# Nontrivial Analytic Varieties

In short, I'd like to know the following:

1. Is there an irreducible analytic variety that has to be defined by at least two distinct sets of holomorphic functions?

2. Are there two irreducible analytic varieties that overlap completely at some open set but are distinct?

Analytic variety of complex manifold $M$ is defined as $V\subset M$ such that each point $x\in V$ has a neighborhood $U$ such that $U \cap V$ is a zero locus of holomorphic functions $f_1 \cdots f_m$ defined at $U$.

This definition had me wondering about examples of analytic varieties that are not simply zero loci of a few functions. Do we truly need an analytic variety to be a patchwork of loci of different set of functions? Also, can one patch be "continued" in at least two different ways?

1. Is there an irreducible analytic variety that can't be described as a zero locus of a single set of functions $\{ f_1 , \cdots f_m\}$? In other words, can we find two distinct sets of functions $\{f_j\}, \{g_j\}$ defined on $U_f,U_g$ which have common zero locus when restricted to $U_f \cap U_g$, but $f_j$ cannot be simultaneously extended to $U_g$ and vice versa.

As far as I know, this cannot happen for $M=\mathbb C$ since we can always find a meromorphic function that has a prescribed divisor.

1. Are there irreducible analytic varieties $V_1 \neq V_2$ such that $\exists U$ such that $U \cap V_1 = U \cap V_2 \neq \emptyset$?

A simple version of this question would ask: are there functions $f, g$ on $\mathbb C^n$ such that $V(f) \neq V(g)$ but $\exists U$ such that $U \cap V(f) = U \cap V(g) \neq \emptyset$?

• If $M$ is compact and connected, there are no nonconstant holomorphic functions on $M$...
– abx
Jan 22, 2017 at 13:33
• The chief example to deal with would be $M=\mathbb C^n$ though, and thus I need noncompact $M$ too. Jan 22, 2017 at 13:42
• It is easy to find examples of curves in $\mathbb{C}^3$ which are not globally defined by 2 equations, see this question and the answer by Hailong Dao.
– abx
Jan 22, 2017 at 14:32
• The answer is "yes" to the first question, if you allow $M$ to be a proper subdomain of $\mathbb{C}^n$. As for the second question, you probably want to put some topological conditions (such as connectedness) to avoid trivial counterexamples: in $\mathbb{C}^2$, $V:=\{\,z=0\,\}$ and $W:=\{\,z(z-1)=0\,\}$ coincide in a common open subset but are different. Jan 22, 2017 at 17:30
• Oops, I see now that in 2) you already assumed the varieties to be irreducible. In this case, I think the answer is "no". If I manage to find a reference I'll post an answer for both. Jan 22, 2017 at 17:33

1. A projective line in the projective plane has to be defined by at least two holomorphic functions on open subsets of the projective plane: there are no nonconstant holomorphic functions defined in a neighborhood of the projective line and vanishing on that line. If there were such a function, we would put that projective line at infinity in an affine chart, and the function would be defined in affine space away from a compact set, and therefore extend to all of the affine space by Hartogs. But then it would be a holomorphic function on the projective plane, so constant.

2. No. If they agree in some open set where they have nontrivial intersection, then their smooth loci agree in that open set. By analytic continuation of local defining functions, and connectivity of the smooth locus of an irreducible variety, their smooth loci agree everywhere, and they are each the Euclidean topology closure of their smooth loci.

1. If $M$ is compact, it's easy to find examples because it doesn't have enough global holomorphic functions. For an example in which the ambient manifold $M$ is not compact: in Fritzsche, Grauert, From holomorphic functions to complex manifolds there is an example of a non-globally definable analytic subset of a domain (connected open) in $\mathbb{C}^2$ (see Chapter I, Section 8, the example on page 36 and Figure 1.5)

Remark: by Cartan's Theorem A, if $M$ is a domain of holomorphy or, more generally, a Stein manifold or a Stein space, the coherent sheaf of ideals $\mathcal{I}_V$ of $V$ is generated by global sections and this implies that $V$ can be globally defined as zero locus of holomorphic functions on $M$.

1. As Ben McKay already said in his answer, all the points of the smooth loci of $V,W\subseteq M$ have to be the same: to see this, use the holomorphic implicit function theorem at smooth points to locally describe $V$ and $W$ as graphs of (vector) holomorphic functions, which have to be (componentwise) equal by the identity principle (since by hypothesis they agree on an open subset). By this local description it's easy to see that the locus $V_{\mathrm{smooth}}\cap W$ is open in $V_{\mathrm{smooth}}$; it's also closed in $V_{\mathrm{smooth}}$ because $V$ and $W$ are closed analytic sets. So $W\cap V_{\mathrm{smooth}}$ is a connected component of $V_{\mathrm{smooth}}$, but -as observed in BenMcKay's answer- the smooth locus of an irreducible analytic variety is connected, so indeed $V_{\mathrm{smooth}}=W_{\mathrm{smooth}}$. Then, $V=W$ by taking closure in the Euclidean topology of $M$.

2. Another (maybe a bit overkill) way to see (2.), I think, is applying Weierstrass preparation theorem around a point in $V\smallsetminus (V\cap W)$. I try to sketch it, assuming $V$ and $W$ of codimension $1$ in $M$ for simplicity. Represent the locally defining function $F$ of $V$ (resp. $G$ of $W$) as $$F(z,z_1,\dots,z_n)=\phi(z,z_1,\dots,z_n)\cdot\sum_{k=0}^df_k(z_1,\dots,z_n)\cdot z^k,$$ respectively $$G(z,z_1,\dots,z_n)=\gamma(z,z_1,\dots,z_n)\cdot\sum_{k=0}^\delta g_k(z_1,\dots,z_n)\cdot z^k$$ where $\phi$ (resp. $\gamma$) are nonzero local holomorphic functions, and $f_k$ (resp. $g_k$) are local holomorphic functions.

Then I think by the hypothesis one can easily show that $d=\delta$, $\phi=\gamma$, and $f_k=g_k$, so the resulting polynomials on the slice with fixed $(z_1,\dots,z_n)$ are the same so, for every fixed $(z_1,\dots,z_n)=\zeta$, we have $$F(z,\zeta)=G(z,\zeta)$$ so locally around every point $V=W$, so $V=W$.