# 7.6: Geometry of Surfaces

- Page ID
- 23341

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)If you've got a surface in your hand, you can find a homeomorphic version of the surface on which to construct hyperbolic geometry, elliptic geometry, or Euclidean geometry. And the choice of geometry is unique: No surface admits more than one of these geometries. As we shall see, of the infinitely many surfaces, all but four admit hyperbolic geometry (two admit Euclidean geometry and two admit elliptic geometry). Thus, if you randomly generate a constant curvature surface for a two-dimensional bug named Bormit, Bormit will no doubt live in a world with hyperbolic geometry.

Chances are, too, that a constant curvature surface cannot be embedded in three-dimensional space. Only the sphere has this nice feature. In fact, if \(X\) is any surface that lives in \(\mathbb{R}^3\text{,}\) like the handlebody surfaces in Figure \(7.5.6\), then it must have at least one point with positive curvature. Why is this? The surface in \(\mathbb{R}^3\) is bounded, so there must be some sphere in \(\mathbb{R}^3\) centered at the origin that contains the entire surface. Shrink this sphere until it just bumps into the surface at some point. The curvature of the surface at this point matches the curvature of the sphere, which is positive.

We have seen that any surface is homeomorphic to a polygonal surface representing \(H_g\) or \(C_g\text{,}\) for some \(g\text{,}\) and we now show that each of these may be given a homogeneous, isotropic and metric geometry (so that it has constant curvature). A polygonal surface can only run into trouble homogeneity-wise where the corners come together. Any point in the interior of the polygon has a nice \(360^\circ\) patch of space about it, as does any point in the interior of an edge. However, if the angles of the corners that come together at a point do not add up to \(360^{\circ}\text{,}\) then the surface has either a cone point or a saddle-point, depending on whether the angle sum is less than or more than \(360^{\circ}\text{.}\) As we saw in Chapter 1 (Exercises \(1.3.5\) and Example \(1.3.7\)), in either case, we will not have a homogeneous geometry; a two-dimensional bug would be able to distinguish (with triangles or circles) a cone point from a flat point from a saddle point.

To smooth out such cone points or saddle points we change the angles at the corners so that the angles *do* add up to \(360^\circ\text{.}\) We now have the means for doing this. If we need to shrink the corner angles, we can put the polygon in the hyperbolic plane. If we need to expand the angles, we can put the polygon in the projective plane.

The standard polygonal representation of \(C_3 = \mathbb{P}^2\# \mathbb{P}^2 \# \mathbb{P}^2\text{,}\) is a hexagon having boundary label \(a_1a_1a_2a_2a_3a_3\text{,}\) as in Figure \(7.6.1\). All six corners of the hexagon come together at a single point. In the Euclidean plane, a regular hexagon has corner angles equal to \(120^{\circ}\text{.}\) To avoid a saddle point when joining the six corners together, shrink the corner angles to \(60^{\circ}\text{.}\) A tiny copy of a regular hexagon in the hyperbolic plane will have corner angles just under \(120^{\circ}\text{.}\) If the hexagon grows so that its vertices approach ideal points, its corner angles will approach \(0^{\circ}\text{.}\) At some point, then, the interior angles will be \(60^{\circ}\) on the nose. (We may construct this precise hexagon as well. See Exercise \(7.6.4\).) If \(C_3\) is built from this hexagon living in the hyperbolic plane, the surface inherits the geometry of the space in which it finds itself; that is, the surface \(C_3\) *admits* hyperbolic geometry, a nice homogeneous, isotropic and metric geometry.

Revisiting Example \(1.3.4\), consider the hexagon with boundary label \(abcabc\text{.}\) The six corners of this polygonal surface come together in groups of two. These corners create cone points because the angle sum of the two corners coming together is less than \(360^{\circ}\) in the Euclidean plane. We can avoid cone points by putting the hexagon in \(\mathbb{P}^2\text{.}\) A small regular hexagon in the projective plane will have corner angles just slightly greater than \(120^{\circ}\text{,}\) but we need each corner angle to expand to \(180^{\circ}\text{.}\) We may achieve these angles by expanding the hexagon until it covers the entire projective plane. In fact, the surface of Example \(1.3.4\) *is* the projective plane, and it admits elliptic geometry.

Each surface admits one homogeneous, isotropic, and metric geometry. In particular, the sphere (\(H_0\)) and projective plane (\(C_1\)) admit elliptic geometry. The torus (\(H_1\)) and Klein bottle (\(C_2\)) admit Euclidean geometry. All handlebody surfaces \(H_g\) with \(g \geq 2\) and all cross-cap surfaces \(C_g\) with \(g \geq 3\) admit hyperbolic geometry.

The following section offers a more formal discussion of how any surface admits one of our three geometries but we present an intuitive argument here.

The sphere and projective plane admit elliptic geometry by construction: The space in elliptic geometry *is* the projective plane, and via stereographic projection, this is the geometry on \(\mathbb{S}^2\text{.}\)

The torus and Klein bottle are built from regular \(4\)-gons (squares) whose edges are identified in such a way that all \(4\) corners come together at a point. In each case, if we place the square in the Euclidean plane all corner angles are \(\dfrac{\pi}{2}\text{,}\) so the sum of the angles is \(2\pi\text{,}\) and our surfaces admit Euclidean geometry.

Each handlebody surfaces \(H_g\) for \(g \geq 2\) and each cross-cap surfaces \(C_g\) for \(g \geq 3\) can be built from a regular \(n\)-gon where \(n \geq 6\text{.}\) Again, all \(n\) corners come together at a single point. A regular \(n\)-gon in the Euclidean plane has interior angle \(\dfrac{(n-2)\pi}{n}\) radians, so the corner angles sum to \((n-2)\pi\) radians. This angle sum exceeds \(2\pi\) radians since \(n \geq 6\text{.}\) Placing a small version of this \(n\)-gon in the hyperbolic plane, the corner angle sum will be very nearly equal to \((n-2)\pi\) radians and will exceed \(2\pi\) radians, but as we expand the \(n\)-gon the corners approach ideal points and the corner angles sum will approach \(0\) radians. Thus, at some point the angle sum will equal \(2\pi\) radians on the nose, and the polygonal surface built from this precise \(n\)-gon admits hyperbolic geometry. Exercise \(7.6.4\) works through how to construct this precise \(n\)-gon.

Of course, one need not build a surface from a regular polygon. For instance, the torus can be built from any rectangle in the Euclidean plane and it will inherit Euclidean geometry. So while the type of geometry our surface admits *is* determined, we have some flexibility where certain geometric measurements are concerned. For instance, there is no restriction on the total area of the torus, and the rectangle on which it is formed can have arbitrary length and width dimensions. These dimensions would have a simple, tangible meaning to a two-dimensional bug living in the surface (and might be experimentally determined). Each dimension corresponds to the length of a geodesic path that would return the bug to its starting point. A **closed geodesic path** in a surface is a path that follows along a straight line (in the underlying geometry) that starts and ends at the same point. Figure \(7.6.2\) shows three closed geodesic paths, all starting and ending at a point near a bug's house. The length of one path equals the width of the rectangle, the length of another equals the length of the rectangle, and the third follows a path that is longer than the first two.

Even in hyperbolic surfaces, where the area of the surface is fixed (for a given curvature) by the Gauss-Bonnet formula, which we prove shortly, there is freedom in determining the length of closed geodesic paths.

If we make three slices in the two-holed torus we obtain two pairs of pants, as indicated in the following figure.

We label our cuts so that we can stitch up our surface later. Match the \(c_i\text{,}\) \(d_i\text{,}\) and \(e_i\) edges to recover the two-holed torus. Any pair of pants can be cut into two hexagons by cutting along the three vertical seams in the pants. It follows that the two-holed torus can be constructed from four hexagons, with edges identified in pairs as indicated below.

The four hexagons represent a cell division of \(H_2\) having \(6\) vertices, \(12\) edges, and \(4\) faces. In the edge identification, corners come together in groups of \(4\), so we need each corner angle to equal \(90^\circ\) in order to endow it with a homogeneous geometry. We know we can do this in the hyperbolic plane. Moreover, according to Theorem \(5.4.6\), there is freedom in choosing the dimensions of the hexagons. That is, for each triple of real numbers \((a,b,c)\) there exists a right-angled hexagon in \(\mathbb{D}\) with alternate lengths \(a,b,\) and \(c\text{.}\) So, there exists a two-holed torus for each combination of six seam lengths (\(a_1,a_2,a_3,b_1,b_2,\) and \(b_3\)).

A surface that admits one of our three geometries will have constant curvature. The reader might have already noticed that the *sign* of the curvature will equal the sign of the surface's Euler characteristic. Of course the *magnitude* of the curvature (if \(k \neq 0)\) can vary if we place a polygonal surface in a scaled version of \(\mathbb{P}^2\) or \(\mathbb{D}\text{.}\) That is, while the type of homogeneous geometry a surface admits is determined by its Euler characteristic (which is determined by its shape), the curvature scale can vary if \(k \neq 0\text{.}\) By changing the radius of a sphere, we change its curvature (though it always remains positive). Similarly, the surface \(C_3\) in Figure \(7.6.1\) has constant curvature \(-1\) if it is placed in the hyperbolic plane of Chapter 5. However, it can just as easily find itself in the hyperbolic plane with curvature \(k = -8\text{.}\) Recall, the hyperbolic plane with curvature \(k \lt 0\) is modeled on the open disk in \(\mathbb{C}\) centered at the origin with radius \(\dfrac{1}{\sqrt{|k|}}\text{.}\) Placing the hexagonal representation of \(C_3\) into this space so that its corner angle sum is still \(2\pi\) produces a surface with constant curvature \(k\text{.}\)

We have finally arrived at the elegant relationship between a surface's curvature \(k\text{,}\) its area, and its Euler characteristic. This relationship crystalizes the interaction between the topology and geometry of surfaces.

The area of a surface with constant curvature \(k\) and Euler characteristic \(\chi\) is given by the formula \(kA = 2\pi \chi.\)

**Proof**-
The sphere with constant curvature \(k\) has radius equal to \(\dfrac{1}{\sqrt{k}}\text{,}\) and area equal to \(\dfrac{4\pi}{k}\text{.}\) Since the sphere has Euler characteristic \(2\), the Gauss-Bonnet formula holds in this case. The projective plane \(\mathbb{P}^2\) with curvature \(k\) has area equal to \(\dfrac{2\pi}{k}\text{,}\) and Euler characteristic equal to \(1\), so the result holds in this case as well. The torus and the Klein bottle each have \(k = 0\) and \(\chi = 0\text{,}\) so in this case the Gauss-Bonnet formula reduces to the true statement that \(0 = 0\text{.}\)

Any surface of constant negative curvature cam be represented by a regular \(n\)-sided polygon with \(n \geq 6\text{.}\) Furthermore, this polygon can be placed in the hyperbolic plane with curvature \(k \lt 0\text{,}\) so that its interior angles sum to \(2\pi\) radians. According to Theorem \(7.4.3\),

\[ kA = 2\pi - (n-2)\pi = 2\pi\bigg(2-\dfrac{n}{2}\bigg)\text{,} \]

where \(A\) is the area of the \(n\)-gon.

Now, for \(g \geq 2\text{,}\) \(H_g\) is represented by a \(4g\)-gon so that

\[ kA = 2\pi(2-2g)=2\pi\chi(H_g)\text{.} \]

For \(g \geq 3\text{,}\) \(C_g\) is represented by a \(2g\)-gon so that

\[ kA=2\pi(2-g) = 2\pi\chi(C_g)\text{.} \]

This completes the proof.

Suppose a two-dimensional cosmologist believes she lives in a surface of constant curvature (because her universe looks homogeneous and isotropic). If she can deduce the curvature of the universe and its total area, she will know its Euler characteristic. That is, by measuring these geometric properties she can deduce the shape of her universe, or, at worst, narrow down the possibilities to two. If \(\chi\) is \(2\) or an odd integer then the cosmologist will know the shape of the universe. If \(\chi \lt 2\) is even, her universe will have one of two possible shapes - one shape orientable, the other non-orientable.

## Exercises

Suppose our intrepid team of two-dimensional explorers from Exercise \(7.4.4\), after an extensive survey, estimates with high confidence that the area of their universe is between \(800,000\) km\(^2\) and \(900,000\) km\(^2\text{.}\) Assuming their universe is homogeneous and isotropic, what shapes are possible for their universe? Can they deduce from this information the orientability status of their universe?

Suppose a certain constant curvature \(3\)-holed torus \(H_3\) has area \(5.2\) km\(^2\text{.}\) What must be the area of a constant curvature \(C_3\) surface so that they have the same curvature?

*Building a hyperbolic surface from pairs of pants.* To construct \(H_g\) for \(g \geq 2\) from pairs of pants, how many do we need? Express your answer in terms of \(g\text{.}\)

Suppose we want to build a regular \(n\)-gon in \((\mathbb{D},{\cal H})\) from the vertices \(v_k = r e^{i \frac{k}{n}2\pi}\) for \(k=0,1,\ldots,n-1\) so that the \(n\) interior angles sum to \(2 \pi\text{.}\) Prove that this is the case when \(r = \sqrt{\cos(\dfrac{2\pi}{n})}\text{.}\)

**Hint**-
It may be helpful to refer to Exercise \(5.4.9\) where we proved this result in the case \(n=8\text{.}\)