Some intuition for Geometric Algebra product contractions and expansions
2020-11-15, updated 2020-11-15 — math ⇦Making zoontycoon2 ultimate edition run on wine on ubuntu. – Revista Contigo⇨
One of the best materials for learning about inner product contractions is, The Inner Products of Geometric Algebra, by Leo Dorst. Here I'll write some intuition to help you remember those results.
First, notice that every product is linear, therefore we can always think about what happens in a product by thinking about what happens with the orthogonal and parallel components. We will use the term "dimension", in a very informal way. For example, if I have the vector \(e_1\) and the vector \(e_2\), and we "sum" their dimensions, we obtain \(e_1 \wedge e_2\). If we have the blade \(e_1 \wedge e_2\) and we remove the dimension \(e^1\), we obtain \(e^2\).
1 \[a \cdot (B \wedge C)\]
Lets start with: \[a \cdot (B \wedge C)\] , where \(B\) and \(C\) are \(k\)-blades with \(k\ge1\). What this formula is saying, intuitively, is that we take 2 blades, \(B\) and \(C\), we add their dimensions, to get a blade of higher dimension, and then we remove the dimensions of \(a\).
For the product to not be null. at most one of \(B\) or \(C\) has the dimension of \(a\). If it is \(B\), then the formula is equivalent to \[(a \cdot B) \wedge C)\] Because removing the dimension of \(a\) from \(B \wedge C\), is equivalent to removing it from \(B\), and then adding the dimension \(C\).
If instead it is \(C\), then, we need to remove the dimension of \(a\) from \(C\), so we will have a term \[\pm? B \wedge (a \cdot C)\], where we need to reason about what should the sign be.
We can split \(C\) into 2 parts, \(a \wedge C'\), therefore we have \[a \cdot(B \wedge a \wedge C') = a \cdot(a \wedge \hat{B} \wedge C') \], where \(\hat{B}\) is the grade inversion of B, i.e. equal to \(-B\), if \(B\) is odd. Therefore,
\[a \cdot (B \wedge C) = (a \cdot B) \wedge C) + \hat{B} \wedge (a \cdot C)\]
In other words, when taking the dot product of a vector with a blade, we can just "put" the vector next to the terms it would cancel, while taking into account grade inversion. For exmple:
\[e_1 \cdot(e_2 \wedge e^1 \wedge e^3) = - e_2 \wedge (e_1 \cdot e^1 \wedge e^3) = -e_2 \wedge e_3\]
2 Contraction
The inner product/ dot product are slightly ambiguous. If we have \((A \cdot B)\), we don't know if we are removing dimensions from \(B\) or from \(A\), i.e. supposing that \(A\) is an \(a\)-blade and \(B\) is a \(b\)-blade, we don't know if \((A \cdot B)\) is an \(a-b\)-blade or \(b-a\)-blade, since we don't know which of them has the higher dimensions. So we define the left contraction inner product \(A \lrcorner B\) to be removing the dimensions of \(A\) from \(B\), i.e. an \(a-b\)-blade (or 0, if \(b>a\).
With this, we can think about: \[(A \wedge B) \lrcorner C =(A \lrcorner (B \lrcorner C)) \] i.e. removing the dimensions of both \(A\) and \(B\) from \(C\) is the same as removing the dimensions of \(B\) from \(C\), and then removing the dimensions of \(A\) from the result.