Chapter I. Classical topology on varieties over $\mathbb{C}$
By a variety over $\mathbb{C}$, we mean a reduced scheme of finite type over $\mathbb{C}$.
Classical topology on an affine variety. If $X$ is an affine complex variety, then $$X = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})) \hookrightarrow \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]) = \mathbb{A}^{n}.$$ Note that $$X(\mathbb{C}) = \{\boldsymbol{a} = (a_{1}, \dots, a_{n}) \in \mathbb{C}^{n} : f_{1}(\boldsymbol{a}) = \cdots = f_{r}(\boldsymbol{a}) = 0\} \subset \mathbb{C}^{n} = \mathbb{A}^{n}(\mathbb{C}).$$ We define the classical topology on $X(\mathbb{C})$ to be the subspace topology of the Euclidean topology on $\mathbb{C}^{n}$. So far, this topology depends on the closed embedding $X \hookrightarrow \mathbb{A}^{n}$ of $\mathbb{C}$-schemes.
Classical topology on an affine variety. If $X$ is an affine complex variety, then $$X = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})) \hookrightarrow \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]) = \mathbb{A}^{n}.$$ Note that $$X(\mathbb{C}) = \{\boldsymbol{a} = (a_{1}, \dots, a_{n}) \in \mathbb{C}^{n} : f_{1}(\boldsymbol{a}) = \cdots = f_{r}(\boldsymbol{a}) = 0\} \subset \mathbb{C}^{n} = \mathbb{A}^{n}(\mathbb{C}).$$ We define the classical topology on $X(\mathbb{C})$ to be the subspace topology of the Euclidean topology on $\mathbb{C}^{n}$. So far, this topology depends on the closed embedding $X \hookrightarrow \mathbb{A}^{n}$ of $\mathbb{C}$-schemes.
Classical topology on affine complex variety does not depend on the choice of a closed embedding. To see this, say $$X = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})) \simeq \mathrm{Spec}(\mathbb{C}[y_{1}, \dots, y_{m}]/(g_{1}, \dots, g_{s}))$$ as $\mathbb{C}$-schemes. This is given by an isomorphism $$\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r}) \simeq \mathbb{C}[y_{1}, \dots, y_{m}]/(g_{1}, \dots, g_{s})$$ of $\mathbb{C}$-algebras. Say $x_i \mapsto G_i(\boldsymbol{y})$ and $y_j \mapsto F_j(\boldsymbol{x})$ under this isomorphism. Then we have two maps $G : \mathbb{C}^{n} \rightarrow \mathbb{C}^{m}$ and $F : \mathbb{C}^{m} \rightarrow \mathbb{C}^{n}$ that when restricted to $V(f_{1}, \dots, f_{r})(\mathbb{C})$ and $V(g_{1}, \dots, g_{s})(\mathbb{C})$, they are set-theoretically inverses to each other. Since these maps are given by polynomials, they are continuous with respect to the Eucleadian topologies coming from $\mathbb{C}^{n}$ and $\mathbb{C}^{m}.$ Thus, we have constructed a homeomorphism $V(f_{1}, \dots, f_{r})(\mathbb{C}) \simeq V(g_{1}, \dots, g_{s})(\mathbb{C}),$ as desired.
Properties. So far $X$ is an affine variety over $\mathbb{C}.$
(1) The classical topology on $X(\mathbb{C})$ is finer than the Zariski topology on $X(\mathbb{C})$ (i.e., the subspace topology $X(\mathbb{C}) \subset X$).
This is quite evident. Every zero set of a set of finitely many polynomials in $n$ variables in $\mathbb{C}^{n}$ is closed in the Euclidean topology of $\mathbb{C}^{n}$, and intersecting such locus with $X(\mathbb{C})$ is another zero set of finitely many polynomials.
(2) If $Z \hookrightarrow X$ is a closed subvariety, then the classical topology on $Z(\mathbb{C})$ is the subspace topology given by $Z(\mathbb{C}) \subset X(\mathbb{C}).$
First, for any $\mathbb{C}$-scheme map $\mathrm{Spec}(\mathbb{C}) \rightarrow Z,$ we get the $\mathbb{C}$-scheme map $$\mathrm{Spec}(\mathbb{C}) \rightarrow Z \hookrightarrow X.$$ This gives us $$Z(\mathbb{C}) \hookrightarrow X(\mathbb{C}).$$ We may choose a closed embedding $$X \hookrightarrow \mathbb{A}^{n}$$ of $\mathbb{C}$-schemes, which gives a closed embedding $Z \hookrightarrow X \hookrightarrow \mathbb{A}^{n}.$ We can choose a presentation of $Z$ by adding more polynomials in the presentation of $X$. Say $$X = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})$$ and $$Z = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r}, f_{r+1}, \dots, f_{s}).$$ Then $$Z(\mathbb{C}) = V(f_{1}, \dots, f_{s})(\mathbb{C}) \subset V(f_{1}, \dots, f_{r})(\mathbb{C}) = X(\mathbb{C}) \subset \mathbb{C}^{n}.$$ The classical topology of $Z(\mathbb{C})$ is given by $\mathbb{C}^{n}$ and it is closed in this sense. Since $X(\mathbb{C})$ is also closed in $\mathbb{C}^{n},$ this implies that the classical topology of $Z(\mathbb{C})$ can be regarded as the subspace $X(\mathbb{C}).$
(3) If $U \hookrightarrow X$ is an affine open subvariety, then the classical topology on $U(\mathbb{C})$ is given by the subspace topology of $X(\mathbb{C}).$
My first thought. We can take $Z := X \setminus U.$ We give the unique reduced closed subscheme structure on $Z.$ Since $Z(\mathbb{C})$ gets a subspace topology $X(\mathbb{C}),$ I thought $U(\mathbb{C}) = X(\mathbb{C}) \setminus Z(\mathbb{C})$ would get one too.
No! But the "classical topology" by definition mean that we are choosing a closed embedding from $U$ to an affine space, and get a topology from there. How do we know that such a topology would match with the topology on $U(\mathbb{C})$ given by that of $X(\mathbb{C})$ and $Z(\mathbb{C})$? This requires a proof.
Oh no, but $U$ is open, how can we find a closed embedding? Well, the key here is that $U$ is an affine scheme! Thus, it is quasi-compact, so $$U = D(g_{1}) \cup \cdots \cup D(g_{s}) = D(g) \subset X,$$ where $g = g_{1} \cdots g_{s}$ with $g_{i} \in \mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})$ if we choose our presentation of $X$ as follows: $$X = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})).$$ Now, note that $$D(g) \simeq \mathrm{Spec}\left(\frac{\mathbb{C}[x_{1}, \dots, x_{n}]}{(f_{1}, \dots, f_{r})}\right)_{g}.$$ For computation, let us write $\bar{g}$ instead of $g$ and say $g \in \mathbb{C}[x_{1}, \dots, x_{n}]$ is a lift of $\bar{g} \in \mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r}).$ Denote $\boldsymbol{x} = (x_{1}, \dots, x_{n}).$ Then we have a $\mathbb{C}$-algebra isomorphism $$\left(\frac{\mathbb{C}[x_{1}, \dots, x_{n}]}{(f_{1}, \dots, f_{r})}\right)_{\bar{g}} \simeq \frac{\mathbb{C}[x_{1}, \dots, x_{n},x_{n+1}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)},$$ given by $x_{i} \mapsto x_{i}.$ The inverse is given by $x_{n+1} \mapsto 1/\bar{g}.$ Thus, we can construct a closed embedding $$U = D(\bar{g}) \simeq \mathrm{Spec}\left( \frac{\mathbb{C}[x_{1}, \dots, x_{n},x_{n+1}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)} \right) \hookrightarrow \mathbb{A}^{n+1}.$$ This implies that the classical topology on $U(\mathbb{C})$ is given by the subspace topology $$U(\mathbb{C}) \simeq V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)(\mathbb{C}) \subset \mathbb{C}^{n+1}.$$ The isomorphism written here is a bit funny precisely because the affine scheme structure of $D(\bar{g})$ is funny. It is more like this isomorphism is declaring the structure on $U(\mathbb{C})$ with the classical topology.
What do I mean? A priori, when we first learn schemes, we don't know whether $D(g)$ is an affine scheme for a given element $g$ in a commutative ring $R$. We understand that $D(g) \simeq \mathrm{Spec}(R_{g})$ is a homeomorphism and declare the affine scheme structure on $D(g)$ using this homeomorphism.
As Zariski topological spaces and also $\mathbb{C}$-schemes (because the reduced structure for a closed subscheme is unique), we have $$U = D(\bar{g}) \simeq D(g) \cap V(f_{1}, \dots, f_{r}) \subset \mathbb{A}^{n},$$ so we may assume $$U = D(g) \cap V(f_{1}, \dots, f_{r}).$$ Similarly, we may assume $$X = V(f_{1}, \dots, f_{r}) \subset \mathbb{A}^{n}.$$ If we follow the $\mathbb{C}$-scheme isomorphism $$V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1) \simeq D(g(\boldsymbol{x})) \cap V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))$$ we get $$(x_{1} - a_{1}, \dots, x_{n} - a_{n}, x_{n+1} - a_{n+1}) \mapsto (x_{1} - a_{1}, \dots, x_{n} - a_{n}),$$ on the sets of $\mathbb{C}$-points. This coincides with the map given by the restriction of $\mathbb{C}^{n+1} \rightarrow \mathbb{C}^{n}$ forgetting the last coordinate. This map is continuous, so the following bijection from left to right is continuous: $$V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)(\mathbb{C}) \simeq D(g)(\mathbb{C}) \cap X(\mathbb{C}) = U(\mathbb{C}),$$ where the right-hand side gets the subspace topology of the classical topology on $X(\mathbb{C}).$ When $g = 0 \in \mathbb{C}[\boldsymbol{x}],$ both sides are empty, so the bijection is a homeomorphism. Otherwise, we can take $(a_{1}, \dots, a_{n}) \mapsto (a_{1}, \dots, a_{n}, 1/g(a_{1}, \dots, a_{n}))$ as the continuous inverse. Thus, the bijection is a homeomorphism, so the classical topology on $U(\mathbb{C})$ coincides with the subspace topology in $X(\mathbb{C}).$
Classical topology on global variety (cf. 5.3.G of Vakil's book). Now let $X$ be any variety over $\mathbb{C}$. For instance, this includes reduced projective schemes over $\mathbb{C},$ which are rarely affine. To define the classical topology on $X(\mathbb{C}),$ let $X = \bigcup_{i \in I}U_{i}$ be an affine open cover. For any distinct $i, j \in I,$ we may have two different subspace topologies on $U_{i}(\mathbb{C}) \cap U_{j}(\mathbb{C}),$ one from $U_{i}(\mathbb{C})$ and another from $U_{j}(\mathbb{C}).$ Both induce the same topology because for any affine open $V \subset U_{i} \cap U_{j},$ we know that the classical topology on $V(\mathbb{C})$ is the same as the subspace topology either from $U_{i}(\mathbb{C})$ or $U_{j}(\mathbb{C}).$ We may use the same argument for the triple intersections, so we can glue $U_{i}(\mathbb{C})$ with the classical topology we studied above to give a unique topology on $X(\mathbb{C}).$ This topology is called the classical topology on $X(\mathbb{C}).$ More specifically, if we fix any affine open cover $X = \bigcup_{i \in I}U_{i},$ any subset $S \subset X(\mathbb{C})$ is open in the classical topology if and only if $S \cap U_{i}(\mathbb{C})$ is open for all $i \in I.$ The classical topology on $X(\mathbb{C})$ does not depend on the choice of an affine open cover of $X$ because if there are two, we may take the common refinement. In particular, for any affine open subset $U \subset X,$ we see that $U(\mathbb{C})$ is open in $X(\mathbb{C}).$
Note that (1) the classical topology on $X(\mathbb{C})$ is finer than the Zariski topology, (2) for any closed subvariety $Z \hookrightarrow X,$ the classical topology on $Z(\mathbb{C})$ coincides with the subspace topology of $X(\mathbb{C}),$ and (3) for any open subvariety $U \hookrightarrow X,$ the classical topology on $U(\mathbb{C})$ coincides with the subspace topology of $X(\mathbb{C}).$ (Proof: take an affine open cover of $X$ and intersect!)
Functoriality (cf. 6.3.N of Vakil's book). Any morphism $X \rightarrow Y$ of $\mathbb{C}$-varieties induces $X(\mathbb{C}) \rightarrow Y(\mathbb{C})$ by $$\mathrm{Spec}(\mathbb{C}) \rightarrow X \rightarrow Y.$$ Due to the description above, it is evident that this is a functor. The source of this functor is the category $\mathbf{Var}_{\mathbb{C}}$ of $\mathbb{C}$-varieties. What is the target category of this functor? Surely, we can start with the category $\textbf{Set}$ of sets. One can check that we can narrow down to the category $\textbf{Top}$ of topological spaces (i.e., the induced morphism $X(\mathbb{C}) \rightarrow Y(\mathbb{C})$ is continuous). Whichever category it sits in we denote this functor by $X \mapsto X^{\mathrm{an}}$ or maybe $$(\pi : X \rightarrow Y) \mapsto (\pi^{\mathrm{an}} : X^{\mathrm{an}} \rightarrow Y^{\mathrm{an}}).$$ Let's call this the analytification functor.
Guess. It seems like the class will soon talk about narrowing the target category down to the category of $\mathbb{C}$-analytic spaces with holomorphic mappings, whatever those mean. Then it will probably discuss GAGA theorems, which seems to be saying that this fuctor becomes something close to a fully faithful functor. (Probably not exactly, but something more interesting!)
In particular, a regular function on $X$ can also get analytified (and we know that it is at least continuous with the classical topology. Indeed, given any $f \in \Gamma(X, \mathscr{O}_{X}),$ the $\mathbb{C}$-algebra map $\mathbb{C}[t] \rightarrow \Gamma(X, \mathscr{O}_{X})$ given by $t \mapsto f$ gives rise to a $\mathbb{C}$-scheme map $X \rightarrow \mathbb{A}^{1},$ which sends all points where $f$ vanishes (not just closed points) are sent to $(t),$ the origin. We then analytify this.
Thought. It would be interesting if this turn out to be (something close to) the way we use to analytify the structure sheaf $\mathscr{O}_{X}.$
Remark. Every point in $X(\mathbb{C}) = X^{\mathrm{an}}$ has a countable basis of open neighborhoods in the classical topology. (I don't think I understand what this means. Perhaps this just means that I can just take an affine Zariski open subset containing that point and take its analytification.)
Analytification commutes with product. This asserts that $$(X \times Y)^{\mathrm{an}} \simeq X^{\mathrm{an}} \times Y^{\mathrm{an}}.$$ Of course, this would only make sense as a homeomorphism so far, unless we think about morphisms in a narrower category than $\textbf{Top}.$
First, note that we get $$(X \times Y)(\mathbb{C}) \simeq X(\mathbb{C}) \times Y(\mathbb{C})$$ in $\textbf{Set}.$ Why? It is because getting a map from $\mathrm{Spec}(\mathbb{C})$ to $X \times Y$ in $\textbf{Var}_{\mathbb{C}}$ is precisely the same as getting a pair of maps to $\mathrm{Spec}(\mathbb{C}) \rightarrow X$ and $\mathrm{Spec}(\mathbb{C}) \rightarrow Y.$ Next, we show that the classical topologies on both side matches. Given any affine opens $U \subset X$ and $V \subset Y,$ we choose closed embeddings $U \hookrightarrow \mathbb{A}^{m}$ and $V \hookrightarrow \mathbb{A}^{n}.$ Then we get the closed embedding $U \times V \hookrightarrow \mathbb{A}^{m + n}.$ The last map is induced categorically, but we check that this is a closed embedding using affine presentations. That is, write $$U = \mathrm{Spec}\left(\frac{\mathbb{C}[x_{1}, \dots, x_{m}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))}\right)$$ and $$V = \mathrm{Spec}\left(\frac{\mathbb{C}[y_{1}, \dots, y_{n}]}{(g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))}\right).$$ We have $$\begin{align*}\left(\frac{\mathbb{C}[x_{1}, \dots, x_{m}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))}\right) & \otimes_{\mathbb{C}} \left(\frac{\mathbb{C}[y_{1}, \dots, y_{n}]}{(g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))}\right) \\ &\simeq \frac{\mathbb{C}[x_{1}, \dots, x_{m}, y_{1}, \dots, y_{n}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))}, \end{align*}$$ so the closed embedding is given by $$U \times V \simeq V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y})) \subset \mathbb{A}^{m+n}.$$ Hence, the classical topology on $(U \times V)(\mathbb{C})$ is given by the subspace topology of $$V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))(\mathbb{C}) \subset \mathbb{C}^{m+n}.$$ Since $$V\left(\begin{array}{c}f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), \\ g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y})\end{array}\right)(\mathbb{C}) = V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))(\mathbb{C}) \times V(g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))(\mathbb{C})$$ in $\mathbb{C}^{m+n},$ we see that the classical topology on $(U \times V)(\mathbb{C})$ is given by the product topology on $U(\mathbb{C}) \times V(\mathbb{C}).$ Since $X \times Y$ can be covered by $U \times V,$ the topology on $(X \times Y)(\mathbb{C})$ can be given by gluing those of $(U \times V)(\mathbb{C}) = U(\mathbb{C}) \times V(\mathbb{C}),$ and since this identity can be seen as the restriction of the bijection $$(X \times Y)(\mathbb{C}) \leftrightarrow X(\mathbb{C}) \times Y(\mathbb{C}),$$ we see that this bijection is a homemomorphism. Thus, the classical topology on $(X \times Y)(\mathbb{C})$ can be given by the product topology on $X(\mathbb{C}) \times Y(\mathbb{C}),$ where $X(\mathbb{C})$ and $Y(\mathbb{C})$ are given the classical topologies.
More thought. I am sure there is more structure to the product identity we have seen than merely as a homeomorphism, but so far at least we know the following. If we consider the analytification functor $\mathrm{an} : \textbf{Var}_{\mathbb{C}} \rightarrow \textbf{Top},$ with the target being just the category of topological spaces with continuous maps, then the following diagram of functors commute:
Of course, one needs to check that the functors commute for the suitable morphisms as well, but this seems to go through without any pain.
Next time. We preview what we will discuss next time.
Theorem. Let $X$ be an irreducible $\mathbb{C}$-variety. If $U \subset X$ is nonempty and Zariski-open, then $U(\mathbb{C}) \subset X(\mathbb{C})$ is dense in the classical topology.
Corollary. Let $X$ be a $\mathbb{C}$-variety. If $U \subset X$ is Zariski-open and Zariski-dense, then $U(\mathbb{C}) \subset X(\mathbb{C})$ is dense in the classical topology.
Proof. Since $X$ is Noetherian, we may write $X = X_{1} \cup \cdots \cup X_{r},$ where $X_{i}$ are distinct irreducible components of $X.$ Write $$U = (U \cap X_{1}) \cup \cdots \cup (U \cap X_{r}).$$ Since $U \subset X$ is dense, taking the closure in $X$ gives $$X = \overline{(U \cap X_{1})} \cup \cdots \cup \overline{(U \cap X_{r})}.$$ This implies that $U \cap X_{i} \neq \emptyset$ for all $i,$ so $U(\mathbb{C}) \cap X_{i}(\mathbb{C})$ is dense in $X_{i}(\mathbb{C})$ for all $i.$ Therefore, we have $U(\mathbb{C})$ dense in $X(\mathbb{C}),$ as we have the finite unions. $\Box$
My first thought. We can take $Z := X \setminus U.$ We give the unique reduced closed subscheme structure on $Z.$ Since $Z(\mathbb{C})$ gets a subspace topology $X(\mathbb{C}),$ I thought $U(\mathbb{C}) = X(\mathbb{C}) \setminus Z(\mathbb{C})$ would get one too.
No! But the "classical topology" by definition mean that we are choosing a closed embedding from $U$ to an affine space, and get a topology from there. How do we know that such a topology would match with the topology on $U(\mathbb{C})$ given by that of $X(\mathbb{C})$ and $Z(\mathbb{C})$? This requires a proof.
Oh no, but $U$ is open, how can we find a closed embedding? Well, the key here is that $U$ is an affine scheme! Thus, it is quasi-compact, so $$U = D(g_{1}) \cup \cdots \cup D(g_{s}) = D(g) \subset X,$$ where $g = g_{1} \cdots g_{s}$ with $g_{i} \in \mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})$ if we choose our presentation of $X$ as follows: $$X = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})).$$ Now, note that $$D(g) \simeq \mathrm{Spec}\left(\frac{\mathbb{C}[x_{1}, \dots, x_{n}]}{(f_{1}, \dots, f_{r})}\right)_{g}.$$ For computation, let us write $\bar{g}$ instead of $g$ and say $g \in \mathbb{C}[x_{1}, \dots, x_{n}]$ is a lift of $\bar{g} \in \mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r}).$ Denote $\boldsymbol{x} = (x_{1}, \dots, x_{n}).$ Then we have a $\mathbb{C}$-algebra isomorphism $$\left(\frac{\mathbb{C}[x_{1}, \dots, x_{n}]}{(f_{1}, \dots, f_{r})}\right)_{\bar{g}} \simeq \frac{\mathbb{C}[x_{1}, \dots, x_{n},x_{n+1}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)},$$ given by $x_{i} \mapsto x_{i}.$ The inverse is given by $x_{n+1} \mapsto 1/\bar{g}.$ Thus, we can construct a closed embedding $$U = D(\bar{g}) \simeq \mathrm{Spec}\left( \frac{\mathbb{C}[x_{1}, \dots, x_{n},x_{n+1}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)} \right) \hookrightarrow \mathbb{A}^{n+1}.$$ This implies that the classical topology on $U(\mathbb{C})$ is given by the subspace topology $$U(\mathbb{C}) \simeq V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)(\mathbb{C}) \subset \mathbb{C}^{n+1}.$$ The isomorphism written here is a bit funny precisely because the affine scheme structure of $D(\bar{g})$ is funny. It is more like this isomorphism is declaring the structure on $U(\mathbb{C})$ with the classical topology.
What do I mean? A priori, when we first learn schemes, we don't know whether $D(g)$ is an affine scheme for a given element $g$ in a commutative ring $R$. We understand that $D(g) \simeq \mathrm{Spec}(R_{g})$ is a homeomorphism and declare the affine scheme structure on $D(g)$ using this homeomorphism.
As Zariski topological spaces and also $\mathbb{C}$-schemes (because the reduced structure for a closed subscheme is unique), we have $$U = D(\bar{g}) \simeq D(g) \cap V(f_{1}, \dots, f_{r}) \subset \mathbb{A}^{n},$$ so we may assume $$U = D(g) \cap V(f_{1}, \dots, f_{r}).$$ Similarly, we may assume $$X = V(f_{1}, \dots, f_{r}) \subset \mathbb{A}^{n}.$$ If we follow the $\mathbb{C}$-scheme isomorphism $$V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1) \simeq D(g(\boldsymbol{x})) \cap V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))$$ we get $$(x_{1} - a_{1}, \dots, x_{n} - a_{n}, x_{n+1} - a_{n+1}) \mapsto (x_{1} - a_{1}, \dots, x_{n} - a_{n}),$$ on the sets of $\mathbb{C}$-points. This coincides with the map given by the restriction of $\mathbb{C}^{n+1} \rightarrow \mathbb{C}^{n}$ forgetting the last coordinate. This map is continuous, so the following bijection from left to right is continuous: $$V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)(\mathbb{C}) \simeq D(g)(\mathbb{C}) \cap X(\mathbb{C}) = U(\mathbb{C}),$$ where the right-hand side gets the subspace topology of the classical topology on $X(\mathbb{C}).$ When $g = 0 \in \mathbb{C}[\boldsymbol{x}],$ both sides are empty, so the bijection is a homeomorphism. Otherwise, we can take $(a_{1}, \dots, a_{n}) \mapsto (a_{1}, \dots, a_{n}, 1/g(a_{1}, \dots, a_{n}))$ as the continuous inverse. Thus, the bijection is a homeomorphism, so the classical topology on $U(\mathbb{C})$ coincides with the subspace topology in $X(\mathbb{C}).$
Classical topology on global variety (cf. 5.3.G of Vakil's book). Now let $X$ be any variety over $\mathbb{C}$. For instance, this includes reduced projective schemes over $\mathbb{C},$ which are rarely affine. To define the classical topology on $X(\mathbb{C}),$ let $X = \bigcup_{i \in I}U_{i}$ be an affine open cover. For any distinct $i, j \in I,$ we may have two different subspace topologies on $U_{i}(\mathbb{C}) \cap U_{j}(\mathbb{C}),$ one from $U_{i}(\mathbb{C})$ and another from $U_{j}(\mathbb{C}).$ Both induce the same topology because for any affine open $V \subset U_{i} \cap U_{j},$ we know that the classical topology on $V(\mathbb{C})$ is the same as the subspace topology either from $U_{i}(\mathbb{C})$ or $U_{j}(\mathbb{C}).$ We may use the same argument for the triple intersections, so we can glue $U_{i}(\mathbb{C})$ with the classical topology we studied above to give a unique topology on $X(\mathbb{C}).$ This topology is called the classical topology on $X(\mathbb{C}).$ More specifically, if we fix any affine open cover $X = \bigcup_{i \in I}U_{i},$ any subset $S \subset X(\mathbb{C})$ is open in the classical topology if and only if $S \cap U_{i}(\mathbb{C})$ is open for all $i \in I.$ The classical topology on $X(\mathbb{C})$ does not depend on the choice of an affine open cover of $X$ because if there are two, we may take the common refinement. In particular, for any affine open subset $U \subset X,$ we see that $U(\mathbb{C})$ is open in $X(\mathbb{C}).$
Note that (1) the classical topology on $X(\mathbb{C})$ is finer than the Zariski topology, (2) for any closed subvariety $Z \hookrightarrow X,$ the classical topology on $Z(\mathbb{C})$ coincides with the subspace topology of $X(\mathbb{C}),$ and (3) for any open subvariety $U \hookrightarrow X,$ the classical topology on $U(\mathbb{C})$ coincides with the subspace topology of $X(\mathbb{C}).$ (Proof: take an affine open cover of $X$ and intersect!)
Functoriality (cf. 6.3.N of Vakil's book). Any morphism $X \rightarrow Y$ of $\mathbb{C}$-varieties induces $X(\mathbb{C}) \rightarrow Y(\mathbb{C})$ by $$\mathrm{Spec}(\mathbb{C}) \rightarrow X \rightarrow Y.$$ Due to the description above, it is evident that this is a functor. The source of this functor is the category $\mathbf{Var}_{\mathbb{C}}$ of $\mathbb{C}$-varieties. What is the target category of this functor? Surely, we can start with the category $\textbf{Set}$ of sets. One can check that we can narrow down to the category $\textbf{Top}$ of topological spaces (i.e., the induced morphism $X(\mathbb{C}) \rightarrow Y(\mathbb{C})$ is continuous). Whichever category it sits in we denote this functor by $X \mapsto X^{\mathrm{an}}$ or maybe $$(\pi : X \rightarrow Y) \mapsto (\pi^{\mathrm{an}} : X^{\mathrm{an}} \rightarrow Y^{\mathrm{an}}).$$ Let's call this the analytification functor.
Guess. It seems like the class will soon talk about narrowing the target category down to the category of $\mathbb{C}$-analytic spaces with holomorphic mappings, whatever those mean. Then it will probably discuss GAGA theorems, which seems to be saying that this fuctor becomes something close to a fully faithful functor. (Probably not exactly, but something more interesting!)
In particular, a regular function on $X$ can also get analytified (and we know that it is at least continuous with the classical topology. Indeed, given any $f \in \Gamma(X, \mathscr{O}_{X}),$ the $\mathbb{C}$-algebra map $\mathbb{C}[t] \rightarrow \Gamma(X, \mathscr{O}_{X})$ given by $t \mapsto f$ gives rise to a $\mathbb{C}$-scheme map $X \rightarrow \mathbb{A}^{1},$ which sends all points where $f$ vanishes (not just closed points) are sent to $(t),$ the origin. We then analytify this.
Thought. It would be interesting if this turn out to be (something close to) the way we use to analytify the structure sheaf $\mathscr{O}_{X}.$
Remark. Every point in $X(\mathbb{C}) = X^{\mathrm{an}}$ has a countable basis of open neighborhoods in the classical topology. (I don't think I understand what this means. Perhaps this just means that I can just take an affine Zariski open subset containing that point and take its analytification.)
Analytification commutes with product. This asserts that $$(X \times Y)^{\mathrm{an}} \simeq X^{\mathrm{an}} \times Y^{\mathrm{an}}.$$ Of course, this would only make sense as a homeomorphism so far, unless we think about morphisms in a narrower category than $\textbf{Top}.$
First, note that we get $$(X \times Y)(\mathbb{C}) \simeq X(\mathbb{C}) \times Y(\mathbb{C})$$ in $\textbf{Set}.$ Why? It is because getting a map from $\mathrm{Spec}(\mathbb{C})$ to $X \times Y$ in $\textbf{Var}_{\mathbb{C}}$ is precisely the same as getting a pair of maps to $\mathrm{Spec}(\mathbb{C}) \rightarrow X$ and $\mathrm{Spec}(\mathbb{C}) \rightarrow Y.$ Next, we show that the classical topologies on both side matches. Given any affine opens $U \subset X$ and $V \subset Y,$ we choose closed embeddings $U \hookrightarrow \mathbb{A}^{m}$ and $V \hookrightarrow \mathbb{A}^{n}.$ Then we get the closed embedding $U \times V \hookrightarrow \mathbb{A}^{m + n}.$ The last map is induced categorically, but we check that this is a closed embedding using affine presentations. That is, write $$U = \mathrm{Spec}\left(\frac{\mathbb{C}[x_{1}, \dots, x_{m}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))}\right)$$ and $$V = \mathrm{Spec}\left(\frac{\mathbb{C}[y_{1}, \dots, y_{n}]}{(g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))}\right).$$ We have $$\begin{align*}\left(\frac{\mathbb{C}[x_{1}, \dots, x_{m}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))}\right) & \otimes_{\mathbb{C}} \left(\frac{\mathbb{C}[y_{1}, \dots, y_{n}]}{(g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))}\right) \\ &\simeq \frac{\mathbb{C}[x_{1}, \dots, x_{m}, y_{1}, \dots, y_{n}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))}, \end{align*}$$ so the closed embedding is given by $$U \times V \simeq V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y})) \subset \mathbb{A}^{m+n}.$$ Hence, the classical topology on $(U \times V)(\mathbb{C})$ is given by the subspace topology of $$V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))(\mathbb{C}) \subset \mathbb{C}^{m+n}.$$ Since $$V\left(\begin{array}{c}f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), \\ g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y})\end{array}\right)(\mathbb{C}) = V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))(\mathbb{C}) \times V(g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))(\mathbb{C})$$ in $\mathbb{C}^{m+n},$ we see that the classical topology on $(U \times V)(\mathbb{C})$ is given by the product topology on $U(\mathbb{C}) \times V(\mathbb{C}).$ Since $X \times Y$ can be covered by $U \times V,$ the topology on $(X \times Y)(\mathbb{C})$ can be given by gluing those of $(U \times V)(\mathbb{C}) = U(\mathbb{C}) \times V(\mathbb{C}),$ and since this identity can be seen as the restriction of the bijection $$(X \times Y)(\mathbb{C}) \leftrightarrow X(\mathbb{C}) \times Y(\mathbb{C}),$$ we see that this bijection is a homemomorphism. Thus, the classical topology on $(X \times Y)(\mathbb{C})$ can be given by the product topology on $X(\mathbb{C}) \times Y(\mathbb{C}),$ where $X(\mathbb{C})$ and $Y(\mathbb{C})$ are given the classical topologies.
More thought. I am sure there is more structure to the product identity we have seen than merely as a homeomorphism, but so far at least we know the following. If we consider the analytification functor $\mathrm{an} : \textbf{Var}_{\mathbb{C}} \rightarrow \textbf{Top},$ with the target being just the category of topological spaces with continuous maps, then the following diagram of functors commute:
$\require{AMScd} \begin{CD} \textbf{Var}_{\mathbb{C}}^{2} @>{\times}>> \textbf{Var}_{\mathbb{C}} \\ @V\mathrm{an}^{2}VV @V\mathrm{an}VV \\ \textbf{Top}^{2} @>{\times}>> \textbf{Top}. \end{CD}$
Of course, one needs to check that the functors commute for the suitable morphisms as well, but this seems to go through without any pain.
Next time. We preview what we will discuss next time.
Theorem. Let $X$ be an irreducible $\mathbb{C}$-variety. If $U \subset X$ is nonempty and Zariski-open, then $U(\mathbb{C}) \subset X(\mathbb{C})$ is dense in the classical topology.
Corollary. Let $X$ be a $\mathbb{C}$-variety. If $U \subset X$ is Zariski-open and Zariski-dense, then $U(\mathbb{C}) \subset X(\mathbb{C})$ is dense in the classical topology.
Proof. Since $X$ is Noetherian, we may write $X = X_{1} \cup \cdots \cup X_{r},$ where $X_{i}$ are distinct irreducible components of $X.$ Write $$U = (U \cap X_{1}) \cup \cdots \cup (U \cap X_{r}).$$ Since $U \subset X$ is dense, taking the closure in $X$ gives $$X = \overline{(U \cap X_{1})} \cup \cdots \cup \overline{(U \cap X_{r})}.$$ This implies that $U \cap X_{i} \neq \emptyset$ for all $i,$ so $U(\mathbb{C}) \cap X_{i}(\mathbb{C})$ is dense in $X_{i}(\mathbb{C})$ for all $i.$ Therefore, we have $U(\mathbb{C})$ dense in $X(\mathbb{C}),$ as we have the finite unions. $\Box$
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