# How to interpret this quote of Lin?

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.

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

• You could try sending a short email to Lin asking this. Jul 27, 2021 at 17:43
• @DeaneYang I appreciate the suggestion; luckily I think I won't have to bother him. A thought came to my mind earlier today, and I believe the sequence below demonstrates the behaviour described in the quote. Jul 27, 2021 at 19:52

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.

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.

• Thank you for your comments! They prompted me to think about the question again, and I think I managed to come up with a sequence that illustrates Lin's quote. What it basically does is rotate a line to cancel with one of the 'spokes' of a triple junction. The example is just a bit more complicated to ensure that the currents are cycles. (That being said, I think this seems like a red herring.) I'd be curious to hear your thoughts! Jul 27, 2021 at 19:48

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.

• This looks about right to me. You can probably simplify it slightly by rotating two of the legs of $S$ to coincide in the limit instead of adding L, but the idea stays the same.
– mlk
Jul 28, 2021 at 6:39
• I'd be interested in a simpler example, but I am skeptical about your suggestion. I don't see how two of the legs could be moved to cancel one another in the limit for a stationary current. Jul 28, 2021 at 15:03
• You need to move all of them of course to keep the solution stationary. But for any choice of $v_1,v_2$ there are $v_3,...,v_6$ that keep them balanced in such a way that there is some uniform distance between each of the latter. So you could have $v_1,v_2$ converge to the same limit vector with their segments oppositely oriented and balance them with a similar sequence of $v_3,...,v_6$ that converge without cancelling.
– mlk
Jul 29, 2021 at 10:29