constant total power out, just like constant DC power into the inverter so no need for storage capacitors / inductors.

Whoa, you're right! Assuming a perfect power factor and perfect balancing on the loads, of course. Your four-phase scheme is very ingenious! I'm sorry I didn't appreciate this at first.

But here's a thing I'm not understanding: yes, the powers sum to a constant. But the way an inverter produces real sinewave power (as opposed to modified square wave) is to PWM an H-bridge to ramp the voltage up and down, filtering it with an inductor (and usually a capacitor, and maybe more than one of each, but those are inessential). An inductor's time integral of voltage, and thus its average voltage drop (disregarding losses to winding resistance, hysteresis, eddy currents, etc.), must be zero to keep its current finite. So the output side of the inductor has the same average voltage as its input side. So, for example, a 25% PWM duty cycle produces 25% of the full-scale output voltage.

However, in your four-phase scheme, when one of the phases (let's say in) is at ±25% output voltage and thus 6.25% peak output power, the other (qu) is at 93.75% output power and thus ±96.8% peak output voltage. This implies that, during that part of the wave, the active in MOSFET needs to be on 25% of the time, while the active qu MOSFET needs to be on 96.8% of the time. That means that between 18.7% and 25% of the time, both MOSFETs are on, so the two phases are actually shorted to each other (though not on the load side of the inductors). Is that okay? I guess it means that the current generated by the solar cells is being shared between the two phases during that time.

Still, it makes me worry that an imbalance of loads or power factors between the four phases could produce some kind of hazardous condition.

You say that the four-phase system doesn't need any energy-storage elements. To a first approximation you're right, again assuming well-balanced, power-factor-corrected loads on the phases: there's no need to store energy harvested close to the zero-crossing for, assuming 50 Hz, the average 5 milliseconds until it can be released close to the peak. 5 milliseconds is a long time, so these storage or filtering elements need to be quite large, 5 millijoules per watt (I think you dropped a factor of ½ in your calculation there, presumably calculating for a full half-cycle instead of a quarter cycle). By contrast, if you're PWMing a sine wave with a PWM frequency of 100 kHz --- a reasonable thing to do with modern IGBTs or power MOSFETs --- the filtering elements only need to store the energy for a maximum of 10 microseconds, and less if the PWM duty cycle is above the minimum. 10 microseconds is 10 microjoules per watt, 500 times smaller. So you only need 0.2% of the energy storage elements you need for what you're describing as typical current systems.

(All that happens if your power factors or loads are imbalanced is that the inverter isn't drawing a consistent amount of current, so potentially the panels become less efficient.)

A thing I'm not sure about is the junction capacitance of photovoltaic cells. A photovoltaic cell is a diode, reverse-biased in normal use, and reverse-biased diodes have a junction capacitance; we'd expect their enormous junctions to have significantly larger junction capacitance than the pF-scale capacitances we see in small-signal diodes. How large is the ½CV² = 50 nJ/microfarad (at V=0.316 V) energy-storage capacity of the PV panel itself? In particular, is it much larger or much smaller than the 5 mJ/W = 5 ms number you'd need for a single-phase inverter to not be wasteful?

As for your questions about the current designs of power-generation inverters and residential inverters, no, I don't know about them. Typically, in both the US and here in Argentina, one side of a normal power outlet is "neutral" (see http://amasci.com/amateur/whygnd.html for the reasons around this) and violating that expectation might cause some problems --- notably, electric shocks from the outside of Edison-screw-type lightbulb sockets.

However, I don't think you need to violate that expectation; you've described a four-phase system, but I think you get the same advantages with a two-phase system, with the phases in quadrature just as you proposed, but with the two phases sharing a neutral wire. The voltage from neutral ("ground") to the in terminal would be V sin(ωt), much as before, while the qu terminal would be at V sin(ωt + ½π), which is to say, V cos(ωt) --- both relative to the neutral wire. In effect you need only a single H-bridge with four MOSFETs (or IGBTs) and two inductors to produce the two voltages, controlled with a scheme slightly different from the usual H-bridge scheme, because it makes sense to have one side of the H-bridge turned on (in, say) while the other side is turned off.