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I am reading a paper that is a little too advanced for me, and it has the following statement:

$$ \text{There is a division ring } K \text{ with involutional automorphism } *: K \rightarrow K$$

I am struggling to find a proper definition of "involutional automorphism" so any help would be really appreciated.

Juan Esteban Arevalo Gomez
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    It might mean $(*)^2 = I$, the identity automorphism. – Robert Lewis Oct 30 '20 at 23:05
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    It is very slightly ambiguous; in some contexts "involutional automorphism" might actually mean an involutional anti-automorphism (reverses the order of multiplication: $(ab)^{\ast} = b^{\ast} a^{\ast}$). For example conjugation on the quaternions has this property. – Qiaochu Yuan Oct 30 '20 at 23:10

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An automorphism is an isomorphism from $K$ to itself. That is, a bijection from $K$ to itself which preserves all the relevant algebraic structure. You might imagine an invertible matrix viewed as a function from a vector space to itself.

An involution is a function which is its own inverse. You might imagine a reflection, or a rotation by $180^\circ$.

So an involutional automorphism is an isomorphism $*$ from $K$ to itself such that $**x = x$.

As a concrete example, consider $\mathbb{C}$ with the complex conjugation operation $z \mapsto \overline{z}$. I will leave it to you to check that this is

  1. A nontrivial field isomorphism from $\mathbb{C}$ to itself (it preserves all the field structure)
  2. Involutional, in the sense that $\overline{\overline{z}} = z$.

As another example, consider the matrix

$$A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

which we view as a function from $\mathbb{R}^2$ to itself. The fact that this is invertible and linear means exactly that it is an automorphism of $\mathbb{R}^2$, and the fact that it is its own inverse ($A^2 = I$) means it is an involution.

I hope this helps ^_^

HallaSurvivor
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