BINARY PRODUCTS, ALGEBRAS, AND DIVISION RINGS

[1998/02/13 - Dave Rusin, rusin@math.niu.edu. Comments welcome.]
[The current version of this document resides at
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Thus is a summary of topics relating to the construction of products
in Euclidean space, generalizing the formulas used to construct the
complex numbers and quaternions, cross products, and so on. The
questions are adapted from questions frequently spotted in the USENET
newsgroup sci.math; the answers include information from some posts
there as well.  In addition, I am relying pretty heavily various
sources, which I will cite, for some of the individual questions.

I would like to recommend the Springer book "Numbers", by Ebbinghaus
et al, as a well-written source for many of the results quoted in
this document. It includes many of the proofs and numerous citations
to the literature.

I am particularly grateful to Daniel Shapiro for bringing me up to date on
Q4 and for prompting a number of needed clarifications.

I am trying not to have this document become a full-scale introduction 
to ring theory -- please use your library if you've not already had
such a course. But I am attempting to check the logical independence of
the many customary axioms used, especially associativity.

Q1. Notation: what are rings, fields, algebras, and so on?
Q2. What is a domain? 
Q3. What is a division ring? division algebra? 
Q3a. What's an associative division ring?
Q3b. What's a non-associative division algebra?
Q4. How do you construct all the division algebras over the real numbers?
Q5. What's the question with answer "n=1, 2, 4, or 8"?
Q6. What makes the real numbers different? (Topology)
Q7. What can you say about finite rings?
Q8. Can you extend the cross product to  N  dimensions?
Q9. What's a Clifford algebra? exterior algebra?
Q10. How does this relate to projective geometry?
Q11. Can you extend the notion of quaternions?

==============================================================================

Q1. Notation: what are rings, fields, algebras, and so on?

The word  ALGEBRA  comes from the first two words (in Arabic) of the
title of a ninth century text on the art of computation; the author is
Al-Khuwarizmi (from whose name we have the word "algorithm").  The
word is now used generally for methods of manipulating mathematical
concepts more general than, say, real numbers. In the US, the word is
used for a secondary-school subject covering the use of variables and
equations. Among mathematicians, the word implies a study of those
mathematical objects defined axiomatically which tend to involve
binary operations such as addition and multiplication. We get
successively more familiar-looking objects by imposing more and more
axioms. Mathematicians also use "algebra" to mean a specific construct;
see below.

To begin, we have the concept of  RINGS:  in the most permissive
definition, these are sets  A  which come equipped with two operations 
denoted "+" and "*" and which satisfy a number of axioms:
(1) For all x and y in  A  we have  x+y=y+x.
(2) For all x,y, and z  in  A  we have x+(y+z)=(x+y)+z.
(3) There is an element  "0"  in  A  with   0+x=x+0=x  for all  x.
(4) For each  x  in  A  there is a  y  with  x+y=0.
    (One can show  y  is then unique; we write  "-x" for this  y.)
(5) For all  x, y, z we have   x*(y+z)=x*y + x*z and  (y+z)*x = y*x + z*x
    (There are occasionally times in which one studies sets in which one 
    or both parts of  (5)  are dropped; we will not do so.)
We often delete the "*" if there is no danger of confusion.

The axioms above include the broadest class of what are called rings.
Usually one or more additional axioms are added; sometimes authors are
careful to distinguish the objects satisfying additional axioms, sometimes
not -- the reader will need to look carefully to check usage.

A  RING WITH (2-sided) MULTIPLICATIVE IDENTITY  is a ring as above
with an  element  "1"  with the property that
	(6) For all  x  in  A,  1*x = x*1 = x.
There are natural rings without multiplicative identities --
the even integers form such a ring. There are rings which have only a
one-sided multiplicative identity. However, in rings which have a left-
and a right-sided multiplicative identity, they are equal  (look at 
1_L * 1_R two ways); similarly, rings which have a multiplicative
identity can have only one. If  A  is any ring, one can make the direct
sum  Z+A  into a ring with multiplicative identity which contains  A  as a 
subring, although this construction may not have nice properties  A  has.
	I have to say -- not being a ring theorist -- that the demand
for rings not having a multiplicative identity is low (and probably
outstripped by supply). This is however a somewhat controversial topic.
A group theorist, number theorist, or algebraic topologist who refers
to a ring almost surely means a ring with a multiplicative identity.
Even some ring theorists insist on it, reserving the name "rng" for
a ring-with-no-I (get it?). But since the point of this document is to 
clarify the distinctions I will define "ring" without requiring an identity.
	Notation varies in other ways: The element  1  is sometimes
called "unity", and  A  a "ring with unity (element)". Sometimes  1  is
called "the unit", and  A  a "ring with unit", but this conflicts with
the definition of units below. There is a class of rings known as
"rings with identity", such as the "rings with polynomial identity" in
which there is an equation of several variables which is to hold for
all choices of elements in the ring -- for example, commutative rings
are rings satisfying the identity xy-yx=0. These will not be part of
our discussion.
	It is interesting to note that perfectly reasonable rings-with-
multiplicative-identity can have sub-(rings-with-multiplicative-identity)
for which the identity element in the smaller ring is different from the
identity element in the larger ring. In one canonical example the
big ring is the set of 2x2 real matrices ("1"=identity matrix) and the small
ring is the set of matrices whose only non-zero entry is in the upper
left corner. Some authors argue, with some persuasion, that this should be
disallowed. In the category of rings-with-multiplicative-identity, (homo-)
morphisms should carry the identity in the domain to the identity in the
co-domain; in particular, a subring (in this category) would have to have
the same identity element as the big ring.

The problem with rings following the broadest definition is that they tend
just to look  like groups. Indeed, our definition of a ring is really just
a pair  (A, f)  consisting of an Abelian group  A  and a group homomorphism
f : A --> Hom(A,A)  (see Q2); the existence of a unit element is sufficient 
(not necessary) to force  f  to be an injection. To force rings to be more
ring-like, one needs to insist that compound products satisfy some additional
axiom(s).

Far and away the most common class of rings are the ASSOCIATIVE RINGS, 
those in which we impose the additional axiom 
	(7) For all x, y, z  in  R,  x*(y*z)=(x*y)*z.
However, there are some extremely important rings which do not satisfy this
axiom. For example, there is the ring in which the underlying set is 
Euclidean space  R^3  with the ordinary addition of vectors, and in which
the product  "*"  is the cross-product. This is an example of what is 
perhaps the most important family of non-associative rings, the  LIE RINGS, 
which replace the axiom (7) with
	(7') For all x, y, z in  R,  x*(y*z)+y*(z*x)+z*(x*y)=0 and x*x=0
For example, this is satisfied by the ring of NxN real matrices in which
A*B is  AB-BA  (where the juxtapositions  AB  and  BA denote ordinary products
of matrices). Other classes of non-associative rings include Jordan rings 
(where  xy=yx and ((xx)y)x=(xx)(yx)) and alternative rings (where   
x(xy)=(xx)y and (yx)x=y(xx) ).

Independent of associativity, it is reasonable to discuss commutativity. A
COMMUTATIVE RING  is one in which
	(8) For all  x, y  in  R,  x*y=y*x.
This axiom is satisfied in many rings commonly studied, but the non-commutative
rings are also of enormous importance. The ring of all square matrices of a 
given size with, say, real entries is the canonical prototype of a 
non-commutative ring. There are disciplines in which it is common to discuss
only commutative rings, so that this adjective is omitted, but even in such
cases the adjective is often retained for emphasis.
	The adjective  "Abelian"  is not used. (All rings are already
Abelian groups under the other binary operation "+".)

Every element of every ring has an additive inverse ("-x"). Our
experience with real numbers suggests not every element can likewise
have a multiplicative inverse, but that we can come close. A  FIELD  is a
commutative, associative ring  A  with multiplicative identity in which
	(9) For every nonzero  x  in  A  there is a  y  with  x*y=y*x=1.
We'll see there do exist non-associative rings satisfying (9); in the
sequel we shall assume all fields are associative.  (I am not aware of
any author who allows a non-associative field.)
	Some authors do not require fields to be commutative, but that
usage is declining. We'll get to these non-commutative analogues in Q3.
We will assume a field is commutative.
	Why not 0, too? Well, if  x  is an element of a ring  A, then
x*0 = x*(0+0) = x*0 + x*0; by uniqueness of  "0", x*0=0. Likewise
0*x=0. Thus if  0  has a multiplicative inverse  Z,  then 1=0*Z=0. But
then for every  x  in  A,  x=1*x=0*x=0. That means  A  has only the
one element, namely  0 . There is just one ring structure one can put
on this ring (0+0=0, 0*0=0). Henceforth, we will assume our rings have
more than one element, to avoid this case. Then there can be no inverse
to  0  in any ring.
	In this document, we'll mostly consider fields like the sets of
real numbers and the sets of complex numbers, but fields like  F_p
(the set of integers modulo a prime  p)  and  R(X) (the set of rational
functions with real coefficients) are heavily used in mathematics.

In any ring with multiplicative inverse (not just fields), a 
UNIT is an element  x  for which an inverse  y  exists with  x*y=y*x=1.
As usual, there are also the concepts of left- and right- inverse, and
examples in which they are distinct from actual inverses. In an
associative ring, if  x  has left and right inverses, they are equal.

Now we come to another use of the word "algebra". An  ALGEBRA 
is a ring  A  which is also a vector space over a field  F,  that is,
there is a field  F  given and another binary operation  "."  which
multiplies elements of  F  with elements of R  to get elements of  A 
subject to these axioms:
(10) for all a,b in  F  and  x  in  A  we have  (ab).x = a.(b.x)
(11) for all a,b in  F  and  x  in  A  we have  (a+b).x=(a.x)+(b.x)
(12) for all a in  F  and  x,y  in  A  we have  a.(x+y)=(a.x)+(a.y)
(13) for all a in  F  and  x,y  in  A  we have  a.(x*y)=(a.x)*y=x*(a.y)
(14) for all  x  in  A  we have  1.x=x and  0.x=0. The  "1" and "0" on the
     left sides of the equations are elements of  F; the  last  "0" is in  A.
Unlike some of the preceding paragraphs, there is in this one
little interest in examining cases in which some axioms hold but not
others. I'm not even sure, offhand, if these are all independent.
	The situation is much easier to understand (and to write!) if
the algebra  A  has a multiplicative identity. Let's assume it has one,
called  I  for now. Then we define a function  f: F --> A  by  f(a) =
a.I  and observe that by the axioms above (and the commutativity of the
field  F)  it follows that  f  is a homomorphism of rings. An elementary fact
about fields implies that the image of  f  is isomorphic to  F except in the
trivial case  A={0}. Thus we may identify  F  with its image  f(F) and
view  F  as a subring of  A  (with the same identity element). It
follows from (13) that  F  is in fact in the center of  A, that is,
ax=xa for all x in  A  when  a  is in  F. (Here  ax  really means
a.x= a.(I*x) = (a.I)*x=f(a)*x, which is  a*x  once we identify  a  with f(a).)
Similarly, it follows that elements of  F  associate with all elements of  A.
	Examples of algebras over the reals include the complex numbers,
the quaternions, and the set of 2x2 real matrices, each with its usual
structure. There are also plenty of example of algebras which are 
infinite-dimensional over  F, for example the ring  A=F[X]  of (formal)
polynomials with coefficients in  F. Among the interesting non-associative
algebras over a field are the Lie algebras.
	The multiplication "*" in an algebra is completely described by
picking a basis for the vector space and giving the products  ei*ej  of
all pairs of basis vectors  ei  and  ej.
	One can of course construct algebras over a ring more general
than a field; we shall not do so, though for example the group algebra
Z[G], an algebra over  Z, is of considerable importance.

There are, of course, many other types of rings which have been
studied and named. Any abstract algebra book will give the 
definitions above and the basic properties that follow from them.

==============================================================================

Q2. What is a domain? 

This concept occurs frequently in, for example, algebraic number theory.
But as this file more or less emphasizes division algebras, I'll restrict
my comments to a few basic ones.

A(n) (INTEGRAL) DOMAIN  is a ring   A  satisfying
	(15) For every  x  and  y  in  A   we have  x*y=0 => x=0 or y=0
That is,  A  has no "zero-divisors" (or "divisors of zero") except  0 itself.
Examples include all fields, as well as the set of integers. Examples of
rings not meeting this condition include the ring of integers mod 6.
(Actually all the texts I have at hand require that  A  also be associative;
almost all ask that it also be commutative, and most require that it have an
identity element different from  0. We won't need to worry about these other 
conditions.)

There is a concept which applies to any ring  A  (as well as to other
algebraic constructs) which we can put to good use here. For any
a  in  A, we define a function  L_a : A --> A  by  L_a(x) = a*x. Observe
that by the distributive law,  L_a  is a homomorphism of the additive
group of  A, so we may view  L_a  as a member of this group  End(A)=Hom(A,A)
of endomorphisms. (End(A)  is also a group under pointwise addition:
(f+g)(x) = f(x)+g(x)  is also a homomorphism  A--> A  if  f  and  g  are.)
Moreover, the distributive law shows that   L_(a+b) = L_a + L_b,  so
that  L  itself can be viewed as a homomorphism from the additive group
of  A  into the group  End(A). We mentioned this view of rings in  Q1.

Now,  End(A)  is more than just a group -- the composition of two
endomorphisms is another one, and with this additional binary operation 
End(A) becomes a ring. It is reasonable to think that  L  is actually
a homomorphism of rings, but this can be seen to be equivalent to the
associative law! (It's also reasonable to think that  L  is an injection,
but this requires extra axioms too; the existence of a unit  1  in  A
is sufficient.)

Of course, all this can be done with right multiplications: we get another
additive map  R : A --> End(A) given by  R_a(x) = x*a; notice that
for an associative ring  A  we have the slightly offbeat R_(a*b)=R_b R_a.
The associative law may be expressed as the condition that the images of
the two maps  L : A --> End(A)  and  R : A --> End(A)  commute (elementwise).

The maps  L  and  R  are known as the "(left and right) regular representation"
of  A, or the  "adjoint representation" (especially for Lie Rings).

If  A  is an algebra over a field  F  then  L_a  and  R_a  are  F-linear
maps, and so lie in the set  Aut_F(A)  of  linear transformations of this
vector space. If  A  is finite-dimensional vector space over  F, Aut_F(A)
may be viewed as the ring of matrices of a certain size, once a basis for
A  is chosen. (This allows a handy characterization of associative,
finite-dimensional F-algebras with  1 : They are up to isomorphism
precisely the sub-algebras of  M_n(F)  for some  n.)

I bring the representations into the story now because one may characterize
domains with them: a ring is a domain iff the endomorphism  L_a  is an
injection for every  a (except  a=0).  Equivalently, it's a domain iff
every  R_a  is an injection except for  a=0. Note that in particular
L_a  is never the zero endomorphism for  a  not  zero, so that  L  is
an isomorphism of  A  with a subgroup of  End(A).

Moreover, if a domain is a finite-dimensional algebra over a field  F, then 
each  L_a and  R_a  is an invertible linear map, so that in particular it is
onto as well; as we shall see in  Q3, this will mean that such a domain
is a division algebra. The example  A=F[X] shows that finite-dimensionality
is essential in this theorem.

There is another connection with division rings which is best handled in
the commutative, associative case: If  A  is such a domain, then there is a
field  K  (unique up to isomorphism)  which contains  A  and all of
whose elements are ratios of elements of  A. This is the classic
construction  of a "quotient field"  used to construct the rational
numbers from the integers; consult any introductory text in ring theory.
Thus, commutative associative domains are precisely the subrings of fields.

I was reminded of some of these topics by Shapiro, who noted:
>Of course a "domain" can be defined as an associative ring with no
>nonzero zero-divisors.  It is interesting to ask whether every domain
>can be embedded into a division ring.  This fails in general...an
>embeddable domain must satisfy the "Ore conditions".  Some authors
>reserve the term "integral domain" for commutative domains.  I think
>that T.Y.Lam's book on "A First Course in Ring Theory" is a good
>reference for all of this.

==============================================================================

Q3. What is a division ring? division algebra? 

Informally, a division ring (once commonly called a skew-field or sfield)
is a ring which looks like a field except that it need not be
commutative; a division algebra is a division ring which is also an
algebra. However, there are several ways of describing the extent to
which a division ring may fail to  be a field.
	Indeed, we have defined a field to be a ring which is commutative,
associate, with multiplicative identity, and with a 2-sided inverse for
every non-zero element. One must decide how many axioms one wishes to
retain despite dropping commutativity. The real sport here seems to be
deciding what is to happen if no associativity is assumed. We will
assume associativity in section  Q3a and return to the non-associative
cases in  Q3b.

It might seem that this description of a division ring would make it
closely allied with the concept of a domain; indeed, in Q2 we have
indicated a little of this. But being a domain is NOT implied by the 
existence of inverses. Consider the following ring, which is a (commutative, 
3-dimensional) algebra (with identity) over the real numbers:
	A= { a 1 + b i + c j | a, b, and c are real },  
where the products of the basis vectors are  1*1=1, 1*i=i, 1*j=j,
i*i=-1, i*j=0, j*j=-1. (Note that  i*(i*j) is not the same as  (i*i)*j.)
One can easily calculate that  (a 1 - b i - c j)/(a^2+b^2+c^2)  is the
inverse of the element  (a 1 + b i + c j )  if the latter is non-zero.
On the other hand,  i*j=0  shows that  A  is not a domain.

Typically, one wishes a division ring to be a domain (so that,
in the opinion of most authors, the ring  A  above is not called a 
division ring). Moreover, some authors seek a definition which does not
require division rings necessarily to have identity elements. To 
meet both goals, we use this definition:

A  DIVISION RING  is a ring  A  such that for all  b  and  all nonzero
a  in  A  there is a unique solution  x  in  A  to the equation  a*x=b
and a unique solution  y  in  A  to the equation  y*a=b.

Equivalently: A  is a division ring iff for each  a  in  A  except  a=0
both endomorphisms  L_a  and  R_a  are isomorphisms.
With this definition, it's easy to check that a division ring must be an
integral domain. The converse is of course not true (e.g. A = the integers)
but as noted in  Q2  the converse does hold if, for example,  A  is
a finite-dimensional algebra over a field  F.

There is some redundancy here. If y*a=b and y'*a=b for distinct  y  and
y', then (y-y')*a=0 and (y-y')*0=0 would violate the uniqueness of
right-solutions. Likewise the uniqueness of  y  implies the uniqueness
of  x. But I believe no two of the four existence/uniqueness
conditions is sufficient to imply the other two.

Usually  x  and  y  are different (as easy matrix examples will show).

A division ring need not have an identity by our definition (although
some authors require it). An example is the set of complex numbers with
ordinary addition but with multiplication  z*w= z.conj(w)  (the complex
conjugate of  w). If however a division ring is associative, this
situation does not occur: for any nonzero  a  find the solution to
a*x=a; then for any  b we have  (a*b)=(a*x)*b=a*(x*b) so by uniqueness,
b=x*b and  x  is a left identity; then a similar argument shows  x  is
also a right identity.
	A nonassociative division ring  D  need not have an identity
element but is closely allied to one that does. The appropriate connection
is "isotopy": every division algebra is isotopic to one with an identity.
	An ISOTOPY  D1 --> D2  between two (division) rings is
a triple of additive isomorphisms  f,g,h  in Hom(D1,D2) with 
f(x) * g(y) = h(x*y)  for all  x, y  in  D1.  If  f=g=h, then  f  is an
isomorphism. If  D1  and  D2  are built on the same additive group  D,
then their left regular representations are related by   h L2_x = L1_f(x) g,
so that in particular two conjugate representations  L: A --> End(A) give
rise to isotopic rings. If  D1  and  D2  are associative with identity
elements, then taking  x=1 and  y=1  shows that  q(x) = g(x)*g(1)^(-1) is
actually an isomorphism between  D1  and  D2.
	We now observe that in any division ring, we can define a new
product  '#' on the same additive group  D  by selecting any two 
elements  u  and  v  in  D  and setting   x#y =  (R_u)^(-1)(x)*(L_v)^(-1)(y).
Then  (xu)#(vy) = x*y and (D,#) is a division ring with identity element vu.  
It's isotopic to the original ring. Of course,  (D,#)  may lose some
other critical property satisfied by  D.  [Thanks to Shapiro for reminding
me of this construction.]

A DIVISION ALGEBRA over a field F is an algebra over F which is a
division ring. This is really not as different from a division ring as
it may appear.  Suppose A is a division ring with a multiplicative
identity.  Let F be the set of all elements a in  A  with the following
properties:
	(1) For all  x, y  in  A,  a*(x*y)=(a*x)*y
	(2) For all  x  in  A,  a*x=x*a.
(If  A  is associative, (1) is unnecessary; if commutative, (2) is; if
both hold, F = A. ) This set is not empty because it includes 0 and
1. This set F can easily be checked to be a field. Then  A  is a
division algebra over F.  Thus although the correspondence is not
perfect, the studies of division algebras and of division rings are
closely allied.

(If  A  is a division ring without identity, one has to form  F  as
the set of endomorphisms  phi: A --> A  having  phi(x*y)=phi(x)*y=x*phi(y)
for all  x  and  y  in  A.  This F  is a commutative, associative ring 
with unit and I think that   A  a division ring makes  F  a field, over
which  A  is then a division algebra.)

==============================================================================

Q3a. What's an _associative_ division ring?

I've separated out this case because it's what many people have in mind
anyway, and it makes for easier reading, fewer counterexamples, and
stronger theorems.

Recall that a  DIVISION RING  is a (now associative) ring  A  such that 
for all  b  and  all nonzero a  in  A  there is 
	a unique solution  x  in  A  to the equation  a*x=b   and
	a unique solution  y  in  A  to the equation  y*a=b.

We have noted in Q3 that such a ring necessarily has a 2-sided identity
element  1. In particular, the equations  a*x=1 and  y*a=1  have a solution
for every nonzero  a; these  x  and  y  are then equal. In this way it is
easy to check that the definition above is equivalent to

A DIVISION RING is an associative ring  A  with identity element such that
every nonzero  a  in  A  has a 2-sided inverse.

In particular, we now see that an  _associative_  ring with identity in 
which every non-zero element has a multiplicative inverse is an
integral domain. Also, it is now clear in the associative case that
a division ring is a field iff it is commutative. 

Again, we can show division rings look like division algebras.
Let F be the set of all elements  a  in  A  with x*a=a*x for every  x  in  A
(the center of  A). Then  F  is also a division ring, now commutative,
and so it a field, over which  A  is an algebra. It is possible that
A  may be infinite-dimensional over  F, but if not, then  A  and  F
lie in a close relationship, as we have noted when studying the
regular representations  A --> M_n(F).

This is the basic tool for studying associative division rings (or
indeed, simple rings) finite-dimensional over their center. There is
quite a detailed understanding which results from this. For example,
one discovers the dimension  [A:F]  is a perfect square, equal
to  [K:F]^2  where  K  is a maximal subfield of  A, and so on. 

One consequence of this is that a division algebra finite-dimensional
over  C  (the complex numbers, or indeed any algebraically closed field)
which has  C  in its center must in fact just be  C.

(The rest of this section is intended for the budding algebraist.)

There is a very pretty theory of finite-dimensional associative
division algebras with center F (a field). The isomorphism classes of
these algebras form a group with identity  F  and product  D1.D2=D3  
defined as follows: D1 tensor_F D2 is isomorphic to the ring of matrices 
of a certain size over a certain division ring whose center is F; 
the size of the matrices and the isomorphism class of the
division ring D3 in this isomorphism are unique.  This group is called
the Brauer Group of F. Within that group are many subgroups, one for
each Galois extension field K of F: let B(K/F) be the set of division
algebras having a maximal (commutative) subfield contained in K;
equivalently, the set of division algebras with D tensor K being full
matrix ring over K.  Then the whole Brauer group  B(F)  is the union of these
B(K/F) in the natural way. Moreover, B(K/F) is isomorphic to the
second cohomology group H^2(Gal(K/F),K^*).

The pity here is that the quaternions give such a poor demonstration 
of this theory. There is only one proper finite-dimensional extension 
field of  R  (namely  C)  and  B(C/R) is the group of  2  elements
{R, H}. However, the situation is more interesting when  F  has lots
of algebraic extensions. For example, if  F  is a local field, 
B(F)  is isomorphic to  Q/Z.  There is also a local-to-global principle
for classifying division algebras over  Q. The entire discussion is
rather number-theoretic.

References: van der Waerden "Algebra II"; Deuring, "Algebra"; Reiner
"Maximal Orders"; Scharlau "Quadratic and Hermitian Forms".

==============================================================================

Q3b. What's a non-associative division algebra?

Recall that at the end of  Q3  we showed every division ring 
can be viewed as an algebra over a field; the phrasing of the question
above reflects that bias.

The point of this section is to demonstrate the construction of all
the (many!) division algebras over a field with a given finite dimension.
A similar construction may be followed in the infinite-dimensional case,
but there are a number of issues which I think are not sufficiently
worked out when the vector space is not finite dimensional.

The construction is straightforward. Suppose  A  is a division algebra over
a field  F  and of finite dimension over  F. Then as we have noted, the
left regular representation  L  of  A  gives an  F-linear embedding of
L  into  End_F(A); once a basis of  A  is chosen,  L  embeds  A  as a
vector subspace of  M_n(F)  of dimension  n. If  A  is a division algebra,
the image  L_a  will be a invertible matrix for every non-zero  a  in  A.

Actually, our definition of a division ring is almost equivalent to this.
We did also require that for every nonzero  a  and every  b  there be a 
unique solution to  y*a = b. But this is redundant in the case  A  is 
finite-dimensional over its center. Indeed, if the equation y*a=b has
no solution, then right-multiplication by  a  is not onto, and so cannot
be one-to-one, and thus c*a=0  for some nonzero  c, which is not possible if
(a is nonzero and) left-multiplication by  c  is injective. But if
right-multiplication is surjective, it must also be injective in the
finite-dimensional case.

Thus the embedding  L : A --> End_F(A)  is all we need check to see if
a ring  A  is a division algebra. In order to check for isomorphisms, we
can proceed as follows. Recall that a  FRAMED  vector space is a vector space 
together with an explicit choice of basis. Likewise we will call a 
division algebra with a preferred basis a  FRAMED DIVISION ALGEBRA.

THEOREM: There is a one-to-one correspondence between the following sets:
(1) Framed division algebras  A  over  F  of dimension  n.
(2) Framed n-dimensional subspaces  V  of  M_n(F)  all of whose elements
    (except the zero matrix) are invertible.
Under this correspondence, a change of basis in the algebra corresponds
to a change of framed subspace as follows: if  X  is the change of
basis matrix, the  i-th matrix   M_i  in the framing is replaced by
X (Sum x_ij M_j ) X^(-1);  in particular, the subspace  V  is replaced
by the subspace  X V X^(-1).

PROOF: Given a basis  {e1, ..., en}  of  A, we let  V  be the set of
all matrices  L_a  for  a  in  A. (We are expressing these linear
maps  L_a  as matrices relative to this basis  {e1, ..., en} of the
vector space.)  As noted above, each of these matrices is nonsingular
(except L_0), and together they form a vector space  V  of dimension  n.
A framing of  V   is the basis  {L_(e1), ..., L_(en)}.
	Conversely, given any framed vector space  V  over  F  of
dimension  n with basis  {M1,...,Mn}  (each  Mi  is a matrix), we form
A = F^n  into an  F-algebra as follows. Let  {e1, ..., en}  be the
standard basis of  F^n. Define a product on  A  by declaring
that  ei*ej  be the  j-th  column of the matrix  Mi, and then extending
bilinearly. If indeed every matrix  M   in the span  V  of the  Mi  is
nonsingular, then the columns of  M  are linearly independent; these
represent the images of  M*ej, so  multiplication by  M  is onto. As noted
above, this means the constructed algebra  A  will be a division algebra.
	It is easy to see that the constructions of these two paragraphs
are inverses of each other, and establish the needed one-to-one
correspondences.
	If the basis of  A  is changed, each of the matrices  L_(ei)
is replaced by  X.L_(ei).X^(-1). We then compute  L_(fi)  as a linear
combination of these to determine the form of the new framing.   QED

This idea is not new to me, and it must be conceded that it doesn't
provide more information about the individual division algebras other
than the size of the set of them. Indeed, I'll quote Shapiro again:

>This Theorem has been observed by a number of authors over the years,
>in various forms, but it does not seem to be at all useful in
>the classification problem.  The major use seems to be to put the
>problem of division algebras into a larger context:  How to find the
>maximal dimension of a linear subspace W of mxn matrices such that
>every element of W - (0) has rank k (or rank \leq k; or rank \geq k).
>Various topological tools can be applied over the reals to get results
>which generalize the 1, 2, 4, 8 theorem.  This is discussed in the
>classic paper Adams-Lax-Phillips, On matrices whose real linear
>combinations are nonsingular, Proc. Amer. Math. Soc. 16 (1965)
>318-322; 17 (1966) 945-947. 
>Also see K.Y.Lam and P.Yiu, Linear spaces of real matrices of
>constant rank, Linear. Alg. Appl. (to appear).  (Probably it has
>appeared by now).

==============================================================================
    
Q4. How do you construct the division algebras over the real numbers?

There are many division algebras which are infinite-dimensional algebras
over the real number field. For example, any extension field of  R(x)  will
do. Then many associative. noncommutative  division algebras central over those
fields may be constructed as mentioned at the end of Q3a.

The real interest is in finite-dimensional division algebras over  R.
There are several well-known examples: the real numbers themselves and
complex numbers are associative and commutative, the quaternions are
associative but non-commutative, and the 8-dimensional octonions are
not associative. This question focuses on algebras of these dimensions
(1, 2, 4, and 8); the next question asks why these dimensions are the 
only ones available.

It is commonly asserted that these four are the only real division algebras.
                  THIS IS NOT TRUE.
We have described in Q3 a 2-dimensional division algebra over  R  which
does not have an identity element, and so is not isomorphic to 
any of these four algebras. Now, we will see that these include the only
associative ones, but the octonions are not associative, so it is
not immediately clear what this claim is supposed to mean.

For general (non-associative) algebras, we may use the classification
theorem of Q3b. That theorem gives a description of all framed finite-
dimensional division algebras; the general linear group  GL(n,F) acts on
the set of all of these in a natural way, and the orbits under this action
correspond to the distinct algebras up to isomorphism. (In English: any
given division algebra will show up more than once in that theorem because
you can choose different bases.)

So just how many distinct division algebras are there of dimension  n?
We can recast the theorem of  Q3b  just a little differently by noting that a
framed subspace of  M_n(R)  is really just a collection of linearly
independent elements of  M_n(R). Thus, the framed subspaces of M_n(R)
lie in one-to-one correspondence with the elements of  M_n(R) x ... x M_n(R)
(n copies). The n-tuples (A1,...,An) in here which correspond to framed
division algebras correspond to the ones on which the function
f(t1,...,tn)  = det( Sum( ti.Ai ) )  has only one zero on its whole
domain  R^n  (namely at  (0,0,...,0)  ).

Note that the restriction of  | f |  to the unit sphere in  R^n  gives a
continuous function on a compact set, which then must attain a minimum
value  M . Thus we have a real-valued function  M  defined on all
n-tuples of matrices in  M_n(R). It is easy to see that  M=0  iff
f has a non-trivial zero on  R^n, so the division algebras correspond
to the set of  n-tuples  for which  M>0. One can show that  M  depends
continuously on  (A1,...,An)  (which may simply be thought of as a point
in  R^(n^3) ), so that the set of division algebras is an _open subet_
of  R^(n^3). In particular, this set, if nonempty, forms a manifold of 
dimension  n^3.

We have already noted that the isomorphism classes of division algebras
now corresponds to the quotient of this space by the action of 
GL(n,R). This last group is a Lie group of dimension  n^2, leaving a
set of dimension  n^3-n^2  to parameterize the set of all isomorphism
classes of division algebras.

For n=1, we already know that  R  is the only division algebra of this
dimension. For  n=2, this already gives us the surprising fact that there
is a 4-parameter family of division algebras of dimension 2 !
(For n=4: 48; for n=8: 448.)

On the other hand, many of these algebras are isotopic, that is, 
the new product may be found from the old one as 
x#y = P((Mx)(Ny)) where  M, N, and  P   are fixed invertible matrices.
Roughly speaking, this can provide at most a 3(n^2)-dimensional family of
framed division algebras starting from any single one.

In particular, for n=4 or 8, this procedure will not exhaust the set 
of division algebras of dimension  n, so that there exist algebras
which are non-isotopic, and so more fundamentally different.

On the other hand, we have already noted that  R  is the only division
algebra of dimension  1. While there is a 4-parameter family of division
algebras of dimension  2, they are indeed all isotopic to the complex field.

I should comment that I don't know a set of internal characteristics
which would allow one to decide whether or not two division algebras
are isomorphic or isotopic. For example, we used the existence of an
identity to distinguish two 2-dimensional algebras. Other internal
signals might include the cardinality of the set of square roots of 1,
and so on.  However, as noted much earlier, this method of
parameterizing the set of division algebras is unlikely to be helpful
for such concerns.

Throughout this section I have assumed that the dimension of the
algebra is 1, 2, 4, or 8, and that the underlying field is the set of
real numbers. I have not used that in any way except to draw on the
existence of already-known division algebras in those dimensions. To
that extent, this procedure is perfectly general, that is, it works
for any field for which we have at least one division algebra of a
given dimension. In Q5 I will discuss the choice of those dimensions.

References: life work of A.A.Albert (ca. 1944-1972)
RH Bruck, TAMS 56 (1944) pp 141-197
J. Marshall Osborn has considered division algebras satisfying various axioms.
H.-D. Ebbinghaus, et al.,"Numbers". (See, "The Dimension of a Division Algebra
	Is 1, 2, 4, or 8")
==============================================================================

Q5. What's the question with answer "n=1, 2, 4, or 8"?

I will resist stupid responses.

One possible response is the Hurwitz problem (1898): For what values of
n is there a product on F^n so that Q(x) Q(y) = Q(x*y) for some
nondegenerate quadratic polynomial Q on F^n?  (Hurwitz was interested
in the case Q(x) = Sum xi^2.) An algebra for which such a  Q  exists is
called a  COMPOSITION ALGEBRA. For a field of characteristic not 2, these
exist iff  n=1, 2, 4, or 8. For some elements of the field c1, c2, c3
the multiplication table may be found as the upper left portion of the
following table

*    |  i0 |  i1   |  i2   |   i3  |  i4   |   i5  |   i6   |     i7   
------------------------------------------------------------------------
i0   |  i0      i1      i2       i3     i4       i5       i6         i7
i1   |  i1   c1 i0      i3   c1  i2     i5   c1  i4  -    i7  -c1    i6
i2   |  i2  -   i3   c2 i0  -c2  i1     i6       i7   c2  i4   c2    i5
i3   |  i3  -c1 i2   c2 i1  -c1c2i0     i7   c1  i6  -c2  i5  -c1c2  i4
i4   |  i4  -   i5  -   i6  -    i7  c3 i0  -c3  i1  -c3  i2  -c3    i3
i5   |  i5  -c1 i4  -   i7  -c1  i6  c3 i1  -c1c3i0   c3  i3   c1c3  i2
i6   |  i6      i7  -c2 i4   c2  i5  c3 i2  -c3  i3  -c2c3i0  -c2c3  i1
i7   |  i7   c1 i6  -c2 i5   c1c2i4  c3 i3  -c1c3i2   c2c3i1   c1c2c3i0

Of course, it is easy to address the situation with degenerate quadratic
forms using this information.

The ring constructed is associative and commutative if  n=1 or 2, associative
but non-commutative when  n=4, and alternative but non-associative when
n=8. (For n>8, the corresponding ring is not even alternative, but does
satisfy some polynomial identities, e.g.  x(yx)=(xy)x, and so is 
power-associative.) The general process used to construct these algebras is 
sometimes referred to as "doubling" the algebra.

When  F  is the field of real numbers, only the sign of the  ci  are
necessary to determine the isomorphism type of the algebra. Taking all
ci = -1  gives respectively the real numbers  R, the complex numbers  C, the
quaternions  H, and the octonions (Cayley algebra)  O   when  n=1, 2, 4, or 8.

Another possible answer is Frobenius' theorem: if  A  is a division algebra
over the reals, it must be  R  or  C  if commutative,  H  if non-commutative
but associative, and (one may show)  O  if non-associative but alternative.
(see Bruck's article in the MAA book "Studies in Modern Algebra" (1963).)

Finally, the question with this answer might be "what are the dimensions
in which a real division algebra exists?"; we shall address this in
the next section. In the course of that discussion we will see a solution
which applies to a number of related problems: for which  n  has the 
unit sphere a trivial tangent bundle; for which  n  is the
sphere in  R^n  a  Lie group  (n=1, 2, or 4), and so on.

Reference: Jacobson's "Basic Algebra I" (1974, WH Freeman, San Francisco)

[Somehow I think there are more answers to  Q5 that I can't quite recall.]

==============================================================================

Q6. What makes the real numbers different? (Topology)

What are the finite dimensions in which a real division algebra are possible?
Somehow we need to use some property of the real numbers. Indeed, for the
complex numbers, there are no division algebras of any finite dimension >1;
for the rational numbers there are division algebras of every dimension.
The tools which turned out to be useful turned out to be topological.

THEOREM: Let  D  be a division algebra finite dimensional over the real
number field  R.  Then  D  is of dimension  1, 2, 4, or  8  over  R.

This result uses the definition of a division algebra in the sense of Q3,
that is, it does not assume  D  is commutative, associative, or has a
multiplicative identity. It does not even assume that elements have
two-sided inverses! (That is, it is sufficient to assume that for
each pair of elements  a  and  b  in  D, there is a unique solution  
x  in  D  to the equation  a*x=b.)

I won't prove this result, of course, but I can give a little of the
flavor of the proof. The main idea is already in  Q3b:  we need to know
that the dimension of a linear subspace of  M_n(R),  whose only
singular member is the zero matrix, is less than  n  for any  n  unless
n=1,2,4, or 8.

Let us consider just the example of  n  odd.  I will show that any subspace
of dimension  2  or more in M_n(R) must contain singular matrices; in 
particular, there is no appropriate subspace of dimension  n  (unless  n=1).

Suppose you had 2 matrices  A  and  B  in  M_n(R), not scalar multiples
of each other. Consider the polynomial  f(x) = det(A + x B). The highest-order
term in this polynomial comes from the products involved in computing
det(xB) = x^n det(B). By assumption, det(B) is not zero, so  f  is indeed
a polynomial of degree  n. Since  n  is odd,  f  must have at least
one real root  r.  (This is where we have used something about the 
topology of  R). Thus  A+rB  is a singular matrix in the vector space.
Since  A  is assumed not to be a multiple of  A,  A+rB  is not the
zero matrix, a contradiction.

At the next level of difficulty, one tries to show that if an
n-dimensional nonsingular subspace exists, then  n  is a power of  2.
There are a number of ways of viewing this problem topologically.
All use some variant of this idea: if  D  is a division algebra over R,
then we have may view the non-zero elements of  D  as special (topological)
subspaces of  R^n  (once a basis of  D  is chosen of course).

For example, if we restrict the multiplication  D x D --> D  to elements
of length one (in this basis), then the product is never zero. It might
not have length one, but we can divide to get a map  f(x,y) = (x*y)/|x*y|.
Since  D  is a real division algebra, the product map is bilinear, and
in particular continuous; then  f  is as well (in fact,  f  is differentiable);
f  is a function  S^{n-1} x S^{n-1} --> S^{n-1}, where  S^{n-1}  is the
n-1-dimensional sphere.

Suppose for simplicity that   D  does have an identity element 1, which
we can assume is the first basis element  e1. Pick another basis
element  e  and consider the elements  1+te for small  t. These
approach  1  as  t --> 0. Taking derivatives at  t=0  gives a vector
tangent to the sphere at  1  and pointing along one of the coordinate
planes in  R^n. The  n-1 vectors so obtained are linearly independent.
For any   x  in  S^{n-1},  compute  f(x, 1+t e); this will
likewise trace out a curve on the sphere, and in the limit as  t -> 0
gives a  tangent vector to the sphere at  x. Informally it is clear
that this gives (for each basis vector) a tangent vector at each point
of the sphere and varying continuously -- this is called a vector field
on the sphere. Doing likewise for each of the other basis vectors gives
n-1  vector fields on the sphere, linearly independent at each point.

A manifold with the property that there exist this many linearly 
independent vector fields is called parallelizable; this argument shows
that any topological group is parallelizable, for example.

Not all manifolds are, though. For example, on the 2-sphere, we can't
even find one non-zero vector field, let alone  2.  (This is the famous
result that "you can't comb the hair on a sphere".) Thus, topologically,
the result we wish to show is that the parallelizable spheres have
dimension one less than a power of  2.

Here is one way to see this if you have some algebraic topology. Observe
projective space would then be parallelizable, i.e. the tangent bundle of
RP^{n-1} is just the product  RP^{n-1} x R^{n-1}. With a trivial tangent 
bundle, the total Stiefel-Whitney class of the tangent bundle in 
H^*(RP^{n-1}, Z/2)  must be 1. But, writing this cohomology ring as a truncated
polynomial ring  Z/2[a] / a^n, one shows that the total Stiefel-Whitney class
is (1+a)^n. Expanding by the binomial theorem, we compute that this polynomial
is just  1  only if  n  is a power of two.

[An algebro-geometric proof of the fact that a real division algebra
has to have dimension a power of 2 is in Shafarevich's Basic Algebraic
Geometry (Springer 1977) IV, par 2, Theorem 3; thanks fo T. Edkedahl for this
reference.]

Shapiro provided some of of the history:
>...I think the problem [parallelizable spheres have dimension  2^N-1]
>was first settled by Stiefel (1940) and Hopf (1940) using two
>different proofs (see Comment. Math. Helv. 13 (1940/41)  201-218 and
>219-239. Stiefel used what are now called Stiefel-Whitney classes,
>while Hopf used what is now called the ring structure of cohomology.

The final step is the proof that the higher-dimensional spheres are
not, in fact parallelizable, so the division algebra must be of dimension
1, 2, 4, or  8  over the real numbers. Continuing the quote above:
>... [n=1, 2, 4, 8] was first
>proved independently by Kervaire and Bott-Milnor in 1958, with proofs
>based on the Bott periodicity Theorem.  Adams proved more general
>results concerning vector fields on spheres which generalize those
>ideas.  Adams' proof led quickly to topological K-Theory, developed in
>the early years by Atiyah, Hirzebruch and others.  That theory
>provides an important tool for analyzing such problems.

[There is also a related paper by Stiefel in Comment Math Helv 8 (1936)
pp. 3-51. The Bott-Milnor paper is in Bull AMS 64 (1958) 87-89 (!) and the
Kervaire paper is in Proc Nat Acad Sci USA 44 (1958) 280-283.]

Adams work includes a discussion of the Hopf invariant. This is an element of
the homotopy group  pi_{2n-3}(S^{n-1}), which is isomorphic to the group of
integers. A map  S^{n-1} x S^{n-1} --> S^{n-1}  of bidegree  d1, d2   leads
to an element of this homotopy group which is  d1*d2  times a generator.
In particular, if  R^n  is a division algebra, the induced map has Hopf
invariant  1. The goal then is to show the non-existence of maps with
Hopf invariant  1  for  n  not  1, 2, 4, or 8.

Adams' results were first proved using secondary operations in 
cohomology, although modern proofs use the (guess what) Adams operations
in K-theory. [Thanks for Geoff Mess for a correction here.]

General references: Milnor and Stasheff, "Characteristic Classes"; 
Hu, "Homotopy Theory"; Husemoller, "K-Theory".

In my personal opinion the resolution of this question about the
existence of real division algebras is a crowning achievement of
20th century mathematics.


Addendum (June 2002): A simpler illustration of the distinct nature of
the real field is in the proof of this more restrictive fact:
   Theorem: There is a COMMUTATIVE division algebra structure on R^n 
            iff  n = 1 or n = 2.
(Commutative, ASSOCIATIVE division algebras may be shown to be fields,
of which there are not many containing  R ! But no associativity is used here.)
A short proof by WIlliam B Gordon may be found in the American Mathematical
Monthly 84 (1977) 28-29. He notes that the squaring map in a commutative
division algebra (of characteristic not 2) is precisely 2-to-1 on the
set of nonzero elements. The key topological aspect needed to contradict
this in R^n (n>2) is a theorem of Hadamard which asserts than an immersion 
between real manifolds must in fact be a homeomorphism (one-to-one) if it 
the inverse images of compact sets are compact. 

==============================================================================

Q7. What can you say about finite rings?

If  A  is a finite ring, it is also a finite Abelian group (under addition)
and so equal to the direct sum of its Sylow p-subgroups. It is easy to see
that each of the latter is a subring, so up to the taking of direct sums,
every ring is a finite  p-group under addition.

Note that a finite ring cannot then be an integral domain unless it is
a finite p-group for a single prime  p.  Moreover, it must be
elementary Abelian (no elements of order  p^2)  and so it is a
vector space over  F_p. Thus, a finite integral domain is an
F_p algebra.

If  A  is a finite integral domain and  a  is a nonzero element of  A,
then since  a*x  is never zero for nonzero  x, a*x must run over all the
different elements of  A  as  x runs over  A; so does  x*a, so that  A
is now a division algebra over  F_p.

If  A  is a finite _associative_ integral domain, then it must be a
division algebra (easy) and in fact a field  (Wedderburn's theorem).

AFAIK, it may be that all the _nonassociative_ finite integral domains are
_isotopic_ to a field, but I'm sure someone will inform me otherwise :).

Finite fields are used extensively in number theory (and elsewhere); there
is (up to isomorphism) precisely one field of each prime-power cardinality.
(Additively, they are vector spaces over the prime fields  Z/pZ; in
particular, the field with  p^r  elements is  NOT  Z/(p^r Z).)

There are interesting non-commutative finite rings (particularly  M_n(F)
for a finite field  F).

Jacobson's "Basic Algebra I" (1974, WH Freeman, San Francisco)

==============================================================================

Q8. Can you extend the cross product to  N  dimensions?

Well, sure, you can extend the definition of a cross product in lots of ways,
but you have to ask what you want the generalization to accomplish. 
If what you intend is to create a product  R^n x R^n --> R^n  which is
bilinear, then you will have made  R^n  into an algebra. What kind of
properties would you like this algebra to have? You have many bilinear 
products to choose from!

One nice property of the ordinary cross product is that if we write
i, j, and k for the basis vectors in R^3, then if we compute a product 
in the quaternions
	(ai+bj+ck) * (di+ej+fk) =  A + Bi + Cj + Dk
then the vector  Bi+Cj+Dk  is just the cross product of the original vectors.
Moreover,  A  is the negative of the dot product (inner product).

So, yes, a "natural" generalization exists: in  R^7 you can define the
"cross product" to be the non-unit portion of the product of two vectors in
the octonions. In dimensions higher than  7  there is no direct analog
of this construction (in view of  Q6).

On the other hand, one useful interpretation of the cross product is
that it provides a perpendicular vector in  R^3. The appropriate
generalization to  R^n is to take not two but  n-1  vectors in  R^n.
The determinant definition of a cross product generalizes to give
another vector in R^n which is perpendicular to all the others, zero if
the others are linearly dependendent, and of magnitude proportional to
the (n-1)-measure of the region "spanned" by the vectors. 
(The corresponding operation in R^2  takes a _single_ vector (x,y) and
produces a vector (-y,x) perpendicular to it. )

Likewise, a valid but unpopular generalization of the cross product is
to take the product of  2  vectors in  R^n  and get, not an  n-tuple,
but an  n(n-1)/2 - tuple consisting of all possible 2x2 determinants of
corresponding entries, which vanishes iff the vectors are collinear.
The corresponding product in  R^2  is  (x,y)*(x',y')=(xy'-yx'), a scalar.
(It's equal to   (x,y) dot U(x',y'),  where  U  is the unary operation
of the preceding paragraph.)

Shapiro forwarded a reference:
>A good survey article on this topic is by Eckmann, Continuous
>solutions of linear equations -- An old problem, its history, and its
>solution, Expo. Math. 9 (1991) 351-365.

Pertti Lounesto writes:
>Your discussion in the section "Q8. Can you extend the cross product 
>to  N  dimensions?" is incomplete.  You consider products of 2 vectors
>in R^7 and products of n-1 vectors in R^n.  If you allow such
>generalizations, then you should, for completeness, study all the
>types of products of k vectors in R^n, 1<k<n.  You would find,
>in addition to your cross products (7,2) and (n,n-1), also a cross
>product (8,3), and nothing else (with an appropriate definition).
>For more information, I can recommend Chapter 7 "The Cross Product"
>of my book "Clifford Algebras and Spinors", CUP, LMS LNS 239, 1997.

==============================================================================

Q9. What's a Clifford algebra? exterior algebra?

Each of these is an associative algebra with identity over a field  F,
and is created from a vector space of dimension n and resulting in an
algebra of dimension 2^n.  As I write this I am reluctant to mention
more than a couple of facts since each has a huge literature
associated with it and I'm not sure which facts are the most salient.

Assume the original vector space has basis  {e1,..., en}

The Clifford algebra is generated by these elements  {e1, ..., en}
with the relations that  ei^2=-1 and  ei*ej= - ej*ei. A basis for
this algebra over the field is the set of products  e_{i1}*...*e_{ik}
where  i1 < i2 < ...  and the sets of subscripts  S = {i1, ..., ik} 
range over all the subsets of  {1,...,n}. These algebras are
semi-simple. The study of Clifford algebras is related to the study
of quadratic forms and their orthogonal groups.

One may show that the "even" subalgebra of the (4-dimensional) Clifford 
algebra over  R^2  is the complex field, and the corresponding algebra
of the  (8-dimensional) algebra over  R^3  is the quaternions. Thus,
arguably, the corresponding (associative!) 8-dimensional subalgebra of
the Clifford algebra over  R^4  is the appropriate "generalization"
of the quaternions.

The exterior algebra is generated by the same  n  elements, and indeed
has the same basis of  2^n elements, but are subject to the relations
ei^2=0 and  ei*ej=-ej*ei. As a consequence, x^2=0  for every element in
the span of the  e_i. These algebras arise from consideration of the
differential forms on a manifold  (e.g. the forms  dx  and  dy  on a plane 
curve) and in multilinear algebra  (the determinant is, roughly, the
last basis element e1*...*en.) The exterior algebra is also known as
the Grassmann algebra associated to the vector space.

Exterior algebras are treated briefly in differential algebra books,
for example Singer and Thorpe, "Lecture Notes on Elementary Topology and
Geometry". For Clifford algebras consider Hestenes and Sobczyk, "Clifford
Algebra to Geometric Calculus". [Thanks to Roger Critchlow for the pointer.]

==============================================================================

Q10. How does this relate to projective geometry?

Given a field  F  we define PROJECTIVE n-SPACE  over  F  to be the set P_n(F)
of lines passing through the origin in the (n+1)-dimensional vector space
F^(n+1). Most of these lines are not "horizontal", and then have a unique
intersection with the hyperplane defined by the equation  x_{n+1}=1, so
for the most part, you can view projective n-space as looking like the
vector space  F^n  (more properly known in this context as "affine space".)

In projective space we define subsets called "lines" (the set of lines through
the origin which all lie on a single plane is called a "line" in projective
space), as well as "planes" and so on. Projective space is cleaner than
affine space: any two elements lie in a single line, and every pair of
lines has a unique point of intersection. Similar statements hold for the
higher-dimensional subsets.

Now, if one takes some care about order of products, one can likewise construct
a space  P_n(D)  for any associative division ring  D. Alternatively, one can
build such spaces axiomatically: a _projective n-space_ is a set with 
collections of subsets known as "lines", "planes", and so on, meeting the 
appropriate intersection properties. For example, a _projective plane_ is a
set  P  given with a collection of subsets called lines such that
	(a) every pair of elements of  P  belongs to a unique line
	(b) every pair of lines intersects in a unique element of  P.
and a non-degeneracy condition (there exist 4 points no three of which are
collinear.)

The great theorem is that every projective n-space is isomorphic to
P_n(D)  for some associative division ring D, unique up to isomorphism;
thus the theory of associative division rings is roughly equivalent to
theory of projective spaces...

Except: the theorem does not hold for  n=2. One needs to add the condition 
that the projective plane be embeddable in a projective  3-space. This
can be checked internally by postulating that the plane be Desarguian, i.e.
that a certain geometric proposition (having to do with intersections of
lines) hold in the plane. I will return to this in a moment.

Within the class of Desarguian planes, we can distinguish the class of
Pappian planes, those in which Pappus' theorem holds. This is a theorem
of "classical" projective geometry which states: given two distinct lines
L1  and  L2,  and distinct points A1, B1, C1 on L1 as well as distinct
points A2, B2, and C2 on L2, then the following three points A3, B3, and
C3  are also collinear: A3, for example, is the intersection of the line
joining B1 and C2 with the line joining C1 and B2; B3 and C3 are defined
similarly.

The important theorem is that a Desarguian plane  P(D)  is Pappian iff 
D  is commutative (i.e., a field).  Personally, I consider this result
to be among the most aesthetically pleasing in mathematics. For
example, it may be combined with the results in  Q7  to show the highly
non-trivial fact that Pappus' theorem holds in every finite
(Desarguian) projective plane.

Among the projective _planes_ we need also be concerned with the
non-Desarguian case. It is possible to begin with any division ring  D  with
an identity element and construct a projective plane  P(D)  from it, although 
a different definition must be used for non-associative rings if we wish to 
create something consistent which agrees with the usual construction for 
associative algebras. Among the results which I will cite are 

Theorem: Two projective planes  P(D) and  P(D')  are isomorphic iff the
division rings  D  and  D'  are isotopic. (In particular, one may as well
assume  P(D)  to be defined for all division rings.)

Theorem: A division ring is alternative iff  P(D)  is a Moufang plane
(i.e., a certain geometric condition is satisfied).

I should also add that there are projective planes which are not isomorphic
to  P(D)  for any division ring  D.  Here, perhaps, is the one place in
this document in which the term "division ring" might be interpreted even
more liberally (although in practice alternative labels are used). It is true
that given any projective plane one may find a set  R  containing elements
called 0 and  1  and a ternary operation  T : R x R x R --> R which satisfies
a number of ring-like axioms (roughly, T(x,y,z) behaves like  xy+z). One
may use  T  to give  R  binary operations  x+y = T(1,x,y)  and x*y=T(x,y,0).
However, this "ring"  R  need not, for example, necessarily even have an
associative law for addition! This leads to concepts such as "quasifield"
(associative addition and one distributive law) and "nearfield" (quasifield
with associative multiplication) among these "ternary rings" (which all
have identity element and invertible left- and right-multiplications).
Ordinarily, the resulting algebraic morass would not merit serious interest
except among those fond of axiomatics, but since the projective planes 
themselves are of interest, these algebraic categories do arise naturally.

A celebrated line of inquiry is the determination of those integers which
can be the number of points n+1 in each line of a projective plane (the
number is a constant); the plane is then said to be of order n. It has
now been shown, with an enormous amount of computer time, that none
exists with n=10.

Most of this information comes from Hughes and Piper, "Projective Planes";
Artin, "Geometric Algebra"; and Dembowski, "Finite Geometries".

==============================================================================

Q11. Can you extend the notion of quaternions? 

Again, it is important to ask what the goal is of the generalization
is. This will clarify what features you want to retain in the
generalization.

We have already addressed the questions of the construction of
division algebras over the reals; the process of doubling an algebra
and the construction of the octonions; the theory of Clifford
algebras; and the existence of a quadratic "norm" (the Hurwitz
problem). 

There is a certain amount of number theory related to the quaternions,
such as Lagrange's proof that every number is the sum of four perfect
squares. I cannot recall many details. (Except I know I studied this
and learned that the "right" ring of integers is actually the set of all
(1/2)(a+bi+cj+dk) where  a,b,c,d all have the same parity.) 
I'm sure someone will write me and refresh my memory.

I am open to addressing other extensions one might hope to find.

==============================================================================

I would appreciate hearing of other topics that fit into this general
thread which are asked often enough that they deserve a home of their own.
Dave rusin@math.niu.edu