There is no unique complex conjugation in C
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Complex numbers can be defined as ordered pairs (x,y) of real numbers x,y in R
with coordinatewise addition and with the product
(x1,y1)(x2,y2) = (x1x2-y1y2,x1y2+y1x2).
By construction, C has a specified subfield isomorphic to the real field R,
namely {(x,0)|x in R}, and a specified automorphism of degree 2, namely the
complex conjugation. The field injection R->C brings extra structure to
the field C, making it an algebra over R.
Considered as a field, with no specified subfield R, C has infinitely many
automorphisms of degree 2. Picking up one fixes a subfield isomorphic to R.
Each such subfield R induces on C a topology, where all the other copies of R
are dense. In other words, there is no distinguished topology on the field C,
considered without a field injection R->C.
In contrast, the real field R has only one automorphism, the identity.
The field C is not isomorphic to the R-algebra C
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There are two distinct objects "C considered as an algebra over R" and
"C without a specified subfield R", which belong to different categories.
In the former category, an automorphism fixes the specified subfield R, while
in the latter category an automorphism permutes an infinite number of
isomorphic copies of R. Thus, in the sense of category theory, the "field C"
is not isomorphic to the "field C with a specified subfield R". The latter
has more structure than just the field structure.
Is there an infinity of automorphisms of C of order 2 swapping i and -i?
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Yes, assuming the standard set theory including the Axiom of Choice.
There are plenty of order 2 automorphisms of C. Let t be a non-real
transcendental number, and take K to be a maximal real-closed subfield
of C containing Q(t) (use Zorn's Lemma). Then C has degree 2 over K,
and there is a corresponding automorphism of C with order 2.
As K <> R this ain't standard complex conjugation.
Do all automorphisms of C, which interchange i and -i, have order 2?
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No, not all automorphisms interchanging i and -i have order 2.
Let eta be a primitive 16-th root of unity. The automorphism
tau of the cyclotomic field Q(eta) taking eta to eta^3 has order 4,
but swaps i and -i. tau can be lifted to an automorphism of the
field of algebraic numbers, and then to an automorphism of C.
It swaps i and -i yet has order at least 4. (Actually it has infinite
order.) Hence given any automorphism of C of order 2 swapping
i and -i there are automorphisms of C fixing i, which are not
products of this automorphism and another of this type.
Are all automorphisms of C, which have finite order, of order 2 (or 1)?
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Yes. If tau is an automorphism of C of finite order m, then C has degree
m over the fixed field K of tau. So m = 1 or 2. In the latter case tau must
swap i and -i.
Is Q the largest subfield of C that is fixed by all automorphisms of C?
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No. Each automorphism of C preserves the cyclotomic field K = Q(alpha)
where alpha = (1+i)/sqrt(2) is a primitive 8-th root of unity.
The automorphism group of K is elementary abelian of order 4, so any
automorphism has order 1 or 2, and its fixed field on K is either K
itself, or a quadratic field. Equivalently, each automorphism of C fixes
one of the following fields pointwise: Q(i), Q(sqrt(2)), Q(sqrt(-2)).
Do all automorphisms of C send the unit circle {z in C | |z|=1} to itself?
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No. Given two algebraically independent transcendentals, there is an
automorphism of C swapping them. Just choose one on the unit circle
and one off it. (NB again such arguments use the Axiom of Choice.)
The automorphism group of C with a specified subfield R is of order two,
namely O(1). When characterized as the algebraically closed field of
characteristic zero having transcendence degree 2^aleph_0 over Q, C has a
very big automorphism group, at least of size 2^(2^aleph_0).
Quaternions
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Quaternions can be defined as ordered pairs (z,w) of complex numbers z,w in C,
but this time the product involves complex conjugation
(z1,w1)(z2,w2) = (z1z2-w1w2*,z1w2+w1z2*).
Alternatively, quaternions are quadruplets (w,x,y,z) of real numbers in R^4
with (a distributive) multiplication being defined for
1 = (1,0,0,0), i = (0,1,0,0), j = (0,0,1,0), k = (0,0,0,1)
by
i^2 = j^2 = k^2 = -1, ij = -ji = k, ki = -ik = j, jk = -kj = i.
The center of the division ring of quaternions H is R. Therefore, there is a
distinguished subring R in H, which makes the ring H an algebra over R.
There is a distinguished anti-involution in the ring H, namely the quaternion
conjugation.
The group of automorphisms of H is isomorphic to SO(3), the group of rotations
of R^3. The group of automorphisms and antiautomorphisms of H is O(3).
Octonions
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Octonions can be defined as pairs (p,q) of quaternions p,q in H, but this time
order of multiplication matters
(p1,q1)(p2,q2) = (p1p2-q2*q1,q2p1+q1p2*) or
(p1,q1)(p2,q2) = (p1p2-q2q1*,p1*q2+p2q1).
In a division algebra D the equations ax = b and ya = b have unique solution
x,y for all non-zero a in D.
The group of automorphisms of O is the exceptional Lie group G_2.
Real division algebras
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The only associative real division algebras are R, C and H.
The only alternative real division algebras are R, C, H and O.
The dimension of a real norm-preserving division algebra is 1, 2, 4 or 8.
The only norm-preserving real division algebras with unity are R, C, H and O.
Does a real norm-preserving division algebra have 1?
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No. Take C with product (z,w) -> zw*.
Is the dimension of a division algebra 1, 2, 4 or 8?
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No. Consider a 3D algebra over Q with basis (1,i,i^2), unity 1 and multiplication
rules i(i^2) = (i^2)i = 3, (i^2)^2 = -6i. Multiplication by x+iy+(i^2)z has
determinant x^3+3y^3-18z^3, which has no rational roots. Thus, we have a division
algebra, 3D over Q.
If each non-zero element of an algebra A has an inverse, is A a division algebra?
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No. Consider a 3D algebra over R with basis (1,i,j), unity 1 and multiplication
rules i^2 = j^2 = -1, ij = ji = 0. Each non-zero element x+iy+jz has an inverse
(x-iy-jz)/(x^2+y^2+z^2), but the algebra has zero-divisors, by definition.
Is an algebra without zero-divisors a division algebra?
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No. Only in finite dimensions an algebra without zero-divisors is a division
algebra. For instance, the tensor algebra and the polynomial algebra do not
have zero-divisors although they are not division algebras, since elements of
positive degree do not have an inverse.