A compact complex manifold $X$ which admits an embedding into $P^n(\mathbb{C})$ (for some $n$) is called a projective algebraic manifold. And by a theorem of Chow, every complex submanifold $V$ of $P^n(\mathbb{C})$ is actually an algebraic submanifold. However, how to determine if the embedded complex submanifold is an irreducible algebraic set?
Is a projective algebraic manifold irreducible algebraic set in $P^n$
algebraic-geometrycomplex-geometry
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A theorem of Demailly and Pǎun (https://arxiv.org/abs/math/0105176) gives a condition for a compact complex manifold to be of Fujiki class $\mathcal{C}$ (that is bimeromoprhic to a Kähler manifold). A manifold is Kähler if and only if it supports a Kähler current. A Kähler current is a real positive $(1,1)$-current $T$, such that $T - \varepsilon \omega > 0$ for a constant $\varepsilon > 0$ and positive hermitian $(1,1)$-form $\omega$.
The homology class of such current is integral if and only if the manifold is projective (this is the standard Kodaira embedding theorem).
A similar theorem for Moishezon manifolds was proved by Popovici (https://arxiv.org/abs/math/0603738). It states, roughly speaking, that a complex manifold is Moishezon if and only if it supports a semi-positive and strictly positive almost everywhere real $(1,1)$-current with integral homology class.
The integrality condition is hardly a differential-geometric one, but you cannot escape it (at least, as far, is we know by now), and I don't really think one might give a more differential-geometric answer.
In a general situation it is, of course, rather hard to verify the Popovici condition. Another "geometric" approach to studying algebraic dimensions of complex manifolds is to study the algebraic reduction. For any complex manifold $X$ there exists a normal projective variety $\overline{X}$ and a meromorphic map $\alpha \colon X \to \overline{X}$, such that any meromorphic function on $X$ can be lifted from $\overline{X}$. The variety $\overline{X}$ is unique up to birational equivalence.
Being Moishezon is equivalent to $\alpha$ being a birational equivalence. More generally, $a(X) = \dim_{\mathbb C}(\overline{X})$.
The answer is essentially yes: since every variety is birational to a projective variety and every projective variety is birational to a smooth projective variety, every variety is birational to a smooth projective variety. And from every smooth projective variety you can get a complex manifold in a canonical way. But the variety and the complex manifold won't be literally birational because of the difference between varieties and manifolds (generic points, eg). But if you're looking for the incarnation of a variety in the world of complex manifolds, the analytification is the thing.
Some references for analytification are https://ncatlab.org/nlab/show/analytification, and also https://www.math.ucdavis.edu/~osserman/classes/248B-W10/analytic-top.pdf , where it's spoken of as putting the analytic topology on a scheme.
One important point to remember here is that there are multiple different definitions of what a "variety" is. Some people believe that varieties over a field k are certain closed subsets of $k^n$ or $\Bbb P^n_k$ with the Zariski topology, other people believe that they're schemes of finite type over $k$. If you believe the first definition, smooth varieties over $\Bbb C$ are really complex manifolds. If you believe the second definition, to each smooth variety over $\Bbb C$, we can canonically associate a complex manifold. The latter viewpoint on what a variety is has been very useful over the past 60-70 years, and there's a canonical way to get from the first to the second in some cases, like varieties in the first sense over $\Bbb C$ with certain adjectives attached. And thinking about these varieties as in the first viewpoint is often a useful reasoning tool, but there are times where you'll have to be careful and explicit about what you're assuming. Especially when talking to people not already well-versed in the field. (This sort of confusion is a big problem for many learners - I went through it my self, for instance.)
Best Answer
Assuming you mean embedding to mean injective on points, this question has nothing to do with the embedding. For a smooth locally noetherian scheme over a field, irreducible components are connected components - these conditions imply that all the local rings of the variety are regular local rings, which in turn implies that each point is on exactly one irreducible component (by the correspondence between minimal primes of the local ring and irreducible components containing the point). So $X$ is irreducible iff it's connected.