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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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For the first one: it's closed under dilation, so contractible. –  Allen Knutson Jun 21 '13 at 3:17
    
If $Hilb^m(C^n)$ the Hilbert scheme of points? –  Mariano Suárez-Alvarez Jun 21 '13 at 3:34
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1 Answer 1

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.

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