The exponents of the Ando-Hiai property for the Wasserstein mean

For positive definite matrices $A$ and $B$ and a weight $t \in (0,1)$, the Wasserstein mean $A \circ_t B$ minimizes $(1-t)d(X,A)^2 + t\,d(X,B)^2$ over positive definite $X$, where $d$ is the Bures-Wasserstein distance. The mean of weight $t$ has the Ando-Hiai property for an exponent $q > 0$ on $m \times m$ matrices if $A \circ_t B \le I$ implies $A^q \circ_t B^q \le I$ for all such $A$ and $B$. Gan and Kim asked whether $A^2 \circ_t B^2 \le I$ implies $A \circ_t B \le I$ for every $t$, which is the property for the exponent $1/2$. They also conjectured that $(A^p \circ_t B^p)^{1/p}$ decreases in weak log-majorization as $p$ decreases, where weak log-majorization compares the products of the $k$ largest eigenvalues for each $k$. We determine the set of exponents with the property for each $m$ and $t$. For $q \le 1$, the ratio of the largest eigenvalue of $A^q \circ_t B^q$ to the $q$th power of that of $A \circ_t B$ is at most $c$, where $c = (1-t)^{2-2q} + t^{2-2q}$ for $1/2 \le q \le 1$ and $c = 1$ for $q \le 1/2$, so every $q \le 1/2$ has the property. For $m \ge 3$ and $1/2 \le q < 1$ the ratio tends to $c$ along an explicit family of $m \times m$ pairs. The exponent $1$ has the property trivially, every $q > 1$ fails and $c > 1$ for $1/2 < q < 1$ so the set is $(0,1/2] \cup \{1\}$ for $m \ge 3$, whereas a separate argument gives $(0, 1]$ for $m \le 2$. Hence Gan and Kim's implication holds for every $t$, whereas they stated it only for weights that depend on $A$ and $B$. The conjecture fails at $k = 2$ for one $2 \times 2$ pair, and at $k = 1$ for $m \ge 3$ because its case $k = 1$ is the property for every $q \le 1$.