How to write down the connection morphism in the long exact sequence in Čech cohomology explicitly in this specific case? 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!
 A: 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?
