I am researching the spectrum of eigenvalues for the divergence operator on Riemannian manifolds and how they deform tensor fields. This is mainly motivated by trying to understand dynamical systems on more complicated geometries, and as a starting point, I am looking into the generalization of eigenfunctions which satisfy $D f = \lambda f$ where $D$ is a differential operator.

Let $(M^n,g)$ be an n-dimensional Riemannian manifold with metric $g$ and $T\in\mathcal{T}M$ a tangent vector field. Suppose the connection $\nabla$ is of Levi-Civita type. In a local orthonormal coordinates basis, the vector field may be written as $T = (f_1(x_1),...,f_n(x_n))$ where $\forall i$ $f_i(x_i)$ are smooth functions.

Suppose $\nabla\cdot T = \xi \cdot T$, where $\xi\in\mathcal{T}^*M$ is a one form and the $\cdot$ denotes an inner product. By $\xi$, we mean the collection of eigenvalues from the divergence operation, i.e., $$ \begin{align}\nabla\cdot T &= \left(\frac{\partial f_1(x_1)}{\partial x_1}+\cdots+\frac{\partial f_n(x_n)}{\partial x_n}\right) = (\xi_1 f_1(x_1)+\cdots+\xi_n f_n(x_n))\\ &= (\xi_1,\cdots,\xi_n)\cdot(f_1(x_1),\cdots,f_n(x_n)) = \xi\cdot T. \end{align}$$

Question 1: The standard ODE question is with regards to uniqueness and existence and the same applies here. Is it possible to solve the above first order linear differential equation using standard ODE arguments and find a solution in local coordinates such as $$\begin{equation} T = T_0\exp(\xi\cdot x) \end{equation}$$ where $T_0$ is an initial tensor field whose divergence is zero, i.e., $\nabla\cdot T_0 = 0$. That is $T_0$ satisfies a continuity equation. Here the vector $x$ is the local coordinate vector.

Question 2: How easily, if at all, can these statements be made coordinate independent?

Question 3: It is well known that a divergence of a vector field can be expressed as the Lie derivative of the metric along the vector field, i.e., $\nabla\cdot X = \frac{1}{2} \text{Tr}_g(\mathcal{L}_X g)$ For instance, see this math stackexchange. What can be said about the metric in the case? Specifically, can the eigenvalues $\xi_i$ provide information about the deformations of the metric or how the metric deforms the tensor field $T$?


1 Answer 1


It is a bit strange to speak of the eigenvalues of a map between two different vector spaces (here, the space of one forms and zero forms on $M$): I would call them singular values, where we seek orthonormal $\{b_i\}$ and $\{\beta_i\}$ for $\Omega^0(M)$ and $\Omega^1(M)$ so that $$\nabla \cdot v = \sum_{i} \sigma_i b_i \beta_i(v)$$ for some singular values $\sigma_i$.

Just as we can probe the singular values of a finite-dimensional rectangular matrix $A$ by examining the spectrum of $A^TA$, we can form the adjoint $-\nabla$ of divergence (since $\langle \nabla\cdot v, f\rangle = -\langle v, \nabla f\rangle$) and look at the spectrum of $-\nabla \cdot \nabla = -\Delta$. Let $f_i$ be the eigenfunctions of the Laplace-Beltrami operator $\Delta$ with $-\lambda_i$ the eigenvalues. Then the one-forms $\lambda_i^{-1/2} df_i$ are orthonormal, and we have $b_i = f_i, \sigma_i = \lambda_i^{1/2}$, and $\beta_i = \lambda_i^{-1/2} df_i$. In your notation, then $\xi= f_idf_i$ and $T=\nabla f_i$.

I don't know about your last question: on the one hand, the singular values are global properties of the divergence operator, so it's not clear that they will give any local information about the relationship between $g$ and $T$. Of course, $\sigma_i$ are just the square roots of the eigenvalues of $-\Delta$, and the spectrum of the Laplace-Beltrami operator is known to be intimately related to the geometry of $M$.

  • $\begingroup$ Could you clarify what you mean by $\langle\nabla\cdot v,f\rangle = -\langle v,\nabla f\rangle?$ In the case that the notation means to integrate $(\nabla\cdot v) f$ over $\mathbb{R}$, what boundary conditions are you assuming? $\endgroup$
    – Dhuality
    Jan 16, 2019 at 19:33
  • $\begingroup$ @Aditya I was taking $M$ to be a compact manifold without boundary; with boundary you need e.g. Dirichlet boundary conditions on $f$. $\endgroup$
    – user7530
    Jan 16, 2019 at 19:56
  • $\begingroup$ Thank you for the clarification! This has provided me with some good insight. $\endgroup$
    – Dhuality
    Jan 16, 2019 at 21:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .