# Geometric visualization of covector?

How could I geometrically visualize a linear functional?

• Draw the hyperplane on which the functional assumes the value $1$. Jul 9 '13 at 12:36

I think the nicest visualization is the set of hyperplanes where it attains integer values (yes, technically one hyperplane, like that one where its value is $1$, suffices to uniquely specify it, but the whole set is IMHO more intuitive, and that's the whole point of visualization, isn't it?). They are the equal value hypersurfaces of the linear function. Of course you'll also have to specially mark which direction is "up" (e.g. by giving the positive value surfaces a different color). The value when applying to a vector is the number of the hyperplane where it ends (if it ends in between, you simply interpolate, or alternatively think of a finer-grained set (like the millimeter lines in between centimeter lines on a ruler). Or alternatively, the value is the number of planes it intersects (again, with interpolation/finer grained subsets).

You immediately see the linearity from the fact that they are equal spaced and parallel; a twice as long vector goes through twice as many hyperplanes. Also, multiplication with a scalar is quite obvious: If you multiply it with $m/n$, you get $m$ planes in the same space you've gotten $n$ planes before. And addition of parallel covectors is basically stuffing the planes from both covectors in the same space, but equally spaced.

Addition of non-parallel covectors is only slightly more involved: You basically have to draw hyperplanes through all the intersections of the hyperplanes of the two added covectors. However you've got to be careful to do it correctly: The zero plane of the sum covector goes through the crossings of positive hyperplanes from one, and negative hyperplanes from the other term.

Note that for this representation you do not need to assume an inner product (if you do so on a space where there's not a natural choice of inner product, your intuition might be misled by giving significance to absolute lengths and angles in your visualization, which are just meaningless in a space without inner product).

• An intermediate "visualization is the set of hyperplanes where it attains" positive integer values, so that one doesn't "also have to specially mark which direction is 'up'". $\;$
– user57159
Sep 25 '14 at 3:10

In the finite dimensional case you always have that a linear functional corresponds to a unique vector through the inner product $\psi(x) = \langle x,y \rangle$ for some $y$, this passes to the infinite dimensional case when you have an inner product space (Riesz Representation Theorem)

• +1 for the general exposition @Maurice: can you find such $y$ for the functionals in my answer? All you need is an orthonormal basis in $\mathbb R^2$... Jul 9 '13 at 12:39
• This is only unique if you have a given inner product. Otherwise it is only unique up to the choice of inner product. Jul 9 '13 at 14:19

I give you an example of geometric visualization: consider, for example, any vector $v=(v_1,v_2)\in\mathbb R^2$ and the linear functionals

$$\varphi_1: \mathbb R^2\rightarrow \mathbb R$$ $$\varphi_2: \mathbb R^2\rightarrow \mathbb R$$

given by $\varphi_1(v_1,v_2)=v_1$ and $\varphi_2(v_1,v_2)=v_2$. You can visualize them as the functionals which provide you with the projections of the given vector $v$ to the $x$-axis, resp. the $y$-axis.

In ${\Bbb{R}}^2$, being $f:{\Bbb{R}}^2\to {\Bbb{R}}$ linear, its level curves are lines perperdicular to the vector given by ${\rm graf} f$.

In ${\Bbb{R}}^3$, being $f:{\Bbb{R}}^3\to {\Bbb{R}}$ linear, its level surfaces are planes perperdicular to the vector given by ${\rm graf} f$.

More details:

I said that the level sets are perpendicular to the gradient is because in the case of the domain is ${\Bbb{R}}^n$ then your covector must be of the form $$f(x_1,x_2,...,x_n)=a_1x_1+a_2x_2+\cdots+a_nx_n,$$ for some constants $a_i$ in $\Bbb{R}$.

Then the gradient also is going to be constant $${\rm grad}f=[a_1,a_2,...,a_n].$$

At a level set $f^{-1}(c):=\{q=(x_1,...,x_n):f(q)=c\}$ consider a point $p\in f^{-1}(c)$ and a curve $\alpha(t)$ such that $$\alpha(0)=p$$ and $$\alpha'(0)=V$$ It is known that $V$ is tangent to the curve $\alpha$ at $p$.

So by differentating $f\circ\alpha(t)=c$ you are going to get (after evaluating at $t=0$) $${\rm grad}f(\alpha(0))\cdot\alpha'(0)=0$$ i.e. $${\rm grad}f(p)\cdot V=0$$ That is $V$ is perpendicular to $[a_1,...,a_n]$.

So, the curve $\alpha$ being in $f^{-1}(c)$ has a tangent $\alpha'$ always perpendicular to $[a_1,...a_n]$ so the curve $\alpha$ is a line in $f^{-1}(c)$.

Being this for each $p$ and each $V$ then the level set $f^{-1}(c)$ is a hyperplane.

With $n=2$ a level line. With $n=3$ a level plane,... et cetera.

Here's a picture of the scalar field pulled back onto the plane it's assigning values to. So what are the basis elements of the covector? The assignment here is $$1 \cdot \mathrm{vert} + 4 \cdot \mathrm{horiz}$$; so I chose basis elements that match the $\vec{x}$'s basis elements. If you changed the 4 and 1 you'd shift the way these level sets "slant"...... (and this would be isomorphic to choosing the same rates-of-increase but changing the direction of rate-of-increase (different basis).

You can also think of moving the covector as changing the weightings of a weighted average (sum, really, but weighted averages are more common in everyday talk).