\ge(c_0+1)w^2(\mcT\cap\mbB_2).
\end{gather}

Let $c_1=c_0+1$. Then the minimum sample size and the restricted eigenvalue are given by

\begin{gather}
n\ge\parenf{\msfC_{RE}^2 c_1\max\parenc{1,\frac{16K_0^2}{\Lambda_{\min}(\Sigma_X)}} w^2(\mcT\cap\mbB_2)}^{1/\gamma}, \quad
\alpha_{RE}=\Lambda_{\min}(\Sigma_X)/2.
\end{gather}

The precondition $t>1/n$ boils down to $n\ge 2/\Lambda_{\min}(\Sigma_X)$.

§.§ Proof of Proposition .

The proof is virtually identical to that of Proposition (<ref>).

§.§ Proof of Proposition .

The proof is virtually identical to that of Proposition (<ref>), the only difference being that we apply the single concentration bound for the deviation term $\fr{X^\top\eta}$ in Proposition (<ref>), instead of Proposition (<ref>).

§.§ Proof of Proposition .

The proof is virtually identical to that of Proposition (<ref>), the only difference being that we apply the single concentration bound for the gram matrix $\fr{X^\top X}$ in Proposition (<ref>), instead of Proposition (<ref>).

§.§ Proof of Proposition .

The proof is virtually identical to that of Proposition (<ref>).

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##  Appendix A Proof of Theorem 1.1

We now prove Theorem 1.1. One can simply implement the algorithms outlined in
this paper on a computer to verify this result (although it needs a little bit
more effort than naively implementing the algorithm because we want to
determine polystability for all primes, which is apriori an infinite set of
computations). However, we will not directly appeal to the algorithms and
instead give an explicit argument. This has a few advantages. First, it
demonstrates the flexibility we actually have in using the ideas developed in
this paper. Second, we want to make the computations as manageable as
possible, i.e., even though we omit many of the computational details, we
intend for it to be hand checkable by the reader with sufficient (but not
unearthly) patience. Indeed, we did these computations by hand. Finally, we
want to illustrate the flavor of combinatorial computations one encounters,
and we hope that a deeper analysis of the combinatorics involved can lead to a
better understanding of polystability for interesting classes of symmetric
polynomials, beyond what we discussed in Section 9.

###### Proof of Theorem 1.1.

First, recall that $h_{3}(x,y,z)$ is the sum of all degree $3$ monomials in
$x,y$ and $z$, so

$h_{3}(x,y,z)=x^{3}+y^{3}+z^{3}+x^{2}y+x^{2}z+xy^{2}+xz^{2}+yz^{2}+y^{2}z+xyz.$

  * •

Case 1: $p\nmid(3=n)$, i.e., $p\neq 3$:

Suppose $h_{3}(x,y,z)$ is not polystable. Then, the optimal parabolic subgroup
must either be $\mathcal{F}=0\subseteq L\subseteq K^{3}$ or
$\mathcal{G}=0\subseteq M\subseteq K^{3}$, where $L={\rm span}(x+y+z)$ and
$M={\rm span}(x-y,y-z)$ by Corollary 8.7.

Suppose $\mathcal{F}$ is the optimal parabolic. Then, a compatible basis is
$\mathcal{B}=(t,q,r)$, where $t=x+y+z,q=y,r=z$. Let $T_{\mathcal{B}}$ be the
corresponding torus. We compute the change of basis:

$h_{3}(x,y,z)=h_{3}(t-q-r,q,r)=t^{3}-2t^{2}q-2t^{2}r+2tq^{2}+3tqr+2tr^{2}-q^{2}r-qr^{2}.$

Recall that we are in the case $p\neq 3$. When $p\neq 2$, the Newton polytope
for $h_{3}(x,y,z)$ with respect to the torus $T_{t,q,r}$ is:

$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$
$\textstyle{t^{3}}$$\textstyle{t^{2}q}$$\textstyle{t^{2}r}$$\textstyle{tq^{2}}$$\textstyle{tqr}$$\textstyle{tr^{2}}$$\textstyle{q^{3}}$$\textstyle{q^{2}r}$$\textstyle{qr^{2}}$$\textstyle{r^{3}}$

The picture on the right gives the dictionary between the monomials and their
weights. Note that the weight of a monomial is just its exponent vector, so
weight of $t^{3}$ is $(3,0,0)$, weight of $tr^{2}$ is $(1,0,2)$, etc. Now, by
Corollary 3.2, we conclude that $h_{3}(x,y,z)$ is $T_{\mathcal{B}}$-polystable
since $(1,1,1)$ is in the relative interior of the Newton polytope.

When $p=2$, the above simplifies to:

$h_{3}(x,y,z)=h_{3}(t-q-r,q,r)=t^{3}+tqr+q^{2}r+qr^{2}.$

Now, the Newton polytope is a convex hull of 4 points, and it is easily seen
that $(1,1,1)$ is again in the interior of the Newton polytope, picture below:

$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$

Hence, we get that the $T_{\mathcal{B}}$ orbit of $h_{3}(x,y,z)$ is closed.
Thus for all $p$ such that $p\nmid n$, $h_{3}(x,y,z)$ is
$T_{\mathcal{B}}$-polystable, so $\mathcal{F}$ cannot be an optimal parabolic
subgroup by Theorem 4.1.

Now, suppose $\mathcal{G}$ is the optimal parabolic subgroup. Then, a
compatible basis is $t=x-y$, $q=y-z$ and $r=z$. Write $\mathcal{B}=(t,q,r)$
and let $T_{\mathcal{B}}$ be the corresponding torus. We compute the change of
basis:

$\displaystyle h_{3}(x,y,z)$ $\displaystyle=h_{3}(t+q+r,q+r,r)$
$\displaystyle=t^{3}+4t^{2}q+5t^{2}r+6tq^{2}+15tqr+10tr^{2}+4q^{3}+15q^{2}r+20qr^{2}+10r^{3}.$

Unless $p=2$ or $p=5$, it is easy to conclude that $h_{3}(x,y,z)$ is
polystable with respect to $T_{\mathcal{B}}$ by computing its Newton polytope
and we leave the details to the reader. On the other hand, when $p=2$, we
have:

$h_{3}(x,y,z)=t^{3}+t^{2}r+tqr+q^{2}r.$

The Newton polyope is:

$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$$\textstyle{\bullet}$

As is evident, the point $(1,1,1)$ is not in the relative interior, so
$h_{3}(x,y,z)$ is not $T_{\mathcal{B}}$-polystable. Thus, it suffices to check
if $w={\rm ess}(h_{3}(x,y,z))$ (with respect to $T_{\mathcal{B}}$) is in the
$\operatorname{SL}_{3}$-orbit of $h_{3}(x,y,z)$. One easily computes
$w=t^{2}r+tqr+q^{2}r=r(t^{2}+tq+q^{2})$, which is reducible. But
$h_{3}(x,y,z)=t^{3}+t^{2}r+tqr+q^{2}r$ is irreducible – think of it as a
polynomial in the variable $t$ with coefficients in the PID $K(q)[r]$ and
apply Eisenstein’s criterion with the prime $r$. Thus $h_{3}(x,y,z)$ and $w$
are not in the same orbit.

Thus, to summarize, for $p=2$, $h_{3}(x,y,z)$ is not $\operatorname{SL}_{3}$
polystable, $\mathcal{G}$ is an optimal parabolic subgroup and
$w=t^{2}r+tqr+q^{2}r=r(t^{2}+tq+q^{2})$ is a point in the boundary of the
$\operatorname{SL}_{3}$ orbit of $h_{3}(x,y,z)$.

Now, the case of $p=5$. In this case, we have

$h_{3}(x,y,z)=t^{3}+4t^{2}q+6tq^{2}+4q^{3}.$

We omit the details, but one can check by a similar analysis as above that
$t^{3}+4t^{2}q+6tq^{2}+4q^{3}$ is actually unstable with respect to
$T_{\mathcal{B}}$. So, $h_{3}(x,y,z)$ is $\operatorname{SL}_{3}$ unstable (and
in particular not polystable) when $p=5$!

  * •

Case 2: The case $p=3$: We will be brief with this case. Suppose
$h_{3}(x,y,z)$ is not polystable, then there are three possible choices for
optimal parabolic. However, the basis $t=x+y+z,q=y-z,r=z$ is compatible with
all possible optimal parabolics. Thus, if we check that $h_{3}(x,y,z)$ is
$T_{\mathcal{B}}$ polystable, where $\mathcal{B}=(t,q,r)$, then we get a
contradiction, so $h_{3}(x,y,z)$ must be polystable.

We compute the change of basis

$\displaystyle h_{3}(x,y,z)$ $\displaystyle=h_{3}(t+2q+r,q+r,r)$
$\displaystyle=t^{3}+7t^{2}q+5t^{2}r+17tq^{2}+25tqr+10tr^{2}+15q^{3}+35qr^{2}+30qr^{2}+10r^{3}$
$\displaystyle=t^{3}+t^{2}q+2t^{2}r+2tq^{2}+tqr+tr^{2}+2qr^{2}+r^{3}.$

We leave it to the reader to check that $h_{3}(x,y,z)$ is
$T_{\mathcal{B}}$-polystable by drawing the Newton polytope.

Thus, we conclude that $h_{3}(x,y,z)$ is $\operatorname{SL}_{3}$ polystable
unless $p=2$ or $p=5$. When $p=2$, it is $\operatorname{SL}_{3}$ semistable,
not $\operatorname{SL}_{3}$ polystable and perhaps most surpisingly, when
$p=5$, it is $\operatorname{SL}_{3}$ unstable! ∎