Difference Between Gluing of Manifolds and that of Schemes

algebraic-geometrymanifolds

I understand that both manifolds and schemes can be defined by gluing of Euclidean spaces and affine schemes respectively. For example, one may regard a manifold $M$ as a limit of a collection $\{U_i\}_{i\in I}$ of open subsets of $\mathbb{R}^n$, where $n=\dim M$, together with another open subsets $U_{ij}$ of $\mathbb{R}^n$ for each $i\neq j\in I$ and embeddings $\phi^{ij}_i:U_{ij}\to U_i$ and $\phi^{ij}_j:U_{ij}\to U_j$. This way we can glue two manifolds $M$, $N$ along another manifold $L$ of same dimension whenever embedding $L\to M$ and $L\to N$ is given.

I am not very good at algebraic geometry, so I am not sure of details, but I think similar things will be possible for schemes: defining schemes as a limit of affine schemes, and gluing two schemes along embeddings.

However, for me, they look very different. For me, it seems quite easier to patching manifolds up, but gluing schemes is very difficult and 'rigid.' For example, there are uncountably many ways to construct $S^2$ by patching up $\mathbb{R}^2$, but the complex projective line $\mathbb{CP}^1$, which is topologically homeomorphic to $S^2$, can only be constructed by glueing two affine lines $\mathrm{Spec} \mathbb{C}[t]$, $\mathrm{Spec} \mathbb{C}[s]$ along the relation $\mathbb{C}[t]_t\to\mathbb{C}[s]_s, t\mapsto 1/s$.

For another example, one can glue two real lines, each of them identified by open intervals $(0,2)$ and $(1,3)$, to make another real line $(0,3)$, but in the world of algebraic geometry, this looks impossible to me.

This is really strange and confusing to me, since gluing of two manifold is quite easy and straightforward, but the gluing of two schemes looks nearly impossible. Am I misunderstanding something? Or are they really different? Can someone give me a good example of gluing two schemes?

Best Answer

However, for me, they look very different. For me, it seems quite easier to patching manifolds up, but gluing schemes is very difficult and 'rigid.' p For another example, one can glue two real lines, each of them identified by open intervals $(0,2)$ and $(1,3)$, to make another real line $(0,3)$, but in the world of algebraic geometry, this looks impossible to me.

I think the difference is that you have a kind of "intuitive" understanding of manifolds, where schemes appear to be more abstract. To glue the both intervals formally, you need to construct a homeomorphism on the parts you want to glue. This is obvious in your case, because both contain the interval $(1, 2)$, and you can glue along the identity.

In the same sense, you can easily glue schemes which "already belong to each other".

This is really strange and confusing to me, since gluing of two manifold is quite easy and straightforward, but the gluing of two schemes looks nearly impossible. Am I misunderstanding something? Or are they really different? Can someone give me a good example of gluing two schemes?

For example take the schemes $X = \mathbb{A}^1 = \text{Spec }k[x]$, and the open set $U = D(x) = \{ p \mid p \neq 0 \}$ (that set-notation should not be taken literally). Then you can glue two copies of $X$ along $U$, and obtain a scheme $Y$ which has the zero-point twice. Think about this like a line with a double-point, a bit similar to glueing two copies of $\mathbb{R}$ along the set $V = \{ p \mid p \neq 0 \}$. Of course the latter is not a manifold, but still a topological space.

$Y$ is a famous example for a scheme that is integral and of finite type over a field $k$, but not separated, so not a variety.

but the complex projective line $\mathbb{CP}^1$, which is topologically homeomorphic to $S^2$, can only be constructed by glueing two affine lines $\text{Spec }\mathbb{C}[t], \text{Spec }\mathbb{C}[s]$ along the relation $\mathbb{C}[t]_t \to \mathbb{C}[s]_s, t \mapsto 1/s$

You could still apply linear transformations, glueing along $$\mathbb{C}[t]_{at + b} \to \mathbb{C}[s]_{cs + d}, t \mapsto \frac{1 - b(cs + d)}{a(cs + d)}$$ for $a, b, c, d \in \mathbb{C}$ and $a, b \neq 0$. This is just an isomorphic renaming.

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