0

I am reading Coxeter's projective geometry book and he says that the isosceles triangle doesn't belong to the projective plane. I see how in projective geometry we solely use the compass as opposed to in Euclidean geometry we use both the compass and straightedge. In projective geometry do we only use the compass for construction and in Euclidean geometry we use just the compass and straightedge for construction but is it the same as in Euclidean geometry where not everything in the plane is constructible? Because since the projective plane is really the Euclidean plane + line at infinity there is nothing stopping from an isosceles triangle existing even though we can't measure anything with a ruler.

user19405892
  • 15,592
  • 2
    It's not that certain triangles themselves don't exist, it's the notion of isosceles itself doesn't exist: It depends on a notion of length, which, like you've noted, doesn't exist in projective geometry. – Travis Willse Mar 05 '16 at 13:28
  • But the isosceles triangle can still exist in the projective plane, right? – user19405892 Mar 05 '16 at 13:30
  • How is that question different from your original question? – Travis Willse Mar 05 '16 at 13:31
  • I mean mathematically it can exist just as it does in the euclidean plane, but we just can't tell if its isosceles or not. Is the concept of not using a striaghtedge not just common to construction in projective geometry but for all of projective geometry? – user19405892 Mar 05 '16 at 13:33
  • I am sorry but the part of sentence (that I imagine you have seen somewhere) "in projective geometry we solely use the compass as opposed to in Euclidean geometry we use both the compass and straightedge" has absolutely NO BASIS. For example, any type of conic section can be transformed in any other type, in particular a circle or an ellpse may be transformed into an hyperbola, etc. – Jean Marie Mar 05 '16 at 15:18

1 Answers1

4

If you fix an embedding of $\mathbb{A}^2(\mathbb{R})$ (affine plane) in $\mathbb{P}^2(\mathbb{R})$ (projective plane), and fix some euclidian structure on $\mathbb{A}^2(\mathbb{R})$, then of course you can take an isoceles triangle in $\mathbb{A}^2(\mathbb{R})$ and consider its image in $\mathbb{P}^2(\mathbb{R})$. It exists.

But it just doesn't make any sense to call it isoceles anymore because there is no notion of length in $\mathbb{P}^2(\mathbb{R})$. For instance, this property won't be preserved by the projective automorphisms.

Captain Lama
  • 25,743