Exterior Algebra Notes #4: The Interior Product

[January 27, 2019]

Vector spaces are assumed to be finite-dimensional and over \(\bb{R}\). The grade of a multivector \(\alpha\) will be written \(\| \alpha \|\), while its magnitude will be written \(\Vert \alpha \Vert\). Bold letters like \(\b{u}\) will refer to (grade-1) vectors, while Greek letters like \(\alpha\) refer to arbitrary multivectors with grade \(\| \alpha \|\).

More notes on exterior algebra. This time, the interior product \(\alpha \cdot \beta\), 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 \(\alpha \cdot \beta\) or \(\iota_{\alpha} \beta\). It subtracts grades (\(\| \alpha \cdot \beta \| = \| \beta \| - \| \alpha \|\)) and, conceptually, does something akin to ‘dividing \(\alpha\) out of \(\beta\)’. 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 \(\| \alpha \| = \| \beta \|\), then \(\alpha \cdot \beta = \beta \cdot \alpha = \< \alpha, \beta \>\).

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

\[\boxed{\< \gamma \^ \alpha, \beta \> = \< \alpha, \gamma \cdot \beta \>} \tag{1}\]

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 \(\< \b{a \^ b}, \b{c \^ d} \>\). We switched from the \(\^\) inner product to the \(\o\) inner product, by writing both sides as tensor products, with the right side antisymmetrized using \(\text{Alt}\):1

\[\begin{aligned} \< \b{a \^ b}, \b{c \^ d} \> &= \< \b{a \o b}, \text{Alt}(\b{c \o d}) \> \\ &= \< \b{a \o b}, \b{c \o d - d \o c} \>_{\o} \\ &= \< \b{a, c} \>_V \< \b{b, d} \>_V - \<\b{a, d} \>_V \< \b{b, c} \>_V \\ &= (\b{a \cdot c}) (\b{b \cdot d}) - (\b{a \cdot d})( \b{b \cdot c}) \end{aligned}\]

Interior products directly generalize inner products to cases where the left side has a lower grade2, (which is why we use \(\cdot\) for both), and can be computed with the exact same procedure:

\[\begin{aligned} \b{a} \cdot (\b{b \^ c}) &= \b{a} \cdot (\b{b \o c - c \o b}) \\ &= (\b{a} \cdot \b{b}) \b{c} - (\b{a} \cdot \b{c}) \b{b} \end{aligned}\]

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

\[\b{a} \cdot (\b{b} \^ \beta) = (\b{a} \cdot \b{b}) \^ \beta - \b{b} \^ (\b{a} \cdot \beta)\]

1.1 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:

\[\beta_{\b{a}} = \text{proj}_{\b{a}} \beta = \frac{1}{\| \b{a} \|^2} \b{a} \^ (\b{a} \cdot \beta) \\ \beta_{\perp \b{a}} = \text{proj}_{\perp \b{a}} \beta = \frac{1}{\| \b{a} \|^2} \b{a} \cdot (\b{a} \^ \beta)\] \[\beta = \beta_{\b{a}} + \beta_{\perp \b{a}} = \frac{1}{\vert \b{a} \vert^2} [\b{a} \^ (\b{a} \cdot \beta) + \b{a} \cdot (\b{b} \^ \beta)]\]

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

\[\b{b}_{\b{a}} = \b{a} (\b{a} \cdot \b{b})\]

That is, we find the scalar coordinate along \(\b{a}\), then multiply by \(\b{a}\) once again. With multivectors, \(\b{a} \cdot \beta\) is not a scalar, so we can’t just use scalar multiplication – so it makes some sense that it would be replaced with \(\^\).3

\[\b{b}_{\b{a}} = \b{a} \^ (\b{a} \cdot \b{b})\]

The classic vector rejection formula is

\[\b{b}_{\perp \b{a}} = \b{b} - \b{b}_{\b{a}} = \b{b} - \b{a} (\b{a} \cdot \b{b})\]

Using the interior product we can write this as

\[\b{b}_{\perp \b{a}} = \b{a} \cdot (\b{a} \^ \b{b}) = (\b{a \cdot a}) \b{b} - (\b{a} \cdot \b{b}) \b{a} = \b{b} - \b{b}_{\b{a}}\]

The multivector version \(\b{a} \^ \beta\) is only non-zero if \(\b{\beta}\) has a component which does not contain \(\b{a}\) – all \(\b{a}\)-ness is removed by the wedge product, leaving something like \(\b{a} \^ \beta_{\perp \b{a}}\). Then \(\b{a} \cdot \b{a} \^ \beta_{\perp \b{a}} = \beta_{\perp \b{a}}\).

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

For example this is the ‘\(\b{x}\)‘-component of a bivector \(\b{b \^ c}\):

\[\begin{aligned} \b{x} \cdot (\b{b \^ c}) &= b_x \b{c} - c_x \b{b} \\ &= b_x (\b{c}_x + \b{c}_{\perp x}) - c_x (\b{b}_x + \b{b}_{\perp x}) \\ &= b_x \b{c}_{\perp x} - c_x \b{b}_{\perp x} \end{aligned}\]

Note that the result doesn’t have any \(\b{x}\) factors in it.

Once more, to be extra explicit about this intuition:

For a unit multivector \(\alpha\), the meaning of \(\alpha \cdot \beta\) is to find \(\beta_{\alpha}\), the ‘\(\alpha\) component’ of \(\beta\).

We can remove the stipulation that \(\alpha\) 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 \(v_{5x}\)? Probably as \(\frac{1}{5} v_x\), so that \(v_x \b{x} = v_{5x} 5 \b{x}\). Likewise, we need to divide through by the magnitude of \(\alpha\), so it’s actually \(\frac{1}{\Vert \alpha \Vert^2} \iota_{\alpha} \beta = \beta_{\alpha}\).

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


1.2 Derivations

If we consider our projection/rejection operations as operators, writing \(L_{\b{a}} \beta = \b{a} \^ \beta\) and \(\iota_{\b{a}} \beta = \b{a} \cdot \beta\), then:

\[\iota_{\b{a}} \circ L_{\b{a}} + L_{\b{a}} \circ \iota_{\b{a}} = \vert \b{a} \vert^2 (\text{proj}_{\b{a}} + \text{proj}_{\perp \b{a}}) = \vert \b{a} \vert^2\]

Since \(\iota^2 = L^2 = 0\), this could also be written as

\[(\iota_{\b{a}} + L_{\b{a}})^2 = \vert \b{a} \vert^2\]

And in fact this works (although the interpretation is trickier) with different vectors for each term:

\[(\iota_{\b{a}} + L_{\b{b}})^2 = \b{a \cdot b}\]

This turns out to be connected to a lot of other mathematics. The short version is that \(\iota\) is, technically, a “graded derivation” on the exterior algebra, which is also codified in the ‘Leibniz’ property:

\[\b{v} \cdot (\alpha \^ \beta) = \b{v} \cdot \alpha + (-1)^{\| \alpha \| } \alpha \^ (\b{v} \cdot \beta)\]

This is the exterior algebra version of \(\p(uv) = (\p u) v + u (\p v)\), and the property that \(\iota L + L \iota = I\) is the equivalent of the fact that \(\p_x x - x \p_x = 1\).

I don’t know much about derivations and abstract algebra yet so that’ll have to wait.


1.3 Duality

If we are keeping track of vector space duality, the left side of an interior product \(\alpha \cdot \beta\) should transform like a dual multivector. (It certainly seems like it should because the left side of an inner product \(\< \alpha , \beta \>\) should.) I have been sloppy about this. I’m hoping to collect all of the duality and metric-tensor related stuff in a later article.

The discussion about projection above seems to me to strongly suggest that we define \(\frac{\iota_{\b{a}}}{\vert \b{a} \vert^2} = \b{a}^{-1}\) as a sort of ‘multiplicative inverse’ of \(\b{a}\) for the \(\^\) operation. It’s not a true inverse, because \(\b{a} \^ \b{a}^{-1} \^ \beta = \beta_{\b{a}}\). Instead of being invertible, dividing and then multiplying pulls out the projection on \(\b{a}\). There is a certain elegance to it anyway.

I sometimes suspect that interior products \(\iota_{\alpha}\), and dual vectors in general, should be considered as negative-grade multivectors, so \(\iota_{\alpha} \in \^^{- \| \alpha \|} V\). Then we could write that \(\alpha \cdot \beta \in \^^{\| \beta \| - \| \alpha \|} V\) even if \(\alpha\) has the higher grade. This is also compelling because it explains why dual vectors transform according to the inverse of a transformation: if \(\alpha \ra A^{\^k}(\alpha)\), of course \(\iota_{\alpha} \ra A^{-\^ k} (\iota_{\alpha})\). I hope to look into it this a later article.


2. More identities

We can use \(\iota\) to prove a few more vector identities. First, note that \(\star\) is just a special case of \(\iota\).5

\[\begin{aligned} \< \alpha \cdot \omega, \star \beta \> &= \< \omega, \alpha \^ \star \beta \> \\ &= \< \omega, \< \alpha, \beta \> \omega \> \\ &= \< \alpha , \beta \> \\ &= \< \star \alpha, \star \beta \> \end{aligned}\]

Since this holds for all \(\alpha, \beta\):

\[\star \alpha = \alpha \cdot \omega\]

(1) implies that \(\iota\) obeys many of the the same rules as \(\^\):

\[\begin{aligned} (\alpha \^ \beta) \cdot \gamma &= \beta \cdot (\alpha \cdot \gamma) \\ &= (-1)^{\| \alpha \| \| \beta \|} \alpha \cdot (\beta \cdot \gamma) \\ \end{aligned}\]

Combining these, we have a way to transform applications of \(\star\):

\[\begin{aligned} \star (\alpha \^ \beta) &= (\alpha \^ \beta) \cdot \omega \\ &= \beta \cdot (\alpha \cdot \omega) \\ &= \beta \cdot (\star \alpha) \\ &= (-1)^{\| \alpha \| \| \beta \|} \alpha \cdot (\star \beta) \end{aligned}\]

Since the cross product is \(\b{a} \times \b{b} = \star (\b{a} \^ \b{b})\):

\[\b{a} \times \b{b} = \b{b} \cdot (\star \b{a}) = - \b{a} \cdot (\star \b{b})\]

This lets us unpack cross product identities. Note that \(\star^2 = (-1)^{(1)(2)} = 1\) in \(\bb{R}^3\).

Here’s the vector triple product:

\[\begin{aligned} \b{a \times (b \times c)} &= \star(\b{a} \^ \star(\b{b \^ c})) \\ &= - \b{a} \cdot \star^2 (\b{b \^ c}) \\ &= - \b{a} \cdot (\b{b \^ c}) \\ &= (\b{a} \cdot \b{c}) \b{b} - (\b{a} \cdot \b{b}) \b{c} \end{aligned}\]

The quadruple product:

\[\begin{aligned} (\b{a \times b}) \times (\b{c \times d}) &= ((\b{a \times b}) \cdot \b{d}) \b{c} - ((\b{a \times b}) \cdot \b{c}) \b{d} \\ &= \star(\b{a \^ b \^ d}) \b{c} - \star(\b{a \^ b \^ c}) \b{d} \end{aligned}\]

The Jacobi Identity:

\[\begin{aligned} 0 &= \b{a \times (b \times c)} + \b{b \times (c \times a)} + \b{c \times (a \times b)} \\ &= -{\star( \b{a} \cdot (\b{b \^ c}) + \b{b} \cdot (\b{c \^ a}) + \b{c} \cdot (\b{a \^ b}))} \\ &= -{\star( (\b{a \cdot b} - \b{b \cdot a}) \b{c} + (\b{b \cdot c - c \cdot b}) \b{a} + (\b{c \cdot a - a \cdot c}) \b{b})} \\ &= 0 \end{aligned}\]

The Jacobi Identity can also be rearranged into the following intriguing form:

\[\begin{aligned} \b{a} \cdot (\b{b \^ c}) &= \b{b} \cdot (\b{a \^ c}) - \b{c} \cdot (\b{a \^ b}) \\ \end{aligned}\]

This is equivalent to \(\b{b \cdot a}) \b{c} - (\b{c \cdot a}) \b{b}\), but it hints at greater structure (which is related to \(\cdot\) being a derivation, above). I know it’s involved in Lie Algebras, but I haven’t been able to find a good purely geometric intuition for what it could mean.

(An alternative proof of the Jacobi identity: the exterior algebra element \(\b{a \^ b \^ c}\) corresponds to the tensor algebra element \(\b{a(bc-cb) + b(ca-ac) + c(ab-bc)}\). The identity follows from contracting any two indexes of this tensor together, since it is antisymmetric in all positions.)


2.1 Matrix Multiplication

One case where the interior product is already being used in mathematics is when multiplying by an antisymmetric matrix. A bivector \(\b{b \^ c}\) can be represented as a tensor product \(\b{b \o c - c \o b}\), which can be treated as an antisymmetric matrix. The interior product \(\b{a} \cdot (\b{b \^ c})\) is then equivalent to matrix multiplication:

\[\begin{aligned} \b{a} \cdot (\b{b \^ c}) &= (\b{a \cdot b}) \b{c} - (\b{a \cdot c}) \b{b} \\ &= \begin{pmatrix} 0 & b_x c_y - b_y c_x & b_x c_z - b_z c_x \\ b_y c_x - b_x c_y & 0 & b_y c_z - b_z c_y \\ b_z c_x - b_x c_z & b_z c_y - b_y c_z & 0 \end{pmatrix} \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} \end{aligned}\]

For instance this is one way of writing a rotation operator which rotates vectors by \(\frac{\pi}{2}\) in the \(\b{bc}\) plane (if \(\b{b}, \b{c}\) are orthogonal unit vectors):

\[R_{\b{bc}} (\b{a}) = \b{a} \cdot (\b{b \^ c})\]

Other articles related to Exterior Algebra:

  1. Oriented Areas and the Shoelace Formula
  2. Matrices and Determinants
  3. The Inner product
  4. The Hodge Star
  5. The Interior Product
  6. All the Exterior Algebra Operations
  1. Recall 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 \(a \cdot b \neq b \cdot a\)?) 

  2. It 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. 

  3. the other candidate would be \(\o\), but we’d like the result to also be a multivector so it makes sense to only consider \(\^\). 

  4. I 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 \((\b{x} \cdot \alpha) \cdot (\b{x} \cdot \beta)\). But that’s complicated and will have to be saved for the future. 

  5. It 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.