1
$\begingroup$

Let $\mathfrak{g}$ denote the Lie algebra of a Lie group $G\leq GL(n)$. The adjoint representation of $G$ is defined as the function $Ad_g:\mathfrak{g}\rightarrow\mathfrak{g}$ that maps each $x\in\mathfrak{g}$ to $Ad_g(x)=gxg^{-1}$, for some $g\in G$. Why does $gxg^{-1}$ fall into $\mathfrak{g}$ ?

EDIT 1: To provide some context, the resources (including this phd thesis, section 2.4) define the Lie algebra $\mathfrak{g}$ of a group $G\leq GL(n)$ as the tangent space at the Identity and more or less limit their study of said lie algebra to computing its generators, before defining the exponential and log, followed by the definition of the adjoint as I gave it in the question. Given only this informations, I don't think the answer to my question is trivial. The definition given in the comment by Tobias Kildetoft (Which says that the Lie algebra in such a case can be defined as all matrices $A\in \mathfrak{gl}(n)$ for which $e^{tA}$ is in $G$) makes the answer to my question obvious, but raises another question, i.e. the equivalence between the two definitions. I don't find that trivial either (as seen in this stackexchange question.). I really want to understand this, so please help me reformulate this question if it is not clear enough.

EDIT 2: As implied from the previous edit, I am from the computer vision community, and this question also seems to be puzzling for the colleagues I have talked to, so I don't think it is useless.

$\endgroup$
9
  • $\begingroup$ This is the wrong definition. The differential of $\Psi_g$, (with $\Psi_g(h)=ghg^{-1}$) is the adjoint representation, and not $\Psi_g$, see here. $\endgroup$ Commented Dec 7, 2015 at 10:35
  • $\begingroup$ How is this wrong? It says after the derivations that $Ad_g(X)=gXg^{-1}$ with $X$ in the lie aglebra... plus, I have seen the definition I gave appeart in the computer vision litterature, e.g in hauke.strasdat.net/files/strasdat_thesis_2012.pdf $\endgroup$
    – Ash
    Commented Dec 7, 2015 at 10:44
  • $\begingroup$ Sorry I meant after the part where it says "taking the derivative of $\psi(...$ $\endgroup$
    – Ash
    Commented Dec 7, 2015 at 10:48
  • $\begingroup$ Ohh, this is only if you already have immersed your Lie group into $GL(n)$. Then it is by definition in the Lie algebra, because it is matrices. $\endgroup$ Commented Dec 7, 2015 at 10:50
  • $\begingroup$ Yes, of course, I'm mainly interested in $Sim3$, I will edit the question. But can you please explain more about why it is in the Lie algebra? $\endgroup$
    – Ash
    Commented Dec 7, 2015 at 10:53

1 Answer 1

1
$\begingroup$

Suppose $V$ is a tangent vector of $G\subseteq\mathrm{GL}(n)$ at the identity. That means there is some smooth path $v(t)$ taking values in $G$ with velocity vector $v(0)=\mathrm{id}_G$ and initial condition $v'(0)=V$. If $g\in G$, then $gv(t)g^{-1}$ is a different path in $G$ with $gv(0)g^{-1}=\mathrm{id}_G$, so it will also yield a velocity vector at the identity by differentiating at $t=0$. Since $\frac{d}{dt}(gv(t)g^{-1})=g v'(t)g^{-1}$, its velocity vector at the identity will be $gVg^{-1}$.

Note that conjugation in a Lie group usually has a natural interpretation in terms of its action on some space. For instance in $\mathrm{SO}(n)$, if $R$ is a rotation by $\theta$ around the oriented axis $\ell$, and $S$ is any other rotation, then $SRS^{-1}$ will be a rotation by $\theta$ around the oriented axis $S\ell$. In one dimension, in either $\mathrm{O}(1)$ or any dihedral group $D_n$ (countable discrete groups are $0$-dimensional Lie groups), let $R$ be any rotation and $F_{\ell}$ a reflection across a line $\ell$. Then $RF_\ell R^{-1}=F_{R\ell}$.

Or, consider the affine group $\mathrm{Aff}_0(1)$, the group of transformations of the real line of the form $f(x)=ax+b$ with $a>0$. Every group element is either a translation or an inward/outward scaling from a unique fixed point. Let $B$ be any group element, with scaling factor $s$. If $A$ is a translation by $h$, then $BAB^{-1}$ is a translation by $sh$, and if $A$ is a scaling in/out from some fixed point $t$, then $BAB^{-1}$ is a scaling by the same factor from fixed point $Bt$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .