Theorem. Let $X$ be any complex variety. If $X$ is connected, then $X^{\mathrm{an}}$ is connected.
Remark. Before we proceed with the proof, note how astonishing this theorem is. For instance, it proves that for any irreducible polynomial $f(x, y) \in \mathbb{C}[x, y],$ the zero set $\{(z, w) \in \mathbb{C} : f(z, w) = 0\} \subset \mathbb{R}^{4}$ is connected in the Euclidean topology. I am not entirely sure how to prove this directly! (Maybe some compactification with Mayer-Vietoris argument will do? I am not sure.)
To assist our proof, we give a separate lemma:
Lemma. Let $X$ be any connected algebraic variety over an algebraically closed field $k$. For any $p, q \in X(k),$ there is an irreducible curve $C \subset X$ as a closed subset such that $p, q \in C.$
We will use this lemma to show Theorem first and then go back to proving the lemma.
Proof that Lemma implies Theorem. There will be several steps.
Step 1. We reduce to the case where $X$ is irreducible.
Suppose that the statement is proven for irreducible components. To show $X^{\mathrm{an}} = X(\mathbb{C})$ is connected, write $X(\mathbb{C}) = U \sqcup V,$ where $U, V$ are clopen sets in $X(\mathbb{C}).$ Then consider the unique irreducible decomposition $$X = X_{1} \cup \cdots \cup X_{r},$$ where $X_{i}$'s are irreducible components. Our assumption says $X_{i}(\mathbb{C})$ are connected, so it is contained in exactly of $U$ and $V,$ so we can eventually write $$X(\mathbb{C}) = (X_{1}(\mathbb{C}) \cup \cdots \cup X_{s}(\mathbb{C})) \sqcup (X_{s+1}(\mathbb{C}) \cup \cdots \cup X_{r}(\mathbb{C}))$$ after some rearrangements, where unionands are equal to $U$ and $V.$ Since each $X_{i}$ is quasi-compact, this necessarily implies that $$X = (X_{1} \cup \cdots \cup X_{s}) \sqcup (X_{s+1} \cup \cdots \cup X_{r}).$$ But then $X$ is connected, so either $X_{1} \cup \cdots \cup X_{s}$ or $X_{s+1} \cup \cdots \cup X_{r}$ must be empty. This implies that either $U$ or $V$ is empty, showing $X(\mathbb{C})$ is connected. Hence, this reduces the problem to the case where $X$ is irreducible. This finishes this step.
Step 2. We reduce to the case where $X$ is an irreducible curve.
Suppose that the statement is proven for irreducible curves. For any $p, q \in X(\mathbb{C}),$ we use Lemma to choose an irreducible curve $C \subset X$ that contains both $p$ and $q.$ Analytifiying the curve, we see that $C(\mathbb{C})$ is connected, so it follows that $X(\mathbb{C})$ is connected (e.g., Munkres Theorem 23.3). This finishes this step (except that we need to prove Lemma later).
Step 3. We reduce to the case where $X$ is a smooth irreducible curve.
Take the normalization $\tilde{X} \twoheadrightarrow X$ (Vakil 9.7.C for surjectivity). Since $X$ is irreducible, we also have $\tilde{X}$ as irreducible (Vakil, beginning of Section 9.7). Normal (Noetherian) curves are regular, so the fact that we are over $\mathbb{C}$ tells us that $\tilde{X}$ is smooth.
Step 4. We reduce to the case where $X$ is a smooth irreducible projective curve.
For this, we will use that there is an open embedding $X \hookrightarrow \bar{X},$ where $\bar{X}$ is a smooth irreducible projective curve (e.g., Hartshorne I.6 Corollary 6.10). Hence, if we assume the statement for the case of smooth irreducible curves, then $\bar{X}^{\mathrm{an}}$ is connected. Now, note that $\bar{X} \setminus X$ is a $0$-dimensional Noetherian scheme, so it must have finitely many points. Thus, we see $\bar{X}(\mathbb{C}) \setminus X(\mathbb{C})$ is also a finite set. As $\bar{X}(\mathbb{C})$ is a connected real manifold of dimension $2,$ taking finitely many points away will keep its connectivity.
Personal remark. Shubhodip Mondal kindly explained how Hartshorne's exposition constructs a desired projective curve, using discrete valuations as points. I have not gone through this in detail, so I cannot explicitly tell in which generality this fact is true, although for the case that we deal with does not require us to understand the proof. As Shubhodip also mentioned, this is can be seen as a special case of Nagata's compactification theorem because proper curves are projective (Vakil 18.7.2).
Remark. We have used that given a real manifold $M$ of dimension $\geq 2,$ if $M$ is connected, for any $p \in M,$ the punctured manifold $M \smallsetminus \{p\}$ is connected. To see this, we may argue by contradiction, so say $M \smallsetminus \{p\} = U \sqcup V$ with nonempty open $U, V \subset M \smallsetminus \{p\} .$ Since $M$ is connected and $M = \bar{U} \cup \bar{V},$ the intersection $\bar{U} \cap \bar{V}$ must be nonempty. For any $q \in \bar{U} \cap \bar{V},$ as any open neighborhood of $q$ in $M$ should intersect both of $U$ and $V,$ which are open in $M,$ we cannot have $q \in U \cup V.$ This implies that $q = p,$ so $\bar{U} \cap \bar{V} = \{p\}.$ Now, take any open neighborhood $W \ni p$ in $M$ that is homeomorphic to an open disk in $\mathbb{R}^{2}.$ We now have $W \smallsetminus \{p\} = (W \cap U) \sqcup (W \cap V),$ and each unionand is nonempty while the left-hand side is connected. This is a contradiction, so $M \smallsetminus \{p\}$ must have been connected.
Step 5. We prove the statement for the case where $X$ is an irreducible smooth projective curve.
For contradiction, let $X^{\mathrm{an}} = U \sqcup V,$ where $U, V$ are nonempty open subsets in the classical topology. Let $g(X)$ be the genus of the curve $X.$ Fix any $p \in U \subset X(\mathbb{C}).$ If $\deg(\mathscr{O}_{X}(np)) = n \geq 2g(X),$ then there is a global section $$s \in \Gamma(X, \mathscr{O}_{X}(np)) = \{t \in K(X)^{\times} : \mathrm{div}(t) + np \geq 0\}$$ that does not vanish at $p$ (Vakil 19.2.11). Thus, we have $$np = \mathrm{div}(s) = q_{1} + \cdots + q_{m} \in \mathrm{Cl}(X),$$ for some finite sequence of points $q_{1}, \dots, q_{m} \in X(\mathbb{C})$ that are not equal to $p.$ (We have $\mathrm{div}(s) \neq np$ as Weil divisors; it is only their divisor classes that match.) Thus, we have some $f \in K(X)^{\times}$ such that $\mathrm{div}(f) = q_{1} + \cdots + q_{m} - np.$ Since the only pole of $f$ is at $p,$ it defines a regular function $f : X \smallsetminus \{p\} \rightarrow \mathbb{A}^{1}.$ This implies that $f : X(\mathbb{C}) \setminus \{p\} \rightarrow \mathbb{C}$ defines a holomorphic function, and since $X$ is projective, its analytification $X^{\mathrm{an}} = X(\mathbb{C})$ is compact. As $p \notin V,$ we may take any connected component $T$ of $V$ (which is a closed subset in $X(\mathbb{C})$ so that it is compact) and realize that $f|_{T}$ is constant. Since $T$ is connected and Hausdorff, it must be an infinite set, so $f$ is constant on infinitely many points. Since $T$ is compact, this implies that $f$ is constant on the whole $X(\mathbb{C}) \smallsetminus \{p\}.$ This is enough to deduce that $f \in \mathbb{C} \subset K(X)^{\times},$ so $\mathrm{div}(f) = 0,$ which is a contradiction as the earlier one had a pole at $p.$ This finishes the proof. $\Box$
Proof of Lemma. We now discuss a proof of Lemma. When $\dim(X) = 1,$ the statement is tautological, so assume $n := \dim(X) \geq 2.$
Step 1. Reduction to the case where $X$ is an irreducible projective $k$-variety.
By Nagata's compactification theorem (e.g., Lütkebohmert's note), our variety $X$ can be regarded as a dense open subset of some complete variety $Y$ over $k.$ Since $X$ is irreducible, this implies that $Y$ must be irreducible as well. Applying Chow's lemma (Vakil 18.9.2 and 18.9.6) to $Y,$ we get a birational surjective $k$-scheme map $\pi : \tilde{X} \twoheadrightarrow X,$ where $\tilde{X}$ is a nonempty open subset of an irreducible projective variety over $k.$ Birationality means that there is a dense open subset $U \subset X$ such that $$\pi : \pi^{-1}(U) \overset{\sim}{\longrightarrow} U.$$ Since the image of any irreducible set under a continuous map is irreducible, this implies that if we take an irreducible curve closed in $\tilde{X},$ the closure of its image is an irreducible curve closed in $X.$ This lets us reduce to the case where $X$ is an open subset of some irreducible projective $k$-variety $\bar{X}$. Then note that proving the statement for $\bar{X}$ is enough, so we have reduced the problem to the case where $X$ is an irreducible projective $k$-variety.
Step 2. Proof for the case where $X$ is an irreducible projective $k$-variety.
Now, fix any two distinct points $p, q \in X(k),$ and consider the blow-up $$\pi : \mathrm{Bl}_{\{p,q\}}(X) \twoheadrightarrow X.$$ Since $X$ is projective over $k,$ so is $\tilde{X} = \mathrm{Bl}_{\{p,q\}}(X)$ (e.g., Hartshorne II Proposition 7.16 or Vakil Theorem 22.3.2 + 17.3.B, both of which use $\pi$ is a projective morphism). Using the universal property of the blow-ups (Section 22.2 of Vakil), we may see that $\pi^{-1}(p)$ and $\pi^{-1}(q)$ are effective Cartier divisors of $\tilde{X},$ so they must have dimension $n - 1.$ (Recall that we are in the case $n \geq 2.$) Since $\tilde{X}$ is projective (of dimension $n$), we have a closed embedding $\pi : \tilde{X} \hookrightarrow \mathbb{P}^{N}$ for some $N \geq n = \dim(X).$ Then we apply Bertini's theorem to choose $n-1$ distinct hyperplanes $H_{1}, \dots, H_{n-1} \subset \mathbb{P}^{N}$ such that
Since $\dim(\pi^{-1}(p)) = \dim(\pi^{-1}(q)) = n - 1,$ both fibers intersect $H_{1} \cap \cdots \cap H_{n-1}$ Since $\pi$ is projective, it is closed, so its image $\pi(C) \subset X$ is a closed subset and an irreducible curve that contain $p$ and $q$. $\Box$
Personal remark about Bertini. When $X$ is smooth, Theorem 8.18 in Hartshorne (note: this theorem seems to use the hypothesis that $k$ is algebraically closed) does the job, but the general version we use is harder. (I need to ask the lecturer again where this is written.) I was told that one can use Hironaka's theorem on the resolution of singularities for the case when the characteristic is $0.$ This will be okay as we only care about the case $k = \mathbb{C}$ in this class, but I am now unsure whether the statement is true for any algebraically closed $k,$ not just characteristic $0$ ones.
Step 3. We reduce to the case where $X$ is a smooth irreducible curve.
Take the normalization $\tilde{X} \twoheadrightarrow X$ (Vakil 9.7.C for surjectivity). Since $X$ is irreducible, we also have $\tilde{X}$ as irreducible (Vakil, beginning of Section 9.7). Normal (Noetherian) curves are regular, so the fact that we are over $\mathbb{C}$ tells us that $\tilde{X}$ is smooth.
Step 4. We reduce to the case where $X$ is a smooth irreducible projective curve.
For this, we will use that there is an open embedding $X \hookrightarrow \bar{X},$ where $\bar{X}$ is a smooth irreducible projective curve (e.g., Hartshorne I.6 Corollary 6.10). Hence, if we assume the statement for the case of smooth irreducible curves, then $\bar{X}^{\mathrm{an}}$ is connected. Now, note that $\bar{X} \setminus X$ is a $0$-dimensional Noetherian scheme, so it must have finitely many points. Thus, we see $\bar{X}(\mathbb{C}) \setminus X(\mathbb{C})$ is also a finite set. As $\bar{X}(\mathbb{C})$ is a connected real manifold of dimension $2,$ taking finitely many points away will keep its connectivity.
Personal remark. Shubhodip Mondal kindly explained how Hartshorne's exposition constructs a desired projective curve, using discrete valuations as points. I have not gone through this in detail, so I cannot explicitly tell in which generality this fact is true, although for the case that we deal with does not require us to understand the proof. As Shubhodip also mentioned, this is can be seen as a special case of Nagata's compactification theorem because proper curves are projective (Vakil 18.7.2).
Remark. We have used that given a real manifold $M$ of dimension $\geq 2,$ if $M$ is connected, for any $p \in M,$ the punctured manifold $M \smallsetminus \{p\}$ is connected. To see this, we may argue by contradiction, so say $M \smallsetminus \{p\} = U \sqcup V$ with nonempty open $U, V \subset M \smallsetminus \{p\} .$ Since $M$ is connected and $M = \bar{U} \cup \bar{V},$ the intersection $\bar{U} \cap \bar{V}$ must be nonempty. For any $q \in \bar{U} \cap \bar{V},$ as any open neighborhood of $q$ in $M$ should intersect both of $U$ and $V,$ which are open in $M,$ we cannot have $q \in U \cup V.$ This implies that $q = p,$ so $\bar{U} \cap \bar{V} = \{p\}.$ Now, take any open neighborhood $W \ni p$ in $M$ that is homeomorphic to an open disk in $\mathbb{R}^{2}.$ We now have $W \smallsetminus \{p\} = (W \cap U) \sqcup (W \cap V),$ and each unionand is nonempty while the left-hand side is connected. This is a contradiction, so $M \smallsetminus \{p\}$ must have been connected.
Step 5. We prove the statement for the case where $X$ is an irreducible smooth projective curve.
For contradiction, let $X^{\mathrm{an}} = U \sqcup V,$ where $U, V$ are nonempty open subsets in the classical topology. Let $g(X)$ be the genus of the curve $X.$ Fix any $p \in U \subset X(\mathbb{C}).$ If $\deg(\mathscr{O}_{X}(np)) = n \geq 2g(X),$ then there is a global section $$s \in \Gamma(X, \mathscr{O}_{X}(np)) = \{t \in K(X)^{\times} : \mathrm{div}(t) + np \geq 0\}$$ that does not vanish at $p$ (Vakil 19.2.11). Thus, we have $$np = \mathrm{div}(s) = q_{1} + \cdots + q_{m} \in \mathrm{Cl}(X),$$ for some finite sequence of points $q_{1}, \dots, q_{m} \in X(\mathbb{C})$ that are not equal to $p.$ (We have $\mathrm{div}(s) \neq np$ as Weil divisors; it is only their divisor classes that match.) Thus, we have some $f \in K(X)^{\times}$ such that $\mathrm{div}(f) = q_{1} + \cdots + q_{m} - np.$ Since the only pole of $f$ is at $p,$ it defines a regular function $f : X \smallsetminus \{p\} \rightarrow \mathbb{A}^{1}.$ This implies that $f : X(\mathbb{C}) \setminus \{p\} \rightarrow \mathbb{C}$ defines a holomorphic function, and since $X$ is projective, its analytification $X^{\mathrm{an}} = X(\mathbb{C})$ is compact. As $p \notin V,$ we may take any connected component $T$ of $V$ (which is a closed subset in $X(\mathbb{C})$ so that it is compact) and realize that $f|_{T}$ is constant. Since $T$ is connected and Hausdorff, it must be an infinite set, so $f$ is constant on infinitely many points. Since $T$ is compact, this implies that $f$ is constant on the whole $X(\mathbb{C}) \smallsetminus \{p\}.$ This is enough to deduce that $f \in \mathbb{C} \subset K(X)^{\times},$ so $\mathrm{div}(f) = 0,$ which is a contradiction as the earlier one had a pole at $p.$ This finishes the proof. $\Box$
Proof of Lemma. We now discuss a proof of Lemma. When $\dim(X) = 1,$ the statement is tautological, so assume $n := \dim(X) \geq 2.$
Step 1. Reduction to the case where $X$ is an irreducible projective $k$-variety.
By Nagata's compactification theorem (e.g., Lütkebohmert's note), our variety $X$ can be regarded as a dense open subset of some complete variety $Y$ over $k.$ Since $X$ is irreducible, this implies that $Y$ must be irreducible as well. Applying Chow's lemma (Vakil 18.9.2 and 18.9.6) to $Y,$ we get a birational surjective $k$-scheme map $\pi : \tilde{X} \twoheadrightarrow X,$ where $\tilde{X}$ is a nonempty open subset of an irreducible projective variety over $k.$ Birationality means that there is a dense open subset $U \subset X$ such that $$\pi : \pi^{-1}(U) \overset{\sim}{\longrightarrow} U.$$ Since the image of any irreducible set under a continuous map is irreducible, this implies that if we take an irreducible curve closed in $\tilde{X},$ the closure of its image is an irreducible curve closed in $X.$ This lets us reduce to the case where $X$ is an open subset of some irreducible projective $k$-variety $\bar{X}$. Then note that proving the statement for $\bar{X}$ is enough, so we have reduced the problem to the case where $X$ is an irreducible projective $k$-variety.
Step 2. Proof for the case where $X$ is an irreducible projective $k$-variety.
Now, fix any two distinct points $p, q \in X(k),$ and consider the blow-up $$\pi : \mathrm{Bl}_{\{p,q\}}(X) \twoheadrightarrow X.$$ Since $X$ is projective over $k,$ so is $\tilde{X} = \mathrm{Bl}_{\{p,q\}}(X)$ (e.g., Hartshorne II Proposition 7.16 or Vakil Theorem 22.3.2 + 17.3.B, both of which use $\pi$ is a projective morphism). Using the universal property of the blow-ups (Section 22.2 of Vakil), we may see that $\pi^{-1}(p)$ and $\pi^{-1}(q)$ are effective Cartier divisors of $\tilde{X},$ so they must have dimension $n - 1.$ (Recall that we are in the case $n \geq 2.$) Since $\tilde{X}$ is projective (of dimension $n$), we have a closed embedding $\pi : \tilde{X} \hookrightarrow \mathbb{P}^{N}$ for some $N \geq n = \dim(X).$ Then we apply Bertini's theorem to choose $n-1$ distinct hyperplanes $H_{1}, \dots, H_{n-1} \subset \mathbb{P}^{N}$ such that
- $C = \tilde{X} \cap H_{1} \cap \cdots \cap H_{n-1}$ is an irreducible curve, and
- none of $\pi^{-1}(p)$ and $\pi^{-1}(q)$ are contained in any of $H_{1}, \dots, H_{n-1}.$
Since $\dim(\pi^{-1}(p)) = \dim(\pi^{-1}(q)) = n - 1,$ both fibers intersect $H_{1} \cap \cdots \cap H_{n-1}$ Since $\pi$ is projective, it is closed, so its image $\pi(C) \subset X$ is a closed subset and an irreducible curve that contain $p$ and $q$. $\Box$
Personal remark about Bertini. When $X$ is smooth, Theorem 8.18 in Hartshorne (note: this theorem seems to use the hypothesis that $k$ is algebraically closed) does the job, but the general version we use is harder. (I need to ask the lecturer again where this is written.) I was told that one can use Hironaka's theorem on the resolution of singularities for the case when the characteristic is $0.$ This will be okay as we only care about the case $k = \mathbb{C}$ in this class, but I am now unsure whether the statement is true for any algebraically closed $k,$ not just characteristic $0$ ones.
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