Can we recover the metric by knowing volume and area?

Question

Does someone know that what is the definition of area of a manifold? and then

Question: Can we recover the metric by knowing its volume and area?

In the simplest case ($n=1$) I don't know whether two simple closed curve of the same area and circumference are congruent or not!! Any explanation, proof, reference or counterexample would be very appreciated.

Answer 1

Certainly you can have two curves with the same perimeter and area. Let one curve be a unit square. Start with a circle with the same perimeter, which has larger area. Now push an indent into the circle, maintaining the perimeter. You can make the indent the proper size to match the area of the square.

Answer 2

Just knowing the integrated area and volume is not sufficient because you can mess with the shape, as Ross shown above. However, more in the spirit of the question, if you know the induced $n$- and $(n-1)$-dimensional Hausdorff measures then you can recover the metric, but it is tedious in practice.

Let's assume $M$ is orientable for simplicity. We can define a (bilinear) pairing $\langle\cdot,\cdot\rangle\colon T_p^*M\times Gr_{n-1}^+(T_pM)\to\mathbb{R}$ which sends $(\alpha,\Pi)$ ($\alpha\in T_p^*M$ and $\Pi$ is an oriented $(n-1)$-plane in $T_pM$) to $\dfrac{\alpha\wedge\mathrm{d}A}{dV(p)}$, where $\mathrm{d}A$ is the area element of $\Pi$ (which we know). Thus we recover the length of $\alpha$ by $\lvert\alpha\rvert=\max_\Pi\langle\alpha,\Pi\rangle$. Now just find $n$ unit-length cotangent vectors $\alpha_1,\dots,\alpha_n$ with $\alpha_1\wedge\dots\wedge\alpha_n=\pm\mathrm{d}V(p)$ and these $\alpha_i$ will give your metric $g(p)=\sum_i\alpha_i\otimes\alpha_i$. If $(\alpha'_1,\dots,\alpha'_n)$ are another set of $n$ unit-length cotangent vectors wedging to give $\mathrm{d}V(p)$, then $\alpha'$ and $\alpha$ are related by an orthogonal matrix, so $g(p)$ is unique.