in $\mu$ with positive leading coefficient, whose roots are $$\frac{p^{2}-5p+8 \pm 4(p-2)\sqrt{4-p}}{p}.$$ So, for $p\geq 5$, there are no real roots, while in the case $p=4$, we have a double root at $\mu=1$.

Similarly, for $k=2p+1$, the vector $v_{p-2}$ is ordered if and only if $$\frac{\mu+p-2}{2(p-1)} \geq c.$$ For $c = c_{\operatorname{vol}}$, we get $$\frac{\mu+p-2}{2(p-1)}\geq \sqrt{\frac{2\mu}{2p+1}},$$ which is equivalent to $f(\mu) = (2p+1)(\mu+p-2)^{2}-8\mu(p-1)^{2}\geq 0$. Again, this is a quadratic polynomial in $\mu$ with positive leading coefficient, whose roots are $$\frac{2p^{2}-5p+6 \pm (p-1)\sqrt{8(4-p)}}{2p+1}$$ So, for $p\geq 5$, there are no real roots while, in the case $p=4$, we have a double root at $\mu=2$.

\[rmk:AutomorphismsTrivialInitialStep\] Observe that the vector $v_{i+1}$ is obtained from $v_{i}$ by applying a Cremona transformation followed by the permutation $\mathfrak{R}:=(1, 2, k+1, k+2, 3, \ldots, k)$. For convenience, we denote the corresponding element of $D_{K}(1, k+1)$ by $\mathfrak{R} \mathfrak{C}$. We can thus write $v_{i}=( \mathfrak{R} \mathfrak{C} )^{i}v_{0}$.

We now differentiate our discussion depending on whether $k$ isodd or even.

The odd case $k=2p+1$
---------------------

This subsection will provide the proof of part $(i)$ of Theorem \[thm:MainTheoremTrivial\].

By Lemma \[lemma:TrivialBundleEvenOddDichotomy\], we know that the vector $$v_{p-2} =
  \left(\lambda_{p-2}\,; \lambda_{p-2}-1,c^{\times 4},(1-c)^{\times (2p-3)} \right)$$ is ordered whenever $k\geq 4$, $\mu\geq 