2Question: A mathematician studying algebraic topology explores symmetries of a surface modeled by a regular polygon with rotational symmetry of order $ n $. If the polygon has 12 sides, what is the sum of all distinct rotational symmetries (in degrees) that map the polygon onto itself?

To solve this problem, we begin by understanding that a regular polygon with 12 sides exhibits rotational symmetry. The rotational symmetry group of a regular $ n $-gon includes all rotations by multiples of $ \frac{360^\circ}{n} $ that map the polygon onto itself.

Given the polygon has 12 sides, the full rotational symmetry is generated by rotations of:

$$\frac{360^\circ}{12} = 30^\circ$$

Thus, the distinct rotational symmetries are:

$$0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ, 180^\circ, 210^\circ, 240^\circ, 270^\circ, 300^\circ, 330^\circ$$

There are 12 such angles, corresponding to the 12th roots of unity in the complex plane, and they form a cyclic group under composition. The sum of these angles can be computed as:

$$\sum_{k=0}^{11} \left(30k\right) = 30 \sum_{k=0}^{11} k = 30 \cdot \frac{11 \cdot 12}{2} = 30 \cdot 66 = 1980$$

Thus, the sum of all distinct rotational symmetries (in degrees) that map the 12-sided polygon onto itself is:

$$\boxed{1980}$$