6
$\begingroup$

Is the any difference between shape operator and second fundamental form for surfaces?

$\endgroup$
10
$\begingroup$

For the sake of clarity I will restrict my answer to the special case of (regular, embedded) surfaces in $\mathbb R^3$. The difference between shape operator and second fundamental form is that the first is an operator, while the second is a bilinear form. Specifically, if $n$ is the outward normal to your surface $S \subset \mathbb R^3$, we may define the shape operator $S$ by the rule $$S(X) = D_X n$$ while the second fundamental form is given by $$II(X,Y) = D_X n \cdot Y$$ Here $D_x n$ is the directional derivative of $n$ in the direction $X$. Note that some authors use a negative sign in the definition of shape operator, and may also use the same name for both objects. If you know about tensors and type change, then the second fundamental form is what you get when you change the type of the shape operator from $(1,1)$ to $(0,2)$.

If you are interested in coordinate formulas and/or other applications, I would recommend Ted Shifrin's book, available on his website: "Differential Geometry: A first Course in Curves and Surfaces".

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think you left out a - in the shape operator. Shape operator is negative of the derivative of the normal. $\endgroup$ – thang Apr 7 '15 at 10:36
  • $\begingroup$ @thang It is a matter of convention, as I mentioned in my answer. $\endgroup$ – treble Apr 7 '15 at 18:18
  • $\begingroup$ oh sorry. i didn't see that part of the text. I have a bad habit of just looking at formulas and numbers... and ignoring everything else. $\endgroup$ – thang Apr 7 '15 at 23:35
  • $\begingroup$ I beg your pardon, let us be precise. Let $f:U\to\mathbb R^3$ be an immersed surface. Let $x$ be a vector field on $U$. Given a Gauss map $N$, we may define the shape operator $S$ by $dN(x)=df(S(x))$. Is this shape operator the one from your answer? $\endgroup$ – superAnnoyingUser May 25 '15 at 10:46
  • $\begingroup$ Using your notation can we show that $D_xN=dN(x)$? $\endgroup$ – superAnnoyingUser May 25 '15 at 12:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy