I am trying to figure out what the next calculation of the "area" (or "volume" in higher dimensional analogues) using Stokes' theorem really means. Here is my thought process:

$2$-dimensional case: given a closed simple piece-wise smooth curve $C$ in $\mathbb{R}^2$ you can find out the area enclosed by $C$ using Green's theorem by choosing an orientation on $C$ and calculating $$\text{Area} = \left\vert \dfrac{1}{2} \oint_C (x dy - y dx) \right\vert .$$ $2$-dimensional case in $\mathbb{R}^n$: When you have a piece-wise smooth non self intersecting map $\phi : S^1 \rightarrow \mathbb{R}^n$ such that $\phi (S^1)$ is contained in a $2$-dimensional plane of $\mathbb{R}^n$, you can again find the area enclosed by $\phi (S^1)$ by choosing an orthonormal coordinate system $(x_1,...,x_n)$, choosing an orientation on $S^1$ and calculating using Stokes' theorem $$\text{Area}^2 = \sum_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint_{\phi (S^1)} (x_i dx_j - x_j dx_i) \right)^2 .$$

Now my question is what is the meaning of calculating the above quantity when $\phi (S^1)$ is not contained in a $2$-dimensional plane. I want to use the same computation as above, but since I want the number I get to be independent on the coordinate system, I'll have to average with respect to all coordinate systems. To be explicit: let $\phi : S^1 \rightarrow \mathbb{R}^n$ be a piece-wise smooth non self intersecting map and fix an orientation on $S^1$. Let $SO_n (\mathbb{R})$ be the special orthogonal group and let $\mu$ be the normalized Haar measure on $SO_n (\mathbb{R})$ (i.e., $\mu (SO_n (\mathbb{R}) ) =1$). Define the "area" (or the "Stokes area" in lack of a better name) bounded by $\phi (S_1)$ as $$\text{"Stokes area"}^2 = \int_{SO_n (\mathbb{R})} \left( \sum_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint_{\gamma.\phi (S^1)} (x_i dx_j - x_j dx_i) \right)^2 \right) d \mu (\gamma) .$$ It is not hard to give analogues in any dimension to get "Stokes volume" (note that the analogue in dimension $1$, i.e., for $S^0$, comes out just the usual Euclidean distance between points).

My questions are:

  1. Is this quantity well-known / studied ? If so, I would very much appreciate a reference.

  2. What does this "Stokes area" represent geometrically (I have no intuition about it)?

  3. Is there a connection (say by some inequality) to the area of a minimal surface enclosed in $\phi (S^1)$?

Edit: by the answer of Will Sawin below, it is clear that we can consider only $$\text{"Stokes area"}^2 = \sum_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint_{\gamma.\phi (S^1)} (x_i dx_j - x_j dx_i) \right)^2 .$$ and that is quantity is always less or equal than the area of a surface enclosed by the curve. The question remains is there is a connection to the infimum of the areas of all the surfaces enclosed by the curve. In the case the curve is contained in a plane, the formula computes this minimum. But what about the general case: is there a constant (maybe depending on the dimension of $\mathbb{R}^n$) $c(n)$ such that $$ \text{"Stokes area"} \geq c(n) (\text{infimum of the areas of surfaces enclosed by the curve}) ?$$

  • 2
    $\begingroup$ Note that all terms in the sum over $i,j$ contribute equally, as $SO_n$ acts transitively on ordered pairs $x_i,x_j$ for $n\geq 3$. For the same reason, the average over $SO_n(\mathbb{R})$ would have vanished if you had omitted the squares. As far as I can tell, your Stokes area is a quadratic average over all 2d planes $P$ of the area of the projection of the curve onto $P$. Once properly normalized, that should be less than the area of a minimal surface. Instead, averaging over $SO_n$ before integrating over $\phi(S^1)$ gives an upper bound to the area of a minimal surface. $\endgroup$ Dec 9, 2014 at 21:10
  • 5
    $\begingroup$ @BrunoLeFloch Sometimes, the projection of the curve onto a $2$-plane can look like a figure $8$. In this case, integrating $xdy-ydx$ over the figure $8$ we can get $0$. $\endgroup$ Dec 9, 2014 at 21:50

3 Answers 3


Yes, this is a lower bound on the area of a surface bounded by the curve. Parameterize the surface and apply Stokes' theorem. Your squared area is:

$$\sum_{1 \leq i < j \leq n} \left( \int_S dx_i dx_j \right)^2 = \int_S \int_S \sum_{1\leq i < j \leq n} dx_i dx_j dy_i dy_j$$

whereas the actual squared area is:

$$ \left( \int_S \sqrt{ \sum_{1 \leq i < j \leq n} \left( dx_i dx_j \right)^2} \right)^2= \int_S \int_S \sqrt{\sum_{1 \leq i < j \leq n} \left( dx_i dx_j \right)^2} \sqrt{\sum_{1 \leq i < j \leq n} \left( dy_i dy_j \right)^2}$$

which is at least as large by Cauchy-Schwartz. Equality occurs exactly when the area element of the surface is always pointed in the same direction in $\wedge^2 \mathbb R^n$ - that is, when the surface is flat.

This also shows that your definition is coordinate-invariant even before you average over $SO_n$.


The answer to my question in the edit is negative. Following the remark of Liviu Nicolaescu, one can always take the figure 8, embed it in the $x_1x_2-$plane and do a small perturbation such it doesn't intersect it self. Doing so you can get a "Stokes area" as small as you like, no matter how big your figure 8 is.


This is not the answer to your question, but a related result.
If $\gamma(x)=(x(s),y(s))$ is a closed curve on the plane, then $$ \mathfrak{D}=\frac{1}{2}\int_0^1(\dot{y}(s)x(s)-\dot{x}(s)y(s))\, ds $$ is the oriented area. In particular if the curve has the shape of the digit $8$, the oriented area equals zero because the curve has opposite orientation in the upper and the lower part of $8$.

In the case of closed curves in $\mathbb{R}^{2n}$ with coordinates $x_1,y_1,\ldots,x_n,y_n$, one can define the symplectic area as the sum of oriented areas of projections onto the $x_iy_i$-planes, $i=1,2,\ldots,n$. More precisely, the symplecic area is $\mathfrak{D}=\mathfrak{D}_1+\ldots+\mathfrak{D}_n$, where $$ \mathfrak{D}_j=\frac{1}{2}\int_0^1(\dot{y}_j(s)x_j(s)-\dot{x}_j(s)y_j(s))\, ds $$ is the oriented area of the projection of the curve on the $x_iy_i$-plane.

Theorem (Isoperimetric inequality). If $\gamma=(x_1,\ldots,x_n,y_1,\ldots,y_n):[0,1]\to\mathbb{R}^{2n}$ is a closed rectifiable curve prametrized by constant speed, then the following isoperimetric inequality is true $$ L^2\geq 4\pi|\mathfrak{D}|, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*) $$ where $L$ is the length of $\gamma$ and $\mathfrak{D}=\mathfrak{D}_1+\ldots+\mathfrak{D}_n$ is defined above. Moreover the equality in (*) holds if and only if $\gamma$ is a circle of the following form: there are points $A,B,C,D\in\mathbb{R}^n$ such that form \begin{equation} \label{Eq1} \gamma(s)=(C+iD)+(1-e^{+2\pi is})(A+iB), \quad \text{when} \quad L^2=4\pi\mathfrak{D} \end{equation} and \begin{equation} \label{Eq2} \gamma(s)=(C+iD)+(1-e^{-2\pi is})(A+iB) \quad \text{when} \quad L^2=-4\pi\mathfrak{D}. \end{equation}

The proof follows from a simle and straightforward adaptation of a standard proof of the isoperimetric inequality due to Hurwitz, see:

P. Hajłasz, S. Zimmerman, Geodesics in the Heisenberg group. Anal. Geom. Metr. Spaces 3 (2015), 325–337.


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