A discrete version of curvature may help highschool students. Take a polyhedron in $R^3$ and define the curvature at a vertex $v$ by $2 \pi K(v) = 2 \pi - \theta_1 - \ldots - \theta_k$, where $\theta_j$ is the angle at $v$ of the j-th 2-face containing $v$. This gives a measure of how sharp that vertex is. For simplicity, assume there are only three faces meeting at $v$. Let the normal vectors to the faces be $n_1$, $n_2$, $n_3$. They are the corners of a geodesic triangle on the sphere, which is an analog of the region spanned by the Gauss map on a surface. By spherical geometry we have that the area of this triangle is $A = \beta_1 + \beta_2 + \beta_3 - \pi$, where $\beta_j$'s are the angles. One then shows that $\beta_j = \pi - \theta_j$, so we have the nice formula $A = 2 \pi K(v)$. I think that similarily one can argue with parallel transport . Define it like this: start with a vector tangent to one face and transport it along a curve in the standard parallel way while inside a face, but when you change face you project it to the next face(see comment by pasquale below). I have not tried the computation, but it should work. Also the global Gauss-Bonnet theorem holds: if you sum the curvature of the vertices on a closed polyhedron the result is the Euler charcteristic (I think this was orginally proved by Descartes, but I'm not sure).
A discrete version of curvature may help highschool students. Take a polyhedron in $R^3$ and define the curvature at a vertex $v$ by $2 \pi K(v) = 2 \pi - \theta_1 - \ldots - \theta_k$, where $\theta_j$ is the angle at $v$ of the j-th 2-face containing $v$. This gives a measure of how sharp that vertex is. For simplicity, assume there are only three faces meeting at $v$. Let the normal vectors to the faces be $n_1$, $n_2$, $n_3$. They are the corners of a geodesic triangle on the sphere, which is an analog of the region spanned by the Gauss map on a surface. By spherical geometry we have that the area of this triangle is $A = \beta_1 + \beta_2 + \beta_3 - \pi$. pi$, where$\beta_j$'s are the angles. One then shows that$\beta_j = \pi - \theta_j$, so we have the nice formula$ A = 2 \pi A = K(v)$. I think that similarily one can argue with parallel transport. Define it like this: start with a vector tangent to one face and transport it along a curve in the standard parallel way while inside a face, but when you change face you project it to the next face. I have not tried the computation, but it should work. Also the global Gauss-Bonnet theorem holds: if you sum the curvature of the vertices on a closed polyhedron the result is the Euler charcteristic (I think this was orginally proved by Descartes, but I'm not sure). 1 A discrete version of curvature may help highschool students. Take a polyhedron in$R^3$and define the curvature at a vertex$v$by$2 \pi K(v) = 2 \pi - \theta_1 - \ldots - \theta_k$, where$\theta_j$is the angle at$v$of the j-th 2-face containing$v$. This gives a measure of how sharp that vertex is. For simplicity, assume there are only three faces meeting at$v$. Let the normal vectors to the faces be$n_1$,$n_2$,$n_3$. They are the corners of a geodesic triangle on the sphere, which is an analog of the region spanned by the Gauss map on a surface. By spherical geometry we have that the area of this triangle is$A = \beta_1 + \beta_2 + \beta_3 - \pi$. One then shows that$\beta_j = \pi - \theta_j$, so we have the nice formula$2 \pi A = K(v)\$. I think that similarily one can argue with parallel transport. Define it like this: start with a vector tangent to one face and transport it along a curve in the standard parallel way while inside a face, but when you change face you project it to the next face. I have not tried the computation, but it should work. Also the global Gauss-Bonnet theorem holds: if you sum the curvature of the vertices on a closed polyhedron the result is the Euler charcteristic (I think this was orginally proved by Descartes, but I'm not sure).