Defining and Interpreting the Geometric Product

Previous section: Introduction to Geometric Algebra and Basic 2D Applications.

Next section: Rotors and Rotations in the Euclidean Plane.


The algebraic properties of vector addition and scalar multiplication are insufficient to characterize the geometric concept of a vector as a directed line segment, because they fail to encode the properties of magnitude and relative direction. That deficiency is corrected by defining suitable algebraic rules for multiplying vectors.


Geometric product:
The product ab for
vectors a, b, c is defined by the rules

          • Associative:                (ab)c = a(bc)

          • Left distributive:         a(b + c) = ab + ac

          • Right distributive:      (b + c)a = ba + ca 

          • Euclidean metric:        a2 = a2,

where a = |a| is a positive scalar (= real number) called the magnitude of a. 

In terms of the geometric product ab we can define two other products, a symmetric inner product

          (1)    a∙b = ½(ab + ba) = b∙a

and an antisymmetric outer product 

          (2)    a∧b = ½(ab − ba) = − b∧a

Adding (1) and (2), we obtain the fundamental formula

          (3)    ab = a∙b + a∧b called the expanded form for the geometric product. 


Our next task 
is to provide geometric interpretations for these three products. 


Exercise: For a triangle defined by the vector equation a + b = c,

a b c

                   

derive the standard Law of Cosines:  a2 + b2 + 2a∙b = c2, and so prove that the inner product a∙b is always scalar-valued. 

{Just multiply c2 = (a + b) (a + b) and use the Euclidean metric above and definition (1) above for the inner product.
  Notice also that
if a and b are perpendicular, using the Pythagorean Theorem we can conclude a∙b is zero (and vice versa).
  Further, 
if a and b are perpendicular, using formulas (1) & (2) above, a∧b = ab = ½(ab − ba), which means ab = − ba.}


Therefore, the inner product can be given the usual geometric interpretation as the length of a perpendicular projection of one line segment on the direction of another.


a b ^ . a b

                                                           


The outer product  a∧b = − b∧a  generates a new kind of geometric quantity called a bivector, that can be interpreted geometrically as directed area in the plane of a and b.

a b a b = a^b b^a -

                

We have shown that the geometric product interrelates three kinds of algebraic entities: scalars (0-vectors), vectors (1-vectors), and bivectors (2-vectors) that can be interpreted as geometric objects of different dimension. Geometrically, scalars represent 0-dimensional objects, because they have magnitude and orientation (sign) but no direction. Vectors represent directed line segments, which are 1-dimensional objects. Bivectors represent directed plane segments, which are 2-dimensional objects. It may be better to refer to a bivector

\[\Large{{\bf{B}} = B{\bf{\hat B}}}\]


as a
directed area, because its magnitude B = |B| is the ordinary area of the plane segment and its direction

\[\Large{{{\bf{\hat B}}}}\]


represents the plane (and orientation) in which the segment lies, just as a unit vector represents the direction of a line. The shape of the plane segment is not represented by any feature of B, as expressed in the following equivalent geometric depictions for B (with 
counterclockwise orientation):

= = = B B B



Prove the following: 


         
Given any non-zero vector a in the plane of bivector B, one can find a vector b such that

                    B = ba = –ab, 

                    B2 = −|B|2 = −a2b2,

                    aB = –Ba,        that is, B anticommutes with every vector in the plane of B. 

                                       
                   
{Looking again at the depictions directly above, it should be clear we can pick a b perpendicular to a
                      and also in the plane of B so that the rectangle depicting b∧a has the magnitude and orientation of B.
     
                      Further, since we've picked
b to be perpendicular to a, by the arguments in the exercise above, 
                      
B = b∧a = ba = – ab. Then, B2 = (–ab) (ba) = – a2b2 = − a2b2 = −|B|2 < 0.
   
                      Finally,  
aB = – aab = − a2b = − baa = – Ba.}


         
Every vector a has a multiplicative inverse: 

\[\Large{{{\bf{a}}^{ - 1}} = \frac{1}{{\bf{a}}} = \frac{{\bf{a}}}{{{a^2}}}}\]

                    that is, geometric algebra makes it possible to divide by vectors.

                    {Simply multiply a on either side by a/a2. As usual, we must assume a≠0.}



Prove the following theorems about the geometric meaning of commutivity and anticommutivity:


          a∙b = 0    ⇔    ab = −ba          Orthogonal vectors anticommute!!          {See the exercise above.}

                          

          a = λb    ⇔    a∧b = 0    ⇔    ab = ba          Collinear vectors commute!!


                    


The problem remains to assign geometric meaning to the quantity ab without expanding it into inner and outer products.



Previous section: Introduction to Geometric Algebra and Basic 2D Applications.

Next section: Rotors and Rotations in the Euclidean Plane.


© David Hestenes 2005, 2014