transformation $\mathfrak{C}$ followed by the permutation $\mathfrak{R}:=(1, 2, k+1, k+2, 3, \ldots, k)$, we get the vector $$v_{1} =
  \left(\lambda_{1}\,; \lambda_{1}-1, c^{\times(k-3)}, (1-c)^{\times 3}\right),$$ which is ordered if and only if $\lambda_{1}-1 = \mu+1-3c \geq c$. Assuming $v_{1}$ ordered, its defect is also $d_{0}=2c-1>0$, so that we can apply another Cremona move $\mathfrak{C}$ and a reordering $\mathfrak{R}$ to get $$v_{2} =
  \left( \lambda_{2}\,; \lambda_{2}-1, c^{\times(k-5)}, (1-c)^{\times 5} \right)$$ provided $k\geq 5$. Clearly, we can repeat this process $n$ times to get a vector $$v_{n} := (\mathfrak{R} \mathfrak{C})^{n}v_{0} =
  \left( \lambda_{n}\,; \lambda_{n}-1, c^{\times(k-2n)}, (1-c)^{\times (2n+1)} \right),$$ as long as $2n\leq k$ and $\lambda_{n-1}-1=\mu+1-c-(n-1)d_{0}\geq c$.

\[lemma:TrivialBundleEvenOddDichotomy\] Given $k\geq 4$, let us write $k=2p$ or $k=2p+1$ depending on the parity of $k$. Choose any $\mu\geq1$ and $c\in(1/2, c_{\operatorname{vol}}]$. Then the following holds:

-   If $k$ is even, the vector $v_{p-3}$ is ordered.

-   If $k$ is odd, the vector $v_{p-2}$ is ordered.

Let $k=2p$. Then the vector $v_{p-3}$ is ordered if and only if $\lambda_{p-3}-1\geq c$, which is equivalent to $$\frac{\mu+p-3}{2(p-2)} \geq c.$$ Clearly, it is sufficient to consider the case $c=c_{\operatorname{vol}}$ for which the previous inequality becomes $$\frac{\mu+p-3}{2(p-2)}\geq \sqrt{\frac{\mu}{p}}.$$ This is equivalent to $f(\mu) = p(\mu+p-3)^{2}-4\mu(p-2)^{2}\geq 0$, which is a quadratic polynomial 