Map to exterior power gives rise to smooth embedding of Grassmannian in projective space?

Question

How do I see that the map$$(x_1, \dots, x_n) \mapsto x_1 \wedge \dots \wedge x_n$$from $V_n(\mathbb{R}^m)$ to the exterior power $\wedge^n(\mathbb{R}^m)$ gives rise to a smooth embedding of $G_n(\mathbb{R}^m)$ in the projective space $$G_1(\wedge^n(\mathbb{R}^m)) \cong \mathbb{P}^{\binom{m}{n} - 1}?$$

Answer 1

If $V\subset \bf R^n$ is a $n$ dimensional subspace, if $e_1,...e_n$ and $f_1,...f_n$ are two bases $D.e_1 \wedge e_2....\wedge e_n= f_1 \wedge f_2....\wedge f_n \not =0$, where $D\in \bf R$ is a determinant of the linear map which send $e_i$ to $f_i$. In particular these multivectors have the same image in the projectif space $P(\wedge ^n(\bf R^m))$. We have a map from the Grassmanian to the projectif space.To check that this map is injective, note that $x\in Vect (e_1,...e_n)$ iff $x\wedge e_1 \wedge e_2....\wedge e_n=0$. For the smoothness, complete the base $e_1,..e_n$ as a base of $\bf R^m$, $e_1,...e_m$ Recall that linear map $\bf R^n\to \bf R^{m-n}$ form a chart around the point $\bf R^n$ as follows : to the linear map $l$ one associate the vector space $c(l)$ generated by $e_i+l(e_i)$. In this chart, the map lift to a map from $L(\bf R^n,\bf R^{m-n})$ to $\wedge ^n\bf R^m$ $l\to (e_1+l(e_1))\wedge (e_2+l(e_2))...\wedge( e_n+l(e_n))$ which is polynomial in the coefficient of $l$ viewed as a matrix.