M algebraic-coding-theory-notes.pdf → algebraic-coding-theory-notes.pdf

M algebraic-coding-theory-notes.tex → algebraic-coding-theory-notes.tex

@@ -1420,6 +1420,66 @@ \end{cor}
 
 $\Longrightarrow$ the MLCT problem (V2) is the same as the packing problem of finding $max_{d-1}(r, q)$.
 
+\subsection{The projective geometry PG(r-1,q)}
+Let the vector space
+$$V(r,q)= \{ (a_1, a_2, \ldots, a_r) ~|~ a_i \in GF(q) \}$$
+
+\begin{defn}{}[Projective Geometry]
+  $PG(r-1, q)$
+  \begin{itemize}
+    \item \emph{points} are the one-dimensional subspaces of $V(r, q)$
+    \item \emph{lines} are the two-dimensional subspaces of $V(r, q)$
+  \end{itemize}
+
+  Each point $P \in PG(r-1, q)$, as a subspace of $V(r,q)$ of dimension $1$, is generated by a single non-zero vector.
+
+  So if $a=(a_1, a_2, \ldots, a_r) \in P$, then $P= \{ \lambda a ~|~ \lambda \in GF(q) \}$.
+
+  $\Longrightarrow$ for $a=(a_1, a_2, \ldots, a_r),~b=(b_1, b_2, \ldots, b_r) \in P$, then
+  $$a=b ~\text{in}}~ PG(r-1,q) ~\text{iff}~ a=\lambda b ~\text{in}~ V(r,q)$$
+  for some non-zero scalar $\lambda$.
+\end{defn}
+
+\begin{lemma}{14.6}
+  in $PG(r-1,q)$,
+  \begin{enumerate}
+    \item number of points is $\frac{q^r-1}{q-1}$
+    \item any two points lie on exactly one line
+    \item each line contains exactly $q+1$ points
+    \item each point lies on $\frac{q^{r-1} -1}{q-1}$ lines
+  \end{enumerate}
+\end{lemma}
+\begin{proof}
+  \begin{enumerate}
+    \item number of points is $\frac{q^r-1}{q-1}$:\\
+      since each of the $q^r-1$ nonzero vectors in $V(r,q)$ has $q-1$ nonzero scalar multiples, the number of points of $PG(r-1,q)$ is $\frac{q^r-1}{q-1}$
+
+    \item any two points lie on exactly one line:\\
+      if $a,b$ distinct points of $PG(r-1,q)$, then the unique line through them consists of the points $\lambda a + \mu b$, where $\lambda,\mu$ are scalars not both zero.
+
+    \item each line contains exactly $q+1$ points:\\
+      in (ii) there are $q^2-1$ choices for the pair $(\lambda, \mu)$, but we're identifying scalar multiples, so the number of distinct points on the line is
+      $$\frac{q^2 -1}{q-1}=q+1$$
+
+    \item each point lies on $\frac{q^{r-1} -1}{q-1}$ lines:\\
+      let $t$ be the number of lines on which a point $P$ lies. Let the set
+      $$X=\{(Q,L) ~|~ Q ~\text{is a point}~ \neq P,~ L~\text{is a line containing both}~ P ~\text{and}~ Q \}$$
+
+      Count the members of $X$ in two ways:
+
+      for each of the $\frac{q^r -1}{q-1} -1$ choices of $Q$, $\exists !$ line containing $P$ and $Q$. Thus,
+      $$|X|=\frac{q^r-1}{q-1} -1 = \frac{q^r-q}{q-1}$$
+
+      On the other hand, for each of the $t$ lines through $P$, there are (by (iii)) $q$ points $Q$ other than $P$ lying on $L$.
+
+      Thus, $|X|=tq$,
+      $$\Longrightarrow tq = \frac{q^r-q}{q-1}, ~~~~ \Longrightarrow t=\frac{q^{r-1}-1}{q-1}$$
+  \end{enumerate}
+\end{proof}
+
+\begin{defn}{}[Projective plane]
+  The $PG(2,q)$ is called the \emph{projective plane} over $GF(q)$.
+\end{defn}
 
 
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