Exercise 11.7

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Try this exercise if you know a little projective geometry. Let $N_s$ be the number of lines in $P_n(F_{p^s})$. Find $N_s$ and calculate $\sum_{s=1}^\infty N_su^s/s$. (The set of lines in projective space form an algebraic variety calles a Grassmannian variety. So do the set of planes three-dimensinal linear subspaces, etc.)

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Write $q = p^s$. The set of lines in $P_n(F_{q})$ is in bijective correspondence with the set of planes of the vector space $F_q^{n+1}$. To count these planes, consider the set $A$ of  linearly independent pairs$(u,v)$ of the space $F_q^{n+1}$, and $B$ the set of planes of $F_q^{n+1}$, and
$$
f 
\left\{
\begin{array}{ccl}
A & 	o B \
(u,v) & \mapsto \langle u, v angle.
\end{array}
ight.
$$
 The set of pre-images of a fixed plane $P$ in $B$ is the set of basis of this plane $P$. Thus, to obtain $N_s$, we divides the number of linearly independent pairs$(u,v)$ of the space by the number of basis of a fixed plane. To build such a pair, we choose first a nonzero vector $u$, and then a vector $v$ not on the line generated by $u$. Therefore
\begin{align*}
N_s &= \frac{(q^{n+1} - 1)(q^{n+1}- q)}{(q^2 - 1)(q^2 -q)}\
&= \frac{(q^{n+1} - 1)(q^{n}- 1)}{(q^2 - 1)(q -1)}.
\end{align*}
\end{proof}

$\bullet$ If $n = 2m+1$ is odd, then
\begin{align*}
N_s &= \frac{q^{2m+2}-1}{q^2-1} \cdot \frac{q^{2m+1}-1}{q-1}\
&=\sum_{k=0}^m q^{2k} \sum_{l = 0}^2 q^l\
&=\sum_{k=0}^m \sum_{l = 0} ^{2m} q^{2k+l}\
&=\sum_{r=0}^{4m} a_r q^r \qquad (r= 2k+l),
\end{align*}
where $a_r$ is the cardinality of the set
$$A_r = \{(k,l) \in \gcro 0, m \dcro 	imes \gcro 0, 2m \dcro \mid 2k+l = r\}.$$
We note that $0 \leq l = r-2k \leq 2m$ gives 
$$
\left\{
\begin{array}{rcl}
\frac{r}{2} - m \leq &k& \leq \frac{r}{2},\
0 \leq &k& \leq m,
\end{array}
ight.
$$
that is
\begin{align}
\max\left( 0,  \frac{r}{2}  - m ight) \leq k \leq \min \left(\frac{r}{2} ,m ight),
\end{align}
 and each such $k$ gives a unique pair $(k,l) = (k,r-2k)$ in $A_r$.

\begin{enumerate}
\item[$-$] If $0 \leq r \leq 2m$, then $(1) \iff 0 \leq k \leq \frac{r}{2}$, thus $a_r = \left \lfloor \frac{r}{2} ight floor + 1$.
\item[$-$] If $2m < r \leq 4m$, then $(1) \iff \frac{r}{2} -m  \leq k \leq m$, thus $a_r = 2m - \left \lceil \frac{r}{2} ight ceil + 1$.
\end{enumerate}
If $n$ is odd, we have proved that
\begin{align*}
N_s &= \sum_{r=0}^{2m} \left (\left \lfloor \frac{r}{2} ight floor + 1 ight) q^r + \sum_{r=2m+1}^{4m} \left(2m+1 - \left \lceil \frac{r}{2} ight ceil ight) q^r\
&=\sum_{r=0}^{n-1} \left (\left \lfloor \frac{r}{2} ight floor + 1 ight) p^{sr} + \sum_{r=n}^{2n-2} \left (n- \left \lceil \frac{r}{2} ight ceil  ight) p^{sr}.
\end{align*}

\bigskip

$\bullet$ If $n = 2m$ is even, then
\begin{align*}
N_s &=\frac{q^{2m}-1}{q^2-1}\cdot  \frac{q^{2m+1}-1}{q-1} \
&= \sum_{k=0}^{m-1} q^{2k} \sum_{l = 0}^{2m} q^l\
&=\sum_{k=0}^{m-1} \sum_{l = 0} ^{2m} q^{2k+l}\
&=\sum_{r=0}^{4m-2} b_r q^r \qquad (r= 2k+l),
\end{align*}
where $b_r$ is the cardinality of the set
$$B_r = \{(k,l) \in \gcro 0, m-1 \dcro 	imes \gcro 0, 2m \dcro \mid 2k+l = r\}.$$
Here $0 \leq l = r-2k \leq 2m$ gives 
$$
\left\{
\begin{array}{rcl}
\frac{r}{2} - m \leq &k& \leq \frac{r}{2},\
0 \leq &k& \leq m -1,
\end{array}
ight.
$$
that is
\begin{align}
\max\left( 0,  \frac{r}{2}  - m ight) \leq k \leq \min \left(\frac{r}{2} ,m-1 ight),
\end{align}
 and each such $k$ gives a unique pair $(k,l) = (k,r-2k)$ in $B_r$.
\begin{enumerate}
\item[$-$] If $0 \leq r \leq 2m-1$, then $(2) \iff 0 \leq k \leq \frac{r}{2}$, thus $b_r = \left \lfloor \frac{r}{2} ight floor + 1$.
\item[$-$] If $2m \leq r \leq 4m-2$, then $(2) \iff \frac{r}{2} -m  \leq k \leq m-1$, thus $b_r = 2m - \left \lceil \frac{r}{2} ight ceil$.
\end{enumerate}
If $n$ is odd, we have proved that
\begin{align*}
N_s &= \sum_{r = 0}^{2m-1}  \left (\left \lfloor \frac{r}{2} ight floor + 1ight) q^r + \sum_{r=2m}^{4m-2} \left ( 2m - \left \lceil \frac{r}{2} ight ceil ight) q^r\
&= \sum_{r = 0}^{n-1}  \left (\left \lfloor \frac{r}{2} ight floor + 1ight) p^{sr} + \sum_{r=n}^{2n-2} \left ( n - \left \lceil \frac{r}{2} ight ceil ight) p^{sr}.
\end{align*}
This is the same formula as in the odd case !
To conclude, for all dimension $n$,
$$N_s = \sum_{r = 0}^{n-1}  \left (\left \lfloor \frac{r}{2} ight floor + 1ight) p^{sr} + \sum_{r=n}^{2n-2} \left ( n - \left \lceil \frac{r}{2} ight ceil ight) p^{sr},$$
therefore
$$\sum_{s=1}^\infty \frac{N_s u^s}{s} = - \sum_{r=0}^{n-1}  \left (\left \lfloor \frac{r}{2} ight floor + 1 ight) \ln(1-p^ru) - \sum_{r=n}^{2n-2}  \left (n - \left \lceil \frac{r}{2} ight ceil  ight) \ln(1 - p^r u)$$
This gives the order of the poles $p^{-r}$ of $Z(u) = \exp\left( \sum_{s=1}^\infty \frac{N_s u^s}{s}  ight)$.

To verify the equality between the two formulas giving $N_s$, we test this equality with a Sage program.
\begin{verbatim}
def N(n,p,s):
     q = p^s
     num = (q^(n+1) - 1)*(q^(n+1) - q)
     den = (q^2 - 1)*(q^2-q)
     return num // den
     
 def M(n,p,s):
    q = p^s
    a = sum((floor(r/2) +1)*q^r for r in range(n))
    b = sum((n - ceil(r/2))*q^r for r in range(n,2*n-1))
    return a+b
    
 N(4,5,3),M(4,5,3)
\end{verbatim}
\begin{center}
(3845707062626, 3845707062626)
\end{center}

Exercise 11.7

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Exercise 11.7

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