There are several possibilities when talking about a trigger that does not fully return when releasing it fully after firing.
When the trigger releases the hammer, it immediately flies forward, detonating the cartridge. It goes all the way forward, and it's face is actually resting on the frame. And, the firing pin is sticking into the cylinder area. And, the hammer block is fully retreated into the side plate. And he hand has to be drawn down, and the cylinder stop reset, and the hammer block slid into place. All of this has to be corrected upon release of the trigger.
As the trigger is released, the trigger rebound slide moves forward. At some point, the bump on the top of the rebound slide contacts the bottom of the hammer, causing it to start to retreat, pulling the firing pin out of the chamber. Simultaneously, as the trigger is being released, it has to push past the cylinder stop. It does this not by droping the cylinder stop, but by pushing it forwards. At some point, the pushing is done, and the cylinder stop moves backward, but not downward or upward. As the trigger is being released, it pulls the hand down - it does not drop by itself. The hand itself has previously been pushed tight against the frame, by the hammer block plunger in the sideplate. The power necessary to pull the hand down is coming from the rebound slide spring. And lastly, the hammer block leaf spring has to completely rise up, letting the hammer block flange slide between the frame and the trigger.
So, a question is - when the trigger stops its retreat, what is the status of the lockwork? Has the firing pin completely retreated? Has the cylinder stop completed its forward movement, and come back to its normal position? If you pull just a bit on the trigger, does the firing pin start to come back into the cylinder area?
You can see that there is a lot of things happenings, that can contribute to a premature stop of the slow trigger retreat.
Study the action carefully of all this activity, and let us know what you think.
Regards, Mike