# equations over (some) lie groups

To be concrete, let $G=SL(n, \mathbb{C}),$ $\phi$ an automorphism of $G.$ Is there a characterization of those $x$ for which there exists a $y$ such that $x = y \phi(y)?$ In the special case, the automorphism group of $G$ is generated by inner automorphisms and $\psi(x) = (x^{-1})^t.$ For inner $\phi,$ we are asking which $x$ satisfy the quadratic equation $x = y a y a^{-1}$ for a fixed a (if $a$ is allowed to vary, it is well-known that every element of a complex semi-simple lie group is a commutator, so that should presumably imply that every element has that form for some $a$ ). For $\psi,$ we want to characterize matrices of the form $y = x (x^{-1})^t$ (by dimension counting this is a proper subset; it is pretty clear that it contains the complex orthogonal group).

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Why is the subset of matrices $y=x(x^{-1})^t$ proper? –  Mark Sapir May 6 '12 at 12:08
It is proper for $n=2$, but not because of the dimension. The set has full dimension 3. It is just the whole $SL(2,\mathbb{C})$ without some subvariety of co-dimension 1. –  Mark Sapir May 6 '12 at 12:22
Well, I was doing this in my head, and observed that at identity all of the symmetric directions are annihilated by the jacobian of $x \rightarrow x \psi(x).$ At a general point, the annihilation condition is that $v^t = x v x^t,$ and it is not clear that this has any nontrivial solutions (for traceless $v$), so this is not clear... –  Igor Rivin May 6 '12 at 13:15
I do not understand the comment. The case $n=2$ is easy, and can be treated using any CAS. –  Mark Sapir May 6 '12 at 13:22
@Igor: To avoid confusion, the definition of $\psi$ in the third lilne of the question should be corrected by removing the factor $x$ on the left. –  Jim Humphreys May 6 '12 at 16:47

For every $x$ the equation $x=yaya^{-1}$ has a solution. Indeed, replace $ya$ by $z$. We get $x=z^2a^{-2}$ or $xa^2=z^2$. But in $SL_n(\mathbb{C})$ every element is a square, so $z$ can be found, hence $y=za^{-1}$ exists.

Update 1. In order to see that every complex non-singular matrix is a square, it is enough to consider a Jordan block $J_n(a)$ ($a$'s on the diagonal, 1 on the next diagonal, the rest are 0). If we denote $b=\sqrt{a}$, then a square root of $J_n(a)$ is a triangular matrix, it has $b$ on the diagonal, $2/b$ on the next diagonal, $-\frac{1}{8b^3}$ on the second diagonal, $\frac{1}{16b^5}$ on the third diagonal, etc. The number on the diagonal # $i$ is equal to $c_i/(b^{2i-1})$ where $c_i$ can be defined by induction, it does not depend on $a$. Also if you represent your matrix as $A=\exp(B)$ for some $B$, then the square root of $A$ is just $\exp(B/2)$. Note that we need $a\ne 0$, so this does not work for singular matrices.

Update 2. About the equation $y=x(x^{-1})^t$. Note that this equation is stable under conjugation by complex orthogonal matrices (i.e. matrices $a$ with $a^t=a^{-1}$).Thus instead of $y$ we can consider $aya^{-1}$ with $a$ orthogonal. Hence we can assume that $y$ is triangular. For $n=2$ this immediately gives:

** A triangular matrix $y\in SL_2(\mathbb{C})$ is of the form $x(x^{-1})^t$ if and only if either $y=1$ or the eigenvales of $y$ are not equal to 1. Thus $y$ is of that form iff either $y=1$ or $y$ is not unipotent.

Update 3 For $n=3$ the description is more complicated. For example, all uni-upper triangular representable matrices $A$ have the form $$\left(\begin{array}{lll} 1 & ca &cb \\\ 0 & 1 & c\\\ 0& 0 & 1\end{array}\right)$$ Hence if $A[3,2]=0$, $A$ must be equal to 1. On the other hand if $c\ne 0$, $A$ is arbitrary.

Update 4. If $y=x(x^{-1})^t$, and $a$ is an eigenvalue of that matrix, then $1/a$ is also an eigenvalue. Indeed, $yv=av$ implies $y^tw=aw$ for some $w$ (since $y$ and $y^t$ have the same eigenvalues), $x^{-1}x^tw=aw$. Hence $x^tw=a xw$. Hence $1/aw=(x^{-1})^txw$, so $1/a$ is an eigenvalue of $(x^{-1})^tx$, hence an eigenvalue of $y$ (since matrices $pq$ and $qp$ have the same eigenvalues). This implies for $n=3$, a matrix $y$ of that form must have eigenvalue 1.

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I might be more jet lagged than I thought, but it seems to me that substituting $y=za$ gives you $x=z a a z a a^{-1} = z a^2 z \neq z^2 a^{-2}.$ –  Igor Rivin May 6 '12 at 11:48
A misprint: you need to substitute $z=ya$ (I fixed that). Then $yaya^{-1}=zza^{-1}a^{-1}=z^2a^{-2}$. –  Mark Sapir May 6 '12 at 11:50
Though since every invertible matrix is a square, as you say, my first question is equivalent to, for every $b,$ the solvability of $x = y b y.$ –  Igor Rivin May 6 '12 at 11:51
Ah, OK, I am too jet lagged to fix typos :) –  Igor Rivin May 6 '12 at 11:52
That's very interesting: I did not see whether you checked whether all non-unipotent matrices are representable, or is that still a conjecture only... –  Igor Rivin May 7 '12 at 10:40