What *is* the working form of a 1-vector in geometric algebra?

Question

Consider the geometric algebra definition of 0-vectors (scalar), 1-vectors (vector), and the inner product.

Let $a$ and $b$ be 1-vectors. Then $a + b = c$, where $c$ is another 1-vector.

Now, consider $c \cdot c$:

$c \cdot c = |a|^2 + |b|^2 + 2(a \cdot b)$

Fantastic, easy.

Now, how do I compute $a \cdot b$?

Do I have to somehow relate $a$ and $b$ down to $\hat i$ and $\hat j$ in order to do a computation? In that case, what is the difference between the normal vector algebra method by which I compute $a \cdot b$?

Example of how it would be done in normal vector algebra:

$\hat{i} \cdot \hat{j} = 0$

$a = \alpha \hat i + \beta \hat j$

$b = \gamma \hat i + \delta \hat j$

$\therefore a \cdot b = \alpha \gamma + \beta \delta$

Answer 1

So I think the real question here is, what do we mean by "coordinate-free"?

One way to put it is that we can do algebraic operations with the objects in questions--vectors and tensors and so on--prior to choosing a set of coordinates or a basis.

Such a system naturally relies upon operations that are independent of the choice of coordinates--in your example, a dot product. What a coordinate-free approach really relies on are these coordinate-independent quantities.

For instance, one coordinate-free approach to using tensors might only talk about tensors as linear maps on collections of vectors and dual vectors, defining the transformation law of tensors to counteract the transformations of their vector and dual vector arguments. Operations like contractions, traces, and so on do not depend on the choice of coordinates, and so they are natural to use.

If you're familiar with differential geometry (e.g. through general relativity), then you might have seen index notation. Originally, index notation referred to components in a basis. So-called "abstract" index notation uses the same general look of things but really uses the idea of tensors as maps that I outlined above. So abstract index notation tries to have the convenience of index notation but the power of a coordinate-free approach all combined in a single notation.

A subtle point with differential geometry is that often we say we're working in some coordinate chart: explicitly, a map from the manifold to a piece of a flat vector space, where the components of vectors in that space are the coordinates of a point on the manifold. Think of how we parameterize a sphere into two coordinates--often, two angles. Even if the sphere is embedded in a 3d ambient space, it only requires two coordinates to describe all the points on it. Though the chart is arbitrary, it is often necessary to acknowledge its presence, as much of differential geometry's calculations are actually done in the space that the chart maps to: the coordinate space.

Geometric algebra's approach to differential geometry is different: the embedded view takes any manifold and puts it in an infinite dimensional vector space. While you can parameterize a manifold, having a prescription for that manifold's unit pseudoscalar at every point gives you access to a lot of information without having to pick a parameterization.

In short: if you're looking for a formalism that allows you to do hard computations (with real quantities, chosen from some example or physical system) without breaking into a basis, you're out of luck. I'm not sure any such system exists, or if it makes sense to consider one. Coordinate-free may not really be "coordinate-free" in that sense, but it does mean you can do a lot of algebra on quantities, simplify results as much as possible, and then break things down into a basis to get hard numbers.

Answer 2

If you want to work coordinate-freely, you have to think carefully about what questions are coordinate-free.

Now, imagine you're a formless blob floating in empty space, but you have a ruler. Suddenly a single vector appears. What can you tell about the vector? You can certainly measure its length with the ruler. As a matter of fact, there's nothing more that you can meaningfully measure: the angle relative to the ruler is meaningless, since you're free to position your ruler as you like.

So for one vector, call it $a$, the only information available is that

1. it is a vector
2. it has a squared length $a^2$

which is exactly what you can extract from a single vector in geometric algebra. Now, you might argue that in $d$-dimensions you have the $d$ components $a_i$ of the vector. However, unless there is a preferred frame of reference, that is not true, since those components depend entirely on your choice of coordinates, except for the one constraint $\sum_{i} a_i^2 = a^2$. In other words, any list of numbers $a_i$ obeying that one constraint are the components of the vector $a$ in some coordinate system. That constraint is exactly equivalent to 2. above.

Now, imagine another vector, call it $b$, appears. Now you can measure the lengths of both vectors, and the angle between them. In addition, they define a plane. So now have the following pieces of geometric information that you can define and measure using your ruler:

1. $a$ is a vector, and it has length squared $a^2$
2. $b$ is a vector, and it has length squared $b^2$
3. there is an angle between $a$ and $b$, call it $\theta_{ab}$
4. there is a plane defined by $a$ and $b$

Geometric algebra provides exactly these pieces of information: 1. and 2. are trivial, 3. can related to the measurements by, for example $a \cdot b = |a||b|\cos(\theta_{ab})$, and 4. is represented by $\frac{a\wedge b}{|a \wedge b|}$. Again, you might argue that you, as a matter of fact, have $2d$ pieces of information in the components $a_i$ and $b_i$. However, that is not true, as any two lists of numbers $a_i, b_i$ obeying the conditions $$ \sum_i a_i^2 = a^2\\ \sum_i b_i^2 = b^2\\ \sum_i a_i b_i = |a||b| \cos(\theta_{ab}) $$ are components of $a$ and $b$ in some coordinate system. So there is no more information there.

Note that you cannot, for example, say anything more about the plane than that it exists, and that $a$ and $b$ are contained in the plane, since there is nothing else to compare it to. So this is indeed all the information available to you, unless more geometric objects (vectors, bivectors, $k$-blades etc) are given for reference.

So to you specific problem: you have two vectors $a$ and $b$, and a third one is known to be related to them by $c = a + b$. So here you have the same pieces of information that can be extracted from $a$ and $b$ as in the previous example, and in addition the relation that fixes $c$ in terms of $a$ and $b$. So you have to measure/assume known/somehow gain access to three independent pieces of information, for example $|a|, |b|$ and $\theta_{ab}$ (or just as well $a \cdot b$ directly) in order to define your system. Once that is done, you can indeed relate the number $c^2 = a^2 + b^2 + 2 a\cdot b$ to the information you have.

Notice that the result you get in components, $a\cdot b = \alpha \gamma + \beta \delta$ does not provide you any more information, as the components $\alpha, \gamma, \beta, \delta$ are completely arbitrary up to the constraints described above. Simply ask yourself this: if you are again that formless blob with a ruler, how do you use that result of yours to arrive at the value of $a\cdot b$? Will you end up essentially measuring $a\cdot b$ anyway while doing that?

There are two points to take out of this:

• using geometric algebra, you can relate any quantity (quantity not being limited to scalars here) that is defined independently of coordinates to any other quantity that is defined independently of coordinates, if you have enough relations to fix them.
• There is a common illusion that when you write something down in general coordinates, you have in some sense solved the problem more thoroughly than when you simply state the maximal number of invariant relations between the given quantities and the result. This is however just an illusion.

Answer 3

You have vector $a, b, c$, and the squared length $a^2, b^2, c^2$. Now if we add the relationship $$ c = a + b$$, we can define $$ 2a\cdot b = c^2 - a^2 - b^2$$. Note the logic flow is from $c^2$ to $a \cdot b$, not from $a\cdot b$ to $c^2$. In this sense, the inner product is index free.