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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).

slackenerny
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    I suppose you do not want to be shown identity maps? – Mariano Suárez-Álvarez Feb 02 '11 at 08:10
  • For what purpose? I'm not sure I understand. There are, of course, lots of equivariant maps in geometry; this idea goes all the way back to Klein's Erlangen program. – Qiaochu Yuan Feb 02 '11 at 11:55
  • @Mariano: you supposed correctly. – slackenerny Feb 02 '11 at 12:46
  • Strictly speaking, the height function is an invariant map, i.e. $h(gx) = h(x)$. An equivariant map would be something where the group acts on the codomain as well, $h(gx) = gh(x)$. (Though you could choose the trivial action of course, where the elements of the codomain are left fixed.) – Greg Graviton Feb 02 '11 at 18:20

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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.

Mariano Suárez-Álvarez
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  • Thanks for those. But can't we find something simpler? I mean, $S^2$ is just a ball and the action of the circle just spins that ball around. I guess my difficulty with your examples lies in imagining the domain in its entirety (or the group, or the codomain for that matter :P) – slackenerny Feb 02 '11 at 12:49
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    Instead of looking for what you call simple, you shoul probably try to get used to dealing with examples you can't immediately imagine: it is the only way to eventually being able to imagine them too! – Mariano Suárez-Álvarez Feb 02 '11 at 16:57
  • Sure, I agree (I am known for being a slacker). I just find it disappointing not being able to come up with an easy example. – slackenerny Feb 02 '11 at 21:47
  • @slackenemy: I don't understand what you consider an easy example. Another: 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. One can go on forever... – Mariano Suárez-Álvarez Feb 02 '11 at 22:01
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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.

Jack Schmidt
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  • thanks, that's the sort of example I was looking for. – slackenerny Feb 04 '11 at 11:49
  • No problem. I think it might even be the case that this is the only (full isometry group) G-equivariant map that is not just the identity map. If you take the group of rotations with common axis through x, then x is fixed by all of them, so f(x) has to be fixed by all of them. So f(x) is either x or -x each time. Since G acts transitively, I think either f(x)=x for every x, or f(x)=-x for every x. For S^1 the argument is different, but I think the answer is the same. For S^1 you have to include reflections in G, and for S^n you have to allow lots of axes of rotation. – Jack Schmidt Feb 04 '11 at 17:39
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I just came across this and I have a really simple example.

Let $f: S^2 \rightarrow S^2$ be a reflection i.e $f(x,y,z)=(-x,y,z)$. Then $f$ is clearly $S^1$-equivariant where the group action is by conjugation.

Hesky Cee
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