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A ring $R$ is said to be a ring with an involution if there exists a mapping $*\colon R \to R$ such that for every $a, b \in R$:

  • $a^{**} = a$,
  • $(a + b)^* = b^* + a^*$,
  • $(ab)^* = b^*a^*$.

Can anyone please explain this definition with an example?

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3 Answers 3

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An involution is a so-called anti-homomorphism (that's what that last condition expresses) that, if applied twice, gives the identity map. Note that your second condition is the same as $(a+b)^* = a^*+b^*$ – addition is commutative!

As a typical example, let $n$ be a natural number, let $A$ be any ring, and consider the ring of $n$-by-$n$ matrices over $A$. The transpose is an involution on that ring. Another classical example: on the ring of Hamilton quaternions, the map $a+bi+cj+dk \mapsto a-bi-cj-dk$ is an involution.

If the ring is commutative, then an involution is just a ring automorphism (an invertible homomorphism from the ring to itself) of order 2.

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An involution is a map which is it's own inverse. For example consider the complex number with the complex conjugation as involution: It's easy to check that complex conjugation satisfies the properties of involution.

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  • $\begingroup$ So this is what I have understood that if we consider an identity map then $3* = 3$. What if we consider $R = Z_6$, what will be the matrix $A*$ if matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$? $\endgroup$
    – Amanda
    Commented Jun 7, 2018 at 10:36
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    $\begingroup$ If by "map" you mean a ring homomorphism, then that's not the usual definition of involution. Unless the ring is commutative, an involution is not a ring homomorphism. $\endgroup$
    – Alex B.
    Commented Jun 7, 2018 at 10:37
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An easy example is $(\mathbb C, *)$ where $*$ is the conjugate operator.

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  • $\begingroup$ Amanda, I know it has been a long time since you asked this question. I was wondering I know a reference about involution for a ring. Thank you in advance $\endgroup$
    – 00GB
    Commented Jul 15, 2022 at 21:16

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