enter image description here

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?

  • 1
    $\begingroup$ The title of your question says “nowhere differentiable”, but it should be “nowhere-zero differentiable”. $\endgroup$ – Hans Lundmark May 19 '18 at 9:32

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.