Let us consider a permutation $\sigma$ of $\{1, \dots, n\}$ that puts the coefficients $\lambda_i$ in increasing order:

$$ \lambda_{\sigma_1} \le \dots \le \lambda_{\sigma_n}. \qquad (B.7) $$

Then

$$ \begin{aligned} I &= \sum_{i \neq j} q_{\sigma_i \sigma_j} \left( 1 + \frac{1}{n} (x_{\sigma_j} - x_{\sigma_i}) \right) \\ &= \sum_{i < j} (q_{\sigma_i \sigma_j} - q_{\sigma_j \sigma_i}) \left( 1 + \frac{1}{n} (x_{\sigma_j} - x_{\sigma_i}) \right) + 2 \sum_{i < j} q_{\sigma_i \sigma_j} \\ &=: I_3 + I_4. \end{aligned} $$

It is clear that $I_4$ is non-negative. We will show that so is $I_3$. By (B.5), it is easy to show that

$$ q_{\sigma_i \sigma_j} - q_{\sigma_j \sigma_i} = \sum_{k=i}^{j-1} (q_{\sigma_k \sigma_{k+1}} - q_{\sigma_{k+1} \sigma_k}) $$

for any $i < j$. Therefore,

$$ \begin{aligned} I_3 &= \sum_{i<j} \sum_{k=i}^{j-1} (q_{\sigma_k \sigma_{k+1}} - q_{\sigma_{k+1} \sigma_k}) \left(1 + \frac{1}{n}(x_{\sigma_j} - x_{\sigma_i})\right) \\ &= \sum_{k=1}^{n-1} a_k (q_{\sigma_k \sigma_{k+1}} - q_{\sigma_{k+1} \sigma_k}) \\ &= \sum_{k=1}^{n-1} a_k (\lambda_{\sigma_{k+1}} - \lambda_{\sigma_k}), \end{aligned} \qquad (B.8) $$

where in the last step we have used (B.5) again and the coefficients $a_k$ are

$$ \begin{aligned} a_k &:= \sum_{\substack{i \le k \\ j \ge k+1}} \left(1 + \frac{1}{n}(x_{\sigma_j} - x_{\sigma_i})\right) \\ &= k(n-k) + \frac{k}{n} \sum_{j=k+1}^{n} x_{\sigma_j} - \frac{n-k}{n} \sum_{i=1}^{k} x_{\sigma_i}. \end{aligned} $$

Bearing in mind that the sum of all the $x_i$ vanishes by hypothesis, we have

$$ a_k = k(n-k) - \sum_{i=1}^{k} x_{\sigma_i}. $$

Thus, $a_k \ge 0$ by hypothesis. Since we have chosen $\sigma$ so that (B.7) takes place, the result follows from the expression (B.8) for $I_3$. ■
**Appendix C. Characterizing the sticking conditions**

Our purpose in this appendix is to give explicit conditions on the weights yielding necessary and sufficient conditions for sticking of particles (3.12) and (3.13) in Subsections 4.3 and 4.1 respectively. The first part is devoted to the supercritical case and the second part will focus on the critical case.

Apart from the linear operators in Definition B.1 we will need the following ones.

**Definition C.1.** For any $n \in \mathbb{N}$ and $i, j \in \{1, \dots, n\}$ we define the linear operator

$$T_{ij}: \text{Skew}_n(\mathbb{R}) \to \mathbb{R}, \quad Y \mapsto \sum_{k=1}^{n} (y_{ik} - y_{jk}).$$

Notice that by definition we get the following relation to the operators in Definition B.1:

$$T_{ij} = \sum_{k=1}^{n} (L_{ik} - L_{jk}).$$

The next result yields a characterization of the sticking condition (3.13).

**Lemma C.1.** For any $n \in \mathbb{N}$ and $M \in \text{Skew}_n(\mathbb{R})$, the following assertions are equivalent:

(1) There exists $Y \in \text{Skew}_n(\mathbb{R})$ such that

$$M = Y \cdot J + J \cdot Y.$$

(2) There exists $Y \in \text{Skew}_n(\mathbb{R})$ such that

$$T_{ij}(Y) \leq m_{ij} \quad \text{and} \quad -T_{ij}(Y) \leq -m_{ij}.$$

(3) We have

$$m_{1i} + m_{ij} + m_{j1} = 0 \qquad (C.1)$$

for all $2 \leq i < j \leq n$.

(4) We have

$$m_{ij} + m_{jk} + m_{ki} = 0 \qquad (C.2)$$

for all $1 \leq i < j < k \leq n$.

*Proof.* First, it is clear that the first two assertions are equivalent. Second, let us briefly show that (C.1) and (C.2) are equivalent. On the one hand, it is clear that (C.1) is a particular case of (C.2). On the other hand, assume that (C.1) holds. Then in particular for $1 \leq i < j < k \leq n$ we have

$$m_{1i} + m_{ij} + m_{j1} = 0,$$

$$m_{1j} + m_{jk} + m_{k1} = 0,$$

$$m_{1k} + m_{ki} + m_{i1} = 0.$$

Taking the sum of these equations we obtain (C.2) by the skew-symmetry of $M$.
Hence, let us prove the equivalence between the second and third assertions. By Lemma B.2, (2) amounts to saying that whenever $\Lambda \in \mathcal{M}_n(\mathbb{R})$ satisfies

$$ \sum_{i,j=1}^{n} \lambda_{ij} T_{ij} = 0, $$

then

$$ \sum_{i,j=1}^{n} \lambda_{ij} m_{ij} \geq 0. $$

Evaluating on the basis $e_i \otimes e_j - e_j \otimes e_i$ we equivalently write the former condition as

$$ \sum_{k=1}^{n} [(\lambda_{ik} - \lambda_{ki}) - (\lambda_{jk} - \lambda_{kj})] = 0. $$

Hence, if we define $p_{ij} = \lambda_{ij} - \lambda_{ji}$ we can conclude that (2) is equivalent to the fact that whenever $P \in \text{Skew}_n(\mathbb{R})$ satisfies

$$ \sum_{k=1}^{n} (p_{ik} - p_{jk}) = 0 \qquad (C.3) $$

for all $i, j \in \{1, \dots, n\}$, then

$$ \sum_{i,j=1}^{n} p_{ij} m_{ij} \geq 0. \qquad (C.4) $$

• Step 1. Here, we characterize the condition (C.3). Taking

$$ x = \left( \sum_{k=1}^{n} p_{jk} \right) j $$

in Lemma 4.4 shows that the matrices $P \in \text{Skew}_n(\mathbb{R})$ fulfilling (C.3) coincide with those in the kernel of the operator $\mathcal{L} = (L_1, \dots, L_n)$. Recall that $\mathcal{L}$ has rank $n-1$. Since $\text{Skew}_n(\mathbb{R})$ is a vector space of dimension $d_1 := n(n-1)/2$, we find that

$$ d_2 := \dim(\ker \mathcal{L}) = \frac{n(n-1)}{2} - (n-1) = \frac{(n-1)(n-2)}{2}. $$

Consider the matrices

$$ P_{ij} := E_{1i} + E_{ij} + E_{j1} = E_{1i} - E_{ij} + E_{ij}, \qquad (C.5) $$

where $E_{ij}$ are the skew symmetric matrices in (B.2). Then

$$ \mathcal{L}(P_{ij}) = \mathcal{L}(E_{1i}) + \mathcal{L}(E_{ij}) + \mathcal{L}(E_{j1}) = (e_1 - e_j) + (e_i - e_j) + (e_j - e_1) = 0. $$

Hence, the subset

$$ \mathcal{P} := \{ P_{ij} : 2 \leq i < j \leq n \} \subseteq \ker \mathcal{L} $$
consists of $(n-1)(n-2)/2$ different elements, which can be ordered according to the lexicographic order of the pairs $(i, j)$. Let us show that they are linearly independent, thus generating the whole kernel. We first consider the basis of skew-symmetric matrices

$$ \mathcal{B} := \{E_{ij} : 1 \le i < j \le n\}, $$

and again we arrange them in lexicographic order. Let $\mathcal{M} \in \mathcal{M}_{d_2 \times d_1}(\mathbb{R})$ be the matrix of coordinates of the elements in $\mathcal{P}$ with respect to the basis $\mathcal{B}$. Then by (C.5) one infers that the $d_2 \times d_2$ identity matrix appears as the submatrix of $\mathcal{M}$ consisting of all the $d_2$ rows but just the last $d_2$ columns. Hence, rank $\mathcal{M} = d_2$ and consequently

$$ \ker \mathcal{L} = \operatorname{span}(\mathcal{P}). $$

• Step 2. Here, we characterize the condition (C.4), which clearly amounts to

$$ \sum_{i,j=1}^{n} p_{ij} m_{ij} = 0 $$

for every $P \in \mathcal{P}$. Taking $P = P_{ij}$ for $2 \le i < j \le n$ we get

$$ \sum_{i,j=1}^{n} p_{ij} m_{ij} = \frac{1}{2}(m_{1i} - m_{i1} + m_{ij} - m_{ji} + m_{j1} - m_{1j}) = m_{1i} + m_{ij} + m_{j1}, $$

and this concludes the proof.

<figure><img src="image_12.png" /></figure>

Finally, we focus on the sticking condition (3.11) in the critical case. The next result gives an explicit characterization that uses similar techniques to those in Lemma B.3.

**Lemma C.2.** For any $n \in \mathbb{N}$ and $M \in \operatorname{Skew}_n(\mathbb{R})$, the following assertions are equivalent:

(1) There exists $Y \in \operatorname{Skew}_n([-1, 1])$ such that $M = Y \cdot J + J \cdot Y$.

(2) There exists $Y \in \operatorname{Skew}_n(\mathbb{R})$ such that

$$ T_{ij}(Y) \le m_{ij}, \quad -T_{ij}(Y) \le -m_{ij} \quad \text{and} \quad L_{ij}(Y) \le 1. $$

(3) We have

$$ \sum_{i,j=1}^{n} q_{ij} + \frac{1}{2} \sum_{i,j=1}^{n} p_{ij} m_{ij} \ge 0 $$

for any $i, j = 1, \dots, n$, and for every $P \in \operatorname{Skew}_n(\mathbb{R})$ and $Q \in \mathcal{M}_n(\mathbb{R}_0^+)$ such that $\sum_{k=1}^n (p_{ik} - p_{jk}) + q_{ij} - q_{ji} = 0$.

(4) The following two conditions hold:

(a) Condition (C.2) holds true.

(b) We have

$$ \sum_{i=1}^{m} \sum_{j=m+1}^{n} m_{\sigma_i \sigma_j} \in [-nm(n-m), nm(n-m)] \quad (C.6) $$

for every permutation $\sigma$ of $\{1, \dots, n\}$ and any $1 \le m \le n$.
*Proof.* Assertions (1) and (2) are apparently equivalent due to the definition of the linear operators involved. Also, (2) and (3) are equivalent by an application of Lemma B.2 that is analogous to that in the proof of Lemma B.3; hence, we skip the proof for simplicity. Therefore, we will only focus on the equivalence with the last assertion. First, assume that for some $Y \in \text{Skew}_n([-1, 1])$ the first assertion holds true, i.e.,

$$m_{ij} = \sum_{k=1}^{n} (y_{ik} - y_{jk}).$$

By Lemma C.1 we arrive at (C.2). Moreover,

$$ \begin{aligned} \sum_{i=1}^{m} \sum_{j=m+1}^{n} m_{\sigma_i \sigma_j} &= \sum_{i=1}^{m} \sum_{j=m+1}^{n} \sum_{k=1}^{n} (y_{\sigma_i \sigma_k} - y_{\sigma_j \sigma_k}) \\ &= (n-m) \sum_{i=1}^{m} \sum_{k=m+1}^{n} y_{\sigma_i \sigma_k} - m \sum_{j=m+1}^{n} \sum_{k=1}^{m} y_{jk} = n \sum_{i=1}^{m} \sum_{k=m+1}^{n} y_{ik}. \end{aligned} $$

Since it is $n$ times the sum of $m(n-m)$ numbers in $[-1, 1]$, the condition (C.6) is also satisfied. Conversely, assume that both (C.2) and (C.6) hold and take any $P \in \text{Skew}_n(\mathbb{R})$ and $Q \in \mathcal{M}_n(\mathbb{R}_0^+)$ such that

$$ \sum_{k=1}^{n} (p_{ik} - p_{jk}) + q_{ij} - q_{ji} = 0 \qquad (C.7) $$

for any $i, j = 1, \dots, n$. Without loss of generality we can assume that $q_{ii} = 0$ for every $i = 1, \dots, n$. Also, define $\lambda_i := \sum_{k=1}^{n} p_{ik}$ and consider a permutation $\sigma$ of $\{1, \dots, n\}$ such that the $\lambda_{\sigma_i}$ are in non-decreasing order,

$$ \lambda_{\sigma_1} \le \dots \le \lambda_{\sigma_n}. \qquad (C.8) $$

Let us split

$$ I := \sum_{i,j=1}^{n} q_{\sigma_i \sigma_j} + \frac{1}{2} \sum_{i,j=1}^{n} p_{\sigma_i \sigma_j} m_{\sigma_i \sigma_j} =: I_1 + I_2. $$

Using (C.2) in the second term we can write

$$ I_2 = \frac{1}{2} \sum_{i,j=1}^{n} p_{\sigma_i \sigma_j} (m_{\sigma_i \sigma_k} - m_{\sigma_j \sigma_k}) $$

for any $k = 1, \dots, n$. Let us take the average with respect to $k$ in the above expression:

$$ \begin{aligned} I_2 &= \frac{1}{2n} \sum_{i=1}^{n} \left( \sum_{k=1}^{n} m_{\sigma_i \sigma_k} \right) \lambda_{\sigma_i} + \frac{1}{2n} \sum_{j=1}^{n} \left( \sum_{k=1}^{n} m_{\sigma_j \sigma_k} \right) \lambda_{\sigma_j} \\ &= \frac{1}{n} \sum_{i=1}^{n} \left( \sum_{k=1}^{n} m_{\sigma_i \sigma_k} \right) \lambda_{\sigma_i} = \frac{1}{n} \sum_{i=1}^{n} \left( \sum_{k=1}^{n} m_{\sigma_i \sigma_k} \right) (\lambda_{\sigma_j} + q_{\sigma_j \sigma_i} - q_{\sigma_i \sigma_j}) \end{aligned} $$
for any $j = 1, \dots, n$, where (C.7) has been used in the last step. Taking the average with respect to $j$ we get

$$ 
\begin{aligned}
I_2 &= \frac{1}{n^2} \sum_{i,j=1}^{n} \left( \sum_{k=1}^{n} m_{\sigma_i \sigma_k} \right) (q_{\sigma_j \sigma_i} - q_{\sigma_i \sigma_j}) \\
&= \frac{1}{2n^2} \sum_{i,j=1}^{n} \left( \sum_{k=1}^{n} (m_{\sigma_i \sigma_k} - m_{\sigma_j \sigma_k}) \right) (q_{\sigma_j \sigma_i} - q_{\sigma_i \sigma_j}) \\
&= \frac{1}{n^2} \sum_{i<j} \left( \sum_{k=1}^{n} (m_{\sigma_i \sigma_k} - m_{\sigma_j \sigma_k}) \right) (q_{\sigma_j \sigma_i} - q_{\sigma_i \sigma_j}).
\end{aligned} 
\quad (C.9) $$

On the other hand,

$$ I_2 = \sum_{j>i} q_{\sigma_i \sigma_j} + \sum_{i<j} (q_{\sigma_i \sigma_j} - q_{\sigma_j \sigma_i}). \quad (C.10) $$

Putting (C.9)–(C.10) together we obtain

$$ I = 2 \sum_{j>i} q_{ij} + \sum_{i<j} \left( 1 - \frac{1}{n^2} \sum_{k=1}^{n} (m_{\sigma_i \sigma_k} - m_{\sigma_j \sigma_k}) \right) (q_{\sigma_i \sigma_j} - q_{\sigma_j \sigma_i}). $$

Finally, notice that for any $i < j$, the condition (C.7) entails

$$ q_{\sigma_i \sigma_j} - q_{\sigma_j \sigma_i} = \sum_{m=i}^{j-1} (q_{\sigma_m \sigma_{m+1}} - q_{\sigma_{m+1} \sigma_m}), $$

and consequently

$$ I = 2 \sum_{j>i} q_{ij} + \sum_{k=1}^{n} a_m (q_{\sigma_m \sigma_{m+1}} - q_{\sigma_{m+1} \sigma_m}), $$

where the coefficients are

$$ 
\begin{aligned}
a_m &= \sum_{i=1}^{m} \sum_{j=m+1}^{n} \left( 1 - \frac{1}{n^2} \sum_{k=1}^{n} (m_{\sigma_i \sigma_k} - m_{\sigma_j \sigma_k}) \right) \\
&= m(n-m) - \frac{1}{n} \sum_{i=1}^{m} \sum_{j=m+1}^{n} m_{\sigma_i \sigma_j}.
\end{aligned} $$

Here, (C.2) has been used again. Since the $a_m$ are all non-negative by (C.6) and the $\lambda_{\sigma_i}$ are ordered as in (C.8), we conclude that $I \ge 0$ and this ends the proof. ■
*Funding.* The work of J. Park was supported by the Basic Research Program through the National Research Foundation of Korea (NRF) funded by the MSIT(NRF-2020R1A4A3079066). The work of D. Poyato (D.P.) and J. Soler (J.S.) has received funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme (grant agreement No. 639638), the MECD (Spain) research grant FPU14/06304 (D.P), the MINECO-Feder (Spain) research grant number RTI2018-098850-B-I00, the Junta de Andalucía (Spain) Projects PY18-RT-2422 & A-FQM-311-UGR18 (D.P., J.S.).

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