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In algebraic geometry, I keep seeing the notation $\mathbb{P}V$ when $V$ is given as a vectorspace. My best guess is that $\mathbb{P}V$ is to mean the projective closure of $V$. But it would be nice to know if this notation is standard, and if so, what the precise definition is so there is no confusion on my part.

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  • $\begingroup$ Thinking about schemes, this ought to be $\operatorname{Proj}(\operatorname{Sym}^\bullet V^\vee)$. To be concrete one often works with $\operatorname{Proj} k[x_1, \dots, x_n]$ and here the $x_i$ should be thought of as the "obvious" dual basis for $(k^n)^\vee$. $\endgroup$ – TTS Sep 27 '13 at 20:17
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I would expect this to mean the projective space defined by $V$: the space nonzero vectors modulo the relation $r\vec{v} \sim \vec{v}$ whenever $r \neq 0$.

So, for example, $\mathbb{P} \mathbb{R}^3$ is the usual real projective plane.

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  • $\begingroup$ Ah... So the 1-dimensional subspaces of V, right? $\endgroup$ – Tac-Tics Sep 27 '13 at 21:09
  • $\begingroup$ Right.${}{}{}{}$ $\endgroup$ – user14972 Sep 27 '13 at 21:49
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The symbol $\mathbb{P}V$ stands for the projectivisation of the vector space $V$. The one-dimensional linear subspaces of $V$ are the elements of $\mathbb{P}V$.

If $V$ is a vector space over the field $\mathbb{K}$, then the projectivisation $\mathbb{P}V$ of $V$ is the quotient space $(V \setminus \{0\}) /\sim$ where, given $u,v \in V \setminus \{0\}$ we have $u \sim v$ if and only if there exists a non-zero $\lambda \in \mathbb{K}$ for which $u = \lambda v$.

For a fixed $v \in V \setminus \{0\}$, the set of all $u \in V \setminus \{0\}$ with $u \sim v$ are of the form $\lambda v$. This set of vectors can be identified with the span of $v$.

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In algebraic geometry, the projective space $\mathbb P(V)$ associated to a vector space $V$ is $\mathrm{Proj}(\mathrm{Sym}(V))$, see EGA, II.4.1.1. In particular, the $k$-rational points of $\mathbb P(V)$ are the linear forms modulo $k^*$: $$ \mathbb P(V)(k) = (V^{\vee} \setminus \{ 0\})/k^*.$$ In other words, they are hyperplanes of $V$. A good reason is that if you want to generalize to projective scheme associated to a coherent sheaf $\mathscr E$ over a scheme $S$, you take $\mathrm{Proj}(\mathscr Sym(\mathscr E))$. If you take $ \mathrm{Proj}(\mathscr Sym(\mathscr E)^{\vee})$ instead, you lose much information from $\mathscr E$ because, for instance, taking the dual kills the torsions of $\mathscr E$.

Edit Another reason over a field. Consider the projective space $\mathrm{Proj}(k[t_0, \dots, t_n])$. It is natural to consider it as $\mathbb P(V)$ where $V=kt_0+\cdots+kt_n$. Now if you have surjective homomorphism $$ \phi: k[t_0, \dots, t_n] \to k[s_0, \dots, s_m]$$ of homogeneous $k$-algebras, you have a surjective linear map $$ \phi_1: V \to E=ks_0+\cdots +ks_m.$$ We know that $\phi$ induces a morphism $$ \mathbb P(E) = \mathrm{Proj}k[s_0, \dots, s_m]\to \mathbb P(V).$$ On the level of rational points, with EGA's definition, it corresponds to $$ E^{\vee} \setminus \{ 0\} /k^* \to V^{\vee} \setminus \{ 0\} /k^* $$ which is correct (given by the injective $\phi_1^*$). But if use the defintion of lines, you need a linear map $V\setminus \{ 0\}\to E\setminus \{ 0\}$ and this is impossible if $\phi_1$ is not injective.

Of course, over a field, people usually do not notice the difference between $V$ and its dual, and this is harmless in general if you only work over fields.

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  • $\begingroup$ +1 for the correct definition. But I don't really agree that the Edit concerning functoriality is a reason for this definition. Namely, with the definition using lines, we have instead that $V \subseteq W$ induces $P(V) \hookrightarrow P(W)$, which is very very useful. The functoriality is just vice versa. $\endgroup$ – Martin Brandenburg Oct 1 '13 at 9:44
  • $\begingroup$ What does Sym mean here? $\endgroup$ – Tac-Tics Oct 3 '13 at 20:29
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    $\begingroup$ Less "correct definition" and more "generalization" and "moving the problem into the framework of schemes". If the question were about an (algebraic) vector bundle (even a vector bundle on a point) -- or one believed that the concept "vector space" is really just a proxy for the notion of vector bundle -- this would clearly the Right Thing. $\endgroup$ – user14972 Oct 3 '13 at 20:38
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    $\begingroup$ @Tac-Tics: Sym is the symmetric tensor algebra over $V$. If B is a basis for $V$, then $Sym(V) \cong k[B]$ -- that is, the polynomial ring in $|B|$ variables. $\endgroup$ – user14972 Oct 3 '13 at 21:10

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