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[SOLVED] Normal Curvature function

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Poirot

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Feb 15, 2012
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Let a regular surface M have shape operator $\begin{bmatrix}-2 & 3\\3 & 6\end{bmatrix}$
In each case find the tangent vector v or explain why it doesn't exist.

(i) a unit vector v such that K(v)=-3 where K is normal curvature function
(ii)unit vector v such that K(v)=10
(iii) a vector v (possibly non unit) such tha K(v)=12

for (i) and (ii) I found a quartic equation and found for (i) there is a solution but in (ii) the solutions were complex. My problem is this was quite hard work, plus for (iii) I would not have the condition that the sum of the squares of the components of v is 1, thus I would have one equation in 2 unknowns. So what to do?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Let a regular surface M have shape operator $\begin{bmatrix}-2 & 3\\3 & 6\end{bmatrix}$
In each case find the tangent vector v or explain why it doesn't exist.

(i) a unit vector v such that K(v)=-3 where K is normal curvature function
(ii)unit vector v such that K(v)=10
(iii) a vector v (possibly non unit) such tha K(v)=12

for (i) and (ii) I found a quartic equation and found for (i) there is a solution but in (ii) the solutions were complex. My problem is this was quite hard work, plus for (iii) I would not have the condition that the sum of the squares of the components of v is 1, thus I would have one equation in 2 unknowns. So what to do?
Disclaimer: I never studied differential geometry, so I may have this all wrong.

As far as I know, the normal curvature function is defined in terms of the shape operator $A$ by the inner product $K(\mathbf{v}) = \langle A\mathbf{v},\mathbf{v}\rangle.$ For a unit vector $\mathbf{v}$, the maximal and minimal values of $K(\mathbf{v})$ are the maximal and minimal principal curvatures, and these in turn are the eigenvalues of $A$.

For the matrix $A = \begin{bmatrix}-2 & 3\\3 & 6\end{bmatrix}$, the characteristic equation is $\lambda^2 - 4\lambda -21=0.$ The eigenvalues are $7$ and $-3$, and the corresponding unit eigenvectors are $$\mathbf{v_{\max}} = \frac1{\sqrt{10}}\begin{bmatrix}1\\3\end{bmatrix},\qquad \mathbf{v_{\min}} = \frac1{\sqrt{10}}\begin{bmatrix}3\\-1\end{bmatrix}.$$ Thus the answer to (i) is $\mathbf{v_{\min}}$, with $K(\mathbf{v_{\min}}) = -3.$ The answer to (ii) is that there is no unit vector $\mathbf{v}$ with $K(\mathbf{v}) = 10$ because the maximum value of $K(\mathbf{v})$ is 7.

For (iii), you are right that the answer will not be unique once the condition that $\mathbf{v}$ should be a unit vector is dropped. One solution would be to use the fact that $K(\alpha\mathbf{v}) = \langle A(\alpha\mathbf{v}),\alpha\mathbf{v}\rangle = \alpha^2\langle A\mathbf{v},\mathbf{v}\rangle = \alpha^2K(\mathbf{v}) $ (for a scalar $\alpha$). You know that $K(\mathbf{v_{\max}}) = 7$, so a suitable multiple of $\mathbf{v_{\max}}$ will give you an answer.
 
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Poirot

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Feb 15, 2012
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Thanks