Gaussian and Mean Curvature

How curved is a surface at a point? Can a single number capture it? A sphere curves the same way in every direction. A saddle curves oppositely in perpendicular directions. A cylinder curves in one direction but not the other. Principal, Gaussian, and mean curvature are the precise numerical answers to "how curved." They are built directly from the Second Fundamental Form and the shape operator. This operator packages a surface's bending into a single symmetric linear map at each point. This matters because curvature is the bridge between two views of a surface. One is the extrinsic picture — how it sits and bends in $\mathbb{R}^3$ . The other is its intrinsic geometry, the geometry an inhabitant of the surface could measure without ever leaving it. Gauss's astonishing discovery, the Theorema Egregium, concerns one particular combination of the two principal curvatures: their product. This product is secretly intrinsic. It depends only on the first fundamental form, not on the embedding. Mean curvature, by contrast, is stubbornly extrinsic. It governs soap films, minimal surfaces, and the physics of surface tension. The big picture is this: at each point of a regular surface, the shape operator (also called the Weingarten map) is a symmetric linear operator on the tangent plane. Being symmetric, it is diagonalizable with real eigenvalues $k_1, k_2$ — the principal curvatures. It also has orthogonal eigenvectors — the principal directions. Gaussian curvature $K = k_1 k_2$ and mean curvature $H = \tfrac{k_1+k_2}{2}$ are the elementary symmetric functions of these eigenvalues. They correspond exactly to the determinant and trace (up to normalization) of the shape operator. This article develops principal curvatures rigorously from the second fundamental form. It proves the diagonalization theorem underlying their existence. It derives the standard coordinate formulas for $K$ and $H$ . It also works two fully computed examples — a torus and a hyperbolic paraboloid. These were chosen precisely because they are asymmetric enough to expose every sign and factor. It closes by previewing why $K$ , unlike $H$ , turns out to be an intrinsic invariant.

Key takeaways

At each point $p$ of a regular surface $S$ , the shape operator $dN_p$ is a symmetric linear map on the tangent plane $T_pS$ .
Its eigenvalues $k_1, k_2$ (with $k_1 \geq k_2$ by convention) are the principal curvatures; the eigenvectors give the principal directions.
Gaussian curvature $K = k_1 k_2 = \det(dN_p)$ ; mean curvature $H = \frac{k_1+k_2}{2} = \frac{1}{2}\operatorname{tr}(dN_p)$ .
In local coordinates with first fundamental form $(E,F,G)$ and second fundamental form $(e,f,g)$ ,

K = \frac{eg-f^2}{EG-F^2}, \qquad H = \frac{eG - 2fF + gE}{2(EG-F^2)}.

The sign of $K$ classifies points as elliptic ( $K>0$ ), hyperbolic ( $K<0$ ), parabolic ( $K=0$ , shape operator singular but nonzero), or planar (shape operator zero).
Key idea: Gaussian and mean curvature are, respectively, the determinant and (half the) trace of the shape operator, and this spectral viewpoint is what makes both quantities well-defined, coordinate-free, and computable.

Where this lives today

Gaussian curvature's intrinsic nature (foreshadowed here, proved in the Theorema Egregium) is the seed of Riemannian geometry, which underlies general relativity's description of spacetime curvature via the Riemann Curvature Tensor. Mean curvature governs minimal surfaces and soap films through the Young–Laplace equation, where pressure difference across a membrane is proportional to $H$ ; mean curvature flow, in which a surface evolves in the direction that most decreases area, is a fundamental tool in geometric analysis and has been used to attack the Poincaré conjecture and to model image processing and denoising. In computer graphics and geometry processing, discrete curvature estimates drive mesh smoothing, remeshing, and fairing algorithms. In cartography, Gauss's theorem explains rigorously why no flat map can perfectly preserve distances on a sphere ( $K_{\text{sphere}} = 1/r^2 \neq 0 = K_{\text{plane}}$ ), a fact underlying every choice of map projection. The classification of points by the sign of $K$ (elliptic/hyperbolic/parabolic) also connects directly to the Gauss-Bonnet Theorem, which integrates $K$ over a closed surface to recover a purely topological invariant, the Euler characteristic.

Ask the tutor: Why does the Theorema Egregium let us conclude $K$ is intrinsic while $H$ is not, given that both come from the same shape operator?

S \subset \mathbb{R}^3

Intuition

Picture standing at a point on a surface and looking in every possible tangent direction. In each direction $v$ , the surface bends toward or away from the normal $N$ with some normal curvature $k_n(v)$ — the curvature of the planar curve cut out by the plane through $N$ and $v$ . As $v$ sweeps around the unit circle in the tangent plane, $k_n(v)$ oscillates between a maximum and a minimum value. These extreme values are exactly the principal curvatures $k_1 \geq k_2$ , and they are achieved in perpendicular directions — a fact that is not obvious a priori but follows from the spectral theorem applied to the symmetric shape operator. Gaussian curvature multiplies these two extremes together. If both principal curvatures have the same sign (the surface bends the same way in every direction, like a sphere or the inside of a bowl), $K>0$ and the point is called elliptic. If they have opposite signs (the surface bends up in one direction and down in the perpendicular direction, like a saddle or the middle of a torus), $K<0$ and the point is hyperbolic. If one principal curvature vanishes, $K=0$ : this is a parabolic point (e.g. anywhere on a cylinder), unless both vanish, giving a planar point (e.g. a flat plane, or the flat point at the "waist" of certain surfaces of revolution). Mean curvature is the average of the two extremes. It measures the surface's tendency to feel a net normal force from surface tension — a physical soap film with no pressure difference across it satisfies $H \equiv 0$ everywhere, a minimal surface. Unlike $K$ , mean curvature genuinely depends on how the surface sits in space: reflecting a surface through a plane, or embedding two isometric surfaces differently in $\mathbb{R}^3$ , can change $H$ while $K$ , remarkably, cannot change at all — this is the content of the Theorema Egregium waiting in the wings.

Figure

Normal curvature

k_n(\theta)

traces an ellipse of extremes: principal curvatures at perpendicular directions, illustrated for

k_1=3,\ k_2=-1

S \subset \mathbb{R}^3

\mathbf{x}: U \to S

Remark on the sign convention. The proof above surfaces a genuine and universal bookkeeping subtlety: whether one defines principal curvatures via $dN_p$ or via $S_p = -dN_p$ is a matter of convention, and different textbooks (do Carmo uses $dN_p$ directly and defines $\mathrm{II}_p(v) = -\langle dN_p(v),v\rangle$ , which makes $K,H$ come out with the sign such that a sphere with inward normal has $K>0$ ; with outward normal the signs of $H$ flip but $K = \det(dN_p)$ is unaffected since $\det(-A) = \det(A)$ in dimension 2). The theorem's content — symmetry, real eigenvalues, orthogonal eigendirections, extremal property — is convention-independent; only the overall sign of $H$ (never of $K$ , in dimension 2) depends on the choice of normal.

K

dN

Worked examples

Example 1: A torus of revolution. Let $0 < r < R$ and parametrize the torus by

\mathbf{x}(u,v) = \big((R + r\cos u)\cos v,\ (R+r\cos u)\sin v,\ r\sin u\big), \qquad u,v \in [0,2\pi).

Here $u$ is the angle around the small (tube) circle and $v$ the angle around the big circle; we take $R=3$ , $r=1$ concretely for a fully numeric check at a specific asymmetric point. Compute the tangent vectors:

\mathbf{x}_u = (-r\sin u\cos v,\ -r\sin u \sin v,\ r\cos u), \qquad \mathbf{x}_v = (-(R+r\cos u)\sin v,\ (R+r\cos u)\cos v,\ 0).

First fundamental form:

E = \mathbf{x}_u\cdot\mathbf{x}_u = r^2, \qquad F = \mathbf{x}_u\cdot\mathbf{x}_v = 0, \qquad G = \mathbf{x}_v\cdot\mathbf{x}_v = (R+r\cos u)^2.

( $F=0$ is easily checked: the two vectors have a $z$ -component of $r\cos u$ and $0$ respectively contributing nothing, and the $x,y$ components form a dot product $r\sin u (R+r\cos u)(\cos v\sin v - \sin v\cos v) = 0$ .) The outward unit normal is

N(u,v) = (\cos u\cos v,\ \cos u \sin v,\ \sin u)

(this is unit length and one checks $N\cdot \mathbf{x}_u = N\cdot\mathbf{x}_v=0$ directly). Second fundamental form via $e = \mathbf{x}_{uu}\cdot N$ , etc. Compute

\mathbf{x}_{uu} = (-r\cos u\cos v,\ -r\cos u\sin v,\ -r\sin u), \quad e = \mathbf{x}_{uu}\cdot N = -r\cos^2 u - r\sin^2 u = -r.

\mathbf{x}_{uv} = (r\sin u\sin v,\ -r\sin u\cos v,\ 0), \quad f = \mathbf{x}_{uv}\cdot N = r\sin u\sin v\cos u\cos v - r\sin u\cos v\cos u\sin v = 0.

\mathbf{x}_{vv} = (-(R+r\cos u)\cos v,\ -(R+r\cos u)\sin v,\ 0), \quad g = \mathbf{x}_{vv}\cdot N = -(R+r\cos u)\cos u.

Now apply the formulas:

K = \frac{eg-f^2}{EG-F^2} = \frac{(-r)\big(-(R+r\cos u)\cos u\big) - 0}{r^2(R+r\cos u)^2} = \frac{r(R+r\cos u)\cos u}{r^2(R+r\cos u)^2} = \frac{\cos u}{r(R+r\cos u)}.

H = \frac{eG - 2fF + gE}{2(EG-F^2)} = \frac{(-r)(R+r\cos u)^2 + 0 + \big(-(R+r\cos u)\cos u\big)(r^2)}{2\, r^2(R+r\cos u)^2} = \frac{-(R+r\cos u)\big[r(R+r\cos u) + r^2\cos u\big]}{2r^2(R+r\cos u)^2}.

Simplify the bracket: $r(R+r\cos u) + r^2\cos u = rR + r^2\cos u + r^2 \cos u = rR + 2r^2\cos u = r(R + 2r\cos u)$ . So

H = \frac{-(R+r\cos u)\cdot r(R+2r\cos u)}{2r^2(R+r\cos u)^2} = -\frac{R+2r\cos u}{2r(R+r\cos u)}.

Numeric check at $u = \pi/3$ , $R=3$ , $r=1$ : $\cos u = 1/2$ , so $R + r\cos u = 3.5$ , and

K = \frac{1/2}{1\cdot 3.5} = \frac{1}{7} \approx 0.1429, \qquad H = -\frac{3 + 1}{2(3.5)} = -\frac{4}{7}\approx -0.5714.

Check via $k_1,k_2$ solving $\lambda^2 - 2H\lambda + K=0$ , i.e. $\lambda^2 + \frac{8}{7}\lambda + \frac17=0$ : discriminant $= \frac{64}{49} - \frac{4}{7} = \frac{64-28}{49} = \frac{36}{49}$ , root $\sqrt{\cdot} = 6/7$ , so $\lambda = \frac{-8/7 \pm 6/7}{2} = \frac{-2/7}{2}, \frac{-14/7}{2} = -\frac17, -1$ . Indeed $k_1k_2 = (-\tfrac17)(-1) = \tfrac17 = K$ ✓ and $\tfrac12(k_1+k_2) = \tfrac12(-\tfrac17-1) = \tfrac12\cdot(-\tfrac87) = -\tfrac47 = H$ ✓. At $u=\pi/3$ (outer-ish region since $\cos u>0$ ), the point is elliptic ( $K>0$ ), consistent with the torus's outer band curving like a sphere in all directions. At $u = 2\pi/3$ : $\cos u = -1/2$ , $R+r\cos u = 2.5$ , giving $K = \frac{-1/2}{2.5} = -0.2 < 0$ : a hyperbolic point on the inner band, as expected, since the tube curves one way while wrapping the other way around the hole.

Example 2: A hyperbolic paraboloid (non-symmetric scaling). Let $S$ be the graph $z = f(x,y) = \dfrac{x^2}{2} - \dfrac{y^2}{3}$ , parametrized by $\mathbf{x}(x,y) = (x,y,f(x,y))$ , and evaluate at the asymmetric point $(x,y) = (1,\ 2)$ . Compute derivatives: $f_x = x$ , $f_y = -\tfrac{2y}{3}$ , $f_{xx}=1$ , $f_{xy}=0$ , $f_{yy} = -\tfrac23$ . At $(1,2)$ : $f_x = 1$ , $f_y = -\tfrac43$ . Tangent vectors: $\mathbf{x}_x = (1,0,f_x) = (1,0,1)$ , $\mathbf{x}_y = (0,1,f_y) = (0,1,-\tfrac43)$ . First fundamental form:

E = 1 + f_x^2 = 1+1 = 2, \qquad F = f_xf_y = (1)(-\tfrac43) = -\tfrac43, \qquad G = 1+f_y^2 = 1 + \tfrac{16}{9} = \tfrac{25}{9}.

EG - F^2 = 2\cdot\tfrac{25}{9} - \tfrac{16}{9} = \tfrac{50}{9}-\tfrac{16}{9} = \tfrac{34}{9}.

For a graph $z=f(x,y)$ , the standard normal is $N = \dfrac{(-f_x,-f_y,1)}{\sqrt{1+f_x^2+f_y^2}}$ , and the second fundamental form coefficients are

e = \frac{f_{xx}}{\sqrt{1+f_x^2+f_y^2}}, \quad f_{\text{(coef)}} = \frac{f_{xy}}{\sqrt{1+f_x^2+f_y^2}}, \quad g = \frac{f_{yy}}{\sqrt{1+f_x^2+f_y^2}}.

(This is a standard consequence of $\mathbf{x}_{xx}=(0,0,f_{xx})$ etc. dotted with $N$ ; we take it as a known formula for graphs, derivable directly from the definitions exactly as in the coordinate corollary above.) At $(1,2)$ : $1+f_x^2+f_y^2 = 1 + 1 + \tfrac{16}{9} = \tfrac{34}{9}$ , so $\sqrt{1+f_x^2+f_y^2} = \sqrt{34}/3$ .

e = \frac{1}{\sqrt{34}/3} = \frac{3}{\sqrt{34}}, \qquad f_{\text{coef}} = 0, \qquad g = \frac{-2/3}{\sqrt{34}/3} = \frac{-2}{\sqrt{34}}.

Now:

K = \frac{eg - f_{\text{coef}}^2}{EG-F^2} = \frac{\left(\tfrac{3}{\sqrt{34}}\right)\left(\tfrac{-2}{\sqrt{34}}\right) - 0}{34/9} = \frac{-6/34}{34/9} = \frac{-6}{34}\cdot\frac{9}{34} = \frac{-54}{1156} = -\frac{27}{578}.

Let's double check numerically: $-6/34 = -0.176470588$ ; dividing by $34/9 = 3.777\overline{7}$ gives $-0.176470588/3.77778 = -0.046713...$ . And $-27/578 = -0.046713...$ ✓. So $K = -27/578 \approx -0.0467$ , confirming a hyperbolic point, exactly as expected for a saddle surface. For $H$ :

H = \frac{eG - 2f_{\text{coef}}F + gE}{2(EG-F^2)} = \frac{\left(\tfrac{3}{\sqrt{34}}\right)\left(\tfrac{25}{9}\right) - 0 + \left(\tfrac{-2}{\sqrt{34}}\right)(2)}{2\cdot 34/9}.

Numerator: $\dfrac{3}{\sqrt{34}}\cdot\dfrac{25}{9} = \dfrac{75}{9\sqrt{34}} = \dfrac{25}{3\sqrt{34}}$ , and $\dfrac{-2}{\sqrt{34}}\cdot 2 = \dfrac{-4}{\sqrt{34}} = \dfrac{-12}{3\sqrt{34}}$ . Sum: $\dfrac{25-12}{3\sqrt{34}} = \dfrac{13}{3\sqrt{34}}$ . Denominator: $2\cdot \tfrac{34}{9} = \tfrac{68}{9}$ .

H = \frac{13/(3\sqrt{34})}{68/9} = \frac{13}{3\sqrt{34}}\cdot\frac{9}{68} = \frac{117}{204\sqrt{34}} = \frac{39}{68\sqrt{34}}.

Numerically: $\sqrt{34}\approx 5.83095$ , $68\sqrt{34}\approx 396.505$ , so $H \approx 39/396.505 \approx 0.09836$ .

Consistency check via $k_1 + k_2 = 2H \approx 0.19672$ and $k_1k_2 = K \approx -0.046713$ : solving $\lambda^2 - 0.19672\lambda - 0.046713=0$ gives discriminant $0.19672^2 + 4(0.046713) = 0.038699+0.186851=0.225551$ , $\sqrt{\cdot}\approx 0.47493$ , roots $\approx \frac{0.19672\pm 0.47493}{2}$ , i.e. $k_1\approx 0.33582$ , $k_2\approx -0.13910$ . Check product: $0.33582\times(-0.13910)\approx -0.046715$ ✓ (matches $K$ to rounding), and sum $\approx 0.19672 = 2H$ ✓. Opposite signs confirm the hyperbolic (saddle) point, matching the geometric picture of $z=x^2/2 - y^2/3$ curving upward in $x$ and downward in $y$ .

A

Scope, edge cases & common traps

Situation	What happens	Why
Umbilic point ( $k_1=k_2$ )	Every direction is principal; $K=k_1^2\ge 0$ actually always $\geq 0$ at umbilics	Shape operator is a scalar multiple of the identity there
Planar point ( $k_1=k_2=0$ )	$K=H=0$ , shape operator is the zero map	Second fundamental form vanishes identically at that point
Parabolic point ( $K=0$ , not planar)	Exactly one $k_i=0$ ; occurs on developable surfaces like cylinders/cones	Shape operator singular but nonzero
Choice of unit normal $N$ vs. $-N$	$K$ unchanged; $H$ changes sign	$\det(-A)=\det A$ in dimension 2, but $\operatorname{tr}(-A) = -\operatorname{tr}A$
Non-orientable surface (e.g. Möbius band)	No globally consistent $N$ , so $H$ is only locally/sign-ambiguously defined	$K$ , being intrinsic, is still globally well-defined (Theorema Egregium) but $H$ needs an orientation choice
Reparametrization of the surface	$K, H, k_1, k_2$ all unchanged	They are properties of the point $p\in S$ and the map $dN_p$ , not of coordinates

Common trapThe classic error: confusing "

K>0

" with "convex" or assuming

K

determines

H

A very common mistake is to assume that Gaussian curvature alone determines mean curvature, or vice versa — but $(K,H)$ are two independent real numbers subject only to the constraint $H^2 \geq K$ (since $H^2 - K = \left(\tfrac{k_1-k_2}{2}\right)^2 \geq 0$ , with equality exactly at umbilic points). A second classic error is asserting that $K>0$ at a point implies the surface is globally convex near that point in some strong topological sense — $K>0$ only says the surface is locally on one side of its tangent plane near $p$ (locally convex), not that the whole surface is a convex body. Finally, students often forget that flipping the choice of unit normal flips the sign of $H$ (and of each $k_i$ ) while leaving $K$ untouched; any formula or claim about $H$ that doesn't specify which normal was chosen is incomplete.

The mathematician behind it

Origins · 1 minCarl Friedrich Gauss and how this was discovered

Carl Friedrich Gauss

1777–1855 · Göttingen, geodetic surveying and the founding of intrinsic differential geometry

Gauss developed the theory of curved surfaces largely while directing the geodetic survey of the Kingdom of Hanover in the 1820s, a practical project measuring the Earth's true shape via triangulation over hilly terrain. This hands-on engagement with real, non-flat surfaces led him to his 1827 treatise Disquisitiones generales circa superficies curvas, in which he introduced the curvature that now bears his name and proved — to his own evident astonishment, he called it "egregium" (outstanding, remarkable) — that it is an intrinsic quantity, measurable by an inhabitant of the surface using only lengths and angles, with no reference to any ambient space.

History

- 1827 — Gauss publishes Disquisitiones generales circa superficies curvas, defining Gaussian curvature via the spherical (Gauss) map and proving the Theorema Egregium.

- 1850s — Bonnet and others develop the second fundamental form and mean curvature systematically, connecting

H

to minimal surfaces and Plateau's soap-film problem.

- 1868 — Riemann's posthumously published habilitation lecture generalizes curvature to arbitrary dimensions, founding Riemannian geometry from Gauss's intrinsic viewpoint.

- 1900s–1930s — Global consequences (Gauss–Bonnet, rigidity theorems) are proved connecting local curvature to global topology.

- 20th–21st century — Mean curvature flow (Huisken and others) and Gaussian curvature bounds become central tools in geometric analysis, general relativity, and computer graphics.

Why are

k_1

and

k_2

always real numbers, never complex?

Because the shape operator $dN_p$ is symmetric (self-adjoint) with respect to the inner product on the tangent plane, a fact proved directly from equality of mixed partial derivatives of a $C^2$ parametrization. Any symmetric real matrix has characteristic polynomial with discriminant equal to a sum of squares — concretely $(a-c)^2+4b^2$ for a $2\times 2$ symmetric matrix $\begin{pmatrix}a&b\\b&c\end{pmatrix}$ — which is automatically nonnegative. Nonnegative discriminant guarantees two real roots (possibly equal). This is a small instance of the general spectral theorem for symmetric operators, and it is precisely why "curvature" can be reported as ordinary real numbers rather than complex or undefined quantities.

Is Gaussian curvature the same thing as "curvature" in the sense of a plane curve's curvature?

No, though they are related by Euler's formula. A plane curve has a single real-valued curvature at each point, describing how fast its tangent turns. A surface point instead has infinitely many normal curvatures, one for each tangent direction, ranging continuously between $k_2$ and $k_1$ as the direction sweeps around. Gaussian curvature $K=k_1k_2$ is a single number built from the two extremes of this whole family, not a curvature of any one curve on the surface (although it does equal the normal curvature times something specific only at umbilic points, where all normal curvatures coincide).

Why does the sign of

K

classify the local shape of the surface?

Because $K=k_1k_2$ is the product of the two extreme normal curvatures, and a product of two reals is positive iff they share a sign. If $k_1,k_2$ have the same sign, the surface bends toward the same side of its tangent plane in every direction (elliptic, locally convex, like a sphere). If they have opposite signs, it bends to opposite sides in perpendicular directions (hyperbolic, saddle-shaped). If one vanishes, the surface is "flat" in exactly one direction (parabolic), and if both vanish it's flat to second order in every direction (planar). This local classification is purely algebraic, following straight from the eigenvalue picture.

Does mean curvature depend on the surface's intrinsic (metric) geometry, or only on how it's embedded?

Mean curvature genuinely depends on the embedding. Two isometric surfaces (identical intrinsic geometry, same distances and angles) can have different mean curvatures if embedded differently in $\mathbb{R}^3$ — for instance, a flat sheet of paper ( $H=0$ when lying flat) rolled into a cylinder acquires $H = 1/(2r) \neq 0$ , even though rolling it preserves all intrinsic distances. Gaussian curvature, by contrast, is invariant here ( $K=0$ for both the flat sheet and the cylinder), which is the essential content of the Theorema Egregium: $K$ is intrinsic, $H$ is not.

What is a minimal surface, and how does it relate to

H

A minimal surface is one with mean curvature identically zero, $H\equiv 0$ , at every point. The name comes from the calculus of variations: such surfaces are critical points (often, but not always, local area minimizers) of the surface area functional among surfaces spanning a fixed boundary curve, which is exactly the mathematical model of a soap film spanning a wire frame with no pressure difference across it. Examples include the plane, the catenoid, and the helicoid; none of these need have $K\equiv 0$ (indeed the catenoid has $K<0$ everywhere except along its central circle), showing $H=0$ is compatible with a wide range of Gaussian curvature behavior.

How do principal curvatures relate to the eigenvalues of the "shape operator matrix"

I^{-1}\mathrm{II}

used in computations?

When computing in a coordinate parametrization, one often forms the matrix $I^{-1}\mathrm{II}$ (first fundamental form inverse times second fundamental form) in the coordinate basis $\{\mathbf{x}_u,\mathbf{x}_v\}$ . This matrix represents the shape operator in a (generally non-orthonormal) basis, so it is not symmetric as a literal array of numbers unless $E=G$ and $F=0$ . However its eigenvalues are still exactly $k_1,k_2$ (up to the sign convention chosen), because eigenvalues, determinant, and trace of a linear operator do not depend on which basis is used to represent it as a matrix — only genuine matrix symmetry (which requires an orthonormal basis) is basis-dependent, not the operator's spectrum.

Can Gaussian curvature be negative everywhere on a closed (compact, boundaryless) surface?

No. This is a consequence of the Gauss–Bonnet theorem: for a closed orientable surface of genus $g$ , $\int_S K\,dA = 2\pi(2-2g)$ , which is strictly positive for a sphere ( $g=0$ ) and can be negative for higher genus, but on any compact surface without boundary there must exist at least one point with $K\geq 0$ — indeed a point of maximal distance from some origin (or the point of maximum "outward bulge") is elliptic, by a direct argument comparing the surface locally to a supporting sphere. So $K<0$ everywhere is impossible on a closed surface, though it is entirely possible on complete non-compact surfaces like the hyperbolic plane model or a catenoid.