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I'm trying to solve Exercise 1.26 from the book "Moduli of Curves" by Harris and Morrison on page 14:

Exercise (1.26) Determine the normal bundle to the rational normal curve $C \subset \mathbb{P}^r =: X$ and show, by computing its $h^0$, that the Hilbert scheme parameterizing such curves is smooth at any point corresponding to a rational normal curve.

The part I can't solve is that one how the dimension of global sections of the normal bundle $h^0= \dim_{\mathbb{C}}H^0(C, N_{C / X})$ imply that the Hilbert scheme $\mathcal{H}$ is smooth at $[C]$.

Since $C$ is rational normal curve $C \cong \mathbb{P}^1$ and the long exact cohomology sequence of the exact sequence $0\rightarrow \mathcal{T}_{C} \rightarrow \mathcal{T}_{X}\otimes\mathcal{O}_{C} \rightarrow \mathcal{N}_{X|C} \rightarrow 0$ gives

$$\operatorname{dim} H^0 (C,N_{C/X}) = \operatorname{dim} H^0 (C,T_X \otimes O_C) -3 = (r+1) \cdot \dim \ H^0 (C,O_C(1))-3$$

(the last one is consequence of Euler sequence). Fine now we know $h^0$. Immediately before it was shown that the tangent space of $\mathcal{H}$ at $[C]$ is

$$T_{[C]}\mathcal{H}= H^0(C, N_{C / X})$$

Per definition a scheme $X$ is smooth at a point $P$ if the dimension of tangent space at this point coinsides with local dimension: ie there exist an open affine subscheme $U \subset X$ with $P \in U$ and $\dim_P T= \dim \ U$.

Working through previous pages I nowhere found informations about local dimension of the Hilbert scheme parametrizing rational normal curve so I have no idea how can I compare the dimension of $T_{[C]}\mathcal{H}= H^0(C, N_{C / X})$ which I calculated above with the local dimension of $\mathcal{H}$.

Can anybody give me some hints how to attack this part of the exercise?

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    $\begingroup$ You can represent the open subscheme of the Hilbert scheme as a GIT quotient. For instance for twisted cubics, one needs to choose 4 homogeneous polynomials of degree $3$ in $X$, $Y$, up to common scaling and then quotient out the $\mathbf{PGL}_2$-action on $X$, $Y$. This gives a $12$-dimensional component of the Hilbert scheme. $\endgroup$
    – Evgeny Shinder
    Commented Jun 15, 2020 at 21:30
  • $\begingroup$ @EvgenyShinder: Could you explain why the Hilbert scheme of rational normal curves can be modeled locally by such quotient? The reason why exactly 4 homogeneous poynomial of degree $3$ do the job isn't clear to me. Or could you give a reference where this constrcution you have described can be looked up? $\endgroup$
    – user267839
    Commented Jun 15, 2020 at 21:49
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    $\begingroup$ At least on the level of points of the moduli space, to specify a rational normal curve of degree $r$, you can first parametrize it as a map from $\mathbf{P}^1$ to $\mathbf{P}^r$ of degree $r$ (Veronese embedding). This amounts to choosing the basis of degree $r$ polynomials in $X$, $Y$. Two maps give the same curve if they differ by linear action on $X$, $Y$. $\endgroup$
    – Evgeny Shinder
    Commented Jun 15, 2020 at 22:05

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