# Transitive embedding of the projective space $\Bbb R P^2$ into the $4$-sphere

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

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

of the 2-dimensional projective space $$\Bbb R P^2$$ into the $$4$$-sphere, that is transitive, i.e. for any two $$x,y\in P^2\Bbb R$$ 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 $$\Bbb R P^n$$: what is the lowest dimensional Euclidean space needed for such an embedding.

• The notation $P\mathbb{R}^2$ makes it look like the projectivization of $\mathbb{R}^2$, i.e. $\mathbb{RP}^1$. Perhaps $\mathbb{RP}^2$ or $\mathbb{P}^2(\mathbb{R})$ are better notations. – Ben McKay Apr 25 at 8:07

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 $$\left=\sum_{ij} A_{ij} B_{ij}$$ is clearly preserved on $$\mathbb{R}^6$$, and $$\left=1$$ in this metric. You take the quotient metric on $$\mathbb{R}^5$$, i.e. project to the orthogonal complement of that 1-dimensional subspace.

• 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 $\mathrm{O}(G)$. – Vít Tuček Apr 25 at 9:29
• Here is a little further discussion of the embedding Ben describes: ldtopology.wordpress.com/2012/07/12/… – Ryan Budney Apr 25 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.