Projective plane and its dual So the projective plane $\mathbb{RP}^2$ is not a vector space.  Is it still isomorphic to its dual?  If not, is there at least an invertible map that takes $\mathbb{RP}^2$ to its dual?
 A: Yes, though the word "dual" is somewhat questionable.  If you mean is $\mathbb{P}(\mathbb{R}^3)$, the standard projective plane, isomorphic to $\mathbb{P}((\mathbb{R}^3)^*)$, the projectivization of the dual, then yes, it follows from the isomorphism of vector spaces.
Much more interestingly, the duality allows you to switch points and lines in theorems, such as the Mystic Hexagon and Brianchon's Theorem.
A: This is a comment more than an answer, mainly in response to Qiaochu's question, but I don't have sufficient reputation to comment.
The dual to a projective plane is the set of all lines in the plane, which itself is a projective plane (as hinted at in Charles Siegel's answer).  This is an important concept in classical projective geometry.
Concretely, the equation for a line has the form $a x + b y + c z = 0$, where $a,b$ and $c$ are some parameters, not all zero, and $x,y,z$ are homogeneous coords. for the points in the proj. plane.  The set of all such lines can thus be thought of as the set of all $(a,b,c)$ not all zero.
But note that simultaneously multiplying $a, b$ and $c$ by a non-zero scalar doesn't change the solution set, i.e. doesn't change the line, so the line should really be thought of as corresponding to the homogeneous coordinates $(a:b:c)$.
Thus the set of all lines is again a projective plane.
