Push-forward of inverse map

Question

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

José Carlos Santos · Answer 1 · 2018-11-15T14:58:55.280

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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.

edited Nov 15 '18 at 14:58

answered Nov 15 '18 at 14:57

José Carlos Santos

427,504

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
If $X$ is not an element of $T_gG$, what is it then? A vector field defined in $G$? Or a left-invariant vector field, perhaps? – José Carlos Santos Nov 15 '18 at 15:18
A vector field in $G$ as it's explained in the question – Vicky Nov 15 '18 at 15:21
Then, there is no reason why $i^*(X|g)$ should be $\pm X|{g^{-1}}$. – José Carlos Santos Nov 15 '18 at 15:25
Why? I'm just getting more confused – Vicky Nov 15 '18 at 15:28
Take a concrete example. Suppose, say that $G$ is $(\mathbb{R},+)$. Suppose that the vector field $X$ is, say, $X(f)(a)=(a+3)f'(a)$. Finally, take $g=1$ and compute $i^*(X|1)$ and $X|{-1}$ (in this case, $i(x)=-x$, of course). They are different. – José Carlos Santos Nov 15 '18 at 15:34
If $i_: T_gG \rightarrow T_{g^{-1}}G$, then $i_(X|g) \in T{g^{-1}}G$. Now recall that $X|{g^{-1}} \in T{g^{-1}}G$, so maybe it is right that $i_*(X|g) = X|{g^{-1}}$ – Vicky Nov 15 '18 at 16:57

Travis Willse · Accepted Answer · 2018-11-15T20:45:45.420

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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.

edited Nov 15 '18 at 20:45

answered Nov 15 '18 at 16:34

Travis Willse

99,363

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 https://math.stackexchange.com/questions/2999933/how-does-push-forward-of-inverse-map-look-like). 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 Willse 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
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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 Willse Nov 15 '18 at 17:15
And how do you what part is in $T_e$ and in $T_g$, because in your formula for $i_*$ appears both? – Vicky Nov 15 '18 at 17:43
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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 Willse Nov 15 '18 at 17:58
Ok, great, got it. Really thanks for your time and your answer – Vicky Nov 15 '18 at 18:11
You're welcome, I'm glad you found it useful. For a clear exposition of some basics about pushforwards, I recommend $\S~$3 of Lee's Introduction to Smooth Manifolds. – Travis Willse Nov 15 '18 at 18:15

Push-forward of inverse map

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