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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Aut{Aut}$EDITED Let $G=\SL_{n,{\mathbb{C}}}$, the special linear group over ${\mathbb{C}}$. Let $H\subset G$ be a finite subgroup. Set $X=G/H$ be the corresponding homogeneous space, it is a complex variety. Let $\tau\in \Aut({\mathbb{C}})$ be an automorphism of the field of complex numbers, not necessarily continuous. Consider the conjugate variety $\tau X$.

We have $\tau X=\tau G/\tau H$. We take the standard $\mathbb{Q}$-form $\SL_{n,\mathbb{Q}}$ of $G$, then we identify $\tau G$ with $G$, and we obtain $\tau X= G/\tau(H)$.

Question 1. Is it always true that $\tau X\simeq X$ as homogeneous spaces of $G=\SL_{n,\mathbb{C}}$?

Question 2. Is it always true that $\tau X\simeq X$ as algebraic varieties over ${\mathbb{C}}$?

Question 3. Is it always true that that the $(\tau X)({\mathbb{C}})\simeq X({\mathbb{C}})$ as $C^\infty$-manifolds?

I expect the answer NO to Question 1.
Let us try to construct a counter-example. We wish to construct a finite subgroup $H\subset G$ such that $G/\tau(H)\not\simeq G/H$, i.e., $\tau(H)$ is not conjugate to $H$ in $\SL(n,\mathbb{C})$. We come to the following question:

Question 4. Let $\rho\colon H\to \SL(n,\mathbb{C})$ be an $n$-dimensional complex representation of a finite group $H$. Is it possible that $\tau(\rho(H))$ is not conjugate to $\rho(H)$ in $\SL(n,\mathbb{C})$ for some $\tau\in \Aut({\mathbb{C}})$?

In the last question by $\tau(\rho(H))$ we mean $\{ \tau(\rho(h))\, |\ h\in H\}$, where $\tau(\rho(h))$ is the matrix with matrix elements $\tau(\rho(h)_{i,j})$.

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  • $\begingroup$ Question 4: the two $3$-dimensional irreducible representations of $\mathfrak{A}_5$ are conjugate under $z\mapsto \bar{z}$, but not under $\operatorname{SL}(3,\Bbb{C}) $. $\endgroup$
    – abx
    Jun 4, 2023 at 12:25
  • $\begingroup$ @abx however, the two 3-dimensional irreducible representations of $\mathfrak A_5$ are also conjugate under the outer automorphism of $\mathfrak A_5$ given by conjugation by a transposition. Thus the image of these two representations are conjugate in $SL_3(\mathbb C)$, although the representations are not. $\endgroup$ Jun 4, 2023 at 15:29
  • $\begingroup$ @Joshua Mundinger: I am a bit confused: this transposition lives in $\mathfrak{S}_5$, which is not contained in $\operatorname{SL}(3,\mathbb{C}) $? $\endgroup$
    – abx
    Jun 4, 2023 at 16:05
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    $\begingroup$ @abx if $\rho_1, \rho_2: \mathfrak A_5 \to SL_3(\mathbb C)$ are the two nonisomorphic irreducible representations of degree 3, then $\rho_1$ and $\rho_2 \circ \varphi$ are conjugate for $\varphi$ the outer automorphism of $\mathfrak A_5$ given by conjugation by a transposition (since their characters are equal). The conjugation gives you an element of $SL_3(\mathbb C)$. $\endgroup$ Jun 5, 2023 at 3:26
  • $\begingroup$ In general, irreducible representations $V$ of an alternating group with irrational character come in the form $V \oplus \overline{V} = Res^{S_n}_{A_n} W$ for some irrep $W$ of $S_n$ and $\overline{V}$ the conjugate representation (note that $W$ always has rational character). It follows from this description that irreps of $A_n$ with Galois-conjugate characters are also conjugate under the outer automorphism of $A_n$ given by conjugation by a transposition. Hence, no example as in my answer can come from $A_n$. $\endgroup$ Jun 5, 2023 at 3:31

1 Answer 1

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$\DeclareMathOperator\SL{SL} \DeclareMathOperator\GL{GL} \DeclareMathOperator\PSL{PSL} \DeclareMathOperator\Aut{Aut} \DeclareMathOperator\Out{Out} \DeclareMathOperator\Ad{Ad} $I can only address Question 1/Question 4. Let me replace $\SL_n$ with $\GL_n$; our $H$ will be simple, so it will provide an example for $\SL_n$ also. If $\rho, \psi: H \to \GL_n(\mathbb C)$ are faithful representations with the same image, then $\psi^{-1} \circ \rho$ is an automorphism of $H$. Hence asking whether $(\tau\rho)(H)$ is conjugate to $\rho(H)$ as a set for faithful $\rho$ is asking whether there is an automorphism $\varphi \in \Aut(H)$ such that $\tau \rho$ is isomorphic to $\rho \varphi$ as a complex representation of $H$. This reduces the problem to character theory. Note also that the action of $\Aut(H)$ on characters factors through $\Out(H)$.

An example is given by $H = \PSL_2(\mathbb F_p)$ for $p > 13$. For any nonsquare $a \in \mathbb F_p^\times$, $$\Out(H) = \left\langle \Ad \begin{pmatrix} a & 0 \\ 0 & 1\end{pmatrix}\right\rangle \cong \mathbb Z/2\mathbb Z.$$ See [1] or [2] for this. Let $$T = \left\{ \begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix} \mid x \in \mathbb F_p^\times \right\} / \{\pm 1\}$$ be the diagonal torus in $H$. If $\zeta$ is a $(p-1)/2$th root of unity, let $\alpha: \mathbb F_p^\times \to \mathbb C^\times$ be a multiplicative character sending a primitive element of $\mathbb F_p^\times$ to $\zeta$. Note that $\alpha(-1) = 1$. There is an irreducible representation of degree $p+1$ given by parabolically inducing the character $\begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix} \mapsto \alpha(x)$ of $T$ up to $H$. Its character $\chi$ satisfies $$ \chi\begin{pmatrix} x & 0 \\ 0 & x^{-1}\end{pmatrix} = \alpha(x) + \alpha(x^{-1}).$$ See e.g. [3] for a description of the characters of $\mathrm{PSL}_2(\mathbb F_p)$. If $\varphi$ is the automorphism of conjugation by $\begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}$, then $\varphi$ fixes $T$ elementwise, so $\chi\varphi$ has the same values on $T$ as $\chi$.

Lemma: If $\zeta$ is a primitive $m$th root of unity, then $\zeta^{-1} + \zeta$ is irrational for $m \notin \{1,2,3,4,6\}$.

This lemma also appears in the theory of root systems.

If $\zeta$ is chosen to be a primitive $(p-1)/2$th root of unity, then $\zeta + \zeta^{-1}$ is irrational for $(p-1)/2 \notin \{1,2,3,4,6\}$, i.e. $p = 11$ or $p > 13$. This provides an example of a character $\chi$ and $\tau \in \Aut(\mathbb C)$ such that $\tau \chi \neq \chi \varphi$ for all $\varphi \in \Out(H)$.


Proof of the lemma: We may assume $m > 2$. The degree $[\mathbb Q(\zeta):\mathbb Q]$ is $\phi(m)$. As $\zeta + \zeta^{-1}$ is real, $[\mathbb Q(\zeta): \mathbb Q(\zeta + \zeta^{-1})] > 1$, but $\zeta$ satisfies the quadratic equation $$\zeta^2+1=(\zeta+\zeta^{-1})\zeta $$ over $\mathbb Q(\zeta + \zeta^{-1})$, so $[\mathbb Q(\zeta):\mathbb Q(\zeta + \zeta^{-1})] = 2$. Hence the degree of $\zeta^{-1} +\zeta$ is $\phi(m)/2$. If $m = p_1^{k_1}\cdots p_\ell^{k_\ell}$, then $\phi(m) = p_1^{k_1 - 1}(p_1 - 1)\cdots(p_\ell^{k_\ell})(p_\ell - 1)$. If $\phi(m) = 2$, then primes $p_i$ appearing must have $p_i - 1$ divides $2$, whence $p_i$ can be only $2$ or $3$. Further, only $m=3$, $m=4$, and $m=6$ are possible.


References

[1] Steinberg, R. (1960). Automorphisms of Finite Linear Groups. Canadian Journal of Mathematics, 12, 606-615. doi:10.4153/CJM-1960-054-6

[2] What is the outer automorphism group of $\operatorname{SL}(2,\mathbb{F}_q)$?

[3] Bonnafé, Representations of $SL_2(\mathbb F_q)$, §5.3, or Fulton and Harris, Representation Theory, §5.2

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  • $\begingroup$ For $p=7$ we have $(p-1)/2=3$. Taking $\zeta_3=\exp 2\pi i/3$ we obtain $\zeta_3+\zeta_3^{-1}=-1$, which is rational. For $p=13$ we have $\zeta_6+\zeta_6^{-1}=1$, which is again rational. However, for $p>13$, indeed the number $\zeta+\zeta^{-1}$ is irrational (do you know a reference for that?) $\endgroup$ Jun 8, 2023 at 14:41
  • $\begingroup$ What is a reference for the assertion that the group ${\rm Out}(H)$ is of order 2 whose nontrivial element is induced by ${\rm Inn}\begin{pmatrix} a & 0 \\ 0 & 1\end{pmatrix}$ for a nonsquare element $a\in{\mathbb F}_p^\times$? $\endgroup$ Jun 8, 2023 at 14:49
  • $\begingroup$ What do you mean by "appear as character values for the diagonal torus in ${\rm PSL}_2(\mathbb F_p)$ at certain principal series representations"? As a character value for some element of the diagonal torus? Which element? What principal series representation? Please kindly add references to your answer! $\endgroup$ Jun 8, 2023 at 16:25
  • $\begingroup$ @MikhailBorovoi Dear Profesor Borovoi, I apologize for the incorrect statement about $\zeta + \zeta^{-1}$. I've added some more explanation and references for the characters and automorphisms of $\mathrm{PSL}_2(\mathbb F_p)$. Please let me know if you have more questions or concerns. $\endgroup$ Jun 8, 2023 at 16:48
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    $\begingroup$ Many thanks for your detailed answer, especially for the proof of the lemma. This proof should be in the Internet! I have edited the statement and proof of this lemma:. Namely, I have added $m=6$. See my first comment. $\endgroup$ Jun 8, 2023 at 18:47

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