There Are Only Four Finite-Dimensional Division Algebras Over the Real Numbers

实数域上有限维可除代数只有四种 苏剑林 · 2014-11-12 · Mathematics algebra matrix ring theory

Today in abstract algebra class, the professor was discussing division rings and gave an example of a non-commutative division ring — the quaternions. He then mentioned that "there are only 4 finite-dimensional division algebras over the field of real numbers," namely the reals themselves, the complex numbers, the quaternions, and the octonions (here a division algebra is just a division ring). This statement sounded somewhat familiar to me, yet somehow off. I recalled reading in some book that if we define hypercomplex number systems over the reals and additionally require multiplicativity of the norm, then indeed only these four exist. But the professor's statement implied that even without requiring multiplicativity of the norm, there are still only four. I naturally assumed the professor had misremembered, and I argued with him for a while. Then, back in my dorm, I looked things up and finally confirmed: there really are only four finite-dimensional division algebras over the reals! Below I'll briefly share my understanding of this problem.

Of course, I cannot possibly give a proof of this proposition here, since the proof is quite involved and I myself have not fully worked through it. But it is possible to get a rough sense of why it should be true. On first seeing this proposition, our reaction might well be: how could there be so few! Let's use some examples to show, informally, that indeed there can't be many.

We're already quite familiar with the complex number system, which is a vector space defined over the reals with basis $\{1,i\}$, and with multiplication defined by

$$1\times i=i \times 1=i,\quad 1^2=1,\quad i^2=-1$$

We can then define division. What we ultimately find is that, except for 0, every element of this vector space has a unique inverse — this is a necessary condition for being a division ring! There's another vector space very similar to the complex numbers, with basis $\{1,j\}$, but with multiplication defined by $j^2=1$ — yet this one is not a division ring!

Where does it go wrong? Let's consider the inverse of $a+bj$, where $a,b$ are not both zero. Then

$$\frac{1}{a+bj}=\frac{a-bj}{(a+bj)(a-bj)}=\frac{a-bj}{a^2-b^2}$$

See the denominator? For the complex numbers, this denominator is $a^2+b^2$, which is nonzero for any $a,b$ not both zero. But here the denominator is $a^2-b^2$, and even if $a,b$ are both nonzero, it can still equal zero, in which case the inverse is undefined. In other words, some nonzero elements of this ring have no inverse, which is exactly why it fails to be a division ring.

Now let's look at the difference between $a^2+b^2$ and $a^2-b^2$: the former is positive definite, while the latter is not. This positive-definiteness of the quadratic form seems to be exactly what guarantees that "not all entries being zero" implies "the result is not zero" — this seems to be the core distinction between the two cases. If this is indeed the key, then we need to figure out the meaning of $a^2+b^2$ and $a^2-b^2$. Let's look at inversion from another angle: suppose the inverse of $a+bj$ is $c+dj$. Then

$$1=(a+bj)(c+dj)=(ac+bd)+(bc+ad)j$$

That is, $ac+bd=1,bc+ad=0$, which in matrix form reads

$$\left( {\begin{array}{c} a&b\\ b&a \end{array}} \right)\left( {\begin{array}{c} c\\d \end{array}} \right) = \left( {\begin{array}{c} 1\\0 \end{array}} \right)$$

So finding the inverse is equivalent to solving this linear system, but this requires

$$\det\left( {\begin{array}{c} a&b\\ b&a \end{array}} \right)=a^2-b^2\neq 0$$

Now we've found the meaning of $a^2-b^2$: it's precisely the determinant of the coefficient matrix that arises when computing the inverse!

Note that the above is purely exploratory reasoning and speculation — it is not a proof, nor a rigorous derivation. What we observe is that a matrix shows up here, and if the determinant of this matrix is positive definite (of course, negative definite works equally well), then the ring can successfully become a division ring. This matrix is determined by the multiplication table of the basis elements. In general, aside from sign differences, every row of this matrix is a rearrangement of the entries in the first row. (In the most general case, it should be a linear combination.)

Now let's analyze when the determinant can possibly be positive definite — take, for instance, the third-order determinant (which corresponds to the existence of a three-dimensional division algebra):

$$\det\left( {\begin{array}{c} a&b&c\\ d&e&f\\ g&h&i\end{array}}\right)=a e i-a f h-b d i+b f g+c d h-c e g$$

How could we make this positive definite? A natural idea is to flip the sign of the negative terms (there are three of them), meaning we'd change the sign of some of the numbers appearing in those terms. But this necessarily involves at least three numbers, and once we flip the signs of those three numbers, it will inevitably affect the other three terms as well, turning what were originally positive terms into negative ones. This is an irreconcilable contradiction. Therefore, a three-dimensional division algebra over the reals cannot exist. It's worth noting that it took Hamilton ten years of struggle before he came to realize this...

However, in the fourth-order determinant, this turns out to be achievable — otherwise the quaternions wouldn't exist either. For odd-order determinants, a similar contradiction to the third-order case always arises; moreover, even orders that contain an odd factor run into an analogous contradiction. Hence, finite-dimensional division algebras over the reals can only exist in dimension $2^n$.

Had this article been written before 1860, it might have offered a hint to friends studying algebra. Of course, 1860 is now far behind us, and as it happens:

In 1861, Weierstrass proved that a finite-dimensional division algebra over the real or complex field satisfying commutativity of multiplication can only be the algebra of real numbers or complex numbers (published in 1884).
In 1870, Dedekind independently obtained the same result (published in 1888).
In 1878, Frobenius (F. G. Frobenius, 1849–1917) proved that the only finite-dimensional division algebras over the reals are the reals, the complex numbers, and the real quaternions.
In 1881, Charles Sanders Peirce also obtained this result independently. In 1958, using methods from algebraic topology, it was proved that finite-dimensional division algebras over the reals — even including non-associative division algebras — can only have dimensions among the four known values: 1, 2, 4, and 8. This shows that the real and complex fields possess quite special properties.

References

http://zh.wikipedia.org/zh/赋范可除代数

http://baike.baidu.com/view/10903855.htm