Birational Invariants Let $X$ be a smooth rational variety of dimension $n$. We have $\dim H^0(X,\Omega_X^p) = \dim H^0(\mathbb{P}^n,\Omega_{\mathbb{P}^n}^p)$ for any $p$. These are Hodge numbers. I know that we can not expect an equality for the sections of the sheaves $\Omega_X^p(k)$. 
However, is it true that if $p < n$ then $\dim H^0(X,\Omega_X^p(k)) = 0$ if and only if $\dim H^0(\mathbb{P}^n,\Omega_{\mathbb{P}^n}^p(k))=0$ ?
 A: No, that is not true, even after the modifications.  Begin with a Veronese surface $X = v_2(\mathbb{P}^2) \subset \mathbb{P}^5$, i.e., $$ X = \{ [x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2] \in \mathbb{P}^5 | [x_0,x_1,x_2]\in \mathbb{P}^2 \} = $$ $$\{ [y_{2,0,0},y_{1,1,0},y_{1,0,1},y_{0,2,0},y_{0,1,1},y_{0,0,2}] \in \mathbb{P}^5 | y_Iy_J - y_Ky_L = 0 \},   $$ where the relations range over all $4$-tuples $(I,J,K,L)$ of elements in $\{(2,0,0),\dots,(0,0,2)\}$ such that $I+J$ equals $K+L$.  Now form the linear birational equivalence, $$ f : X \dashrightarrow \mathbb{P}^2, \ [y_I] \mapsto [y_{1,1,0},y_{1,0,1},y_{0,1,1}]. $$  The maximal domain of definition of $f$ is $v_2(U)$ for the open subscheme $$U=\mathbb{P}^2 \setminus \{ [1,0,0], [0,1,0], [0,0,1] \}.$$  The rational transformation $f$ is usually called the "standard Cremona transformation".  In particular, $f^*\mathcal{O}_{\mathbb{P}^2}(1)$ is isomorphic to $v_2^*\mathcal{O}_{\mathbb{P}^5}(1)$, and this is isomorphic to $\mathcal{O}_{\mathbb{P}^2}(2)|_U$.  In particular, even though $h^0(\mathbb{P}^2,\Omega^1_{\mathbb{P}^2}(1))$ equals $0$, $h^0(X,\Omega^1_X\otimes_{\mathcal{O}_X} f^*\mathcal{O}_{\mathbb{P}^2}(1))$ is nonzero.  
The point is, because the group of birational automorphisms of every rational variety is so large, the invertible sheaves $f^*\mathcal{O}_{\mathbb{P}^d}(k)$ can take on many different values.  
