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.'
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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.
Best Answer
I feel that the thing to do is to start off by thinking about real projective spaces. There are also some tricks to help understand projective spaces in general.
One trick is to "lift" back to the underlying vector space. For example, to come to grips with the idea that any two projective lines in $\mathbb P^2(\mathbb R)$ meet in one point, we can think about what happens in the vector space $\mathbb R^3.$ A projective line is a linear collection of projective points, which themselves are described by lines through the origin in $\mathbb R^3.$ So, a projective line is described by a plane through the origin in $\mathbb R^3.$ Translating what "two lines in $\mathbb P^2(\mathbb R)$ meet in a point" means to their representatives in $\mathbb R^3,$ the statement becomes that any two planes through the origin intersect in a unique line. Can you visualize this?
Another trick is to remember that projective space decomposes (non-uniquely) into disjoint pieces: $\mathbb P^n = \mathbb A^n \sqcup\mathbb P^{n-1}.$ Often times, we can ignore the fact that we're working with projective space, if the behaviour we're interested in takes place entirely within the $\mathbb A^n$ piece.
For example, the projective plane is what we get by adding the projective line to the affine plane $\mathbb A^2$ (in our case $\mathbb A^2 = \mathbb R^2,$ but without the vector space "structure", i.e., we don't add points of affine space, or multiply them by scalars). Remember from your notes, the projective line $\mathbb P^1(\mathbb R)$ can be thought of as the circle $S^1$ with antipodal points identified. If we put $S^1$ inside $\mathbb R^2$ as the unit circle centred at the origin, then every line through the origin defines a unique point of $\mathbb P^1,$ by its intersection with $S^1.$ This means that the projective plane is what we get by adding to the affine plane one point for each direction (i.e. slope!) that a line can have in $\mathbb R^2.$ If we are interested in the intersection of two lines in $\mathbb R^2$ that we know have different slopes, then we can ignore projective geometry. But, if two lines are parallel, then they do not intersect in $\mathbb R^2,$ but they do define the same point of $\mathbb P^1,$ which means that they intersect in the $\mathbb P^1$ portion of $\mathbb P^2$ ("at infinity").
Regarding visualizing complex geometry as real geometry, well, it's limited, because visualizing geometry past 4 dimensions isn't really possible. So, as soon as we think about spaces of complex dimension two or more, we're stuck. I think it's better to understand the real picture where we can, and then try to accept that the complex numbers arise in a way that preserves (and simplifies) the essential algebraic properties of the real numbers. When you study algebraic geometry, you will learn that geometric properties (in the situations we can visualize) can be expressed by algebraic conditions (vanishing of polynomials, etc.). The algebraic conditions can make perfect sense over different fields and in any dimension, but you will still only be able to visualize the geometry in $\mathbb R^1,\mathbb R^2,\mathbb R^3$ (maybe also over finite fields...). This doesn't mean we cannot draw diagrams to help our understanding, but those diagrams will no longer be so closely linked to the literal solutions to our equations, if that makes sense. But it's surprising how far one can get with "meaningless" pictures, so I encourage you to draw them in the real case, and to keep them in mind over other fields!
Anyway, those notes you are reading look very nice, I would certainly recommend you to keep reading them, and things will become clearer the more you do.