n-2\alpha(n),\;\;\;\;em(M^{2n})\geq 4n-2
\alpha(n)+1.
\end{eqnarray*}
\end{conjecture}

\section{Manifolds $M^{dn}$}

\subsection{Construction}\label{cons}

Let $P^n$ be a simple polytope such that $\chi (P^n)=n$ and
$h:\{F_1, \dots,F_m\}\rightarrow\{1,\ldots,n\}$ be its proper
coloring. Denote by $\mathcal{F}_j$ the set $h^{-1} (j)$. Every
vertex $V\in P^n$ is the intersection of $n$ differently colored
facets. Take an arbitrary vertex $V=H_1\cap\cdots\cap H_n$, where
the facet $H_i$ is colored by color $i$. Assign to each facet
$H_i$ the vector
$\lambda_i=(\underbrace{0,\dots,0}_{i-1},\underbrace{1,\dots,1}_{n-i+1})^t$
and vectors $\lambda_F=(\underbrace{0,\dots,0}_{i-1}, 1,
\underbrace{0,\dots,0}_{n-i})^t$ to remaining facets $F\in
\mathcal{F}_i\backslash \{H_i\}$. The corresponding matrix
$\Lambda$ clearly induces the characteristic map, since $\det
{\Lambda}_{(v)}=1$ for every vertex $V\in P^n$. Define
$M^{dn}=M(P^{n}, \Lambda)$ as the manifold constructed from the
characteristic pair $(P^{n}, \Lambda)$.

\subsection{Cohomology ring}

Let $u_1,\dots,u_n$ be Poincar\'e duals to characteristic
submanifolds over the facets $H_1,\dots,H_n$ respectively. For
every facet $F$ of $P^n$ distinct from $H_1,\dots,H_n$, let $v_F$
denotes the Poincar\'e dual to the characteristic submanifold over
$F$.

The cohomology ring of the manifold $M^{dn}$ is determined by the
Stanley-Reisner ideal $\mathcal{I}$ of $P^n$ and the ideal
$\mathcal{J}$ which is generated by linear forms
\begin{equation} \label{j1}
\theta_i:=\sum_{j=1}^i u_j+\sum_{F\in
{\mathcal{F}_i\setminus\{H_i\}}} v_F, i=1, \dots, n.
\end{equation}
Recall that $$H^*(M^{dn}; \mathbb{Z}_2)=\mathbb{Z}_2[u_1, \dots,
u_n, v_F |F\in {\mathcal{F}\setminus\{H_1,\dots,
H_n\}}]/(\mathcal{I}+\mathcal{J}).$$

From the coloring of $P^n$ we easily deduce

\begin{proposition}\label{p1} For facets $F$, $G\in {\mathcal{F}_i\setminus\{H_i\}}$, $F\neq G$
corresponding classes satisfy the following relation in the
cohomology ring $H^*(M^{dn}; \mathbb{Z}_2)$
\[v_F v_G=u_i v_F=u_i v_G=0.\]
\end{proposition}

Proposition \ref{p1} and (\ref{j1}) together imply

\begin{proposition}\label{p2} The following equalities hold in $H^*(M^{dn};
\mathbb{Z}_2)$
\begin{equation}\label{r1}
u_1^2=0, \, u_2^2=u_1 u_2, \dots, u_n^2=u_1 u_n+\cdots+u_{n-1}
u_n.
\end{equation}
\end{proposition}

Let $F$ be an arbitrary facet of $P^n$ and $v_F$ the corresponding
class over $F$, even if $F$ is one of $H_1$, $\dots$, $H_n$.

\begin{lemma}\label{lema1}
Let $k$ be a positive integer and $h$ a proper coloring map of the
polytope $P^n$. Then the class $v_F^k$ is either trivial or equal
to the degree $k$ homogenous polynomial
$Q_k^F(u_1,\ldots,u_{h(F)-1},v_F)$ whose monomials are square
free.
\end{lemma}

\begin{proof} Let $h(F)=i$. We prove the claim by induction on $k$. For
$k=1$ it is trivial. Assume that $v_F^k=Q_k^F (u_1,\ldots,u_{i-1},
v_F)$. By multiplying with $v_F^k$ the relation
$$\sum_{j=1}^{i-1} u_j + \sum_{F\in
{\mathcal{F}_i\setminus\{H_i\}}} v_{F}=0$$ and by using
Proposition \ref{p1} we get
\[v_F^{k+1}=(u_1+\cdots+u_{i-1}) Q_k^F (u_1, \dots, u_{i-1},
v_F).\] If $v_F^{k+1}=0,$ the claim follows directly. In the
opposite case it is obvious that $(u_1+\cdots+u_{i-1}) Q_k^F (u_1,
\dots, u_{i-1}, v_F)$ is the degree $k+1$ homogenous polynomial in
variables $u_1$, $\dots$, $u_{i-1}$, $v_F$. However, from
relations (\ref{r1}) follows $(u_1+\cdots+u_{i-1}) Q_k^F (u_1,
\dots, u_{i-1}, v_F)=Q_{k+1}^F (u_1, \dots, u_{i-1}, v_F)$, where
all monomials of the polynomial $Q_{k+1}^{F}$ are square free.
\end{proof}

From lemma \ref{lema1} immediately follows

\begin{corollary}\label{posl1}
Each class  $v_{F_{i_1}}^{r_1}\cdots v_{F_{i_k}}^{r_k}$ is either
trivial or equal to the degree $r_1+\dots+r_k$ homogenous
polynomial in variables $u_1$, $\dots$, $u_{n}$, $v_{F_{i_1}}$,
$\dots$,  $v_{F_{i_k}}$ whose monomials are square free.
\end{corollary}

\begin{proposition}\label{p3} The class $u_1 \dots u_n$ is the fundamental
cohomology class in the ring $H^{dn} (M^{dn}; \mathbb{Z}_2)$.
\end{proposition}

\begin{proof} Denote by $\mathcal{F}_V$ the set of facets containing a vertex $V\in
P^{n}$. Let $V^{\ast}=\prod_{F\in\mathcal{F}_V}v_F$. We show that
$V^{\ast}=V'^{\ast}$ for each two vertices $V,V'\in P^{n}$. For
this it is sufficient to suppose that vertices are connected by
the edge $VV'$. In this case we have
$\mathcal{F}_{V'}=\mathcal{F}_V\setminus\{G\}\cup\{G'\}$ for
unique facets $G$ and $G'$ which are colored by the same color. If
$h(G)=h(G')=i$, we multiply by
$\prod_{F\in\mathcal{F}_V\cap\mathcal{F}_{V'}}v_F$ the relation
$$\sum_{j=1}^{i-1} u_j + \sum_{F\in \mathcal{F}_i} v_{F}=0.$$
From Proposition \ref{p1} we obtain
$$(u_1+\ldots+u_{i-1})\prod_{F\in\mathcal{F}_V\cap\mathcal{F}_{V'}}v_F+V^{\ast}+V'^{\ast}=0.$$
In the case $i=1$ we get the required equality
$V^{\ast}=V'^{\ast}$. If $i>1$, for each $1\leq j<i$, we have by
Propositions \ref{p1} and \ref{p2}
$$u_j\prod_{F\in\mathcal{F}_V\cap\mathcal{F}_{V'}}v_F=\left\{\begin{array}{cc} 0,
&\mbox{if} \ H_j\notin\mathcal{F}_V\cap\mathcal{F}_{V'} \\
(u_1+\cdots
u_{j-1})\prod_{F\in\mathcal{F}_V\cap\mathcal{F}_{V'}}v_F, &
\mbox{if} \
H_j\in\mathcal{F}_V\cap\mathcal{F}_{V'}\end{array}\right..$$ Then
by induction follows
$u_j\prod_{F\in\mathcal{F}_V\cap\mathcal{F}_{V'}}v_F=0,
j=1,\ldots,i-1$ which again gives $V^{\ast}=V'^{\ast}$.

Suppose now that $u_1\cdots u_n=0$. From the so far proven part
follows that $V^{\ast}=0$ for each vertex $V\in P^{n}$. Moreover
in this case all $dn$-dimensional classes vanish since by
Corollary \ref{posl1} square free monomials of degree $n$ linearly
generate $H^{dn} (M^{dn}; \mathbb{Z}_2)$. But it contradicts the
known fact that $H^{dn} (M^{dn}; \mathbb{Z}_2)\simeq\mathbb{Z}_2$.
\end{proof}

\subsection{Stiefel-Whitney classes}

Proposition \ref{p1} implies that the total Stiefel-Whitney class
could be expressed as
\[w(M^{dn})=\prod_{i=1}^{n} \prod_{F\in \mathcal{F}_i} (1+v_F)=\prod_{i=1}^{n}(1+\sum_{F\in \mathcal{F}_i}v_F).\]
By applying (\ref{j1}) we obtain

\[w (M^{dn})=(1+u_1)(1+u_1+u_2)\cdots(1+u_1+u_2+\cdots+u_{n-1}).\]

In order to prove the main theorem \ref{main}, we are going to use
another set of generators $t_1,\dots,t_n$ which are defined by
\begin{eqnarray*}\label{tgen}
t_i=\sum_{j=1}^{i}u_j, i=1,\ldots,n.
\end{eqnarray*}
Consequently we have
\begin{equation}\label{sw3} w(M^{dn})=(1+t_1)\cdots(1+t_{n-1}).
\end{equation}
By Proposition \ref{p2} the classes $t_1^2$, $t_2^2+t_1
t_2,\ldots,t_n^2+t_{n-1}t_n$ vanish. Let $\mathcal{T}_n$ be the
ideal generated by these classes. By Proposition \ref{p3} it is
easily seen that the class $t_1 t_2 \cdots t_n$ also represents
the fundamental class.

\section{Proof of Theorem \ref{main}}

In order to prove Theorem \ref{main} it is sufficient to show that
the top dual Stiefel-Whitney class $\overline{w}_{d(n-1)}(M^{dn})$
is nontrivial.

\subsection{Dual Stiefel-Whitney classes}

Stiefel-Whitney classes and dual Stiefel-Whitney classes are
related by
$$w(M^{dn})\cdot \overline{w}(M^{dn})=1.$$
From the relation (\ref{sw3}) we obtain

\begin{lemma}\label{swd}
The total Stiefel--Whitney class $\overline{w}(M^{dn})$ is
expressed by
$$\overline{w}
(M^{dn})=(1+t_1)(1+t_2+t_2^2)\cdots(1+t_{n-1}+\cdots+t_{n-1}^{n-1}).$$
\end{lemma}

\begin{proof}
The statement follows from the fact that $t_k^{k+1}=0$ for all
$k=1,\ldots,n-1$. In fact $t_k^{k+1}=(u_1+\cdots+u_k)^{k+1}$ is
the sum of monomials of the form $\sum u_{i_1}^{r_1}\cdots
u_{i_j}^{r_j},$ where $1\leq i_1<\cdots<i_j\leq k$ and $r_1+\cdots
r_k=k+1$. In its turn, by Corollary \ref{posl1} and Proposition
\ref{p2} this is a sum of homogeneous polnomials of the degree
$k+1$ of the form $\sum Q_{k+1}(u_1,\ldots,u_k)$. Since monomials
of each polynomial in the sum is square free, we have
$Q_{k+1}(u_1,\ldots,u_k)=0$.
\end{proof}

We want to determine the highest nontrivial dual $\overline{w}_k
(M^{dn})$. For small $n$, we could calculate $\overline{w}
(M^{dn})$ directly.

\begin{example}\begin{itemize}
\item[(1)] $\overline{w} (M^{d \cdot 2})=1+t_1$, \item[(2)]
$\overline{w} (M^{d\cdot 3})=1+(t_1+t_2)$, \item[(3)]
$\overline{w} (M^{d\cdot 4})=1+(t_1+t_2+t_3)+t_1 t_3+t_1 t_2 t_3$,
\item[(4)] $\overline{w} (M^{d\cdot 5})=1+(t_1+t_2+t_3+t_4)+(t_1
t_3+t_1 t_4+t_2 t_4)+(t_1 t_2 t_3+t_2 t_3 t_4)$.
\end{itemize}
\end{example}

\subsection{The class $\overline{w}_{d (n-1)}(M^{dn})$}

Consider arbitrary two manifolds $M^{dn}$ and $M^{d(n+1)}$
constructed as in subsection \ref{cons} over simple polytopes
$P_1^{n}$ and $P_2^{n+1}$, properly colored in $n$ and $n+1$
colors, respectively.

\begin{lemma}\label{sub} The ring $\mathbb{Z}_2[t_1,\ldots,t_n]/\mathcal{T}_n$
is a subring of the cohomology ring $H^{\ast}(M^{dn};
\mathbb{Z}_2)$ which is generated by elements $t_1,\ldots,t_n$.
\end{lemma}
\begin{proof} It is necessary to prove that $\mathcal{T}_n$
is the ideal of all relations among elements $t_1,\ldots,t_n$ in
$H^{\ast}(M^{dn}; \mathbb{Z}_2)$. The rings $\mathbb{Z}_2 [t_1,
\dots, t_n]/\mathcal{T}_n$ and $\mathbb{Z}_2 [u_1, \dots,
u_n]/\mathcal{U}_n$ are isomorphic, where $\mathcal{U}_n$ is the
ideal generated by elements $u_1^{2}$ and
$u_i^{2}+(u_1+\ldots+u_{i-1})u_i, i=2,\ldots,n$ from Proposition
\ref{p2}. It is sufficient to show that $\mathcal{U}_n$ is the
ideal of all relations among $u_1,\ldots,u_n$ in $H^{\ast}(M^{dn};
\mathbb{Z}_2)$.

By Corollary \ref{posl1} any homogeneous polynomial in
$H^{\ast}(M^{dn}; \mathbb{Z}_2)$ is expressed as a sum of square
free monomials. So let we have a relation
$$ \sum_{j=1}^{m}u_{I_j}=0,$$ where
$I_j=\{a_{j1}<\ldots<a_{jk}\}\subset\{1,\ldots,n\}, j=1,\ldots,m$
and $u_{I}=\prod_{i\in I}u_i$. It is easy to convince that
$u_1\cdots u_{i-1}u_i^{2}=0, i=1,\ldots,n$. Let us order sets
lexicographically $I_1<\ldots<I_m$. Let $I_m=\{a_1<\ldots<a_k\}$
and $d_i=a_i-a_{i-1}, i=1,\ldots,k$ where $a_0=0$. Define
following elements
$$U_i=\left\{\begin{array}{cc} u_{a_{i-1}+1}\cdots u_{a_i-1}, & d_i>1
\\ 1, & d_i=1\end{array}\right.,$$ where $i=1,\ldots,k$. By multiplying
the relation with $U_1\cdots U_k,$ we get $u_1\cdots u_{a_k}=0,$
which contradicts the fact that
 $u_1\cdots u_n$ is the fundamental class.
\end{proof}

By abbreviating all terms in the identity in Lemma \ref{swd} we
obtain that each class $\overline{w}_{dk}(M^{dn})$ is expressed in
the ring $\mathbb{Z}_2[t_1,\dots,t_n]/\mathcal{T}_n$ by square
free homogeneous polynomials $\overline{W}_{k}(t_1,\dots,t_n)$ of
the degree $k$. Note that the inclusion
$$i:\mathbb{Z}_2[t_1, \dots, t_n]\rightarrow\mathbb{Z}_2[t_1,
\dots, t_n, t_{n+1}]$$ induces a natural monomorphism

\[i^*:\mathbb{Z}_2[t_1, \dots, t_n]/\mathcal{T}_n\rightarrow
\mathbb{Z}_2[t_1, \dots, t_n, t_{n+1}]/\mathcal{T}_{n+1},\] which
by Lemma \ref{sub} allows us to consider the total Stiefel-Whitney
class $\overline{w}(M^{dn})$ as an element of the ring
$H^*(M^{d(n+1)}; \mathbb{Z}_2)$. Thw total Stiefel-Whitney classes
$\overline{w} (M^{dn})$ and $\overline{w} (M^{d(n+1)})$ satisfy
the following relation in $H^*(M^{d(n+1)}; \mathbb{Z}_2)$
$$\overline{w} (M^{d(n+1)})=\overline{w}(M^{dn})(1+t_n+\dots+t_n^n).$$
Explicitly
\begin{equation}\label{jed2}
\overline{w}_{dk}(M^{d(n+1)})=\overline{w}_{dk}(M^{dn})+t_n\overline{w}_{d(k-1)}(M^{dn})+\dots+t_n^k,
k=0,\ldots, n.
\end{equation}
Recall that $\overline{w}_{dn}(M^{dn})=0$ (see \cite{10}), which
implies
\begin{equation}\label{jed3}
\overline{w}_{dn}(M^{d(n+1)})= t_n
\overline{w}_{d(n-1)}(M^{dn})+\cdots+t_n^n=t_n\overline{w}_{d(n-1)}(M^{d(n+1)}).
\end{equation}

We use the same trick as in \cite{1}. Define numbers $\sigma^k_n,
0\leq k\leq n-1$ as follows
\[
\sigma^k_n=\overline{W}_{dk}(\underbrace{1,\dots,1}_n)\pmod 2.
\]
By (\ref{jed2}) and (\ref{jed3}), we have
$\sigma^k_{n+1}=\sum_{i=0}^{k} \sigma^i_n $ for every
$k=1,\ldots,n-1$ and $\sigma^{n}_{n+1}=\sigma^{n-1}_{n+1}$. By
definition of $\sigma^k_n$, if $\sigma^k_n=1$, then
$\overline{w}_{dk}$ is the sum of an odd number of linearly
independent square free monomials, which implies
$\overline{w}_{dk}(M^{dn})\neq 0. $

An easy mathematical induction shows that
 $$ \sigma^k_n\equiv\bn{n+k}{k}\pmod{2}. $$
Particularly, if $n=2^r$, we have
\[
\sigma^{n-1}_n
\equiv\bn{2^r+(2^r-1)}{2^r-1}\equiv\bn{2^{r+1}-1}{2^r-1}\equiv
1\pmod{2}.
\]
Consequently,
\[
\overline{w}_{d(n-1)}(M^{dn})=t_1 t_2\cdots t_{n-1}\neq 0.
\]
Therefore by Theorem \ref{imeem} we obtain the required bounds
\[imm(M^{dn})\geq d(2n-1), em(M^{dn})\geq d(2n-1)+1.\] For the
small cover $M^{n}$, Whitney's theorem implies $imm(M^{n})=2n-1$
and $em(M^{n})=2n$. The quasitoric manifold $M^{2n}$ is
orientable, so it can be embedded into $\mathbb{R}^{4n-1}$. From
Lemma \ref{swd} follows
$\overline{w}_2(M^{2n})=t_1+t_2+\cdots+t_{n-1},$ which implies
that the characteristic class $\overline{w}_2(M^{dn})\cdot
\overline{w}_{2n-2}(M^{2n})$ vanishes. If $n\geq 3$, the result of
Massey \cite[Theorem V]{11} yields
 $$imm(M^{2n})=4n-2 ,$$ which finishes the proof of Theorem \ref{main}.

\section{Proof of Theorem \ref{exist}}

Let $n=2^{r_1}+2^{r_2}+\dots+2^{r_t}$, $r_1>r_2>\dots>r_t\geq 0$
be the binary representation of $n$ and let $m_i=2^{r_i}$ for
$i=1,\dots,t$ and $m_0=0$. Let $P^n$ be a simple $n$-polytope such
that $$P^n= P_1^{2_{r_1}}\times \cdots\times P_t^{2^{r_t}},
$$ where each $P_i^{2_{r_i}}$ is $2^{r_i}$-colored simple
$2^{r_i}$-polytope. It is obvious that polytope $P^n$ is
$n$-colored.

In subsection \ref{cons} we constructed manifolds $M^{d 2^{r_i}}$
over polytopes $P_i^{2_{r_i}}$. It follows from
\cite[Proposition~4.7]{2} that
$M^{dn}=M^{d2^{r_1}}\times\dots\times M^{d 2^{r_t}}$ is a
$G_d^{n}$-manifold over the polytope $P^n= P_1^{2_{r_1}}\times
\cdots\times P_t^{2^{r_t}}$. The total Stiefel--Whitney class of
the manifold $M^{dn}$ can be easily determined using the following
formula (see \cite[pp.\,27,\,54]{4})
\[
w(M^{dn})=w(M^{d 2^{r_1}})\cdots w(M^{d 2^{r_t}})\in
H^*(M^{dn})\cong H^*(M^{d 2^{r_1}})\otimes\dots\otimes H^*(M^{d
2^{r_t}}).
\]
The corresponding dual total Stiefel-Whitney class is expressed as
\begin{equation}\label{dpr} \overline{w}(M^{dn})=\overline{w}(M^{d 2^{r_1}})\cdots
\overline{w}(M^{d 2^{r_t}}).\end{equation} Let $\rank w M):=\max
\{k | w_k (M)\neq 0\}$. Thus, from formula (\ref{dpr}) we have

$$\mathrm{rank} \ \overline{w}(M^{dn})=\sum_{i=1}^t\mathrm{rank} \ \overline{w}(M^{d2^{r_i}})=\sum_{i=1}^t d(2^{r_i}-1)=nd -\alpha (n) d.$$
In this way, Theorem \ref{exist} is a consequence of Theorems
\ref{imeem} and \ref{skewteo}.

\section*{Aknowledgements}

The authors express their special thanks to T. E. Panov and A. A.
Gaifullin for attention to this work, as well as to reviewers for
helpful comments.

config: arxiv


TITLE: Generators of Poincare Groups
QUESTION [9 upvotes]: How can I determine the generators of the Poincare Group, $P(1,3)$ explicitly?
Here $P(1,3)$ means a matrix Lie group.

REPLY [17 votes]: The general idea.
Let's restrict the discussion to matrix Lie Groups for simplicity.  Determining the generators of a given Lie group $G$ simply means (by definition) determining a basis for its Lie algebra $\mathfrak g$.  Here's a standard method for finding such a basis:

Recall that the Lie algebra $\mathfrak g$ of a matrix Lie group $G$ is defined as the set of all matrices $X$ for which $e^{s X}$ is an element of $G$ for all real numbers $s$.
Use the properties of elements of $G$ to constrain the elements $X$; the resulting admissible elements $X$ are precisely the elements of the Lie algebra $\mathfrak g$ which is a vector space of matrices.
Determine a basis for this resulting vector space.

An example: $\mathrm{SO}(2)$.
Elements of the rotation group in two dimensions are precisely those $2\times 2$ real matrices $R$ for which
\begin{align}
  R^tR = I, \qquad \det R = 1.
\end{align}
Now, suppose that $X$ is an element of its Lie algebra $\mathfrak{so}(2)$, then $e^{sX}$ is an element of $\mathrm{SO}(2)$ which means that
\begin{align}
  e^{s X} (e^{sX})^t = I, \qquad \det (e^{sX}) = 1
\end{align}
for all $s\in\mathbb R$.  Now we use the facts that $(e^M)^t = e^{M^t}$ and $\det e^M = e^{\mathrm{tr}M}$ to assert that
\begin{align}
  e^{s(X + X^t)} = I, \qquad e^{s(\mathrm{tr} X)} = 1
\end{align}
which implies that
\begin{align}
  X = -X^t, \qquad \mathrm{tr} X = 0.
\end{align}
Thus, the Lie algebra $\mathfrak{so}(2)$ is given by the set of all real, antisymmetric, traceless $2\times 2$ matrices.  This is a one-dimensional vector space of matrices whose general element can be written as a scalar multiple of
\begin{align}
  J = \begin{pmatrix}
        0 & 1 \\
        -1 & 0 \\
      \end{pmatrix}
\end{align}
which is therefore the generator we were looking for.
The Poincare group.
The same procedure can we used to determine an explicit matrix form for the generators of the Poincare group.  Since, in that case, the Poincare group can be written as a semi-direct product of the Lorentz group $\mathrm{SO}(3,1)$ with the four-dimensional translation group $\mathbb R^4$, one often finds the generators of the Lorentz and translation groups individually for convenience.
When finding the generators of the Lorentz group, you simply need to use the analogous properties to $RR^t = I$ and $\det R = 1$ that define its elements.  For example, in this case, the defining property is that the elements of $\mathrm{SO}(3,1)$ preserve the Minkowski scalar product;
\begin{align}
  \Lambda^\mu_{\phantom\mu\alpha} \Lambda^\nu_{\phantom\nu\beta}\eta^{\alpha\beta} = \eta^{\mu\nu} \tag{$\star$}
\end{align}
where $\eta^{\mu\nu} = \mathrm{diag}(-1,+1,+1,+1)$.  This can be written in matrix form, and then the same procedure used above for $\mathrm{SO}(2)$ can be used to find the Lie algebra and generators.
How it's done in practice by physicists.
In order to determine the algebra as outlined above using the invariance condition $(\star)$, physicists will often write a Lorentz group element as
\begin{align}
  \Lambda^\mu_{\phantom\mu\nu} = \delta^\mu_\nu + \omega^\mu_{\phantom\mu\nu}
\end{align}
where $\omega = (\omega^\mu_{\phantom\mu\nu})$ is an "infinitesimal" matrix.  Note that this is simply the same thing as writing out the matrix exponential to first order in the parameter $s$ in the procedure above, and identifying the first order term as the Lie algebra element.  Then, one can plug this expression into the invariance condition $(\star)$ and determined what properties $\omega$ needs to obey by comparing the terms that are first order in $\omega$.  This is equivalent to the procedure outlined above for $\mathrm{SO}(2)$, but it's often more computationally convenient.
If you go through this computation for the Lorentz group, you should find that
\begin{align}
  \omega_{\mu\nu} = -\omega_{\nu\mu}
\end{align}
where $\omega_{\mu\nu} = \eta_{\mu\alpha}\omega^\alpha_{\phantom\alpha\nu}$.


TITLE: How can I find Sobolev generalized derivative of function $f(t)$
QUESTION [0 upvotes]: Given following function
$$
f(t) :=
\begin{cases}
 0 & \text{ if } t \in (-\infty ,1]  \\
 2x -2& \text{ if } t \in (1, 2]  \\
 2& \text{ if } t \in (2, 3)  \\
 8 - 2x& \text{ if } t \in (3, 4]  \\
 0 & \text{ if } t \in (4, \infty)
\end{cases}
$$
How can I find Sobolev generalized derivative of function $f(t)$

REPLY [1 votes]: In general, if $f$ is continuous and piecewise continuously differentiable (say on the intervals $(-\infty,t_1],[t_1,t_2],\dots,[t_{n-1},t_n],[t_n,\infty)$), then
$$
g(t)=\begin{cases}f'(t)&\text{if }t\neq t_k\\0&\text{otherwise}\end{cases}
$$
is a weak derivative of $f$. This can be easily seen by splitting up the integral in the definition of weak derivatives and integrating by parts in each of the pieces. I leave it up to you to apply this result to the function at hand.


TITLE: Stone-Cech compactification of the separable Hilbert space
QUESTION [1 upvotes]: Where can I read about the Stone-Cech compactification of the separable Hilbert space?

REPLY [7 votes]: Hilbert space $l^2$ is homeomorphic to product space $\mathbb R^\infty$, but Stone-Cech compactification of that space is not very nice either.  Probably the best thing is to read about the Stone-Cech compactification in general.  Check (haha) your library for The Stone-Cech Compactification by Russel Walker.


TITLE: Assuming that 5 cards are drawn with replacement, what is the chance that there are 3 cards of rank Ace and 2 cards of rank King?
QUESTION [0 upvotes]: I have the answer, but there is one step that I don't understand.
So the total sample space is $52^5$, as 5 cards are drawn, one card is drawn each time from 52 cards.
3 ace out of 4, that is $4^3$ and 2 king out of 4 thats $4^2$. But the final answer is $(5C3*4^3*4^2)/52^5$
Where does the $5C3$ comes from?

REPLY [0 votes]: There are 5 ways in which cards can be drawn. Therefore multiply by 5!. And since 3 cards and 2 cards are of the same rank divide by 2! and 3!. Altogether they combine to form 5C3.


TITLE: Neat way to prove $\sin(\alpha+\beta)$ using complex exponential
QUESTION [1 upvotes]: I am supposed to prove that $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$ using complex exponentials:
$$
\begin{align}
\sin\theta&=-\frac{1}{2}i(e^{i\theta}-e^{-i\theta})\\
\cos\theta&=\frac{1}{2}(e^{i\theta}+e^{-i\theta})
\end{align}
$$
The proof that I have done is very long and messy and essentially I am showing that LHS and RHS are the same thing. I was wondering if there is a neater proof?

REPLY [5 votes]: Hint
Take the imaginary part of
$$e^{i(\alpha+\beta)}=e^{i\alpha}e^{i\beta}$$


TITLE: Mathematical induction condition "p(k)$\Rightarrow$p(k+1)" for the divisibility by a prime number
QUESTION [0 upvotes]: "

Mathematical induction If p(n) is a statement involving the natural number n such that:
p(1) is true, and
p(k)$\Rightarrow$p(k+1) for any arbitrary natural number k, then p(n) is true for any natural number n.

Example 12 Prove by mathematical induction that
1+2+3+...+n= n(n+1)/2  for every natural number n
[Proof] Here p(n) represents the statement 1+2+3+...+n= n(n+1)/2
In particular p(1) represents 1 = (1ㆍ2)/2, which obviously is a true statement. Therefore, condition (1) for mathematical induction is satisfied.
To prove that condition (2) is satisfied, we assume that p(k), which is "1+2+3+...+k= k(k+1)/2" is true. Then add k+1 to both sides of this equality. We have
1+2+3+...+k+(k+1) = k(k+1)/2 + (k+1)
                           = k(k+1)/2+ 2(k+1)/2
                           = (k+1)(k+2)/2
which shows that p(k+1) is true. We have now shown that the condition (1) and (2) of mathematical induction are satisfied. Therefore, by the principle of mathematical induction,
1+2+3+...+n= n(n+1)/2  is true for every natural number n"
Source: Set Theory by You-Feng Lin, Shwu-Yeng T. Lin
In the below proof, I have a question.
I see "P(2) is true" is used instead of "p(1) is true" for the divisibility by a prime. But unlike from the above proof and the definition, why P(i) is divisible by a prime number for all integers i from 2 through k, rather than for some integer k≥2?

Source: Discrete Mathematics with Applications: Susanna S. Epp

REPLY [0 votes]: It uses so-called "strong induction". If it is not explicitly mentioned in the book what this is, it is a variation on the principle of mathematical induction. Specifically, it is the following principle:
If $P(n)$ is some property that some natural number $n$ might have, and it is true that

$P(a)$ is true for some $a \in \mathbb{N}$
$P(k)$ is true for all $a \leq k < n$ implies that $P(n)$ is true

then $P(n)$ is true for all natural numbers greater than or equal to $a$.
This principal follows from the "normal" version of induction:
Suppose that we have the property $P$ with the conditions above satisfied. Let $Q$ be the predicate:
$$ Q(n) = ((k \geq a \text{ and } k \leq n) \implies P(k)) $$
In other words, $Q(n)$ is true if either $n$ is less than $a$, or if $P(k)$ is true for all $k$ from $a$ to $n$.
You can then check that $Q(1)$ is true, and that from the conditions on $P$, that we have that $Q(n) \implies Q(n + 1)$. So by "normal" induction, $Q(n)$ is true for all natural numbers $n$. But $Q(n)$ implies $P(n)$ for all $n \geq a$, so we then also have that $P(n)$ is true for all $n \geq a$.
In your case, we are proving by induction the property $Q(n)$, where $Q(n)$ is given by the statement
"If $n \geq 2$, then all natural numbers from $2$ to $n$ have a prime factor."