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How can I solve this problem? I know that the curvature $K$ is given by $\det(dN_p)$ for any point $p$ in the surface. Here I will write capital $V$'s for the vectors in the problem statement. We should show that $$\langle d(fN)(V_1)\times d(fN)(V_2),fN\rangle=f^3\det(dN_p)$$I know that $$V_1=\alpha_1'(s)=x_uu_1'(s)+x_vv_1'(s)$$ $$V_2=\alpha_2'(s)=x_uu_2'(s)+x_vv_2'(s)$$ for curves $\alpha_1,\alpha_2$ mapping into $V$ from $R^2$. Therefore I think I can write $dfN(V_i)$ which I assume means $d(f \circ N)(V_i)$ as $$dfN(V_i)=dN(V_i)dfN(V_i)=(N_uu_i'(s)+N_vv_i'(s))dfN(V_i)$$ by the chain rule but obviously this doesn't make sense. I feel like I completely misunderstand something very fundamental, perhaps it should be $$dfN(V_i)=df(dN(V_i))=df(N_uu_i'(s)+N_vv_i'(s))$$ Furthermore I think that $N_uu_i'(s)+N_vv_i'(s)$ lies in the tangent plane and so $V_1,V_2$ are made up of the basis vectors $N_u,N_v$ so we can rewrite $$N_uu_i'(s)+N_vv_i'(s)=a_iV_1+b_iV_2$$ which would make sense given the hint in the book. Then can write $$dfN(V_i)=df(a_iV_1+b_iV_2)$$ assuming the second way I wrote the differential is correct. But this is probably completely wrong and I can't see how I could carry on from here anyway. Can anybody help me out?

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    $\begingroup$ The title of your question says “nowhere differentiable”, but it should be “nowhere-zero differentiable”. $\endgroup$ – Hans Lundmark May 19 '18 at 9:32
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If $f_i=df\ v_i,\ N_i=dN\ v_i$, then $(N,N_i)=0$ so that

$$(d(fN)v_1\wedge d(fN) v_2, fN) = ((f_1N+fN_1)\wedge (f_2N+fN_2),fN)=f^3 (N_1\wedge N_2, N) $$

Here $(N_1\wedge N_2, N)$ is a Gaussian curvature.

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  • $\begingroup$ From what I understand you used that $d(fN)(v_i)=df(v_i)N(v_i)+f(dN(v_i))$, but I don't understand this step, could you explain it? $\endgroup$ – Dan May 19 '18 at 16:39
  • $\begingroup$ $d$ is Leibniz then $d(fN)=(df)N+fdN$ $\endgroup$ – janmarqz May 19 '18 at 17:03

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