I was reading a book on differential geometry and after the intro to the concepts of regular singular points I came across an example under it:
The set $M:=\{(x^2,y^2,z^2,yz,zx,xy)|x,y,z\in \mathbb R, x^2+y^2+z^2=1\}$ is a smooth 2-manifold in $\mathbb R^6$. To see this, define an equivalence relation on the unit sphere $S^2\subset\mathbb R^3$ by $p\sim q$ iff $q=\pm p$. The quotient space (the set of equivalence classes is called the real projective plane and is denoted by $$\mathbb RP^2:=S^2/\{\pm1\}.$$ It is equipped with the quotient topology, i.e. a subset $U\subset\mathbb RP^2$ is open, by denfinition, iff its premiage under the obvious projection $S^2\to\mathbb RP^2$ is an open subset of $S^2$. Now the map $f:S^2\to\mathbb R^6$ defined by$$f(x,y,z):=(x^2,y^2,z^2,yz,zx,xy)$$ descends to a homeomorphism from $\mathbb RP^2$ onto $M$. An atlas on $M$ is given by the local smooth parametrizations $$\Omega\to M:(x,y)\mapsto f(x,y,\sqrt{1-x^2-y^2}),$$$$\Omega \to M:(x,z)\mapsto f(x,\sqrt{1-x^2-z^2},y,z),$$$$\Omega\to M:(y,z)\mapsto f(\sqrt{1-y^2-z^2},y,z),$$defined on the open unit disc $\Omega\subset \mathbb R^2$. We remark the following.
(a) $M$ is not the primage of a reguar value under a smooth map $\mathbb R^6\to \mathbb R^4.$
(b) $M$ is not diffeomorphic to a submanifold of $\mathbb R^3$.
(c) The projection $\Sigma:=\{(yz,zx,xy)|x,y,z\in\mathbb R,x^2+y^2+z^2=1\}$ of $Z$ onto the last three coordinates is called the Roman surface and was discovered by Jakob Steiner. The Roman surface can also be represented as the set of solutions $(\xi,\eta,\zeta)\in \mathbb R^3$ of the equation $\eta^2\zeta^2+\zeta^2\xi^2+\xi^2\eta^2=\xi\eta\zeta.$ Roman surface is not a submanifold of $\mathbb R^3$
Exercise: Prove this. Show that $M$ is diffeomorphic to a submanifold of $\mathbb R^4$
My questions are:
(A) Is $\mathbb RP^2$ a unit sphere? The forms of "local smooth parametrizations" suggests so. However, by definition $\mathbb RP^2$ is the collection of equivalence classes on the unit sphere $S^2\subset\mathbb R^2$ under the relation $q=\pm p$ If $q=-p$ means that $q$ and $p$ are vectors with opposite direction and the same magnitude, then $\mathbb RP^2$ should be a semiphere.
(B)How can I prove claims (a) and (b), I tried and I didn't make so much progress.
(C)Why is Roman surface not a submanifold of $\mathbb R^3$? My answer is that at the origin an intersection between any open sphere and the surface is not "nice" enough, but how can one rigorously show that any M open neighborhood of the origin is not diffeomorphic to a subset of $\mathbb R^2$?
(D)In which way does the author show that $M$ is a smooth 2-manifold in $\mathbb R^6$? He has showed that $M$ is homeomorphic to the unit sphere, but this doesn't prove the claim.
(E) Are there any linear structures on $\mathbb R^6,\mathbb R^3, \mathbb RP^2$ etc?
Please help!