Why is this a good picture of a covector?

Question

I'm reading a book about applied differential geometry and the author says: "suppose $V$ is a finite dimensional vector space. For a given covector $\omega \in V^\ast$, the set $\hat{\omega}$, of vectors such that $\omega(v)=1$, that is $\hat{\omega}=\{v\in V : \omega(v)=1\}$ form a hyperplane in the vector space, and provide a faithful representation for $\omega$."

Now, why is this a good representation of a covector? I've heard that a covector can be thought of as a family of hyperplanes and it's value on a vector are then the number of hyperplanes crossed. This seems to relate to this idea, but how? Is this the idea that "if $v\in \hat{\omega}$, then $v$ crosses $\omega$ once if $\dfrac{1}{k}v\in \hat{\omega}$ then $v$ crosses $\omega$ $k$-times"?

I'm really failing to get the geometrical intuition behind those objects.

Thanks very much in advance.

Answer 1

First of all, the book Gravitation, by Misner, Thorne, and Wheeler has an extensive discussion of the intuition behind covectors, differential forms, etc, complete with excellent diagrams. Since their audience is physicists, they focus on fostering geometric intuition. Section 2.5 should help.

Here are three ways to think above covectors. (I will assume throughout that $\omega\neq 0$.) First, by definition, $\omega\in V^*$ is a linear function $\omega:V\rightarrow \mathbb{R}$. Let's let $V$ be two dimensional, just to make things easy to visualize. You know that you can gain insight into function by graphing it, but also by drawing level-curves for it. Well, you can graph $\omega$ in $\mathbb{R}^3$, letting $V$ be the $xy$-plane. You get a plane passing through the origin. If you draw level curves in the xy-plane, say for all integer values, you'll get parallel lines. That is, if you plot $\omega(v)=c$ for all integer values of $c$, you'll get parallel lines in the xy-plane.

If $\omega$ is "gently sloped", then the lines will be spaced far apart. If $\omega$ is "steeply sloped", then the lines will be spaced close together. In particular, if we plot the level curves for $2\omega$, the lines will be space precisely twice as close together.

This brings us to the second way of looking at $\omega$. $\mathbb{R}^2$ has an inner product defined on it. If we've identified $V$ with $\mathbb{R}^2$, that means we can talk about the dot product $v\cdot w$ for $v,w\in V$. (Of course for a general vector space we can't do this, but we're trying to develop geometric intuition, so let's assume this.)

Pick $v\in V$. The function $w\mapsto v\cdot w$ is linear, thus an element (call it $\omega_v$) of $V^*$. In fact, every element of $V^*$ is of this form. See if you can convince yourself that $v$ is perpendicular to the parallel lines (level curves) for $\omega_v$.

Finally, we have the representation you mentioned. This is simply picking one of the level curves for $\omega$, namely the one with $c=1$. The level curves are evenly spaced and all parallel, and the one for $c=0$ must pass through the origin. So if you are given the level curve $\omega(v)=1$, that's enough to determine the whole family of level curves.

These ideas are all useful for thinking about differential forms, and the intuition for differential forms reflects back to the more basic concept of covectors. Given a smooth function $f:\mathbb{R}^2\rightarrow \mathbb{R}$, the gradient $\nabla f$ determines a covector at each point: if we pick a vector $w$ at a point $p$, then the directional derivative of $f$ in the direction $w$ is $\nabla f\cdot w$. But this just means that $\nabla f$ determines a linear function on the space of all vectors "with tails at" the point $p$. So it's a covector. Now if you plot level curves of $f$, you'll get some kind of contour plot. If you "blow up" (i.e., magnify) the plot near $p$, you'll get curves that are approximately straight, parallel, and evenly spaced. Those represent the covector determined by $\nabla f$ at $p$. (We may need to choose our level curves more fine-grained than $f(x)=c$ for integer $c$, if we want to see detail under high magnification.)

One caveat: as just noted, $\nabla f$ determines a differential form, and hence a covector at each point $p\in\mathbb{R}^2$. But that's not the definition of a differential form. Without going into fine points, a differential form is any way of smoothly assigning covectors to each point. If the form can be described as $\nabla f$ for some function $f$, then the form is said to be exact.

Answer 2

Let's take $\omega = dz$. I believe the text means to say that the set of vectors for which $dz(v)=0$ can be identified with $\omega$. Here this gives all vectors with $z$-component trivial namely $\hat{\omega} = \{ \langle a,b,0 \rangle \ | \ a,b \in \mathbb{R} \}$. So the dual vector $dz$ is naturally identified with the set of vectors which it annihilates. Now, a deeper question is this, can we identify that set of vectors as being tangent to some surface in $\mathbb{R}^3$. For $dz$ the answer is clearly yes: the horizontal vectors form tangents to $z=c$. However, in general, the situation is more perilous. It is not generally true that the kernels of a oneform are tangent to some family of surfaces which foliate the space. I think the discussion you may be seeking is found in the excellent text on differential forms by David Bachman http://arxiv.org/abs/math/0306194. See pages 86-88 or so in the pdf which is available through the link.