Alan ... in wa2ise's simulation, once you account for the fact that you have 3 lights operating, and figure it in turns of real power used, a diode will in fact economize, even in a three phase system. You can only get so much power to be dissipated by each load, and for resistive loads, the diode cuts that in half. Actually, it's probably just a bit more than half because the diode won't start conducting immediately when the zero crossover transitions to forward voltage (resulting in less overlap and more current diverted to the neutral), and because of the lower operating temperature of the filaments resulting in more current.
The problem is that because of the timings of the intermittent (half cycle) current, much or most of the return current goes by way of the same shared common neutral conductor. If the diodes were to conduct only during the center 66% of the half cycle (33% of the whole cycle), then at no time could any line-to-line current flow happen, and all of the current will be forced through the neutral. Regular dimmers can cause this problem, too. Because regular dimmers do (there may be a few cheap ones that don't) conduct in both directions, the dimmer would have to be adjusted so it is conducting only 33% of the time for this effect to be at its worst.
So if you have a multi-wire branch circuit on a common neutral, protected at 20 amps (North American standards), with 15 amps of load (for example, 30 60-watt incandescent bulbs) on each phase, you would expect essentially zero current on the neutral (ignoring slight bulb wattage differences due to manufacturing variations). Now put these diode type power reducers in place on each bulb, all in the same polarity. Ignoring the effects of the reduced filament temperature (which would lower resistance and raise the current, slightly), you now have 7.5 amps of load on each phase. During the highest part of the conducting cycle of each phase, neither of the other phases are conducting. This might be as little as 33% of the conducting time, but it is during the current peaks, so it will be most of the total averaged current. This could be an average of say, 6 amps per phase. This 6 amps conducts through the respective phase and the neutral. And this happens in sequence for all three phases, giving a net average on the neutral of 18 amps (at 180 Hz in North America).
That's at least within the designed in tolerance-error forgiving deratings that NEC requires. But, that narrows the forgivable tolerance errors in other aspects (for example, the 60-watt bulbs might actually be 67 watts, due to variations in manufacturing that batch they all came from).
But someone might be thinking "I now only have 7.5 amps of load on each phase of this MWBC ... so I can double up and join another branch in with it and still be well within ratings". So just double all the figures I gave above and that someone would end up with 36 amps on the neutral which is likely to be AWG #12 because it is a 20 amp circuit. Don't forget I-squared-R means you are dissipating 4+ times as much power in that neutral as it should be in a normal worst case (all loads within rating on one phase). Each phase conductor will still be around 15 to 16 amps, which should never trip the breaker. So the 36 amp neutral current will continue at that level until something bad happens.
Three phase power can easily be a problem with non-linear harmonic loads like this. But the diodes operating in half-wave mode (conducting in one direction only) can also pose a similar risk to two pole single phase circuits with a shared neutral. For that to happen, (most of) the diodes would have to be polarized one way on phase A and the other way on phase B. So with 30 60-watt bulbs on each pole of a two-wire shared neutral circuit, these diodes would cause all the current on the neutral (because there's zero conduction overlap time between poles at 180 degrees). Ignoring the slight rise in current due to the lower filament temperatures, that's the same as the worse case (all loads on one pole) and not yet a hazard. But the apparent reduced load on each pole could lead people to double up the loads, resulting in double the current (30 amps) on the neutral. It would be almost as bad as the three phase scenario.
The single phase scenario would have very little DC current on the neutral. if both poles are equivalent windings on the same transformer core, then the effect should still be an alternating magnetic field. But if the 120/240 system is derived from two separate 120 volt transformers in series, each transformer will be dealing with DC current. They won't be happy. Fortunately in North America, a setup like that is unheard of because 120/240 transformers are "everywhere", so the primary hazard is someone doubling up the neutral.
The three phase scenario will have lots of DC current. Making sure that half the diodes are one way, and half the other way, on every phase, should at least eliminate the direct current issue. I'd just avoid these things on a big scale. They could be OK for a few lights. But a smaller wattage bulb would still do better (more efficient in terms of usable light per watt because they would be running at design temperature). If you have a large scale lighting system and need to save power, you should already be doing something other than incandescent.
See NEC 210.4(A)FPN and 310.10 FPN(2) and of course 310.15(B)(4)(c).
You can run the calculations yourself for European standards on a 400/230 volt three phase system and a rare 460/230 volt 2-pole single phase system. A single pole single phase system won't have the same kinds of problem. It can still pose problems to the source transformer, especially if it is three phase and customers on other phases are doing the same thing, in significant amount.
