# Riemannian Manifolds II

### The Structural Equations

In the previous blog post I introduced a bit on Riemannian manifolds and differential geometry from the perspective of vector fields and their respective operators; despite the fact that I did not get a chance to go into too many examples, many of you will come to see that geometry can be very difficult to compute using vector fields alone. Indeed, vector fields do not produce nicely graded algebraic structures which could allow them to be multiplied or differentiated in a fashion similar to differential forms. Therefore, we first wish to further generalize the notion of a connection so that it is applicable to differenital forms.

It's been a little while since I originially introduced fibre bundles in my second post, so I'll go ahead and reiterate over the definitions in order to clear up any confusion. Whenever we are given some topological spaces (not necessarily manifolds) $$E$$ and $$M$$, as well as a continuous surjection $$\pi : E \to M$$, we say that the tuple $$(E, M, \pi)$$ is a fibre bundle if there is a local trivialization. By local trivialization, we mean that at each point $$x \in E$$ there is a neighborhood $$U$$ of $$\pi(x)$$ such that $$\pi^{-1}(U)$$ is isomorphic to the direct product of $$U$$ with some space $$F$$ known as the fibre.

I realize this is a bit of an abstract definition with a lot of moving parts, so let me simplify things a bit: we originally defined the tangent bundle $$TM$$ to be the space

$$TM = \coprod_{p\in M} T_pM = \bigcup_{p \in M} (\{ p\} \times T_pM)$$

Therefore, we associate to each point $$x \in M$$ a linear space $$T_xM$$ which represents all tangent vectors at that point — effectively, this doubles the degrees of freedom in our space. Using similar notation to classical mechanics, we may represent the coordinates of $$TM$$ by $$(x, v)$$ (where $$v$$ is a tangent vector in $$T_xM$$), giving us the obvious projection map $$\pi : TM \to M$$ defined by $$\pi(x, v) = x$$. Connecting $$TM$$ to our terminology above, if we let $$E = TM$$ one can easily see that $$(TM, M, \pi)$$ is a fibre bundle. The local trivialization comes from the fact that, given $$(x, v) \in TM$$, our local chart $$U$$ centered at $$x$$ satisfies $$\pi^{-1}(U) \simeq U \times \mathbb{R}^n$$.

In a similar fashion, we could represent the coordinates of our cotangent bundle $$T^*M$$ by $$(x, \alpha)$$ for some $$\alpha \in (T_xM)^*$$. By defining a projection map $$\pi : T^*M \to M$$ similarly by $$\pi(x, \alpha) = x$$, we can see that the tuple $$(T^*M, M, \pi)$$ is a fibre bundle as well. Thus, we may think of fibre bundles as generalizations of spaces such as the tangent and cotangent bundle over a manifold $$M$$ — our local trivialization merely says that the locally Euclidean condition is preserved by $$\pi$$.

I'm using a bit more generality than necessary here since we may actually assume more about our class of bundles related to $$TM$$ and $$T^*M$$. Indeed, at each point $$x \in M$$, our tangent plane $$T_xM$$ is simply a vector space isomorphic to $$\mathbb{R}^n$$ (where $$n$$ is the dimension of our manifold) — thus, we say that a fibre bundle $$(E, M, \pi)$$ is a vector bundle if each $$\pi^{-1}(x)$$ is a vector space.

In addition, the reader may recall that we defined a section $$\sigma : M \to E$$ to be a continuous right inverse of our projection map $$\pi$$, so that $$\pi(\sigma(x)) = x$$ for all $$x \in M$$ — in other words, sections associate every point $$x \in M$$ to an element $$(x, \omega_x)$$ of our vector space $$\pi^{-1}(x)$$. We denote the set of all smooth sections of a vector bundle $$(E, M, \pi )$$ by $$\Gamma(E)$$. For example, we saw that the archetypal examples of sections were vector fields when $$E = TM$$ and differential forms when $$E = T^*M$$, so we have that $$\Gamma(TM) = \mathfrak{X}(M)$$ and $$\Gamma(T^*M) = \Omega^1(M)$$.

From elementary linear algebra, we know that each vector space $$\pi^{-1}(x) = E_x$$ has a basis — however, when our basepoint $$x \in M$$ begins to vary, we begin to wonder whether we can encompass all bases in a smooth section. Indeed, over local trivializations we may simply pull back the standard basis for $$\mathbb{R}^n$$, thus allowing us to define a frame as a collection of sections $$e_1, \dots, e_n : M \to E$$ such that at each $$x \in M$$ we have that $$e_1(x), \dots, e_n(x)$$ is a basis for the vector space $$E_x$$.

Since we have generalized our vector fields $$\mathfrak{X}(M)$$ to sections over a vector bundle $$\Gamma(E)$$, we wish to further extend the analogy to connections as well. Originally, we defined an affine connection to be a map $$\nabla : \mathfrak{X}(M) \times \mathfrak{X}(M) \to \mathfrak{X}(M)$$ that satisfies

1. $$\nabla_{fX}Y = f \nabla_X Y$$ for all $$f \in C^\infty(M)$$ and $$X, Y \in \mathfrak{X}(M)$$
2. $$\nabla_X(fY) = (Xf) Y + f \nabla_XY$$ for all $$f \in C^\infty(M)$$ and $$X, Y \in \mathfrak{X}(M)$$

It's fairly easy to generalize a connection to sections $$s \in \Gamma(E)$$ in our first property above — that is,

$$\nabla_{fs}r = f \nabla_s r$$

for all $$f \in C^\infty(M)$$ and $$r, s \in \Gamma(E)$$. However, when we look at the second property we get stuck at the definition of $$sf$$ — how does a section of $$\Gamma(E)$$ act on a smooth function? This behavior was never defined, even for specific examples like when $$s = \omega$$ is a differential form! In fact, the only types of sections we defined to act on smooth functions were vector fields $$X \in \mathfrak{X}(M)$$. For this reason, we are unable to fully generalize a connection to a map $$\nabla : \Gamma(E) \times \Gamma(E) \to \Gamma(E)$$ and are thus left with a map $$\nabla: \mathfrak{X}(M) \times \Gamma(E) \to \Gamma(E)$$.

Ultimately, not much has changed from the previous blog post — we may still define the curvature and torsion using the same equations as before:

\begin{align} R(X, Y)s &= \nabla_X\nabla_Y s - \nabla_Y\nabla_X s - \nabla_{[X, Y]}s \\ T(X, Y) &= \nabla_XY - \nabla_YX - [X, Y] \end{align}

Now suppose $$e_1, \dots, e_n$$ is a frame over our vector bundle $$E$$ — by linearity we may use this frame to express any section $$s \in \Gamma(E)$$ as $$s = \sum_i \omega_i e_i$$ where the $$\omega_i \in \mathbb{R}$$ are our coefficients. Given a vector field $$X \in \mathfrak{X}(M)$$, we know that $$\nabla_Xs$$ is also a section over $$E$$ so it may also be represented by a sum $$\sum_i \omega_i(X, s) e_i$$ where the coefficients $$a_i \in \mathbb{R}$$ are now dependent on both $$X$$ and $$s$$. By linearity, we don't even need to consider all possible sections $$s$$, but instead the frame elements $$e_1, \dots, e_n$$. Hence, we write

$$\nabla_Xe_j = \sum_i \omega_j^i(X) e_i$$

so that our coefficients $$\omega_j^i$$ are dependent on $$X$$ alone.

At this point, one should notice that since our $$\omega_j^i$$'s are dependent on $$X$$, they live in the set $$\textrm{Hom}\,(TM, \mathbb{R}) = T^*M$$ and thus give rise to a finite set of differential forms known as the connection forms. Depending on the author, the reader may also see these referred to in the context of linear algebra as the connection matrix $$\omega = [\omega_j^i]$$.

Similar to our connection $$\nabla$$ above, we may represent the torision section $$T(X, Y)$$ as a linear combination $$T(X,Y) = \sum_i \tau_i(X, Y) e_i$$. Since the coefficients $$\tau_i(X, Y)$$ are solely dependent on $$X$$ and $$Y$$, we have that our $$\tau_i$$ are indeed $$2$$-forms on $$M$$. This brings us to the first important topic covered by this blog post: the structural equations.

##### First Structural Equation
Let $$e_1, \dots, e_r$$ be a smooth frame for the tangent bundle $$TM \to M$$ and let $$\theta^1, \dots, \theta^r$$ be the dual frame, so that $$\theta^i(e_j) = \delta_i^j = \begin{cases} 1 & i = j \\ 0 & i \neq j \end{cases}$$ Then we have the following equality: $$\tau_i = d\theta^i + \sum_j \omega_j^i \wedge \theta^j$$

Lastly, recall that our curvature tensor $$R(X,Y)$$ is not actually a section, but instead an endomorphism on $$\Gamma(E)$$ — thus, we must supply a section $$s$$ in order to get a curvature section $$R(X, Y)s$$ back. As done previously, we simply look at the frame elements in order to represent each section $$R(X, Y)e_j$$ as a linear combination $$R(X, Y)e_j = \sum_i \Omega_j^i(X, Y)e_i$$. Here, the coefficients $$\Omega_j^i$$ are solely dependent on the vector fields $$X, Y \in \mathfrak{X}(M)$$, so we have a family of $$2$$-forms $$\Omega_j^i$$. Much like the curvature matrix, since our curvature forms $$\Omega_j^i$$ are not just parameterized by the $$e_i$$ frame sections, but also the input sections $$e_j$$, we are able to find a curvature matrix given by $$\Omega = [\Omega_j^i]$$. This leads us into the second of our two structural equations:

##### Second Structural Equation
Let $$\nabla$$ be a connection over a vector bundle $$E \to M$$ of rank $$r$$ with frame $$e_1, \dots, e_r$$. Then $$\Omega_j^i = d\omega_j^i + \sum_k \omega_k^i \wedge \omega_j^k$$

To wrap this section up, we turn our attention to the amount of effort it takes to compute all these differential forms. Given an arbitrary $$n$$-dimensional manifold, there is an unfortunately large number of forms $$\omega_j^i$$ which we have to compute in order to get the full connection matrix $$\omega = [\omega_j^i]$$. To be precise, we have to compute $$n^2$$ connection forms $$\omega_j^i$$ since, for a fixed index $$1 \leq i \leq n$$, we have to iterate over all $$1 \leq j \leq n$$. However, we could cut that number in half if we knew some nice properties of our matrix $$\omega$$ such as symmetry or skey symmetry — in fact, that is precisely what this next theorem tells us!

###### Theorem
Let $$E \to M$$ be a Riemannian bundle in the sense that our metric $$g$$ now assigns an inner product $$\langle\,,\,\rangle_p$$ to each fibre $$E_p$$. Additionally, let $$\nabla$$ be a connection on $$E$$. Then the matrices $$\omega = [\omega_j^i]$$ and $$\Omega = [\Omega_j^i]$$ are skew-symmetric if and only if $$\nabla$$ is compatible with the metric.

### What is a Geometry?

Before we continue any further in our study of differential geometry, it may be helpful to some readers that I reiterate over the significance of material from my last blog post. We first defined the notion of a directional derivative $$D : \mathfrak{X}(M) \times \mathfrak{X}(M) \to \mathfrak{X}(M)$$ to be the Lie algebra homomorphism defined by

$$D_XY = \sum_i (Xb_i) \frac{\partial}{\partial x_i}$$

for $$Y = \sum_i b_i \frac{\partial}{\partial x_i}$$. In essence, this measures how much a tensor changes under the one parameter group of transformations generated by a vector field $$X$$. For example, we can see how $$Y$$ is transformed into $$D_XY$$ by $$X$$ in the following picture:

As the reader has hopefully realized by this point, not everything behaves as nicely as it does on flat Euclidean space $$\mathbb{R}^n$$. We ultimately generalized the notion of a directional derivative through the idea of a connection $$\nabla : \mathfrak{X}(M) \times \Gamma(E) \to \Gamma(E)$$ over some vector bundle $$E \to M$$ — when $$E = TM$$, we referred to this as an affine connection. Indeed, the Riemannian connection was a special case which allowed us to transport tangent vectors in a way that agreed with both the Riemannian metric and Lie algebra properties.

The reader may notice that the derivative of a vector field $$V(t)$$ need not be parallel to the surface at all points, like the directional derivative may lead you to believe. Indeed, our directional derivative and connections have derivational properties which tell us a huge amount about our surface through things like curvature; however, these derivations never actually deviate from our underlying frame $$e_1, \dots e_n$$. Indeed, we would expect that given a vector field $$X = \sum_i a_i e_i$$, a derivative of our vector field looks something like

$$\frac{\partial X}{\partial x_k} = \sum_i \frac{\partial a_i}{ \partial x_k} e_i + a_i \frac{\partial e_i}{\partial x_k}$$

When $$\frac{\partial e_i}{\partial x_k} = 0$$ for all $$i, k$$, we would have a derivation identical to our directional derivative — unfortunately, this phenomenon should only occur in flat Euclidean space. Over an arbitrary manifold, we can decompose $$\frac{\partial e_i}{\partial x_k}$$ into its tangential (i.e. frame) components, as well as its normal components

$$\frac{\partial e_i}{\partial x_k} = \left( \sum_j \Gamma^j_{ki} e_j + N^j_{ki} n_j \right)$$

where the $$n_j \in \Gamma(NM)$$ form a smooth frame over the normal bundle denoted $$NM$$. I will not go into details since we have discussed nothing of the normal bundle, but it turns out that our tangential components $$\Gamma_{ki}^j$$ are intrinsic (i.e. indpendent of embedding) to our manifold while the normal components $$N^j_{ki}$$ are not. For this reason, if we want to define a derivation that represents a characterization of $$M$$ and not the embedding, we must exclude any information about the normal bundle.

For a fixed curve $$c : [a, b] \to M$$ and affine connection $$\nabla$$, we formally define what is known as the covariant derivative to be a map

$$\frac{D}{dt} : \Gamma( TM\vert_{c(t)} ) \to \Gamma( TM\vert_{c(t)} )$$

which satisfies $$\frac{De_i}{dt} = \nabla_{c'(t)} e_i$$. Precisely, if $$X = \sum_i a_i(t) e_i \in \Gamma(TM\vert_{c'(t)})$$ is a vector field along the curve, then

$$\frac{D X}{dt} = \sum_i \frac{\partial a_i}{dt} e_i + a_i(t) \nabla_{c'(t)} e_i$$

In this case, we have that our tangential components are given by $$\nabla_{e_i}e_j = \sum_k \Gamma^k_{ij} e_k$$ and are known as the Christoffel symbols.

###### Theorem
An affine connection $$\nabla$$ on $$M$$ is torsion free if and only if the Christoffel symbols $$\Gamma^k_{ij}$$ are symmetric in $$i$$ and $$j$$.

Remember from your first calculus class you (hopefully) learned that the derivative may be used to find the minimum or maximum of some function $$f$$. This idea was later extended in multivariable calculus, where the gradient $$\nabla f$$ was shown to indicate the direction of steepest ascent / descent along the surface $$Z(f) = \{ \overrightarrow{x} \in \mathbb{R}^n \mid f(\overrightarrow{x}) = 0 \}$$. In general, maps with derivation-like properties are often used for minimization in applications ranging from optimization to analysis — for the covariant derivative, this is no different.

Before we jump into the thick of things, what is it that we really want to minimize? Well to answer that question, we have to go back and look at the definition of our covariant derivative $$\frac{D}{dt}$$ — indeed, we have that the covariant derivative is uniquely defined in terms of an affine connection $$\nabla$$ over a smooth curve $$c : [a, b] \to M$$. When the connection $$\nabla$$ is a Riemannian connection, it generalizes the directional derivative and is compatible with our Riemannian metric tensor $$g$$ so we may use $$\frac{D}{dt}$$ to measure change in distance (with respect to $$g$$) per unit time.

Alright, so at this point we have that the covariant derivative is used on a Riemannian manifold $$M$$ to minimize the distance travelled by a path $$c : [a, b] \to M$$. What can we expect this to look like? In flat (Euclidean) space, we have that the closest distance between two points $$P$$ and $$Q$$ is simply the straight line given by $$P + \lambda (Q - P)$$. Any other path $$\gamma$$ that travels from $$P$$ to $$Q$$ at a constant speed must have some curvature, and thus a nonzero accecleration vector $$\gamma''(t)$$. In particular, since we are in flat space the acceleration vector will have zero normal components and non-zero tangential components.

On the other hand, suppose we want to find the shortest path between $$P$$ and $$Q$$ in curved space (i.e. over some Riemannian manifold $$M$$) — in this case, every path must have some curvature induced by the surface. Now suppose we get in a car with no steering wheel and drove from point $$P$$ to $$Q$$ — the car would not turn due to the lack of steering wheel, so the acceleration would solely have non-zero normal components. However, if we were to draw some path that sidewinds around (so that it is not the shortest path from $$P$$ to $$Q$$), the acceleration vector would additionally have a non-zero tangential component associated to it.

As I mentioned above, when we defined our covariant derivative $$\frac{D}{dt}$$ we decided to throw away the normal components $$\sum_k N^k_{ij}n_k$$ since they were dependent on the choice of embedding. However, this implies that the covariant derivative only accounts for the tangential acceleration — therefore, a curve $$\gamma : I \to M$$ is a shortest path if and only if $$\frac{D\gamma'(t)}{dt} = 0$$. We refer to such a path $$\gamma$$ as a geodesic. For those interested in etymology, the word geodesic comes from the Greek words γη (Earth) + δαιεσθαι (distribute, divide), since geodesy was originally the study of measuring the Earth.

An Important Caveat: I haven't been entirely truthful in saying that geodesics and shortest paths are one in the same; as many of you may know, the differential is a local operator as it requires a coordinate neighborhood. For this reason, the requirement that a path $$\gamma$$ is a geodesic if it satisfies $$\frac{D\gamma'(t)}{dt} = 0$$ only tells us that $$\gamma$$ is locally the shortest path. For example, consider the $$2$$-dimensional sphere $$S^2$$:

Both the red path and the green path are geodesics (in particular, geodesics on $$S^2$$ are arcs of great circles); however, only the green path is the shortest path between $$P$$ and $$Q$$.

##### The Geodesic Equations
Let $$M$$ be a manifold with connection $$\nabla$$. A parameterized curve $$c(t)$$ is a geodesic if and only if, relative to any chart $$(U, x_1, \dots, x_n)$$, we have that $$y_k'' + \sum_{i,j} \Gamma^k_{ij}\, y_i'y_j' = 0 \ \ \ \ \ \ \ \ \ \forall k = 1, \dots, n$$ where $$y_i = x_i \circ c$$.

Since the coordinates $$y_i$$ in the theorem above exist on the real line, a generalized version of the Picard-Lindelof theorem tells us that there exists a unique solution to the ODE.

Let's look at what we have done in this section so far — we formulated a new kind of derivative to measure the tangential change of some smooth frame over a curve, and used that to generalize the notion of a straight line between two points. On the Euclidean plane, straight lines play a pivotal role in the study of geometry — in fact, Euclid himself originally proposed five axioms which he believed to serve as a basis for all Euclidean geometry. These axioms, also known as Euclid's Postulates, stated the following:

1. Given two points $$P$$ and $$Q$$, there exists a line segment joining $$P$$ and $$Q$$.
2. Any straight line segment can be extended indefinitely.
3. Given a straight line segement $$l$$, there exists a circle with $$l$$ as its radius and one endpoint as its center.
4. All right angles are congruent.
5. (Parallel Postulate) Given a line $$l$$ and a point $$P \notin l$$, there exists exactly one line $$l'$$ which passes through $$P$$ and is parallel to $$l$$.

So what does this have to do with our Riemannian manifolds? Great question — we defined a geodesic to be the generalization of a line on a Riemannian manifold $$M$$, so a good starting point for defining geometry on $$M$$ would be to replace the word "line" in the axioms above with "geodesic". We never really defined what it meant for geodesics to be parallel, but it shouldn't come as much of a surprise to know that two geodesics are considered parallel if they do not intersect. In this case, we may generalize the parallel postulate to the following:

1. (Generalized Parallel Postulate) Given a geodesic $$\gamma$$ on a Riemannian manifold $$M$$ and a point $$P \in M$$ not on $$\gamma$$, there exists exactly one geodesic $$\gamma'$$ which passes through $$P$$ and does not intersect $$\gamma$$.

One of the most significant results in the field of differential geometry is that the parallel postulate does not hold on an arbitrary Riemannian manifold. That is, there are numerous other geometries which are just as logically consistent as Euclidean geometry that may have infinitely many lines passing through a given point considered "parallel" to a straight line.

### Hyperbolic Geometry

We will now explore the first and most popular example of a non-Euclidean geometry, as originally discovered by mathematicians Janos Bolyai, Nikolai Lobachevsky, and Carl Fredrich Gauss — as the title likely gives away, this geometry is referred to as hyperbolic geometry. There are multiple spaces which are standard models for hyperbolic geometry such as the Poincaré disk, the Poincaré half-plane, and Minkowski space; however, we will restrict our attention to the Poincaré half-plane in order to minimize the number of coordinate charts we are working with.

To begin, we look at the half plane

$$\mathbb{H}^2 = \{ (x, y) \in \mathbb{R}^2 \mid y \gt 0 \}$$

Fortunately for us, this manifold only has one coordinate chart since $$\mathbb{H}^2$$ is an open set of $$\mathbb{R}^2$$. We next make $$\mathbb{H}^2$$ into a Riemannian manifold by equipping it with a Riemannian metric

$$g(x, y) = \frac{dx \otimes dx + dy \otimes dy}{y^2}$$

To give the reader a bit of an understanding, the standard Euclidean metric tensor on $$\mathbb{R}^2$$ is given by $$g(x,y) = dx \otimes dx + dy \otimes dy$$ — thus, our Riemannian metric for $$\mathbb{H}^2$$ depends on our position along the $$y$$-coordinate. In other words, as $$y$$ approaches $$0$$, our metric becomes scaled much larger, so what we would consider as a unit vector must be much smaller than normal to compensate. Conversely, as $$y$$ grows large, our metric "shrinks" so that a unit vector must look quite large to compensate.

Alas, no matter how much I try to talk through a concept many readers will find that visual aids lead to the best understanding. In my experience, I have found the mathematician Jos Leys' embeddings of M.C. Escher's tilings into the hyperbolic plane the most helpful — I encourage readers to take a look at the full gallery on his website. ###### M.C. Escher's (no. 35) ###### M.C. Escher's Angel-Devil (no. 45) ###### A Hyperbolic Tessellation of M.C. Escher's E35 by Jos Leys ###### A Hyperbolic Tessellation of M.C. Escher's E45 by Jos Leys

Going back to our equations, it is easy to check that our Riemannian metric $$g$$ yields the following orthonormal frame over the tangent bundle $$TM$$:

$$e_1 = y \frac{\partial}{\partial x} \hspace{3em} e_2 = y\frac{\partial}{\partial y}$$

Recall that, given a smooth frame $$e_1, \dots, e_n$$, we defined the dual frame $$\theta^1, \dots, \theta^n$$ to satisfy

$$\theta^i(e_j) = \delta^i_j = \begin{cases} 1 & i = j \\ 0 & i \neq j \end{cases}$$

We already know that our covectors $$dx_i$$ satisfy $$dx_i(\frac{\partial}{\partial x_j}) = \delta^i_j$$, so all that remains is to eliminate the coefficient $$y$$. Thus

$$\theta^1 = \frac{1}{y} \,dx \hspace{3em} \theta^2 = \frac{1}{y} \, dy$$

Since we wish to represent $$\mathbb{H}^2$$ as a Riemannian manifold, our Riemannian connection $$\nabla$$ must be torsion free by definition — in other words, $$\tau_i = 0$$ for $$i = 1, 2$$. By the first structural equation, this tells us that

\begin{align} 0 &= \tau_1 = d\theta^1 + \omega^1_1 \wedge \theta^1 + \omega^1_2 \wedge \theta^2 \\0 &= \tau_2 = d\theta^2 + \omega^2_1 \wedge \theta^1 + \omega^2_2 \wedge \theta^2 \end{align}

Since we are assuming $$\nabla$$ is a Riemannian connection, it must be compatible with the metric — by our theorem above the matrices $$\omega = [\omega^i_j ]$$ and $$\Omega = [\Omega^i_j]$$ are skew-symmetric. Hence, $$\omega^1_1 = \omega^2_2 = 0$$ and $$\omega^2_1 = -\omega^1_2$$. Since all of $$\mathbb{H}^2$$ is encompassed by a single coordinate chart, we may represent the 1-form $$\omega^1_2$$ as a linear combination

$$\omega^1_2 = a\,dx + b\,dy$$

Therefore, our equations above simplify to the following:

\begin{align} 0 &= \frac{1}{y^2}\,dx \wedge dy + (a\,dx + b\,dy) \wedge \frac{1}{y}\,dy \\&= \Big( \frac{1}{y^2} + \frac{a}{y} \Big) \, dx \wedge dy \\ 0 &= 0 - (a\,dx + b\,dy) \wedge \frac{1}{y}\,dx \\&= \frac{b}{y}\,dx \end{align}

Putting the two equations together, we can see that $$a = \frac{-1}{y}$$ and $$b = 0$$ so that

$$\omega = \begin{pmatrix} 0 & \frac{-1}{y}\,dx \\ \frac{1}{y} \, dx & 0 \end{pmatrix}$$

Fortunately, in two dimensions the curvature forms are fairly easy to compute — by skew-symmetry, $$\Omega^1_1 = \Omega^2_2 = 0$$ and $$\Omega^2_1 = - \Omega^1_2$$. In addition, the term

$$\sum_k \omega^i_k \wedge \omega^k_j$$

from the second structural equation always contains a diagonal entry, so it disappears. Thus, we need only compute $$\Omega^1_2 = d\omega^1_2 = \frac{1}{y^2} \,dy \wedge dx = \frac{-1}{y^2}\,dx\wedge dy$$ to see that

$$\Omega = \begin{pmatrix} 0 & \frac{-1}{y^2} dx \wedge dy \\ \frac{1}{y^2} dx \wedge dy & 0 \end{pmatrix}$$

and

$$K = \Omega^1_2(e_1, e_2) = \frac{-1}{y^2} (dx \wedge dy)\left( y \frac{\partial}{\partial x}, y \frac{\partial}{\partial_y} \right) = -1$$

Now that we know the intrinsic (Gaussian) curvature of $$\mathbb{H}^2$$, we wish to examine the geometry of the half space model via our geodesics. In order to identify the geodesics on $$\mathbb{H}^2$$, we must first compute all the Christoffel symbols with respect to our connection $$\nabla$$. Since our orthonormal frame is given by $$e_1 = y \, \partial_1, e_2 = y \,\partial_2$$, we may represent the standard basis $$\partial_1, \partial_2$$ by

$$\partial_1 = \frac{1}{y}\,e_1 \hspace{3em} \partial_2 = \frac{1}{y}\,e_2$$

For a fixed vector field $$X \in \mathfrak{X}(\mathbb{H}^2)$$, we may explicitly compute

\begin{align} \nabla_X \partial_1 &= \nabla_X \left( \frac{1}{y} \,e_1 \right) \\&= X\left( \frac{1}{y} \right)\,e_1 + \frac{1}{y}\nabla_Xe_1 \\&= \frac{-1}{y^2} (Xy) (y\,\partial_1) + \frac{1}{y} \omega^2_1(X)e_2 \\&= \frac{-1}{y} (Xy) \partial_1 + \frac{1}{y^2}\,dx(X) (y\,\partial_2) \\&= \frac{-1}{y} (Xy) \partial_1 + \frac{1}{y} \,(Xx) \partial_2 \\ \nabla_X \partial_2 &= \nabla_X \left( \frac{1}{y}\,e_2 \right) \\&= X \left( \frac{1}{y} \right)\,e_2 + \frac{1}{y}\nabla_X e_2 \\&= \frac{-1}{y^2}(Xy) (y\, \partial_2) + \frac{1}{y} \omega^1_2(X)e_1 \\&= \frac{-1}{y}(Xy) \partial_2 - \frac{1}{y^2}\,dx(X)(y\,\partial_1) \\&= \frac{-1}{y} (Xx) \partial_1 - \frac{1}{y} \,(Xy) \partial_2 \end{align}

By replacing $$X$$ with the basis elements $$\partial_1$$ or $$\partial_2$$, we get the following:

\begin{align} \Gamma^1_{12} &= \Gamma^1_{21} = \Gamma^2_{22} = \frac{-1}{y} \\ \Gamma^2_{11} &= \frac{1}{y} \\ \Gamma^k_{ij} &= 0 \hspace{3em} \textrm{otherwise} \end{align}

Let $$(\mathbb{H}^2, \tilde{x}, \tilde{y})$$ be the coordinate chart on $$\mathbb{H}^2$$ and let $$\gamma : I \to \mathbb{H}^2$$ be some curve on the Poincaré half-plane with constant speed. If we let $$x = \tilde{x} \circ \gamma$$ and $$y = \tilde{y}\circ \gamma$$ denote the coordinates on $$\mathbb{R}^2$$, we may use the geodesic equations to see that $$\gamma$$ is a geodesic if and only if it satisfies

\begin{align} x'' - \frac{2}{y} x'y' = 0 \\ y'' + \frac{1}{y} (x')^2 - \frac{1}{y}(y')^2 = 0 \end{align}

Since we required that our curve have constant speed, it follows that

$$\langle c'(t), c'(t) \rangle = \frac{(x')^2 + (y')^2 }{y^2} = \lambda$$

We may even reparameterize so that $$\lambda = 1$$, which gives us $$(x')^2 + (y')^2 = y^2$$.

Next consider two cases: $$x'(t)$$ is not identically zero and $$x'(t)$$ is the zero function. In the first case, we may divide our first differential equation by $$x'$$ so that we have

$$\frac{x''(t)}{x'(t)} = 2 \frac{y'(t)}{y(t)}$$

Integrating, we get that $$\ln(x') = 2\ln(y) + C = \ln(y^2) + C$$ and thus $$x' = k\,y^2$$ for some $$k \in \mathbb{R}$$. If we plug this into the third differential equation (i.e. the constant speed), we get that

$$k^2y^4 + (y')^2 = y^2$$

which yields $$y' = \pm y \sqrt{ 1 - k^2y^2 }$$. We therefore wind up with the following expression:

$$\frac{dx}{dy} = \pm \frac{ky^2}{y\sqrt{1 - k^2y^2} } = \pm \frac{ky}{\sqrt{ 1 - k^2y^2}}$$

Thus, our problem simply reduces down to freshman calculus — this equation can be solved using simple $$u$$-substitution with $$u = 1 - k^2y^2$$. In the end, the reader will wind up with the following solution:

$$x = \mp \frac{1}{k} \sqrt{1 - k^2y^2} + C$$

which is equivalent to $$(x - C)^2 + y^2 = \left( \frac{1}{k} \right)^2$$.

In the second case, when we assume $$x'(t)$$ is identically zero, we are unable to divide any equations by $$x'$$ for simplification. However, there is ultimately no need to, since $$x'(t) \equiv 0$$ tells us that our $$x$$ coordinate is constant and thus the curve $$\gamma$$ is a vertical line. Therefore, we come to the conclusion that geodesics on $$\mathbb{H}^2$$ are vertical lines and half-circles with center on the $$x$$-axis.

This image of the half-plane $$\mathbb{H}^2$$ divided along its geodesics is an incredibly important concept in the study of projective and arithmetic geometry, known as the Dedekind tessellation. I will not go into too much detail, but the Dedekind tesselation shows a relationship between a tiling of $$\mathbb{H}^2$$ with hyperbolic triangles and $$PSL(2, \mathbb{Z})$$ acting on $$\mathbb{H}^2$$ — it additionally allows one to decompose the hyperbolic plane into fundamental domains.

The main significance of $$\mathbb{H}^2$$ that I want to convey to the reader is that we can look at some point $$P$$ on the intersection of two circles $$C_1$$ and $$C_2$$ and choose a vertical line $$l$$ that is not touching these circles in order to break Euclid's parallel postulate. Indeed, since $$l \cap C_1 = l \cap C_2 = \emptyset$$, we have that $$C_1$$ and $$C_2$$ are both parallel to the vertical line $$l$$ — however, this would imply that there is more than one geodesic passing through $$P$$ which is parallel to $$l$$. In fact, it turns out that for any point $$P$$, there are infinitely many geodesics of $$\mathbb{H}^2$$ passing through $$P$$!

I have run out of material to cover in this blog post, but I applaud anyone who has read this far — most of my writing is subpar and I closely adapt many of the results from Loring Tu's Differential Geometry. However, I can assure you that we will begin to examine results not covered in the vast majority of differential geometry texts in the upcoming blog posts. Thanks again for reading! 😁