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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: math.stackexchange.com/questions/205489/… –  Andrew Oct 18 '12 at 21:41
    
@Andrew Yes, I do. But for examples curves can have cusps which make them not differentiable. But they are still topological manifolds. –  k.stm Oct 19 '12 at 6:37

1 Answer 1

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. –  Brad Oct 18 '12 at 21:46
    
Oops, I should have read more carefully. At least the specific example still applies! –  user123123 Oct 18 '12 at 21:50
    
Can you prove you end up with a space homeomorphic to an 'X' shape, i.e. to $(-1,1) \times \{0\} \cup \{0\} \times (-1,1) \subseteq \mathbb{R}^2$? –  k.stm Oct 19 '12 at 17:18
    
Also, the inverse function theorem doesn't work as converse, does it? –  k.stm Oct 20 '12 at 9:09

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