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I encountered the below quote defining the gaussian curvature not mentioning normals in the context of unfamiliar differential forms formalism. How does one reconcile this with the definition of the gaussian curvature that I am familiar with which mentions normals which is $$K = \det dN = \det\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$$ where $$N_u=a_{11}\mathbf{x}_u + a_{21}\mathbf{x}_v$$ $$N_v=a_{12}\mathbf{x}_u + a_{22}\mathbf{x}_v$$

Let $S$ be a surface in $R^3$. A frame field on $S$ is a choice of vector fields $\{\mathbf{E_1,E_2}\}$ such that at each point $p$ of $S$, $\mathbf{E_1}(p)$ and $\mathbf{E_2}(p)$ form an orthonormal basis for $T_p S$.

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We define $\Omega(V)=(\nabla_V \mathbf{E_1})\cdot \mathbf{E_2}$. Since $(\nabla_V \mathbf{E_1})\cdot \mathbf{E_2}=d\mathbf{E_1}(V)\cdot\mathbf{E_2} = d\mathbf{E_1}\cdot \mathbf{E_2}(V)$, it is sometimes more convenient to write $$\Omega=d\mathbf{E_1}\cdot \mathbf{E_2}$$ We now define a numerical measure of the curvature of $S$ at each point, which does not depend on a choice of tangent vector at that point.

Definition 4. The gaussian curvature at each point of $S$ is defined to be the number $$K=-d\Omega(\mathbf{E_1},\mathbf{E_2})$$

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The answer to your question is the Mainardi-Codazzi equation. Letting $\mathbf E_3$ be the unit normal, write $$\omega_{ij} = d\mathbf E_i\cdot \mathbf E_j \quad\text{for } i,j=1,2,3.$$ Note that your $\Omega = \omega_{12}$. Note also that $\omega_{ij}=-\omega_{ji}$ because $\mathbf E_i\cdot\mathbf E_j$ is a constant function.

Note that $d\mathbf E_i = \sum \omega_{ij}\mathbf E_j$, so, differentiating again, we have $$0 = \sum d\omega_{ij}\mathbf E_j - \sum\omega_{ij}\wedge d\mathbf E_j = \sum d\omega_{ik}\mathbf E_k - \sum\omega_{ij}\wedge\omega_{jk}\mathbf E_k.$$ Since $\{\mathbf E_k\}$ is linearly independent, we deduce that $$d\omega_{ik} = \sum\omega_{ij}\wedge\omega_{jk}.$$ In particular, \begin{equation}d\omega_{12} = \omega_{13}\wedge\omega_{32} = -\omega_{13}\wedge\omega_{23}. \tag{$\star$} \end{equation} (Equation ($\star$) is one formulation of the Mainardi-Codazzi equation.) Writing \begin{align*} \omega_{13} &= h_{11}\omega_1 + h_{12}\omega_2 \\ \omega_{23} &= h_{21}\omega_1 + h_{22}\omega_2, \end{align*} we see that $\omega_{13}\wedge\omega_{23} = (h_{11}h_{22}-h_{12}h_{21})\omega_1\wedge\omega_2$. The matrix $\big[h_{ij}\big]$ is the matrix of $-dN$ with respect to the basis $\{\mathbf E_1,\mathbf E_2\}$. Since the determinant of a linear map is independent of the basis one uses to express its matrix, we deduce that $$d\omega_{12} = -\det(dN)\omega_1\wedge\omega_2.$$ Evaluating on $\mathbf E_1,\mathbf E_2$, we see that, in fact, $K = \det(dN)$.

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  • $\begingroup$ I don't understand your notation. What is $\omega_1$ with only one subscript? And what does $dE_i = \sum \omega_{ij}E_j$ mean given that the left side is a vector of one-forms and on the left side $\omega_{ij}$ is a one form and $E_j$ is a vector? $\endgroup$ – user782220 Sep 30 '18 at 22:36
  • $\begingroup$ $\omega_1,\omega_2$ are the dual basis to $\mathbf E_1,\mathbf E_2$. I was assuming you were reading O'Neill, who does the moving frames approach methodically. To understand my notation better, take a look at pp. 101-106 of my differential geometry text. $\endgroup$ – Ted Shifrin Sep 30 '18 at 22:46

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