# Is the decomposition $M = AA^T$ unique for $M,A \in \operatorname{SL}(n,\mathbb{Z})$?

I'm trying to extend the univariate discrete gaussian kernel $f(k|t)=e^{-t}I_k(t)$ to $\mathbb{Z}^d$. In analogy to the multivariate normal, let $X=AK + b$ where $K$ is a vector of iid discrete gaussian variables with unit variance. The pdf of $K$ is then $$P(K) = e^{-d}\prod_i^d I_{k_i}(1)$$ and it is a valid distribution on $\mathbb{Z}$. By substitution $$P(X) = e^{-d}\prod_i^d I_{[A^{-1}(X-b)]_i}(1)\equiv I(A^{-1}(X-b)).$$ Because $A$ and $A^{-1}$ must map $\mathbb{Z}\to\mathbb{Z}$, $\det{A} = 1$ such that $A\in \operatorname{SL}(d,\mathbb{Z})$ (note $P(K) = P(-K)$). Since $\operatorname{E}[K] = 0$ and assuming $$e^{-t}\sum_{k=-\infty}^{\infty}k^2I_k(t)=t$$ (I've only been able to verify this numerically), then $\operatorname{E}[X] = b$ and $$\Sigma_X=\operatorname{E}[(AK+b)(b^T+K^TA^T)-bb^T]=AA^T$$ just like the continuous case.

I'm curious if, like the continuous case, this distribution is uniquely defined by the matrix $\Sigma$ or whether it is instead defined by the unique matrix $A$. Hence the question of whether $M = AA^T$ is a unique decomposition on $\operatorname{SL}(d,\mathbb{Z})$.

EDIT: I realized that $M = AA^T = LL^T$ where $A\neq L$ is not unique on $\operatorname{GL}(d,\mathbb{R})$, which probably makes this question moot; there are several matrices $A$ that produce random vectors from a given distribution.