# An invitation to differential geometry?

For the sake of context: I've just finished a master degree in Mathematics and my goal is to get a Ph.D. in Complex Analysis. In the years I spent studying my undergraduate mathematics degree I had always avoided the Geometry courses because the subject (or maybe the way it was taught to me) seemed... tedious. I remember to think "Euclidean Geometry, manifolds, Riemann metrics, curvature, geodesics... Ok, so what?". That thought, that inability to appreciate the beauty of Geometry ---which I can tell it exists by the many people in the mathematical community who know way more than me about it and claim so--- in the same way I distinctly see it (specially) in Complex Analysis has always bothered me.

A few days ago I was reading Rudin's Real and Complex Analysis and reached the following theorem:

Theorem: If $\varphi$ is convex on $(a,b)$, then $\varphi$ is continuous on $(a,b)$.

PROOF The idea of the proof is most easily conveyed in geometric language. Those who may worry that this is not "rigorous" are invited to transcribe it in terms of epsilons and deltas.

Suppose $a<s<x<y<t<b$. Write $S$ for the point $(s,\varphi(s))$ in the plane, and deal similarly with $x,y,$ and $t$. Then $X$ is on or below the line $SY$, hence $Y$ is on or above the line through $S$ and $X$; also, $Y$ is on or below $XT$. As $y\to x$, it follows that $Y\to X$, i.e., $\varphi(y)\to\varphi(x)$. Left-hand limits are handled in the same manner, and the continuity of $\varphi$ follows.

The instant I read "in geometric language" I started frowning, but I continued reading. After drawing a picture and give it some thought I was thinking, much to my surprise, that it couldn't exist a more beautiful proof of this theorem. I couldn't like more that "three-lines-Lipschitz" clean argument; (!!) Today I had a similar "incident" working out the proof that the (angle preserving) set of isometries of the Poincaré disk model coincides exactly with the Mobius transformations of the unit disk, so I have decided to give a hard try to Geometry.

I want to begin with Differential Geometry which, being closer to my comfort zone, I expect to be a good choice. I have already chosen from which books I will study (Spivak's A Comprehensive Introduction to Differential Geometry), so my question is

Question: Could you please share some example of a geometric argument, geometric result, geometric idea, or even a geometric calculation in the realm of Differential Geometry which ---in the same spirit as the two examples I gave--- you find exceptionally beautiful or enlightening and explain why?

It would be a good source of motivation to keep studying and learning Geometry to me (and maybe others with a similar problem), so I would be sincerely grateful to hear from you. =)

[Sorry for the extension, I couldn't find a shorter way to accurately explain what I am looking for.]

• This is a really good question; sadly I am not sure it is a good fit for the SE model. I'm certainly not going to be the one to cast the first close vote, but it might happen. You may have luck by asking in chat, especially if @TedShifrin is around :) – Eric Stucky Nov 4 '16 at 3:13
• Not really in the same spirit as the question but since you are interested in complex analysis, the theory of several complex variable relies on a lot of notions from geometry. – ClassicStyle Nov 4 '16 at 3:18
• @EricStucky Thank you for replying, and foor the reference too. I was aware that some might think that this is "primarily-opinion-based" and that that could be a problem sooner or latter. However I feel that even if I can't express the question without using the words "you find" or "beautiful", it has so much meaning as asking for recomendations for a book or something (which are relatively frequent questions here at MSE). Anyway, I hope to get some answers before that time. – user378947 Nov 4 '16 at 3:26
• @ClassicStyle Yes I know. Even with the one variable case there are lots of scenarios when Geometry comes to play. That actually is the reason I was dealing with Geometry in the second example. – user378947 Nov 4 '16 at 3:29
• – Martín-Blas Pérez Pinilla Jan 9 '17 at 17:46

$\newcommand{\Reals}{\mathbf{R}}$Theorem: Let $S^{2} \subset \Reals^{3}$ be the unit sphere centered at the origin, and $n = (0, 0, 1)$ the north pole. The stereographic projection mapping $\Pi:S^{2} \setminus\{n\} \to \Reals^{2}$ is conformal.

Proof: Fix a point $p \neq n$ arbitrarily, and let $v$ and $w$ be arbitrary tangent vectors at $p$. The plane containing $n$ and $p$ and parallel to $v$ cuts the sphere in a circle $V$ tangent to $v$. Similarly, the plane containing $n$ and $p$ and parallel to $w$ cuts the sphere in a circle $W$ tangent to $w$. The circles $V$ and $W$ form a digon with vertices $p$ and $n$. By symmetry, the angle $\theta$ they make at $p$ is equal to the angle they make at $n$. Let $v'$ and $w'$ be the tangent vectors at $n$ obtained from $v$ and $w$ by reflection across the plane of symmetry that exchanges $p$ and $n$.

Because the tangent plane to the sphere at the north pole is parallel to the $(x, y)$-plane, the image $\Pi_{*}v$ of $v$ under $\Pi$ is parallel to the vector obtained by translating $v'$ along the ray from $n$ through $p$ to the $(x, y)$-plane. Similarly, $\Pi_{*}w$ is parallel to the vector obtained by translating $w'$.

It follows at once that the angle between $\Pi_{*}v$ and $\Pi_{*}w$ is equal to the angle between $v'$ and $w'$, which is equal to the angle between $v$ and $w$.

Conformality of stereographic projection allows the holomorphic structure on the Riemann sphere to be visualized in three-dimensional Euclidean geometry, providing a basic link between complex analysis and differential geometry.

In coordinates, stereographic projection and its inverse are given by $$\Pi(x, y, z) = \frac{(x, y)}{1 - z},\qquad \Pi^{-1}(u, v) = \frac{(2u, 2v, u^{2} + v^{2} - 1)}{u^{2} + v^{2} + 1}.$$ It's possible to calculate the induced Riemannian metric on the plane: $$g = \frac{4(du^{2} + dv^{2})}{(u^{2} + v^{2} + 1)^{2}}.$$ Conformality is encoded in $g$ being a scalar function times the Euclidean metric. While this analytic argument has its own elegance, the geometric argument above is essentially obvious.

• Wow!! Thank you for your answer, I won't be able to forget that visual proof of the conformality of the stereographic projection. This is exactly the kind of answers I was looking for, and the Complex Analysis taste is delightful!! =) – user378947 Jan 5 '17 at 13:53

Here is an interesting result on the study of curves in $\mathbb R^3$. The proof is analytic, and will be paraphrased from Do Carmo's book "Differential Geometry of Curves and Surfaces," which is a standard reference. While much of differential geometry has to do with manifolds of higher dimension than just curves, there are many notions that I don't think I would be comfortable trying to distill down into a rather short argument or explanation of some theorem without really losing the beauty of the statement. There are many experts on differential geometry here, and I will let them have the pleasure of explaining some of these results. It's worth noting that much of differential geometry is real-analytic, rather than complex. Hopefully you still find this result interesting. First, some terminology:

Let $\alpha(s)$ be a curve parameterized by arc length. The derivative of $\alpha$ is a tangent vector $t$, and since $\alpha$ is parameterized by arc length, it is unit length. This means that $|\alpha''(s)|$ measures the rate of change of the angle which neighboring tangents make with the tangent at $s$. This value is denoted by $\kappa$ and is the 'curvature' of $\alpha$. If the curvature at a point of a curve is non-zero, then there is a unit vector $n$ in the direction of $\alpha''$ defined by $\alpha''(s) = \kappa (s) n(s)$, which is perpendicular to the tangent line (this follows from differentiating $\alpha' \cdot \alpha' = 1$). This vector $n$ is the normal vector, and the plane spanned by the tangent and normal is called the 'osculating plane'.

In geometry, we often times want coordinates adapted to our particular situation. In particular, when dealing with curves, it is desirable to have a set of linearly-independent vectors that live at each point of a curve, called a 'moving frame.' The third and final vector of this frame is found by taking the cross product of the tangent with the normal, to produce the 'binormal.' This frame is very special and is called the 'Frenet frame.'

There is a final notion we need - the derivative of the binormal. This is called the 'torsion.' Similar to curvature, this measure how much the curve pulls and twists away from the osculating plane. It is denoted by $\tau$

Now we are ready to state the claim:

THEOREM: Given differentiable functions on an interval $k(s) >0$ and $\tau(s)$, there exists a regular parameterized curve $\alpha$ such that $s$ is the arc length, $k(s)$ is the curvature of $\alpha$ and $\tau$ is the torsion of $\alpha$. Moreover, any other curve with the same conditions is isometric to $\alpha$, in the sense that there is some rigid motion (i.e. element of $O(3)$ and/or a translation) which maps the other curve onto $\alpha$.

We will sketch the proof, which uses the theory of ordinary differential equations.

From the definitions, it is not hard to show that the Frenet frame can be given by the following equations.

$$\frac{dt}{ds} = \kappa n$$

$$\frac{dn}{ds} = -\kappa t - \tau b$$

$$\frac{db}{dt} = \tau n$$

Now, really since these are all vector quantities, they are each actually a linear function of three variables - the coordinates of the vectors at each point. And so the Frenet frame gives a linear differential system on $I \times \mathbb R^n$.

This system might sound like a hot mess, but we do have an existence theorem that can handle it. In particular, since the system is linear, the usual local existence result from analysis can handle this system on the whole interval. It follows that given these functions, and the initial conditions we use to create the orthonormal frame at one point, we can solve the system of ODEs without any issues. However, we don't know that the solutions remain orthonormal, and this is key if we want our curve to be essentially defined by the Frenet frame we are trying to prescribe.

Now to check orthnormality, we will use the Frenet equations to check that the quantities $\langle t,n \rangle, \langle t, b \rangle, \langle n, b \rangle, \langle t ,t \rangle, \langle n, n \rangle, \langle b, b\rangle$ are all either $0$ or $1$, respectively.

From these expressions, we can simply differentiate each of these expressions, and then a relatively easy computation shows that the desired result is true.

Now we must actually obtain the curve. This is straight forward enough. Set $\alpha(s) = \int t(s) ds$. This ensures that $\alpha'(s) =t (s)$ and that $\alpha''(s) = \kappa n$, so that $\kappa (s)$ actually is the curvature at each point. That we have succeeded in prescribing the torsion is a little harder. We can see that $\alpha '''(s) = \kappa ' n - \kappa^2t - \kappa \tau b$ from the product rule and the definition. Then we have to use another (computational) fact, which is also not super hard to prove, that $\frac{-\langle \alpha' \times \alpha'', \alpha''' \rangle}{\kappa^2} = \tau$, and this shows that $\alpha$ is the claimed curve.

The uniqueness part is thankfully easier. The Frenet frames being linearly independent sets and in fact orthonormal means that we can just rotate one frame and translate so that origins coincide, and then use the uniqueness part of our existence/uniqueness theorem for ordinary differential equations, to show that after this change, the resulting solutions, and hence curves, coincide.

In hindsight, I feel like this argument might be rather terse. The basic theory of curves and surfaces has many computations in it, most of which are very simple, but sometimes a little long, but are always satisfying to work out, in my opinion. I encourage you to hunt down a copy of Do Carmo's book, and take a pass at it. The book has a good number of illustrations, but spares few if any details on mathematical rigor. The analyst in you will be pleased that proofs are not distilled into pictures, with the actual details left to the reader - you will find a complete telling of all the details of elementary differential geometry in this text.

• You missed an equal sign in "$\alpha(s)=\int t(s)ds$". I think I have not enough rep. to edit it. – user378947 Nov 5 '16 at 19:54
• @mathbeing I see what you have pointed out, and I seem to be having a hard time making that equal sign appear. When I click 'edit,' it is there in the LaTeX but I am not sure what is causing it not to appear.... – Alfred Yerger Nov 5 '16 at 19:57

The one I really like is the geometric intuition behind the generalized Stokes' Theorem:

$$\int_Md\omega=\int_{\partial M}\omega$$

It's none other than the familiar notions of the Kelvin-Stoke's Theorem, Divergence Theorem, Green's Theorem, and the Fundamental Theorem of Calculus and also a generalization of them. It unites all of them under a common framework and significantly generalizes them at the same time (from $\mathbb{R}^n$ to a (nice enough) arbitrary manifold). The geometric intuition for this theorem is the same as it is for multiple integration: You can solve the integral iteratively, sweeping out the region to be integrated over 1 variable at a time.

• +1. I always found Stokes' theorem very beautiful. Btw, I edited your answer to make it more pleasant to see. I hope you don't mind :) – Xam Jan 7 '17 at 5:10
• But, shouldn't the intuition for this theorem really be the "telescoping sum" phenomenon that occurs in the intuitive derivations of the fundamental theorem of calculus and also the divergence theorem and the classical Stokes' theorem? You chop up a surface or manifold into tiny pieces, compute the "flow" over each tiny piece, add up all the contributions from all the pieces and notice that beautiful cancellation occurs, leaving us with only boundary terms. – littleO Jan 9 '17 at 18:38
• That's an equivalent way of stating it, but I very much liked the intuition I got in Vector Calculus for multiple integration as imagining that you are sweeping out the region to be integrated over one variable at a time. – Justin Benfield Jan 10 '17 at 22:18

$3$ nice facts, which I personally like very much:

$(1)$ Any compact surface (without boundary) which is embedded in $\mathbb{R}^3$ has an elliptic point, i.e a point where the Gaussian curvature is positive.

This is intuitively plausible, once you understand that positive curvature amounts to "the space is folded (or collapsed) into itself" - think of a sphere for instance - when you start moving from the north pole along two great circles they diverge at first, but then gets closer after some point until they meet again at the south pole.

So, if $S$ is a compact surface we have two options:

• It has a non-empty boundary (like the closed unit disk- which is flat)

• It has no boundary. In that case, since it's compact, it cannot "run away" forever - hence at some point it must "turn onto iteslf" (fold back)- and at this point the curvature will be positive.

$(2)$ If a plane intersects a surface at one point only, than this plane is the tangent plane to the surface at the point of intersection. (easily visualisable).

$(3)$ The affect of curvature on geodesics. Geodesics in spaces of positive curvature diverge more slowly than straight lines in Euclidean space, while Geodesics in spaces of negative curvature diverge faster than straight lines in Euclidean space. The awesome thing is that this is an infinitesimal fact, i.e the divergence of geodesics emanating from a point is governed by the sectional curvature at that point.

You've probably noticed that many of the nice theorems in complex analysis relate the values of a function on the boundary of a domain with its values in the interior, and that this is reminiscent of classic theorems in calculus like Greene's, Gauss's, and Stokes'. In fact this is not a coincidence, and complex analysis has a nice presentation using the generalized Stokes' theorem.

Define the $\mathbb{C}$-linear operator $\frac{\partial}{\partial x}$ on the space of $C^\infty$ function $\mathbb{C}\to\mathbb{C}$ by

$$\frac{\partial}{\partial x}f = \frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}$$

and define $\frac{\partial}{\partial y}$ similarly. They satisfy the product rule, eg.

$$\frac{\partial}{\partial x}(fg) = \frac{\partial f}{\partial x}g + f\frac{\partial g}{\partial x}$$

Note the complex multiplication. Linear combinations of derivations are again derivations (eg $z\frac{\partial}{\partial x} + w\frac{\partial}{\partial y}$), and thus $\big\{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\big\}$ is a basis for a (complex) vector space of derivations, with dual basis $\{dx,dy\}$.

We can choose an alternative basis $\{dz, d\bar{z}\}$

$$dz=dx+idy$$

$$d\bar{z}=dx-idy$$

and dual basis $\big\{\frac{\partial}{\partial z}, \frac{\partial}{\partial \bar{z}}\big\}$ (sometimes called the Wirtinger derivatives)

$$\frac{\partial}{\partial z} = \frac{1}{2}\Bigg(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\Bigg)$$

$$\frac{\partial}{\partial \bar{z}} = \frac{1}{2}\Bigg(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\Bigg)$$

This notation is intentionally suggestive: some simple algebra together with the Cauchy-Riemann equations shows that if $f$ is holomorphic then $\frac{\partial f}{\partial z}=f'(z)$; and $\frac{\partial f}{\partial \bar{z}}=0$ if and only if $f$ is holomorphic.

For our purposes, a differential $1$-form $\omega$ is a choice of smooth functions $g(z)$ and $h(z)$

$$\omega=g(z)dz + h(z)d\bar{z}$$

Given a differential form with complex coefficients, we can rewrite it in terms of the $\{dx,dy\}$ basis and split it up into real and imaginary parts $\omega_r + i\omega_r$. If we want to integrate a differential form, we can then integrate the real and imaginary parts separately, just as for contour integrals. We can define the exterior derivative by $d\omega = d\omega_r + id\omega_r$. Using the $\{dz,d\bar{z}\}$ basis, $d\omega$ can be computed

$$d\omega = \Bigg(\frac{\partial h}{\partial z}-\frac{\partial g}{\partial \bar{z}}\Bigg)dz\wedge d\bar{z}$$

In particular, let us associate to a smooth function $f$ the form $\omega=f(z)dz$; if $f$ is holomorphic, $d\omega=-\frac{\partial f}{\partial \bar{z}}dz\wedge d\bar{z}=0$. Speaking loosely, holomorphic differential 1-forms are closed.

Recall that Stokes' theorem relates the integral of a form on the boundary to the integral over the interior

$$\int_{\partial M}\omega=\int_Md\omega$$

By applying this to $\omega_r$ and $\omega_i$ separately, Stokes' theorem is valid for differential forms with complex coefficients.

Putting all of this together, Stokes' theorem says that for compact submanifolds $M$ of $\mathbb{C}$ on which $f$ is holomorphic,

$$\int_{\partial M}f(z)dz=0$$

1. It's not hard to check that for a smooth curve $\gamma:[a,b]\to\mathbb{C}$, the usual definition of a contour integral $\int_a^b f(\gamma(t))\gamma '(t)dt$ is equivalent to $\int_\gamma f(z)dz$. So if $\gamma$ is the boundary of a disk $D$,

2. Choosing $M$ to be $\bar{D}$, we get Cauchy's theorem for a disk: $\int_{\partial \bar{D}}f(z)dz = 0$.

3. Choosing $f(z)=\frac{g(z)}{z-\xi}$ for $g$ holomorphic, $f$ is holomorphic except at a point. Thus we can apply Stokes' theorem by integrating over $D$ minus a small open disk around $\xi$ (a cookie with a hole - see the picture). There are two boundary components whose integrals are thus equal; the outer one gives $\int_{\partial \bar{D}}\frac{g(z)}{z-\xi}dz$ and the integral around the inner one goes to $2\pi ig(\xi)$ as $D'$ gets smaller. In other words, we get Cauchy's integral formula for a disk: $g(\xi) = \frac{1}{2\pi i}\int_{\partial \bar{D}}\frac{g(z)}{z-\xi}dz$. 1. The residue theorem follows by a similar argument: remove a small disk around each singularity and integrate over the remaining region (a cookie with multiple holes). The integral around the outside boundary component equals the sum of the rest of the integrals, each of which is equal to the corresponding residue (times $2\pi i$).

Finally, relaxing the requirement that $f$ be holomorphic, Cauchy's theorem and integral formula become respectively

$$\int_{\partial \bar{D}} f(z)dz = -\int_D \frac{\partial f}{\partial \bar{z}}dz\wedge d\bar{z}$$

$$f(\xi) = \frac{1}{2\pi i}\int_{\partial \bar{D}}\frac{f(z)}{z-\xi}dz - \frac{1}{2\pi i}\int_E\frac{\partial f}{\partial \bar{z}}\frac{1}{z-\xi}dz\wedge d\bar{z}$$

where $E$ is the cookie with a hole. Recalling that $\frac{\partial f}{\partial \bar{z}}=0$ if and only if $f$ is holomorphic, we see both theorems now have an error term that measures how much $f$ deviates from holomorphicity. If $f$ is holomorphic, we recover the original theorems. Thus I like to say, holomorphic functions have so many nice properties because they correspond to closed differential forms.

• Thank you for this great answer, and particularly for taking the care to make it suitable for my current knowledge of DG. Just in case you wonder, the reasons I didn't choose this answer for the bounty over Andrew's are that, coincidentally, I am already very familiar with Wirtinguer derivatives and the $\bar\partial$ operator (my master's thesis is about the Corona Theorem) as well as the complexified Stokes version and its consequences; and that (to me at least) it reads like a motivation of the study of DG almost solely because of the consequences of Stokes' theorem. – user378947 Jan 10 '17 at 19:17
• Ha yea in all honesty the only motivation I could find when I started Riemannian geometry proper was general relativity. All the indices and Christoffel symbols and connection stuff seemed so dry to me, and still does, even though I loved eg. Lee's Smooth Manifolds. However I recently started Walschap's book because I wanted to learn about characteristic classes, and I find the subject much more interesting the way he presents it. – juan arroyo Jan 10 '17 at 20:08
• For example, he starts by considering connections on arbitrary vector bundles, and defines a connection to be a distribution on the tangent bundle of the vector bundle, which right away kind of piqued my interest. His book has a very different feel from the other RG books I've read, like less analytic and more topological, and it's a bit more advanced and faster and has a different focus. I wouldn't recommend it to a neophyte but it might be good for someone like you who is already a bit familiar with the basic concepts of RG. – juan arroyo Jan 10 '17 at 20:09
• Dry is definitely the word to describe what they feel like to me, and that surely subjective appreciation is one of the things I want to change first. Thank you again, for your comments and for the recommendation! :) – user378947 Jan 10 '17 at 22:02

Shall avoid equations in an answer because most of my motivation to study DG is by observation of large deformations,in a non-linear formulation regime.

When I first read a two line statement of Egregium Theorem I was sure the technicalities of isometry in surface theory are hidden deep under several meters of concrete.

But it all changed when I realized that I can visually understand it by simply holding some of it in my hands for experimentation.. a plastic hemisphere can be squeezed together to realize all its spherical surface equivalents (spindle, cusped segment endings).. it was just flabbergasting.

Continuing on this experimentation trail.. if we take a larger thinner plastic hemisphere and squeeze on it further it is seen to result in an non axi-symmetrical surface shaped somewhat like a huge grain of wheat. There is no closed formula or code to find this shape as of now known to me. All this comes described under a single metric, the first fundamental form.

Seeing catenoid-helicoid isometry pictured in E.Kreyzig's book ( Differential and Riemannian geometry), I cut out one segment from a narrow waisted water bottle. A generator is cut and pulled apart to twist it .. is nothing but the manner in which a pseudosphere goes to Dini surface.

Experience and intuition comes first, mathematical analysis, explanations next. When you can rely in what you see or can geometrically sense you have got the necessary inspiration already, as seeing is believing to sweat it out later through odes and pdes.

Shapes of corals are visually shown to belong to hyperbolic geometry (before defining its shape) by Daina Taimina.. Basically adding more material on a woollen disk periphery in comparison to central parts. Rim annular area at boundary in a disk grows, like in crochet knitting/ embroidery to warp the new added surface. A more elementary model was hand made much earlier by Beltrami in cloth/ paper mache.. But there is no simple ode/pde formula, one cannot easily find a computational direction to be able to describe it. The beauty is visible, connections are teasing, however calculation would be rewarding.

Pressurized or not, soap bubble surfaces give a shape description by H =const, or H = 0 for DeLaunay minimal surface areas respectively.

It is amazing, that simple inversions in complex analysis can send elliptic geometry to flat and hyperbolic geometries. The bilinear (Moebius) mapping video shows how a grid and the surface it is written on gets trans/de/formed.

Gauss-Bonnet theorem connects 3D and 2D (line rotation and solid angles ) differential geometry and topology framed together in a single famous equation...

Even the history of hyperbolic geometry is fascinating. Some 18 centuries after Euclid, it got moving once again. A young gutsy army officer puts in effort even when his father who is a teacher pre-warned him about possible fruitless research outcome, but nevertheless went ahead to succeed (Bolyai) ..