10
$\begingroup$

Let $G$ be a Lie group, $g\in G$ and $L_g$ be left translation by $g$. I want to compute the differential $dL_g|_0$ of $L_g$ at $0$.

Attempt:

Let $v\in T_0G$ be a tangent vector at $0$. Let $x:\mathbb R^n\to G$ be chart around $0$, such that $x(0)=0$. Let $a:\mathbb R\to \mathbb R^n:t\to vt$ be a straight line with gradient $v$.

Then $dL_g(v)=v(L_g)=\frac{d}{dt}|_0[L_g(x(a(t)))]=\frac{d}{dt}|_0[gx(vt)]=gx'(0)v$

As $g$ is a group element, and $x'(0)v$ is a tangent vector, we cannot multiply these together. Therefore, the last term above is meaningless. How should we remedy this problem? How can we compute the matrix of $dL_g|_0$?

EDIT: Thank you for your comment, Jorkug. I am worried because by following the definitions I arrive at the meaningless symbol $gx'(0)v$.

$\endgroup$
5
  • 2
    $\begingroup$ To my experience, they usually do not compute it explicitly and call it the left translation of a vector. To write the matrix explicitly though, one would have to compute the derivatives of the multiplication on the group. I will also notice, that locally any group is a matrix subgroup, so your formula will make sense if understood locally and up to an isomorphism. $\endgroup$
    – Jorkug
    Commented Mar 23, 2015 at 5:43
  • $\begingroup$ What are you asking for, precisely, with "how can we compute the matrix of $\left.dL_g\right\rvert_0$? Are you asking for the matrix of the transformation in some coordinates $T_e G \cong \mathbb R^n \to \mathbb R^n \cong T_g G$? $\endgroup$
    – epimorphic
    Commented Jun 16, 2015 at 18:57
  • $\begingroup$ Yes, we need to work in a coordinate system in order to speak of the matrix of a linear operator. $\endgroup$ Commented Jun 16, 2015 at 19:01
  • $\begingroup$ That's a rather peculiar request because the matrix depends on the choice of charts at both the identity and $g$. Say if $F \colon M^m \to N^n$ is smooth, $p \in M$, and $A$ the matrix of $dF_p$ in some pair of coordinates at $p$ and $F(p)$, then for any $B \in (GL_n)(A)(GL_m)$ there exists a pair of coordinates such that $B$ is the matrix of $dF_p$. $\endgroup$
    – epimorphic
    Commented Jun 16, 2015 at 19:31
  • 1
    $\begingroup$ Here, if the chart at $g$ is just the left-translation $L_g x$ of your chart $x \colon \mathbb R^n \to G$ at the identity, then the matrix of $dL_g$ at the identity will just be $I_n$. For arbitrary coordinates, the chart could be any element of $GL_n$. $\endgroup$
    – epimorphic
    Commented Jun 16, 2015 at 19:31

1 Answer 1

13
+50
$\begingroup$

$\newcommand{\Reals}{\mathbf{R}}$To be invariant about it, if $v$ is a tangent vector at $e$, then the one-parameter subgroup $x(t) = \exp(tv)$ is a curve through $e$ with $x'(0) = v$, and $$ dL_{g}(v) = \frac{d}{dt}\bigg|_{t=0} L_{g}\exp(tv), $$ just as you say.

To get more detailed and/or explicit information, you need to invoke specifics about the group structure. If $G$ is a Lie subgroup of $GL(n, \Reals)$, the group operation is the restriction of matrix multiplication. For each $g$ in $G$, left multiplication $L_{g}(x) = gx$ is the restriction of a linear transformation $\Reals^{n \times n} \to \Reals^{n \times n}$, and may be identified with its differential: $$ dL_{g}(v) = gv. $$

For example, if $G = SO(n)$, then $g$ is a rotation matrix, a tangent vector $v$ at the identity is a skew-symmetric matrix, and $$ dL_{g}(v) = gv $$ is an ordinary matrix product, viewed as an element of $T_{g} SO(n)$.

$\endgroup$
6
  • $\begingroup$ Thank you, sir. What did I do wrong that led me to a meaningless symbol? $\endgroup$ Commented Jun 16, 2015 at 16:07
  • 1
    $\begingroup$ I suppose it comes down to your final equality; by the chain rule, the derivative of$$t \mapsto gx(vt) = (L_{g} \circ x)(vt)$$ at $t = 0$ is$$dL_{g}\bigl(x(0)\bigr) dx(0) v.$$(Incidentally, our "$v$"s mean slightly different things; I used $v$ as a tangent vector to $G$ at $e$, which you'd denote $x'(0)v$. In that respect, your formula is correct if $G$ is a matrix group and $dL_{g}\bigl(x(0)\bigr)$ is identified with multiplication by $g$.) $\endgroup$ Commented Jun 16, 2015 at 17:00
  • $\begingroup$ It's worth noting that $dL_g(x(0)) dx(0) v$ is just $dL_g(v)$, which is what we started with. Both of you are denoting as $v$ both an element of $T_e G$ and the corresponding element of $T_0 \mathbb R^n$. $\endgroup$
    – epimorphic
    Commented Jun 16, 2015 at 18:14
  • 1
    $\begingroup$ I might be misreading a bit. Perhaps "I used $v$ as a tangent vector to $G$ at $e$" refers to the entire answer, while the $dx(0) v$ in the comment is just you emulating OP's slightly amorphous notation. $\endgroup$
    – epimorphic
    Commented Jun 16, 2015 at 18:23
  • 4
    $\begingroup$ Can you elaborate on why we can identify left multiplication in $G$ with left multiplication in its tangent space? $\endgroup$
    – mi.f.zh
    Commented Feb 2, 2020 at 11:24

You must log in to answer this question.

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