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Let $X$ be an affine variety over $\mathbb{C}$. Let $x\in X$.

If $x$ is non-singular, then $x$ is locally holomorphic (in the Euclidean topology). See here for a relevant MO post.

By results of Milnor if $x$ is singular, then it is not locally $C^1$ (and hence not smooth in any sense).

However, singular points in $X$ may be locally Euclidean (of class $C^0$) as the cuspidal cubic ($x^2=y^3$) shows. However, this example is not normal (normalization is desingularization in dimension 1, so this can't happen for normal curves).

In dimension 2, a theorem of Mumford shows if a surface is normal then all singularities are topological singularities (every neighborhood of $x$ in the Euclidean topology is not homeomorpic to a ball).

This theorem does not generalize to higher dimensional varieties (Egbert Brieskorn, Beispiele zur Differentialtopologie von Singularitäten, Invent. Math. 2 (1966), 1–14). In fact, germs of links can get very complicated in general: see here.

Here is my question (modified after Jason's comment):

If $X=\mathbb{C}^n/\!/ G$ where $G$ is a reductive affine algebraic group (acting rationally), what are conditions on $X$ so that Mumford's theorem holds when $\dim X\geq 3$?

Note that $X$ will be normal since $\mathbb{C}^n$ is smooth, and normality is preserved by GIT quotients.

Remark: By the Luna Slice Theorem, in this situation, one can come up with a local model around a singularity that is of the form $V/\!/Stab(x)$ where $V$ is smooth. So one can then try to show the smooth locus of this local model is not what one would expect if the local model was Euclidean (comparing homology groups, for example).

Relevant MO Links (following Jason Starr's Comments):

  1. Algebraic varieties which are topological manifolds
  2. Is there a topological Chevalley-Shephard-Todd Theorem?

Remark: The above links and Jason's comments show that there are examples where $X$ is a topological manifold yet has algebraic singularities in dimensions higher than 3.

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    $\begingroup$ For the disconnected icosahedral group $\Gamma$ with linear action $\sigma:\Gamma \to \textbf{GL}(\mathbb{C}^2\times \mathbb{C})$ equal to the direct sum of the standard representation and a trivial representation, the quotient $(\mathbb{C}^2\times \mathbb{C})/\Gamma$ is a normal, topological manifold that is singular. Now choose a faithful representation $\rho:\Gamma \hookrightarrow \textbf{GL}_n(\mathbb{C})$. This is the same as the quotient of $\textbf{GL}_{n}(\mathbb{C}) \times (\mathbb{C}^2\times \mathbb{C})/(\rho,\sigma)(\Gamma)$ by its left action of $\textbf{GL}_n(\mathbb{C})$. $\endgroup$
    – Jason Starr
    Commented Mar 19, 2019 at 12:32
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    $\begingroup$ mathoverflow.net/questions/211100/… $\endgroup$
    – Jason Starr
    Commented Mar 19, 2019 at 12:33
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    $\begingroup$ Note that the quotient $\textbf{GL}_n(\mathbb{C})\times (\mathbb{C}^2\times \mathbb{C})/(\rho,\sigma)(\Gamma)$ is a smooth, affine variety, and the geometric quotient by $\textbf{GL}_n(\mathbb{C})$ is a singular, normal variety that is topologically a manifold. $\endgroup$
    – Jason Starr
    Commented Mar 19, 2019 at 12:55
  • $\begingroup$ @JasonStarr Thanks! This definitely gives me a counter-example and references. I edited the question in response to your example, as I still would like to have some set of conditions to guarantee the quotient is not a topological manifold at singular points. But maybe the point is that this example is so basic, that general conditions are not reasonable to expect. $\endgroup$
    – Sean Lawton
    Commented Mar 19, 2019 at 13:38

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