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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!

Leafar
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  • MathJax is not full-blown TeX so for example the enumerate environment is not supported. The mark up language does support lists. Check out the tool bar on top of the edit window and pick either the numbered list or the bulleted list. Play with it for a while - you'll figure out how it works soon enough! – Jyrki Lahtonen Nov 24 '14 at 22:04
  • And welcome to the site! – Jyrki Lahtonen Nov 24 '14 at 22:05

1 Answers1

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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.

  1. 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.
  2. $G(2,4)$ 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. $G(1,3)=\mathbb P(V^\vee)$.
  3. 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 $V$.
Brenin
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  • We denote by $Gr(p,V)$ all the subspaces of the vector space $V$, for any vector space $V$, not necessairly $\mathbb{C}$. When $V = \mathbb{K}^n$ we write $Gr(p,n)$.
    • – Leafar Nov 25 '14 at 00:36
  • $2.+3.-$ You mean $\mathbb{P}(V) := Gr(1,V)$. Can you explain what is $V^{\vee}$? I don't understand too the notation that you used to describe the points in the third topic. – Leafar Nov 25 '14 at 00:41
  • Maybe you'd prefer $V^*$ instead? There's just some fiddling with dual vector spaces. – Hoot Nov 25 '14 at 00:59
  • Sorry, I used $\mathbb C$ as my "default" ground field but of course you can change it to any field (paying the price that you do not deal with a complex manifold anymore, if that is ok for you). – Brenin Nov 25 '14 at 10:01
  • Dear Brenin: I don't think there is a canonical isomorphism $\mathbb P(V)\cong \mathbb P(V^\vee)$ and I dont understand how you can describe a hyperplane by a map $ V\to H\to 0 $ . – Georges Elencwajg Nov 25 '14 at 10:40
  • @GeorgesElencwajg: you are right, I modified my answer accordingly. Now it does not look canonical anymore. – Brenin Nov 25 '14 at 15:18