in projective space. However, if we took the representative of the first point to be (2,0) instead, the midpoint would become (1,1/2) which goes into projective space as [(2, 1)]. Taking different representatives gives us a different midpoint.

The trouble with that view is that we're trying to impose the vector-space structure of R^2 onto the projective structure of P^2, and it doesn't work.

I'm not aware of any sensible way of defining a midpoint that behaves as you'd expect, but I'm decidedly not an expert.

Reply 2

number23

OP

15

Original post by Smaug123

It's undefined if you were to take the intuitive way of defining it. Consider the points [(1,0)] and [(0,1)] viewed as being in the 2d projective

\mathbb{R}

-space. In the vector space

\mathbb{R}^2

, the midpoint would be

(\dfrac{1}{2}, \dfrac{1}{2})

, which is

[(1,1)]

in projective space. However, if we took the representative of the first point to be (2,0) instead, the midpoint would become (1,1/2) which goes into projective space as [(2, 1)]. Taking different representatives gives us a different midpoint.

The trouble with that view is that we're trying to impose the vector-space structure of R^2 onto the projective structure of P^2, and it doesn't work.

I'm not aware of any sensible way of defining a midpoint that behaves as you'd expect, but I'm decidedly not an expert.

true, in that case it wont work

is there a way to just find distances in projective space?

Reply 3

theOldBean

15

Original post by number23

true, in that case it wont work

is there a way to just find distances in projective space?

Disclaimer: I am not an algebraic geometer. However, I suspect that the answer is 'yes, but it is not a useful notion.' (Algebraic geometry is really about the Zariski topology which does not arise from a metric.)

When we define affine space over a field K, we do it by saying

A^n=K^n

, except that we 'forget the vector space structure', and so the origin is no longer a special point. For certain fields, like

K=R,C

, we can still view

A

as a metric space, but what if

K=F_p

? Really the best we can do here is the 'discrete metric' where two points are zero distance apart if they are the same point, and distance 1 apart otherwise.

Things get even 'worse' from this perspective when we take the appropriate quotient to get into projective space.

P1 can be viewed as a line, together with a single point at infinity i.e. a circle. It would be meaningful here to define distance as 'how far you need to walk along the circle'.

P2 is the set of lines through the origin in 3D space. Here I suppose you could consider a unit sphere and define the distance between two lines to be 'how far you need to walk along the sphere', but you still need to be slightly careful about antipodal points. (i.e. given a line, it intersects the sphere in two diametrically opposite points; which do you choose?)

The reason projective space is 'nice' is because
e.g. Two lines always meet (in P2 at least!); there is no analogue of 'parallel lines' like in

R^n

e.g. Projective space is compact wrt the Zariski topology. (To think about: Why do we pay special attention to compact manifolds?)
and not because there is some canonical 'distance function' on it (to my knowledge).

I'm not sure if this is exactly the answer you were looking for, but hopefully it helps!

(edited 9 years ago)

How could you define the midpoint of two points in projective space?

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