First a little bit of context: in engineering, the finite element method (FEM) is commonly used to find approximate solutions of PDEs. It requires a mesh of the domain of interest, usually into triangles/quadrangles in two dimensions or tetrahedra/hexahedra in three dimensions. In the following, the domain I'm interested in is in $\mathbb{R}^2$. Adaptive meshing aims at adapting the mesh based on some error indicator obtained from a first computed solution. For PDEs with a pronounced anisotropic behaviour, anisotropic adaptive meshing aims at generating distorted triangulations to better capture the physics (e.g. shocks in gas dynamics). To do this, a Riemannian metric, distinct from the canonical Euclidean metric, is defined on $\mathbb{R}^2$ based on an error indicator. Then a triangulation is constructed with the goal of creating triangles close to an equilateral triangle when edges length are measured in this metric, thus the triangulation is stretched with respect to the Euclidean metric. We usually abuse the notation and simply call "metric" the matrix representation of the metric tensor, thus a symmetric and positive-definite 2x2 matrix. This matrix $M$ has a diagonalization of the form $M = R\Lambda R^T$.
To define the metric tensor, I consider its components in the canonical basis of $\mathbb{R}^2$ through the following diagonalized form: \begin{equation} M = (\boldsymbol{v}_1 ~~\boldsymbol{v}_2) \begin{pmatrix} \lambda_1 & 0\\0 & \lambda_2 \end{pmatrix}(\boldsymbol{v}_1 ~~\boldsymbol{v}_2)^T, \end{equation} where for any $x \in \mathbb{R}^2$ the $\boldsymbol{v}_i(x)$ are orthonormal vectors and $\lambda_i(x) > 0$. The eigenvalues are related to the Euclidean sizes $h_i$ of the stretched triangles by the relation $\lambda_i = h_i^{-2}$. It is my understanding however that we cannot talk about eigenvectors/eigenvalues of the metric tensor, see e.g. here: https://physics.stackexchange.com/questions/382390/eigenvectors-and-eigenvalues-of-a-metric
I have two distinct, but related questions:
- Are the eigenvectors $\boldsymbol{v}_i$ and eigenvalues $\lambda_i$ of the matrix $M$ of intrinsic significance? Again, my understanding is "no", because they're just eigenvectors and eigenvalues of some representation of the metric tensor, in some basis, and only the sign of the eigenvalues is invariant. Does it change anything that this is in the canonical basis?
- For a given eigenvector, say $\boldsymbol{v}_1$, I consider the curve $x(t)$ such that $x(0) = x_0$ and it is everywhere tangent to $h_1\boldsymbol{v}_1 = \lambda_1^{-\frac{1}{2}}\boldsymbol{v}_1$. By construction, it has unit norm $\Vert \lambda_1^{-\frac{1}{2}}\boldsymbol{v}_1 \Vert = (\lambda_1^{-\frac{1}{2}}\boldsymbol{v}_1)^T M (\lambda_1^{-\frac{1}{2}}\boldsymbol{v}_1) = 1$. Is this curve geodesic with respect to the metric that I have defined? To answer this, I tried the following:
- Find basis vectors $\boldsymbol{g}_1, \boldsymbol{g}_2$ such that $M_{ij} = \boldsymbol{g}_i \cdot \boldsymbol{g}_j$ and write the tangent vector in this basis as $\lambda_1^{-\frac{1}{2}}\boldsymbol{v}_1 = V^1\boldsymbol{g}_1 + V^2\boldsymbol{g}_2$
- Write the second derivative $x''(t) = \frac{d}{dt}(V^1\boldsymbol{g}_1 + V^2\boldsymbol{g}_2)$ using the chain rule, to make the Christoffel symbols appear in the derivatives of the basis vectors. I end up with \begin{equation} \left( \frac{dV^k}{dt} + \Gamma^k_{ji}V^j\frac{dx^i}{dt} \right) \boldsymbol{g}_k, \end{equation} which, if I'm not mistaken, is the covariant derivative of $\lambda_1^{-\frac{1}{2}}\boldsymbol{v}_1$ along the curve, and what is expected when computing $x''(t)$.
- Then I think I need to show that this quantity is 0 if the curve is geodesic, but I'm not sure how to proceed. I tried computing the derivative $\frac{dV^k}{dt}$, but this involves derivatives of eigenvectors, and the only expression I know of involves the pseudo-inverse of $M - \lambda_kI$.
I have little background in differential geometry, so please feel free to point out if there are big nonsense or misunderstandings in what I said.
Thanks in advance!