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 $\dim H^0(\mathbb{P}^n,\Omega\_{\mathbb{P}^n}^p(k))=0$ ?



Birational Invariants
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. 


