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Is there a vector version of The Elements?

by Bipolarity
Tags: elements, vector, version
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Bipolarity
#1
Dec8-12, 12:31 AM
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I have noticed that many theorems in classical geometry can be proven using vectors. Thus, I am naturally inclined to believe that pretty much every theorem in the Elements can be proven using vectors (using the vector space axioms and the vector definitions of norm, angle, and parallel, rather than the axioms, postulates and definitions adopted by Euclid).

First of all, is this possible? What would be the vector form of Euclid's 5th (parallel) postulate be? After all, I know that Euclid's 5th postulate is the basis of Euclidean geometry, so certainly a construction of the Elements that uses vectors must have some vector analog of the 5th postulate, otherwise it would produce non-Euclidean geometries?

If it is possible, wouldn't it be circular? Don't some of the vector space axioms follow from Euclidean geometry? Or are the vector space axioms chosen to be consistent with the axioms and theorems of Euclidean geometry, in which case it does not really matter?

And finally, if it is possible, has it been done? What's the name of the text in that case? I am particularly interested in the usage of vectors to prove theorems involving circles and tangents from Euclidean geometry.

I am only speculating, so I deeply apologize if my idea sounds dumb. In such case I would appreciate an explanation of why this (construction) is not possible.

Thanks!

BiP
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jedishrfu
#2
Dec8-12, 12:58 AM
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here's the closest book I found on the topic:

http://www.amazon.com/Vector-Space-A.../dp/0486404528

but I don't think there is such a book since vector concepts are founded on more general principles than Euclidean geometry.
lavinia
#3
Dec8-12, 10:55 AM
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I think that a 2 dimensional real vector space must satisfy the Parallel postulate if you take lines to mean translations of 1 dimensional subspaces.

This is interesting because it shows that the idea of a vector space provides a model of Euclidean geometry and that the primitive idea of line in Euclid's elements does not determine the plane as a vector space. Very cool.

micromass
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Dec8-12, 12:49 PM
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Is there a vector version of The Elements?

Quote Quote by lavinia View Post
I think that a 2 dimensional real vector space must satisfy the Parallel postulate if you take lines to mean translations of 1 dimensional subspaces.

This is interesting because it shows that the idea of a vector space provides a model of Euclidean geometry and that the primitive idea of line in Euclid's elements does not determine the plane as a vector space. Very cool.
If you take enough axioms for your geometry, then you will eventually obtain that your geometry is a vector space.

In his book "Geometry: Euclid and beyond", Hartshorne takes some very natural axioms which are close to the axioms of Euclid. Then he constructs an order field F and he proves that your geometry is isomorphic to the vector space [itex]F^2[/itex].

If you're content with "vector spaces" over just division rings or non-ordered fields, then even less axioms are needed. The parallel postulate always seems necessary though. I would be interested to know if a geometric structure that does not satisfy the parallel postulate can also be given an algebraic structure.
Fredrik
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Dec8-12, 06:05 PM
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Quote Quote by Bipolarity View Post
I have noticed that many theorems in classical geometry can be proven using vectors. Thus, I am naturally inclined to believe that pretty much every theorem in the Elements can be proven using vectors (using the vector space axioms and the vector definitions of norm, angle, and parallel, rather than the axioms, postulates and definitions adopted by Euclid).
...
If it is possible, wouldn't it be circular?
I haven't seen all the details worked out, but I have no doubt that it's possible. For example, let X be a 2-dimensional vector space over ℝ. We can define the term "point" by saying that we call the members of V points. We can define the term "line" by saying that a set ##L\subset X## is a line if there exist ##x,v\in X## such that ##v\neq 0## and L is the range of the map ##C:\mathbb R\to X## defined by ##C(t)=x+tv## for all ##t\in\mathbb R##. Then we can start proving theorems, like "Let K and L be lines in X. If K intersects L, then either K=L or K intersects L at exactly one point". Angles can be defined by
$$\cos\theta=\frac{x\cdot y}{|x|\,|y|}.$$ It would surprise me a lot if we can't prove all of Euclids axioms in this framework.

Micromass confirmed something else that I've always suspected, that you can start with something very similar to Euclid's axioms and use them to define a vector space structure on the set (i.e. to define an addition operation and a scalar multiplication operation that satisfy the vector space axioms).

This isn't any more circular than the proof of any "if and only if" statement in mathematics.
mathwonk
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Dec9-12, 11:26 PM
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with euclid's axioms plus archimedes axiom and dedekind's axiom one can prove that the only possible such plane is indeed R^2. In the other direction, if one has the usual axioms for R, then all Euclid's axioms and theorems are true in R^2, and can be proven there along lines suggested by Fredrick and micro mass. There is no need for a 5th postulate or any other geometric postulate once you are in R^2, since all the Euclidean postulates are theorems there. I.e. R^2 is a model for Euclidean geometry, a space where all Euclidean axioms hold. The existence of R^2 proves that Euclid's axioms are consistent, i.e. not self contradictory. Since Archimedes and Dedekind's axioms are not included in Euclid's, leaving them out allows many more models for Euclidean geometry, modeled as micromass said, and Hartshorne proves, on different number systems from R. Of course rather than Euclid's own axioms, it is prudent to use a more careful and precise version due to Hilbert, as Hartshorne does.
lavinia
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Dec10-12, 07:26 AM
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Quote Quote by micromass View Post

If you're content with "vector spaces" over just division rings or non-ordered fields, then even less axioms are needed. The parallel postulate always seems necessary though. I would be interested to know if a geometric structure that does not satisfy the parallel postulate can also be given an algebraic structure.
What about projective planes?
What about vector addition on a torus(project vector addition in the plane onto a torus)? no I guess this does satisfy the parallel postulate.
mathwonk
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Dec12-12, 08:51 PM
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a projective plane is a surface of constant positive curvature and a torus is a surface of constant zero curvature, like euclid's plane. to check euclid's parallel postulate you should state it. the version stated by euclid is true also on the sphere for instance, since it ays that under certain conditions two lines meet, and that is always true on a sphere and projective plane. (although euclid's postulate speaks of sides of a line which do not make sense on a projective plane.) in fact it also holds on a torus, i.e. two lines meet if there is a third line transverse to both and making angles with the two original lines adding to less than a straight angle on one side of the transversal. you may be amused to read euclid's axioms and try to see which one is not true on a torus, if any.


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