Why is there no decent multiplication in 3D space?
We have all learnt to add and multiply real numbers at school and therefore
we can easily enumerate the properties of these operations:
(a) Addition is associative, commutative
and has a neutral element. Moreover, for each element a there is an opposite
element a.
(b) Multiplication is associative,
commutative and has a neutral element 1. Moreover, for every nonzero element
a there is an inverse element a^{1}.
(c) Multiplication distributes over
addition.
Besides the set R of all real numbers, we have become acquainted
with the set N of all natural numbers, the set Z of all integers
and the set Q of all rational numbers. Even if addition and multiplication
can be performed within each of these sets, only Q and R
satisfy all the requirements (a)  (c). The mathematician would say shortly
that R and Q are fields, whereas N and Z are
not.
Can you provide an example of a field included
between Q and R?
The set C of all complex numbers is an example of a field greater
than R. Let's recall that the complex numbers are expressions of
the form a+bi with a and b in R. We add and multiply such numbers
just like we do with polynomials, with one additional relation: i^{2}=1.
It is not hard to check that operations so defined have all the properties
(a)  (c), i.e. C is a field.
For example:
 (2+3i)+(4+2i)=6+5i,
 (2+3i)(4+2i)=8+16i+6i^{2}=2+16i.
Every complex number being basically a pair of real numbers (a,b), the
set C can be identified with the real plane R^{2 }equipped
with addition and multiplication of points. Under this interpretation the
operations can be expressed by the formulas (a,b)+(c,d)=(a+c,b+d) and (a,b)(c,d)=(acbd,ad+bc).
Let's observe that such an operation of addition can be easily generalized
to the threedimensional space R^{3 }by adding triples like
vectors , i.e coordinatewise: (a,b,c) + (d,e,f) = (a+d,b+e,c+f). But then
a natural question arises: can the threedimensional space R^{3
}be also equipped with multiplication in such a way that the two operations
together satisfy (a)  (c)? We might also require that this multiplication
be consistent with the multiplication by scalars, welldefined in R^{3}.
In other words, that (tv) w=v (t w) for all
v, w in R^{3} and t in R. We will call such
a multiplication bilinear.
Coordinatewise multiplication is not good. The
product of two nonzero numbers may be equal to zero. Indeed, (1,0,0)(0,1,0)=(0,0,0).
This does not happen in any field. Can you say why?
We will now prove the following
Theorem. There is no multiplication
in R^{3} which satisfies conditions (a)
 (c) together with the coordinatewise addition.
Let's suppose the contrary, i.e. let R^{3} have such
a multiplication. It will be convenient to think of elements of R^{3}
as vectors starting in 0=(0,0,0). It follows from (b) that one of
the vectors is the multiplication unit ; let's call it 1. Let L
be the straight line extending this vector, i.e.
For a vector v in (R^{3 }\ L) let P(v)
denote the plane determined by the vectors 1 and v. Observe
first that the following holds:
Lemma 1. Let v in R^{3}
\ L. If its square v^{2} lies on the plane P(v),
then the plane is a subset closed with respect to addition, subtraction,
multiplication and inverse.
In this case we say that the subset P(v)
is a subfield of
R^{3}.
Proof: Clearly a plane containing
the point 0 is closed with respect to addition and subtraction of
vectors. Every element of P(v) is of the form s1+t v for
some real numbers s and t. When two such elements are multiplied, the result
becomes a combination of the vectors 1, v and v^{2}.
Each of them being in P(v), the product is there, too. It remains
to prove that if w is in P(v) and w is nonzero, then
the inverse of w belongs to P(v). If w=t 1,
then w^{1}=t^{1} 1 is an element of P(v).
In the opposite case w elongs to P(v) \ L and therefore P(w)=P(v).
We know already that P(v) is closed with respect to multiplication,
so w^{2 }is in P(v)=P(w), i.e. w^{2}=s1+t
w for some real numbers s and t. If s=0, then w^{2}=tw
and consequently w (wt 1)=0. This, however,
is not possible, since both factors are nonzero. Therefore s is nonzero,
which implies w (s^{1}wts^{1} 1)=1.
Hence w^{1}=s^{1}wts^{1}1
belongs to P(v).
And now we will see that the case treated by Lemma 1 ... never occurs.
Lemma 2. If v in R^{3
}\ L, then v^{2} does not belong to P(v).
Proof: Suppose v^{2 }belongs
to P(v) and take any vector w in R^{3}
\ P(v). Then the three vectors 1, v and w
generate the entire space R^{3}. In particular, the vector
wv can be represented as a combination of the three: wv=r
1+s v+t w. But then w(vt 1)=r
1+s v and so
We know by Lemma 1 that the expression on the right hand side represents
an element in P(v). Hence w is in P(v), contrary to
the assumption.
Now we can return to the proof of the theorem. Take an arbitrary vector
v which belongs to R^{3 }\ L. By Lemma 2 we infer
that the vectors 1, v and v^{2} generate the
space R^{3}. Thus the vector v^{3} can be
represented as their combination: v^{3}=r 1+s v+t
v^{2}. Consider the function f:R >R such
that f(x)=x^{3}rsxtx^{2}. Since f(x)<0 for large
negative numbers and f(x)>0 for large positive numbers, there is a real
number c such that f(c)=0, so we can write f(x)=(xc)(a+bx+x^{2}).
Substituting the vector v for x we get
The assumptions and Lemma 2 imply that both factors are nonzero. This
is a contradiction, which completes the proof.
Note for the connoisseurs: We have not used commutativity
of the multiplication in our proof. Hence the space R^{3}
does not even have the structure of a noncommutative algebra
with division.
Zbigniew Marciniak
