Question

Prove s² is a smooth manifold using a smooth A.

Answer 1

If A is a smooth manifold, then the square of A, denoted as s², is also a smooth manifold due to the preservation of smooth structures in the Cartesian product operation.

This conclusion stems from the property that the product of two smooth manifolds results in a smooth manifold, ensuring that the square of A maintains its smoothness when A itself is a smooth manifold.

Step-by-step explanation:

To establish the smoothness of s², where (A) is a smooth manifold, we leverage the property that the product of two smooth manifolds yields a smooth manifold. The square of (A), denoted as s², essentially corresponds to the product manifold
`\(A * A\)`. Given that (A) is smooth, the product manifold inherits its smooth structure, affirming that s² is indeed a smooth manifold.

This assertion is grounded in the foundational principles of differential geometry, where operations on smooth manifolds maintain their smoothness. By treating (A) as a smooth manifold, we extend this property to the Cartesian product
`\(A * A\)`, demonstrating that the square of (A) preserves its smooth nature.

Understanding such properties is pivotal in advanced mathematical contexts, especially in differential geometry, where the manipulation of smooth manifolds plays a central role in various mathematical analyses.