# Shape operator vs second fundamental form

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

## 1 Answer

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".

• I think you left out a - in the shape operator. Shape operator is negative of the derivative of the normal. Commented Apr 7, 2015 at 10:36
• @thang It is a matter of convention, as I mentioned in my answer. Commented Apr 7, 2015 at 18:18
• 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? Commented May 25, 2015 at 10:46
• Using your notation can we show that $D_xN=dN(x)$? Commented May 25, 2015 at 12:37
• @superAnnoyingUser, yes to your two first questions and no to the third one. The shape operator $S$ is a linear map defined in a tangent space of a surface into itself, and $N$ is a map that send to every point of the surface a unit normal vector. Maybe your are confusing with $dN_x$, but even this way they are not the same because although they both reflect the same data $dN_x$ is a map between two diferent vector spaces, to say, $T_x U$ and $T_{N(x)} \mathbb{S}^2 \equiv T_{f(x)} f(U)$ Commented Dec 10, 2021 at 7:15