Third Fundamental Form and How to Find it
In this post, we attempt to relate the Third Fundamental Form with the First and Second Fundamental Form .
Finding Characteristic Polynomial
At each point \(x\) of a surface, the tangent space is an inner product space.
\begin{align} (S_xv, w) = (dn(v), w) \end{align}
Now I will use without proof that \(S\) can be written as:
\[S = \begin{pmatrix} L & M \\ M & N \end{pmatrix}\]Now we attempt to find its Characteristic Polynomial. Recall that it is written as:
\[p_A(\lambda) = \det(A - \lambda I)\]For the Shape Operator, we have:
\[A - \lambda I = \begin{pmatrix} L & M \\ M & N \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} L - \lambda & M \\ M & N - \lambda \end{pmatrix}\]Now,
\[p_S(\lambda) = \det \begin{pmatrix} L - \lambda & M \\ M & N - \lambda \end{pmatrix}\]Resulting in
\[p_S(\lambda) = \lambda^2 - (L + N)\lambda + (LN - M^2)\]It can be easily shown that it is equivalent to:
\[p_S(\lambda)=\lambda^2-2H \lambda +(LN - M^2)\]Cayley-Hamilton Theorem and Simplifications
By the Cayley-Hamilton theorem, the shape operator \(S\) satisfies its own characteristic polynomial:
\[S^2 - 2H S + KI = 0\]Applying it to a vector and with some simplifications:
\[\begin{align} (S^2 - 2H S + KI)(v) &= 0 \\ S^2(v)-2HS(v)+K(v) &=0 \\ S^2(v) \cdot w - 2H S(v) \cdot w + K v \cdot w &= 0 \\ \mathrm{III}(v, w) - 2H \mathrm{II}(v, w) + K\mathrm{I}(v, w) &= 0 \end{align}\]Using:
\[I(v, w) = v \cdot w\] \[\mathrm{II}(v, w) = S(v) \cdot w\] \[\mathrm{III}(v, w) = S^2(v) \cdot w = S(v) \cdot S(w)\]Exercises for the Reader
- Show that the shape operator is self-adjoint.
- Hence show that \(S^2(v) \cdot w = S(v) \cdot S(w)\) for the Third Fundamental Form (Hint: \((S(v), w) = (v, S^*(w))\))
- How does the shape operator contain information about the curvature of the surface? (Hint: Eigenvalues)
- State and prove Cayley-Hamilton Formula and show it works using a matrix as an example.
- Why is the Third Fundamental Form not that useful?
