In the intro we say "But for other homogeneous spaces, inverting \varphi_x may have no relationship with any DLP, and may potentially be harder to solve, even for a quantum computer". But Couveignes notes that we can always use BSGS under certain conditions on G and X, which I think are met in the isogeny paradigm, and which I'm going to wildly generalize and say are probably met whenever X is a cryptographically useful HHS, too. So the classical difficulty is at most O(\sqrt{#X}) time and space, which isn't harder than the DLP. Classic isogeny-finding algorithms are just doing the Pollard version of this, I suppose. My point is that
Maybe HHS inversion is indeed "harder to solve" than the DLP on a quantum computer, but
It can't be harder to solve in an absolute sense.
I can't think of a way of expressing this in the intro without everything spiralling out of control, but we should make sure we've fixed this statement before submitting.
In the intro we say "But for other homogeneous spaces, inverting \varphi_x may have no relationship with any DLP, and may potentially be harder to solve, even for a quantum computer". But Couveignes notes that we can always use BSGS under certain conditions on G and X, which I think are met in the isogeny paradigm, and which I'm going to wildly generalize and say are probably met whenever X is a cryptographically useful HHS, too. So the classical difficulty is at most O(\sqrt{#X}) time and space, which isn't harder than the DLP. Classic isogeny-finding algorithms are just doing the Pollard version of this, I suppose. My point is that
I can't think of a way of expressing this in the intro without everything spiralling out of control, but we should make sure we've fixed this statement before submitting.