The GaussBonnet Theorem (for orientable surfaces without boundary) states that for surface $M$, with Gaussian curvature at a point $K$, we have $$\int_M K\ dA=2\pi\chi(M).$$ Right now, this just says to me that the integral of something I don’t care about is equal to the ratio of a circumference to radius of a (Euclidean) circle times something I do care about. I get why normal curvature is interesting. I get Gaussian curvature is important in the Theorema Egregium and classification of surfaces of constant curvature. In the former, it helps classify surfaces up to isometry and show that developable surfaces are ruled. In the latter, it tells us that Riemannian manifolds with transitive isometry groups locally have a structure from a very small selection. But that’s just Gaussian curvature giving us other things. I don’t understand why I should like it for its own sake, and as such, why I should care about GaussBonnet. I get that it links differential geometry and topology, but maybe there are other ways to do this, and this one feels a little convoluted, and topology is already linked to differential manifolds, and thus Riemannian manifolds, through PoincareHopf.
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7Well, GaussBonnet itself gives you a reason to care about curvature: curvature is a local geometric quantity that can be used to compute a global topological invariant that you care about. – Eric Wofsey Jan 21 '22 at 02:38

8It's not clear to me why you want a reason to care about curvature "for its own sake". I would think that having *applications* of it would be *more* compelling as a reason to care about it. – Eric Wofsey Jan 21 '22 at 02:45

That’s reasonable. I guess I’m just more of a fan of relationships than computation. We like to feel that because mathematics is (modulo philosophical disputes) objective, mathematical taste is as well. But of course it’s not. – Thomas Anton Jan 21 '22 at 02:51

10Do you care about differential geometry at all? What kind of math do you care about? – Moishe Kohan Jan 21 '22 at 02:51

3PoincareHopf connect vector fields to topology, GaussBonnet connects metrics to topology. You should think of them as equivalent theorems. I think I can understand what you are trying to say because undergraduate courses in Surfaces usually make it sound like the most important Theorem in mathematics. I think there are also some historical reasons behind this. Note that when special cases for this theorem were proved (1840), topological spaces were not even defined so I bet that it got more and more popular until 1930's or so. Note that 90% of undergraduate material is from that era – Aitor Iribar Lopez Jan 21 '22 at 03:02

1It seems like you’re asking about why gaussian curvature is intrinsically important? Not every mathematical quantity may have this I think. But in this case, the gaussian curvature is a number which immediately describes the way the surface is curved around a point. Positively or negatively curved, and how sharply. It seems more interpretable as “shape” than the metric data. Is this not good enough? – Keshav Jan 21 '22 at 04:45

2I should add though that for me Gauss Bonnet is the first step in a general principle linking differential geometry to topology, as you mention. See “chern gauss bonnet” and “chern weil theory” for more direct extensions of this idea. These results are very useful and very clearly the same idea as gauss bonnet, so the intuition builds off of that. – Keshav Jan 21 '22 at 04:50

So you can rigorously explain why every twodimensional map projection is "wrong" in some way. – Dan Jan 22 '22 at 00:04

2Growing up science fiction exposes you to the idea of warped space. As a child you are left wondering what that means. How can you tell if the space you are living in is not flat  what would be different? Most numerical quantities you are exposed to do not tell you about this. If you were a dot, whose universe was a planar curve, you would not care about the derivative of the curve  it would not make anything different for you. When I went university and learnt that there was a numerical invariant that describes a space itself, not just how it sits in some other space, it was profound. – tkf Jan 22 '22 at 05:49

1While I don't want to convince you to care about something you just don't care about, e.g. curvature, I'll add that since you're a fan of relationships you should be a fan of this theorem. It describes a profound and useful relationship between the geometry of a surface (the left hand side) and the topology of the surface (the right hand side). – Lee Mosher Jan 22 '22 at 14:07
2 Answers
In retrospect, this post got quite long. Also, the level varies greatly  sorry! Feel free to ask any questions. I guess I am quite fond of this topic, even though I am not as knowledgeable on it as other people on here. Anyway, hopefully this is helpful to someone :)
In my opinion this theorem is one of the crown jewels of mathematics.
There are a number of ways to interpret or generalize this statement. For instance, since $2\pi \chi(M)$ is a purely topological quantity, this tells us that stretching or deforming $M$ through smooth isotopies will not change the integral of the Gaussian curvature. So, we could take a sphere $S^2(r)$ of radius $r$ in $\Bbb{R}^3$. Then the Gaussian curvature is everywhere $\frac{1}{r^2}$. The integral is $$ \int_{S^2(r)}\frac{1}{r^2}\mathrm{vol}=\frac{4\pi r^2}{r^2}=4\pi=2\pi\chi(S^2(r)). $$ Now, we can deform this sphere $S^2(r)$ as we wish through isotopies and we see that the GaussBonnet theorem gives us a "conservation law":
if we deform our surface $M$ in a volumepreserving manner such a way that the curvature in one region becomes greater, then the curvature in another region must become smaller.
Drawing a picture illustrates this idea:
This is a nice intuition for at least some of what the theorem is saying. However, there is another interpretation. Indeed, we have a classical topological invariant $\chi(M)$, and with some work we were able to find a "geometric quantity" to integrate to return $\chi(M)$, establishing a connection between a global geometric quantity and a topological quantity. This can be interpreted as $$ \int_M\frac{K}{2\pi}\mathrm{vol}=\chi(M). $$ Now, the natural objects of integration on manifolds are differential forms and indeed we have integrated a $2$form $\frac{K}{2\pi}\mathrm{vol}$ to return $\chi(M)$. This is a closed $2$form and defines a cohomology class $\omega\in H^2(M,\Bbb{R})$. So, we arrive at a natural question:
Given a compact $n$manifold $M$, can we find a cohomology class $\omega \in H^n(M,\Bbb{R})$ so that $\int_M\omega=\chi(M)$?
For $n=2$, the result is GaussBonnet. If $n$ is odd, it is an easy consequence of Poincaré Duality that $\chi(M)=0$, so the answer is yes but for stupid reasons: simply take $0\in H^n(M,\Bbb{R})$. For $n$ even, the question is more interesting. We next note that we had more structure in the case of the GaussBonnet theorem. Indeed, we were using the Riemannian structure coming from our manifold being in $\Bbb{R}^3$. So, we can equip $M$ with a Riemannian structure $(M,g)$. Now that we have done that, we would like to introduce a notion of curvature. Curvature is  in some sense  a second order phenomenon, so we need to introduce a notion of a second derivative. The method of doing this is to introduce an affine connection on $TM$. This is an operator $$\nabla:\mathfrak{X}(M)\times \mathfrak{X}(M)\to \mathfrak{X}(M)$$ written as $\nabla(X,Y)=\nabla_XY$. This has some natural properties:
 $\nabla$ is $\Bbb{R}$linear in both $X$ and $Y$
 For any $f\in C^\infty(M)$, $\nabla_{fX}Y=f\nabla_XY$.
 For any $f\in C^\infty(M)$, $\nabla_X(fY)=X(f)Y+f\nabla_XY$, which we call the Leibniz property.
The standard example of this is the directional derivative of a vector field $Y\in \mathfrak{X}(\Bbb{R}^3)$ with respect to another vector field $X$, written in multivariable calculus by $D_XY$. Anyway, associated to this $\nabla$ are a pair of tensors: $$ T(X,Y)=\nabla_X Y\nabla_YX [X,Y] $$ called the torsion and $$ R(X,Y)Z=\nabla_X\nabla_YZ\nabla_Y\nabla_X Z\nabla_{[X,Y]}Z $$ called the curvature. There is a theorem which says that on a Riemannian manifold $(M,g)$, there is a unique connection $\nabla$ satisfying $T(X,Y)\equiv 0$ and $\nabla_Xg(Y,Z)=g(\nabla_X Y,Z)+g(Y,\nabla_X Z)$. This connection is called the LeviCivita connection. The virtue of this is that there is a canonical connection associated to a Riemannian manifold, which will turn out to be quite important.
With this in hand, we can take our Riemannian manifold $(M,g)$ with its LC connection $\nabla$ and study the properties of its curvature. (By the way, you can recover the Gaussian curvature from this mysterious $R$ in the case of a surface, but I won't explain how.) If we take a trivialization of the tangent bundle by a local frame $(e_1,\ldots, e_n)$, we can describe the data of the connection using the equations: $$ \nabla_X e_j=\sum_i \omega^i_j(X)e_i $$ where $\omega_j^i$ are differential $1$forms. Furthermore, we can describe the data of the curvature using $$ R(X,Y)e_j=\sum_i \Omega^i_j(X,Y)e_i $$ for all $1\le i \le n$, where the $\Omega^i_j$ are $2$forms. In particular, we get matrices $\omega=(\omega^i_j)$ and $\Omega=(\Omega^i_j)$ describing the connection and the curvature, respectively, in a local trivialization. We would like to assemble from these guys a differential form that we can integrate to get back $\chi(M)$. The bridge is provided by the ChernWeil homomorphism, which allows us to construct cohomology classes from these matrices of differential forms. The statement is
Theorem (ChernWeil) Suppose $E$ is a rank $r$ vector bundle with connection $\nabla$. Suppose $P$ is a $\mathrm{GL}(r,\Bbb{R})$invariant polynomial on $\mathfrak{gl}(r,\Bbb{R})$ of degree $k$. Then, the $2k$form $P(\Omega)$ on $M$ is globally defined, closed, and $[P(\Omega)]\in H^{2k}(M,\Bbb{R})$ is independent of the connection.
The definition of a connection on $E$ is analogous to the above. A polynomial on $\mathfrak{gl}(r,\Bbb{R})$ means a polynomial that takes an $r\times r$ matrix $X=(x_j^i)$ as its input. Anyway, this theorem tells us that we can construct cohomology classes explicitly from the information of the curvature of a connection on a vector bundle on $M$. Moreover, the cohomology class is independent of the choice of connection. One way to interpret this is that we are getting a way to construct distinguished representatives of our cohomology classes on $M$. This is a motif that appears frequently in geometry. This lets us state our final theorem.
(ChernGaussBonnet) Let $(M,g)$ denote a compact Riemannian manifold of dimension $2n$. Then $$ \int_M \frac{1}{(2\pi)^n}\mathrm{Pf}(\Omega)=\chi(M). $$
Here, $\mathrm{Pf}$ denotes the Pfaffian, which is a certain polynomial defined on $\mathfrak{so}(2n,\Bbb{R})$ characterized (up to a sign) by $\mathrm{Pf}^2=\det$ as polynomial functions. I have swept a lot of details under the rug here  but the moral of the story is that the presence of a Riemannian metric gives a version of the above ChernWeil theorem for $SO(2n)$invariant polynomials on $\mathfrak{so}(2n,\Bbb{R})$.
Lastly, here is a high level explanation of what we are doing here. In topology, one associates to a vector bundle $E\to M$ characteristic classes, which are cohomology classes in $H^*(M)$ that are invariants of the bundle. I won't get too much into this, but there is a characteristic class of oriented real vector bundles called the Euler class, $e(E)$. It is suggestively named because in the case where $E=TM$, $e(E)\in H^n(M,\Bbb{R})$ (viewed for instance as a de Rham cohomology class) satisfies $$ \int_M e(E)=\chi(M). $$ So, we have really constructed an explicit form $\mathrm{Pf}(\Omega)$ on $M$ representing this cohomology class $e(E)$, and that is the content of the theorem. It turns out that characteristic classes of oriented bundles rank $r$ can be defined as cohomology classes of the classifying space $BSO(r)$. It can be proven that the cohomology of $BG$ (for $G$ a compact Lie group) is $H^*(BG,\Bbb{R})\cong \Bbb{R}[\mathfrak{g}]^G$, i.e. the invariant polynomials on the Lie algebra $\mathfrak{g}$. So, the ChernWeil homomorphism in its most general form expresses characteristic classes of a (say) oriented real bundle of rank $r$ in terms of polynomials of the curvature. So, in this setup the GaussBonnet theorem is a special case of the fact that the polynomial $\mathrm{Pf}$ computes the Euler class.
If this is of interest to you, I started learning about this in Tu's Differential Geometry: Connections, Curvature, and Characteristic classes, which is a clear and pleasant read for someone with a basic knowledge of manifolds and de Rham cohomology.
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While I may still not consider it a "crown jewel", after thorough digestion, this answer goes a long way. – Thomas Anton Jan 23 '22 at 06:06
I think one does not need to care about the notion of curvature or even Euler characteristics to find GaussBonnet at least surprising.
A manifold is just a topological space. But when you make it into a Riemannan manifold, you add some additional structure. And it is reasonable to ask "in how many different ways can I do this?". Visually, a Riemannian structure on a, say, topological sphere makes it into some more or less deformed potatoe shape. And this can get pretty wild up to a point where it does not embed into $\Bbb R^3$ anymore. So are there any limits at all?
Note the following two things about GaussBonnet.
 Whatever curvature is, it is something that is defined purely in terms of the Riemannian structure, and it is completely determined by it.
 When you integrate it over the manifold, it does not depend on the Riemannian structure.
Isn't that surprising? We just uncovered that Riemannian structures cannot be chosen completely arbitrary.
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