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$\DeclareMathOperator\GL{GL}$Let $ H $ be an irreducible finite subgroup of $ \GL(n,\mathbb{C}) $. Define $ N^r(H) $ inductively by $$ N^{r+1}(H)=\{ g \in \GL(n,\mathbb{C}): g H g^{-1} \subset N^r(H) \} $$ starting from $ N^1(H):= H $.

Note that, since $ H $ is finite, the "semi-normalizer" $ N^2(H) $ is just the normalizer of $ H $ (see comment from LSpice). And thus is closed under inverses. But for $ r \geq 3 $ the set $ N^r(H) $ is generally not a group.

My question: Is $ N^r(H) $ closed under inverses?

The answer appears to be yes for $ n=2 $ (and is yes trivially for $ n=1 $ since then $ N^r(H)=\GL(1,\mathbb{C}) $ for all $ r \geq 2 $).

Context:

This construction is used for a naturally occurring object called the Clifford hierarchy which appears in quantum computing. For example, the (single qubit) Clifford hierarchy is the case where $ n=2 $ and $ H= \left\langle \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \right\rangle $ is the single qubit Pauli group.

For $ H $ infinite there are many counterexamples. There is a counterexample with reducible $ H $ due to Arturo Magidin in this MathSE question with $ H = \left\{\left.\left(\begin{array}{cc} 1 & m\\ 0 & 1\end{array}\right)\in G\ \right|\ m\in\mathbb{Z}\right\} $ where $ x = \left(\begin{array}{cc} 2 & 0\\ 0 & 1\end{array}\right) $ is in $ N^2(H) $ but $ x^{-1} $ is not.

And there is a counterexample for infinite $ H $ with $ H $ irreducible given by Moishe Kohen in the answer to this MathSE question

More generally we can consider representations over fields other than $ \mathbb{C} $.

Some thoughts: If $ H $ is projectively self-normalizing (meaning the normalizer is just $ H $ with scalar matrices) then clearly the hierarchy $ N^r(H) $ is closed under inverses because $ N^r(H)=<H,\mathbb{C}^\times I > $ for all $ r \geq 2 $. This applies to all projectively self-normalizing subgroups $ H $ of $ GL(n,\mathbb{C}) $, including all $ H $ such that $ H/Z(H) $ is a maximal finite subgroup of $ PGL(n,\mathbb{C}) $ (since irreducible finite subgroups always have a projectively finite normalizer). For example all subgroups for $ n=1 $. And the subgroups $ H=2.A_5 $ and $ H=2.S_4 $ for $ n=2 $.

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    $\begingroup$ I think the norm (as opposed to the normaliser 😄) is that it is better to avoid editorialising, like "interesting", "difficult", etc., in the title (and preferably in the body), so I edited out "interesting". $\endgroup$
    – LSpice
    Nov 30 at 22:59
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    $\begingroup$ @LSpice +1 good point, you are kind and wise as always! $\endgroup$
    – Ian Gershon Teixeira
    Dec 1 at 0:47
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    $\begingroup$ @LSpice I really do mean $ N^2(H) $ fails to be closed under inverses! The subtle thing here is that there is a containment not an equality in the condition $ gHg^{-1} \subset N^r(H) $ so if $ H $ is infinite then even for $ r=1 $ the containment $ gHg^{-1} \subset H=N^1(H) $ can be proper! So although when $ H $ is finite then $ N^2(H) $ is the normalizer, when $ H $ is infinite then $ N^2(H) $ is just some set that (properly) contains the normalizer. $\endgroup$
    – Ian Gershon Teixeira
    Dec 4 at 22:43
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    $\begingroup$ Re, ah, I see. I was thinking, as in @YCor's now-deleted answer, of the algebraic case, where one can show with finite invariants (dimension and number of components) that what I guess should be called the "semi-normalizer" $N^2(H)$ is actually the normalizer; but now I take your point. For example, one could take $\tilde H$ to be the integer-indexed sequences of reals and $H$ to be the subgroup of $\tilde H$ that vanishes on negative integers, and put $G = \langle\sigma\rangle \ltimes \tilde H$, where $\sigma$ is the right shift. $\endgroup$
    – LSpice
    Dec 4 at 22:48
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    $\begingroup$ Can we understand a simple case? E.g. take $H= S_4$ acting on $\mathbb{R}^3 \simeq \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4: x_1+x_2+x_3+x_4 = 0\}$ by permuting variables (how's that called? Standard?). I was wondering if one can 'classify' sets $X$ that satisfy $g H g^{-1} \subset X$ iff $g^{-1} H g \subset X$ $\endgroup$
    – Andrea Marino
    Dec 5 at 8:49

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