moduli space of k3 surfaces

I consider the moduli space of k3 surfaces, defined by N:={(X,$\phi$)}/$\sim$, where X is a k3 surface and $\phi:H^2(X,\mathbb{Z})\rightarrow \Lambda:=-2E_8\oplus 3U$ is an isometry considering the intersection product on $H^2(X,\mathbb{Z})$.

$(X,\phi)\sim (X',\phi')$ iff exists $\psi:X \rightarrow X'$ biholomorphic such that $\phi'=\phi \psi^*$.

I know that N is a complex variety of dimension 20 (i studied it in Huybrechts' work http://arxiv.org/abs/1106.5573), since it can be covered by the subspaces Def(X) (universal deformation of X).

The thing i'm not getting is why a priori N is not an Hausdorff space, since Def(X) seem to be Hausdorff

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I agree. The sentence "Firstly, $M$ is a complex manifold but in general it is not Hausdorﬀ," to be terribly strange. What definition of complex manifold omits Hausdorff (or doesn't have it as a consequence)? Maybe this just means that there is also a natural Zariski topology on the space which is definitely not Hausdorff. –  Matt Dec 31 '12 at 19:07