I want to try and identify a geometric structure I thought up while doing some weird stuff with making things walk on the surface of a 3D model and trying to incorporate backface culling into the surface geometry itself. See, in computer graphics each side of a polygon or triangle are considered separate entities and so I specifically desired to capture this within the geometry I constructed. Below, I will describe the different properties I know of to see if anyone can identify it as anything previously studied.

The Structure Itself

Let us define a special triangular mesh. Let's just call it a "half-triangle mesh" since I don't know what else to call it. In this context we define a half-triangle mesh to be a collection of half-triangles and we define a half-triangle to be an ordered triplet of points. These points technically form the vertices of Euclidean triangle in space.

Now comes the somewhat weird part. We can say that a half-triangle only has one side. If we look at it from a geodesic perspective and a physics perspective, from the side in space where the points are in clockwise order, there is nothing on that side. Literal emptiness. The geodesics will behave as if that triangle isn't there. However, from the other side, the half-triangle does exist and geodesics extending onto that triangle will behave as if it is there. Think of it like a one way window.


Now I did technically say a bit about the geodesics, but let me be more rigorous. When triangles facing the right direction form a surface like in a triangular mesh or this image geodesics behave like you would expect. In fact, two half-triangles with the same three points but facing in opposite directions form a regular triangle.

The unusual case is if we had a shape like in the T-surface within the below thing I found on google images. If all the protrusions are formed by normal triangles, then under my system the blue line is a geodesic/straight line.

a geodesic on a T shape

Whereas if the surface was formed by half-triangles and the unseen backsides did not have any half-triangles, then by extending the green lines according to the allowed rule set forth in Euclid's second postulate which states "lines may be extended infinitely in either direction" we get the following:

geodesics on a T shape without backsides

I hope the image makes sense. Basically, when there is no side from where the line is coming from, the line just ignores it and passes by it (hence the action of the blue line). Furthermore because there is no backside for the red line to wrap around onto, it just ends. These are just two examples.

Note, that this also means that by extending backwards along the blue line, the blue line can essentially grow a perpendicular segment that is part of the line. This essentially means that the lines in this system can split and branch wherever a lone half-triangle intersects a plane on the side that exists. I feel that this is in particular an important property of this geometric system.

Let's presume that this concept can be extended to 'surfaces' in general. Not just polygons. I dare not attempt it myself, but I'm sure the polygonal case is clear enough for people to get an idea of how that might extend. What sort of 'surface' geometry am I doing it and what is it classified under?

  • $\begingroup$ At no point did I say your question is wrong. I am correcting a questionable claim in the motivation of your question. If I wanted to show that your question was wrong, I would do so in an answer. $\endgroup$
    – user856
    Jul 6, 2017 at 17:19
  • $\begingroup$ @Typhon: Some definitions incorporate self-conflicting assumptions, which makes the definitions incorrect/faulty. This is the case with your definition as well. There is no such thing as "one-sided triangle". Instead, what you have, is a locally Euclidean geometry with discontinuities at zone boundaries. (Similar structures exist in nature, too, by the way.) Instead, you might wish to reword your question to asking that if you define certain boundaries to be one-way, what would be the overall geometry be classified as? (I don't know, I'm not a mathematician.) $\endgroup$ Jul 6, 2017 at 18:09
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    $\begingroup$ @Typhon: Will do. I should point out that my use of "discontinuity" above is from an outside perspective: from within the geometry, the discontinuities are undetectable. (That is, the geometry is locally Euclidean at every point.) (Even Portal, the game series which uses exactly such a discontinuity as the key game mechanic, uses a colored "ring" around the discontinuity to make them stand out.) $\endgroup$ Jul 6, 2017 at 18:46
  • $\begingroup$ @Typhon: Unfortunately, as a non-mathematician, I don't have anything better to offer than "locally Euclidean geometry at all points, with regions connected in a way that may violate Euclidean geometry" or something similar. (In your case, there is one way to detect zone boundaries corresponding to "one-way triangles": by searching for points where $(\vec{p} + \vec{t}) - \vec{t} \ne \vec{p}$, with $\vec{t} \to 0$ at the boundary.) $\endgroup$ Jul 6, 2017 at 19:40
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    $\begingroup$ For what it's worth, even a single "half triangle" is not a surface in the customary sense. A surface (viewed intrinsically) doesn't have "sides" at all, and saying "a curve lies on one side" has no standard meaning. (If you like, we're forced to use indelible markers on tissue paper.) Only the image of a surface under an embedding into a three-dimensional ambient space has "sides" in the customary sense. Here again, however, there's no standard meaning to the notion that "a curve on an embedded surface lies on one side of the surface". $\endgroup$ Jul 6, 2017 at 22:10

3 Answers 3


We can drop a dimension, since you can assume that locally the structure is the product of $\Bbb{R}$ and another geometric structure that I will describe.

This structure is a topological space together with a continuous action of the semigroup $\Bbb{R}_+$.

It is very similar to a switch in a train track structure. There are two semigroup actions of $\Bbb{R}_+=\{x\in\Bbb{R}$, s.t.\ $x\ge 0\}$, corresponding to moving forward and backwards. There can be no group action of $\Bbb{R}$, since that would imply time reversibility.

Consider three copies of the half line, blue, red and green, so the set is $S=\{b,r,g\}\times \Bbb{R}_+$.

Then the first semigroup action is given by $\rho_+:S\times \Bbb{R}_+\to S$ defined as \begin{eqnarray*} \rho_+((b,x),t)&=&\left\{ \begin{array}{ccc} (b,x-t)&if& t\le x\\ (g,t-x)&if& t>x \end{array}\right.\\ &&\\ \rho_+((r,x),t)&=&\left\{ \begin{array}{ccc} (r,x-t)&if& t\le x\\ (g,t-x)&if& t>x \end{array}\right.\\ &&\\ \rho_+((g,x),t)&=&(g,x+t)\\ \end{eqnarray*} and the second semigroup action is given by $\rho_-:S\times \Bbb{R}_+\to S$ defined as \begin{eqnarray*} \rho_-((b,x),t)&=&(b,x+t)\\ &&\\ \rho_-((r,x),t)&=&(r,x+t)\\ &&\\ \rho_+((g,x),t)&=&\left\{ \begin{array}{ccc} (g,x-t)&if& t\le x\\ (r,t-x)&if& t>x \end{array}\right.\\ \end{eqnarray*} The interpretation in terms of the structure described in the OP is the following: On the bottom you have the blue half line on the left and the green half line on the right, and the red half line is standing vertical on the joint. However, the joint is represented by the three different endpoints. enter image description here The interpretation of the actions of $\Bbb{R}_+$ is that when you travel away from the joint, nothing special happens, but if you enter the joint from the blue half line, you don't see the red line and continue on the green line. If you came from the red line, you also continue on the green line, but if you come from the green line, the you must continue on the red line. Note that the actions $\rho_+$ and $\rho_-$ of $\Bbb{R}_+$ are not compatible, so they do not induce an action of $\Bbb{R}$.

You can also provide $S$ with a topology in order to make the action continuous: Take as basis for the topology all open intervals in $\Bbb{R}_+$ for each of the three halflines, add the open-closed-intevals on the blue line of the form $(b,[0,x))$ (containing the endpoint) and add the unions of red and green open-closed-intevals $(r,[0,x))\cup (g,[0,y))$. With other words, if an open set contains the blue endpoint it has to contain some blue end-interval, if it contains the red endpoint, it has to contain some red end-interval AND some green end-interval, and if it contains the green endpoint, it has to contain some green end-interval AND some red end-interval.

With this topology both actions are continuous.

The topology is not Hausdorff, in particular it comes not from a metric structure (but there is a pseudometric structure with the distance between the endpoints being zero) the point $(b,0)$ is closed, but the points $(r,0)$ and $(g,0)$ are not closed (the union of both points $\{(r,0),(g,0)\}$ is closed.

If you take the quotient space and collapse the three endpoints (identifying them as only one point), then the action is no longer well defined.

In general topological spaces with continuous actions of $\Bbb{R}$ which are also called flows, is very well studied, for example in the theory of differential equations. However, if you only have a semigroup action, or two actions that are not symmetric, there is very few literature.

If you now return to dimension two, you are dealing with oriented geodesics, I didn't find anything specifically referring to your particular problem, but Thurston has worked in related areas.


In the 2dim case you would have the semigroup $\Bbb{R}_+\times\Bbb{R}$ acting on three halfplanes (with two different actions).

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    $\begingroup$ Actually the colors do not correspond to your picture, but to the picture of the three triangles in the answer of nominal animal. $\endgroup$
    – san
    Jul 19, 2017 at 4:24

This is not an answer, but an extended comment. It might help understand the situation better.

OP has three triangles, $ABD$ (blue), $BDC$ (red), and $BDE$ (green). In the following picture, the vertices are labeled using letters, and edges with $1$, $2$, or $3$: Three triangles The "real" side of the triangle is the one where the edges are numbered in counterclockwise order.

That is, $ABD$ (blue) and $BDE$ (green) have their real sides towards us, while $BDC$ (red) has its real side away from us.

If we consider the two-dimensional (and "one-sided") surface on those three "one-sided triangles", the geometry within each triangle is strictly Euclidean, but there are discontinuities at the edges of the 3D triangles. If we draw a directed graph of how a point on one triangle can pass to another triangle, by walking over an edge, we get: Directed graph of edge connections

The problem is that a 2D point in the $BDE$ (green) triangle traveling to the edge $1$ experiences a discontinuity at the edge. When it passes to the surface on the $ABD$ (blue) triangle, suddenly the entire halfspace (half-plane) behind it snaps to the $BDC$ (red) triangle!

If there were no such discontinuities, the geometry on the surface could be considered Euclidean.

A particular issue is that it is undefined which half-plane the 2D point observes, if it travels from the surface on the $BDE$ (green) triangle to the edge $1$, and then along the edge $1$: we cannot mathematically determine it based on the location of the point alone.

In computer programs, this problem is avoided by always having the point exist on the surface of exactly one triangle, and only changing to another triangle when the point passes over the edge. The discontinuous "snapping" happens only when the point goes outside the current triangle, onto a new one.

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    $\begingroup$ @Typhon: I meant, if we travel along the line segment 1 towards $D$, we cannot mathematically determine whether we see the plane $BCD$ (red) or $BDE$ (green) on our right side, using our location on the surface alone. The discontinuity is one problem, but defining when and how the discontinuity occurs, using just the point location alone, makes it a very difficult, ambiguous problem, mathematically. The question you pointed out shows that in fact, there might be more than one discontinuity at the same edge, making it even more of a problem. $\endgroup$ Jul 14, 2017 at 8:35
  • $\begingroup$ @Typhon: Right. Oh: I mentioned I looked at using an additional coordinate $\theta$ for the points on the surfaces, with $\theta$ constant except at an edge; the coordinates $(u, v, \theta)$ being shorthand for $(u, v, \cos\theta, \sin\theta)$, so that $\vec{p}_1 \cdot \vec{p}_2 = u_1 u_2 + v_1 v_2 + \cos(\theta_1 - \theta_2)$, and $(\vec{p}_1 - \vec{p}_2)\cdot(\vec{p}_1 - \vec{p}_2) = (u_1 - u_2)^2 + (v_1 - v_2)^2 + 2 - 2 \cos(\theta_1 - \theta_2)$. Now, $\theta$ "warps" distances so that $(u, v , \theta_1)$ and $(u , v , \theta_2)$ are separated by $\sqrt{2 - 2 \cos(\theta_1 - \theta_2)}$. $\endgroup$ Jul 14, 2017 at 20:04
  • $\begingroup$ @Typhon: However, $\theta$ can also be used to "choose" which triangle the point is upon, simply by assigning each triangle a $\theta$ of their own. However, my math-fu is not strong enough to tell me whether this makes any mathematical sense, especially whether such embeddings have any other use, so that they might have been defined or at least explored before. Finally, other than the properties of the 4D vector dot product, I haven't investigated whether vector algebra on them makes any sense. But it might be interesting! $\endgroup$ Jul 14, 2017 at 20:12

$\newcommand{\Reals}{\mathbf{R}}$This may not be an answer, but is too lengthy for a comment.

For simplicity, let's push the dimension down by one. Consider the plane $\Reals^{2}$ with the "half-interval cut" $C = \{0\} \times (-1, 1)$, and assume:

  1. The map $\gamma_{+}(t) = (t, 0)$, with $t$ real, is a continuous path. (Light can travel left-to-right through the cut.)

  2. The map $\gamma_{-}(t) = (-t, 0)$, with $t \leq 0$, is a continuous path.

  3. There exists no $a > 0$ such that $\gamma_{-}(t) = (-t, 0)$, with $t \leq a$, is a continuous path. (Light traveling right-to-left through $C$ "hits $C$ and stops/is absorbed/ceases to exist".)

I read the question as

How are we to describe such a geometric structure using the language/concepts and machinery of topology?

If the preceding properties capture the intent of a "half-triangle", one seemingly insurmountable obstacle to modeling with a topological space is: The map $\gamma_{-}$ is formally the composition of the continuous "time reversal" map $t \mapsto -t$ with the continuous map $\gamma_{+}$, but there exists no neighborhood of $0$ on which $\gamma_{-}$ is continuous (or even defined).

This violates the fact that in topology, a composition of continuous maps is continuous.

  • $\begingroup$ If items 1-3 really capture the properties of "one-sides obstacles", then it does look to me that topological spaces are not a natural framework. As for what happens in computer graphics, there's a camera-dependent mapping from space (the scene) to a plane, and a one-sided triangle is an obstacle for light rays only in one direction. At the moment, I'm inclined to say "one-sidedness" is a property of the camera (and the scene), not of the scene itself. [...] $\endgroup$ Jul 13, 2017 at 10:09
  • $\begingroup$ A bit more formally, it looks as if one-sidedness cannot be described mathematically by "photographing" a single topological space (a scene containing one-sided triangles), but can be described as a mapping from camera-and-scene to a plane. $\endgroup$ Jul 13, 2017 at 10:12
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    $\begingroup$ I explored the notion of using 4D coordinates $(u, v, \cos\theta, \sin\theta)$ to describe the "surface"; with the intent that $\theta$ represented a direction perpendicular to the surface, and all Cartesian axes used. That way the "triangles" are actually 3D subspaces of 4D Euclidean space partially projected to 3D, and the triangles have a "phase angle" of $\theta$ with respect to the 3D triangle surface. Or, in another words, the "one-sided triangles" reside on specific side of the 3D surface. Initially it looked like it works... but my math-fu is so weak I believe I just generated garbage. $\endgroup$ Jul 13, 2017 at 12:54
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    $\begingroup$ @Typhon: Games do. The floor is not always flat. (Think of games where you can push or move objects; in particular those where you can push a crate on relatively flat terrain, but not steep uphill.) $\endgroup$ Jul 17, 2017 at 20:37
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    $\begingroup$ @Typhon: Oh no, I was definitely referring to weird-physics puzzle-type games only. Yes, these games do use normal two-sided triangles, but the way they are "glued" together at the edges (or surfaces for volumes), is the key similarity to your one-sided triangles I was thinking of. For example, you can glue say 3 triangles together to form a continuous surface for a 2D point to travel on. The surface is locally Euclidean, but has very weird boundary conditions (depending on how the edges connect). The more triangles you use, the weirder the constructs can be, with "local" loops and whatnot. $\endgroup$ Jul 18, 2017 at 3:12

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