Let $(M,g)$ be a manifold with metric $g$ parametrized by the mapping $S$ and parametric domain $\Omega$. The sobolev space of order one with respect to the $L_2(M)$-norm $H^1_2(M)$ is defined as follows: Let $C_2^1 = \{ u \in C^\infty(M) s.t. \forall j = 0,1: \int_M |\nabla^j u |^2 dv(g) < \infty \}$, where $\nabla^0 u = u$ and $\nabla u $ is the ordinary gradient in local coordinates (i.e. in $\Omega$). Note that with above metric, we have $|\nabla u|^2 = (\nabla u)^T g \nabla u$. The Sobolev space $H^1_2(M)$ is the completion of $C_2^1$ with respect to the norm \begin{align} \left \| u \right \|_{H^1_2} = \sum_{j = 0}^1 \left(\int_M | \nabla^j u |^2 dv(g) \right)^{1/2}. \end{align}

I have a subtle problem with this definition. As an example, suppose that $\Omega = [0,1]$ and \begin{align} u(x) = \left \{ \begin{array}{ll} 1-2x & \quad 0 \leq x \leq 1/2 \\ 0 & \quad 1/2 < x \leq 1 \end{array} \right. . \end{align} Suppose $M$ is just the line-segment $\Omega$ mapped onto another line in $\mathbb{R}^2$. Then, I believe, $u \in H_2^1(M)$ should be true. Now suppose that $[0,1]$ is mapped onto a closed circle (via the mapping $S(x) = (r \cos(2 \pi x),r \sin(2 \pi x)$)), then I feel like $u \notin H_2^1(M)$, this is because on the circle, the function $u$ contains a `jump' at the point where the the boundaries of $\Omega$ connect.

The problem that I am having is that $\left \| u \right \|_{H_2^1} < \infty$ even when $M$ is a circle. The integral is carried out in $\Omega$ and there the discontinuity does not exist. I believe that the reason that $u$ is not in the sobolev space on the circle must be that it is not contained in the completion of $C_2^1$ in that case. However, I have no idea how to demonstrate that.

More specifically, I'd like to demonstrate that functions that contain discontinuities on $M$, in general, are not contained in $H_2^1(M)$, also when $M$ is an n-dimensional manifold.

Can somebody help me ? Thank you in advance !

  • $\begingroup$ What is a "mapping operator" and a "parametric domain"? $\endgroup$ – levap Dec 8 '15 at 14:34
  • $\begingroup$ $s|_\Omega$ gives you $M$. Sorry, I am really not an expert in this field of mathematics. $\endgroup$ – Yamamoto Dec 8 '15 at 14:37
  • $\begingroup$ No problem, but it will help if you provide some details and/or a reference about what is $s$ and $\Omega$ because the definition of Sobolev spaces on manifolds I'm familiar with doesn't use this kind of terminology at all. $\endgroup$ – levap Dec 8 '15 at 14:39
  • $\begingroup$ Also, the point of Sobolev spaces is to allow less regular functions, and in particular, functions with discontinuities so it is unlikely you will be able to show that such functions don't belong to $H_2^1(M)$. There are discontinuous functions that belong to $H_2^1(\Omega)$ where $\Omega$ is an open subset of $\mathbb{R}^n$ with $n > 1$ and also to the Sobolev spaces on manifolds using the definitions I'm familiar with. $\endgroup$ – levap Dec 8 '15 at 14:41
  • $\begingroup$ @levap: What the OP wanted to say is that $M$ has a global chart $s^{-1} : M \to \Omega$. $\endgroup$ – Alex M. Dec 8 '15 at 14:44

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