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Fix an integer $k$. Let $X=G/P$ be a complex rational homogeneous variety. I assume here $G$ is a simply connected semi simple complex Lie group and $P=P_k$ is a maximal parabolic subgroup defined by dropping the $k$-th simple root (Assume we have chosen and fixed a Borel subgroup to avoid ambiguity, and we use Bourbaki convention about the order of simple roots).

I want to compute the cohomology of the homogeneous bundle $T_X\otimes L^{-\lambda_k}$ over $X$. Here, $T_X$ is the tangent bundle while $L^{-\lambda_k}$ is the line bundle corresponding to the 1-dimensional $P$-representation with character induced by $\lambda_k$. For example, if $X$ is a Grassmannian variety embedded into a projective space using Plücker embedding, then $L^{-\lambda_k}$ is just $\mathcal O(-1)$.

My strategy is to use Borel Weil Bott theorem. But $T_X$ is not irreducible in general (i.e. The $P$-representation $\mathfrak{g/p}$ is not irreducible), so is not $T_X\otimes L^{-\lambda_k}$. Hence, I have to find a $P$-representation filtration of $\mathfrak{g/p}$, say $0\subset s_1\subset s_2\subset\ldots\subset s_r=\mathfrak{g/p}$ with quotients $T_i$ irreducible $P$-representations. We shall use the same notation for the homogeneous vector bundles corresponding to $s_i, T_i$. My plan is: STEP I. compute the cohomology of $T_i\otimes L^{-\lambda_k}$ using Borel Weil Bott theorem for all $i$ since they are irreducible. STEP II. Using the filtration and step I to get the cohomology of $T_X\otimes L^{-\lambda_k}$.

STEP I is easily done by prudent computation, and STEP II can be done in most cases. But I meet some difficulties in step II for some special cases: I need to write the connection morphism down explicitly in these cases. I will use the following example to demonstrate my dilemma here.

From now on, let $G$ be the simply connected Lie group of type $B_l$ and $P=P_2$ be a maximal parabolic subgroup defined by dropping the second simple root. The the filtration of the tangent bundle of $X=G/P$ is $0\subset s_1\subset s_2=T_X$, and hence we have a short exact sequence $$ 0\to s_1\to T_X\to s_2/s_1\to 0$$ Tensoring $L^{-\lambda_2}$, we get another short exact sequence which we simply write as $$0\to s_1(-1)\to T_X(-1)\to s_2/s_1(-1)\to 0$$from which we have a long exact sequence. According to my computation using Borel Weil Bott theorem $$H^q(s_1(-1))=\mathbb C, q=1; 0, q\ne 1$$ $$H^q(s_2/s_1(-1))=\mathbb C, q=0; 0, q\ne 0$$

Hence, my long exact sequence looks like $$ 0\to H^0(T_X(-1))\to \mathbb C \stackrel{\delta}{\to} \mathbb C \to H^1(T_X)\to 0$$ Therefore, to determine what I want: $H^q(T_X(-1))$, I need to write down $\delta$ explicitly. Maybe there are other methods that can determine the cohomology of $T_X(-1)$ directly without $\delta$, so any idea is welcome and appreciated!

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Let me show you a more geometric way to compute the desired cohomology for types $BCD$. I will illustrate it for $X=IGr(k, V)$, where $\dim V=2n$ and $3\leq k\leq n$.

Let $Y$ denote the Grassmannian $Gr(k,V)$, let $i:X\to Y$ be the natural closed embedding. Under $i$ the variety $X$ gets identified with the zero subscheme of a regular section $\omega\in H^0(Y, \Lambda^2U^*)\simeq \Lambda^2V^*$, where $U$ will denote the tautological both on $X$ and $Y$. In particular, the normal bundle $N_{X/Y}$ is isomorphic to $\Lambda^2U^*$.

Now, consider the short exact sequence $$ 0\to T_X(-1)\to i^*T_Y(-1)\to N_{X/Y}(-1)\to 0. $$ BBW on $X$ tells you that $H^\bullet(X, N_{X/Y}(-1))=0$. Thus, $$ H^\bullet(X, T_X(-1))\simeq H^\bullet(X, i^*T_Y(-1)) \simeq H^\bullet(Y, T_Y(-1)\otimes i_*\mathcal{O}) $$ I clame that the latter cohomology groups vanish.

Let us recall that $T_Y\simeq U^*\otimes V/U$, and use the Koszul resolution $$ 0\to \det \Lambda^2U\to\cdots\to\Lambda^2\Lambda^2U\to\Lambda^2U\to\mathcal{O} \to i_*\mathcal{O}\to 0. $$ Once you replace $i_*\mathcal{O}$ with its resolution, you get a spectral sequence with terms of the form $$ H^i(Y, U^*\otimes V/U(-1)\otimes \Lambda^t\Lambda^2U) \simeq \mathrm{Ext}^i(U^\perp, \Lambda^{k-1}U\otimes\Lambda^t\Lambda^2U).$$ It remains to show that for any irreducible summand $\Sigma^\alpha U\subset \Lambda^{k-1}U\otimes \Lambda^t\Lambda^2U$ one has $$ \mathrm{Ext}^\bullet(U^\perp, \Sigma^\alpha U)=0. $$ Here is the lazy way to show the latter: it is quite easy to see that every summand appearing in the decomposition of $\Lambda^t\Lambda^2 U$ is of highest weight $\gamma$ with $\gamma_1\leq k-1$ (actually, the decomposition is well known). Thus, from Pieri's formulas we get that $\alpha_1\leq k\leq 2n-k$. If one thinks of $\alpha$ as of a Young diagram, it is inscribed in the standard $k\times (2n-k)$ rectangle and has $|\alpha|=2t+k-1>1$ boxes.

Now, let $\beta$ be a Young diagram inscribed in the $(2n-k)\times k$ rectangle. It is a good exercise to check the following: $$ \mathrm{Ext}^i(\Sigma^\beta U^\perp, \Sigma^\alpha U)= \begin{cases} \mathsf{k}, & \text{if } \alpha=\beta^T \text{ and } i=|\alpha|,\\ 0 & \text{otherwise}. \end{cases} $$ (Actually, these are elements of dual exceptional collections in $D^b(Y)$ constructed by Kapranov.)

Finally, in our case $\beta$ consists of a single box. Thus, $|\alpha|>|\beta|$ and $\alpha\neq\beta^T$.

Teaser. Remark that when $k=2$, the variety $X$ is a hyperplane section of $Y$, and $N_{X/Y}(-1)\simeq \mathcal{O}$. What happens in the spectral sequence?

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