Are Absolutely Maximally Entangled states maximally entangled under all entanglement monotones?

Question

In Ref. [1] absolutely maximally entangled (AME) states are defined as:

An $\textrm{AME}(n,d)$ state (absolutely maximally entangled state) of $n$ qudits of dimension $d$, $|\psi\rangle \in \mathbb{C}^{\otimes n}_d$, is a pure state for which every bipartition of the system into the sets $B$ and $A$, with $m = |B| \leq |A| = n − m$, is strictly maximally entangled such that $$ S(\rho_B) = m \log_2 d. $$

As the name would have you believe, does this mean that an $\textrm{AME}(n,d)$ state is maximally entangled across all entanglement monotones (for fixed $n$ and $d$)?

[1]: Helwig, Wolfram, et al. "Absolute maximal entanglement and quantum secret sharing." Physical Review A 86.5 (2012): 052335.

Answer 1

There is no a priori reason why an entanglement monotone should have this property. I think the particular issue is that, restricting to pure states, when you start to look at multiparticle states, you get inequivalent states. For example, for 3 qubits, the W state and GHZ state are inequivalent, and both are maximally entangled representatives for their SLOCC equivalence class (stochastic local operations and classical communication). There's no particular reason that an entanglement monotone should always favour GHZ (the AME(3,2)) over W-state. Indeed, they don't. Looking through the list of common multipartite entanglement monotones on Wikipedia, the one that stands out to me is the Schmidt Measure. It has value $\log_2(3)$ for a W state and one a value 1 for the GHZ state, so the AME is not maximally entangled by this measure.