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A lower bound of the number of conjugacy classes in the automorphism group of a simple Lie algebra $\mathfrak{s}$, of finite dimension over an arbitrary field $\mathbb{F}$, can be the size of the image of the automorphism group by the trace.

I think this map $\operatorname{tr}:\operatorname{Aut}(\mathfrak{s})\longrightarrow\mathbb{F}$ must be surjective. However, I have no clue. (For splitable Lie algebras, there is an explicit way).

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    $\begingroup$ For the real field and a compact Lie group, the trace is bounded, hence not surjective. $\endgroup$
    – YCor
    Commented Apr 22, 2022 at 5:55
  • $\begingroup$ Thanks for the counterexample. Instead, in characateristic zero or for an infinite field, $\operatorname{im}\operatorname{tr}(\operatorname{Aut}(\mathfrak{s}))$ is it infinite? $\endgroup$ Commented Apr 22, 2022 at 22:02
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    $\begingroup$ Over a perfect field, $\mathrm{Aut}(\mathfrak{s})^0$ is Zariski dense in the underlying algebraic group (Rosenlicht), so if the trace had a finite image, it would be constant and also constant over the algebraic closure. But this would contradict the assertion for split $\mathfrak{s}$. $\endgroup$
    – YCor
    Commented Apr 22, 2022 at 23:09
  • $\begingroup$ Dear @YCor, if you repost your comment as an answer for the infinity of the image of the trace, and so for the existence of infinity conjugacy classes in $\operatorname{Aut}(\mathfrak{s})$. I will accept it as an anwer. $\endgroup$ Commented Apr 23, 2022 at 23:26

2 Answers 2

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  1. If $\mathfrak{s}$ is $K$-anisotropic, where $K$ is a real or $p$-adic field (this is equivalent to $\mathfrak{s}$ not containing $\mathfrak{sl}_2$, and also to the corresponding group be compact), then the trace is bounded, hence non-surjective.

  2. Suppose that $\mathrm{Aut}(\mathfrak{s})^0$ is Zariski-dense in the underlying algebraic group (this is automatic if $K$ is perfect, by Rosenlicht's theorem). In this case, by Zariski density, if $\mathrm{Aut}(\mathfrak{s})^0$ has infinitely many traces (resp. characteristic polynomials), then the same holds over the algebraic closure. So we can reduce (in this case) to the algebraically closed case.

  3. In the algebraically closed case, if the number of characteristic polynomials of $\mathrm{Aut}(\mathfrak{s})^0$ is finite, it is (by connectedness) reduced to one, and hence it follows that every element in $\mathrm{Aut}(\mathfrak{s})^0$ is unipotent. In characteristic zero, this is not possible. I'm not sure about the modular case.

The same conclusion holds for the number of traces, although the characteristic zero is then used in a stronger way.

  1. If $\mathfrak{s}$ is the Lie algebra of a simple algebraic group, the restrictions on the characteristic should be dropped (still needing Rosenlicht), to infer that the number of characteristic polynomials is infinite.
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  • $\begingroup$ Unipotent elements have the same trace as the identity. $\endgroup$ Commented Apr 24, 2022 at 19:00
  • $\begingroup$ So you accepted an incorrect answer :) anyway I'll fix it in some way. $\endgroup$
    – YCor
    Commented Apr 25, 2022 at 7:47
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    $\begingroup$ I accepted it before I recognize the minor problem with unipotents. I was trying to fix it and I lefted accepted because it holds the key argument. $\endgroup$ Commented Apr 25, 2022 at 15:48
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If $\mathfrak{s}$ is a simple Lie algebra over an algebraically closed field of characteristic zero or a classical simple Lie algebra over an algebraically closed field of characteristic $p>3$. Then the trace over the automorphism group $\operatorname{Aut}(\mathfrak{s})$ is surjective.

Let $\mathfrak{h}$ be a Cartan subalgebra and $\Phi$ its root system, we have the decomposition $$\mathfrak{s}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Phi}\mathfrak{s}_\alpha$$ where $\mathfrak{s}_\alpha=\{x\in\mathfrak{s}\vert~ [h,x]=\alpha(h)x\text{ for all }h\in\mathfrak{h}\}$ is of dimension $1$. For a simple root $\alpha_i\in\Delta$, we set $\Lambda_{i}=\{\beta=\sum_{j=1}^l m_j(\beta){\alpha_j}\in\Phi\vert~m_i(\beta)\neq0\}$ and $n_i$ its size. Let $c_i$ be a nonzero scalar, with respect to the Chevalley basis $\{x_\alpha,h_i\vert~\alpha\in\Phi\text{ and }1\leq i\leq\dim\mathfrak{h}\}$, we define the automorphism $\varphi_i$ as follows, \begin{align*} \left\{\begin{array}{ll} \varphi_i(h)=h,&\text{for all } h\in\mathfrak{h},\\ \varphi_i(x_\beta)=c_i^{m_i(\beta)} x_\beta,&\text{for all } \beta=\sum_{j=1}^lm_j(\beta)x_{\alpha_j}. \end{array} \right. \end{align*} Hence \begin{align*} \operatorname{tr}(\varphi_i)&=\dim\mathfrak{h}+\sum_{\beta\in\Delta}c_i^{m_i(\beta)}\\ &=\dim\mathfrak{h}+\vert\Delta\vert-n_i+\sum_{\beta\in\Lambda_i}c_i^{m_i(\beta)}\\ &=\dim\mathfrak{s}-n_i+\sum_{\beta\in\Lambda_i}c_i^{m_i(\beta)} \end{align*} Let $\Delta=\{\alpha_1,\cdots,\alpha_l\}$ be a basis of $\Phi$, with the ordering follows Dynkin diagrams in Bourbaki. From Hasse diagrams of poset of positive root systems and by induction, one obtains

  • $\mathrm{A}_l~(l\geq1)$ : for all $1\leq i\leq l$, $$\operatorname{tr}(\varphi_i)=\dim\mathfrak{s}+i\left(l+1-i\right)\left(c_i+c_i^{-1}-2\right).$$

  • $\mathrm{B}_l~(l\geq2)$ : for all $1\leq i\leq l$, \begin{align*} \operatorname{tr}(\varphi_i)=\dim\mathfrak{s}-i\left(4l+1-3i\right)+i\left(2(l-i)+1\right)\left(c_i+c_i^{-1}\right)+\dfrac{(i-1)i}{2}\left(c_i^2+c_i^{-2}\right). \end{align*}

  • $\mathrm{C}_l~(l\geq3)$ : for all $1\leq i\leq l-1$, \begin{align*} \left\{ \begin{array}{ll} \operatorname{tr}(\varphi_i)=\dim\mathfrak{s}-i\left(4l+1-3i\right)+2i\left(l-i\right)\left(c_i+c_i^{-1}\right)+\dfrac{i(i+1)}{2}\left(c_i^2+c_i^{-2}\right),\\ \operatorname{tr}(\varphi_l)=\dim\mathfrak{s}+\dfrac{l(l+1)}{2}\left(c_l+c_l^{-1}-2\right). \end{array} \right. \end{align*}

  • $\mathrm{D}_l~(l\geq3)$ : for all $1\leq i\leq l-2$, \begin{align*} \left\{ \begin{array}{ll} \operatorname{tr}(\varphi_i)=\dim\mathfrak{s}-i\left(4l-1-3i\right)+2i(l-i)\left(c_i+c_i^{-1}\right)+\dfrac{(i-1)i}{2}\left(c_i^2+c_i^{-2}\right),\\ \operatorname{tr}(\varphi_j)=\dim\mathfrak{s}+\dfrac{(l-1)l}{2}\left(c_j+c_j^{-1}-2\right),~\text{ for }j=l-1\text{ or }l. \end{array} \right. \end{align*}

  • $\mathrm{E}_6$ :
    \begin{align*} \left\{ \begin{array}{ll} \operatorname{tr}(\varphi_i)=46+16\left(c_i+c_i^{-1}\right),\text{ for }i=1\text{ or }6,\\ \operatorname{tr}(\varphi_i)=28+20\left(c_i+c_i^{-1}\right)+5\left(c_i^2+c_i^{-2}\right),\text{ for }i=3\text{ or }5,\\ \operatorname{tr}(\varphi_2)=36+20\left(c_2+c_2^{-1}\right)+\left(c_2^2+c_2^{-2}\right),\\ \operatorname{tr}(\varphi_4)=20+18\left(c_4+c_4^{-1}\right)+9\left(c_4^2+c_4^{-2}\right)+2\left(c_4^3+c_4^{-3}\right). \end{array} \right. \end{align*}

  • $\mathrm{E}_7$ :
    \begin{align*} \left\{ \begin{array}{ll} \operatorname{tr}(\varphi_1)=67+32\left(c_1+c_1^{-1}\right)+\left(c_1^2+c_1^{-2}\right),\\ \operatorname{tr}(\varphi_2)=49+35\left(c_2+c_2^{-1}\right)+7\left(c_2^2+c_2^{-2}\right),\\ \operatorname{tr}(\varphi_3)=39+30\left(c_3+c_3^{-1}\right)+15\left(c_3^2+c_3^{-2}\right)+2\left(c_3^3+c_3^{-3}\right)\\ \operatorname{tr}(\varphi_4)=27+24\left(c_4+c_4^{-1}\right)+18\left(c_4^2+c_4^{-2}\right)+8\left(c_4^3+c_4^{-3}\right)+3\left(c_4^4+c_4^{-4}\right),\\ \operatorname{tr}(\varphi_5)=33+30\left(c_5+c_5^{-1}\right)+15\left(c_5^2+c_5^{-2}\right)+5\left(c_5^3+c_5^{-3}\right),\\ \operatorname{tr}(\varphi_6)=49+32\left(c_6+c_6^{-1}\right)+10\left(c_6^2+c_6^{-2}\right),\\ \operatorname{tr}(\varphi_7)=79+27\left(c_7+c_7^{-1}\right). \end{array} \right. \end{align*}

  • $\mathrm{E}_8$ :
    \begin{align*} \left\{ \begin{array}{ll} \operatorname{tr}(\varphi_1)=&92+64\left(c_1+c_1^{-1}\right)+14\left(c_1^2+c_1^{-2}\right),\\ \operatorname{tr}(\varphi_2)=&64+56\left(c_2+c_2^{-1}\right)+28\left(c_2^2+c_2^{-2}\right)+8\left(c_2^3+c_2^{-3}\right),\\ \operatorname{tr}(\varphi_3)=&52+42\left(c_3+c_3^{-1}\right)+35\left(c_3^2+c_3^{-2}\right)+14\left(c_3^3+c_3^{-3}\right)+7\left(c_3^4+c_3^{-4}\right),\\ \operatorname{tr}(\varphi_4)=&36+30\left(c_4+c_4^{-1}\right)+30\left(c_4^2+c_4^{-2}\right)+20\left(c_4^3+c_4^{-3}\right)\\ &\hspace{.88cm}+15\left(c_4^4+c_4^{-4}\right)+6\left(c_4^5+c_4^{-5}\right)+5\left(c_4^6+c_4^{-6}\right),\\ \operatorname{tr}(\varphi_5)=&40+40\left(c_5+c_5^{-1}\right)+30\left(c_5^2+c_5^{-2}\right)+20\left(c_5^4+c_5^{-3}\right)+10\left(c_5^4+c_5^{-4}\right)+4\left(c_5^5+c_5^{-5}\right),\\ \operatorname{tr}(\varphi_6)=&54+48\left(c_6+c_6^{-1}\right)+30\left(c_6^2+c_6^{-2}\right)+16\left(c_6^3+c_6^{-3}\right)+3\left(c_6^4+c_6^{-4}\right),\\ \operatorname{tr}(\varphi_7)=&82+54\left(c_7+c_7^{-1}\right)+27\left(c_7^2+c_7^{-2}\right)+2\left(c_7^3+c_7^{-3}\right),\\ \operatorname{tr}(\varphi_8)=&134+56\left(c_8+c_8^{-1}\right)+\left(c_8^2+c_8^{-2}\right). \end{array} \right. \end{align*}

  • $\mathrm{F}_4$ :
    \begin{align*} \left\{ \begin{array}{ll} \operatorname{tr}(\varphi_1)=22+14\left(c_1+c_1^{-1}\right)+\left(c_1^2+c_1^{-2}\right),\\ \operatorname{tr}(\varphi_2)=12+12\left(c_2+c_2^{-1}\right)+6\left(c_2^2+c_2^{-2}\right)+2\left(c_2^3+c_2^{-3}\right),\\ \operatorname{tr}(\varphi_3)=12+6\left(c_3+c_3^{-1}\right)+9\left(c_3^2+c_3^{-2}\right)+2\left(c_3^3+c_3^{-3}\right)+3\left(c_3^4+c_3^{-4}\right),\\ \operatorname{tr}(\varphi_4)=22+8\left(c_4+c_4^{-1}\right)+7\left(c_4^2+c_4^{-2}\right). \end{array} \right. \end{align*}

  • $\mathrm{G}_2$ :
    \begin{align*} \left\{ \begin{array}{ll} \operatorname{tr}(\varphi_1)=4+2\left(c_1+c_1^{-1}\right)+\left(c_1^2+c_1^{-2}\right)+2\left(c_1^3+c_1^{-3}\right),\\ \operatorname{tr}(\varphi_2)=4+4\left(c_2+c_2^{-1}\right)+\left(c_2^2+c_2^{-2}\right). \end{array} \right. \end{align*}

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