# Singular affine real varieties are no manifolds?

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?

-
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

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.
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