Extremal curves and surfaces in symmetric tensor fields

Abstract

The visualization of symmetric second-order tensor fields in two or three dimensions is still a challenging task, particularly if global structures of the data are desired. One approach is tensor field topology which provides structures characterizing the behavior of the eigenvector fields. Another widely used approach is analyzing tensor fields by means of scalar invariants, i.e., quantities invariant with respect to changes of the coordinate system. In this case, the selection of the relevant invariants might be difficult. Thus, we propose an approach which analyzes the complete invariant part of the tensor. We define extremal points for tensor fields in a mathematically rigorous way, which form curves for two-dimensional and surfaces for three-dimensional tensor fields. We propose a way to compute extremal curves or surfaces from a suitable set of two or three invariants, respectively. We also show that commonly used sets of invariants lead to the same extremal points. Consequently, extremal points contain minima and maxima of most invariants used in tensor field analysis and they are linked to the tensor field topology by containing the degenerate points. Moreover, we show that each extremal point is an extremum or a saddle of a certain invariant. The method is demonstrated on synthetic datasets as well as on stress tensor fields from structure simulations.

Appendix

The rows of the Jacobian matrices of $\bar{I}$, $\bar{K}$, $\bar{R}$ are the gradients of the respective invariant functions, i.e.,

$$\begin{aligned}&J\bar{I} = \left[ \begin{array}{ccc} \nabla \bar{I}_1&\nabla \bar{I}_2&\nabla \bar{I}_3 \end{array} \right] ^\top , \\&J\bar{K} = \left[ \begin{array}{ccc} \nabla \bar{K}_1&\nabla \bar{K}_2&\nabla \bar{K}_3 \end{array} \right] ^\top ,\\&J\bar{R} = \left[ \begin{array}{ccc} \nabla \bar{R}_1&\nabla \bar{R}_2&\nabla \bar{R}_3 \end{array} \right] ^\top . \end{aligned}$$

The gradients of the invariant functions depending on the eigenvalues $\lambda _1, \lambda _2, \lambda _3$ are given by

$$\begin{aligned} \nabla \bar{I}_1= & {} \nabla \bar{K}_1 = \left[ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right] , \quad \nabla \bar{I}_2 = \left[ \begin{array}{c} \lambda _2 + \lambda _3 \\ \lambda _1 + \lambda _3 \\ \lambda _1 + \lambda _2 \end{array} \right] ,\\ \nabla \bar{I}_3= & {} \left[ \begin{array}{c} \lambda _2 \lambda _3 \\ \lambda _1 \lambda _3 \\ \lambda _1 \lambda _2 \\ \end{array} \right] , \quad \nabla \bar{K}_2 = \left[ \begin{array}{c} \frac{1}{ \Vert \tilde{T} \Vert } \tilde{\lambda }_1 \\ \frac{1}{ \Vert \tilde{T} \Vert } \tilde{\lambda }_2 \\ \frac{1}{ \Vert \tilde{T} \Vert } \tilde{\lambda }_3 \\ \end{array} \right] ,\\ \nabla \bar{R}_2= & {} \left[ \begin{array}{c} \frac{1}{\sqrt{6}} \Vert T \Vert ^{-3} \Vert \tilde{T} \Vert ^{-1} ( {{\mathrm{tr}}}T ) (\lambda _1 \lambda _2 + \lambda _1 \lambda _3 - \lambda _2^2 - \lambda _3^3 ) \\ \frac{1}{\sqrt{6}} \Vert T \Vert ^{-3} \Vert \tilde{T} \Vert ^{-1} ( {{\mathrm{tr}}}T ) (\lambda _2 \lambda _1 + \lambda _2 \lambda _3 - \lambda _1^2 - \lambda _3^3 ) \\ \frac{1}{\sqrt{6}} \Vert T \Vert ^{-3} \Vert \tilde{T} \Vert ^{-1} ( {{\mathrm{tr}}}T ) (\lambda _3 \lambda _1 + \lambda _3 \lambda _2 - \lambda _1^2 - \lambda _2^3 ) \end{array} \right] ,\\ \nabla \bar{K}_3= & {} \nabla \bar{R}_3 = \left[ \begin{array}{c} \sqrt{6} \Vert \tilde{T} \Vert ^{-5} (\lambda _2 - \lambda _3)^2(\lambda _1 - \lambda _2)(\lambda _1 - \lambda _3) \\ \sqrt{6} \Vert \tilde{T} \Vert ^{-5} (\lambda _1 - \lambda _3)^2(\lambda _2 - \lambda _1)(\lambda _2 - \lambda _3) \\ \sqrt{6} \Vert \tilde{T} \Vert ^{-5} (\lambda _1 - \lambda _2)^2(\lambda _3 - \lambda _1)(\lambda _3 - \lambda _2) \end{array} \right] . \end{aligned}$$