About Plücker embedding I'm doing a work about Plücker embedding and I need some help about a few topics.
I'm going to list them:
$1-$ I know that Plücker embedding is well-defined and is injective.
However, Plücker embedding is called an embedding and not an injection.
Why do we call Plücker embedding an embedding?
$2-$ Some papers say that $Gr(2,4)$ is the simplest grassmannian that is not a projective space. Why $Gr(2,3)$ is considered as a projective space?
$3-$ Related to the previous topic, we know that $Gr(1,n)$ is a projective space, and we have a bijection between $Gr(1,n)$ and $Gr(n-1,n)$ given by $W \mapsto W^{\perp}$. Is this map an isomorphism so we can conclude that $Gr(n-1,n)$ is a projective space?
Thanks for helping!
 A: Notation (added). By $V^\vee$ I mean the dual of $V$, and by $\mathbb P(V)$ I mean the set of hyperplanes of $V$: $$\mathbb P(V)=\textrm{Proj Sym}(V)=(V^\vee\setminus\{0\})/k^\times.$$ I think this is quite standard in algebraic geometry, but if you do not like it or the dual confuses you, no problem: call $\mathbb P(V)$ the set of lines and exchange the rôle of quotients and kernels in 3. below.



*

*As far as I know, injection only refers to injectivity on points, which is quite a poor condition. The actual fight could nevertheless sussist between immersion and embedding. (See here.) As a map between complex manifolds (if $V$ is complex), it is correct to call $G(k,V)\to \mathbb P(\wedge^kV)$ an embedding. As algebraic varieties (over any field), one usually talks about closed immersion.

*$G(2,4)$ parameterizes $2$-planes in $k^4$, and this is not a projective space. Instead, $G(2,3)$ are the hyperplanes in $V=k^3$, i.e. $G(2,3)=\mathbb P(V)$, and this is dual to the set of lines in $V$, i.e. $G(1,3)=\mathbb P(V^\vee)$.

*As for $n=3$ above, you always have, for $V=k^n$, the following isomorphic Grassmannians: $$G(n-1,n)=\mathbb P(V),\qquad G(1,n)=\mathbb P(V^\vee).$$ An isomorphim $\mathbb P(V)\cong \mathbb P(V^\vee)$ is given by letting correspond a hyperplane $$H:a_0x_0+a_1x_1+\dots+a_{n-1}x_{n-1}=0$$ to the point $$(a_0,a_1,\dots,a_{n-1})\in \mathbb P(V^\vee).$$ As Georges points out in his comment, any isomorphism would depend on the choice of a basis of $V$.

