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The curve $C \colon x^3 + x^2 = y^2$ is a singular affine variety with a node at zero. How would one show that as an real affine variety $C \subseteq \mathbb{A}_\mathbb{R}^2 = \mathbb{R}^2$ it is no topological manifold?

More generally: How can someone tell and show which singular affine varieties over the reals given by equations are topological manifolds (when equipped with the subspace topology)?

I tried to find a property of a node which makes it impossible to have euclidian neighbourhoods (like having a closed neighbourhood which is the finite union of closed non-neighbourhoods), but this seems way too complicated – how could one even show that this curve has this property at the node an euclidian space hasn't?

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    You might find this relevant: http://math.stackexchange.com/questions/205489/is-it-possible-to-compute-if-an-algebraic-variety-is-a-differential-manifold2012-10-18
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    @Andrew Yes, I do. But for examples curves can have cusps which make them not differentiable. But they are still topological manifolds.2012-10-19

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Clearly, it must be a one-dimensional manifold if it is a manifold. Try and sketch the graph of it as a function - there's a singularity at $0$. Take a neighbourhood of 0, and you have a space homeomorphic to an 'X' shape. If you remove $0$, you end up with $4$ disjoint, path-connected segments, while if you remove any point in the real line, you end up with two disjoint segments. Thus, the variety isn't locally homeomorphic to $\mathbb{R}$ at $0$, and so it can't be a manifold.

More generally, for a singular variety, you could look at using the inverse function theorem to show that there is no local diffeomorphism at the singularity.

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    In the more general situation, the original poster was asking about the question of whether a given variety is a *topological* manifold (i.e. is locally homeomorphic to Euclidean space) rather than whether it is a smooth submanifold of the ambient space.2012-10-18
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    Oops, I should have read more carefully. At least the specific example still applies!2012-10-18
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    Can you _prove_ you end up with a space homeomorphic to an 'X' shape, i.e. to $(-1,1) \times \{0\} \cup \{0\} \times (-1,1) \subseteq \mathbb{R}^2$?2012-10-19
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    Also, the inverse function theorem doesn't work as converse, does it?2012-10-20