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I was doing some computation to find the principal curvatures on a helicoid and came accross the apparent contradiction that the eigenvectors I calculated appeared to be non-orthogonal. This problem is dealt with in this question here : Showing directly that the principal directions are going to be orthogonal.

The accepted answer shows that, even though the calculated eigenvectors may not seem be orthogonal intrinsically, they are indeed orthogonal when taking into account the metric form $G$.

However, this answer does not deal with the apparent contradiction between the two following facts

  1. The Weingarten operator is symmetric.
  2. This operator should be represented by the matrix $G^{-1}H$, which is not symmetric.

My question is thus the following : is the second claim false ? How do we solve this apparent contradiction.

ged
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  • To see the orthogonality more directly, you should write the eigenvectors as $3$-dimensional vectors that happen to be tangent to the surface. I also recommend avoiding using coordinates in this situation. Instead, fix an orthonormal basis of the tangent space and write the first and second fundamental form with respect to that basis. – Deane May 05 '23 at 14:56
  • So a symmetric operator needs not to be represented by a symmetric matrix in every base ? I think I see, it should be symmetric only in an orthonormal basis, right ? – ged May 05 '23 at 16:30
  • No, it should be symmetric with respect to any basis. But if the two eigenvalues are different, then the eigenvectors are orthogonal – Deane May 05 '23 at 17:13
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    No, the matrix is only symmetric, in general, when you write the linear map as a matrix with respect to an orthogonal basis where the vectors have the same length. – Ted Shifrin May 05 '23 at 17:21
  • Sorry, I misspoke. The matrix associated with the Weingarten operator need not be symmetric. – Deane May 05 '23 at 19:35

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