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Let $X\subset \mathbb{C} \mathbb{P}^{N}$ be a smooth, algebraic (locally closed) complex submanifold of $\mathbb{C} \mathbb{P}^N$ of complex dimension $k$. More concretely, $X$ is of the following type $$ X := \{ p \in \mathbb{C} \mathbb{P}^N: \phi_1(p) =0, \phi_2(p) \neq 0 \} $$ where $\phi_1$ and $\phi_2$ are sections of some holomorphic vector bundle and whenever $\phi_2(p) \neq 0$, $\phi_1$ is transverse to the zero set.

Let $\overline{X}$ be the closure of $X$ inside $\mathbb{C}\mathbb{P}^N$.

1) 1) Is it true that $\overline{X}$ is an algebraic variety?

2) Is it true that the ``dimension'' of $\overline{X}-X$ is strictly less than the dimension of $X$?

3) In particular does $\overline{X}$ always define a homology class $$ [\overline{X}] \in H_{2k}(\mathbb{C} \mathbb{P}^N, \mathbb{Z}) $$
the basic idea being that the singularities of $\overline{X}$ are of complex codimension one, hence real codimension two.

Everything is over complex numbers. Note that although $X$ may not be connected, I am assuming that every connected component of $X$ has the same dimension $k$.

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    $\begingroup$ Sorry, did you mean smooth algebraic (locally closed) submanifold? Then the answer to all the questions is yes. $\endgroup$
    – Serge Lvovski
    Commented Jan 25, 2013 at 16:27
  • $\begingroup$ Thank you for the counter example, in my case I am interested only in algebraic (locally closed) sub manifold. Is there a reference for this fact if X is algebaric, ie locally closed? You are saying in that case the answer is yes. $\endgroup$
    – Ritwik
    Commented Jan 25, 2013 at 17:24
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    $\begingroup$ $\overline{X}-X$ may not be irreducible: Consider a section $\phi_1\in H^0(\mathbb{P}^2,O(2))$ defining a smooth conic $C$, and let $\phi_2\in H^0(\mathbb{P}^2,O(1))$ define a line $L$ intersecting $C$ transversely. Then $X$ is $C$ with two points $p,q$ removed, but $\overline{X}-X=${p,q}, which is not an algebraic variety. $\endgroup$
    – J.C. Ottem
    Commented Jun 10, 2013 at 15:11
  • $\begingroup$ Actually that was a typo. I meant to ask if $\bar{X}$ is an algebraic variety. $\endgroup$
    – Ritwik
    Commented Jun 11, 2013 at 4:07
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    $\begingroup$ If $\Phi_1$ and $\Phi_2$ are algebraic, then the answer to all your questions is yes, as Serge Lvovski pointed out above. You can think of $\bar X$ as the variety defined by $\{\Phi_1=0\}$ minus all irreducible components contained in $\{\Phi_2=0\}$. It follows that every irreducible component of $\bar X$ has complex dimension $k$. One way to see that $\bar X$ defines a homology class is to use the existence of a Whitney stratification with all strata being complex varieties. $\endgroup$
    – Brett Parker
    Commented Jun 11, 2013 at 4:51

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1) No. Suppose that $X$ is the set of pairs $(z,w)\in\mathbb C^2$ s.t. $w=e^z$. Then the closure of $X$ in $\mathbb{CP}^2$ is union of $X$ and the line at the infinity (it follows from, say, the big Picard theorem). This is not an algebraic variety.

2) No. In the example above, $\dim(\bar X\setminus X)=\dim X=1$.

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  • $\begingroup$ I have edited the question to indicate what $X$ can be. $\endgroup$
    – Ritwik
    Commented Jan 25, 2013 at 17:30

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