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The shape operator at acts on the tangent space of a manifold at $p$:

$S_p: T_pM \rightarrow T_pM$

By: $$S_p(v) = D_v(n(p))$$

That is, it sends the normal vector at $p$ to it's directional derivative in the direction of the input vector. The proof that this map is symmetric follows easily from the fact that partial derivatives commute, but I was hoping that somebody here could tell me why the geometric reason why this map is symmetric w.r.t. an inner product on the tangent space:

$$S_p(V) \cdot U = V \cdot S_p(U)$$

The eigenvalues of this map are the principle curvatures and the direction of the eigenvectors are the principle directions.

HeroZhang001
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PhysicsIsHard
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    I don’t really have a good answer, but the reason why the second fundamental form is symmetric (and hence why the shape operator is self-adjoint) is because the connection being used (Levi-Civita) is torsion-free. Btw, it’s not quite what you said but having a torsion-free connection is equivalent to the second covariant derivatives of functions being symmetric (i.e Hessian of functions being symmetric), not just mixed partials being equal. Anyway, I don’t have a very good geometric explanation to tie it all together to explain why the vanishing of torsion implies symmetry of the 2nd FF. – peek-a-boo May 11 '24 at 03:13
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    partly because legitimately visualizing torsion itself is a challenge for me… – peek-a-boo May 11 '24 at 03:14
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    Perhaps a geometric explanation comes from considering the osculating quadric surface to the surface at the point. So can we see it “geometrically” in that case? – Ted Shifrin May 11 '24 at 03:20
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    Here is a way to describe @TedShifrin's answer: You can assume that $p$ is the origin and $T_pM$ is the hyperplane $x^n=0$. In that case, $M$ near $p$ is the graph of a function $f$, i.e., $$ M = { (x^1, \dots, x^{n-1}, f(x^1, \dots, x^{n-1}) } \subset \mathbb{R}^n. $$ $M$ is well approximated at $p$ by the quadric hypersurface $$ x^n = \frac{1}{2}h_{ij}x^ix^j, $$ where $h_{ij} = \partial^2_{ij}f(0).$ That partials commute implies that $h_{ij}=h_{ji}$. The second fundamental form at $p$, with respect to the coordinates $(x^1,\dots, x^{n-1})$ is given by this symmetric matrix. – Deane May 11 '24 at 16:17
  • @TedShifrin Thank you so much for the great differential geometry notes! The exercises are so challenging, If you know of any solution guide available I'd appreciate it! Peace – PhysicsIsHard May 12 '24 at 13:48
  • No solutions available, but you’ll find a number of the more challenging questions discussed on this very site. (There are plenty of standard questions in the notes, too, you know. And I assume you know that there are some answers/solutions at the back.) – Ted Shifrin May 12 '24 at 15:26

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