Subset of symmetric matrices is submanifold Let $Sym(3)\cong\mathbb{R}$ denote the set of real symmetric $3\times 3$-matrices. Let $$M:=\{ P\in Sym(3)|\ P^2=P,\ \operatorname{tr}{P}=1 \}.$$
I asked to show that $M$ is a submanifold diffeomorphic to the projective space $\mathbb{R}P^2$.

My effort: I'm familiar with two types of definition for a submanifold:


*

*Definition of an immersed manifold and

*A subset $N$ of a manifold $M$ is called a submanifold if for all $p\in N$ there exists a chart $(U,\phi)$ of $M$ with $p\in U$ such that $\phi(N\cap U)=(\mathbb{R}\times 0)\cap \phi(U)$.


I want to use the second one. We can choose a global chart given by $(Sym(3),\phi)$, where $$\phi:\begin{pmatrix}a&d&f\\d&b&e\\f&e&c\end{pmatrix}\mapsto(a,b,c,d,e,f).$$
In this case we have: $\phi(Sym(3))=\mathbb{R}^6$ and $N\cap Sym(3)=N$. This idea ends in analyzing the conditions $\operatorname{tr}P=1$ and $P^2=P$. For example we can set $a=1-b-c$.
Before I continue: Is there an easier way to solve the problem?
Furthermore: I have no idea how to show that $M$ is diffeomorphic to $\mathbb{R}P^2$.
Your help is greatly appreciated.
 A: There is a completely different way to think about this problem. 
Note that your set is an orbit of the group compact $SO(3)$ acting on Sym(3) by conjugation (a matrix in $M$ is conjugate via an orthogonal matrix to the diagonal matrix with eigenvalue $(0,0,1)$. This action is certainly $C^{\infty}$.
Now an orbit of a point $x_0$ under the action of a compact group $K$ on a manifold is always a submanifold (this follows from the constant rank theorem), diffeomorphic to $K/K_{x_0}$, where $K_{x_0}$ is the stabilizer of $x_0$. Here $x_0$ is the diagonal matrix with diagonal $(0,0,1)$, and its stabilizer is $O(2)$, the subgroup of $SO(3)$ which fixes the horizontal plane, or equivalently, the vertical line. This is the same isotropy group as the isotropy group of the vertical line viewed as a point in $RP^2$, with the natural action of $SO(3)$ on the projective plane.
This explanation immediately generalizes to the set of symmetric projectors of trace $k$ in $Sym(n)$ and proves that this set is a submanifold diffeomorphic to the Grassmanian.
A: At first, we are going to show that this set of matrices is exactly a set of all lines in $\mathbb{R}^3$. Let us define some tautological map, $p: \mathbb{R}P^2 \to M$, every line $l$ is defined by its projector in it: $p_{l} : \mathbb{R}^3 \to l$ that is by definition a linear mapping which is self-adjoint, (in standart euclidean structure in $\mathbb{R}^3$ it is symmetricity property) its image is exactly $l$, and $p_{l}^{2} = p_{l}$, and trace of $p_{l}$ is 1. Geometrically, it is just orthogonal projection onto $l$. (Not so easy to see is condition about trace, hint: write a matrix of $p_{l}$ in basis with first vector collinear to $l$ ) 
Now we actually have a plenty of ways to prove that $M$ is truly a submanifold of $Sym(3)$. First one is most straight-forward, you actually have to evaluate differential of following mapping: $E: Sym(3) \to (Sym(3) \times \mathbb{R}), E(P) = (P^2 - P, tr(P))$. If $D(E) := \frac{dE}{dP}$ is non-zero on $M$ then by implicit function theorem you are done. (Hint: to evaluate a differential you will have to take some pertrubation of $P: P + \epsilon A$ and evaluate $lim_{\epsilon \to 0} \frac{E(P+\epsilon A) - E(P)}{\epsilon}$). Second one is little unfair, you can say that $p_{l}$ mapping is in fact diffeomorphism and by that you are done.
