Is the definition $a\cdot b= 0.5(ab+ba)$ in geometric algebra justified, or is it mostly arbitrary?

Question

According to one of the basic axioms of geometric algebra, the square of a vector with itself is a scalar. For two vectors $a$ and $b$, this results in $ab+ba = (a+b)^2-a^2-b^2$. Therefore $ab+ba$ must be a scalar. When then "define"

$$ a \cdot b = \frac{1}{2}(ab+ba)$$

The dot product is connected to geometry because it relates to the cosine of the angle between the vectors $a$ and $b$. Could we have just as easily "defined" the dot product to be something even more complicated, like $ab+ba = 3.5\lVert a\rVert \lVert b\rVert \log(\cos(\sin([\textrm{angle between} a \textrm{ and } b])))$ (perhaps notwithstanding the infinity in the logarithm)? This formula also respects the fact that $ab+ba$ is a scalar. Is the cosine relationship actually required?

Answer 1

A geometric algebra is a structure as follows: You have a vector space $V$, then you have an inner product (i.e. a non-degenerate, symmetric bilinear form) $V\ni a,b\mapsto a\cdot b$, and then the geometric algebra is some algebra $A$ that contains scalars in $\mathbb{C}$ and vectors in $V$ (and bi-vectors and tri-vectors etc.), such that for $a,b\in V$ we have

1. $aa$ (the algebra multiplication) $= a\cdot a 1_\mathbb{C}$ (the bilinear form).

We also have:

2. $a\cdot b$ must be bilinear: $a\cdot(b+c)=a\cdot b+a\cdot c$ etc.
3. $a\cdot b$ is symmetric: $a\cdot b$=$b\cdot a$

Now we can derive the conclusion: Let $a,b\in V$ be two vectors. We can do two things:

1. We can take the inner product of any two vectors, and this is a scalar: $(a+b)\cdot(a+b)$ is a scalar, $a\cdot a$ is a scalar, $a\cdot b$ is a scalar, $b\cdot b$ is a scalar.
2. We can take the algebra-multiplication of two vectors. Only when multiplying a vector with itself, will this be a scalar.

So using the inner product, we have $(a+b)\cdot(a+b) = a\cdot a + a\cdot b + b\cdot a + b\cdot b$, where everything is a scalar. Furthermore, $b\cdot a = a\cdot b$ by symmetry. On the other hand, using the algebra product we have $(a+b)(a+b) = aa + ab + ba + bb$, where $ab$ and $ba$ are not scalars, but bi-vectors.

Using that $aa=a\cdot a1_\mathbb{C}$, we obtain that $$a\cdot a1_\mathbb{C}+b\cdot b1_\mathbb{C}+2a\cdot b1_\mathbb{C}=(a+b)\cdot(a+b)1_\mathbb{C} $$$$= (a+b)(a+b)=aa + ab + ba + bb = a\cdot a1_\mathbb{C}+b\cdot b1_\mathbb{C} + ab+ba$$ Therefore $$a\cdot b1_\mathbb{C} = \frac 12(ab+ba).$$

This follows from the fact that $\cdot$ is an inner product.

When you replace $\cdot$ with any other function that is bilinear and symmetic, everything still works out. However, your funny function is not bilinear, so thing will not work out as smoothly.

Answer 2

There isn't actually that much choice in what to call such a symmetric product of vectors. Consider first the product of two vectors $ a = \sum_i x_i \mathbf{e}_i $ and $ b = \sum_i y_i \mathbf{e}_i $, assuming that $ \left\{ { \mathbf{e}_1, \cdots \mathbf{e}_N } \right\} $ is an orthonormal basis. $$\begin{aligned} a b &= \sum_{1 \le i,j \le N} x_i \mathbf{e}_i y_j \mathbf{e}_j \\ &= \sum_{i = j} x_i y_i \mathbf{e}_i^2 + \sum_{i \ne j} x_i y_j \mathbf{e}_i \mathbf{e}_j \\ &= \sum_{i = 1}^N x_i y_i + \sum_{i \ne j} x_i y_j \mathbf{e}_i \mathbf{e}_j \\ &= \sum_{i = 1}^N x_i y_i + \sum_{1 \le i < j \le N} \left( { x_i y_j - x_j y_i } \right) \mathbf{e}_i \mathbf{e}_j \\ &= \sum_{i = 1}^N x_i y_i + \sum_{1 \le i < j \le N} \begin{vmatrix} x_i & x_j \\ y_i & y_j\end{vmatrix} \mathbf{e}_i \mathbf{e}_j.\end{aligned}$$ Reversing the products, we find $$\begin{aligned} b a &= \sum_{i = 1}^N y_i x_i + \sum_{1 \le i < j \le N} \begin{vmatrix} y_i & y_j \\ x_i & x_j\end{vmatrix} \mathbf{e}_i \mathbf{e}_j \\ &= \sum_{i = 1}^N x_i y_i - \sum_{1 \le i < j \le N} \begin{vmatrix} x_i & x_j \\ y_i & y_j \end{vmatrix} \mathbf{e}_i \mathbf{e}_j.\end{aligned}$$

Forming the symmetrization, there isn't any choice about the specific value of this scalar term $$ \frac{1}{{2}} \left( { a b + b a } \right) = \sum_{i = 1}^N x_i y_i,$$ and we are forced to identify the symmetric sum as the dot product.

Incidentally, we also have no choice about the specific value of the grade-2 portion of the vector product, which we may similarily pick out using an antisymmetrical sum $$ \frac{1}{{2}} \left( { a b - b a } \right) = \sum_{1 \le i < j \le N} \begin{vmatrix} x_i & x_j \\ y_i & y_j \end{vmatrix} \mathbf{e}_i \mathbf{e}_j.$$ We can then define the grade-2 part of the vector product as the wedge product. From the coordinate expansion above, it is clear that this has the desired properties $ a \wedge a = 0, a \wedge b = - b \wedge a $. It's also possible to show that this has the $ \sin\theta $ (parallelogram area) property that we desire of the wedge product.