In this post, 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?
