## G-equivariant embeddings of manifolds

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This is my first post, and I plan on sporadically writing some in the future. I’m Peter, a third year grad student at Cornell, and I talk to Greg pretty often, so I thought I’d write down some of the things I say. This first post won’t be long or deep, but it’s kind of cute, and the trick behind it is useful in many other situations, so I decided to share it. Let’s say we have a finite group $G$ acting on a compact manifold $M$. The Whitney embedding theorem says that we can embed $M$ into $\mathbb{R}^k$ for sufficiently large $k$, and what I want to show in this post is that you can do this in a $G$-equivariant way, i.e. there is an embedding $\phi:M \to \mathbb{R}^k$ and an injective homomorphism $f:G \to O(\mathbb{R}^k)$ such that $\phi(g\cdot x) = f(g)\cdot \phi(x)$. I guess the moral of the story is that compact manifolds are really just “nice” subsets of Euclidean space, and a compact manifold with a finite group action is really nothing but a “nice” subset of Euclidean space that is preserved by the action of a finite group of permutation matrices. Also, if one were to summarize the moral of the trick used, one might say “averaging over the elements of a group makes things equivariant,” and this idea comes up almost uncountably many times in many areas of mathematics (of course, averaging takes on different meanings in different situations).

The notation for this is a little messy, but the idea isn’t too hard. First we define an embedding $M \to V$ and then we show that this embedding is $G$-equivariant. Pick a finite covering of $M$ by coordinate charts $\{U_i, \phi_i:U_i \to \mathbb{R}^d\}_{1 \leq i \leq r}$ and a partition of unity $\{\lambda_i:M \to \mathbb{R}\}_{1 \leq i \leq r}$ which is subordinate to the cover of $M$. To make the notation a little more clear later, we can break up the $\phi_i$ into their components $\phi_i = (\phi_{i,1},\ldots,\phi_{i,d})$. Now we let $e_{i,j,k}$ be the standard basis for $\mathbb{R}^{rnd}$ with $1 \leq i \leq r$, $1 \leq j \leq n = \left\vert G \right\vert$, and $1 \leq k \leq d = \dim M$. (To simplify things, I’ll use the same bounds for each index letter throughout the post.) We also let $e'_{i,j}$ be the standard basis for $\mathbb{R}^{rn}$. Now let $V = \mathbb{R}^{rn(d+1)}$ be the vector space spanned by $e_{i,j,k}$ and $e'_{i,j}$. Then we can define our embedding $\varphi:M \to V$ with the following formula:

$\varphi(x) = \sum_{i,j,k} \lambda_i(g_j(x))\phi_{i,k}(g_j(x)) e_{i,j,k} + \sum_{i,j}\lambda_i(g_j(x))e'_{i,j}$

(In this formula we are defining $\lambda_i\phi_i = 0$ if $\lambda_i = 0$ even if $\phi_i$ is undefined.) Since all the functions involved are smooth, this is obviously a smooth map, so to show it is an embedding it is sufficient to show it is a homeomorphism. Since $M$ is compact and its image is Hausdorff, it is sufficient to prove that $\varphi$ is one-to-one. Let $x,y \in M$ and suppose that $\varphi(x) = \varphi(y)$. Then since the $\lambda_i$ are a partition of unity, there is some $i$ such that $\lambda_i(x) \not= 0$. By letting $g_j = id_G$, we see that $\lambda_i(x) = \lambda_i(y)$, which shows that $x,y \in U_i$. Then from the definition of $\varphi$ we see that $\lambda_i(x)\phi_{i,k}(x) = \lambda_i(y)\phi_{i,k}(y)$ for all $k$. However, the $\phi_i$ are injective and $\lambda_i(x) = \lambda_i(y)$, so $x = y$. (By the way, this is exactly the same proof that is in Conlon’s book Differentiable Manifolds, just with the other group elements added in. If $g_{j_0} = id_G$, then you can take this embedding and project all the points onto the subspace spanned by $\{e_{i,j_0,k},e'_{i,j_0}\}$ and you will get his embedding.)

Now we want to show that there is an orthogonal action of $G$ on $V$ that makes the embedding $\varphi$ $G$-equivariant. This is intuitively obvious since we put the $g_j$ in the formula specifically for this purpose, but at first glance it seems like the action of $G$ on $V$ will be a right action, so I’ll be a bit pedantic just to be careful. (Tim Goldberg helped me out with this part. Until just now I’ve always been confused by the fact that acting on coordinates and acting on basis vectors are “dual” in the same sense that left and right actions are “dual”, and also in the same sense that acting with $g$ and $g^{-1}$ are “dual”.) Since we’ve fixed an ordering $g_1,\ldots,g_n$ of our group elements, given any $g \in G$ we know that $g_j g = g_{\sigma^{-1}_g(j)}$ for some permutation $\sigma \in S_n$. Then we can define a map $f:G \to S_n$ by $g \mapsto \sigma_g$, and the question is whether this is a homomorphism or an anti-homomorphism. Intuitively, we’ve “dualized” twice, once by multiplying on the right and the other by inverting, so it should be a homomorphism. Formally,
$\begin{array}{lll} g_{\sigma^{-1}_{gh}(j)} &=& g_j (gh)\\ &=& (g_j g)h\\ &=& (g_{\sigma^{-1}_g(j)})h\\ &=& g_{\sigma^{-1}_h(\sigma^{-1}_g(j))} \end{array}$
which shows that $\sigma^{-1}_{gh} = \sigma^{-1}_h(\sigma^{-1}_g)$. This is the same as $f(gh)^{-1} = (f(h)f(g))^{-1}$, so $f$ is a homomorphism. Now $S_n$ acts (on the left) on $V$ in the obvious way, by $e_{i,j,k} \mapsto e_{i,\sigma(j),k}$ and $e'_{i,j} \mapsto e'_{i,\sigma(j)}$. This gives a left action of $G$ on $V$, i.e. a homomorphism $f:G \to O(V)$.

Now let’s try to obtain the formula $\varphi(g \cdot x) = f(g)\cdot \varphi(x)$. To simplify the formulas, I’ll just show this for the second term in the formula (which I’ll write as $\varphi'$), which will make it obvious for the first term. We have
$\begin{array}{lll} \varphi'(g \cdot x) &=& \sum\lambda_i(g_j\cdot (g\cdot x))e'_{i,j}\\ &=& \sum\lambda_i((g_jg)\cdot x))e'_{i,j}\\ &=& \sum\lambda_i(g_{\sigma^{-1}_g(j)}\cdot x)e'_{i,j}\\ &=& \sum\lambda_i(g_j\cdot x)e'_{i,\sigma_g(j)}\\ &=& f(g) \cdot \varphi'(x)\end{array}$
(In the fourth equality we’ve used the fact that if we apply $\sigma_g$ to all the index positions where $j$ goes, then the sum is unchanged since we are summing over all elements of $G$.) I should mention that this embedding doesn’t help with any kind of visualization since if the group and manifold are both nontrivial then $\dim(V) \geq 4$. It does give a nice perspective on compact manifolds with finite group actions, and maybe it helps explain some of the nice properties they enjoy. I don’t know of any uses of this, but I’d be interested in hearing of one if you find one.

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### 7 Responses to “G-equivariant embeddings of manifolds”

1. This week in the arXivs… « It’s Equal, but It’s Different… Says:

[...] G-equivariant embeddings of manifolds [...]

2. No wai Says:

For an $n$-dimension manifold in the non-equivariant case, there is a beautiful ‘sharp’ bound on what the smallest possible $k$ is – it is $2n-b(n)$ where $b(n)$ denotes the number of 1′s in the binary expansion of $n$. This is the famous immersion theorem of Cohen. I wonder if there is an equivariant version!

3. onaka Says:

サイト訪問しました。

4. Peter Says:

I’m not sure about an equivariant version of that bound, I haven’t actually seen the equivariant embedding written anywhere, so I’m not sure how much it’s been studied (I’m also not sure how much equivariant immersions have been studied). Such a bound might depend in subtle ways on the group action, so it’s probably a pretty hard question in general.

5. John Baez Says:

Please tell the other folks at this blog to post more stuff! It’s been ages!

6. Peter Says:

looks like greg just did. I plan on writing a couple posts over the break, too

7. Chris Says:

Peter May pointed me to an excellent reference that describes equivariant embeddings in G-representations: the paper by Palais in Borel’s Seminar on Transformation Groups. Mostow’s Annals paper from 1957 also proves many results about equivariant embeddings.