This is, indeed, an energy dissipation problem. The motor has kinetic energy. Newton says that without some opposing force being applied, the motor will spin forever.
I've already mentioned some of the forces that are contributing to the motor stopping: bearing friction, windage (different source of friction), PN losses, transmission losses, etc. If the motor has more energy than these losses can account for, then its energy will be converted into something else. In this case, that "something else" is DC Link voltage due to the inverse-voltage protection diodes across the IGBT's that act as a rectifier when the motor regenerates electricity.
This energy is going to be fed into the capacitors on the DC Link bus. There is nowhere else for it to go. There is no way for the drive to stop or even otherwise mitigate the accumulation of this energy as voltage on the DC bus, except by giving it an alternate path, such as a braking resistor.
Let's go back to basic induction: When a magnetic field has relative motion (perpendicular) to a conductor, an electro-motive force (EMF) is produced in the conductor. The spinning motor has a magnetized rotor, and its stator windings are perpendicular to this rotating magnetic field, by design. This is going to produce an EMF in the Stator windings.
The Drive's input rectifier, assuming it is a diode converter front end, cannot allow this force to be passed to the incoming line. Therefore, the only place for the energy to go is the DC link. However, the DC link is being supplied by the incoming AC (rectified). Assuming a 460V drive, then you'll have an absolute maximum of about 650 VDC on the DC bus, assuming peak charging of the DC link capacitors.
The Motor's EMF is either going to be voltage or current, or a combination of the two. Since the motor is (for the first few milliseconds) not generating enough EMF to overcome the incomer line, the EMF becomes a voltage on the motor terminals and the drive output terminals. Once this voltage exceeds the level necessary to forward bias the IGBT-protection diodes, then current will flow from the motor onto the DC link. This regenerated motor voltage MUST exceed 460 VAC in order to forward bias the inverter's diodes. This increased voltage will "pump up" the DC link voltage unless it is dissipated. That's what the chopper and braking resistor do; they dissipate the motor's regenerated energy in the form of heat.
If the drive is providing a "deceleration torque," then it is providing power to the motor. Torque times speed is power. Speed is changing (due to the deceleration), so you have torque and (delta) speed, which is power. This power is opposing the energy from the motor. Where does the motor energy go?
The only places for the energy from the motor to go are:
1. Back onto the line, to be used elsewhere. This is not the case, as the front end converter is not bi-directional.
2. Be converted into heat in the motor. This can be the case, as the drive can provide a constant DC to the stator windings. This DC will provide a high flux level, which provides only a braking torque to the rotor. The energy in the motor will be converted to heat in the rotor windings. (See Lens' Law or Flux braking on an internet search for more information.) Any "switching" of the inverter devices to accomplish the negative torque will, likewise, cause some rotor heating.
3. Be converted into heat elsewhere, such as a braking resistor as already discussed.
4. Have the current path for the EMF interrupted. This will allow the rotor's magnetic field to dissipate, causing no more regenerated EMF, and therefore no more braking. The motor will coast to a stop with only friction losses causing it to slow. Any drive power supplied to the motor in this case will serve to re-magnetize the rotor, and give us the other three scenarios once again.
I see no other place for the motor's energy to go. Any other method of braking must account for the rotational energy's conversion. The friciton brakes that have been discussed convert the rotational energy into heat, either in the drum or the disk.