What's a very simple example of an equivariant map?

Question

If we take $S^1$ acting on $S^2$ by rotation, then the height function $h: S^2\to R$ is an example of an invariant map (or an equivariant map where the action on $R$ is the trivial one).

I'm looking for an easy example of an equivariant map which is as simple as this one (and which can be visualized).

Answer 1

If you dig straight down into the Earth, where do you end up? What if we take into account the Earth's rotation?

A silly example of a G-equivariant map f from X to Y takes X=Y={ (x,y) : x,y in R s.t. x²+y² = 1 } to be the unit circle, and G to be the full group of rotations (and reflections, if you'd like) of X. Taking f(x,y) = (−x,−y) to be the "antipodal" or opposite point of x turns out to be G-equivariant, since the "opposite" of a point is defined intrinsically. This f is not invariant, but is equivariant.

You can do the same thing for any ball or sphere, and G can be any subgroup of the isometry group. For instance take X=Y=S² to be the sphere, and take G to be a group of rotations with a common axis of rotation. If you want to know where you get after drilling straight down into the earth, you could ask before the Earth rotates or after, but it wouldn't really matter, since the opposite point rotates at the same rate.

Answer 2

It is very easy to come up with those. A few examples:

• The function which maps each $k$-tuple of linearly independent vectors in $\mathbb C^n$ to the $k$-dimensional subspace they span is equivariant with respect to the obvious actions of $\mathrm{GL}(\mathbb C^n)$ on its domain and codomain;

• The function mapping each triangle in $\mathbb R^2$ to its area, a real number, is equivariant with respect to the action of $\mathrm{GL}(\mathbb R^2)$ acting in the obvious way on the domain, and by multiplication by the determinant on the codomain;

• the function which takes a finite word in the letters $a$, $b$ and returns the word with the same letters but with all $a$s put before all the $b$s, is equivariant with respect to the action of $\mathbb Z_2$ on the domain by interchaning $a$s and $b$s in each word, and on the codomain, by interchanging $a$s and $b$s and reflecting the word backwards.

• the function which maps vertices to the diagonal to which it belongs is equivariant for the group of symmetries of a cube acting in the obvious way on the domain and codomain of the function.