Everyone knows (citation needed) that (at fixed rim diameter) tyres with smaller section require less effort to move around (at least on a paved road).
TL; DR. Why?
A bit of context.
I always thought this phenomenon is due to the size of the contact patch between each tyre and the road. Notably, the area of this patch decreases when you decrease the tyre cross section. [This assumes that a tyre with smaller cross section will have a higher minimum pressure, which is generally true]. Friction is proportional to the contact area, QED.
However, I recently considered that friction (as almost everyone knows) would generate a torque that causes the wheel to spin faster [note 1]. So I had to come up with an alternative explanation. My best attempt is this. If tyre A has a smaller cross section than tyre B, tyre A will have a higher manufacturer pressure than tyre B, and therefore will be less subject to deforming. In terms of conservation of energy, deforming a tyre continuously requires a fair amount of energy, and it's here that our precious kinetic energy goes when we stop pedalling and our expensive toys (or in my actual case, inexpensive toy) come to a sad halt. So tyre A sucks less energy than tyre B, and therefore requires less effort. How can we put this in terms of forces? There must be an asymmetry in the forces near the contact point, causing a torque that slows down the spin of the wheel. Can you describe this asymmetry?
[Note 1. Friction is a force applied at the contact patch, with direction opposite to the direction of motion. Therefore, friction generates a torque that spins the wheel faster. For instance, in absence of any friction, a wheel would slide seamlessly without rolling.]
[EDIT. I apolgise for the sloppy formulation of the question. I edited to add clarifications where necessary. I opted for adding text rather than removing because some comments would otherwise look out of place]