Exterior Algebra Notes #4: The Interior Product

January 27, 2019

Vector spaces are assumed to be finite-dimensional and over $R$ . The grade of a multivector $α$ will be written $∣ α ∣$ , while its magnitude will be written $∥ α ∥$ . Bold letters like $u$ will refer to (grade-1) vectors, while Greek letters like $α$ refer to arbitrary multivectors with grade $∣ α ∣$ .

More notes on exterior algebra. This time, the interior product $α \cdot β$ , with a lot more concrete intuition than you’ll see anywhere else, but still not enough.

I am not the only person who has had trouble figuring out what the interior product is for. This is what I have so far…

1. The Interior Product

The last main tool of exterior algebra is the interior product, written $α \cdot β$ or $ι_{α} β$ . It subtracts grades ( $∣ α \cdot β ∣ = ∣ β ∣ - ∣ α ∣$ ) and, conceptually, does something akin to ‘dividing $α$ out of $β$ ’. It’s also called the ‘contraction’ or ‘insertion’ operator. We use the same symbol as the inner product because we think of it as a generalization of the inner product: when $∣ α ∣ = ∣ β ∣$ , then $α \cdot β = β \cdot α = ⟨ α, β ⟩$ .

Its abstract definition is that it is adjoint to the wedge product with respect to the inner product:

⟨ γ \land α, β ⟩ = ⟨ α, γ \cdot β ⟩ (1)

In practice this means that it sort of ‘undoes’ wedge products, as we will see.

When we looked at the inner product we had a procedure for computing $⟨ a \land b, c \land d ⟩$ . We switched from the $\land$ inner product to the $\otimes$ inner product, by writing both sides as tensor products, with the right side antisymmetrized using $Alt$ :¹

¹Recall that we basically elect to antisymmetrize one side because if we did both we would need an extra factor of $1/ n!$ for the same result. It might be that there are abstractions of this where you do need to do both sides (for instance if $a \cdot b \neq = b \cdot a$ ?) ↩

⟨ a \land b, c \land d ⟩ = ⟨ a \otimes b, Alt (c \otimes d)⟩ = ⟨ a \otimes b, c \otimes d - d \otimes c ⟩_{\otimes} = ⟨ a, c ⟩_{V} ⟨ b, d ⟩_{V} - ⟨ a, d ⟩_{V} ⟨ b, c ⟩_{V} = (a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c)

Interior products directly generalize inner products to cases where the left side has a lower grade²

²It is probably possible to generalize to either side having the lower grade, but it’s not normally done that way. I want to investigate it sometime. ↩

, (which is why we use

\cdot

for both), and can be computed with the exact same procedure:

a \cdot (b \land c) = a \cdot (b \otimes c - c \otimes b) = (a \cdot b) c - (a \cdot c) b

A general formula for the interior product of a vector with a multivector, which can be deduced from the above, is

a \cdot (b \land β) = (a \cdot b) \land β - b \land (a \cdot β)

2. Projection

The intuitive meaning of the interior product is related to projection. We can construct the projection and rejection operators of a vector onto a multivector with:

β_{a} = proj_{a} β = \frac{1}{∣ a ∣ ^{2}} a \land (a \cdot β) β_{⊥ a} = proj_{⊥ a} β = \frac{1}{∣ a ∣ ^{2}} a \cdot (a \land β)

β = β_{a} + β_{⊥ a} = \frac{1}{∣ a ∣ ^{2}} [a \land (a \cdot β) + a \cdot (b \land β)]

To understand this, recall that the classic formula for projecting onto a unit vector is:

b_{a} = a (a \cdot b)

That is, we find the scalar coordinate along $a$ , then multiply by $a$ once again. With multivectors, $a \cdot β$ is not a scalar, so we can’t just use scalar multiplication – so it makes some sense that it would be replaced with $\land$ .³

³the other candidate would be $\otimes$ , but we’d like the result to also be a multivector so it makes sense to only consider $\land$ . ↩

b_{a} = a \land (a \cdot b)

The classic vector rejection formula is

b_{⊥ a} = b - b_{a} = b - a (a \cdot b)

Using the interior product we can write this as

b_{⊥ a} = a \cdot (a \land b) = (a \cdot a) b - (a \cdot b) a = b - b_{a}

The multivector version $a \land β$ is only non-zero if $β$ has a component which does not contain $a$ – all $a$ -ness is removed by the wedge product, leaving something like $a \land β_{⊥ a}$ . Then $a \cdot a \land β_{⊥ a} = β_{⊥ a}$ .

The correct interpretation of $a \cdot β$ , then, is a lot like what it means when $β = b$ : it’s finding the ‘ $a$ -component’ of $β$ . It’s just that, when $β$ is a multivector, the ‘ $a$ -coordinate’ is no longer a scalar.

For example this is the ‘ $x$ ‘-component of a bivector $b \land c$ :

x \cdot (b \land c) = b_{x} c - c_{x} b = b_{x} (c_{x} + c_{⊥ x}) - c_{x} (b_{x} + b_{⊥ x}) = b_{x} c_{⊥ x} - c_{x} b_{⊥ x}

Note that the result doesn’t have any $x$ factors in it.

So, we might things up like this: For a unit multivector $α$ , the meaning of $α \cdot β$ is to find $β_{α}$ , the ‘ $α$ component’ of $β$ .

We can remove the stipulation that $α$ be a unit multivector, but it requires being a bit careful. To illustrate why, consider an example with just vectors. What should be the value of $v_{5 x}$ ? Probably as $\frac{1}{5} v_{x}$ , so that $v_{x} x = v_{5 x} 5 x$ . Likewise, we need to divide through by the magnitude of $α$ , so it’s actually $\frac{1}{∥ α ∥ ^{2}} ι_{α} β = β_{α}$ .

Unfortunately the rejection formula doesn’t work if $α$ is a multivector. It’s still true that $α \cdot β$ gives the ‘ $α$ -coordinate’ of $β$ , if there is one. But we can only use $β_{⊥ α} = β - \frac{1}{∥ α ∥ ^{2}} α \land (α \cdot β)$ . The problem is that there are cases where both $α \land β = α \cdot β = 0$ , such as for $x \land y$ and $y \land z$ .⁴

⁴I think there’s a way to make it work. It looks something like: for each basis multivector of lower grade, remove it from both sides, like $(x \cdot α) \cdot (x \cdot β)$ . But that’s complicated and will have to be saved for the future. ↩

3. More identities

We can use $ι$ to prove a few more vector identities. First, note that $⋆$ is just a special case of $ι$ .⁵

⁵It is easier to use the $\cdot$ notation for inner products, since after all they are a special case of interior products. But sometimes I use $⟨, ⟩$ anyway when it makes things clearer. ↩