# The topology of $y^2=(x-1)(x-2)(x-3)(x-4)$

Andreas Gathmann's lecture notes on algebraic geometry start by considering the curve $C_n=\{(x,y): y^2 = (x-1)(x-2)\cdots(x-2n)\} \subset \mathbb{C}^2$.

He claims that the topology of this curve is the same as that obtained by removing from two copies of $\mathbb{C}$ the real intervals $[1,2],...,[2n-1,2n]$ and gluing the two copies along those intervals in a reverse manner, identifying opposite sides, as indicated by his following diagram: I hope to understand this through some clarifications:

1. After gluing, what corresponds to the two points of $C_n$ having value $x=1.7$? It seems we lost them in the process of removing the intervals.

Answered: they are present in the A part of each sheet and not in B, and after the gluing they are the horizontal perimeter of the circular hole

2. Why do we need to glue the intervals in opposite sides? (sides A and B in the picture have non-compatible labels in the two copies, so the gluing is supposedly more contrived than is shown to the right).

Answered: We don't glue A to A. We glue instead A to B, to match the part with the crossing points to the part without them. This gives us the picture shown on the right.

3. In what exact sense does it characterize the topology of the curve? Do we say that $C_n$ as a topological space with the euclidean topology is homeomorphic to the glued space? Can we say what are all the (euclidean) open sets in $C_n$? (In other words, what did we achieve?)

4. Does this example even have any reference or bearing to the usual Zariski topology that we give to algebraic curves? What is the Zariski topology on the curve $C_n$?

• Because A 's neibor is surface,but B's neibor is hole. Commented May 6, 2016 at 15:43
• Hint: the bottom surface is a copy of the first after reversal. This is why $A_{top}$ faces $B_{bottom}$ and you glue these together. The two points $x=1.7$ are still there and they can serve as a crossing points between the top and bottom sheets, but changing the sign of the imaginary part (given the reversal).
– user65203
Commented May 20, 2016 at 12:55
• This is about the usual topology of the complex curve (or a real surface) - not Zariski topology. Commented Jun 2, 2016 at 7:22

Here is a summary of the answers.

1. After gluing, what corresponds to the two points of $C_n$ having value $x=1.7$? It seems we lost them in the process of removing the intervals.

Answered: they are present in the A part of each sheet and not in B, and after the gluing they are the horizontal perimeters of the circular holes.

2. Why do we need to glue the intervals in opposite sides? (sides A and B in the picture have non-compatible labels in the two copies, so the gluing is supposedly more contrived than is shown to the right).

Answered: We don't glue A to A. We glue instead A to B, to match the part with the crossing points to the part without them. This gives us the picture shown on the right.

3. (I don't know)

4. Does this example even have any reference or bearing to the usual Zariski topology that we give to algebraic curves? What is the Zariski topology on the curve $C_n$?

Answer: The zariski (affine) topology on every irreducible curve is the cofinite topology, so the only closed sets in the curve are the finite subsets and the curve itself.

We show that the curve is indeed irreducible: it is equivalent to the irreducibility of the polynomial $y^2-(x-1)...(x-2n)$ in $\mathbb{C}[x,y]$. It is true since otherwise if both factors have degree 1 in $y$ they must be $(y-p(x))(y+p(x))$ and we have p(x)^2=(x-1)...(x-2n), and if they are with degrees 2 and 0 in $y$ then they are $p(x)(q(x)y^2+h(x))$ and then $p, q$ must be constants. Thus we arrived at a contradiction.