# Why is there no decent multiplication in 3-D 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 non-zero element a there is an inverse element a-1.

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: i2=-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+6i2=2+16i.

Every complex number being basically a pair of real numbers (a,b), the set C can be identified with the real plane R2 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)=(ac-bd,ad+bc). Let's observe that such an operation of addition can be easily generalized to the three-dimensional space R3 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 three-dimensional space R3 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, well-defined in R3. In other words, that (tv) w=v (t w) for all v, w in R3 and t in R. We will call such a multiplication bilinear.

Coordinatewise multiplication is not good. The product of two non-zero 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 R3 which satisfies conditions (a) - (c) together with the coordinatewise addition.

Let's suppose the contrary, i.e. let R3 have such a multiplication. It will be convenient to think of elements of R3 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.

L={ t . 1 : t being a real number }.

For a vector v in (R3 \ 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 R3 \ L. If its square v2 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 R3.

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 v2. 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 w2 is in P(v)=P(w), i.e. w2=s1+t w for some real numbers s and t. If s=0, then w2=tw and consequently w (w-t 1)=0. This, however, is not possible, since both factors are non-zero. Therefore s is nonzero, which implies w (-s-1w-ts-1 1)=1. Hence w-1=-s-1w-ts-11 belongs to P(v).

And now we will see that the case treated by Lemma 1 ... never occurs.

Lemma 2. If v in R3 \ L, then v2 does not belong to P(v).

Proof: Suppose v2 belongs to P(v) and take any vector w in R3 \ P(v). Then the three vectors 1, v and w generate the entire space R3. In particular, the vector wv can be represented as a combination of the three: wv=r 1+s v+t w. But then w(v-t 1)=r 1+s v and so

w=(r 1+s v)(v-t 1)-1.

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 R3 \ L. By Lemma 2 we infer that the vectors 1, v and v2 generate the space R3. Thus the vector v3 can be represented as their combination: v3=r 1+s v+t v2. Consider the function f:R ->R such that f(x)=x3-r-sx-tx2. 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)=(x-c)(a+bx+x2). Substituting the vector v for x we get

0=f(v)=(v-c 1)(a1+b v+v2).

The assumptions and Lemma 2 imply that both factors are non-zero. 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 R3 does not even have the structure of a non-commutative algebra with division.

Zbigniew Marciniak