# Trivial extension for the tangent bundle of Grassmannian of planes in $\mathbb C^5$

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?

• You need Borel-Bott-Weil theorem. Nov 15, 2021 at 13:38
• All the cohomology vanishes for $d=-4,-3,-2,-1,1$. And the only value where $H^1\ne 0$ is $d=0$. Nov 16, 2021 at 16:05
• @JakeLevinson could you provide more details for the proof that if $d \ne 0$ then $H^1 \ne 0$? Nov 16, 2021 at 16:15
• Because in that case, we have a non-splitting short exact sequence on $X$ given by $$0 \to \mathcal O \to N \to T_X \to 0.$$ I want to understand the most I can about $N$ Nov 16, 2021 at 16:23

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.