I was reading through Harris's Algebraic Geometry book, and was slightly perplexed by the following paragraph:

"Note that the set of hyperplanes in a projective space $\mathbb{P}^{n}$ is again a projective space, called the dual projective space and denoted $\mathbb{P}^{n^{*}}$. Intrinsically, if $\mathbb{P}^{n} = \mathbb{P}V$ is the projective space associated to a vector space V, the dual projective space $\mathbb{P}^{n^{*}} = \mathbb{P}(V^{*})$ is the projective space associated to the dual space $V^{*}$. More generally, if $\Lambda \cong \mathbb{P^{k}} \subset \mathbb{P^{n}}$ is a $k$-dimensional linear subspace, the set of (k+1)-planes containing $\Lambda$ is a projective space $\mathbb{P^{n-k-1}}$, and the space of hyperplanes containing $\Lambda$ is the dual projective space $(\mathbb{P^{n-k-1}})^{*}$. Intrinsically, if $\mathbb{P^{n}} = \mathbb{P}V$ and $\Lambda = \mathbb{P}W$ for some $(k+1)$-dimensional subspace $W \subset V$, then the space of $(k+1)$-planes containing $\Lambda$ is the projective space $\mathbb{P}(V/W)$ associated to the quotient, and the set of hyperplanes containing $\Lambda$ is naturally the projectivization $\mathbb{P}((V/W)^{*}) = \mathbb{P}(Ann(W)) \subset \mathbb{P}(V^{*})$ of the annihilator $Ann(W) \subset V^{*}$ of $W$."

I understand that in general, $V^{*}$ is defined as the set of linear functionals $f \colon V \to R$ and it makes sense that hyperplanes, but I can't picture intuitively how the hyperplanes would be associated with the dual projective space. In addition, I was having trouble trying to understand geometrically how the last sentence works.

  • $\begingroup$ Think about a necessary and sufficient condition for two linear functionals to define the same hyperplane. $\endgroup$ – Zhen Lin Sep 23 '14 at 0:37
  • $\begingroup$ Would two linear functionals have to define the same point in $\mathbb{P^{n}}$ $\endgroup$ – Sky123 Sep 23 '14 at 0:56
  • $\begingroup$ If you take the kernel of a (non-zero) functional, you get a subspace of dimension $n-1$, and so a hyperplane in projective space. Now two functionals with the same kernel must be proportional to each other, but those are exactly the functionals that are identified in $\mathbb{P}(V^*)$. $\endgroup$ – Prometheus Sep 23 '14 at 2:19
  • $\begingroup$ Oh alright thanks, is there a similar way to picture the last sentence regarding the projective space of a quotient? $\endgroup$ – Sky123 Sep 23 '14 at 2:40

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