Generalized Sides
This was inspired by a conversation in the new Mathblr Discord server, come join if it sounds like fun!
So a polygon is a 2-dimensional shape that has a certain, finite number of sides, right? These are the parts of the boundary of the shape that are straight line segments in the plane. It seems like intuitively there are more kinds of shapes that can be said to have a certain number of sides, though. The boundary of a Reuleaux triangle, for example, contains no straight line segments but it looks like it has three ‘sides’, for some definition thereof. You might even ask how many sides a circle has, or a teardrop shape. Let’s try to generalize.
The rest of this post is written in the online math typesetting format KaTeX, so viewing it on the Tumblr dashboard is not ideal. View the post in a browser on my Tumblr page to render the math properly.
A (planar) curve \(C\) is a subset of \(\R^2\) such that there is at least one continuous function \(\gamma \colon S^1 \to \R^2\) such that \(C\) is the image of \(\gamma\), where \(S^1\) is the unit circle in \(\R^2\). We wish to define a notion of the amount of ‘sides’ and ‘vertices’ that this curve has.
The curve \(C\) is called differentiable in the point \(x \in C\) if there is some parametrisation of \(C\) such that this parametrizing function can be linearly approximated near \(x\). EDIT: if there is some homeomorphism from a real open interval to an open \(C\)-neighbourhood of \(x\) that can be linearly approximated near \(x\). The exact specifics are a little technical, and ultimately unnecessary; as long as you have an intuitive idea of what it means for a curve to be smooth at certain points it’s fine.
We will call a connected open subset \(A \subset C\) a smooth segment of \(C\) if for every \(a \in A\) we have that \(C\) is differentiable in \(a\). \(A\) is additionally called a side of \(C\) if for every smooth segment \(B\) we have that \(A \subset B\) implies \(A = B\). A vertex of \(C\) is then a point not contained in the union of all sides of \(C\).
The following three facts are true about sides and vertices:
1. If some \(x \in C\) has a \(C\)-neighbourhood that is a smooth segment, then there is some side \(E\) with \(x \in E\).
2. Any two distinct sides are disjoint.
3. If the set of vertices is finite and non-empty, and there is a parametrisation \(\gamma\) of \(C\) that is injective, then the number of sides equals the number of vertices.
Proof. This is the longest proof I’ve written on this blog so far. If you’re ever lost, I’d recommend drawing your favourite \(n\)-sided polygon, marking its vertices, then drawing a circle with \(n\) distinct points marked on it, and finally drawing a big arrow from the circle to the polygon labelled \(\gamma\).
Let \(C\) and \(x \in C\) be such that \(U\) is a smooth segment that contains \(x\). Define \(E(x) := \bigcup_{V \in \mathcal{V}} V\), where \(\mathcal{V}\) is the collection of all smooth segments \(V\) of \(C\) such that \(x \in V\).
We know that \(U \subset E(x)\), so it is not empty. \(E(x)\) is a union of open sets, so it is itself open. It is also a union of connected sets that each contain \(x\), so \(E(x)\) is connected. For every \(a \in E(x)\) it holds that \(a\) is contained in some smooth segment, so \(C\) is differentiable in \(a\). It follows that \(E(x)\) is a smooth segment. Assume that \(B\) is a smooth segment such that \(E(x) \subset B\). Then \(x \in B\), so \(B \subset E(x)\). We conclude that \(E(x)\) is a side of \(C\). Let \(A,B\) be sides of \(C\) such that there is some \(x \in A \cap B\). Then \(A,B\) are smooth segments that contain \(x\), so \(A,B \subset E(x)\). Because \(A\) and \(B\) are sides, it follows that \(A = E(x) = B\). We conclude that any two distinct sides of \(C\) are disjoint. Let \(V\) be the set of vertices, and \(\mathcal{E}\) the set of sides. Assume that \(C\) is such that \(V\) is finite and non-empty, and let \(\gamma \colon S^1 \to \R^2\) be an injective parametrisation of \(C\). \(C\) is precisely the image of \(\gamma\), so there is an inverse mapping \(\gamma^{-1} \colon C \to S^1\). Note that any continuous bijection from \(S^1\) into a Hausdorff space is a homeomorphism, so \(\gamma\) and \(\gamma^{-1}\) are additionally open maps between \(S^1\) and \(C\). For the sake of notational brevity, whenever \(x \in C\) or \(A \subset C\), define \(x_\circ := \gamma^{-1}(x)\) and \(A_\circ := \gamma^{-1}(A)\) respectively. Any finite subset of \(S^1\) is discrete, for every vertex \(v \in V\) we have that \(\{v_\circ\}\) is an isolated point. For every \(v \in V\) there is some angle \(\varepsilon(v) > 0\) such that \(B_{\varepsilon(v)}(v_\circ) \cap V_\circ = \{v_\circ\}\), where the metric on \(S^1\) is taken to be the smallest angle between two points. Let \(f \colon V \to \mathcal{E}\) be given by
\[ f(v) = E \left( \gamma \left( v_\circ + \frac{\varepsilon(v)}{2} \right) \right), \]
where \(+\) on \(S^1\) is taken as counterclockwise addition of angles. First we need to show that \(f\) is well-defined: we must show that \(C\) is differentiable at \(p := v_\circ + \varepsilon(v)/2\). Define \(A := B_{\varepsilon(v)/2}(p)\). We know that \(A \cap V_\circ = \emptyset\), and \(A\) is connected, so \(A \cap (S^1 \setminus V_\circ) = A\) is a connected open set containing \(p\) that is itself contained in the inverse image of some smooth segment of \(C\), hence \(\gamma(A)\) is a connected open subset of \(C\) containing \(\gamma(p)\) on which \(C\) is differentiable. We conclude that \(f\) is well-defined. Now we prove that \(f\) is bijective. \(V\) is not empty, so let \(v,w \in V\) be such that \(f(v) = f(w)\). By definition of \(f\) there is some connected subset \(K \subset S^1 \setminus V_\circ\) such that \(v_\circ + \varepsilon(v)/2\) and \(w_\circ + \varepsilon(w)/2\) are both in \(K\). We have that the counterclockwise open arcs \((v_\circ,v_\circ+\varepsilon(v)/2)\) and \((w_\circ,w_\circ+\varepsilon(w)/2)\) are subsets of \(S^1 \setminus V_\circ\), so we have \(w_\circ \notin (v_\circ, v_\circ + \varepsilon(v)/2)\) and \(v_\circ \notin (w_\circ, w_\circ + \varepsilon(w)/2)\). Hence the two arcs are either disjoint or equal. If the two arcs are equal, then \(v = w\) and we are done. If the arcs are disjoint, then there is no connected subset of \(S^1\) that contains \(v_\circ + \varepsilon(v)/2\) and \(w_\circ + \varepsilon(w)/2\) but not one of \(v_\circ\) and \(w_\circ\). Because \(K\) must satisfy this, it follows that \(v = w\), so \(f\) is injective. Now let \(E \in \mathcal{E}\) be some side of \(\gamma\). Because \(E_\circ\) is a connected open subset of \(S^1 \setminus V_\circ\), and \(V_\circ\) is non-empty, it follows that \(E_\circ = (a,b)\) for some (not necessarily distinct) \(a,b \in S^1\), where \((a,b)\) is, again, the counterclockwise open arc from \(a\) to \(b\). We prove that \(a,b \in V_\circ\). Assume that \(a \notin V_\circ\). Then there is some \(A \subset S^1\) containing \(a\) such that \(\gamma(A)\) is a smooth segment of \(C\). Then \(E \cup \gamma(A) = \gamma((a,b)) \cup \gamma(A)\) is a smooth segment strictly larger (because it contains \(\gamma(a)\)) than \(E\), which is a contradiction. We conclude that \(a \in V_\circ\). The proof for \(b\) is exactly analogous. Then, because \(a + \varepsilon(\gamma(a))/2 \in E\), it follows that \(E = f(\gamma(a))\), and we conclude that \(f\) is surjective, hence bijective. We have constructed a bijection between \(V\) and \(\mathcal{E}\), so the number of vertices equals the number of sides, QED.
Note that nowhere in this proof we actually used the definition of what it means for \(C\) to be differentiable in some point \(x\). In fact, if we were to replace differentiability with an arbitrary property \(\Phi(x)\), these three facts we’ve just proven would still hold true. The only thing we need is that a ‘\(\Phi\)-side’ is connected, open, and is a maximal such set on which \(\Phi(x)\) holds for all \(x\).
However, if we wanted to determine what kinds of planar transformations preserve the number of sides and vertices, we would need to use the proper definition. It is my hope that a set being a side is preserved under a diffeomorphism of \(C\), but this post is already way too long so I leave that to any curious readers :)
