Minimal area of non-planar lattice curves

Question

Consider a $\mathbb{Z}^d$ lattice whose edges connect nearest-neighbor sites only, i.e. a $d$-dimensional hypercubic grid. Let $C$ be a closed curve along such edges. In general, for $d>2$ such curves are non-planar. If an elementary square is given area 1, is there any formula that computes the minimal area bounded by a generic lattice curve $C$? For example, the following curve has minimal area of 2:

Edit: Precisely, what I would like to know is if it is possible to decide whether a generic curve $C$ bounds an even or odd area. For example, the curve above has area 2, which is even. I suspect that parity is a property of the curve and is preserved in non-minimal areas, but I don't know how to prove it.

Answer 1

For any $\mathbb Z_2$-valued function $f$ on the set of segments $[v,v+e_i]$ for $v\in \mathbb Z^3$ and $i=1,2,3$, and any $S$ that bounds $C$ $$ \sum_{P\in S}\sum_{edge\in P} f(edge) $$ is independent of the choice of $S$ (it is equal to the sum of $f$ over the segments of $C$).

Thus to prove the independence of parity of the area from the choice of $S$, it suffices to find a function on the set of (directed) segments $[v,v+e_i]$ such that the sum of it over edges of any size one lattice square is $1\in \mathbb Z_2$.

One way of thinking about such $f$ is the following homological algebra construction. If you naively tensor three complexes, then the corresponding squares will commute rather than anticommute. However, it is possible to change some of the signs to make the total complex to be an actual complex. The edges that needed to be altered are the ones that will have $f=1$.

For a more elementary argument, we will declare $f([v,w])$ to be $1$ if and only if $[v,w]$ is equal modulo $2$ to one of the following: $$ [(0,0,0),(1,0,0)],~[(0,0,1),(0,1,1)],~[(0,1,1),(1,1,1)], $$ $$ [(1,0,0),(0,0,0)],~[(0,1,1),(0,0,1)],~[(1,1,1),(0,1,1)], $$ I will now argue that every lattice square contains exactly one such edge.

Note that the set above is preserved by translations by elements of $2\mathbb Z^3$. It is also preserved by symmetries across lattice planes, as these have an effect of switching $[v,w]$ and $[w,v]$ modulo $2$ or preserving $[v,w]$ modulo $2$. Every lattice square can be moved into one of the six facets of the standard cube $[0,1]^3$ by the above shifts and symmetries. It remains to observe that each of the faces of the standard cube contains exactly one edge from the first line and none from the second line.

Answer 2

This is not a full answer, but at least it can be proved that the parity is kept using few basic properties of homology of cubical complexes. (See also the last paragraph "After edit" discussing how to determine this parity.)

Let $S_1$ and $S_2$ be two surfaces bounded by $C$. The symmetric difference $S$ of $S_1$ and $S_2$ is a cycle in $\mathbb Z_2$-chain complex of the grid. [If you are not familiar with homology, think of $S$ as a multicomponent possibly self-intersecting $2$-surface with empty (!) boundary.] The fact that $S_1$ and $S_2$ have the same parity is equivalent with showing that the number of squares forming $S$ is even.

Since $S$ is a cycle and the homology of the cubical grid is trivial, it is also a boundary of some $3$-chain over $\mathbb Z_2$. That is, there are cubes $Q_1, \dots, Q_k$ such that the boundary of the union $Q$ of these cubes is $S$. Start removing these cubes from $Q$ one by one. (That is, make the symmetric differences.) In each single removal, the parity of the number of squares of the boundary does not change, since a single cube has an even number of squares. After removing all the cubes, the boundary is empty. Therefore the boundary of $Q$ has to consist of an even number squares.

After edit: This homology approach should essentially also answer whether a given curve $C$ bounds an even or an odd area. It is just necessary to find a 2-chain $S'$ such that the boundary of $S'$ is $C$, perhaps not a minimal one. (In some bounded subcomplex of the grid containing $C$ in order to make the computations finite.) This should be then just some matrix manipulation.

Answer 3

You might look at this work by John Sullivan: "Discrete minimal surfaces in the cubic lattice," and his lecture on the topic here:
               
Incidentally, another version of the Schwartz $P$-surface is at Alan Schoen's link: