Let $(X\_n)\_{n\geq 1}$ be a sequence of random variables defined on the $d-$simplex ($d\geq 1$) : $\Sigma\_d=\big\lbrace\boldsymbol{x}\in\mathbb{R}\_+^d,\,\sum\_{1\leq i\leq n} x\_i=1\big\rbrace$. Assuming that there exists $\alpha\in\Sigma\_d$ such that for $n\geq 1$, $\mathbb{E}[X\_n]=\alpha$, and that the sequence of covariance matrices $\text{Cov}(X\_n)$ converges to the matrix with coefficients $M\_{ij}=\alpha\_i(\delta\_{ij}-\alpha\_j)$ ($i,j\in \{1,...,d\}$ and $\delta\_{ij}$ is the Kronecker symbol), does the following weak convergence holds: $$\lim\limits\_{n\rightarrow\infty}\,\mathbb{P}\_{X\_n}\longrightarrow \sum\limits\_{1\leq i\leq d} \alpha\_i \delta\_{e\_i},$$
where $(e\_i)\_{1\leq i\leq d}$ is the canonical base of $\mathbb{R}^d$, and $\delta\_{x}$ the Dirac measure for $x\in\mathbb{R}^d$?


The specific case for which every $X\_n$ follow a Dirichlet law was solved: [Weak convergence of Dirichlet distributions to a "multi-Bernoulli" distribution](https://mathoverflow.net/questions/374314/weak-convergence-of-dirichlet-distributions-to-a-multi-bernoulli-distribution) .



Weak convergence to a "multi-Bernoulli" distribution
$\newcommand\Ga\Gamma\newcommand\R{\mathbb R}$This answer is similar to the one linked by the OP.


Indeed, let $a:=\alpha$ and $(X\_{n,1},\dots,X\_{n,d})$.


We have $EX\_{n,1}=a\_1$ and $Var\,X\_{n,1}\to(1-a\_1)a\_1$, whence $EX\_{n,1}^2\to a\_1$ and $E(1-X\_{n,1})X\_{n,1}\to0$. So, for each $t\in(0,1)$,
$$P(t\le X\_{n,1}\le1-t)=E1(t\le X\_{n,1}\le1-t)
\le\frac{E(1-X\_{n,1})X\_{n,1}}{(1-t)t}\to0$$
and hence
$$P(t\_{n,1}\le X\_{n,1}\le1-t\_{n,1})\to0\tag{1}$$
for sequence $(t\_{n,1})$ in $(0,1)$ converging to $0$. Passing to subsequences if needed, without loss of generality we may assume that $P(X\_{n,1}>1-t\_{n,1})\to p$ for some $p\in[0,1]$. Hence, by (1), $P(X\_{n,1}<t\_{n,1})\to1-p$, which implies
$$a\_1=EX\_{n,1}=EX\_{n,1}\,1(0\le X\_{n,1}<t\_{n,1})+EX\_{n,1}\,1(t\_{n,1}\le X\_{n,1}\le1-t\_{n,1})
+EX\_{n,1}\,1(1\ge X\_{n,1}>1-t\_{n,1})\to0+0+p,$$
so that $p=a\_1$ and $P(X\_{n,1}>1-t\_{n,1})\to a\_1$.


Similarly, for each $j\in[d]:=\{1,\dots,d\}$ and some sequence $(t\_{n,j})$ in $(0,1)$ converging to $0$,
$$P(X\_{n,j}>1-t\_{n,j})\to a\_j,$$
whence
$$P(X\_{n,j}\le 1-t\_{n,j}\ \forall j\in[d])\to1-\sum\_{j=1}^d a\_j=0.$$


So, for any continuous function $f\colon\R^d\to\R$,
$$Ef(X\_n)=\sum\_{j=1}^d Ef(X\_n)1(X\_{n,j}>1-t\_{n,j}) \\ 
+Ef(X)1(X\_{n,j}\le 1-t\_{n,j}\ \forall j\in[d]) \\ 
\to\sum\_{j=1}^d f(e\_j)a\_j+0,$$
where $e\_j$ is the $j$th standard basis vector of $\R^d$; here we used the implications $$X\_{n,j}>1-t\_{n,j}\iff1\ge X\_{n,j}>1-t\_{n,j}\\ 
\implies0\le X\_{n,i}<t\_{n,j}<1-t\_{n,i}\ \forall i\in[d]\setminus\{j\},$$
which hold for all large enough $n$; these implications imply that the events $\{X\_{n,j}>1-t\_{n,j}\}$ are pairwise disjoint in $j$.


Thus, your conjecture holds.


