Geometric interpretation of symmetries of a manifold

Question

I have been doing some reading about Lie groups and algebras and stumble across the phrase that those theories "measure" the symmetries of the manifolds, like an analog of the groups that measure the symmetries of objects like a triangle, a polygon etc. I am trying to understand what that means. I think I would understand better the above with 2 examples

1. So lets take a 2-d manifold-a sphere. Does this mean that we have infinite symmetries? (how do we compute them mathematically meaning if someone ask how many symmetries a sphere has how do we answer that mathematically)
2. can you give an example of a manifold that has a finite symmetry

How can I think about the symmetries of a manifold? Thinking about an axes' rotation is probably wrong, so how do we think it geometrically?

Answer 1

The key phrase "those theories 'measure' the symmetry of the manifolds, like an analog of the groups that measure the symmetries of objects like a triangle, a polygon etc." isn't too clear.

Maybe you have in mind starting with something like the symmetries of a regular $n$-gon. That gives rise to the dihedral groups as an automorphism group. So now maybe you want to take a manifold, consider its automorphism group, and conclude that automorphism group has the additional structure of a Lie group? This program is subtle and depends on what sort of manifolds and automorphisms you're considering, e.g. see here.

A much simpler analogue is to instead start with a group $G$ and to think of each $g \in G$ as giving rise to the "symmetry" given by conjugation, $x \mapsto gxg^{-1}$, an inner automorphism of $G$. The same thing holds for Lie Groups. The simplest example to keep in mind is when $G = \mathrm{GL}_n(\mathbb{R})$, in which case every invertible matrix $A$ gives rise to a "smooth symmetry" $X \mapsto AXA^{-1}$. One often considers left translations in Lie groups, $X \mapsto AX$, which has much the same flavor at this level of specificity.

For your sphere example, you can think of it as a homogeneous space, for what that's worth.

Answer 2

It's essential to understand just what kind of structure you are looking at the symmetries of. If a smooth manifold M is given a Riemannian metric g, then it is a theorem of Myers and Steenrod that its isometry group (the group of self-diffeomorphisms of M that preserve its metric) is a Lie group. Call this Lie group Isom(M,g).

Indeed, all Lie groups whose dimension is positive contain a continuum of points, so they are certainly infinite.

For a circle, the only invariant of a Riemannian metric is its circumference. So its isometry group will be isomorphic to the group O(2), which topologically consists of two circles. (Think of the set of rotations by any angle, and the set of reflections about any line through the center of the circle.)

For a round sphere Sⁿ in Rⁿ⁺¹, its isometry group is called O(n+1) and has dimension equal to n(n+1)/2.

Then certainly the dimension dim(Isom(M,g)) is a "measure" of how symmetrical (M,g) is.

Among surfaces, a round sphere has an isometry group of dimension equal to 3. A flat torus (the quotient of the plane by a lattice, such as R²/Z², has an isometry group whose dimension is 2. A compact surface of genus greater than or equal to 2, on the other hand — no matter how symmetrical it is — cannot have an isometry group of positive dimension; instead its isometry group is a finite group.

Answer 3

When speaking of symmetries of a polygon, there's an implicit assumption that a symmetry is a Euclidean motion. When speaking of manifolds, it's crucial to be explicit about the structure being preserved.

A sphere, for example, may be viewed as a Riemannian manifold with a round metric, in which case a "symmetry" is an isometry, and the group of isometries is isomorphic to $O(3)$, the group of real orthogonal $3 \times 3$ matrices under matrix multiplication. If we fix an orientation in addition, the symmetry group is isomorphic to $SO(3)$.

A sphere may also be viewed as the Riemann sphere, equipped with a conformal structure and an orientation (equivalently, since we're on a real surface, a holomorphic structure), in which case a "symmetry" is a Möbius transformation and the group of symmetries is isomorphic to $PGL(2, \mathbf{C})$, the complex projective general linear group of fractional linear transformations.

Instead, one might equip the sphere with:

• A smooth structure and the area form of the round metric, in which case a "symmetry" is an area-preserving diffeomorphism;
• A smooth structure, in which case a "symmetry" is a diffeomorphism;
• A topological structure, in which case a "symmetry" is a homeomorphism;
• The structure of an uncountable set, in which case a "symmetry" is a bijection.

In these examples (which by no means exhaust all possibilities), there are infinitely many symmetries, so counting symmetries is not a distinguishing invariant. These symmetry groups are, however, mutually non-isomorphic.

In the other direction, we might equip the sphere with a Riemannian metric having the same symmetry as a polyhedron (think of the solution set of $x^{4} + y^{4} + z^{4} = 1$ with the metric induced from Euclidean three-space, a "rounded cube"), which would have finite symmetry group. The extreme here is a generic Riemannian metric, which has precisely one symmetry, the identity map.