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Frobenius theorem (real division algebras)

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In mathematics, more specifically in abstract algebra, the Frobenius theorem, proved by Ferdinand Georg Frobenius in 1877, characterizes the finite dimensional associative division algebras over the real numbers. The theorem proves that the only associative division algebra which is not commutative over the real numbers is the quaternions.

If D is a finite dimensional division algebra over the real numbers R then one of the following cases holds

[edit] Proof

The main ingredients for the following proof are the Cayley-Hamilton Theorem and the Fundamental theorem of algebra.

We can consider D as a finite-dimensional R-vector space. Any element d of D defines an endomorphism of D by left-multiplication and we will identify d with that endomorphism. Therefore we can speak about the trace of d, the characteristic and minimal polynomials. Also, we identify the real multiples of 1 with R. When we write a \le 0 for an element a of D, we tacitly assume that a is contained in R. The key to the argument is the following

Claim: The set V of all elements a of D such that a^2 \le 0 is a vector subspace of D of codimension 1.

To see that, we pick an aD. Let m be the dimension of D as an R-vector space. Let p(x) be the characteristic polynomial of a. By the fundamental theorem of algebra, we can write

 p(x)= (x-t_1)...(x-t_r) (x-z_1)(x - \overline{z_1}) ... (x-z_s)(x - \overline{z_s})

for some real ti and (non-real) complex numbers zj. We have 2s + t = m. The polynomials x^2 - 2\operatorname{Re}(z_i) + |z_i|^2 = (x-z_i)(x-\overline{z_i}) are irreducible over R. By the Cayley-Hamilton Theorem, p(a)=0 and because D is a division algebra, it follows that either ati = 0 for some i or that a^2 - 2\operatorname{Re}z\,a + |z|^2=0, z = zj for some j. The first case implies that a ∈ R. In the second case, it follows that x^2 - 2\operatorname{Re}z\,x + |z|^2 is the minimal polynomial of a. Because p(x) has the same complex roots as the minimal polynomial and because it is real it follows that

 p(x)=(x^2 - 2\operatorname{Re}z\,x + |z|^2)^k

and m=2k. The coefficient of x2k − 1 in p(x) is the trace of a (up to sign). Therefore we read from that equation: the trace of a is zero if and only if \operatorname{Re}z = 0, that is a2 = − | z | 2.

Therefore V is the subset of all a with tr a = 0. In particular, it is a vector subspace (!). Moreover, V has codimension 1 since it is the kernel of a (nonzero) linear form. Also note that D is the direct sum of R and V (as vector spaces). Therefore, V generates D as an algebra.

Define now for a,b \in V:\ B(a,b):= -ab - ba. Because of the identity (a + b)2a2b2 = ab + ba, it follows that B(a,b) is real and since a^2 \le 0, B(a,a)>0 if  a \neq 0. Thus B is a positive definite symmetric bilinear form, in other words, an inner product on V.

Let W be a subspace of V which generated D as an algebra and which is minimal with respect to that property. Let e1,...,en be an orthonormal basis of W. These elements satisfy the following relations:

e_i^2 =-1, e_i e_j = - e_j e_i.

If n=0, then D is isomorphic to R.

If n=1, then D is generated by 1 and e1 subject to the relation e_1^2 = -1. Hence it is isomorphic to C.

If n=2, we have seen above above that D is generated by 1,e1,e2 subject to the relations e_1^2 = e_2^2 =-1,\ e_1 e_2 = - e_1 e_2 and (e1e2)(e1e2) = − 1. These are precisely the relations for H.

If n > 2, the D cannot be a division algebra! Assume that n > 2. Put u: = e1e2en. It is easy to see that u2 = 1 (this only works if n > 2). Therefore 0 = u2 − 1 = (u − 1)(u + 1) implies that u= ±1 (because D is still assumed to be a division algebra). But if u= ±1, then e_n = \mp e_1 e_2 and so e1,e2,...en − 1 generates D. This contradicts the minimality of W.

Remark: The fact that D is generated by e1,...en subject to above relation can be interpreted as the statement that D is the Clifford algebra of Rn. The last step shows that the only real Clifford algebras which are division algebras are Cl0, Cl1 and Cl2.

Remark: As a consequence, the only commutative division algebras are R and C. Also note that H is not a C-algebra. If it were, then the center of H has to contain C, but the center of H is R. Therefore, the only division algebra over C is C itself.

[edit] Pontryagin variant

If D is a connected, locally compact division ring, then either D = R, or D = C, or D = H.

[edit] References

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