Is there an embedding (i.e. injective continuous map)

$$\phi:\Bbb R \Bbb P^2\hookrightarrow S^4\subseteq\Bbb R^5$$

of the projective plane $\Bbb R\Bbb P^2$ into the $4$-sphere, that is transitive, i.e. for any two $x,y\in \Bbb R\Bbb P^2$ there is an orthogonal transformation $T\in\mathrm{O}(\Bbb R^5)$ that fixed the image $\mathrm{im}(\phi)$ set-wise, and has $Tx=y$?

Is $\Bbb R^5$ the lowest dimensional space in which such an embedding is possible, or do we need even more dimensions?

I may ask the same question for $\Bbb R\Bbb P^n$: what is the lowest dimensional Euclidean space needed for such an embedding.


Yes. You take each vector $v \in \mathbb{R}^3$ to the vector $v \cdot v \in \operatorname{Sym}^2\mathbb{R}^3=\mathbb{R}^6$. This takes each unit vector $v$ to the same place as $-v$. So it descends to $S^2/\pm 1=\mathbb{RP}^2$. If we identify each element of $\operatorname{Sym}^2\mathbb{R}^3$ with a symmetric matrix by identifying $\sum a_{ij} e_i \cdot e_j$ with $A=(a_{ij})$, then the action of $\operatorname{SO}(3)$ is by conjugation on these symmetric matrices. The element $e_1 \cdot e_1 + e_2 \cdot e_2 + e_3 \cdot e_3$ is identified with the identity matrix. The action preserves identity matrix, so preserves a 1-dimensional subspace in $\mathbb{R}^6$. We quotient out that subspace to get an action of $\operatorname{SO}(3)$ on $\mathbb{R}^5$. This is well known in representation theory as the unique 5-dimensional irreducible representation of $\operatorname{SO}(3)$. The metric $\left<A,B\right>=\sum_{ij} A_{ij} B_{ij}$ is clearly preserved on $\mathbb{R}^6$, and $\left<v\cdot v,v\cdot v\right>=1$ in this metric. You take the quotient metric on $\mathbb{R}^5$, i.e. project to the orthogonal complement of that 1-dimensional subspace.

  • 2
    $\begingroup$ You mean preserving the trace of the symmetric matrix, not $x^2 + y^2 + z^2$. (What are $x, y, z$ anyway?) The only detail left is for which scalar product $G$ on $\mathrm{Sym}^2 \mathbb{R}^3$ the conjugation by matrices from $\mathrm{O}(3)$ belongs to $\mathrm{O}(G)$. $\endgroup$ Apr 25 '19 at 9:29
  • 2
    $\begingroup$ Here is a little further discussion of the embedding Ben describes: ldtopology.wordpress.com/2012/07/12/… $\endgroup$ Apr 25 '19 at 19:19

Since you may ask about $\mathbb{RP}^n$, I may point out that you can embed $\mathbb{RP}^n$ into $\mathbb{R}^{(n+2)(n+1)/2-1}$ with a transitive action of $O(n+1)$.

From e.g. Exercise 5-C of Milnor-Stasheff, $\mathbb{RP}^n$ is the space of symmetric idempotent $(n+1)\times (n+1)$ matrices with trace $1$. This lives inside $\mathbb{R}^{(n+2)(n+1)/2-1}$ as the space of symmetric $(n+1)\times (n+1)$ matrices with trace $1$. $O(n+1)$ acts by conjugation on this space as isometries, and preserving the subspace of idempotent matrices.

An exceptional case is $\mathbb{RP}^3 \cong SO(3)$. The group $SO(3) \subset \mathbb{R}^9$ as $3\times 3$ matrices. But we can do a bit better: we may think of $SO(3)$ as pairs of orthogonal unit vectors $(v_1,v_2)\in (\mathbb{R}^3)^2$. This gives an embedding of $SO(3)\subset S^5\subset \mathbb{R}^6$ with a transitive group action. I believe that this special embedding exists since $so(4)=so(3)\oplus so(3)$.

One might be able to construct similar smaller dimensional embeddings using fibrations $S^1\to \mathbb{RP}^{2n+1}\to \mathbb{CP}^n$ and $\mathbb{RP}^3\to \mathbb{RP}^{4n+3}\to \mathbb{HP}^n$, but I haven't checked if they give smaller embeddings with isometric actions.

However, $\mathbb{RP}^{2n}$ is not a fibration (for the same reason that $S^{2n}$ is not a fibration). Hence, if we have an embedding $\mathbb{RP}^{2n}\subset \mathbb{R}^k$ and a transitive action by isometries $G\leq O(k)$, then the representation of the compact group $G$ must be irreducible. If not, then there is a splitting $\mathbb{R}^k=\mathbb{R}^{k_1}\times \mathbb{R}^{k_2}$ which is invariant under $G$. In this case, we get $v=(v_1,v_2)\in \mathbb{RP}^n\subset \mathbb{R}^k$, $v_i\in \mathbb{R}^{k_i}$, and $\mathbb{RP}^n = G\cdot v \to G\cdot v_1$. Hence we have a fibration $\mathbb{RP}^{2n} \to G\cdot v_1$, a contradiction unless $G\cdot v_1=\mathbb{RP}^{2n}$, in which case $k$ was not minimal.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.