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I'm looking for a comprehensive note/paper/chapter of a book which discusses the Hilbert Scheme of Points of Riemann sphere ($\mathbb{P}_{\mathbb{C}}^1$) (maybe via a less abstract, more constructive approach?)

For the case of $\mathbb{C}^2$ the Hilbert scheme of $n$ points, $\mathrm{Hilb}^n(\mathbb{C}^2)$, turns out to be smooth and symplectic. I'm wondering if $\mathrm{Hilb}^n(\mathbb{P}^1_{\mathbb{C}})$ also shares those nice properties. More generally what does $\mathrm{Hilb}^n(\mathbb{P}^1_{\mathbb{C}})$ look like as a scheme? I'm specially interested in knowing about morphisms $\mathrm{Hilb}^n(\mathbb{P}_\mathbb{C}^1)\to \mathbb{A}^1_\mathbb{C}$. Or maybe more generally rational functions instead of functions?

I'm a rookie in algebraic geometry so I'm hoping that a simple special case like Riemann sphere, can be treated more directly, rather than the usual abstract approach (which I still have trouble understanding). At the end of the day understanding $\mathrm{Hilb}^n(\mathbb{P}_{\mathbb{C}}^1)$ is all I need for pushing my theoretical physics research further.

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    $\begingroup$ I find it quite interesting that this Hilbert scheme has applications to theoretical physics. Could you perhaps tell us a little about the relationship, maybe by linking to some document? $\endgroup$ Jan 19 '17 at 22:08
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    $\begingroup$ Dear Georges, the connection is with Quantum Hall states and is my own findings so there is no document of it anywhere yet. Quantum Hall states turn out to be polynomial functions (roughly speaking) over a Riemann surface (2D nature is crucial). They are a many body system, so in a sense a function on Hilbert scheme of $n$ points is (again roughly speaking) a wavefunction (section) of the $n$ particles. These particles are bosons so the wavefunction should be symmetric. This translates into unordered $n$-tuple of points. $\endgroup$
    – Hamed
    Jan 20 '17 at 4:04
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    $\begingroup$ I'm very close to publication, once the preprint is out I'll definitely ping a message here since you're interested. The connection however to Hilbert schemes will be very brief, I plan to dig deeper for my next work. The work starts with Hilbert schemes, goes into binary invariants, then connects them regular graphs and stays graph theoretic until the end. Quantum Hall states are extremely rich mathematically actually. $\endgroup$
    – Hamed
    Jan 20 '17 at 4:10
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    $\begingroup$ Dear Hamed, thank you very much for your explanations. I'm impressed by the way you are pioneering a connection between Quantum Hall states (of which I have heard but know nothing about) and such a very abstract mathematical concept like Hilbert scheme. I wish you much well-deserved success in your research and I am looking forward to the ping that you kindly promised. $\endgroup$ Jan 20 '17 at 8:23
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$Hilb^n(\mathbb{P}^1) \cong \mathbb{P}^n$.

EDIT (an explanation). A subscheme $Z$ of length $n$ is determined by its ideal $I_Z$ and its embedding $I_Z \hookrightarrow \mathcal{O}$. Every ideal on a smooth curve is invertible, hence $I_Z \cong \mathcal{O}(-n)$, and its embedding is determined by an element of $Hom(\mathcal{O}(-n),\mathcal{O}) = H^0(\mathbb{P}^1,\mathcal{O}(n)) = k^{n+1}$. Thus $Hilb^n(\mathbb{P}^1) \cong \mathbb{P}(k^{n+1}) \cong \mathbb{P}^n$.

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  • $\begingroup$ An explanation of why this is so would surely benefit the OP and others... $\endgroup$ Jan 19 '17 at 18:19
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    $\begingroup$ @MarianoSuárez-Álvarez: You are right, of course. $\endgroup$
    – Sasha
    Jan 19 '17 at 18:28
  • $\begingroup$ Thank you. I'm wondering if this isomorphism can be thought of as follows too: From any set of $n$ points on $(x_i:y_i)\in \mathbb{P}^1$ one can construct a homogeneous polynomial $f\in \mathbb{C}[X,Y]$ such that the homogeneous roots of $f$ are exactly these $n$ points. The coefficients of $f$ now can be thought of the coordinates of a point in $\mathbb{P}^n$. The correspondence in fact goes both ways... $\endgroup$
    – Hamed
    Jan 19 '17 at 18:39
  • $\begingroup$ @Hamed: Yes, precisely. In my description, the morphism $O(-n) \to O$ is given by multiplication with such $f$. $\endgroup$
    – Sasha
    Jan 19 '17 at 19:15
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The Hilbert scheme of $n$ points on $\mathbb P^1$ is the space of unordered families of $n$ non-necessarily distinct points of $\mathbb P^1$, and we will see that it is isomorphic to $\mathbb P^n$.
To understand this concretely it is best to interpret $\mathbb P^n$ as the projective space associated to the vector space $\mathbb C[x,y]_n$ of homogeneous polynomials of degree $n$ in two variables .
In other words, the polynomial $a_0x^n+a_1x^{n-1}y+\cdots+a_ny^n$ is seen as the point $[a_0:a_1:\cdots:a_n]\in \mathbb P^n$
The required isomorphism then comes from the morphism $F:(\mathbb P^1)^n\to \mathbb P^n$ sending an $n$-tuple $([u_i:v_i])_{i=1}^n$ to the point corresponding to the polynomial $\Pi_{i=1}^n (v_i x-u_iy)\in \mathbb C[x,y]_n$.
Dividing out by the action of the symmetric group $S_n$ on $(\mathbb P^1)^n$ we get the required isomorphism $$f:\operatorname {Hilb}^n(\mathbb P^1)=(\mathbb P^1)^n/S_n=\operatorname {Sym}^n(\mathbb P^1)\stackrel {\cong}{\to} \mathbb P^n$$

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  • $\begingroup$ How does one know that Hilb is just the symmetric power in this case? Usually it isn't, no? $\endgroup$ Jan 19 '17 at 19:37
  • $\begingroup$ @Mariano: No, in general it isn't. A priori one has a Hilbert-Chow morphism $\operatorname {Hilb}^n(\mathbb P^1)\to \operatorname {Sym}^n(\mathbb P^1)$ which is birational, as can be seen by looking at the space of $n$ distinct points. However both the source and the target are smooth and the morphism is bijective (because there is only one fat point of given size on a smooth curve). Hence (by Zariski's Main Theorem) the Hilbert-Chow morphism is an isomorphism. Giving complete proofs would require quite some space.but I just wanted to show how concrete the Hilbert scheme was here. $\endgroup$ Jan 19 '17 at 20:36
  • $\begingroup$ I was curious, because I've never seen a really elementary argument for this. ;-) $\endgroup$ Jan 19 '17 at 21:14
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Here is another, extremely geometric, vision of the Hilbert scheme $\mathcal H=\mathcal H^{[n]}$ of $n$ points in the projective line $\mathbb P^1_{x:y}$, in other words the Hilbert scheme of the subschemes $S\subset \mathbb P^1$ with Hilbert polynomial the constant polynomial $n\in \mathbb Q[T]$.
Like all hypersurfaces of any projective space, each $S=S_a\subset \mathbb P^1_{x:y}$ has an equation, namely $f(a;x,y)= a_0x^n+a_{n-1}x^{n-1}y+\cdots+a_ny^n=0$ and the Hilbert scheme is the hypersurface $$ \mathcal H=V(f(a;x,y)) \subset \mathbb P^n_a\times \mathbb P^1_{x:y} $$ endowed with its projection $$p:\mathcal H\to \mathbb P^n_a:([a_0:a_1:\cdots:a_n],[x:y])\mapsto [a_0:a_1:\cdots:a_n]$$ so that $p^{-1}(a)=S_a\subset \mathbb P^1$, the subscheme with equation $a_0x^n+a_{n-1}x^{n-1}y+\cdots+a_ny^n=0$.
The Hilbert scheme $\mathcal H$ is smooth and connected but its fibres over $\mathbb P^n$ are not: the non-smooth fibres $S_a$ of $p$ correspond to those $a\in \mathbb P^n$ such that the discriminant $$\operatorname {discr} \: (a_0x^n+a_{n-1}x^{n-1}y+\cdots+a_ny^n)\in \mathbb C[a_0,\cdots,a_n]$$ is zero.
This is in line with the fact that $\mathbb P^n$ being simply connected cannot have an étale connected covering of degree two and thus $p:\mathcal H\to \mathbb P^n$ must have a non-empty ramification locus.

Remarks
1) The construction shows that $\mathcal H$ is projective so that all morphisms to $\mathbb A^1_\mathbb C$ are constant, which answers one of Hamed's questions.
2) The connectednes of $\mathcal H$ is an extremely special case of Hartshorne's brilliant Ph.D: he proved that every Hilbert scheme (corresponding to some fixed Hilbert polunomial) is connected
3) The smoothness of $\mathcal H$ is a low-tech calculation using only the implicit function theorem of advanced calculus.
These calculations can easily be generalized to the case of the Hilbert scheme of hypersurfaces of $\mathbb P^N$.
Unfortunately these calculations cannot be found in the literature (as far as I know), and they could be the theme of another question...

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    $\begingroup$ Hi Georges, three questions: 1) In this example you're not fixing the whole Hilbert polynomial, but only the leading term, so is this moduli space called the Hilbert scheme? 2) Can't smoothness be checked simply by taking the Jacobian of the defining equation after using the Segre embedding? 3) When you talk about the Hilbert scheme of hypersurfaces, do you mean all hypersurfaces? Cheers. $\endgroup$
    – user347489
    Jan 21 '17 at 7:48
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    $\begingroup$ Dear @user347489: 1) But yes, I'm fixing the whole Hilbert polynomial ! Only for subschemes consisting of $n$ points, that polynomial is just the constant $n$, so that it coincides with its leading term. 2) Yes, in principle this should work but I don't know if the computations are manageable. 3) I mean the Hilbert scheme of all hypersurfaces of fixed degree $n$ in some fixed projective space $\mathbb P^N$. These hypersurfaces all have the same Hilbert polynomial (of degree $N-1$) , namely $P(t)=\binom {t+N}{N}-\binom {t+N-n}{N}$ (Hartshorne, page 52, line before last). $\endgroup$ Jan 21 '17 at 8:40
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    $\begingroup$ Great! I assume in that case $$ H=V(f(a,x_0,\dots, x_N))\subset\mathbb{P}^{{N+d\choose N}-1}\times\mathbb{P}^N, $$ and everything else you say works verbatim? Would you mind sketching how would you check smoothness using the implicit function theorem? Thanks! $\endgroup$
    – user347489
    Jan 21 '17 at 9:09
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    $\begingroup$ Dear @user347489, yes you are absolutely right and your notation $d$ for the common degree of the hypersurfaces is better than the $n$ of this thread. As for the proof of smoothness, it is a bit too technical for a comment. I encourage you to ask a new, carefully thought out question about this on our site and hope for the best :-) $\endgroup$ Jan 21 '17 at 9:55

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