The Third Fundamental Form and How to Find It

In this post we relate the third fundamental form of a surface to the first and second fundamental forms, using nothing more than the Cayley–Hamilton theorem.

The shape operator and its characteristic polynomial

At each point \(x\) of a surface, the tangent space is an inner product space. The shape operator (or Weingarten map) \(S_x\) is the linear map on the tangent space defined by

\[(S_x v, w) = -(d\mathbf{n}(v), w),\]

where \(\mathbf{n}\) is the unit normal. (Some authors drop the minus sign; it only changes the sign of \(H\).)

In an orthonormal basis of the tangent space, \(S\) is represented by a symmetric matrix, which we write as

\[S = \begin{pmatrix} L & M \\ M & N \end{pmatrix}.\]

(In a general coordinate basis \(S = \mathrm{I}^{-1}\mathrm{II}\), where \(\mathrm{I}\) and \(\mathrm{II}\) are the matrices of the first and second fundamental forms. The orthonormal basis keeps the algebra simple.)

The characteristic polynomial of a matrix \(A\) is \(p_A(\lambda) = \det(A - \lambda I)\). For the shape operator,

\[p_S(\lambda) = \det \begin{pmatrix} L - \lambda & M \\ M & N - \lambda \end{pmatrix} = \lambda^2 - (L + N)\lambda + (LN - M^2).\]

The trace of \(S\) is the sum of the principal curvatures, \(2H\), where \(H\) is the mean curvature, and its determinant is their product, the Gaussian curvature \(K\). Therefore

\[p_S(\lambda)=\lambda^2-2H \lambda +K .\]

Cayley–Hamilton and the third fundamental form

By the Cayley–Hamilton theorem, \(S\) satisfies its own characteristic polynomial:

\[S^2 - 2H S + K\,\mathrm{Id} = 0 .\]

Apply this to a tangent vector \(v\) and take the inner product with another tangent vector \(w\):

\[\begin{aligned} S^2(v) - 2H\,S(v) + K v &= 0, \\ S^2(v) \cdot w - 2H\, S(v) \cdot w + K\, v \cdot w &= 0 . \end{aligned}\]

In terms of the fundamental forms,

\[\mathrm{I}(v, w) = v \cdot w, \qquad \mathrm{II}(v, w) = S(v) \cdot w, \qquad \mathrm{III}(v, w) = S(v) \cdot S(w) = S^2(v) \cdot w,\]

this becomes

\[\boxed{\mathrm{III} - 2H\, \mathrm{II} + K\,\mathrm{I} = 0 .}\]

The third fundamental form is therefore determined by the first two, which is why it is rarely used.

Exercises

• Show that the shape operator is self-adjoint.
• Hence show that \(S^2(v) \cdot w = S(v) \cdot S(w)\). (Hint: \((S v, w) = (v, S^* w)\).)
• How does the shape operator encode the curvature of the surface? (Hint: eigenvalues and eigenvectors.)
• State and prove the Cayley–Hamilton theorem, and check it on an explicit \(2 \times 2\) matrix.
• Why is the third fundamental form not very useful in practice?

Changes on 25 October 2026

What was wrong. The post wrote the shape operator as the matrix of the second fundamental form without saying that this only holds in an orthonormal basis. In a general coordinate basis it is \(\mathrm{I}^{-1}\mathrm{II}\).

What changed. The assumption is now stated, together with the sign convention.