Compute the derivative of Plucker Embedding Let V be an n-dimensional vector space over $\mathbb{R}$, and 
$$\Psi: G(k,V)\rightarrow \mathbb{P}(\Lambda^k V)$$ be the Plucker embedding, where 
$$L=span \{u_1, ..., u_k\} \mapsto \Psi(L)=[u_1 \wedge u_2 \wedge ... \wedge u_k].$$
(a) How can I show that this map is smooth?
(b) How can I compute its derivative? ($\Psi_*:Hom(L,L^\perp)\rightarrow Hom(det(L), det(L)) $)
Any help is appreciated!
 A: I guess this is some serious algebraic geometry stuff, so I wouldnt pretend I know it quite well. I will instead only offer my personal opinion from differential geometry point of view and would hope there are better answers to come. 
(a) It is standard knowledge that $G(k,V)$ is a $k\times(n-k)$ dimensional manifold. Assume without loss of generality that after change of basis, an element $G$ of $G(k,V)$ looks like:
$$
G=\left[\begin{array}{cccccc}
1 &  &             & g_{1,k+1} & \cdots & g_{1,n}\\
  & \ddots &  &  \vdots & & \vdots \\
 & & 1 & g_{k,k+1}& \cdots & g_{k,n}
\end{array}\right]
$$
with $g_{i,k+j}$'s being its coordinates. Now consider the map:
$$
\tilde\Psi:G(k,V)\to \mathbb R^{{n\choose k}} 
$$
that takes the value of all $k\times k$ minors of $G$, i.e. $\tilde\Psi$ is $\Psi$ without taking the quotient (or just think of homogeneous coordinates). The determinant map are all polynomial functions and are obviously smooth. Therefore $\tilde\Psi$ is smooth, and so is $\Psi$.
(b) To compute the derivative of this map, it is enough to compute the derivatives of the minors w.r.t. the entries $g_{i,k+j}$'s, and is a well known result in linear algebra.
Hope my naive reply helps.
