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Rolling a sheet into a circular cylinder bends it without changing its internal lengths and angles. Fitting that same sheet smoothly onto a sphere requires distortion of its first fundamental form. Gaussian curvature distinguishes these situations: it measures the intrinsic curvature of a smooth surface, even though its simplest definition uses how the surface bends in Euclidean space.

The Gaussian curvature at a point is the determinant of the shape operator, or the product of the two principal curvatures:Here $I=Ed{u}^2+2Fdudv+Gd{v}^2$ is the first fundamental form and $II=ed{u}^2+2fdudv+gd{v}^2$ is the second fundamental form, with $S=-dN$. Regularity makes $EG-F^2>0$. Reversing the unit normal negates $e,f,g$ and both principal curvatures, leaving Gaussian curvature unchanged. ^[ref]

The sign describes bending at the point: positive Gaussian curvature means the principal curvatures have the same nonzero sign, as at an elliptic point; negative Gaussian curvature means opposite signs, as on a saddle; zero Gaussian curvature means at least one principal curvature vanishes. Zero Gaussian curvature at a single point does not imply that a neighborhood is a plane. ^[ref]

Four worked models make the signs concrete:

The patches use a common length unit, with $R=1$ for the sphere and circular cylinder. The marked red and blue curves have principal directions at the black point; the arrow is the chosen unit normal. For physical lengths, principal curvature and mean curvature have units of inverse length, while Gaussian curvature has units of inverse squared length. Rescaling a surface by $\lambda >0$ divides its principal curvatures by $\lambda$ and its Gaussian curvature by ${\lambda }^2$. Writing the saddle as $h=(x^2-y^2)/(2L)$ restores an explicit length scale: $K(0)=-1/L^2$.

The Theorema Egregium says that Gaussian curvature depends only on the first fundamental form and its first two coordinate derivatives. A local isometry therefore preserves Gaussian curvature, even if the shape operator and the principal curvatures change. ^[ref]

Here is the outline, rather than a full proof. The Gauss formula splits second derivatives of a parametric surface into tangential Christoffel symbols and the second fundamental form. The Christoffel symbols are determined by the first fundamental form and its first derivatives. Commuting the next partial derivatives gives the Gauss equation, which identifies $\det S$ with the intrinsic sectional curvature computed from the Riemann curvature tensor. Thus the apparently extrinsic product $k_1{k}_2$ is determined by the Riemannian metric. ^[ref]

One can see the distinction directly by rolling a plane strip with coordinates $(u,v)$ into the circular cylinder $X(u,v)=(R\cos (u/R),R\sin (u/R),v)$. Both have $I=d{u}^2+d{v}^2$, so the map is a local isometry, and is one-to-one on an open strip of width less than $2\pi R$. Extending $u$ over all real numbers wraps around periodically; it does not give a global isometry between the entire plane and the entire circular cylinder. A nonempty open patch of a sphere cannot be flattened by a local isometry into a plane, because its Gaussian curvature $1/R^2$ cannot become zero. The obstruction concerns preservation of the first fundamental form, rather than a particular drawing or choice of seam.

Adjoin ${\mathbb{P}}^1(\mathbb{Q})$ to the complex upper half-plane. Ordinary points retain their usual neighborhoods. At a rational boundary point $a=\sigma \infty$, a neighborhood basis consists of $a$ together with $\sigma \{z\in \mathbb{H}:\mathrm{Im}z>Y\}$, with $Y$ arbitrarily large. Different choices of $\sigma \in S{L}_2(\mathbb{Z})$ change it by a translation and sign, so give the same neighborhood system. Give $X(1)$ the quotient topology. The modular group acts properly discontinuously on $\mathbb{H}$; its modular cusp neighborhoods descend to punctured neighborhoods with their modular cusp point filled in. This quotient is Hausdorff: interior orbits can be separated by proper discontinuity, and a sufficiently high modular cusp neighborhood avoids any prescribed compact interior neighborhood.

Away from fixed points of the effective group $PS{L}_2(\mathbb{Z})$, a small disk is disjoint from its nonidentity translates, so its coordinate gives a holomorphic quotient coordinate. At an elliptic fixed point ${\tau }_0$, useIts effective stabilizer rotates $u$ by an $e$th root of unity, where $e=2$ at $i$ and $e=3$ at $\rho$. The quotient coordinate is $w=u^e$, not $u$. All rational boundary points lie in one modular cusp class. At infinity the stabilizer acts by $\tau \mapsto \tau +1$, andis a punctured-disk coordinate; adjoining $q=0$ fills in the modular cusp. These charts and their holomorphic transition maps define the complex structure of the compactified modular curve.

For compactness use the standard regionIts part with $\mathrm{Im}\tau \le Y$, $Y>1$, is compact: the imaginary part is at least $\sqrt{3}/2$. The quotient of its upper tail, with its two vertical sides identified and infinity added, is the closed disk $|q|\le e^{-2\pi Y}$. These two compact images cover $X(1)$, including the boundary-edge identifications. Thus $X(1)$ is a compact Riemann surface.

The $j$ function also extends over the modular cusp with one simple pole, since $j=q^{-1}+O(1)$. Its local expressions at elliptic points descend to holomorphic functions of $u^e$. By Question 1 it maps the remaining quotient bijectively onto $\mathbb{C}$, so its extension is a degree-one holomorphic map $X(1)\to {\mathbb{P}}^1$. A degree-one map of compact Riemann surfaces is biholomorphic: any local ramification would already give degree at least two. Therefore $X(1)$ is the Riemann sphere; this will also identify the base genus in Question 5.

A weight-$2k$ modular form gives an invariant tensor differentialbecause $d(\gamma \tau )=(c\tau +d{)}^{-2}d\tau$ cancels its automorphy factor. Assume first $k\ge 0$. At an ordinary quotient point this differential is holomorphic. At an elliptic point of order $e$, use $u$ and $w=u^e$ from part (a), and write the pullback as $A(u)(du{)}^k$, with $A$ holomorphic. Its invariance says $A(\zeta u){\zeta }^k=A(u)$, so a nonzero term $a_r{u}^r$ has $r+k\equiv 0(\mathrm{mod}e)$. Sincethe exponent in the descended coefficient is the integer $(r-k(e-1))/e$. With $r\ge 0$, it is at least $-\lfloor k(e-1)/e\rfloor$. At the modular cusp, $d\tau =dq/(2\pi iq)$ and $f$ has a power series with nonnegative exponents, so ${\omega }_f$ has a pole of order at most $k$.

Consequently the linear map $f\mapsto {\omega }_f$ injects $M_{2k}$ into the meromorphic sections of $K^{\otimes k}$ with poles bounded byThis section space is $H^0(X(1),K^{\otimes k}(D))$. It is finite-dimensional on a compact Riemann surface by the Riemann-Roch theorem. Concretely choose one nonzero meromorphic section of the line bundle; dividing by it identifies these sections with a Riemann-Roch space with a fixed finite divisor bound. This proves the requested finite-dimensionality by compact-surface theory.

In fact the injection is an isomorphism. If a descended coefficient has order at least $-\lfloor k(e-1)/e\rfloor$, its pulled-back exponent $ed+k(e-1)$ is nonnegative; the modular cusp pole bound likewise gives a holomorphic Fourier expansion of a modular form. These pullbacks therefore recover modular forms. Since $X(1)$ is the Riemann sphere, $\mathrm{deg}K=-2$, and a line bundle of degree $d$ on the sphere has $\max (0,d+1)$ sections. HenceFor negative $k$, the valence formula excludes a nonzero holomorphic modular form because all its orders would be nonnegative but their sum would be $k/6<0$; the space is zero. The mechanism behind finite-dimensionality is modular forms as meromorphic differentials on X(1), with explicit elliptic and modular cusp pole bounds.

Reduction modulo $N$ has kernel $\Gamma (N)$ and is onto $S{L}_2(\mathbb{Z}/N\mathbb{Z})$. To see surjectivity, use the Chinese remainder theorem to reduce to $\mathbb{Z}/p^a\mathbb{Z}$. In that local ring a unimodular column has a unit entry. Elementary row operations move a unit to the upper position and clear the lower one; diagonal matrices $\mathrm{diag}(u,u^{-1})$ are products of elementary matrices as well. Explicitly, with $E_{12},E_{21}$ denoting elementary shears,and multiplying this by the same matrix with $u=-1$ gives $\mathrm{diag}(u,u^{-1})$. Thus elementary matrices generate the special linear group over each local ring. Chinese remaindering their parameters gives generation over the product ring, and every such parameter lifts to an integer. This proves the reduction is onto.

Over ${\mathbb{F}}_p$, there are $p^2-1$ choices for a nonzero first column and $p$ choices for a second column with determinant one, so $|S{L}_2({\mathbb{F}}_p)|=p(p^2-1)$. Each reduction from modulus $p^{a+1}$ to $p^a$ has kernel represented by $I+p^{a}B$, with $B$ modulo $p$ and $\mathrm{tr}B=0$, of size $p^3$. Hence, for $N>1$,For $N=1$ the index is one. Since $-I\in \Gamma (N)$ exactly when $N=1,2$,

For $N\ge 2$ there are no nontrivial effective elliptic stabilizers. Indeed $A=I+NB\in \Gamma (N)$ with determinant one has $\mathrm{tr}B=-N\det B$, and thereforeA noncentral elliptic element of $S{L}_2(\mathbb{Z})$ has trace $0$ or $\pm 1$, none congruent to $2$ modulo $N^2$ for $N\ge 2$. The central element $-I$ at level two acts trivially and is not an elliptic point of the effective quotient. Thus ${\nu }_2={\nu }_3=0$ for $N\ge 2$, while at level one both numbers are one.

The subgroup $\Gamma (N)$ is normal, so all its modular cusps have the same effective width as infinity. That width is $N$: the shear $T^h$ belongs to $\pm \Gamma (N)$ exactly when $N\mid h$; for $N\ge 3$ its diagonal entries exclude the minus sign, and for $N=1,2$ that sign changes nothing. Since modular cusp widths sum to $\mu$,For $N\ge 3$ the resulting principal congruence modular curve genus isFor clarity, the small levels, including all exceptional terms, areFor $N>6$, every factor in the genus correction is positive, so $g>1$. Therefore the genus-zero levels are exactly $N=1,2,3,4,5$.

Suppose $S$ lies in the closed ball of radius $R$ centred at $c$. By compactness, the continuous function $q\mapsto |q-c|$ attains its maximum $r\le R$ at some $p\in S$. The sphere of radius $r$ about $c$ is a supporting sphere tangent to $S$ at $p$.

Put $N=(p-c)/r$. For any unit vector $v\in T_{p}S$, choose a surface curve $\alpha$ with $\alpha (0)=p$ and ${\alpha }^{\prime }(0)=v$. Since $|\alpha (t)-c{|}^2$ has a local maximum at zero,The scalar $⟨{\alpha }^{\prime \prime }(0),N⟩$ is the normal curvature in direction $v$, so every normal curvature is at most $-1/r$ with this choice of unit normal. Applying this to principal directions shows that both principal curvatures satisfy $k_i\le -1/r$. ThereforeThis proves both assertions, including the existence of an elliptic point on every compact regular surface.