How to interpret this quote of Lin?

Question

I recently stumbled across a quote of Fang-Hua Lin that I have trouble understanding [1, page 42].

It is a well-known fact that a weakly converging sequence of stationary integral currents may have a limit which is not a stationary current.

Question. How should I interpret this quote? What does Lin mean by a 'stationary current', and which sequence demonstrates this 'well-known fact'?

My initial guess would be that an integral current $T$ is 'stationary' if the varifold $\lvert T \rvert$ obtained by forgetting orientations is stationary. If I am not mistaken, this should mean that $\partial T = 0$? However my impression is that a flat limit $T$ would be 'stationary' in this sense of the word.

[1] F.-H. Lin. Mapping problems, fundamental groups and defect measures. Acta Math. Sin. 15 (1999), 25-52.

Answer 1

The following counterexample is due to Leon Simon. Take $\mathbb{R}^2$ with coordinate label $(x,y)$. Define a current $T$ supported in $x$-axis $\cup$ $y$-axis of the shape $-\lrcorner+\ulcorner.$ To be precise, $\lrcorner$ trace the negative $x$-axis, then the non-negative $y$-axis, and the minus sign means reversing orientation. $\ulcorner$ traces the negative $y$-axis and then the non-negative $x$-axis. By construction, $T$ is a cycle with no boundaries and can be realized as the boundary of two quadrants suitably oriented.

Let $T_a$ be the unique line parallel to the $x$-axis and passing through $(0,a).$ Then $T_a+T$ is stationary for all $a\not=0$ both as varifolds and currents. However, $T_0+T$ equals $\llcorner+\ulcorner$ and is not stationary.

Answer 2

This is not a full answer, since I do not know the counterexample Lin refers to, but I can offer some explanations and guesses which are too long for a comment:

You can define a first variation for currents similarly to that for varifolds by considering $\frac{d}{ds}|_{s=0} E((h_s)_\# T)$ for some smooth enough family of diffeomorphisms $h_s$ with $h_0 = \operatorname{id}$ (and $h_s = \operatorname{id}$ outside a compact set). Here $E$ is the energy you want to consider (i.e. mass in the case of minimal surfaces) and $h_\#$ denotes the pushforward. This should coincide with $\lvert T\rvert$ being a stationary varifold though.

Regarding the boundary, you can fix it, by setting $\partial T = R$ and correspondingly only allowing variations with $h_s = \operatorname{id}$ on $\operatorname{supp} R$. Otherwise, if you are not additionally penalizing the mass of the boundary, $\partial T=0$ is the only one that makes sense in the context of minimal surfaces.

I don't know the counterexample Lin refers to, but I think it would need to involve cancellation of parts with opposite orientation (otherwise $\lvert\lim_{k\to\infty} T_k\rvert = \lim_{k \to \infty} \lvert T_k\rvert$ which is known to be stationary). Additionally I think it should not be a sequence of minimal currents, otherwise the limit will be minimal and thus stationary again.

Answer 3

I believe the following sequence demonstrates the failure of flat limits to be stationary. This would be consistent with the natural interpretation of the quote, meaning: a current $T$ is called stationary if the varifold $\lvert T \rvert$ is.

(A quick side remark before the construction: on second thought whether $\partial T = 0$ or not seems seems unrelated to the stationarity of $\lvert T \rvert$. For example, a triple junction has boundary as a current, but is stationary as a varifold. A slightly simpler variant of the example below would see the current $S$ replaced with a triple junction.)

That being said, the currents in the constructed sequence $(T_n \mid n \in \mathbf{N})$ are one-dimensional cycles in the unit disc $D \subset \mathbf{R}^2$: $\partial T_n = 0$ for all $n$. They converge weakly as currents to another cycle, say $T_n \to T$ as $n \to \infty$. Most important: $\lvert T_n \rvert$ is stationary for all $n$, but $\lvert T \rvert$ is not.

To construct the sequence, let $\{ v_1,\dots,v_6 \} \subset \partial D$ be unit vectors with \begin{equation} v_1 + \cdots + v_6 = 0, \end{equation} but which do not match up into antipodal pairs. For example \begin{equation} -v_1 \not \in \{ v_1,\dots,v_6 \}. \end{equation}

• Let $S \in I_1(D)$ be the current supported in the union of the segments $\{ t v_i \mid 0 \leq t \leq 1 \}$, oriented so that $\partial S = 0.$ The associated varifold $\lvert S \rvert$ is stationary by construction.

• Let $L$ be the current supported in the segment $\{ tv_1 \mid -1 \leq t \leq 1 \}$, which we orient in the opposite direction. In other words \begin{equation} \{ t v_1 \mid 0 < t \leq 1 \} \cap \mathrm{spt} \, (S + L) = \emptyset. \end{equation} This too has $\partial L = 0$ and $\lvert L \rvert$ stationary.

The orientations are chosen so as to ensure that the current $T := S + L$ is not stationary; this is because $-v_1 + \cdots + v_6 = -2v_1 \neq 0$.

Next we consider a sequence of positive angles $\theta_n \to 0$. We use these angles to rotate $L$, forming a sequence of currents \begin{equation} R_{\theta_n \#} L \to L \text{ as $n \to \infty$.} \end{equation} As long as $\theta_n$ is small enough that $\mathrm{spt} \, R_{\theta_n \#} L \cap \mathrm{spt} \, S = \{ 0 \}$, the cycles $T_n := S + R_{\theta_n \#} L$ are stationary. However \begin{equation} T_n = S + R_{\theta_n \#}L \to S + L = T \text{ as $n \to \infty$} \end{equation} in the current topology, which was pointed out above is not stationary.