1
$\begingroup$

I was doing a spot of light reading (crystallography), when the term "convex" polyhedron came up in a a section (very prominently) in conjunction with something else called the "Euler characteristic".

The Wikipedia article (linked above) on the "Euler characteristic" is written in the same vein as the book I'm using... but try as I might, I can't seem to understand it :/

enter image description here

Excerpt from the section (in the Wiki article) in question.

I suppose my inability to fully comprehend the section (relevant excerpt included above) is due to my shortcomings as a high-school student; so following the other Wiki links wasn't much help...and neither was Google.


My questions:

Can someone explain (in a way a high-school student such as myself would understand),

1) What is a "convex" polyhedron?

2) What's this "Euler characteristic" all about? Does it have any sort of physical implications (by "physical", I mean "geometry" of the solid)?


(Including my thoughts on this...in case it helps a potential answerer in tailoring an answer best suited to my needs/understanding)

Looking at the figures/models/diagrams provided in the Wiki article, has led me to believe that "convex" polyhedra must be the same as (or at least, very similar to) the idea of a "regular" polyhedra (identical faces, sides and dihedral angles) that we've learned when we were younger...except the corresponding Wiki article on "convex" polyhedra does not reflect this simplicity; so drawing such an equivalence appears to be wrong.

I have not studied "Topology" (at least, not as a field of mathematics). Obviously, I'm quite capable of looking at 3D object and classifying it as a "cube", a "sphere", some sort of "prism", etc...but my knowledge of "topology" pretty much ends there. I'm sorry.

I was tempted to ask the question ("What's a convex polyhedron?", "What's the Euler characteristic?") separately. However, seeing as they're intimately linked (at least, from the crystallography perspective), I felt it more prudent to put these questions in a single post.

$\endgroup$
  • 2
    $\begingroup$ Convex for a shape means roughly that any two points are connected by a straight path that lies within the boundaries of the shape. As an example take a crescent moon shape, you can draw a line between two points that has parts of the line outside the shape. A convex polyhedra need not be regular. $\endgroup$ – Triatticus Sep 20 '17 at 7:27
  • 1
    $\begingroup$ It's honestly the most intuitive way to think about it I think, especially from a physics point of view $\endgroup$ – Triatticus Sep 20 '17 at 7:31
  • 1
    $\begingroup$ When I wrote within I had considered the shape to be the interior plus it's boundary to be included, should have clarified. $\endgroup$ – Triatticus Sep 20 '17 at 7:38
  • 2
    $\begingroup$ Any two arbitrary points, essentially it's a consideration over ALL pairs of points. $\endgroup$ – Triatticus Sep 20 '17 at 7:41
  • 2
    $\begingroup$ There's no "roughly" needed: a set $S$ is convex if for every two points $x, y$ in $S$, the line segment from $x$ to $y$ is also in $S$. This includes the boundary, if $S$ includes its boundary. $\endgroup$ – Robert Israel Sep 20 '17 at 7:42
1
$\begingroup$

Let's start with a precise definition of a convex polyhedron: I assume you are interested in bounded convex polyhedra (also called convex polytopes). A formal definition of a 3-dimensional convex polytope $P$ is that it is a bounded subset of $R^3$ (the 3-dimensional Euclidean space) which equals the intersection of finitely many closed half-spaces and is not contained in a plane. A closed half-space in $R^3$ can be defined algebraically as the solution set of a linear inequality $$ ax + by + cz + d\ge 0, $$ where $a, b, c, d$ are given numbers (such that $(a,b,c)\ne (0,0,0)$). In other words, $P$ is the solution set of a system of linear inequalities in three variables; these inequalities are called the defining inequalities of $P$.

Each convex polytope $P$ has faces of various dimensions, they are traditionally called vertices, edges and faces. These can be defined in several ways. One definition is that, say, a vertex of $P$ is a point $v$ of $P$ which is the solution set of a system of equations and inequalities, obtained from the defining inequalities by converting some of them into equations. Same for faces of other dimensions.

One verifies that each polytope has only finitely many vertices, edges and faces. The numbers of these are usually denoted $V, E$ and $F$. Euler's formula is $$ V- E + F =2. $$ One way to appreciate this formula is to notice that it helps you to do "numerology" of polytopes. Suppose, for instance, that you have a simple polytope, i.e. where every vertex has three edges coming out of it. ("Generic" polytopes have this property: If you perturb randomly the linear inequalities defining $P$, the new polytope becomes simple.) Given this, you already know that $$ 2E=3V $$ (every vertex defines three oriented edges). Now, plug this into Euler's formula and you get: $$ F= 2 - V + \frac{3}{2} V= 2 + \frac{1}{2} V. $$ I doubt that you would be able to come up with this equation without knowing Euler's formula.

Euler's formula has other applications and leads to some interesting advanced mathematics.

$\endgroup$
0
$\begingroup$

As other have said, a convex shape is one where the straight line connecting any two points is entirely within the shape.

The Euler characteristic tells you something about the topological shape of the solid in question. By topological I mean you can distort things without changing the Euler characteristic, but no cutting or gluing. So a tetrahedron or a cube are essentially the same shape - you can make either out of a ball of putty just by flattening appropriate places - and have the same Euler characteristic. But if you imagine drilling a square hole through a cube, this would make a different type of solid - instead of a ball, it's sort of like a donut - and this has a different Euler characteristic (you should find $F+V-E=0$, not $2$).

$\endgroup$
  • 1
    $\begingroup$ "The Euler characteristic tells you something about the topological shape of the solid in question. By topological I mean you can distort things without changing the Euler characteristic." Thanks for taking the time to post an answer. However, the part of your answer I quoted is tautological. So I'm still at square one :( Thanks anyways though! ) $\endgroup$ – paracetamol Sep 20 '17 at 9:05
  • $\begingroup$ Take a look at (math.stackexchange.com/questions/1828482/…) $\endgroup$ – Jean Marie Sep 21 '17 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.