In [this post](https://mathoverflow.net/q/397649), an answer claims that the normal bundle of the Veronese decomposes into a filtration, such that the associated graded is $$\bigoplus\_{i=2}^d S^i T,$$ where $T$ is the tangent bundle. But I thought it was actually $$\bigoplus\_{i=2}^d S^i T \otimes \mathcal{O}(i-1)$$... I attempted to prove this in the following way:


We know that the normal bundle sits in the exact sequence


$$0 \to V \otimes \mathcal{O}(1) \to S^d V \otimes \mathcal{O}(d) \to \mathcal{N} \to 0.$$


Given the Euler exact sequence
$$0 \to \mathcal{O} \to V \otimes \mathcal{O}(1) \to T \to 0,$$ symmetrizing yields the exact sequence
$$0 \to S^{d-1}V \otimes \mathcal{O} \to S^dV \otimes \mathcal{O}(1) \to S^d T \to 0.$$


Iterating this and twisting, we obtain exact sequences
$$0 \to S^{d-i}V \otimes \mathcal{O}(d-i) \to S^{d-i+1}V \otimes \mathcal{O}(d-i+1) \to S^{d-i+1}T \otimes \mathcal{O}(d-i) \to 0$$


Then we have a filtration of $S^d V \otimes \mathcal{O}(d)$ via


$$V \otimes \mathcal{O}(1) \subset S^2V \otimes \mathcal{O}(2) \subset \cdots \subset S^{d-1}V \otimes \mathcal{O}(d-1) \subset S^d V \otimes \mathcal{O}(d)$$


such that quotienting out by $V \otimes \mathcal{O}(1)$ gives us the filtration


$$S^2 T \otimes \mathcal{O}(1) \subset \cdots \subset \frac{S^{d-1}V \otimes \mathcal{O}(d-1)}{V \otimes \mathcal{O}(1)} \subset \frac{S^d V \otimes \mathcal{O}(d)}{V \otimes \mathcal{O}(1)} = \mathcal{N}$$.


But then the associated graded is
$$\bigoplus\_{i=2}^d S^i T \otimes \mathcal{O}(i-1).$$ Is this correct?



Normal bundle of veronese as iteration extension of symmetric powers
Your third exact sequence is incorrect --- the correct form is
$$
0 \to S^{d-1}V \otimes \mathcal{O}(d-1)
 \to S^{d}V \otimes \mathcal{O}(d)
 \to S^dT 
 \to 0.
$$


