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An involution semigroup or $\star$-semigroup is a unary semigroup $\langle S,{\cdot}\,,{}^\star\rangle$ that satisfies the equations $$ (x^\star)^\star = x \quad \text{and} \quad (xy)^\star = y^\star x^\star. $$ Common examples include $\langle M_n(\mathbb{R}),{\cdot}\,,{}^T\rangle$ and $\langle M_n(\mathbb{C}),{\cdot}\,,{}^*\rangle$. An inverse $\star$-semigroup is a $\star$-semigroup that satisfies the equations $$xx^\star x = x \quad \text{and} \quad xx^\star yy^\star = yy^\star xx^\star. {\tag1\label{1}}$$

Using standard methods from linear algebra, it is quite easy to show:

  • Every periodic sub-$\star$-semigroup of $\langle M_n(\mathbb{C}),{\cdot}\,,{}^*\rangle$ is an inverse $\star$-semigroup.

In other words, every periodic sub-$\star$-semigroup of $\langle M_n(\mathbb{C}),{\cdot}\,,{}^*\rangle$ satisfies the equations \eqref{1}. Recall that a $\star$-semigroup $\langle S,{}\cdot\,,{}^\star\rangle$ is periodic if $(\forall a \in S)(\exists m \geq 1) \ a^{2m} = a^m$.

A colleague suggested that this may be a special case of some known general result, probably in operator theory. Therefore, I would like to ask if anyone is aware of a reference for the above result.

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    $\begingroup$ $A \approx B$ just means the equation $A = B$. I have replaced every $\approx$ by $=$. $\endgroup$
    – E W H Lee
    Commented Sep 22, 2022 at 16:21
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    $\begingroup$ As the guilty colleague, to not replicate work the point here is that every element of $S$ is a partial isometry due to the periodicity and it is an old result of Halmos and Wallen that the product of two partial isometries $a,b$ on a Hilbert space is another partial isometry iff $a^*abb^*=bb^*aa^*$ and so the question is basically has it been observed before that a periodic $\ast$-semigroup acting on a Hilbert space is automatically a $\ast$-semigroup of partial isometries. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 22, 2022 at 17:54
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    $\begingroup$ @BenjaminSteinberg could you be more precise? Do you mean that $a,aa^*, a^*a$ periodic implies $a$ partial isometry, or something of this flavor? $\endgroup$
    – YCor
    Commented Sep 22, 2022 at 18:03
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    $\begingroup$ @Ycor, I was saying the ordinary algebra is not semisimple and that is why being closed under * is important. That's why I didnt say *-algebra $\endgroup$
    – Benjamin Steinberg
    Commented Sep 22, 2022 at 18:05
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    $\begingroup$ @YemonChoi Thanks; I asked but neither of them was aware of a reference. $\endgroup$
    – E W H Lee
    Commented Oct 3, 2022 at 21:04

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