Describe $\mathcal{N}_{G(\mathbb{P}^1,\mathbb{P}^k)\mid G(\mathbb{P}^1,\mathbb{P}^n)}$ Consider $1 \leq k < n$ positive integers, and denote by $G(\mathbb{P}^k,\mathbb{P}^n)$ the Grassmannian of $\mathbb{P}^k$'s in $\mathbb{P}^n$. It is well known $G(\mathbb{P}^k,\mathbb{P}^n)$ admits a structure of projective variety in $\mathbb{P}^N$, where $N=\binom{n+k}{2}-1$.
It is moreover known that, denoting by $\mathcal{N}_{\mathbb{P}^k\mid \mathbb{P}^n}$ the normal bundle of $\mathbb{P}^k$ in $\mathbb{P}^n$, it holds the following
$$\mathcal{N}_{\mathbb{P}^k\mid \mathbb{P}^n} =\mathcal{O}_{\mathbb{P}^k}(1)^{\oplus n-k}.$$
I was wondering if there exists a similar results if we consider a Grassmannian instead of the projective space, that is
$$?? \text{ } \mathcal{N}_{G(\mathbb{P}^1,\mathbb{P}^k)\mid G(\mathbb{P}^1,\mathbb{P}^n)}=\mathcal{O}_{G(\mathbb{P}^1,\mathbb{P}^k)}(1)^{n-k} \text{ }??$$
I've put the double question mark to stress this may be completely nonsense, but it's my  guess (nothing rigorous).
I was wondering if there exists such a similar description, I've thought about it for a while but didn't managed to find anything. Thanks in advance!
 A: Here are two comments:

*

*The normal bundle  $N_{G(\mathbb P^1,\mathbb P^k)|G(\mathbb P^1,\mathbb P^n)}$ that you are studying should have rank $2(n-k)$, since $\dim G(\mathbb P^1,\mathbb P^n)=2(n-1)$ and $\dim G(\mathbb P^1,\mathbb P^k)=2(k-1)$.


*To talk about $\mathcal{O}_{G(\mathbb P^1,\mathbb P^k)}(1)$, one needs a projective embedding. It is natural to consider the Plucker embedding of $G(\mathbb P^1,\mathbb P^n)$.
Let me do the simplest example when $k=2$ and $n=3$. In this case, $G(\mathbb P^1,\mathbb P^2)$ is a codimension two Schubert cycle of $G(\mathbb P^1,\mathbb P^3)$ and under the Plucker embedding, $G(\mathbb P^1,\mathbb P^3)\cong Q$ is a smooth quadric hypersurface in $\mathbb P^5$, and $G(\mathbb P^1,\mathbb P^2)\cong \mathbb P^2$ is a plane. So we have an exact sequence
$$0\to N_{\mathbb P^2|Q}\to N_{\mathbb P^2|\mathbb P^5}|_{\mathbb P^2}\to N_{Q|\mathbb P^5}|_{\mathbb P^2}\to 0.$$
Since $N_{\mathbb P^2|\mathbb P^5}|_{\mathbb P^2}\cong \mathcal{O}_{\mathbb P^2}(1)^{\oplus 3}$ and $N_{Q|\mathbb P^5}|_{\mathbb P^2}\cong \mathcal{O}_{\mathbb P^2}(2)$, $N_{\mathbb P^2|Q}$ is a rank two bundle with first Chern class $c_1=H$, where $H$ is a hyperplane and $c_2=1$ due to multiplicative identity of Chern polynomial $c(N_{\mathbb P^2|Q})(1+2x)=(1+x)^3$. However, I don't know if the bundle splits.
