Differential of the inversion of Lie group

Question

Let $G$ be a Lie group and $\iota \colon G\to G$ denote the inversion. If $e$ is the identity of $G$, prove that: $$\text d \iota _e = -\text {id} _{T_e G}.$$ I understand that the differential at $e$ should be an involution, since $$\iota \circ \iota = \text{id}_G$$ implies that $$\text d \iota _e \circ \text d \iota _e=\text d (\iota \circ \iota)_e=\text d(\text {id} _G)_e=\text {id} _{T_e G}.$$ However I don't know how to conclude the thesis from this. Any hint?

Answer 1

From your approach, you can only say the eigenvalue is either $1$ or $-1$, therefore impossible to answer your question.

If you use the curve definition of tangent vector, the result can't be more obvious: $$ \iota: g(t)\mapsto g^{-1}(t), g(0)=e. $$ Then by chain rule: $$ d\iota_e(\dot g(0))=\dot g^{-1}(0). $$ However, $g(t)g^{-1}(t)=e$ and therefore by Leibniz rule: $$ \dot g(0)=-\dot g^{-1}(0). $$ and consequently, $$ d\iota_e(\dot g(0))=-\dot g(0), \forall \dot g(0)\in T_eG. $$

Note that a Lie group $G$ is necessarily a symmetric space with the inversion symmetry $S_g$ at $g\in G$ given by $S_g: h\mapsto gh^{-1}g$. So that, every inversion symmetry is involutive and in particular, at $g=e$, $S_g$ coincides with the group inversion. This roughly says you should expect to have a involutive map in the first place.

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Honest check of Leibniz rule:

If $G$ is a matrix Lie group, group multiplication is the matrix multiplication and we are done. For a general Lie group, the situation is similar: $\forall a,b\in G$ parameterized by local coordinates $a=a(x^1,\dots,x^n)$ and $b=b(y^1,\dots,y^n)$, their product is smooth and a map from $(x^1,\dots,x^n;y^1,\dots,y^n)$ to $a\cdot b\in G$ is well defined: $$ a\cdot b(x^1,\dots,x^n;y^1,\dots,y^n)=a(x^1,\dots,x^n)\cdot b(y^1,\dots,y^n) $$ Note that $$ \begin{split} \frac{d}{dt}a\cdot b(x^1(t),\dots,x^n(t);y^1(t),\dots,y^n(t))&=\sum_{i=1}^n\frac{\partial a\cdot b}{\partial x^i}\frac{dx^i}{dt}+\sum_{i=1}^n\frac{\partial a\cdot b}{\partial y^i}\frac{dy^i}{dt}\\ &=\frac{d}{dt}(R_b(a(t)))+\frac{d}{dt}(L_a(b(t)))\\ &=R_{b*}\dot a(t)+L_{a*}\dot b(t) \end{split} $$ In our case, we take $a=g, b=g^{-1}$ and the rest follows.

Answer 2

Based on the fact that $\iota$ is an involution, you cannot conclude that $d\iota_{e} = -\text{id}_{T_{e}G}$. For example, think of the involution $R$ of $\mathbb{R}^3$ which consists in rotation by $\pi$ around the $z$-axis. This has differential at the origin given by \begin{equation} \left[ {\begin{array}{ccc} -1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array} } \right]. \end{equation}

A nice way to see that $d\iota_{e} = -\text{id}_{T_{e}G}$ is to make use of 1-parameter subgroups. Given an element $X \in T_{e}G$, we have the 1-parameter subgroup of $G$ generated by $X$: \begin{equation} \Phi_{X}(t) = \text{exp}(tX). \end{equation} (Note that we are identifying the tangent space $T_{e}G$ with $\text{Lie}(G)$, the Lie algebra of $G$, and that exp is the exponential map: exp$: \text{Lie}(G) \to G$.)

Since $\Phi_{X}'(0) = X$, and that \begin{equation}\iota \circ \Phi_{X}(t) = \text{exp}(tX)^{-1} = \text{exp}(-tX) = \Phi_{-X}(t), \end{equation} we have that \begin{equation} d\iota_{e}(X) = d\iota_{e} \circ d(\Phi_{X})_{0}(\frac{d}{dt}) = d(\iota \circ \Phi_{X})_{0}(\frac{d}{dt}) = \Phi_{-X}'(0) = -X.\end{equation}

Note that making use of the 1-parameter subgroup often comes in handy in problems like this involving differentials of maps on Lie groups.

Answer 3

It can be sooo simple... :)

Applying the multiplication infinitesimally: $$[\alpha]+\mathrm{d}\iota[\alpha]=[\alpha]+[\iota\circ\alpha]=\mathrm{d}\mu[(\alpha,\alpha^{-1})]=[\mu(\alpha,\alpha^{-1})]=[e]=0$$ (Note that all equality signs in this line are for real.)