In an article I am reading it states "the space of algebraic curves of a given degree $d$ is compact". It seems to take this as a basic fact, as there is no explanation on this. I was wondering could someone please explain what this means?
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Well, it depends on the topology under consideration. So, have they said anything about what topology they are using on this set of curves? – Arthur Jul 21 '19 at 22:03
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Can't you just send $[a_{0,0}:\ldots:a_{d,0}]$ to $\sum_{m=0}^d \sum_{l=0}^{d-m} a_{m,l} X^m Y^l Z^{d-m-l}= 0$ ? – reuns Jul 21 '19 at 22:24
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@Arthur They said nothing... – Johnny T. Jul 22 '19 at 08:47
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Let $\mathbb{K}$ denote $\mathbb{R}$ or $\mathbb{C}$.
The vector space of homogeneous polynomials in $\mathbb{K}[X,Y,Z]$ of degree $d$ is isomorphic to $\mathbb{K}^{ {d+2} \choose {2}}$. For example, if $d=2$ it is the span of $\{X^2,Y^2,Z^2,XY,XZ, YZ\}$. Now two homogeneous polynomials define the same algebraic curve if they are scalar multiples of each other. So the space of algebraic curves is basically $\mathbb{P}^{{{d+2} \choose {2}} -1}(\mathbb{K})$. Under the standard quotient topology both the complex and real projective space are compact spaces.
I am quite sure this is what they mean.
Parthiv Basu
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2Also under the Zariski topology they are compact. In fact, they are Noetherian topological spaces. – Parthiv Basu Jul 22 '19 at 01:40
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1@JohnnyT. When removing the non-irreducible polynomials it is not compact anymore (as the irreducible polynomials are dense in all the degree $d$ homogeneous polynomials), same when quotienting by isomorphism classes (in the same way that $SL_2(Z) \setminus H$ is not compact) – reuns Jul 22 '19 at 17:11