1
$\begingroup$

In the book on Linear Algebra that I am using, the author defines a line in an arbitrary vector space $V$, given direction $ 0 \neq d \in V $ and passing through $ p \in V$ as $ l(p;d)= \lbrace v\in V| \;\exists t \in \mathbb R,v=p+td\rbrace $.

He then proceeds to show that $ l(p;d)=l(q;d) $ iff $ (q-p) $ is a multiple of $d $. and further that any two distinct points determine a unique line. He later defines 2 lines $ l(p;d_1) , l(q;d_2)$ as parallel if $d_1= \alpha d_2$ for some $ \alpha \in \mathbb R$.

He finally proves Euclid's parallel postulate using these tools. My question is how does this setup still allow for a Non-Euclidean Geometry? Does it have to do anything with the fact that the definition of line here uses a "$ t \in \mathbb R$" which inhibits the validity of this definition? If so, how does one define a straight line using an arbitrary field? I hope I have been able to convey my thoughts precisely enough to warrant an answer.

  • 3
    We can change the field, but that does not get us to hyperbolic geometry. There are various suitable definition of line. A common one is to define line segment as a *geodesic*, that is, a curve that minimizes distance. In the geometry of the surface of the sphere, such a geodesic is a segment of a "great circle," which is why in flying from North America to Europe one travels so far North.2012-11-18
  • 0
    So the "t" in the definition can be from some other field than $R$?? I don't know how much sense that makes as $R$ has some special properties. Also, is this geometrical setup complete? I mean as the definition involves an arbitrary vector space, how does one arrive at a Non-Euclidean Geometry?2012-11-18
  • 0
    One does not arrive at something like hyperbolic geometry through a linear algebra definition of line. One can reach projective geometry, however. True, the definition of line is most useful when the underlying field is the reals, or, occasionally, the complex numbers.2012-11-18
  • 0
    I am sorry if I am sounding a bit dumb, but what I was trying to understand is this. The need for a Non-Euclidean Geometry was borne out of problems with Euclid's parallel postulate, which here is however proved as a theorem. I wanted to know what is the subtle problem in the definition of a line and parallelism here that still gives scope for the existence of such a geometry. Here line and notion of parallelism are the axiomatic concepts, so there should be something lacking there??2012-11-18
  • 0
    The geometry of $\mathbb{R^n}$, with lines defined as above, *is* Euclidean. There are many other possible definitions of line, and many ways to put a "metric" (distance function) even of $\mathbb{R}^2$.2012-11-18
  • 0
    Ok, I guess I need to read along a bit more to get a better picture. Thanks a lot for your patience.2012-11-18
  • 1
    **Added:** You may find the [Wilipedia article](http://en.wikipedia.org/wiki/Hyperbolic_geometry#Models_of_the_hyperbolic_plane) geometry interesting. I have jumped to the "models of the hyperbolic plane" part.2012-11-18
  • 0
    @AndréNicolas Please consider expanding your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-09-14

0 Answers 0