# $\mathbb{C}P^1\times…\times \mathbb{C}P^1/S_m=\mathbb{C}P^m$

Let $$X:=\mathbb{C}P^1\times...\times \mathbb{C}P^1$$ be the product of $$m$$ copies of $$\mathbb{C}P^1$$ and $$S_m$$ acts on $$X$$ by permuting the factors. Then why is $$X/S_m=\mathbb{C}P^m$$?

• The standard hint is to use the elementary symmetric functions on $n$ variables. – Ted Shifrin Jan 16 at 21:11

Consider the line bundle $$\mathcal{O}(n)$$ on $$\mathbb{CP}^{1}$$ whose sections are homogeneous polynomials of degree n in 2 variables. Then if we projectivise the space of sections (i.e. consider $$\mathbb{P}(H^{0}(\mathbb{P}^{1},\mathcal{O}(n)))$$ we get a projective space of dimension $$n$$. Furthermore, each element of this projective space gives you $$n$$ unordered points on $$\mathbb{P}^{1}$$ since we can factorize a homogeneous polynomial in $$2$$ variables as a product of degree one polynomials.
• Just to make this explicit, take $$\big([a_1:b_1], [a_2:b_2], \ldots, [a_m: b_m]\big) \mapsto [c_0: c_1: \ldots: c_m]$$ where $c_i$ is the coefficient of $x^iy^{m-i}$ in $\prod_i (a_ix + b_iy)$. – hunter Jan 16 at 21:35