Computation of the global sections of a normal sheaf

Question

Let $Y\subset X=\mathbb{P}^r$ be the image of the Veronese embedding $\mathbb{P}^1\rightarrow\mathbb{P}^r$. I want to calculate $dim$ $H^{0}(C,\mathcal{N}_{Y|X})$, where $\mathcal{N}_{Y|X}$ is the normal sheaf of $Y$ in $X$. My idea was to use the short exact sequence $0\rightarrow \mathcal{T}_{Y} \rightarrow \mathcal{T}_{X}\otimes\mathcal{O}_{Y}\rightarrow\mathcal{N}_{Y|X}\rightarrow 0$ but I'm not really sure about the way to proceed. Can someone give me some help? Thank you in advance!

Answer 1

It seems like you are almost there, so let's fill in the remaining details.

As you said in the comments, passing to the cohomology long exact sequence associated to your short exact sequence gives

$$\operatorname{dim} H^0 (Y,N) = \operatorname{dim} H^0 (Y,T_X \otimes O_Y) -3.$$

So we need to calculate the space of sections of $T_X \otimes O_Y$. As @IrfanKadikoylu suggests, we have have the Euler sequence

$$0 \rightarrow O_X \rightarrow O_X(1)^{\oplus r+1} \rightarrow T_X \rightarrow 0;$$

we can restrict this to $Y$ to get the exact sequence

$$0 \rightarrow O_Y \rightarrow (O_X(1)_{|Y})^{\oplus r+1} \otimes O_Y \rightarrow T_X \otimes O_Y \rightarrow 0.$$

(Make sure you understand why this is still exact!)

Now take cohomology; since $H^1(Y,O_Y)=0$, all that remains is to calculate the space of sections of the middle term. But $Y$ is a curve of degree $r$ in $X$, so $O_X(1)_{|Y} = O_Y(r)$, whose sections you understand.