Let $X$ be the 6-dimensional Grassmannian of 2-planes in a 5-dimensional vector space $V$, namely $X=G(2,5)$.
I want to compute $$ H^1(X,\Omega_X \otimes L) $$ where $\Omega_X$ is the cotangent bundle over $X$ and $L$ is a line bundle over $X$, that is $L=\mathcal O(d)$ for some $d \in \mathbb Z$. In particular, I am interested for which $d$ we have $H^1(X,\Omega_X \otimes L)=\{0\}$.
In the book of Okonek, this computation is made in the case of $X=\mathbb P(V)$. Does it exists a reference for the case of Grassmannians?
The weights corresponding to $\Omega_X = Q^* \otimes S$ on $G(2,5)$ are $(0,0,-1)$ (for the $Q^*$) and $(1,0)$ (for the $S$). Twisting by $d$ adds $(d,d,d)$ to the $Q$ weight since $\det(Q) = O(1)$. Concatenating these and applying Borel-Weil-Bott, we get that the cohomology all vanishes if there is a repeated value in $$(4,3,2,1,0) + (d,d,d-1,1,0) = (d+4,d+3,d+1,2,0).$$ If all five values are distinct, the unique nonvanishing $H^i$ is given by $i=$ the length of the permutation needed to sort the list into descending order (number of inversions). If $d\leq-2$, there is more than one inversion. If $d=-1,1$ there is a repeated entry. If $d=0$ there is exactly one inversion and no repeats (this is the case where $H^1$ is nonvanishing; it is the $GL_5$ representation of weight $(0,0,0,0,0)$ after sorting the list and then subtracting out the $(4,3,2,1,0)$ — the trivial one-dimensional representation). If $d\geq2$ there are no inversions.
Comment: Borel-Weil-Bott is basically the procedure above (in full generality it applies to flag varieties). I would suggest looking up the statement and then using the calculation I’ve sketched to help understand how it works.