I'd like to understand a bit more the following problem.

Suppose a potato shape (say a 3D volume bounded by a 2D surface), and define $n$ the normal vector to its surface, with components $n_i$. Define the object

$$\left\langle \dfrac{n_{i}n_{j}}{1+\alpha_{sr}n_{s}n_{r}}\right\rangle =g_{ij}$$

where repeated indices are summed: for instance in 2D one has $\alpha_{sr}n_{s}n_{r}=\alpha_{11}n_{1}^{2}+\left(\alpha_{12}+\alpha_{21}\right)n_{1}n_{2}+\alpha_{22}n_{2}^{2}$. The notation $\left\langle \cdots\right\rangle $ defines the angular averaging over the surface of the potato: for instance in 2D it will be $\left\langle \cdots\right\rangle =\int_{0}^{2\pi}d\phi\left[\cdots\right]/2\pi$ provided the vector $n$ is $\phi$-dependent. I believe this defines a metric $g_{ij}$. Am I correct on this point ? So the problem should be a problem of differential geometry. Otherwise please make a comment, such that I can modify the question accordingly.

I'd like to know the condition on the matrix $\alpha_{ij}$ which guarantee that $g_{ij}$ is diagonal.

Any help is warm welcome, as well as any remark to improve this question.

A few comments: It seems the question is not trivial. For instance, supposing a circle as the potato, one has $n_{1}=\cos t$ and $n_{2}=\sin t$ only, and the generic $$\alpha=\left(\begin{array}{cc} \alpha_{11} & \alpha_{12}\\ \alpha_{21} & \alpha_{22} \end{array}\right)$$ and so in this particular example one has

$$g_{ij}=\int_{0}^{2\pi}\dfrac{dt}{2\pi}\left[\dfrac{n_{i}n_{j}}{1+\alpha_{11}\cos^{2}t+\left(\alpha_{12}+\alpha_{21}\right)\cos t\sin t+\alpha_{22}\sin^{2}t}\right]$$

or in matrix form

$$g=\int_{0}^{2\pi}\dfrac{dt}{2\pi}\dfrac{\left(\begin{array}{cc} \cos^{2} t & \cos t\sin t\\ \cos t\sin t & \sin^{2} t \end{array}\right)}{1+\alpha_{11}\cos^{2}t+\left(\alpha_{12}+\alpha_{21}\right)\cos t\sin t+\alpha_{22}\sin^{2}t}$$

which I can not solve for generic $\alpha_{ij}$. It nevertheless seems to me that $\alpha$ needs to be at least antisymmetric to impose a diagonal $g_{ij}$. For instance, suppose $\alpha_{11}=\alpha_{22}$ and $\alpha_{12}=-\alpha_{21}$ one has clearly

$$\left\langle \dfrac{n_{i}n_{j}}{1+\alpha_{11}}\right\rangle =\dfrac{\left\langle n_{i}n_{j}\right\rangle }{1+\alpha_{11}}=\dfrac{\int_{0}^{2\pi}\dfrac{n_{i}n_{j}}{2\pi}dt}{1+\alpha_{11}}=\dfrac{\delta_{ij}}{1+\alpha_{11}}$$

In contrary, one has for instance $$\alpha=\left(\begin{array}{cc} 1 & 1\\ 1 & 1 \end{array}\right)\Rightarrow g=\dfrac{1}{6}\left(\begin{array}{cc} \sqrt{3} & 3-2\sqrt{3}\\ 3-2\sqrt{3} & \sqrt{3} \end{array}\right)$$

Clearly, $\alpha$ being an antisymmetric matrix is a sufficient condition (actually, $\alpha_{22}$ and $\alpha_{11}$ needs to be related as well, since not all values are allowed for these two parameters), but is it necessary ? So I'd like to know if this problem is already known, and eventually solved.

  • $\begingroup$ I'm having trouble understanding your notation. Some questions: 1. Is a "potato" a region in space bounded by a smooth, closed surface? 2. What do the indices on the normal vector signify? (You write "$n_{i}$ is the normal vector", but then refer to $n_{j}$, $n_{r}$, and $n_{s}$.) 3. Are you summing over $r$ and $s$ in the denominator? 4. Does angular averaging refer to integrating something (what...?) over the surface of the potato? Thank you for any clarification. :) $\endgroup$ May 22, 2015 at 12:21
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    $\begingroup$ @user86418 Yes to all your question : the normal vector to the surface bounding the potato-shape is $n$, its components are $n_i$. I call it a potato since I was explicitly thinking of a 3D object. The angular averaging is over the angle-parametrizing the surface. Summation over repeated indices is implied. Isn't the example a bit clarifying ? Anyways, I'll put complements to the definition, you're right. Thanks for your comment. $\endgroup$
    – FraSchelle
    May 22, 2015 at 12:29
  • $\begingroup$ @user86418 More details added. Thanks again for your remarks. $\endgroup$
    – FraSchelle
    May 22, 2015 at 12:41
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    $\begingroup$ Thank you for the extra details. :) I now realize I'm confused about deeper issues: If you average over the potato, don't you just get a single matrix? (I was initially thinking you were hoping to construct a metric on the potato itself.) In any case, it looks as if conditions on $\alpha_{ij}$ that guarantee diagonality will depend on the specific shape. (For the circle, incidentally, $\alpha_{21} = -\alpha_{12}$ suffices, because if $f$ is continuous, then $\int_{0}^{2\pi} f(\cos\theta) \sin\theta\, d\theta = 0$.) $\endgroup$ May 22, 2015 at 18:31
  • $\begingroup$ @user86418 I've difficulty to understand your previous remark. Do you mean $g$ is by no means a metric, and it's just a matrix ? It's clear it's a matrix, a few questions were: i) may I assume $g$ to be a metric ?, and ii) if so, does it help me to resolve the problem ? Thanks for showing the antisymmetry of $\alpha$ in 2D is a sufficient condition. Is it necessary ? I'm not sure that the shape of the potato determine $\alpha$, which in turns determine $g$. I would naively expect that the topology of the potato lead to a class of $\alpha$, which class should helps calculating $g$. Thank you $\endgroup$
    – FraSchelle
    May 26, 2015 at 1:51


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