Tensor #36. Visualization of Forms: Arrows and Stacks텐서 #36. 형식의 시각화: 화살표와 스택

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 37 of the playlist, Visualization of Forms: Arrows and Stacks. Up to the previous post we treated vectors and forms as maps and built everything abstractly. From this post on, over several posts, we draw those objects as pictures.

How to treat visualization

The lecture opens with a caveat. Visualization is another name for intuition, and intuition is a wish to understand things easily. But the maps, dual spaces, metrics fixed by an inner product, and manifolds and coordinate systems we have built so far are all abstract. Just as intuition for the wave function in quantum mechanics comes after learning the rules, this series followed the order of learning the abstract rules first and attaching intuition afterward. Index gymnastics was originally a crutch that hid the abstraction, but we explained the abstraction up front, so now we may draw pictures.

There are two reference books. One is Misner, Thorne, and Wheeler’s Gravitation (MTW). It has many figures and sidebars drawing forms and vectors, but it is vast for a first read. The other is Gabriel Weinreich’s Geometrical Vectors, a thin book of about 120 pages that pushes visualization all the way. It does not use the words form or wedge product, so we must translate it into the language of this series. Both start in three dimensions, and so does this post.

Arrows

Weinreich calls an element of the vector space $\V$ an arrow. It is a three-dimensional vector space over the reals with basis $e_1, e_2, e_3$. We give no inner product. So even if we draw coordinate axes, saying they are perpendicular has no meaning, and they are just lines drawn along the basis directions. An arrow takes the form $a = a^\mu e_\mu$ and is fixed by three components and the basis.

Rotating the coordinate system leaves the arrow as it is and changes only the components. A more notable change is stretching or shrinking the size of the basis. If we change a basis measured in centimeters to meters, the basis vectors become 100 times longer, and the components that describe the same arrow shrink to one hundredth. This is because the components move opposite to the basis, so the product $a^\mu e_\mu$ stays the same. Classical tensor textbooks call such components a contravariant vector. This series has simply called it a vector, and arrow is Weinreich’s name.

Stacks

An element of the dual space $\V^*$ is called a stack. It is three-dimensional with basis $e^1, e^2, e^3$. A stack belongs to a vector space, so it is also a vector, and it lives in the dual space of the arrows, so it is also a covector. Whichever name we use, it is an element of a different space from the arrows.

The picture is a pile of planes. The basis stack $e^1$ is an infinite set of planes with one plane for each unit length along $e_1$, and a small arrowhead marks the direction. In two dimensions the set of planes becomes a set of lines. The planes can be numbered like $\dots, -1, 0, 1, 2, \dots$, and we imagine planes with fractional numbers between them as well. If length and direction fix an arrow, density and direction of the planes fix a stack. If the basis unit is a centimeter, the basis stack has one plane per centimeter.

We draw $e^2, e^3$ the same way. That the sets of planes seem to overlap at a point is only a matter of the drawing. They are different stacks, so overlapping is no problem.

Q. Do we draw this picture inside the space we live in?
A. It is not drawn there. Every point of space carries a tangent space, and an arrow is drawn inside the tangent space of that point. A stack is drawn inside the cotangent space of the same point. We have a habit of drawing vector fields as arrows in real space, but an arrow belongs to the tangent space at a point, which is a different thing from the point of space itself.

Adding stacks

A stack is also an element of a vector space, so we can add stacks. Take $e^1 + e^3$. Seen from the side, the two sets of planes form a grid of lines. Joining the diagonals across the cells of that grid gives a new set of planes, and its arrowhead points between the two arrowheads. Joining the opposite diagonals gives $e^3 - e^1$. (This explanation reads the picture off the computation that the planes of the sum have the form $x + z = n$ and was not covered in the lecture.) The result of such an addition is still a single stack. The lattice or egg-crate shape comes from a different operation than the sum, and that is the subject of the next post.

The electric field is a stack

Consider a parallel-plate capacitor. The displacement $d$ that expresses the distance between the plates is an arrow. If we change the unit from 1 centimeter to meters, the number for the displacement shrinks from 1 to 0.01. The actual distance between the plates is unchanged. For the same voltage the electric field is $E = V/d$, so the number that was $V/(1\,\text{cm})$ in centimeters becomes $V/(0.01\,\text{m})$ in meters and grows 100 times. Where an arrow would shrink, the electric field grew.

In mathematics this is a test and no mere convention. Things that change differently when we stretch, shrink, or rotate the coordinate system are different kinds of objects. Under rotation it reacts exactly like an arrow, so the difference shows when the size changes. The electric field is a stack. When the basis changes to meters, a set of planes with one plane per centimeter has 100 planes per meter, and this density is the component of the stack. The component grows in the same direction as the basis, so it is covariant.

The picture also lets us read the voltage. If we regard the planes of the stack as surfaces of equal voltage (equipotential surfaces), the voltage between two places is the number of planes the displacement arrow between them crosses. (This interpretation is added for understanding and was not covered in the lecture. The lecture explains the number of planes crossed in general in the next section.)

Q. Was it wrong to draw the electric field as an arrow?
A. In problems with only rotation, drawing it as an arrow causes no discrepancy. But when the unit changes, the components grow and do not shrink like an arrow’s. We use an arrow only because the picture is simple, and the rule by which the components change belongs to a stack.

The relation between arrows and stacks

If there is a metric $g$, we put an arrow $a$ into one slot to get $g(a, \cdot)$, which is a dual vector waiting for another vector, that is, a stack. So the two are often paired, but they are not the same object. There is a temptation to draw a stack with a single arrow whose length is the stack’s density, a short arrow for a sparse stack and a long arrow for a dense one. But when the basis size changes, the arrow components shrink and the stack components grow, so the two arrows change differently. To understand the geometry correctly we must use the picture of stacked planes.

The dual space mapping

Now we read in pictures the rule we have used all along, that a dual vector takes a vector and gives a real number. For an arrow $a$ and a stack $b$, Weinreich defines a dot product $a \cdot b$. It is the dual space mapping, which differs from an inner product. It is defined without an inner product, and in components it is $a^\mu b_\mu$.

The value of the dual space mapping is the number of planes of the stack that the arrow crosses. Lay the arrow on the stack and count the planes. Crossing four and a half planes gives 4.5, and crossing two planes gives 2. If the arrow is parallel to the planes it crosses none, so the value is 0, and then we say the arrow is contained in the stack. Where the arrow is placed does not matter. What matters is only that both objects come from the tangent space and cotangent space of the same point. A stack in the cotangent space of another point cannot take this arrow.

This value does not change when the basis size changes. Stretching the basis grows the stack’s components and shrinks the arrow’s components, so they cancel. This point is made at the start of the next lecture, and it becomes the test for every operation we build from now on, whether it is invariant under changes of coordinates.

Q. What fixes the spacing of the planes of a stack?
A. The components. In two dimensions, if the stack components are $(\alpha_1, \alpha_2)$, the planes have the form $\alpha_1 x + \alpha_2 y = n$, and when we can measure distance the spacing between neighboring planes is $1/\sqrt{\alpha_1^2 + \alpha_2^2}$. The larger the components, the denser the planes. (This formula is added for understanding and was not covered in the lecture.)

In the toy below, drag the arrow with the blue point and the stack with the green point. The planes crossed are drawn bold, and when you change the basis size the components change oppositely but the number of planes crossed stays the same.

Summary

What we confirmed in this post is the following.

• An arrow is an element of the vector space, and its components shrink when the basis size grows. In this series it is a vector.
• A stack is an element of the dual space, drawn as a pile of planes. Its components are the density of the planes and grow along with the basis size.
• The electric field grows when the unit changes, so it must be read as a stack. The displacement is an arrow.
• The dot product of an arrow and a stack is the dual space mapping, which differs from an inner product. It is read as the number of planes the arrow crosses, and it is invariant under changes of coordinates.

In three dimensions the dimensions of the $p$-forms for $p = 0, 1, 2, 3$ are $1, 3, 3, 1$. We have drawn 1-forms and 1-vectors, so what remains are 2-forms, 2-vectors, 3-forms, and 3-vectors. In the next post we continue drawing those pictures.