7.4 The outer product
7.4.3 Orthogonal bases
GA works in any number of dimensions, and anticipating the need to embrace a large number of dimensions we require a notation for the extended orthogonal axial systems. Conventionally, i and j represent the unit basis vectors forR2, and i, j and k represent the unit basis vectors forR3.
Geometric algebra 61
If we continue with this notation the alphabet cannot support very high-dimensional spaces. An alternative convention is to use e1, e2, e3,. . .ento represent the orthogonal unit basis vectors.
Using this notation we define two vectors inR2as
a=a1e1+a2e2 (7.8)
b=b1e1+b2e2. (7.9)
We can now state the outer product as
a∧b=(a1e1+a2e2)∧(b1e1+b2e2) (7.10) which expands to
a∧b=a1b1(e1∧e1)+a1b2(e1∧e2)+a2b1(e2∧e1)+a2b2(e2∧e2). (7.11) Substituting the following observations
e1∧e1=e2∧e2=0 and e2∧e1= −e1∧e2 (7.12) we obtain
a∧b=a1b2(e1∧e2)−a2b1(e1∧e2) (7.13) simplifying, we obtain
a∧b=(a1b2−a2b1)(e1∧e2). (7.14) The scalar terma1b2−a2b1 in Eq. (7.14) looks familiar — in fact, it is the magnitude of the imaginary term of Eq. (3.17), the value of which equalsabsinθ, which is the area of the parallelogram formed bya andb. So in this context, the outer producta∧b is a scalar area multiplying the unit bivector e1∧e2, which just means that the area is associated with the plane defined by e1∧e2. Figure 7.6 illustrates this relationship.
Figure7.6.
Now let’s computeb∧a:
b∧a=(b1e1+b2e2)∧(a1e1+a2e2)
which expands to
b∧a=a1b1(e1∧e1)+a2b1(e1∧e2)+a1b2(e2∧e1)+a2b2(e2∧e2). (7.15) Substituting the following observations
e1∧e1=e2∧e2=0 and e2∧e1= −e1∧e2 (7.16) we obtain
b∧a=a2b1(e1∧e2)−a1b2(e1∧e2). (7.17) Simplifying, we obtain
b∧a= −(a1b2−a2b1)(e1∧e2) (7.18) which confirms thatb∧a= −a∧b.
Now let’s consider the outer product inR3:
a=a1e1+a2e2+a3e3 (7.19)
b=b1e1+b2e2+b3e3. (7.20) The outer product is
a∧b=(a1e1+a2e2+a3e3)∧(b1e1+b2e2+b3e3) (7.21) which expands to
a∧b=a1b1(e1∧e1)+a1b2(e1∧e2)+a1b3(e1∧e3)+a2b1(e2∧e1)+a2b2(e2∧e2) +a2b3(e2∧e3)+a3b1(e3∧e1)+a3b2(e3∧e2)+a3b3(e3∧e3). (7.22) Substituting
e1∧e1=e2∧e2=e3∧e3=0 (7.23) and
e2∧e1= −e1∧e2 e1∧e3= −e3∧e1 e3∧e2= −e2∧e3 (7.24) we obtain
a∧b=a1b2(e1∧e2)−a1b3(e3∧e1)−a2b1(e1∧e2)
+a2b3(e2∧e3)+a3b1(e3∧e1)−a3b2(e2∧e3). (7.25) Simplifying, we obtain
a∧b=(a1b2−a2b1)e1∧e2+(a2b3−a3b2)e2∧e3+(a3b1−a1b3)e3∧e1. (7.26) You may be wondering why the unit basis bivectors in Eq. (7.26) have been chosen in this way, especially e3∧e1. This could easily be e1∧e3. To understand why, refer to Fig. 7.7, which shows a right-handed axial system and where each orthogonal plane is defined by its associated unit basis bivectors.
Geometric algebra 63
X Y
Z
e1 e2
e3
1 2
e ∧e
2 3
e ∧e
3 1
e ∧e
Figure7.7.
Figure 7.7 also shows the orthogonal alignment of the Cartesian axes with the unit basis bivectors:
thex-axis is orthogonal to e2∧e3
they-axis is orthogonal to e3∧e1
thez-axis is orthogonal to e1∧e2
and if Eq. (7.26) is rearranged in this sequence we obtain
a∧b=(a2b3−a3b2)e2∧e3+(a3b1−a1b3)e3∧e1+(a1b2−a2b1)e1∧e2. (7.27) Now let’s look at a definition of the cross product. We begin by declaring two vectors using the conventional orthogonal unit basis vectors i, j and k:
a=a1i+a2j+a3k (7.28)
b=b1i+b2j+b3k. (7.29)
The cross product is
a×b=(a1i+a2j+a3k)×(b1i+b2j+b3k) (7.30) which expands to
a×b=a1b1(i×i)+a1b2(i×j)+a1b3(i×k)+a2b1(j×i)+a2b2(j×j)
+a2b3(j×k)+a3b1(k×i)+a3b2(k×j)+a3b3(k×k). (7.31) The magnitude of the cross product isabsinθ, which means that
i×i=j×j=k×k=0. (7.32)
Therefore,
a×b=a1b2(i×j)+a1b3(i×k)+a2b1(j×i)
+a2b3(j×k)+a3b1(k×i)+a3b2(k×j). (7.33) Because the cross product is antisymmetric
j×i= −i×j k×j= −j×k i×k= −k×i. (7.34)
Substituting these relationships:
a×b=a1b2(i×j)−a1b3(k×i)−a2b1(i×j)
+a2b3(j×k)+a3b1(k×i)−a3b2(j×k). (7.35) Collecting up like terms:
a×b=(a2b3−a3b2)j×k+(a3b1−a1b3)k×i+(a1b2−a2b1)i×j. (7.36) If we place Eqs. (7.27) and (7.36) together and substitute the e notation for i, j and k, we obtain
a∧b=(a2b3−a3b2)e2∧e3+(a3b1−a1b3)e3∧e1+(a1b2−a2b1)e1∧e2 (7.37) a×b=(a2b3−a3b2)e2×e3+(a3b1−a1b3)e3×e1+(a1b2−a2b1)e1×e2. (7.38) In the cross product, the terms(a2b3−a3b2),(a3b1−a1b3)and(a1b2−a2b1)are the components of an orthogonal vector, whereas in the outer product they become signed areas projected onto the planes defined by the unit bivectors e2∧e3, e3∧e1and e1∧e2. And in spite of there being such similarity between the two equations, it would be dangerous to conclude thata∧b≡a×b.
What Hamilton had proposed was that
e2×e3=e1 e3×e1=e2 e1×e2=e3 (7.39) which is fine forR3, but is ambiguous for higher dimensions. So, in GA we substitute the outer product for the cross product and introduce the concept of a directed area, which holds for any number of dimensions.
Before we reveal the imaginary nature of the outer product in the next chapter, consider the scenario shown in Fig. 7.8. Two vectorsa andbare shown forming a parallelogram created by their outer producta∧bwith parallel projections of the parallelogram projected onto the three orthogonal planes. The projections will normally be parallelograms, but under some conditions they could collapse to a line. Whatever happens, at least one will be a parallelogram.
We define two vectors as
a=a1e1+a2e2+a3e3 (7.40)
b=b1e1+b2e2+b3e3. (7.41) Starting with the plane containing e1and e2, which is defined by e1∧e2, the projections ofaand barea′′′andb′′′, respectively, where
a′′′=a1e1+a2e2 (7.42)
b′′′=b1e1+b2e2. (7.43)
Therefore,
a′′′∧b′′′=(a1e1+a2e2)∧(b1e1+b2e2)
=a1b1(e1∧e1)+a1b2(e1∧e2)+a2b1(e2∧e1)+a2b2(e2∧e2)
a′′′∧b′′′=(a1b2−a2b1)e1∧e2 (7.44)
which is the last term in Eq. (7.27).
Geometric algebra 65
Similarly, we can show that
a′∧b′=(a2b3−a3b2)e2∧e3 (7.45) a′′∧b′′=(a3b1−a1b3)e3∧e1. (7.46) Thus we see that instead of creating a new vector, the outer product projects the parallelogram onto the three orthogonal planes to create three new bivectors, whose area is positive or negative.
The cross product, however, takes these areas and uses them to form a vector, which happens to be orthogonal to the original parallelogram.
Figure7.8.
To illustrate this concept, consider two vectorsaandb
a=a1e1+a2e2+a3e3 (7.47)
b=b1e1+b2e2+b3e3 (7.48)
where
a1=1 a2=0 a3=1
b1=1 b2=1 b3=0 (7.49)
which makes
a=e1+e3 b=e1+e2. (7.50)
Using Eq. (7.26)
a∧b=(a1b2−a2b1)e1∧e2+(a2b3−a3b2)e2∧e3+(a3b1−a1b3)e3∧e1
a∧b=(1)e1∧e2+(−1)e2∧e3+(1)e3∧e1. (7.51) The signed area on the plane e1∧e2is+1 and is shown in Fig. 7.9. The projected area is shown crosshatched.
Figure7.9.
Similarly, the signed area on the plane e2∧e3is−1 and is shown in Fig. 7.10. Note that the direction of the projected area opposes the direction of e2∧e3.
Figure7.10.
And the signed area on the plane e3∧e1is+1, and is shown in Fig. 7.11.
Figure7.11.
Now let’s compute the magnitude of the bivectora∧b.
To begin with, we need to know the angle betweena andb, which is revealed using the dot product:
θ=cos−1
a1b1+a2b2+a3b3
ab
θ=cos−1 1
√2√ 2
=60◦. (7.52)
Geometric algebra 67
Therefore,
a∧b = absin 60◦ a∧b =√
2√ 2
√3 2 =√
3. (7.53)
The next question to pose is whether this value is related to the other three areas? Well the answer is “yes”, and for a very good reason:
a∧b2=(a1b2−a2b1)2+(a2b3−a3b2)2+(a3b1−a1b3)2 (7.54) therefore,
√32=(1)2+(−1)2+(1)2=3. (7.55) Remember, that the cross product uses these coefficients as Cartesian components of the axial vector and satisfy the Pythagorean rule:
a2=a21+a22+a32. (7.56) To prove that this holds, we need to show that Eq. (7.54) is correct.
Expanding the LHS of Eq. (7.54):
a∧b2= a2b2sin2θ= a2b2(1−cos2θ )
a∧b2= a2b2− a2b2cos2θ. (7.57) From the dot product
cos2θ =(a1b1+a2b2+a3b3)2
a2b2 . (7.58)
Therefore,
a∧b2= a2b2−(a1b1−a2b2−a3b3)2
a∧b2=(a12+a22+a32)(b21+b22+b23)−(a1b1−a2b2−a3b3)2 and we obtain
a∧b2=(a12b22−2a1a2b1b2+a22b12)+(a22b32−2a2a3b2b3+a32b22) +(a32b21−2a3a1b3b1+a12b23)
a∧b2=(a1b2−a2b1)2+(a2b3−a3b2)2+(a3b1−a1b3)2. (7.59) Therefore, Eq. (7.54) is correct.
Now, as
a∧b = absinθ (7.60)
a∧b2= a2b2sin2θ (7.61)
and
absin2θ=(a1b2−a2b1)2+(a2b3−a3b2)2+(a3b1−a1b3)2 (7.62) therefore
θ=sin−1 √
(a1b2−a2b1)2+(a2b3−a3b2)2+(a3b1−a1b3)2 ab
. (7.63)
Substituting the values for the above example:
θ=sin−1 √
√ 3 2√
2
=60◦. (7.64)
The beauty of the outer product is that it works in any number of dimensions. For example, we can create two vectors inR4as follows:
a=a1e1+a2e2+a3e3+a4e4 (7.65) b=b1e1+b2e2+b3e3+b4e4 (7.66) and form their outer product:
a∧b=(a1e1+a2e2+a3e3+a4e4)∧(b1e1+b2e2+b3e3+b4e4). (7.67) This explodes into
a∧b=a1b1(e1∧e1)+a1b2(e1∧e2)+a1b3(e1∧e3)+a1b4(e1∧e4) +a2b1(e2∧e1)+a2b2(e2∧e2)+a2b3(e2∧e3)+a2b4(e2∧e4) +a3b1(e3∧e1)+a3b2(e3∧e2)+a3b3(e3∧e3)+a3b4(e3∧e4) +a4b1(e4∧e1)+a4b2(e4∧e2)+a4b3(e4∧e3)+a4b4(e4∧e4) and collapses to
a∧b=(a1b2−a2b1)(e1∧e2)+(a2b3−a3b2)(e2∧e3)+(a3b1−a1b3)(e3∧e1)
+(a1b4−a4b1)(e1∧e4)+(a2b4−a4b2)(e2∧e4)+(a3b4−a4b3)(e3∧e4) (7.68) which resolves the outer product into six bivectors.
Geometric algebra 69
These bivectors arise because there are six ways of making 2-tuples from four axes:
4C2= 4!
(4−2)!2! =6. (7.69)
In five dimensions there are 10 bivectors.
5C2= 5!
(5−2)!2! =10. (7.70)
As a final example, let’s consider two vectors inR4and compute their outer product. The vectors are
a=e1+e3+e4 (7.71)
b=e1+e2+e4. (7.72)
Then
a =√
3 b =√
3 (7.73)
and the separating angleθis
θ=cos−1 2
3
≃48.19◦. (7.74)
Similarly,
θ=sin−1 √
5 3
≃48.19◦. (7.75)
Substituting the vectors into Eq. (7.68):
a∧b=(1)(e1∧e2)+(−1)(e2∧e3)+(1)(e3∧e1)+(−1)(e2∧e4)+(1)(e3∧e4). (7.76) Therefore,a∧bis given by
a∧b = absinθ=√ 3√
3 sin 48.19◦≃2.2361. (7.77) Finally, let’s show that theR4equivalent of Eq. (7.54) still holds:
a∧b2= |a1b2−a2b1|2+ |a2b3−a3b2|2+ |a3b1−a1b3|2 + |a1b4−a4b1|2+ |a2b4−a4b2|2+ |a3b4−a4b3|2
2.23612=(1)2+(−1)2+(1)2+(0)2+(−1)2+(1)2=5. (7.78)