# How is the “surface measure” on a manifold defined?

Let

• $k,n\in\mathbb N$ with $k\le n$
• $M$ be a $k$-dimensional $C^1$-submanifold of $\mathbb R^n$
• $\Omega\subseteq\mathbb R^k$ be open, $\phi:\Omega\to M$ be a global chart of $M$ and $$g_\phi(x):=\det\left(\left({\rm D}\phi(x)\right)^\ast {\rm D}\phi(x)\right)\;\;\;\text{for }x\in U$$

Now, let $\lambda^k$ denote the Lebesgue measure on $\mathcal B(\mathbb R^k)$, $\sqrt{g_\phi}{\rm d}\left.\lambda^k\right|_U$ denote the measure with density $\sqrt{g_\phi}$ with respect to $\left.\lambda^k\right|_U$ and $$S_M:=\phi_\ast\left(\sqrt{g_\phi}{\rm d}\left.\lambda^k\right|_U\right)$$ the pushforward measure of $\sqrt{g_\phi}{\rm d}\left.\lambda^k\right|_U$ with respect to $\phi$. $S_M$ is a measure on $\mathcal B(M)$ and we're able to verify that $S_M(A)$ is the surface area of $A\in\mathcal B(M)$.

I'm searching for a generalization of the concept described above for more general $M$. In particular, the $M$ I've got in mind is the finite union of triangles in $\mathbb R^3$. Two different triangles can be assumed to intersect only along one common side or at one common vertex and two different sides can intersect only at one common vertex.

I've searched the internet for a couple of hours, but couldn't find anything (Actually, I wasn't even able to find the construction of $S_M$ above anywhere). So, is there any good textbook on that topic?

Notice that your $S_M$ depends not only on $M$, but also on $\phi$; to define an area form intrinsic to $M$ you need to equip $M$ with a metric, impose additional assumptions (for instance that the metric is pulled back from the ambient space), or specify a particular $\phi$.
A typical construction for the second approach for triangle meshes is to observe that one can construct an affine chart $\phi$ on each triangle (with vertices $\mathbf{p}_i$) from a canonical unit triangle $\mathcal{T} = 0 \leq u,v, u+v \leq 1$ in the plane by the triangle's barycentric coordinates: $$\phi(u,v) = \mathbf{p}_1 + u(\mathbf{p}_2-\mathbf{p}_1) + v(\mathbf{p}_2-\mathbf{p}_1).$$
It is easy to show that the Euclidean metric in $\mathbb{R}^3$ pulls back to a constant metric $d\phi\otimes d\phi$ on $\mathcal{T}$, so that your desired area measure is, analogously to your formula, $$S_{\mathrm{im}\ \phi} = \phi_*\left(\|d\phi\|\,d\lambda_{\mathcal{T}}\right) = \phi_*\left(\sqrt{\det \left[\begin{array}{cc}\|(\mathbf{p}_2-\mathbf{p}_1)\|^2 & (\mathbf{p}_2-\mathbf{p}_1)\cdot (\mathbf{p}_3-\mathbf{p}_1)\\(\mathbf{p}_3-\mathbf{p}_1)\cdot (\mathbf{p}_2-\mathbf{p}_1) & \|(\mathbf{p}_3-\mathbf{p}_1)\|^2\end{array}\right]}\,d\lambda_{\mathcal{T}}\right).$$
• I've considered $M\subseteq\mathbb R^n$ as being equipped with the metric induced by $\mathbb R^n$. And we can show that $S_M$ is independent of the choice of $\phi$. – 0xbadf00d Jun 28 '18 at 17:17