# Tangent Map of an Isometry and the Shape Operator

I have spent a lot of time on this problem and would appreciate some help. Please bear with me.

Let $$M$$ be a surface and $$F\colon\mathbb{R}^3\rightarrow\mathbb{R}^3$$ an isometry. Denote with $$F_\ast$$ the tangent map of $$F$$. Let $$U$$ be a unit normal vector field on $$\mathcal{V}\subset M$$ and let $$\overline{U}=F_\ast U$$ be the unit normal vector field on $$F(\mathcal{V})$$. Let $$S$$ be the shape operator on $$M$$ determined by $$U$$ and $$\overline{S}$$ the shape operator of $$F(M)$$ determined by $$\overline{U}$$. Let $$p\in M$$ and $$v\in T_pM$$. Show that $$F_\ast S_p(v_p)=\overline{S}_{F(p)}(F_\ast v_p).$$

Here is my attempt. First the left-hand side: the shape operator $$S$$ acting on a tangent vector $$v_p$$ is the covariant derivative of $$U$$ at $$p$$ in the direction of $$v$$. So if $$\alpha$$ is a curve such that $$\alpha(0)=p$$ and $$\alpha'(0)=v_p$$ then $$S_p(v_p) =-\nabla_{v_p}U = \frac{d}{dt}U(\alpha(t))\big|_{t=0}.$$ If we call $$U'\colon\mathbb{R}^3\rightarrow\mathbb{R}^3$$ the map that sends points to the vector part of $$U$$ (that is, ignoring the base point) then $$S_p(v_p) = -DU'(\alpha(0))\alpha'(0) ~\text{ at } p =J_{U'}(p)v ~\text{ at } p,$$ where $$DU'$$ is the derivative of $$U'$$ which is also the Jacobian matrix $$J_{U'}$$. Next, the tangent map $$F_\ast$$ sends tangent vectors $$v_p$$ to tangent vectors $$J_F(p)v$$ at $$F(p)$$. Then $$F_\ast S_p(v_p) = F_\ast\left(-J_{U'}(p)v ~\text{ at } p\right) = -J_F(p)J_{U'}(p)v ~\text{ at } F(p).$$ Now for the right-hand side: the shape operator $$\overline{S}$$ acting on a tangent vector $$w_q$$ is the covariant derivative of $$\overline{U}$$ at $$q$$ in the direction of $$w$$. Following our steps above, $$\overline{S}_q(w_q) = -D\overline{U}'(q)w ~\text{ at } q.$$ Since $$F_\ast v_p$$ is $$J_F(p)v$$ at $$F(p)$$, we have $$\overline{S}_{F(p)}(F_\ast v_p) = -D\overline{U}'(F(p))J_F(p)v ~\text{ at } F(p).$$ But what is $$\overline{U}'$$? By definition $$\overline{U}=F_\ast U$$, so at every point $$p\in M$$, $$\overline{U}(p)=F_\ast U(p) = J_F(p)U'(p)$$ at $$F(p)$$. So we can say $$F_\ast U$$ is the function $$p\mapsto J_F(p)U'(p)$$. Then $$\overline{S}_{F(p)}(F_\ast v_p) = -D\big(J_FU'\big)(F(p))J_F(p)v ~\text{ at } F(p).$$ From here, I don't know how to relate the two sides to each other. Part of my trouble is all the varying notation, which may have caused a mistake above. I would appreciate any guidance.