# Blow up the diagonal of a symmetric product space

Consider the complex blow-up of the diagonal $$\triangle\subset Sym^2(\mathbb{C}^n)$$. It is known that the blown-up space is smooth when $$n=1,2$$. I was wondering if that is still smooth for $$n\geq3$$? Moreover, what if $$\mathbb{C}^n$$ is replaced by other complex variety? I'd be grateful if anyone could show me some hint or reference...

• If true for $\Bbb{C}^n$, it will be true for any manifold -- this is a local question.
– abx
Jul 25, 2020 at 7:49
• Another way to see it: a quotient of a smooth space by a free action is still smooth. Jul 25, 2020 at 20:00
• @DmitryVaintrob Do you mean blow up the product space first and then do the S_2-action? I don't see how that is free could you explain more? Thank you! Jul 26, 2020 at 15:55
• @EricYuan, good point! What I wrote above is incorrect (e.g. in the one-dimensional case the blow-up doesn't change the variety) Jul 26, 2020 at 18:04

Consider the quotient map $$\pi:\mathrm{X}\times\mathrm{X}\rightarrow\mathrm{S}^2\mathrm{X}=(\mathrm{X}\times\mathrm{X})/\mathrm{S}_2$$, and a point $$p$$ of $$\mathrm{X}\times\mathrm{X}$$. Then the tangent space of $$\mathrm{S}^2\mathrm{X}$$ at $$\pi(p)$$ equals the tangent space of $$\mathrm{T}_p(\mathrm{X}\times\mathrm{X})/(\mathrm{S}_2)_p$$ at the origin (see for example here), where $$(\mathrm{S}_2)_p$$ is the stabiliser of $$p$$. For $$\mathrm{X}=\mathbf{A}^n$$ you can also write down explicit equations for the blow up; another possibility is to prove that the blow up is isomorphic to the Hilbert scheme of two points $$\mathrm{Hilb}^2\mathrm{X}$$.