So I have been mulling the following question over in my head for awhile now, and want to see if anyone else might have any ideas.
Begin with $M$ a manifold and suppose that $M$ has an antipodal map $\alpha:M\rightarrow M$, i.e. $\forall m\in M$ one has:

$\alpha(m)\neq m$
$\alpha(\alpha(m)) = m$

Let $A^n$ be the usual antipodal map on $S^n$: $A^n(s) = -s$. What I am wondering is if there is always an embedding $e:M\hookrightarrow S^n$ (for some $n$, not necessarily of minimal dimension), such that $e\circ\alpha = A^n|_{e(M)}$, that is that the antipodal map on the submanifold extends to the entire sphere.

One thing I suspect may be true is that when $M$ has more than one class of antipodal map, the necessary dimension of the sphere may depend on which class of maps I start with.

As an example, if I start with $M = \mathbb{T}^2$ with coordinates $(\theta,\phi)$, then I can easily define two classes of antipodal maps:

$\alpha_1(\theta,\phi) = (\theta+\pi,\phi)$
$\alpha_2(\theta,\phi) = (\theta+\pi,\phi+\pi)$

I have been able to convince myself with mental pictures that I can embed $\mathbb{T}^2$ into $S^3$ such that $\alpha_2$ coincides with $A^3$ (although I could even be mistaken about this), but I cannot convince myself that there is an embedding such that $\alpha_1$ coincides with $A^3$, so perhaps for this class of antipodal map, one must embed in a higher dimensional sphere.

Anyone have any insights or know of any results in this direction?

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    $\begingroup$ Your question is typically answered in a 1st course on Lie groups. Take a compact manifold $M$ with a compact Lie group action $G \times M \to M$. Then there exists an embedding of $M$ in some Euclidean space, and a representation of $G$ on that Euclidean space such that the inclusion $M \to \mathbb R^k$ is equivariant. So this reduces your question to the study of order $2$ elements of the orthogonal group. $\endgroup$ Commented Jan 26, 2011 at 18:32
  • $\begingroup$ Oh, I'm stupid. $\endgroup$ Commented Jan 26, 2011 at 18:37
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    $\begingroup$ So I believe the answer to your question is yes. A linear involution of Euclidean space does not have to be an antipodal map but it fixes a subspace, and it's the antipodal map on the orthogonal subspace. This is enough to answer your question. $\endgroup$ Commented Jan 26, 2011 at 18:41
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    $\begingroup$ You don't have a fixed point on $M$, so the embedding misses the origin. $\mathbb R^n \setminus \{0\}$ is $O_n$-equivariantly isomorphic to a subspace of $S^n$. Consider a diffeomorphism $\mathbb R^n\setminus \{0\} \simeq S^{n-1} \times (0,\infty)$. $\endgroup$ Commented Jan 26, 2011 at 19:07
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    $\begingroup$ Couldn't one let $N$ be the orbit space of $\Bbb Z_2$ acting on $M$, then we can embed $N$ in some $\Bbb RP^j$. Lastly we can lift this to an embedding of double covers to get a smooth embedding of $M$ in $S^j$ which is equivariant. $\endgroup$
    – John Klein
    Commented Jan 26, 2011 at 20:31

1 Answer 1


Let me elaborate on my comment above. Suppose $M$ is a manifold equipped with a smooth $\Bbb Z_2$ action that is also free. Then there is an equivariant smooth embedding $M \to S^j$, for some $j$, where we give the sphere the antipodal action. If we let $N$ be the orbit space of this action on $M$, then by Whithney's "easy" embedding theorem, we can embed $N$ smoothly in $\Bbb RP^j$, where $j = 2\dim M + 1$. Now pass to double covers To obtain an equivariant smooth embedding of $M$ in $S^j$.

This argument also works when $M$ is a manifold with free $S^1$-action, where we give the odd spheres the usual free action whose orbits give the complex projective spaces.

More generally, suppose $G$ is a finite group acting freely and smoothly on $M$. Suppose $V$ is a free orthogonal $G$-representation. Then the above can be generalized to show that there is an equivariant smooth embedding from $M$ into $S(jV)$ where the latter is the unit sphere of $j$-copies of $V$, $j$ large.

  • $\begingroup$ There must be something I am misunderstanding. I presume that the image of the embedding of $N$ will (or at least might) be contained in a ball in $\mathbb{R}P^j$. Given this, you get a map 2-1 from $M$ to $\mathbb{R}P^j$ which lifts to a map 2-1 to S^j$. Does this make sense? $\endgroup$ Commented Jan 26, 2011 at 21:22
  • $\begingroup$ The embedding $N \to S^j$ lifts to a map of double covers by covering space theory. The map of double covers is automatically an embedding. $\endgroup$
    – John Klein
    Commented Jan 26, 2011 at 21:30
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    $\begingroup$ You need to start by a map from $N\to \mathbb R P^\infty$ classifying the double cover $M\to N$. This factors through $\mathbb R P^j$ with $j = 2\dim M + 1$, can be perturbed to an embedding, and if $M$ is a non-trivial-cover, then the map $N\to \mathbb R P^j$ is non-trivial on fundamental groups, and so the image cannot be contained in a ball. $\endgroup$
    – Martin O
    Commented Jan 26, 2011 at 21:33
  • $\begingroup$ Thanks Martin, I thought that the starting point was a "random" embedding. $\endgroup$ Commented Jan 26, 2011 at 21:40
  • $\begingroup$ Like Alessandro, I was worried about which double cover one would lift to, but I think Martin's comment about bringing in the classifying space of $Z/2Z$ cleared that up. $\endgroup$
    – ARupinski
    Commented Jan 26, 2011 at 21:47

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