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prove-Gauss-curvature-K-is-intrinsic-by-showing-K-determinant-1-2-E-vv-F-uv-G-uu-1-2-E-u-F-u-1-2-E-v-F-v-1-2-G-u-E-F-1-2-




Question Number 225610 by Lara2440 last updated on 06/Nov/25
prove  Gauss curvature K is intrinsic by showing  K=(( determinant (((−(1/2)E_(vv) +F_(uv) −G_(uu) ),((1/2)E_u ),(F_u −(1/2)E_v )),((            F_v −(1/2)G_u ),(   E),(      F)),((                (1/2)G_v ),(   F),(     G)))− determinant (((    0),((1/2)E_v ),((1/2)G_u )),(((1/2)E_v ),(    E),(    F)),(((1/2)G_v ),(     F),(   G))))/((EG−F^( 2) )^2 ))  E,F,G is First Fundametal form of metric tensor.
$$\mathrm{prove} \\ $$$$\mathrm{Gauss}\:\mathrm{curvature}\:{K}\:\mathrm{is}\:\mathrm{intrinsic}\:\mathrm{by}\:\mathrm{showing} \\ $$$${K}=\frac{\begin{vmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}{E}_{{vv}} +{F}_{{uv}} −{G}_{{uu}} }&{\frac{\mathrm{1}}{\mathrm{2}}{E}_{{u}} }&{{F}_{{u}} −\frac{\mathrm{1}}{\mathrm{2}}{E}_{{v}} }\\{\:\:\:\:\:\:\:\:\:\:\:\:{F}_{{v}} −\frac{\mathrm{1}}{\mathrm{2}}{G}_{{u}} }&{\:\:\:{E}}&{\:\:\:\:\:\:{F}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{G}_{{v}} }&{\:\:\:{F}}&{\:\:\:\:\:{G}}\end{vmatrix}−\begin{vmatrix}{\:\:\:\:\mathrm{0}}&{\frac{\mathrm{1}}{\mathrm{2}}{E}_{{v}} }&{\frac{\mathrm{1}}{\mathrm{2}}{G}_{{u}} }\\{\frac{\mathrm{1}}{\mathrm{2}}{E}_{{v}} }&{\:\:\:\:{E}}&{\:\:\:\:{F}}\\{\frac{\mathrm{1}}{\mathrm{2}}{G}_{{v}} }&{\:\:\:\:\:{F}}&{\:\:\:{G}}\end{vmatrix}}{\left({EG}−{F}^{\:\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${E},{F},{G}\:\mathrm{is}\:\mathrm{First}\:\mathrm{Fundametal}\:\mathrm{form}\:\mathrm{of}\:\mathrm{metric}\:\mathrm{tensor}. \\ $$

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