Wait. My intuition says that the mass of ice cannot possibly have a gravitational influence on nearby water big enough to create the effects described. Back of envelope calculation: surface area of Antarctica is on the order of 10M km^2, with an average thickness of 1 km (a high estimate), means 10M km^3 of ice; some further calculations from volume to mass multiply out to 10^19 kg of ice. (Note that we're assuming that 1kg of ice is 1 cubic deciliter, as with liquid water; this is definitely wrong but let's go with it since this is just an order-of-magnitude estimate). The mass of the Earth is on the order of 10^24 kg, so this is 10^-5 of the mass of the earth.
Can you get really get 50m of local displacement from 10^-5 Earth masses?
Can you get really get 50m of local displacement from 10^-5 Earth masses?