1
$\begingroup$

If I define the inverse map in a Lie group $G$ as,

$$i: G \rightarrow G,\quad i(g) = g^{-1}, \forall g \in G \tag1$$

I think that the associated push-forward would be,

$$i_*: T_gG \rightarrow T_{g^{-1}}G, \quad i_*(X|_g) = X|_{g^{-1}} \equiv -X|_g, \forall X \in \mathfrak{X}(G) \tag2$$

Where, $\mathfrak{X}(G)$ is the set of tangent vector fields in $G$

Is Eq. (2) the action of $i_*$ or it is wrong? The part which I'm not very sure is $i_*(X|_g) = X|_{g^{-1}}$. I have this doubt because in other posts (e.g. Pushforward of Inverse Map around the identity? or Differential of the inversion of Lie group) it is handle $T_eG$ and not $T_gG$ in general.

$\endgroup$
2
  • $\begingroup$ What does $-g$ mean in an arbitrary Lie group? Did you mean $g^{-1}$? $\endgroup$ Commented Nov 15, 2018 at 14:52
  • $\begingroup$ yes, sorry for the mistake $\endgroup$ Commented Nov 15, 2018 at 14:52

2 Answers 2

Reset to default
1
$\begingroup$

In general the proposed equation makes no sense, as quantities on the two sides live in different tangent spaces: For $X \in T_g G$, we have $-X_g \in T_g G$ but $X_{g^{-1}} \in T_{g^{-1}} G$, and these spaces coincide only if $g^2 = e$.

On the other hand, unwinding definitions gives $$i = R_{g^{-1}} \circ i \circ L_{g^{-1}}$$ (here, $L_h$ and $R_h$ respectively denote left and right multiplication by $h$), so differentiating gives $$T_g i = T_e R_{g^{-1}} \circ T_e i \circ T_g L_{g^{-1}},$$ and using that $T_e i \cdot Y = - Y$ gives what you wanted in your equation (2): $$\boxed{T_g i \cdot X = T_e R_{g^{-1}} (-T_g L_{g^{-1}} \cdot X) = -T_g (L_{g^{-1}} \circ R_{g^{-1}}) \cdot X }.$$ For a general Lie group this is already fully simplified, but for a linear Lie group $G \leq GL(n, \Bbb R)$ the usual matrix identifications give $$\boxed{T_A i \cdot X = -A^{-1} X A^{-1}} .$$

Remark The proposed equation is wrong for a different reason, too, namely that the value of the vector field $X$ at one point on $G$ need not be related it its value at any other point. Such a relation is forced, however, if we restrict our attention to left- or right-invariant vector fields.

$\endgroup$
8
  • $\begingroup$ I don't see how you get by differentiating the equationwith the tangent spaces. Could you develop it? (Actually I've made a new post for that because I found that equation but I dindn't undertand it. The post is math.stackexchange.com/questions/2999933/…). I've made some calculus but I don't obtain that formula $\endgroup$ Commented Nov 15, 2018 at 16:40
  • $\begingroup$ I'm not sure exactly what you mean. Using the star notation for pushforward, differentiating the identity $i = R_{g^{-1}} \circ i \circ L_{g^{-1}}$ gives $i_* = (R_{g^{-1}})_* \circ i_* \circ (L_{g^{-1}})_*$. The idea here is to rewrite $i$ so that where $i_*$ appears on the r.h.s., it's the pushforward at $1_G$, for which we already know the identity you wrote down. $\endgroup$ Commented Nov 15, 2018 at 16:46
  • $\begingroup$ I don't see how to do the differentiation. Deriving respect to some parameter? I don't see it. In that post I wrote what I get $\endgroup$ Commented Nov 15, 2018 at 16:48
  • 1
    $\begingroup$ The relevant fact is the chain rule for pushforwards: The pushforward of a composition is the composition of the pushforwards, i.e., $(F \circ G)_* = F_* \circ G_*$. $\endgroup$ Commented Nov 15, 2018 at 17:15
  • 1
    $\begingroup$ That's written in my answer. On the right-hand side, $T_g L_{g^{-1}}$ maps $T_g G$ to $T_{L_{g^{-1}}(g)} G = T_e G$, then $T_e i$ maps $T_e G$ to $T_e G$, and finally $T_g R_{g^{-1}}$ maps $T_e G$ to $T_{R_{g^{-1}}(e)} G = T_{g^{-1}} G$. $\endgroup$ Commented Nov 15, 2018 at 17:58
1
$\begingroup$

No, that doesn't make since. You are defining a map from $T_gG$ into $T_{g^{-1}}G$. It maps $X(\in T_gG)$ into what? Since $X\in T_gG$, $X\notin T_{g^{-1}}G$, and therefore it makes no sense to assert that $i^*(X)=-X$. Of course, there is one exception to the assertion “Since $X\in T_gG$, $X\notin T_{g^{-1}}G$”, which is precisely when $g=e$. But that's the only exception.

$\endgroup$
11
  • $\begingroup$ Then, how should Eq. (2) be written? $\endgroup$ Commented Nov 15, 2018 at 14:58
  • $\begingroup$ I don't think that there is a simple answer to that question. $\endgroup$ Commented Nov 15, 2018 at 15:00
  • $\begingroup$ Is $i_*(X|_g) = X|_{g^{-1}}$ right or is this wrong too? $\endgroup$ Commented Nov 15, 2018 at 15:08
  • $\begingroup$ It is worst than wrong: it makes no sense (by the reason that I explained in my answer). $\endgroup$ Commented Nov 15, 2018 at 15:12
  • $\begingroup$ I didn't say $X \in T_gG$. $X|_g \in T_gG$ or at least this is the way I learnt it in class, so $X|_{g^{-1}}$ should belong to $T_{g^{-1}}G$ $\endgroup$ Commented Nov 15, 2018 at 15:14

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.