[proofplan] The Theorema Egregium asserts that the Gaussian curvature $K$, defined extrinsically as $\frac{LN - M^2}{EG - F^2}$, is in fact computable from the first fundamental form $g = (g_{ij})$ alone. The strategy is to give an explicit formula — the **Brioschi formula** — that expresses $K$ as a rational function of $g_{ij}$, $\partial g_{ij}$, and $\partial^2 g_{ij}$. We do this in two steps. First, we re-express the Gaussian curvature in terms of the Riemann curvature tensor of the induced metric: the Gauss equation for a surface in $\mathbb{R}^3$ reads $R_{1212} = LN - M^2$, where $R_{ijkl}$ is built entirely from $g_{ij}$ via the Levi-Civita connection. Hence $K = R_{1212} / \det g$. Second, we expand $R_{1212}$ using the standard formula for the Riemann tensor in terms of Christoffel symbols and obtain Brioschi's closed-form expression in $g_{ij}$ and their partials. Isometry invariance is then immediate: an isometry $\Phi: (\Sigma_1, g_1) \to (\Sigma_2, g_2)$ pulls back $g_2$ to $g_1$, hence pulls back the Brioschi expression for $K_2$ to the Brioschi expression for $K_1$, so $K_2 \circ \Phi = K_1$. [/proofplan] [step:Set up the Gauss formula relating the surface's connection to the ambient flat connection] Let $\Sigma \subseteq \mathbb{R}^3$ be a smooth surface and let $X: U \to \Sigma$ be a local parametrisation, where $U \subseteq \mathbb{R}^2$ is open with coordinates $(u_1, u_2) = (u, v)$. Write $\partial_i := \partial/\partial u_i$ for $i = 1, 2$, and let $X_i := dX(\partial_i) \in \mathbb{R}^3$ denote the coordinate tangent vectors. The first fundamental form has components \begin{align*} g_{ij} &= \langle X_i, X_j \rangle_{\mathbb{R}^3}, \qquad i, j \in \{1, 2\}, \end{align*} so that $g_{11} = E$, $g_{12} = g_{21} = F$, $g_{22} = G$, and $\det g = EG - F^2 > 0$. Let $N: U \to \mathbb{R}^3$ be a smooth choice of unit normal to $\Sigma$, and let $h_{ij}$ be the components of the second fundamental form, \begin{align*} h_{ij} &= \langle \partial_i \partial_j X, N \rangle_{\mathbb{R}^3}, \end{align*} so that $h_{11} = L$, $h_{12} = h_{21} = M$, $h_{22} = N_{\mathrm{II}}$ (we write $N_{\mathrm{II}}$ to avoid clash with the unit normal). Let $\overline{\nabla}$ denote the Levi-Civita connection of the Euclidean metric on $\mathbb{R}^3$ (which is just componentwise differentiation), and let $\nabla$ denote the Levi-Civita connection of $g$ on $\Sigma$. The **Gauss formula** decomposes the ambient derivative of a tangent vector field into tangent and normal parts: \begin{align*} \overline{\nabla}_{X_i} X_j &= \nabla_{X_i} X_j + h_{ij}\, N. \end{align*} Both this decomposition and the identification $\nabla_{X_i} X_j = \sum_{k=1}^2 \Gamma_{ij}^k\, X_k$ (where $\Gamma_{ij}^k$ are the [Christoffel symbols](/page/Christoffel%20Symbol) of $\nabla$) hold because $\nabla$ is the unique torsion-free, metric-compatible connection on $(\Sigma, g)$, and the tangent component of $\overline{\nabla}_{X_i} X_j$ is precisely such a connection on $(\Sigma, g)$. [/step] [step:Derive the Gauss equation $R_{1212} = LN - M^2$ via curvature of the ambient flat connection] We compute $R(X_1, X_2) X_2$ using the chapter's sign convention \begin{align*} R(X_1, X_2) &= \nabla_{[X_1, X_2]} - [\nabla_{X_1}, \nabla_{X_2}]. \end{align*} Since $X_1$ and $X_2$ are coordinate vector fields, $[X_1, X_2] = 0$. Hence \begin{align*} R(X_1, X_2) X_2 &= -\bigl(\nabla_{X_1} \nabla_{X_2} X_2 - \nabla_{X_2} \nabla_{X_1} X_2\bigr). \end{align*} We now compute $R(X_1, X_2) X_2$ a second way, using the Gauss formula and the flatness of the ambient connection $\overline{\nabla}$. The Riemann tensor of $\overline{\nabla}$ on $\mathbb{R}^3$ is identically zero, and $[X_1, X_2] = 0$, so \begin{align*} 0 &= \overline{\nabla}_{X_1} \overline{\nabla}_{X_2} X_2 - \overline{\nabla}_{X_2} \overline{\nabla}_{X_1} X_2. \end{align*} We expand each term using Gauss. For the first: \begin{align*} \overline{\nabla}_{X_2} X_2 &= \nabla_{X_2} X_2 + h_{22}\, N, \\ \overline{\nabla}_{X_1}\bigl(\nabla_{X_2} X_2 + h_{22}\, N\bigr) &= \overline{\nabla}_{X_1} \nabla_{X_2} X_2 + (\partial_1 h_{22})\, N + h_{22}\, \overline{\nabla}_{X_1} N. \end{align*} Using Gauss again on $\overline{\nabla}_{X_1} \nabla_{X_2} X_2$ (with $\nabla_{X_2} X_2 = \sum_k \Gamma_{22}^k X_k$ a tangent field, so its ambient derivative decomposes as tangent plus $h_{ij} \cdot$ coefficient $\cdot N$): \begin{align*} \overline{\nabla}_{X_1} \nabla_{X_2} X_2 &= \nabla_{X_1} \nabla_{X_2} X_2 + \sum_{k=1}^2 \Gamma_{22}^k\, h_{1k}\, N. \end{align*} The **Weingarten map** $-\overline{\nabla}_{X_i} N$ is a tangent vector (because $\langle N, N \rangle_{\mathbb{R}^3} = 1$ implies $\langle \overline{\nabla}_{X_i} N, N \rangle_{\mathbb{R}^3} = 0$); we do not need its explicit components, only that $\overline{\nabla}_{X_1} N$ is tangent. Combining, \begin{align*} \overline{\nabla}_{X_1} \overline{\nabla}_{X_2} X_2 &= \nabla_{X_1} \nabla_{X_2} X_2 + h_{22}\, \overline{\nabla}_{X_1} N + \biggl(\partial_1 h_{22} + \sum_{k=1}^2 \Gamma_{22}^k\, h_{1k}\biggr) N. \end{align*} By symmetry (interchange the indices $1 \leftrightarrow 2$ in the first slot — the second slot is fixed at $X_2$ throughout, but we are computing the term $\overline{\nabla}_{X_2} \overline{\nabla}_{X_1} X_2$, so the right computation interchanges the *outer* index, replacing the first $X_1$ with $X_2$, and applying Gauss to $\overline{\nabla}_{X_1} X_2 = \nabla_{X_1} X_2 + h_{12} N$): \begin{align*} \overline{\nabla}_{X_2} \overline{\nabla}_{X_1} X_2 &= \nabla_{X_2} \nabla_{X_1} X_2 + h_{12}\, \overline{\nabla}_{X_2} N + \biggl(\partial_2 h_{12} + \sum_{k=1}^2 \Gamma_{12}^k\, h_{2k}\biggr) N. \end{align*} Subtracting and using $\overline{\nabla}_{X_1} \overline{\nabla}_{X_2} X_2 - \overline{\nabla}_{X_2} \overline{\nabla}_{X_1} X_2 = 0$: \begin{align*} \nabla_{X_1}\nabla_{X_2} X_2 - \nabla_{X_2}\nabla_{X_1} X_2 + \bigl(h_{22} \overline{\nabla}_{X_1} N - h_{12} \overline{\nabla}_{X_2} N\bigr) + (\text{normal terms})\, N = 0. \end{align*} Take the inner product with $X_1$ on both sides. The term involving $N$ vanishes because $\langle X_1, N \rangle_{\mathbb{R}^3} = 0$. The terms $\langle \overline{\nabla}_{X_i} N, X_1 \rangle_{\mathbb{R}^3}$ are computed by differentiating $\langle N, X_1 \rangle_{\mathbb{R}^3} = 0$: \begin{align*} 0 &= \partial_i \langle N, X_1 \rangle_{\mathbb{R}^3} = \langle \overline{\nabla}_{X_i} N, X_1 \rangle_{\mathbb{R}^3} + \langle N, \overline{\nabla}_{X_i} X_1 \rangle_{\mathbb{R}^3} = \langle \overline{\nabla}_{X_i} N, X_1 \rangle_{\mathbb{R}^3} + h_{i1}, \end{align*} so $\langle \overline{\nabla}_{X_i} N, X_1 \rangle_{\mathbb{R}^3} = -h_{i1}$. Therefore \begin{align*} \bigl\langle h_{22} \overline{\nabla}_{X_1} N - h_{12} \overline{\nabla}_{X_2} N,\; X_1\bigr\rangle_{\mathbb{R}^3} = -h_{22} h_{11} + h_{12} h_{21} = -(LN_{\mathrm{II}} - M^2). \end{align*} On the other hand, $\langle \nabla_{X_1}\nabla_{X_2} X_2 - \nabla_{X_2}\nabla_{X_1} X_2,\, X_1 \rangle_g = -R_{1212}$ by the chapter sign convention, since \begin{align*} R_{1212} = g(R(X_1, X_2) X_2, X_1) = -g\bigl(\nabla_{X_1}\nabla_{X_2}X_2 - \nabla_{X_2}\nabla_{X_1}X_2,\, X_1\bigr). \end{align*} Adding the two contributions to zero gives $-R_{1212} - (LN_{\mathrm{II}} - M^2) = 0$, i.e. \begin{align*} R_{1212} = -(LN_{\mathrm{II}} - M^2). \tag{Gauss} \end{align*} (Note. Different texts adopt different sign conventions for $R$; the chapter's convention $R = -\nabla \circ \nabla$ flips the sign of $R_{ijkl}$ compared with conventions where $R(X,Y)Z = \nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z - \nabla_{[X,Y]}Z$. What matters is that *up to a fixed overall sign*, $R_{1212}$ equals $\pm(LN_{\mathrm{II}} - M^2)$, and the sign cancels in the formula $K = R_{1212}/\det g$ once one uses the same convention to define $K$ from $R$. We work consistently with the chapter's convention throughout.) [guided] The overarching strategy is to exploit the fact that $\nabla$ on $\Sigma$ is the *projection* of the ambient flat connection $\overline{\nabla}$ onto the tangent bundle — this is the content of the Gauss formula. Because $\overline{\nabla}$ is flat (its Riemann tensor vanishes identically on $\mathbb{R}^3$), iterating $\overline{\nabla}$ on a tangent field must give a symmetric expression. When we decompose into tangent and normal parts via Gauss, the symmetry forces a relation between the curvature $R$ of $\nabla$ and the second fundamental form $h$. That relation is the Gauss equation. We start by computing $R(X_1, X_2) X_2$ from the chapter's sign convention \begin{align*} R(X_1, X_2) &= \nabla_{[X_1, X_2]} - [\nabla_{X_1}, \nabla_{X_2}]. \end{align*} Why does $[X_1, X_2] = 0$ matter here? Because $X_1$ and $X_2$ are *coordinate* vector fields on $U$, partial-derivative operators with respect to $u$ and $v$, and partial derivatives commute. This kills the first term in the formula and leaves \begin{align*} R(X_1, X_2) X_2 &= -\bigl(\nabla_{X_1} \nabla_{X_2} X_2 - \nabla_{X_2} \nabla_{X_1} X_2\bigr). \end{align*} Now we need a *second* expression for the same quantity, derived from the ambient picture, that we can compare against. The strategy: replicate the iterated covariant derivative $\nabla_{X_1}\nabla_{X_2}X_2 - \nabla_{X_2}\nabla_{X_1}X_2$ inside $\overline{\nabla}$, which we know is identically zero on $\mathbb{R}^3$. Since $\mathbb{R}^3$ is flat and $[X_1, X_2] = 0$, \begin{align*} 0 &= \overline{\nabla}_{X_1} \overline{\nabla}_{X_2} X_2 - \overline{\nabla}_{X_2} \overline{\nabla}_{X_1} X_2. \end{align*} This zero is the engine of the proof: we will expand each term using Gauss, peel off the tangent and normal contributions, and read off the relationship. Let's expand the first term. By the Gauss formula, \begin{align*} \overline{\nabla}_{X_2} X_2 &= \nabla_{X_2} X_2 + h_{22}\, N. \end{align*} We now apply $\overline{\nabla}_{X_1}$ to this expression. Since $\overline{\nabla}$ is just componentwise differentiation in $\mathbb{R}^3$, it acts as a derivation, so \begin{align*} \overline{\nabla}_{X_1}\bigl(\nabla_{X_2} X_2 + h_{22}\, N\bigr) &= \overline{\nabla}_{X_1} \nabla_{X_2} X_2 + (\partial_1 h_{22})\, N + h_{22}\, \overline{\nabla}_{X_1} N. \end{align*} The middle term comes from differentiating the scalar coefficient $h_{22}$, and the last from differentiating the unit normal $N$. We still have $\overline{\nabla}_{X_1} \nabla_{X_2} X_2$ to handle. The point is that $\nabla_{X_2} X_2 = \sum_k \Gamma_{22}^k X_k$ is a *tangent* field (this is the formula for the Levi-Civita connection in coordinates), so Gauss applies again — once for each tangent component. Differentiating $\sum_k \Gamma_{22}^k X_k$ ambiently produces a tangent piece (the intrinsic covariant derivative) plus a normal piece coming from the $h_{1k}$-coefficients on each $\overline{\nabla}_{X_1} X_k$: \begin{align*} \overline{\nabla}_{X_1} \nabla_{X_2} X_2 &= \nabla_{X_1} \nabla_{X_2} X_2 + \sum_{k=1}^2 \Gamma_{22}^k\, h_{1k}\, N. \end{align*} A key observation we exploit repeatedly: $\overline{\nabla}_{X_1} N$ is a *tangent* vector. Why? Because $N$ has unit length, so $\langle N, N\rangle_{\mathbb{R}^3} = 1$ identically; differentiating gives $2\langle \overline{\nabla}_{X_1} N, N\rangle_{\mathbb{R}^3} = 0$, so $\overline{\nabla}_{X_1} N \perp N$, hence is tangent to $\Sigma$. We do not need its explicit components yet — only that it is tangent. Combining, \begin{align*} \overline{\nabla}_{X_1} \overline{\nabla}_{X_2} X_2 &= \nabla_{X_1} \nabla_{X_2} X_2 + h_{22}\, \overline{\nabla}_{X_1} N + \biggl(\partial_1 h_{22} + \sum_{k=1}^2 \Gamma_{22}^k\, h_{1k}\biggr) N. \end{align*} The same expansion, with $1 \leftrightarrow 2$ in the *outer* derivative (the inner $X_2$ is fixed, but Gauss is applied to $\overline{\nabla}_{X_1} X_2 = \nabla_{X_1} X_2 + h_{12} N$ this time), gives \begin{align*} \overline{\nabla}_{X_2} \overline{\nabla}_{X_1} X_2 &= \nabla_{X_2} \nabla_{X_1} X_2 + h_{12}\, \overline{\nabla}_{X_2} N + \biggl(\partial_2 h_{12} + \sum_{k=1}^2 \Gamma_{12}^k\, h_{2k}\biggr) N. \end{align*} Subtract the second from the first and set the result equal to zero (since $\overline{\nabla}$ is flat): \begin{align*} \nabla_{X_1}\nabla_{X_2} X_2 - \nabla_{X_2}\nabla_{X_1} X_2 + \bigl(h_{22} \overline{\nabla}_{X_1} N - h_{12} \overline{\nabla}_{X_2} N\bigr) + (\text{normal terms})\, N = 0. \end{align*} The equation has tangent contributions (the first bracket and the Weingarten pieces) and a normal contribution (the $N$-multiplied terms). To extract a scalar relation, we project onto the tangent direction $X_1$ — this is exactly the projection that builds $R_{1212}$ out of $R(X_1, X_2)X_2$, and it kills the normal terms in one stroke since $\langle X_1, N\rangle_{\mathbb{R}^3} = 0$. Take the inner product (Euclidean, but for tangent vectors this coincides with $g$ by the definition of the induced metric) with $X_1$. The Weingarten contributions $\langle \overline{\nabla}_{X_i} N, X_1\rangle_{\mathbb{R}^3}$ are computed by differentiating the orthogonality $\langle N, X_1\rangle_{\mathbb{R}^3} = 0$: \begin{align*} 0 &= \partial_i \langle N, X_1 \rangle_{\mathbb{R}^3} = \langle \overline{\nabla}_{X_i} N, X_1 \rangle_{\mathbb{R}^3} + \langle N, \overline{\nabla}_{X_i} X_1 \rangle_{\mathbb{R}^3} = \langle \overline{\nabla}_{X_i} N, X_1 \rangle_{\mathbb{R}^3} + h_{i1}, \end{align*} so $\langle \overline{\nabla}_{X_i} N, X_1\rangle_{\mathbb{R}^3} = -h_{i1}$. Substituting into the bracket, \begin{align*} \bigl\langle h_{22} \overline{\nabla}_{X_1} N - h_{12} \overline{\nabla}_{X_2} N,\; X_1\bigr\rangle_{\mathbb{R}^3} = -h_{22} h_{11} + h_{12} h_{21} = -(LN_{\mathrm{II}} - M^2), \end{align*} using $h_{11} = L$, $h_{22} = N_{\mathrm{II}}$, $h_{12} = h_{21} = M$. The remaining tangent contribution is the iterated covariant derivative paired with $X_1$. Why does this give $R_{1212}$? Because $R_{1212} = g(R(X_1, X_2)X_2, X_1)$ is the only independent component of the $(0,4)$ Riemann tensor on a $2$-surface (everything else is determined by skew-symmetries), and from the chapter's sign convention combined with $[X_1,X_2] = 0$, \begin{align*} R_{1212} = g(R(X_1, X_2) X_2, X_1) = -g\bigl(\nabla_{X_1}\nabla_{X_2}X_2 - \nabla_{X_2}\nabla_{X_1}X_2,\; X_1\bigr). \end{align*} Adding the two contributions to zero, \begin{align*} -R_{1212} - (LN_{\mathrm{II}} - M^2) = 0, \end{align*} which rearranges to the Gauss equation \begin{align*} R_{1212} = -(LN_{\mathrm{II}} - M^2). \tag{Gauss} \end{align*} A note on signs: different texts adopt different sign conventions for $R$. The chapter's convention $R = \nabla_{[\cdot,\cdot]} - [\nabla, \nabla]$ flips the sign of $R_{ijkl}$ compared with the convention $R(X,Y)Z = \nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z - \nabla_{[X,Y]}Z$ used elsewhere. The sign cancels in the formula $K = R_{1212}/\det g$ once one uses the same convention to define $K$ from $R$. We work consistently with the chapter's convention throughout. [/guided] [/step] [step:Solve for $K$ in terms of $R_{1212}$ and the metric determinant] The Gaussian curvature is defined extrinsically as \begin{align*} K &= \frac{LN_{\mathrm{II}} - M^2}{EG - F^2} = \frac{LN_{\mathrm{II}} - M^2}{\det g}. \end{align*} Combining with the Gauss equation $R_{1212} = -(LN_{\mathrm{II}} - M^2)$ from Step 2, \begin{align*} K &= -\frac{R_{1212}}{\det g}. \tag{$*$} \end{align*} The right-hand side depends only on $g_{ij}$ and the Riemann tensor of $g$. Since the Riemann tensor of $g$ is built entirely from $g_{ij}$ and its first and second partial derivatives via Christoffel symbols (next step), formula $(*)$ already establishes that $K$ is intrinsic. The remaining task is to expand $R_{1212}$ explicitly to obtain the Brioschi formula. [/step] [step:Expand $R_{1212}$ in Christoffel symbols and derive the Brioschi formula] The Christoffel symbols of $\nabla$ are determined by $g$ via \begin{align*} \Gamma_{ij}^k &= \frac{1}{2} \sum_{\ell=1}^2 g^{k\ell}\bigl(\partial_i g_{j\ell} + \partial_j g_{i\ell} - \partial_\ell g_{ij}\bigr), \end{align*} where $(g^{k\ell})$ is the inverse matrix of $(g_{k\ell})$. This formula is standard for the unique torsion-free metric-compatible connection. The Riemann tensor in coordinates is \begin{align*} R^\ell_{\;jik} = \partial_k \Gamma^\ell_{ij} - \partial_i \Gamma^\ell_{kj} + \sum_{m=1}^2 \bigl(\Gamma^\ell_{km} \Gamma^m_{ij} - \Gamma^\ell_{im} \Gamma^m_{kj}\bigr), \end{align*} and the fully covariant tensor is $R_{\ell j i k} = \sum_{p=1}^2 g_{\ell p}\, R^p_{\;j i k}$. With the chapter's sign convention $R(X,Y) = \nabla_{[X,Y]} - [\nabla_X, \nabla_Y]$, applied to coordinate fields where $[X_i, X_j] = 0$, the resulting expression for $R_{1212}$ is a polynomial in $\Gamma_{ij}^k$, $\partial_i \Gamma_{jk}^\ell$, and contractions with $g_{ij}$. Substituting the formula for $\Gamma_{ij}^k$ in terms of $g_{ij}$ and $\partial_k g_{ij}$, and simplifying, one obtains the **Brioschi formula**: \begin{align*} R_{1212} = \frac{1}{\det g}\Biggl[ \det\!\begin{pmatrix} -\tfrac{1}{2}\partial^2_{vv} g_{11} + \partial^2_{uv} g_{12} - \tfrac{1}{2}\partial^2_{uu} g_{22} & \tfrac{1}{2}\partial_u g_{11} & \partial_u g_{12} - \tfrac{1}{2}\partial_v g_{11} \\ \partial_v g_{12} - \tfrac{1}{2}\partial_u g_{22} & g_{11} & g_{12} \\ \tfrac{1}{2}\partial_v g_{22} & g_{12} & g_{22} \end{pmatrix} \\ - \det\!\begin{pmatrix} 0 & \tfrac{1}{2}\partial_v g_{11} & \tfrac{1}{2}\partial_u g_{22} \\ \tfrac{1}{2}\partial_v g_{11} & g_{11} & g_{12} \\ \tfrac{1}{2}\partial_u g_{22} & g_{12} & g_{22} \end{pmatrix} \Biggr]. \end{align*} Combined with $(*)$, we have an explicit closed-form expression for $K$ as a rational function of $g_{11}, g_{12}, g_{22}$ and their first and second partial derivatives in $u, v$: \begin{align*} K = -\frac{R_{1212}}{\det g} = (\text{rational expression in } g_{ij}, \partial g_{ij}, \partial^2 g_{ij}). \tag{Brioschi} \end{align*} This shows directly that $K$ is **intrinsic**: it depends only on the first fundamental form $g$, not on the second fundamental form $h$ or on the embedding $\Sigma \hookrightarrow \mathbb{R}^3$. [guided] The strategy of this step is to take the abstract relation $K = -R_{1212}/\det g$ from the previous step and unpack it into a fully explicit formula in $g_{ij}$ and its partial derivatives. We do this by tracing the dependency chain: $R_{1212}$ depends on Christoffel symbols and their derivatives; Christoffel symbols depend on $g_{ij}$ and $\partial g_{ij}$. Composing these expressions gives a closed-form rational function of $g_{ij}, \partial g_{ij}, \partial^2 g_{ij}$ — the Brioschi formula. The first ingredient is the formula for the Christoffel symbols of the Levi-Civita connection. The defining requirements (torsion-free, metric-compatible) determine $\Gamma_{ij}^k$ uniquely from $g$ as \begin{align*} \Gamma_{ij}^k &= \frac{1}{2} \sum_{\ell=1}^2 g^{k\ell}\bigl(\partial_i g_{j\ell} + \partial_j g_{i\ell} - \partial_\ell g_{ij}\bigr), \end{align*} where $(g^{k\ell})$ is the inverse matrix of $(g_{k\ell})$. Why this combination? The metric-compatibility identity $\partial_i g_{jk} = g(\nabla_{X_i} X_j, X_k) + g(X_j, \nabla_{X_i} X_k)$ together with torsion-freeness $\Gamma_{ij}^k = \Gamma_{ji}^k$ allows one to symmetrise three cyclic permutations and solve for $\Gamma$; the result is the formula above. In dimension $2$, the inverse $g^{k\ell}$ is computed via the cofactor identity $g^{k\ell} = \frac{1}{\det g}\,\mathrm{cof}(g)^{k\ell}$, so $\Gamma_{ij}^k$ is a rational function of $g_{ij}$ and $\partial g_{ij}$ with $\det g$ in the denominator. The second ingredient is the coordinate formula for the Riemann tensor. With the chapter's sign convention $R(X,Y) = \nabla_{[X,Y]} - [\nabla_X, \nabla_Y]$, applied to coordinate fields where $[X_i, X_j] = 0$, one expands $\nabla_{X_i}\nabla_{X_j} X_k$ in Christoffel symbols and reads off the components: \begin{align*} R^\ell_{\;jik} = \partial_k \Gamma^\ell_{ij} - \partial_i \Gamma^\ell_{kj} + \sum_{m=1}^2 \bigl(\Gamma^\ell_{km} \Gamma^m_{ij} - \Gamma^\ell_{im} \Gamma^m_{kj}\bigr). \end{align*} The fully covariant version is obtained by lowering the upper index against $g$: \begin{align*} R_{\ell j i k} = \sum_{p=1}^2 g_{\ell p}\, R^p_{\;j i k}. \end{align*} Already at this stage we see that $R_{1212}$ is a polynomial in $\Gamma_{ij}^k$, $\partial \Gamma_{ij}^k$, and $g_{ij}$, hence — after substituting the Christoffel formula — a rational function of $g_{ij}, \partial g_{ij}, \partial^2 g_{ij}$ with $\det g$ in the denominator. The remaining task is bookkeeping: substitute the Christoffel formula into the Riemann formula, expand all products, collect terms, and use the identities $\det g \cdot g^{k\ell} = \mathrm{cof}(g)^{k\ell}$ and the symmetries of $g_{ij}$ in dimension $2$ to package the result into determinants. Why determinants? Because each $g^{k\ell}$ contributes a $\frac{1}{\det g}$ factor, and the bilinear pairings of $\partial g$ terms against $g_{ij}$ entries naturally arrange themselves as $3\times 3$ minors. The mechanical but lengthy expansion produces the **Brioschi formula**: \begin{align*} R_{1212} = \frac{1}{\det g}\Biggl[ \det\!\begin{pmatrix} -\tfrac{1}{2}\partial^2_{vv} g_{11} + \partial^2_{uv} g_{12} - \tfrac{1}{2}\partial^2_{uu} g_{22} & \tfrac{1}{2}\partial_u g_{11} & \partial_u g_{12} - \tfrac{1}{2}\partial_v g_{11} \\ \partial_v g_{12} - \tfrac{1}{2}\partial_u g_{22} & g_{11} & g_{12} \\ \tfrac{1}{2}\partial_v g_{22} & g_{12} & g_{22} \end{pmatrix} \\ - \det\!\begin{pmatrix} 0 & \tfrac{1}{2}\partial_v g_{11} & \tfrac{1}{2}\partial_u g_{22} \\ \tfrac{1}{2}\partial_v g_{11} & g_{11} & g_{12} \\ \tfrac{1}{2}\partial_u g_{22} & g_{12} & g_{22} \end{pmatrix} \Biggr]. \end{align*} Notice that the $(1,1)$-entry of the first determinant collects all the second-derivative terms, while the lower-right $2 \times 2$ blocks are exactly the metric matrix $(g_{ij})$ itself; the cross terms encode the first derivatives. This compact form is no accident — it reflects the antisymmetric pairing of indices in the Riemann tensor combined with the symmetric structure of $g$. Combining with the relation $(*)$ from Step 3, \begin{align*} K = -\frac{R_{1212}}{\det g} = (\text{rational expression in } g_{ij}, \partial g_{ij}, \partial^2 g_{ij}). \tag{Brioschi} \end{align*} This is the structural payoff: *every* term in the resulting expression is a function of $g_{ij}$ and its first and second partial derivatives. There is no $h_{ij}$ anywhere, no unit normal $N$, no embedding data — exactly the content of Theorema Egregium. Two why-discussions worth flagging. First: what if one tried to define a "curvature" via $LN_{\mathrm{II}} - M^2$ alone, without dividing by $\det g$? Then it would not be parametrisation-invariant, since $h_{ij}$ transforms as a $(0,2)$-tensor whose Jacobian factor cancels only when paired with the $\det g^{-1}$ that converts $(0,2)$-tensor components into a scalar density of weight zero. The denominator $\det g$ is what makes $K$ a genuine scalar function on $\Sigma$. Second: why does this formula give isometry invariance for free (the content of Step 5)? Because once $K$ is expressible from $g_{ij}$ alone via Brioschi, any map $\Phi: (\Sigma_1, g_1) \to (\Sigma_2, g_2)$ satisfying $\Phi^* g_2 = g_1$ pulls back the inputs of the Brioschi formula on $\Sigma_2$ to those on $\Sigma_1$, hence pulls back $K_2$ to $K_1$. As a celebrated consequence, the sphere of radius $R$ has $K = 1/R^2$ everywhere, while the plane has $K \equiv 0$, so no piece of a sphere can be isometric to a piece of the plane — the orange-peel principle. [/guided] [/step] [step:Conclude isometry invariance] Let $\Phi: (\Sigma_1, g_1) \to (\Sigma_2, g_2)$ be a smooth isometry, meaning $\Phi$ is a diffeomorphism with $\Phi^* g_2 = g_1$. Pick a parametrisation $X_1: U \to \Sigma_1$ and let $X_2 := \Phi \circ X_1: U \to \Sigma_2$, which is a parametrisation of $\Sigma_2$ on the same parameter domain $U$. The pullback condition $\Phi^* g_2 = g_1$ means the components $g_{ij}$ of $g_1$ in $X_1$ coordinates equal the components $g_{ij}$ of $g_2$ in $X_2$ coordinates: explicitly, for $i, j \in \{1, 2\}$, \begin{align*} (g_1)_{ij} = (X_1)^* g_1\bigl(\partial_i, \partial_j\bigr) = (X_2)^* g_2\bigl(\partial_i, \partial_j\bigr) = (g_2)_{ij}. \end{align*} By the Brioschi formula, $K_1$ and $K_2$ are the same rational function of $g_{ij}, \partial g_{ij}, \partial^2 g_{ij}$ — and these inputs agree on the two sides. Therefore \begin{align*} K_2 \circ \Phi = K_1. \end{align*} Combined with Steps 3 and 4, this completes the proof of Theorema Egregium: $K$ is an intrinsic invariant of the Riemannian metric $g$. [/step]