MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.