# Existence of parallels in axiomatic plane geometries

• B
Gold Member

## Summary:

Is the existence of parallels provable in Euclidean and Hyperbolic plane geometry?

## Main Question or Discussion Point

Parallels exist in both Euclidean and Hyperbolic geometry. Yet each includes a separate postulate that declares the number of parallels to a line in a plane through a given point. But it seems that if both geometries have parallels then their existence - as opposed to how many of them - should be provable without a parallel postulate. One should be able to derive them from the other postulates, the postulates that describe points and lines and their intersections and how they separate the plane.

My question is: Is there a proof of the existence of parallels without Euclid's Fifth Postulate or its equivalents such as Playfair's Axiom? Same question for Hyperbolic geometry.

The following postulates describe simple properties of the plane that it seems one would have to use to prove the existence of parallels without a parallel postulate.

- Given two points in a plane there is a unique line that contains both of them.

- Two lines that intersect intersect in exactly one point

- A point on a line separates the line into two rays - or half lines. The two rays intersect in the point.

- A line separates the plane into two half planes. These two half planes intersect in the line.

- Let two lines $L_1$ and $L_2$ intersect in the point $P$. Let $R_1$ and $R_2$ be the two rays on $L_1$ determined by $P$. Then $R_1$ lies completely in one of the half planes determined by $L_2$ and $R_2$ lies completely in the other.

Note: These postulates exclude Elliptic geometry since, there, two lines intersect in two points. One might try identifying each pair of intersection points to get only one point but then a line will not separate the plane into two disjoint half planes. If one pictures this geometry as a sphere with great circles as the lines then identifying opposite poles creates the projective plane. In the projective plane the projections of the great circles do intersect in a single point but they do not separate the plane into two pieces. This is because the inverse image of the projective plane minus the projection of the circle is two hemispheres. But these are identified in the projective plane. Also on a sphere opposite poles do not determine a unique line. So it seems that an axiomatic version of this geometry would have to say something like if two points are not "opposite" then they determine a unique line but if they are "opposite" they do not. It would be interesting to derive spherical geometry axiomatically.

Geometric intuition says that two perpendicular lines to a given line $L$ must be parallel. Why is this? Suppose they are not. Then they must intersect in one of the half planes and not there other. But which half plane would that be? The lines make right angles to $L$ in both half planes so there is no difference in the way they enter into the half planes. One would think therefore if they intersected in one they would have to intersect in the other. There seems to be a principle of symmetry here that is implicit in the intuition but is not stated in the postulates. Still one would think that this symmetry is intrinsic to the idea of these two geometries and is independent of any parallel postulate.

So is there a proof or is some additional postulate dealing with symmetry needed?

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mathman
Parallel postulate is needed. Without it a geometry with no parallels is possible (spherical).

• bhobba
Gold Member
@mathwonk do you have any thoughts on this?

pbuk
Gold Member
Note: These postulates exclude Elliptic geometry since, there, two lines intersect in two points.
No they don't, look up the axioms of elliptic geometry.

Geometric intuition says that two perpendicular lines to a given line $L$ must be parallel.
But this is geometric intuition applied to a Euclidean plane - if instead we have a sphere mapped in the conventional way, then any two lines of longitude intersect the equator at right angles, however they meet at the two poles. And here is the crucial point you are missing - in the geometry of a sphere (elliptical geometry), the two poles are actually a single "point". You can see that this is the case by looking at the first three postulates you quote; if antipodes were distinct points then each of these is violated.

Gold Member
No they don't, look up the axioms of elliptic geometry.If you had read mIf
This is from the Wikipedia artice

"Elliptic plane
The elliptic plane is the real projective plane provided with a metric:"

I explained why the projective plane is excluded in the postulates that I listed. Take a look.

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Gold Member
But this is geometric intuition applied to a Euclidean plane -
It is the same or the hyperbolic plane. It is also true for the sphere but not true for the projective plane.

pbuk
Gold Member
I think we are talking at cross-purposes, and I don't think the Wikipedia articles on this subject are very helpful. As others have said it is not possible to prove the parallel postulate from the first four postulates. We can see this because if you assume the parallel postulate is not true opposite you still have a consistent geometry which obeys the first four postulates.

Bear in mind that when we write "two lines that intersect intersect in exactly one point" we are defining what a point is; the fact that when you look at great circles on a sphere you see two intersections does not mean that this axiom is violated, it means that you need to change your concept of a "point".

http://mathworld.wolfram.com/EuclidsPostulates.html
http://mathworld.wolfram.com/ParallelPostulate.html
http://mathworld.wolfram.com/Non-EuclideanGeometry.html

Gold Member
I think we are talking at cross-purposes, and I don't think the Wikipedia articles on this subject are very helpful. As others have said it is not possible to prove the parallel postulate from the first four postulates. We can see this because if you assume the parallel postulate is not true opposite you still have a consistent geometry which obeys the first four postulates.

Bear in mind that when we write "two lines that intersect intersect in exactly one point" we are defining what a point is; the fact that when you look at great circles on a sphere you see two intersections does not mean that this axiom is violated, it means that you need to change your concept of a "point".

http://mathworld.wolfram.com/EuclidsPostulates.html
http://mathworld.wolfram.com/ParallelPostulate.html
http://mathworld.wolfram.com/Non-EuclideanGeometry.html
The problem with the projective plane is that it is not separable by a line. One of the postulates is that a line separates the plane into two half planes.

On the other hand if you accept the separation postulate for a moment then what about the question of symmetry that ended my post? That is what I am really talking about.

BTW: If you do not use the separation postulate then the argument about symmetry that I suggested does not hold since there are not two half planes.

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Infrared
Gold Member
This doesn't fully answer your question, but I think it covers the case of Euclidean, elliptic/projective, hyperbolic geometry. Let's suppose that our plane is a Riemannian 2-manifold, and the lines are geodesics.

Suppose for contradiction that your axioms are met, but there are no parallels, so every two lines meet in exactly one point. Let $L$ be a line, and $p$ be a point not on $L$. Then, for every nonzero $v\in T_pM$, the geodesic through $p$ with initial tangent vector $v$ meets $L$ exactly once by assumption. So, we have a map $T_pM-\{0\}\to L$. Scaling $v$ does not affect the geodesic, and since $M$ is $2$-dimensional, this gives a map $\mathbb{RP}^1=S^1\to L$. This map is onto (since any two points are connected by a geodesic) and one-to-one (since if two lines pass through $p$ and the same point on $L$, they must agree by your second axiom). I think this map is also continuous, so we have a continuous bijection $S^1\to L$. Since $S^1$ is compact and $L$ is Hausdorff, it is a homeomorphism, so $L\cong S^1$.

Now let $P_1$ and $P_2$ be points in one of the regions bounded by $L$. Let $L'$ be the line passing through $P_1$ and $P_2$. Since $L$ and $L'$ are homeomorphic to circles and $P_1$ and $P_2$ are on the same side of $L$, in order for $L$ and $L'$ to meet exactly once, they must be tangent at some point. But two geodesics that are tangent at a point are equal. Contradiction.

• Klystron
pbuk
Gold Member
The problem with the projective plane is that it is not separable by a line. One of the postulates is that a line separates the plane into two half planes.
I think it would be better to stop talking about projective plane/elliptic geometry because you are restricting your interest to Euclidean and hyperbolic spaces. Some of the statements you have made about elliptic geometry are not correct and this is adding unnecessary confusion.

On the other hand if you accept the separation postulate for a moment then what about the question of symmetry that ended my post? That is what I am really talking about.
Are you suggesting that we can just assume symmetry? I don't think we can, in mathematics we can't just make an assumption because it looks intuitive, we either have to prove it as a theorem or we have to introduce it as another postulate/axiom (e.g. "lines that meet in one half plane must meet in the other"*). And we already know that we can replace the parallel postulate with e.g. the triangle rule or any one of a number of equivalents so where has this got us?

* EDIT - I am not suggesting that this is a sufficient condition to be equivalent to the parallel postulate, just asserting the existance of such conditions.

Gold Member
I think it would be better to stop talking about projective plane/elliptic geometry because you are restricting your interest to Euclidean and hyperbolic spaces. Some of the statements you have made about elliptic geometry are not correct and this is adding unnecessary confusion.
I agree there is confusion. But what specifically did I say that was incorrect?

Gold Member
This doesn't fully answer your question, but I think it covers the case of Euclidean, elliptic/projective, hyperbolic geometry. Let's suppose that our plane is a Riemannian 2-manifold, and the lines are geodesics.

Suppose for contradiction that your axioms are met, but there are no parallels, so every two lines meet in exactly one point. Let $L$ be a line, and $p$ be a point not on $L$. Then, for every nonzero $v\in T_pM$, the geodesic through $p$ with initial tangent vector $v$ meets $L$ exactly once by assumption. So, we have a map $T_pM-\{0\}\to L$. Scaling $v$ does not affect the geodesic, and since $M$ is $2$-dimensional, this gives a map $\mathbb{RP}^1=S^1\to L$. This map is onto (since any two points are connected by a geodesic) and one-to-one (since if two lines pass through $p$ and the same point on $L$, they must agree by your second axiom). I think this map is also continuous, so we have a continuous bijection $S^1\to L$. Since $S^1$ is compact and $L$ is Hausdorff, it is a homeomorphism, so $L\cong S^1$.

Now let $P_1$ and $P_2$ be points in one of the regions bounded by $L$. Let $L'$ be the line passing through $P_1$ and $P_2$. Since $L$ and $L'$ are homeomorphic to circles and $P_1$ and $P_2$ are on the same side of $L$, in order for $L$ and $L'$ to meet exactly once, they must be tangent at some point. But two geodesics that are tangent at a point are equal. Contradiction.
I love this proof.

Technically all of this differential topology is not included in the axioms. Still the proof seems to show that existence of parallels depends only on the separation axioms. Really very nice.

Interestingly in elliptic geometry the lines are circles.

• Infrared
Gold Member
Are you suggesting that we can just assume symmetry? I don't think we can, in mathematics we can't just make an assumption because it looks intuitive, we either have to prove it as a theorem or we have to introduce it as another postulate/axiom (e.g. "lines that meet in one half plane must meet in the other"*). And we already know that we can replace the parallel postulate with e.g. the triangle rule or any one of a number of equivalents so where has this got us?
I am suggesting that there is an underlying principle of symmetry that depends only on the separation axioms.

Symmetry is not equivalent to the parallel postulate. It merely guarantees the existence of parallels. It does not say how many.

BTW: The symmetry principle also works on the sphere. A great circle splits the sphere into two disjoint hemispheres. However, once one takes the quotient topology by identifying antipodal points the projections of the circles do not separate the the quotient space i.e. the projective plane.

If you read through @Infrared 's post #9 it seems to say that a model for the separation axioms is a smooth two dimensional manifold whose geodesics are smooth 1 dimensional submanifolds. In this model no parallel postulate is assumed nor any specific metric. Yet the existence of parallels is proved. It does not say how many parallels. I am not sure yet if there is a symmetry principle used in the proof. Will have to think about that. He uses the theorem that the only connected one manifolds without boundary are the line and the circle.

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Infrared
Gold Member
- A point on a line separates the line into two rays - or half lines. The two rays intersect in the point.
I realized that I don't use that axiom, and in fact it is inconsistent with having lines that are homeomorphic to circles (and adjusting this would also mean changing your last axiom). The axiom that lines can't meet more than once already rules out spherical geometry, so maybe this unnecessary?

He uses the theorem that the only connected one manifolds without boundary are the line and the circle.
I'm probably missing something, but where exactly did I use this?

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Gold Member
I realized that I don't use that axiom, and in fact it is inconsistent with having lines that are homeomorphic to circles (and adjusting this would also mean changing your last axiom). The axiom that lines can't meet more than once already rules out spherical geometry, so maybe this unnecessary?

I'm probably missing something, but where exactly did I use this?
Right. So it seems either that the ray postulate is not necessary or the circles contradict it.

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Gold Member
I'm probably missing something, but where exactly did I use this?
You are right. You did not use it.

TeethWhitener
Gold Member
This doesn't fully answer your question, but I think it covers the case of Euclidean, elliptic/projective, hyperbolic geometry. Let's suppose that our plane is a Riemannian 2-manifold, and the lines are geodesics.

Suppose for contradiction that your axioms are met, but there are no parallels, so every two lines meet in exactly one point. Let $L$ be a line, and $p$ be a point not on $L$. Then, for every nonzero $v\in T_pM$, the geodesic through $p$ with initial tangent vector $v$ meets $L$ exactly once by assumption. So, we have a map $T_pM-\{0\}\to L$. Scaling $v$ does not affect the geodesic, and since $M$ is $2$-dimensional, this gives a map $\mathbb{RP}^1=S^1\to L$. This map is onto (since any two points are connected by a geodesic) and one-to-one (since if two lines pass through $p$ and the same point on $L$, they must agree by your second axiom). I think this map is also continuous, so we have a continuous bijection $S^1\to L$. Since $S^1$ is compact and $L$ is Hausdorff, it is a homeomorphism, so $L\cong S^1$.

Now let $P_1$ and $P_2$ be points in one of the regions bounded by $L$. Let $L'$ be the line passing through $P_1$ and $P_2$. Since $L$ and $L'$ are homeomorphic to circles and $P_1$ and $P_2$ are on the same side of $L$, in order for $L$ and $L'$ to meet exactly once, they must be tangent at some point. But two geodesics that are tangent at a point are equal. Contradiction.
This is going to be a stupid question. I'm not a mathematician, so the language is going to be rough, at the very least; I'm just trying to wrap my head around the argument. At least intuitiviely, isn't the geodesic through $p$ with initial tangent vector $v$ equivalent to the geodesic through $p$ with initial tangent vector $-v$ (I don't know what else to call it)? Wouldn't that mean that the map is onto but not one-to-one?

Infrared
Gold Member
My map is from $\mathbb{RP}^1=(T_pM-\{0\})/\mathbb{R^\times}$ to $L$. Note that $v$ and $-v$ represent the same element of $\mathbb{RP}^1.$

Edit: If you're not familiar with projective space, here's a way to restate the argument without it. Choose coordinates so that $\{e^{i\theta}\}$ is the unit circle in $T_pM$, and consider the arc $C=\{e^{i\theta}:0\leq\theta\leq\pi\}.$ Every geodesic based at $p$ is equivalent (can be reparameterized) to one with initial tangent vector in $C$, and the only two tangent vectors in $C$ that give equivalent geodesics are $1=e^{0}$ and $-1=e^{i\pi}$. A map from an interval (arc) to a space where the two endpoints of the arc are mapped to the same point is equivalent to a map from a circle to the space. This map (from the circle) is bijective because now we've counted all the directions exactly once.

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• TeethWhitener
mathman
On a sphere every pair of lines meet at exactly two points. Where did exactly one point come from?

pbuk
Gold Member
One of our definitions is that two lines meet at exactly one point. We therefore interpret a "point" on a sphere as the pair of antipodal points where two great circles cross; the resulting geometry is consistent.

See post #7 and http://mathworld.wolfram.com/SphericalGeometry.html

• dextercioby
Gold Member
On a sphere every pair of lines meet at exactly two points. Where did exactly one point come from?
The projective space its the quotient space of the sphere obtained by identifying each pair of its antipodal points to a single point. Since two great circles intersect in two antipodal points, their projections in the projective plane intersect in one point.

When you identify antipodal points on a circle you still end up with a circle. Think of twisting a rubber band so that it overlaps itself twice. So in the projective plane great circles project to circles. You get a geometry where lines are circles and each pair of lines intersect in exactly one point.

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Gold Member
Note: Why a line does not separate the projective plane into two half planes.

Choose a line in the projective plane. Its preimage in the sphere is a great circle. Think of this great circle as the equator. It is surrounded by a cylindrical strip bounded by the Tropics of Cancer and Capricorn.

Since this strip is symmetrical around the equator all of its antipodal points lie inside of it. A point in the Southern half of the strip is antipodal to a point in the Northern half. It is identified to itself in the projective plane where identification turns it into a Mobius stip. So every line in the projective plane is surrounded by an embedded Mobius strip. One knows that when one removes the equator of a Mobius strip it does not fall into two pieces but just into one connected cylindrical band.

More simply the equator on the sphere splits the sphere into two hemispheres but these become one connected region in the projective plane.

.

• jim mcnamara
mathwonk
Homework Helper
Euclid proves in Prop. I. 31 the existence of parallel lines, without the use of the parallel postulate, i.e. in any "neutral" geometry (also called a Hilbert plane), in particular also in hyperbolic geometry. The proof uses only Prop. I. 28 (equivalent to I.27), which do not use the parallel postulate, which is first used in Prop. I.29.

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jim mcnamara
Mentor

• lavinia, mathwonk and Infrared
Gold Member
Following on @Infrared's proof, I thought a sketch of a proof that does not use calculus might go like this. ( Sorry, I do not know how to include diagrams.)

In a plane, given a line $L$, a point $O$ on it, and a line $L_{OA}$ perpendicular to it, draw a line perpendicular to $L_{OA}$ through $A$ and assume that there are no parallels. This line intersects $L$ in a point $B$.

$B$ lies in one of the half planes $H_1$ determined by $L_{OA}$. In the other half plane $H_2$ a point $P$ on $L$ determines a ray that makes some angle $∠OAP$ with $L_{OA}$. Let $α$ be the least upper bound of all of these angles.

Now draw a line through $A$ that makes the angle $α$ with $L_{OA}$ in the same half plane $H_2$. Call this line $L_{AX}$ where $X$ is the point where the line intersects the original line $L$. $X$ can not be in the half plane $H_2$ since then a larger angle than $α$ could be found. One does this by selecting a point on the ray in $L$ whose base point is $X$ and which does not contain the point $O$.

So $X$ must lie in the half plane $H_1$, the same half plane as the intersection point $B$ of $L_{AB}$ with $L$. Further $X$ can not lie on the line segment between $O$ and $B$ but must be in the half plane determine by $L_{AB}$ that does not contain $O$. (For the moment the case that $X$ and $B$ are the same will be excluded.)

Now consider the angles formed by the constructed lines at the point $A$. One has the 90 degree angle $∠OAB$; the angle $α$ , and a residual angle $∠BAX$ . These angles all together add up to 180 degrees.

* The lines through $A$ that are inside $α$ all intersect $L$ in the half plane $H_2$. The lines inside $∠OAB$ all intersect $L$ on the line segment $OB$. The lines through $∠BAX$ all intersect $L$ on the line segment $BX$.

No line intersects $L$ at a point on the ray that starts at $X$ and does not contain $B$.

But one can always draw such a line.

Note:

- If $X$ and $B$ are the same then one gets the same situation as in * but with the residual angle $∠BAX$ equal to zero.

To flesh this out, one needs a few ideas and facts. All of these - I think - can be defined in terms of the separation axioms.

- An idea of betweeness for three points on a line. This by itself excludes circles as candidates for lines.
- An idea of a line passing through the interior of an angle.
- The assumption that rays always contain at least two points.

I think this line of proof differs from Infrared's only in not using the continuity for the mapping of the projective line at $A$ onto the line $L$.

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