Abstract functional equation

by math_explorer, Jan 26, 2013, 2:05 AM

An anti-homomorphism from a ring $R$ to a ring $R'$ is a homomorphism $\eta$ (what a great letter) between the additive groups that also satisfies $\eta(1) = 1$ and $\eta(ab) = \eta(b)\eta(a)$ for any $a, b \in R$ .

A Jordan homomorphism from a ring $R$ to a ring $R'$ is a homomorphism $\eta$ between the additive groups that also satisfies $\eta(1) = 1$ and $\eta(aba) = \eta(a)\eta(b)\eta(a)$ for any $a, b \in R$ .

Why do people have so many things named after them!?

If $R'$ is a domain, then the Jordan homomorphism is either a homomorphism or an anti-homomorphism.
This is so literally a functional equation...

Proof.
Let $P(a, b)$ denote the statement $\eta(aba) = \eta(a)\eta(b)\eta(a)$ .

$P(a, 1) \Longrightarrow \eta(a^2) = \eta(a)^2 \qquad \forall a \in R$
$P(a, a) \Longrightarrow \eta(a^3) = \eta(a)^3 \qquad \forall a \in R$

Hmm qq is a substring of \qquad. /random

Subtracting $P(a, b)$ and $P(c, b)$ from $P(a+c, b)$ yields

$\eta(abc + cba) = \eta(a)\eta(b)\eta(c) + \eta(c)\eta(b)\eta(a)$

(since $\eta$ is an additive group homomorphism)
Let this statement be $Q(a, b, c)$ .

So, for any $x, y \in R$ we have

$\begin{align*}&(\eta(xy) - \eta(x)\eta(y))(\eta(xy) - \eta(y)\eta(x)) \\ =& \eta(xy)^2 - \eta(xy)\eta(y)\eta(x) - \eta(x)\eta(y)\eta(xy) + \eta(x)\eta(y)\eta(y)\eta(x) \\ =& \eta(xyxy) - \eta(xy^2x + xyxy) + \eta(x)\eta(y^2)\eta(x) \\ =& \eta(xy^2x) - \eta(xy^2x) \\ =& 0 \end{align*}$

using $Q(xy, y, x)$ and $\eta(y)^2 = \eta(y^2)$ and of course the additive-group-homomorphism properties.
(darn I'm not sure what the best way to typeset this without overflowing width is)

So (this is the only place where we use the condition that $R'$ is a domain) one of the factors in the original expression is 0, i.e. either $\eta(xy) = \eta(x)\eta(y)$ or $\eta(xy) = \eta(y)\eta(x)$ for any $x, y \in R$ (*).

In fact, the single condition (*) plus the additive-group-homomorphism properties are sufficient to prove that $\eta$ is either a homomorphism or an anti-homomorphism.

To prove this, we need to show it works like a homomorphism on all pairs $xy$ with $\eta(xy) = \eta(x)\eta(y)$ , or like an anti-homomorphism on all pairs $xy$ with $\eta(xy) = \eta(y)\eta(x)$ . This is somewhat an abuse of English but I think it gets the point across better than equation after equation. We provide two ways to finish.

The Straightforward Bash

Claim. For any $a, b, c$ , either $\eta$ acts as a homomorphism or as an anti-homomorphism on both pairs $ab$ and $ac$ . Similar remarks hold for $ba$ and $ca$ .

Proof of claim. If this is not true, then the only other possibility for (*) is that $\eta$ acts as a homomorphism on one pair and as an anti-homomorphism on the other.

Suppose (WLOG) $\eta(ab) = \eta(a)\eta(b)$ and $\eta(ac) = \eta(c)\eta(a)$ . Consider $\eta(a(b+c)) = \eta(ab) + \eta(ac)$ .

If $\eta(a(b+c)) = \eta(a)\eta(b+c) = \eta(a)\eta(b) + \eta(a)\eta(c)$ then $\eta(ac) = \eta(a)\eta(c)$ too.
Otherwise if $\eta(a(b+c)) = \eta(b+c)\eta(a) = \eta(b)\eta(a) + \eta(c)\eta(a)$ then $\eta(ab) = \eta(b)\eta(a)$ too.

Both contradict our initial assumption. Thus $\eta$ acts as a homomorphism on both or as an anti-homomorphism on both.
-- end claim --

Now: suppose $\eta$ works on some pair as only a homomorphism and on some other pair as only an anti-homomorphism, that is: $\eta(ab) = \eta(a)\eta(b) \neq \eta(b)\eta(a)$ while $\eta(cd) = \eta(d)\eta(c) \neq \eta(c)\eta(d)$ .
From what we just derived, this forces $\eta$ to work as both a homomorphism and an anti-homomorphism on the pairs $ad$ and $bc$ . Thus:

$\begin{align*} \eta(ad) &= \eta(a)\eta(d) = \eta(d)\eta(a) \\ \eta(cb) &= \eta(c)\eta(b) = \eta(b)\eta(c) \\ \eta(a(b+d)) &= \eta(a)(\eta(b) + \eta(d)) = \eta(a)\eta(b+d) \\ &\neq (\eta(b) + \eta(d))\eta(a) = \eta(b+d)\eta(a) \\ \eta(c(b+d)) &\neq \eta(c)(\eta(b) + \eta(d)) = \eta(c)\eta(b+d) \\ &= (\eta(b)+\eta(d))\eta(c) = \eta(b+d)\eta(c) \end{align*}$

i.e. $\eta$ works on $a(b+d)$ as only a homomorphism, and on $c(b+d)$ as only an anti-homomorphism, impossible.

Thus $\eta$ is fully a homomorphism or fully an anti-homomorphism (it may be partially both, on elements that commute). Q.E.D.

Slick Group Theory!

Observe:
1. $\eta(a)\eta(1) = \eta(a)1' = \eta(a1)$
2. $\eta(a)\eta(b) = \eta(ab)$ and $\eta(a)\eta(c) = \eta(ac)$ together imply $\eta(a)\eta(b + c) = \eta(a(b+c))$ (additive homomorphism properties + distributivity)
3. $\eta(a)\eta(b) = \eta(ab)$ implies $\eta(a)\eta(-b) = \eta(a(-b))$ since obviously $-1$ commutes with everybody multiplicatively just like $1$

(Firefox, how can you not think "multiplicatively" is a word?!)

Therefore, for any $a$ , the set of all $b$ such that $\eta(a)\eta(b) = \eta(ab)$ is a subgroup of the additive group of $R$ .

Similar remarks hold if you fix $b$ and collect all $a$ , or if you look at pairs with $\eta(b)\eta(a) = \eta(ab)$ , or both.

Some quick group-theory lemmata:

Lemma 1. The intersection of any number of subgroups of a group is a subgroup. Proof. Just verify all the axioms trivially.

Lemma 2. If the union of two subgroups $H$ and $L$ of a group $G$ is $G$ itself then one of the subgroups is improper. Proof: If $H$ is included in $L$ or vice versa, the result is clear; otherwise, take $h \in H \setminus L$ and $\ell \in L \setminus H$ and observe that $h\ell$ is in neither $H$ nor $L$ , a contradiction.

Therefore:
Fix $a$ and collect $B_1 := \{ b \in R \mid \eta(ab) = \eta(a)\eta(b) \}$ and $B_2 := \{ b \in R \mid \eta(ab) = \eta(b)\eta(a) \}$ .

From (*), $B_1 \cup B_2 = R$ ; also, as we discussed, $B_1$ and $B_2$ are subgroups of the additive group of $R$ . Then by Lemma 2, either $B_1 = R$ or $B_2 = R$ (**).

Now collect $A_1 := \{ a \in R \mid \eta(ab) = \eta(a)\eta(b) \; \forall b \in R \}$ and $A_2 := \{ a \in R \mid \eta(ab) = \eta(b)\eta(a) \; \forall b \in R \}$ .

Note that $A_1 = \bigcap_{b \in R} \{ a \in R \mid \eta(ab) = \eta(a)\eta(b) \}$ , which is a subgroup of $R$ from our initial discussion and Lemma 1. Similarly, $A_2$ is a subgroup of $R$ too. Then (**) says that for any $a \in R$ , either $a \in A_1$ or $a \in A_2$ , i.e. $A_1 \cup A_2 = R$ . By Lemma 2 either $A_1 = R$ , whence $\eta$ is a homomorphism, or $A_2 = R$ , whence $\eta$ is an anti-homomorphism. Q.E.D.

This post has been edited 1 time. Last edited by math_explorer, Mar 26, 2015, 5:36 AM
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abstract algebra

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Abstract functional equation

by math_explorer, Jan 26, 2013, 2:05 AM

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