# Push-forward of inverse map

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.

• What does $-g$ mean in an arbitrary Lie group? Did you mean $g^{-1}$? – José Carlos Santos Nov 15 '18 at 14:52
• yes, sorry for the mistake – Vicky Nov 15 '18 at 14:52

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.

• 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 – Vicky Nov 15 '18 at 16:40
• 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. – Travis Nov 15 '18 at 16:46
• 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 – Vicky Nov 15 '18 at 16:48
• The relevant fact is the chain rule for pushforwards: The pushforward of a composition is the composition of the pushforwards, i.e., $(F \circ G)_* = F_* \circ G_*$. – Travis Nov 15 '18 at 17:15
• 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 $T_{R_{g^{-1}}(e)} G = T_{g^{-1}} G$. – Travis Nov 15 '18 at 17:58

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.

• Then, how should Eq. (2) be written? – Vicky Nov 15 '18 at 14:58
• I don't think that there is a simple answer to that question. – José Carlos Santos Nov 15 '18 at 15:00
• Is $i_*(X|_g) = X|_{g^{-1}}$ right or is this wrong too? – Vicky Nov 15 '18 at 15:08
• It is worst than wrong: it makes no sense (by the reason that I explained in my answer). – José Carlos Santos Nov 15 '18 at 15:12
• 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 $T_{g^{-1}}G$ – Vicky Nov 15 '18 at 15:14