where is the closed path in tesla coil?
For DC circuits, we need a closed path (complete circuit) in order to create a circular current.   If the path is open, then energy cannot be transferred.
And for low frequency AC circuits ...same thing.   At 60Hz, circuitry is much like DC systems, but the polarity keeps reversing.   A closed path is like a flywheel, or like a drive belt.  An open circuit path is like a drive belt with a brake being set.   If one part of the &quot;drive belt&quot; is nailed down and cannot move, then there can be no current in the entire loop.
At high frequencies the rules are very different.   Analogy:  with our flywheel or drive-belt with-a-brake, we can't move the belt, even if we yank back and forth  ...yet sound waves easily travel along the belt!  After all, sound is just AC belt-motion.  If we yank at constant DC, or even at 60Hz, the belt can't move, but instead if we tried to yank on the belt 1000 times per second, it would move quite a bit.
To create currents in open circuits, besides employing high frequency, instead we could also yank really hard, in other words, use a high-voltage system.  Even with the &quot;brakes&quot; applied, our flywheel (or drive-belt) does move significantly when the forces are enormous.   So, high-voltage and high-frequency will act like high-pressure sound waves at high wattage.   The &quot;brake&quot; cannot stop the vibration passing along the rest of the circuit.  Unlike with DC, an open circuit acts compressible.
In detail:  the space around a circuit always acts like a capacitor.   But for low voltage DC, and for low-freq AC, the capacitance is totally insignificant (a few picofarads.)  For 100VAC at 60Hz, the magnitude of current in the dielectric of the wire is roughly I=V*(2piFC),  I = 100(2pi60*1e-11) = 0.4 microamps.     If you plug in a piece of lamp-cord to 120VAC socket, but with no lamp attached, it will draw less than a microamp.
Now try the same for a typical tesla coil: 100 kilovolts, and 100kHz  I = 1e5*(2pi1e5*1e-11) = 0.63 amps!    The current in the surrounding dielectric is quite enormous.  We can't just ignore it like we did for DC and 60Hz circuitry.
So, answer:  the Tesla Coil secondary is like a large capacitor plate, and the surface of the Earth is the other plate, as a capacitor of a few tens of picofarads.  We're all standing inside the capacitor's dielectric!   Install a resistor in series with the metal conductors (such as a glass globe full of low-pressure argon,) and we'll find a significant current through the gas, and quite large voltage across it.  The gas will break down and glow.   Choose the right resistor (by varying the gas pressure,) and the wattage of our wireless lamp could be as much as 100kV * 0.63A = 63 kilowatts!   Well, less than that, for z-match max power transfer.  Maybe ~30kW at most.  (grin.)
In other words, when frequency is high and voltage is large, then meters-thick capacitor-dielectrics behave as good conductors.  An open switch, if it has a couple of picofarads across it, behaves as a closed switch for Tesla coil signals.   With a Tesla Coil as a power supply, we could build entire circuits out of ceramic rods (try barium titanate, or perhaps PZT.)    Or, hang your wireless lighting-globe near the TC terminal, and the few picofarads of empty space on either side of the lamp will provide a good power-line for completing the circuit and lighting the lamp.