Dorst L., Fontijne D., Mann S. Geometric Algebra for Computer Science. An Object Oriented Approach to Geometry
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SECTIONB.2 EQUIVALENCE OF THE IMPLICIT AND EXPLICIT CONTRACTION DEFINITIONS 593
=
k
i=1
(−1)
i−1
(x · a
i
)
y
1
· a
1
··· y
1
· a
i−1
y
1
· a
i+1
··· y
1
· a
k
.
.
.
.
.
.
.
.
.
.
.
.
y
k−1
· a
1
··· y
k−1
· a
i−1
y
k−1
· a
i+1
···y
k−1
· a
k
=
k
i=1
(−1)
i−1
(x · a
i
)(y
k−1
∧···∧y
1
) ∗ (a
1
∧ a
2
∧···∧
˘
a
i
∧···∧a
k
)
= (y
k−1
∧···∧y
1
) ∗
k
i=1
(−1)
i−1
(a
1
∧ a
2
∧···∧(x · a
i
) ∧···∧a
k
)
where the notation
˘
a
i
means that the term a
i
is omitted from the outer product.
Since this must hold for any blade (y
k−1
∧···∧y
1
), we apply (B.2), and we have
proved (3.16) when we write the inner product between vectors as contractions.
It is now easy to establish a recursive formula for a k-blade A
k
:
xA
k
= x(a ∧ A
k−1
) ≡ (xa) ∧ A
k−1
− a ∧ (xA
k−1
).
(B.4)
Note that this formula even works for k = 1. Applying this repeatedly, we get
(3.10), which completes its proof.
Proof of consistency of (3.11): The proof of (3.11) shows how nicely all our products
intertwine:
X ∗
(
(A ∧ B)C
)
=
(
X ∧ (A ∧ B)
)
∗ C
=
(
(X ∧ A) ∧ B
)
∗ C
= (X ∧ A) ∗ (BC)
= X ∗
(
A(BC)
)
This is valid for arbitrary X; therefore, by the orthogonality property of the scalar
product (B.2), we must have (3.11).
All the above shows that the contraction operation defined by equations (3.7) through
(3.11) is consistent with (3.6). For a space with a nondegenerate metric, in which both of
these definitions define the contraction, they are therefore equivalent.
The explicit 5-part definition allows us to also compute the contraction for a space with
a degenerate metric, without getting a contradiction with the now-incomplete implicit
definition. The explicit definition is therefore the one we should use in general. But of
course, (3.6) still holds for any X in any metric. That property, which shows how the
contraction interleaves with the outer product and the scalar product, remains the main
algebraic motivation for introducing the contraction product. (Mathematically, it makes
the contraction the adjoint of the outer product relative to the scalar product a natural
construction in the treatment of algebras.)
594 CONTRACTIONS AND OTHER INNER PRODUCTS APPENDIX B
B.3 PROOF OF THE SECOND DUALITY
Since the second duality relationship is so important, we prove that it follows from our
definitions. We repeat the statement of the second duality (from (3.21)):
(AB)C = A ∧ (BC) when A ⊆C
Its proof is straightforward and rather instructive if you are interested in the mathematics
of using the algebra; but you will not miss anything essentially geometrical if you skip it.
We prove (3.21) by a four-step induction, each result forming the basis for the next.
0. As step 0 of the induction, if at least one of A, B or C is a scalar, this is covered
by either of the cases “A is a scalar” or “A is not a scalar, but B is.” Both are
trivial.
1. As step 1 of the induction, we have the straightforward point of departure for
vectors:
c ∧ ( bc) = ( cb)c,
(B.5)
which is basically two ways of writing the scalar product of the scalar b ∗ c = bc =
b · c with the vector c.
2. Next, we prove from this step 2:
a ∧ ( bC) = (ab)C if a ⊆C. (B
.6)
This is
done by induction: assume that a ∧ (bC
k−1
) = (ab)C
k−1
if a ⊆C
k−1
.
Then form C
k
= c
k
∧ C
k−1
and try to prove validity for that. First, note that a ⊆C
k
is equivalent to a ∧ C
k−1
= 0 for some blade C
k−1
. Then we obtain:
a ∧ ( bC
k
) = a ∧ (b(a ∧ C
k−1
))
= a ∧ ((ba) ∧ C
k−1
− a ∧ (bC
k−1
))
= a ∧ (ba) ∧ C
k−1
= (ab)C
k
.
Induction from (B.5) then proves (B.6).
3. Then we prove step 3:
a ∧ ( BC) = (aB)C for a ⊆C, (B.7)
SECTIONB.4 PROJECTION AND THE NORM OF THE CONTRACTION 595
through assuming a ∧ (B
l−1
C) = (aB
l−1
)C for a ∧ C = 0, and deriving for
B
l
= b
l
∧ B
l−1
:
(aB
l
)C =
= (a(b
l
∧ B
l−1
))C
= (ab
l
)(B
l−1
C) − (b
l
∧ (aB
l−1
))C
= (ab
l
)(B
l−1
C) − b
l
((aB
l−1
)C)
= (ab
l
)(B
l−1
C) − b
l
(a ∧ (B
l−1
C))
= a ∧ (b
l
(B
l−1
C)) [by developing the last term]
= a ∧ ((b
l
∧ B
l−1
)C)
= a ∧ (B
l
C),
and induction from (B.6) now proves (B.7).
4. Finally, we prove the actual second duality in step 4:
A ∧ (BC) = (AB)C for A ⊆C, (B.8)
through assuming A
m−1
∧ (BC) = (A
m−1
B)C for A
m−1
⊆C, and deriving for
A
m
= a
m
∧ A
m−1
:
A
m
∧ (BC) = (a
m
∧ A
m−1
) ∧ (BC)
= a
m
∧ ((A
m−1
B)C)
= ((a
m
(A
m−1
B))C by (B.7)
= (a
m
∧ A
m−1
)B)C
= (A
m
B)C
and induction from (B.6) proves the result.
The required containment of A into C enters this proof already in step 2, to prevent a
proliferation of extra terms that would make the statement less simple and therefore less
usable.
B.4 PROJECTION AND THE NORM OF
THE CONTRACTION
We claimed in Section 3.3 that the norm of the contraction AB is proportional to the
norm of B times the norm of the projection of A onto B. Let us combine the proof of that
statement with a proof of the projection formula P
B
[A] = (AB)B
−1
itself. It is all based
on the scalar product definitions of the norm (3.4), the angle (3.5), the left contraction
(3.6) and the right contraction (3.18), plus usage of duality and the reversion properties
of the scalar product in (3.3).
596 CONTRACTIONS AND OTHER INNER PRODUCTS APPENDIX B
We first express the norm of AB in something resembling the projection:
AB
2
= (AB) ∗ (AB)
= (AB) ∗ (
B
A)
= B ∗
(AB)
∧ A
=
B(AB)
∗ A
=
(AB)
B
∗
A.
We can express the norm of (AB)
B back in terms of the norm of AB:
(AB)
B
2
=
(AB)
B
∗
(AB)
B
=
(AB)
B
∗
B(AB)
= B ∗
(AB)
∧
(AB)
B
= B ∗
(AB)
(AB)
B
= AB
2
B
2
.
Now, defining P ≡
(AB)
B
/
(B ∗
B) and denoting the cosine of P and A as cos(P, A)
we rewrite these results as
AB
2
= PAB
2
cos(P, A)
and
P
2
= AB
2
/
B
2
.
It follows that
P = A cos(P, A),
and since P was proved to be contained in B in Section 3.3, it is the projection of A onto
B. (Its idempotence is proved in Section 3.6.) With that meaning of P, we interpret the
second result as the desired expression for the norm of AB.
C
SUBSPACE PRODUCTS
RETRIEVED
C.1 OUTER PRODUCT FROM GEOMETRIC PRODUCT
We prove that the outer product defined by (6.10) and (6.11) is consistent with the familiar
outer product we had before. We repeat the statements for convenience:
a ∧ B =
1
2
(aB+
Ba), (C.1)
B ∧ a =
1
2
(Ba+ a
B), (C.2)
where
B = (−1)
grade(B)
B is the grade involution of B introduced in Section 2.9.5. (Writing
the equations in this form makes it easier to lift them to general multivectors B.)
We check the earlier defining properties of Section 2.10.
•
If B is a scalar β,wegeta ∧ β = β a = β ∧ a.
•
If B is a vector, we obtain antisymmetry, as we saw above.
•
To demonstrate associativity, let us develop (a ∧ B) ∧ c and a ∧ (B ∧ c).
(a ∧ B) ∧ c =
1
4
(aBc+ ac
B +
B
ca+
cBa)
a ∧ ( B ∧ c) =
1
4
(aBc+
Bac+ c
a
B + cB
a)
597
598 SUBSPACE PRODUCTS RETRIEVED APPENDIX C
The two are equivalent if ac
B − c
a
B =
Bac −
B
ca. But this can be simplified
using (6.9) to (a ·c)
B =
B (a ·c). That holds by the commutativity of scalars under
the geometric product, so the two expressions are indeed equivalent. In particular,
this demonstrates the associativity law for vectors:
(a ∧ b) ∧ c = a ∧ (b ∧ c).
By applying it recursively we get general associativity for blades.
•
Linearity and distributivity over addition trivially hold by the corresponding prop-
erties of the geometric product. They permit extending the results to general mul-
tivectors B.
This demonstrates that the two equations above indeed identify the same outer product
structure that we had before, at least when one of the factors is a vector. Since we have
associativity, this can be extended to general blades, and by linearity to general multivec-
tors. Only the case of two scalars is formally not yet included.
C.2 CONTRACTIONS FROM GEOMETRIC PRODUCT
We prove (6.12) and (6.13), repeated for convenience:
aB =
1
2
(aB−
Ba), (C.3)
Ba =
1
2
(Ba− a
B). (C.4)
We focus on the formula for the left contraction aB, and show consistency with the earlier
definition of (3.7) through (3.11). The right contraction Ba is completely analogous.
•
If B is a scalar β,wegetaβ = 0.
•
If B isavectorb, we obtain symmetry and equivalence to the classical inner product
a · b of (6.9).
•
To demonstrate the distribution of the contraction over an outer product, we
first show
a(bC) =
1
2
(abC+ b
Ca) =
1
2
(abC+ baC− baC+ b
Ca)
= (a · b) C − b (aC).
(C.5)
Similarly, you may derive that
a(
Cb) = (a
C) b + C (a · b). (C.6)
These equations are important by themselves, for they specify the distributivity of
the contraction over the geometric product. Adding them produces a(b ∧ C) =
(a · b) ∧ C − b ∧ (aC). This is essentially (B.4), and we can pick up the rest of the
proof from there: applied repeatedly, this equation gives us the desired result (3.10)
for blades.
SECTIONC.3 PROOF OF THE GRADE APPROACH 599
•
If we follow the same procedure we used to demonstrate associativity of the outer
product, we get a surprise. That derivation does not show the associativity of the
contraction (fortunately!), but the associativity of a “sandwich” combination of the
two contractions:
(aB)c = a(Bc).
For vectors, this becomes the rather trivial (ab)c = a(bc), which holds because
both sides are zero.
•
That leaves us with no tools to prove the remaining defining equation (3.11) for the
earlier contraction: A(BC) = (A ∧ B)C. In fact, we cannot even compute both
sides simultaneously, since we always need a vector as an argument to use (6.12).
As with the outer product, the proposed equations do not define the contraction, and do
therefore not prove it to be identical with the earlier contraction from Chapter 3; but we
have at least shown that they are consistent with it.
C.3 PROOF OF THE GRADE APPROACH
We prove that the subspace products of blades can be retrieved as certain selected grades
of the geometric product, as stated in (6.19) through (6.22). We repeat those statements
for convenience:
A
k
∧ B
l
≡A
k
B
l
l+k
(C.7)
A
k
B
l
≡A
k
B
l
l−k
(C.8)
A
k
B
l
≡A
k
B
l
k−l
(C.9)
A
k
∗ B
l
≡A
k
B
l
0
(C.10)
We show the correctness of the equation for the outer product using the earlier derived
(6.10). Wr ite A
k
as A
k
= a
k
∧ A
k−1
. Then:
A
k
B
l
l+k
= (a
k
∧ A
k−1
) B
l
l+k
=
1
2
a
k
A
k−1
B
l
+
A
k−1
a
k
B
l
l+k
=
1
2
a
k
A
k−1
B
l
+
A
k−1
a
k
B
l
+
A
k−1
B
l
a
k
−
A
k−1
B
l
a
k
l+k
= a
k
∧ (A
k−1
B
l
) +
A
k−1
(a
k
B
l
)
l+k
= a
k
∧ (A
k−1
B
l
)
l+k
[other term at most of grade l+k−2]
= a
k
∧A
k−1
B
l
l+k−1
.
.
.
= A
k
∧B
l
l
= A
k
∧ B
l
This proof does not work when A
k
is a scalar α of grade 0. But then αB
l
l
= αB
l
= α ∧B
l
,
so the result still holds.
600 SUBSPACE PRODUCTS RETRIEVED APPENDIX C
The definition (C.8) of the left contraction, in terms of geometric product and grade
operator, is also consistent with the earlier definition of (3.7) through (3.11), as we show
onebyone.
•
The scalar properties (3.7) and (3.8) are straightforward:
A
k
α = A
k
α
−k
= 0ifk = 0
αA
k
= αA
k
k
= αA
k
•
The inner product correspondence property (3.9) is rather obvious: ab
0
=
a ∗ b = a · b.
•
The property (3.10), which would read a(b ∧B
l
) = (ab)B
l
−b∧(aB
l
) for a blade
B
l
of grade l, takes some more work:
(ab)B
l
− b ∧ (aB
l
) = (a · b)B
l
− b(aB
l
)
l
= (a · b)B
l
− b ∧ (aB
l
) − b(aB
l
)
l
= (a · b)B
l
− b ∧ (aB
l
)
l
[last term of too low grade]
= a(b ∧ B
l
)
l
= a (b ∧ B
l
) − a ∧ (b ∧ B
l
)
l
= a (b ∧ B
l
)
l
[last term of too high grade]
= a(b ∧ B
l
)
•
In contrast to the algebraic methods of Section 6.3.1, we now have in (C.8) a com-
plete definition of the contraction and are therefore also able to prove the final con-
traction property (3.11).
(A
k
∧ a)B
l
= (a ∧
A
k
)B
l
l−k−1
=
1
2
a
A
k
)B
l
+ A
k
aB
l
l−k−1
=
1
2
A
k
aB
l
− A
k
B
l
a + A
k
B
l
a + a
A
k
B
l
l−k−1
= A
k
(aB
l
) + a ∧ (
A
k
B
l
)
l−k−1
= A
k
(aB
l
)
l−k−1
(lowest grade last term |l − k| + 1, too high)
= A
k
(aB
l
),
and recursion over the factors of A
k
then proves the general result (A ∧ A
)B =
A(A
B).
So we see that the symbol A
k
B
l
l−k
, which we proposed as the grade-based alternative
definition of the left contraction inner product, has all the desired properties. Therefore
it is algebraically identical to the contraction of Chapter 3.
SECTIONC.3 PROOF OF THE GRADE APPROACH 601
For the right contraction we could repeat this proof, or just show that its correspondence
(3.19) with the left contraction holds:
A
k
B
l
= A
k
B
l
k−l
= (
B
l
A
k
)
k−l
=
B
l
A
k
k−l
= (B
l
A
k
)
.
Since that is universal for all blades (and can be extended for multivectors), the right con-
traction defined by (C.9) must be identical to our earlier rig ht contraction from Chapter 3.
The scalar product of blades of equal grade should be consistent with how it can be defined
as a special case of the contraction: A
k
∗B
k
= A
k
B
k
= A
k
B
k
0
. For nonequal grades, the
scalar part of A
k
B
l
is zero (as it should be), since the lowest grade present always exceeds
zero when k = l. Therefore, (C.10) for the scalar product is correct.
D
COMMON EQUATIONS
Not all equations are equal, and some occur more often than others in the practice of
geometric algebra. We have collected the equations we use most for convenient reference.
We specify their sources so that you may find the context or a similar equation if the one
we give does not quite satsify your needs.
Some equations have been made somewhat more robust for practical use (such as using
an absolute value under a square root to guarantee validity even with round-off errors).
603