0$ and~$b$. Conversely, if $[v',\,w']=[v,\,w]$,
then $[v',\,w'] = [a\,v,\,\,a\,w+b\,v]$ for some real numbers
$a>0$ and $b$.
It follows that a Finsler norm on~$\bbS$ determines and is determined by a
function~$F$ on~$\bigl(V\setminus\{\bold0\}\bigr)\times V$ that satisfies
$$
F\bigl(v,\,w\bigr) = F\bigl(a\,v,\,\,a\,w+b\,v\bigr)
\qquad\text{for all real~$a>0$ and $b$}
\tag4
$$
as well as the homogeneity condition~
$$
F\bigl(v,\,c\,w\bigr) = c\,F\bigl(v,\,w\bigr)
\qquad\text{for all~$c\ge0$.}
\tag5
$$
\proclaim{Theorem 10}
Let~$C\subset\bbC\bbP^2$ be a conic without real points and let~$Q$ be a
normalized quadratic form on~$V\otimes\bbC$ so that
$C = \{\,\lb v\rb\,\mid\,Q(v)=0\}$. Let the inner product of two vectors~$v$
and $w$ with respect to~$Q$ be denoted $v\cdot w$. Set
$$
F_C(v,w)
= \Re\left[\,
\frac
{\sqrt{\,(w\cdot w)(v\cdot v)-(w\cdot v)^2\,}-i\,(v\cdot w) }
{(v\cdot v)}
\,\right].
\tag6
$$
Then~$F_C$ defines the Finsler norm of the Finsler structure on~$\bbS$ with
linear geodesics and $K=1$ whose canonical section~$\sigma:\bbS^*\to\bbC\bbP^2$
has its image equal to~$C$.
\endproclaim
\remark{Remark}
Before beginning the proof (which will be a calculation), it probably is
a good idea to explain that the quantity inside the radical in the formula
for~$F_C$ can never be a negative real number and the branch of the complex
square root being used is the one satisfying~$\sqrt1=1$ and having
the negative real axis as its branch locus. Formula~(6) then defines
a function on~$\bigl(V\setminus\{\bold0\}\bigr)\times V$ satisfying (4)
and~(5). It will be seen below that this function is positive and smooth away
from points of the form~$(v,\,c\,v)$. (These points represent the zero
section of~$T\bbS$.)
By the remarks after Proposition~4, the normalized $Q$ is only determined
by~$C$ up to a positive real multiple. However, replacing~$Q$ by~$r\,Q$
for any positive~$r$ does not affect the resulting function~$F_C$.
\endremark
\demo{Proof}
Let~$V_{(2)}\subset V\times V$ be the set of pairs of linearly independent
vectors in~$V$. This is a connected, open subset of~$V\times V$.
Let $(v,w)$ be an element of~$V_{(2)}$. Since~$C$ has no real points and
no real tangents, the complexified line spanned by~$v$ and $w$ must
intersect~$C$ transversely in two non-real points.
Let~$p=\lb\,\alpha\,v+\beta\,w\,\rb$ be such a point.
Since it is not real, neither~$\alpha$ nor~$\beta$ can vanish, nor can
the ratio~$\alpha/\beta$ be real. By multiplying~$\alpha$ and~$\beta$
by an appropriate scalar, it can be assumed that~$\beta = ib$ for some
non-zero real number~$b$. Then~$\alpha$ cannot be pure imaginary, so by
dividing both~$\alpha$ and $\beta$ by the real part of~$\alpha$, it can be
assumed that~$\alpha = 1{+}ia$ for some real number~$a$. Thus, such a~
$p\in C$ can be uniquely written in the form
$$
p = \lb\,(1{+}ia)\,v + ib\,w\,\rb
$$
for some real numbers~$a$ and~$b\not=0$.
Now, there are two such points,
$$
\align
p_1 &= \lb\,(1{+}ia_1)\,v + ib_1\,w\,\rb = \lb\,v + i(b_1\,w+a_1\,v)\,\rb\,,\\
p_2 &= \lb\,(1{+}ia_2)\,v + ib_2\,w\,\rb = \lb\,v + i(b_2\,w+a_2\,v)\,\rb\,.\\
\endalign
$$
Since~$\tau(p_i)=[\,v\w(b_i\,w+a_i\,v)\,]=[\,b_i\,v\w w\,]$ and these two
points must represent the two different oriented lines spanned by~$v$~and~$w$,
it follows that~$b_1$ and $b_2$ are always of different signs. Thus, it
can be arranged that~$b_1<0