My first question is whether $m$-th symmetric product of $\mathbb{C}^{n}$ is simply connected, where $n\geq 3$.
The second question is whether $Hilb^{m}(\mathbb{C}^{n})$ is simply connected, where $n\geq 3$.
If not, how about the case when $m\gg1$.
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$\begingroup$ For the first one: it's closed under dilation, so contractible. $\endgroup$– Allen KnutsonCommented Jun 21, 2013 at 3:17
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$\begingroup$ If $Hilb^m(C^n)$ the Hilbert scheme of points? $\endgroup$– Mariano Suárez-ÁlvarezCommented Jun 21, 2013 at 3:34
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1 Answer
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Re your first question: If $\pi_1(X)=H_1(X,{\mathbb Z})=0$, then any $m$-th symmetric product of $X$ is simply connected; this is a special case of Theorem 1.1 in this paper by Kallel and Taamallah.
answered Jun 21, 2013 at 3:18