I was wondering about this question since I learned about rolling motion in the chapter on rotational mechanics. I was unable to come to a solid conclusion due to the reasons mentioned below.
The following diagram shows a ball on a frictionless inclined plane and the forces acting on it:
The forces acting on the ball are shown in red and are the normal contact force $N$ and the gravitational force of attraction $mg$. I qualitatively determined the torque of these forces about two axes - one passing through the centre of mass of the ball of uniform density, and the other passing through the point of contact of the ball and the inclined plane. Both of these axes are perpendicular to the screen.
When the axis passes through the centre of the ball, the torque exerted by $mg$ is zero as its line of action meets the axis. Further, the torque exerted by $N$ is also zero due to the same reason. There are no other forces. So, net torque about this axis is zero, and this tempts us to conclude the ball slides down the inclined plane.
When the axis passes through the point of contact, the torque exerted by $N$ is zero but the torque exerted by $mg$ is non-zero. This means the ball must roll i.e., it rotates while moving down the inclined plane. This conclusion is contradictory to the former case.
So, what exactly will happen to a ball kept on a frictionless inclined plane - will it slide or roll?
The following diagram is a visual interpretation of my question (if the terms slide and roll confuses the reader) where the red arrow denotes the orientation of the ball:
Image Courtesy: My own work :)
Please Note: The question - Ball Rolling Down An Inclined Plane - Where does the torque come from? discusses the case of ball rolling on an inclined plane where friction is present. Since the question - Rolling in smooth inclined plane is marked as duplicate of the former, and has no sufficient details, I planned to ask a new question with additional information.