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Why is the shape operator $A: \mathbb R^2 \to \mathbb R^2 $self adjoint regarding the first fundamental form? I see it is written as a fact everywhere but cannot find a proof for it.

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  • $\begingroup$ See Wolfgang Kuhnel's book on Differential Geometry (third edition), page 67. $\endgroup$ – Isomorphism Mar 7 '18 at 10:43
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I already found out myself that it is equivalent to the symmetry of the 2nd fundamental form.

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