[proofplan] The proof proceeds by localising the global identity to a triangulation. Fix a smooth triangulation $\mathcal{T}$ of $M$ whose closed faces are *geodesic triangles* contained in coordinate charts admitting positively oriented orthonormal frames. On each such triangle $T$, the Cartan structure equations and [Stokes' Theorem](/theorems/1530) reduce $\int_T K\, dA$ to an integral of the connection $1$-form along $\partial T$, which the geodesic equation rewrites as the variation of the angle that the unit tangent makes with the frame. The Hopf Umlaufsatz then yields the local angle excess identity $\int_T K\, dA = \sum_{i=1}^3 \alpha_i - \pi$, where $\alpha_1, \alpha_2, \alpha_3$ are the interior angles. Summing this identity over all faces, the angles at every vertex assemble into $2\pi$ and we obtain $\int_M K\, dA = 2\pi V - \pi F$. The combinatorial identity $3F = 2E$ for triangulations finally converts this into $2\pi(V - E + F) = 2\pi \chi(M)$. [/proofplan]

[step:Set up a geodesic triangulation of $M$ adapted to oriented orthonormal frames]Because $M$ is a smooth compact surface, it admits a smooth triangulation. By taking iterated barycentric subdivisions we may assume each closed face is contained in a single normal coordinate chart of $g$ of radius smaller than the injectivity radius. We then replace each edge by the unique minimising geodesic between its endpoints (which lies inside the chart by [Local Existence and Uniqueness of Geodesics](/theorems/1557)); the resulting decomposition $\mathcal{T} = (V, E, F)$ is a *geodesic triangulation* whose closed faces are geodesic triangles. Denote the sets of vertices, edges and (closed) faces by $V$, $E$, $F$ respectively, and write $|V|, |E|, |F|$ for their cardinalities. For each closed face $T \in F$, fix an open neighbourhood $U_T \supset T$ contained in a coordinate chart on which $TM$ trivialises, and choose a positively oriented orthonormal frame \begin{align*} (e_1^{(T)}, e_2^{(T)}) : U_T &\to TM \times TM, \\ p &\mapsto (e_1^{(T)}(p), e_2^{(T)}(p)), \end{align*} i.e. $g_p(e_i^{(T)}(p), e_j^{(T)}(p)) = \delta_{ij}$ for all $p \in U_T$ and $(e_1^{(T)}(p), e_2^{(T)}(p))$ agrees with the orientation of $M$. Let $(\theta^1_T, \theta^2_T)$ denote the dual coframe, so that $dA|_{U_T} = \theta^1_T \wedge \theta^2_T$.[/step]

[guided]We must place ourselves in a setting where the connection $1$-form is globally defined and [Stokes' theorem](/theorems/1530) applies. Two requirements drive the construction. **First**, we need each face small enough to fit inside a chart that supports an oriented orthonormal frame; otherwise the connection $1$-form $\omega$ introduced below depends on a non-trivial choice and changes by a global obstruction between charts. **Second**, we need the edges to be geodesics so that the angle function $\phi$ along $\partial T$ (introduced later) satisfies the clean relation $d\phi = \omega|_{\partial T}$. A smooth triangulation exists by the classical theorems of Cairns–Whitehead. To meet the size requirement, subdivide (barycentrically) until each closed face lies in a normal coordinate chart of radius less than the injectivity radius — this is possible because $M$ is compact, hence has a positive injectivity radius, and barycentric subdivision shrinks faces. To meet the geodesic requirement, replace each combinatorial edge by the unique minimising geodesic connecting its endpoints; existence and uniqueness of this geodesic come from [Local Existence and Uniqueness of Geodesics](/theorems/1557) applied inside the normal chart. For each closed face $T$, pick a (slightly larger) open neighbourhood $U_T \supset T$ on which the tangent bundle trivialises and choose a positively oriented orthonormal frame $(e_1^{(T)}, e_2^{(T)})$. Such a frame exists because trivialisation provides a smooth section of $TM \times TM$, and Gram–Schmidt then produces an orthonormal one; orientability of $M$ allows the choice to match the orientation. Write $(\theta_T^1, \theta_T^2)$ for the dual coframe. By definition of $dA$, on $U_T$ we have $dA = \theta_T^1 \wedge \theta_T^2$.[/guided]

[step:Identify the curvature $2$-form with the exterior derivative of the connection $1$-form] Fix a face $T \in F$ and write $(e_1, e_2) = (e_1^{(T)}, e_2^{(T)})$, $(\theta^1, \theta^2) = (\theta^1_T, \theta^2_T)$. By the [Fundamental Theorem of Riemannian Geometry](/theorems/1552), the Levi-Civita connection $\nabla$ exists and is unique. Define the connection $1$-form $\omega_T$ as the smooth section \begin{align*} \omega_T \in \Gamma(T^*U_T) = \Omega^1(U_T) \end{align*} (i.e. a smooth $1$-form on $U_T$) characterised by the requirement \begin{align*} \nabla e_1 = -\omega_T \otimes e_2, \qquad \nabla e_2 = \omega_T \otimes e_1. \end{align*} The antisymmetric form of this assignment is forced by metric compatibility $\nabla g \equiv 0$ applied to $g(e_i, e_j) = \delta_{ij}$. This sign convention is chosen so that a geodesic unit tangent with angle $\phi$ relative to $(e_1,e_2)$ satisfies $d\phi = \gamma^*\omega_T$ along each smooth geodesic edge. We claim: [claim:Cartan curvature identity in two dimensions] On $U_T$, \begin{align*} d\omega_T = K \, dA. \end{align*} [/claim] [proof] By the [Local Formula for Curvature](/theorems/1540), the connection forms $\omega^i_j$ are defined by \begin{align*} \nabla e_j = \sum_{i=1}^2 \omega^i_j \otimes e_i. \end{align*} Our convention gives \begin{align*} \omega^1_2 = \omega_T, \qquad \omega^2_1 = -\omega_T, \qquad \omega^1_1 = \omega^2_2 = 0. \end{align*} The same local curvature formula defines the curvature $2$-forms $\Omega^i_j$ by the second Cartan structure equation \begin{align*} \Omega^i_j = d\omega^i_j + \sum_{k=1}^2 \omega^i_k \wedge \omega^k_j. \end{align*} For $(i, j) = (1, 2)$, \begin{align*} \sum_{k=1}^2 \omega^1_k \wedge \omega^k_2 = \omega^1_1 \wedge \omega^1_2 + \omega^1_2 \wedge \omega^2_2 = 0, \end{align*} so $\Omega^1_2 = d\omega_T$. The same theorem identifies Gaussian curvature in an oriented orthonormal frame by \begin{align*} \Omega^1_2 = K \, \theta^1 \wedge \theta^2 = K\, dA. \end{align*} Therefore $d\omega_T = K\, dA$. [/proof] [/step]

[step:Convert the curvature integral on a triangle to a boundary integral of the angle function]Let $T \in F$ be a geodesic triangle with vertices $p_1, p_2, p_3$ and interior angles $\alpha_1, \alpha_2, \alpha_3 \in (0, \pi)$ measured by $g$ at the respective vertices. Orient $\partial T$ as the positively oriented boundary induced from the orientation of $T$, and parametrise it by arc length $s \in [0, L]$ where $L = \mathrm{length}(\partial T)$, writing $\gamma: [0, L] \to M$ for the parametrisation (with $\gamma(0) = \gamma(L)$). Because $\gamma'(s)$ is a unit vector in $T_{\gamma(s)} M$, there exists a continuous lift $\phi: [0, L] \setminus \{s_1, s_2, s_3\} \to \mathbb{R}$ (where $s_1 < s_2 < s_3$ are the times at which $\gamma$ traverses the vertices) such that \begin{align*} \gamma'(s) = \cos \phi(s) \, e_1(\gamma(s)) + \sin \phi(s) \, e_2(\gamma(s)). \end{align*} On each open geodesic arc, the geodesic equation $\nabla_{\gamma'} \gamma' = 0$ expanded in the frame $(e_1, e_2)$ gives \begin{align*} \nabla_{\gamma'} \gamma' = \bigl(\dot \phi(s) - \omega_T(\gamma'(s))\bigr) \bigl(-\sin\phi(s) \, e_1 + \cos\phi(s) \, e_2\bigr), \end{align*} so $\dot\phi(s) = \omega_T(\gamma'(s))$, i.e. $d\phi = \gamma^*\omega_T$ along each smooth arc. By [Stokes' Theorem](/theorems/1530) applied to the smooth $1$-form $\omega_T$ on the closed disk $T \subset U_T$, \begin{align*} \int_T K \, dA = \int_T d\omega_T = \int_{\partial T} \omega_T = \sum_{i=1}^3 \int_{e_i} d\phi, \end{align*} where $e_1, e_2, e_3$ are the three (open) geodesic arcs of $\partial T$ and the last equality uses $d\phi = \omega_T$ on each arc.[/step]

[guided]We want to relate $\int_T K\, dA$ to angles at vertices. The Cartan identity $d\omega_T = K\, dA$ already turns $\int_T K\, dA$ into $\int_{\partial T} \omega_T$ via Stokes, so the task reduces to interpreting $\omega_T$ along $\partial T$. The geodesic equation is what makes this clean. Pick a unit-speed parametrisation $\gamma$ of $\partial T$ — possible because $\partial T$ is piecewise smooth and each smooth piece is a geodesic arc, hence has well-defined arc length. The unit tangent $\gamma'(s) \in T_{\gamma(s)}M$ is a unit vector, so it lies on the unit circle of $T_{\gamma(s)}M$ which we parametrise by angle relative to $e_1$. Lifting this angle through the covering $\mathbb{R} \to S^1$ gives a continuous function $\phi$ on the smooth pieces (at vertices the tangent jumps, so $\phi$ has jumps there). Why does $\dot\phi = \omega_T(\gamma')$ hold? Differentiate $\gamma' = \cos\phi \, e_1 + \sin\phi\, e_2$ covariantly along $\gamma$: \begin{align*} \nabla_{\gamma'} \gamma' &= -\sin\phi\, \dot\phi\, e_1 + \cos\phi\, \nabla_{\gamma'} e_1 + \cos\phi\, \dot\phi\, e_2 + \sin\phi\, \nabla_{\gamma'} e_2 \\ &= -\sin\phi\, \dot\phi\, e_1 - \cos\phi\, \omega_T(\gamma')\, e_2 + \cos\phi\, \dot\phi\, e_2 + \sin\phi\, \omega_T(\gamma')\, e_1 \\ &= \bigl(\dot\phi - \omega_T(\gamma')\bigr)\bigl(-\sin\phi\, e_1 + \cos\phi\, e_2\bigr), \end{align*} where we used $\nabla e_1 = -\omega_T \otimes e_2$ and $\nabla e_2 = \omega_T \otimes e_1$. On a geodesic arc the LHS vanishes, and the vector $-\sin\phi\, e_1 + \cos\phi\, e_2$ is the unit normal to $\gamma'$, in particular nonzero, so the scalar coefficient must vanish: $\dot\phi(s) = \omega_T(\gamma'(s))$, i.e. $\gamma^* \omega_T = d\phi$ as $1$-forms on each open arc. The hypotheses of [Stokes' Theorem](/theorems/1530) are satisfied: $T$ is a compact smoothly embedded $2$-manifold with corners, $\omega_T$ is a smooth $1$-form on a neighbourhood $U_T \supset T$, and the corner set of $\partial T$ has measure zero. Stokes gives $\int_T d\omega_T = \int_{\partial T} \omega_T$, and pulling back to the parametrisation, \begin{align*} \int_{\partial T} \omega_T = \sum_{i=1}^3 \int_{e_i} \gamma^* \omega_T = \sum_{i=1}^3 \int_{e_i} d\phi. \end{align*} Here a *simple piecewise smooth Jordan curve* in $\mathbb{R}^2$ means a continuous injective map $\eta: [0, L] \to \mathbb{R}^2$ with $\eta(0) = \eta(L)$ that is smooth except at finitely many parameter values — this is the regularity class to which the chart image $\psi(\partial T)$ belongs.[/guided]

[step:Apply the Hopf Umlaufsatz to obtain the local angle excess formula]The lift $\phi$ extends continuously to the smooth arcs but jumps at each vertex. Let $\theta_i \in (-\pi, \pi)$ denote the *exterior angle* of $T$ at the $i$-th vertex, defined as the signed angular jump of $\phi$ as $\gamma$ traverses that vertex, oriented to match the orientation of $T$. With the convention that interior angles $\alpha_i$ are the standard $(0, \pi)$-valued angles measured by $g$, \begin{align*} \theta_i = \pi - \alpha_i. \end{align*} [claim:Hopf Umlaufsatz on a geodesic triangle] The total angular variation of $\gamma'$ along $\partial T$ equals $2\pi$: \begin{align*} \sum_{i=1}^3 \int_{e_i} d\phi + \sum_{i=1}^3 \theta_i = 2\pi. \end{align*} [/claim] [proof] The closed triangle $T$ is contained in a normal coordinate chart $\psi: U_T \to \psi(U_T) \subseteq \mathbb{R}^2$. Inside this chart the frame $(e_1, e_2)$ trivialises the tangent bundle, so the lift $\phi: [0, L] \to \mathbb{R}$ (allowing jumps at $s_1, s_2, s_3$) is well-defined modulo $2\pi$ and the difference $\phi(L^-) - \phi(0^+) + \sum_i \theta_i \in 2\pi\mathbb{Z}$ since $\gamma'(0) = \gamma'(L)$. To compute which integer multiple of $2\pi$, use the oriented frame to define the bundle trivialisation \begin{align*} \tau: TU_T &\to U_T \times \mathbb{R}^2 \end{align*} by writing each vector as $a_1 e_1 + a_2 e_2$. The unit tangent map of $\partial T$ in this trivialisation is the map $s \mapsto (\cos\phi(s),\sin\phi(s))$, with the vertex jumps recorded by the angles $\theta_i$. The coordinate chart $\psi$ sends $\partial T$ to a positively oriented simple piecewise smooth Jordan curve in $\mathbb{R}^2$ (i.e. the image of a continuous injective map $\eta: [0,L] \to \mathbb{R}^2$ with $\eta(0)=\eta(L)$, smooth except at finitely many points). By the planar Hopf Umlaufsatz, the Euclidean unit tangent map $s \mapsto \dot\eta(s)/|\dot\eta(s)|$ of this Jordan curve, viewed as a map $S^1 \to S^1$, has *degree* $1$ — where the degree of a continuous map $f: S^1 \to S^1$ is the unique integer $n$ such that any continuous lift $\tilde f: \mathbb{R} \to \mathbb{R}$ along the covering $t \mapsto e^{it}$ satisfies $\tilde f(t + 2\pi) - \tilde f(t) = 2\pi n$. Passing from the Euclidean coordinate trivialisation to the frame trivialisation multiplies the tangent-direction map by a loop in $GL^+(2,\mathbb{R}) := \{A \in \mathbb{R}^{2\times 2} : \det A > 0\}$, the group of orientation-preserving real linear automorphisms of $\mathbb{R}^2$. This loop extends to a continuous map on the disk $T$ (because the frame is defined on all of $U_T \supset T$), and any continuous map from a disk to $GL^+(2,\mathbb{R})$ is null-homotopic when restricted to the boundary circle, so the loop has degree $0$. Hence the tangent direction in the frame trivialisation still has degree $1$, and the total angular variation is $2\pi$. [/proof] Combining this with the previous step, \begin{align*} \int_T K\, dA = \sum_{i=1}^3 \int_{e_i} d\phi = 2\pi - \sum_{i=1}^3 \theta_i = 2\pi - \sum_{i=1}^3 (\pi - \alpha_i) = \sum_{i=1}^3 \alpha_i - \pi. \end{align*}[/step]

[guided]The Cartan computation reduced everything to $\sum_i \int_{e_i} d\phi$, but $\phi$ is multivalued unless we cut at the vertices. The Hopf Umlaufsatz quantifies the total turning of $\gamma'$ as it goes around $\partial T$ once. The exterior angle $\theta_i$ is the jump of $\phi$ at the $i$-th vertex; geometrically, it is the angle you must rotate the incoming tangent through to align it with the outgoing tangent. With the convention that the interior angle $\alpha_i \in (0, \pi)$ is the geodesic angle between the two incoming edges measured from inside $T$, we have $\theta_i = \pi - \alpha_i \in (0, \pi)$ for a convex triangle. Why is the total turning exactly $2\pi$? In Euclidean $\mathbb{R}^2$ this is a planar topology fact: as you traverse a simple closed curve once counterclockwise, the unit tangent makes one full turn — quantitatively, the *degree* of a continuous map $f: S^1 \to S^1$ is the integer $n$ such that any continuous lift $\tilde f: \mathbb{R} \to \mathbb{R}$ along $t \mapsto e^{it}$ satisfies $\tilde f(t+2\pi) - \tilde f(t) = 2\pi n$, and the planar Umlaufsatz states this integer is $1$ for the unit tangent of a positively oriented simple closed curve. On a Riemannian surface, $\gamma'$ lives in the (non-trivial) circle bundle of unit tangent vectors, but the frame $(e_1, e_2)$ trivialises this bundle over the chart $U_T$, so the angle function $\phi$ relative to that frame is well-defined modulo $2\pi$ on each smooth piece. The total signed variation $\sum_i \int_{e_i} d\phi + \sum_i \theta_i$ is an integer multiple of $2\pi$ because $\phi(L^-) - \phi(0^+) + \sum \theta_i \in 2\pi\mathbb{Z}$ (the tangent returns to itself). To pin down the integer, compare the frame trivialisation with the ordinary coordinate trivialisation in the chart. The chart image $\psi(\partial T)$ is a positively oriented simple piecewise smooth Jordan curve in $\mathbb{R}^2$, so the planar Hopf Umlaufsatz says its Euclidean unit tangent has degree $1$. The change from Euclidean coordinate directions to the oriented orthonormal frame $(e_1,e_2)$ is a loop in $GL^+(2,\mathbb{R}) = \{A \in \mathbb{R}^{2\times 2}: \det A > 0\}$ along $\partial T$ that extends over the disk $T$ because the frame is defined on all of $U_T$. Any loop in $GL^+(2,\mathbb{R})$ that extends to a continuous map on the disk $T$ is null-homotopic in $GL^+(2,\mathbb{R})$ (just contract along the extension), hence has rotational degree $0$, so it does not change the tangent degree. Therefore the total turning measured by $\phi$, including the vertex jumps, is $2\pi$. Substituting $\sum_i \theta_i = 3\pi - \sum_i \alpha_i$: \begin{align*} \int_T K\, dA = 2\pi - (3\pi - \sum_i \alpha_i) = \sum_i \alpha_i - \pi. \end{align*} This is the **angle excess formula**: the curvature integrated over a geodesic triangle equals the amount by which the sum of its interior angles exceeds $\pi$.[/guided]

[step:Sum the local identity over all faces and collect angles at vertices]Summing the angle excess formula $\int_T K\, dA = \sum_{i=1}^3 \alpha_i - \pi$ over all $T \in F$, \begin{align*} \int_M K\, dA = \sum_{T \in F} \int_T K\, dA = \sum_{T \in F} \sum_{i=1}^3 \alpha_i(T) - \pi |F|, \end{align*} where $\alpha_i(T)$ is the $i$-th interior angle of $T$. Re-indexing the angle sum by vertices: for each vertex $v \in V$, the triangles meeting at $v$ tile a neighbourhood of $v$, so their interior angles at $v$ sum to $2\pi$. Therefore \begin{align*} \sum_{T \in F} \sum_{i=1}^3 \alpha_i(T) = \sum_{v \in V} 2\pi = 2\pi |V|. \end{align*} Hence \begin{align*} \int_M K\, dA = 2\pi |V| - \pi |F|. \end{align*}[/step]

[guided]We have a local statement on each triangle and want a global statement on $M$. The faces of the geodesic triangulation cover $M$ and overlap only on lower-dimensional sets (edges and vertices), so the integral over $M$ is the sum of integrals over the faces. This step is just additivity of Lebesgue/Riemannian integration over a finite measurable partition. Each triangle contributes three angle terms, one per vertex of that triangle. Reorganise this double sum by collecting angles at each vertex $v \in V$. The triangles meeting $v$ have their interior angles at $v$ filling a complete neighbourhood of $v$ — concretely, because the triangulation is a smooth triangulation of a surface without boundary, the link of $v$ is a topological circle. In a normal chart around $v$, intersecting the incident faces with a sufficiently small metric circle produces a piecewise smooth simple closed curve whose arcs occur in the cyclic order of the incident triangles. Thus the angles at $v$ partition the full tangent circle and sum to $2\pi$. So $\sum_{v} 2\pi = 2\pi |V|$, and putting everything together, \begin{align*} \int_M K\, dA = 2\pi |V| - \pi |F|. \end{align*}[/guided]

[step:Convert the count $2\pi |V| - \pi |F|$ into $2\pi \chi(M)$]For any triangulation, each face has exactly three edges and each edge lies in exactly two faces (because $M$ has no boundary), so $3 |F| = 2 |E|$, equivalently $\pi |F| = 2\pi |E| - 2\pi |F|$, equivalently \begin{align*} 2\pi |V| - \pi |F| = 2\pi(|V| - |E| + |F|). \end{align*} Since the Euler characteristic of $M$ equals the alternating face count of any smooth triangulation, $|V| - |E| + |F| = \chi(M)$. Combining with the previous step, \begin{align*} \int_M K\, dA = 2\pi \chi(M), \end{align*} which is the claimed identity.[/step]

[guided]The right-hand side $2\pi |V| - \pi |F|$ is a triangulation-dependent count. To turn it into the topological invariant $\chi(M)$, we use the combinatorial identity $3|F| = 2|E|$, which holds because each triangular face contributes three edge-incidences and each edge of a closed triangulated surface is shared by exactly two faces. We verify the target identity \begin{align*} 2\pi |V| - \pi |F| = 2\pi(|V| - |E| + |F|) \end{align*} by rearranging the right-hand side: \begin{align*} 2\pi(|V| - |E| + |F|) - (2\pi|V| - \pi|F|) = -2\pi|E| + 3\pi|F| - \pi|F| \cdot 0 = \pi(3|F| - 2|E|) = 0, \end{align*} where the last equality uses $3|F| = 2|E|$. Hence $2\pi |V| - \pi |F| = 2\pi(|V| - |E| + |F|)$. Finally, the alternating sum $|V| - |E| + |F|$ is the Euler characteristic $\chi(M)$ of the topological surface underlying any triangulation — this is the simplicial definition, agreeing with the singular/cellular definition by the standard isomorphism between simplicial and singular homology. So \begin{align*} \int_M K\, dA = 2\pi(|V| - |E| + |F|) = 2\pi \chi(M), \end{align*} completing the proof.[/guided]