In geometric algebra, what is the dot product of a vector and a scalar? what is the wedge product of a vector and a scalar?

Question

I am watching a series on geometric calculus by Alan Mcdonald and in the first episodes he states that for any vector u and multivector M: $uM = u \cdot M + u \land M$

This doesn't really take into account what happens with the 0-grade i.e scalar part of M, let's call it s.

For the identity to be true either $u \cdot s$ or $u \wedge s$ has to be zero, and from what i read previously in some instances $u \land s = us$, which implies $u \cdot s = 0$, which kind of makes intuitive sense i guess?

Answer 1

The general dot product formula for two k-vectors $ a_r, b_s $, of grades r and s respectively, is typically defined as a grade selection of the following sort:

$$a_r \cdot b_s={\left\langle{{ a_r b_s }}\right\rangle}_{{\left\lvert {r - s} \right\rvert}}.$$

Similarly, the wedge product of the same k-vectors is defined as

$$a_r \wedge b_s={\left\langle{{ a_r b_s }}\right\rangle}_{r + s}.$$

Both of these can be extended to multivectors by decomposing that multivector into component k-vectors.

By these definitions, if $u$ is a scalar, and $v$ is a k-vector, we have

$u \cdot v = u v = u \wedge v,$

which also holds if $v$ is a multivector (and $u$ still a scalar.)

Answer 2

The issue is that the "dot product" isn't theoretically nice. There are two definitions I've seen.

In terms of the geometric product, the first is $$ A_r\cdot B_s = \langle A_rB_s\rangle_{|r-s|} $$ where $A_r$ is an $r$-vector and $B_s$ an $s$-vector, and $\langle{-}\rangle_k$ is projection onto the $k$-vector part. This definition fails the identity you cite for exactly the reason you give.

The second definition (which I will disambiguate using a fat dot) is $$ A_r\bullet B_s = \begin{cases} \langle A_rB_s\rangle_{|r-s|} &\text{if }(r\ne0\text{ and }s\ne0)\text{ or }(r = s = 0), \\ 0 &\text{otherwise}. \end{cases} $$ This definition is really just tailor made to make your identity work; the explicit split on grade makes it more difficult to work with.

The theoretically best incarnation of the inner product is the (left) contraction $$ A_r\mathbin\rfloor B_s = \langle A_rB_s\rangle_{s-r} $$ where we define the $k$-vector part for negative $k$ to be zero. This satisfies your identity: $$ uM = u\mathbin\rfloor M + u\wedge M. $$ To demonstrate this products theoretical niceties, one way to define it is via the equation $$ \langle(A\wedge B)C\rangle_0 = \langle A(B\mathbin\rfloor C)\rangle_0 $$ for all multivectors $A, B, C$. In other words, it is the adjoint of the exterior product under the scalar product. It also has a simple geometric interpretation: if $$ [A] = \{v \in \mathbb R^n \;:\; v\wedge A = 0\} $$ for a blade $A$ is the subspace it represents, and $B$ is another blade, then $$ [A\mathbin\rfloor B] = \begin{cases} V &\text{if }[A]\cap[B]^\perp \ne \{0\}, \\ [A]^\perp\cap[B] & \text{otherwise}. \end{cases} $$ The first case is equivalent to $A\mathbin\rfloor B = 0$ and we say that $[A]$ is partially orthogonal to $[B]$. The above characterization is true regardless of the grades of $A$ and $B$, whereas a similar statement using $\cdot$ would not be.

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It is indeed true that $$ M\wedge a = a\wedge M = aM $$ for any scalar and multivector $M$. One way to see this simply is if the wedge product is defined in terms of the geometric product $$ A_r\wedge B_s = \langle A_rB_s\rangle_{r+s}. $$