# The Unapologetic Mathematician

## (Hyper-)Surface Integrals

The flip side of the line integral is the surface integral.

Given an $n$-manifold $M$ we let $S$ be an oriented $n-1$-dimensional “hypersurface”, which term we use because in the usual case of $M=\mathbb{R}^3$ a hypersurface has two dimensions — it’s a regular surface. The orientation on a hypersurface consists of $n-1$ tangent vectors which are all in the image of the derivative of the local parameterization map, which is a singular cube.

Now we want another way of viewing this orientation. Given a metric on $M$ we can use the inverse of the Hodge star from $M$ on the orientation $n-1$-form of $S$, which gives us a covector $\nu(p)\in\mathcal{T}^*_pM$ defined at each point $p$ of $S$. Roughly speaking, the orientation form of $S$ defines an $n-1$-dimensional subspace of $\mathcal{T}^*_pM$: the covectors “tangent to $S$“. The covector $\nu(p)$ is “perpendicular to $S"$, and since $S$ has only one fewer dimension than $M$ there are only two choices for the direction of such a vector. The choice of one or the other defines a distinction between the two sides of $S$.

If we flip around our covectors into vectors — again using the metric — we get something like a vector field defined on $S$. I say “something like” because it’s only defined on $S$, which is not an open subspace of $M$, and strictly speaking a vector field must be defined on an open subspace. It’s not hard to see that we could “thicken up” $S$ into an open subspace and extend our vector field smoothly there, so I’m not really going to make a big deal of it, but I want to be careful with my language; it’s also why I didn’t say we get a $1$-form from the Hodge star. Anyway, we will call this vector-valued function $dS$, for reasons that will become apparent shortly.

Now what if we have another vector-valued function $F$ defined on $S$ — for example, it could be a vector field defined on an open subspace containing $S$. We can define the “surface integral of $F$ through $S$“, which measures how much the vector field flows through the surface in the direction our orientation picks out as “positive”. And we measure this amount at any given point $p$ by taking the covector at $p$ provided by the Hodge star and evaluating it at the vector $F(p)$. This gives us a value that we can integrate over the surface. This evaluation can be flipped around into our vector field notation and language, allowing us to write down the integral as

$\displaystyle\int\limits_SF\cdot dS$

because the “dot product” $\left[F\cdot dS\right](p)$ is exactly what it means to evaluate the covector dual to $dS(p)$ at the vector $F(p)$. This should look familiar from multivariable calculus, but I’ve been saying that a lot lately, haven’t I?

We can also go the other direction and make things look more abstract. We could write our vector field as a $1$-form $\alpha$, which lets us write our surface integral as

$\displaystyle\int\limits_S\langle\alpha,\nu\rangle\omega_M=\int\limits_S\alpha\wedge*\nu$

where $\omega_M$ is the volume form on $M$. But then we know that $*\nu=\omega_S$ the volume form on $S$. That is, we can define a hypersurface integral of any $1$-form $\alpha$ across any hypersurface $S$ equipped with a volume form $\omega_S$ by

$\displaystyle\int\limits_S\alpha\wedge\omega_S$

whether or not we have a metric on $M$.

October 27, 2011 - Posted by | Differential Geometry, Geometry

## 3 Comments »

1. You have a problem with this blog that I noticed long ago but I forgot to signal. At the top-right side of every page, where it says “This is mainly an expository blath”, the link doesn’t work (I had to read that linked article through the Internet Archive).

Comment by Andrei | October 27, 2011 | Reply

2. […] and work — of late, but I think we can get started again. We left off right after defining hypersurface integrals, which puts us in position to prove the divergence […]

Pingback by The Divergence Theorem « The Unapologetic Mathematician | November 22, 2011 | Reply

3. […] , where the differential can take us straight from a line integral over a -dimensional region to a surface integral over an -dimensional […]

Pingback by The Classical Stokes Theorem « The Unapologetic Mathematician | November 23, 2011 | Reply