A general result of Lie Theory is that every Lie group homomorphism $\Phi: G\rightarrow H$ induces a Lie algebra homomorphism $\phi: \frak{g} \rightarrow \frak{h}$.

Which Lie algebra homomorphism is induced by left (or right)-translations:

$L_g h = gh$ for $h,g \in G$, which is a map $G \rightarrow G$ ?

A first idea would be looking at the corresponding pushforward: $L_{ g \star} $ is a map between tangent vectors at $g$ and $h$ respectively. For $ X \in T_g G$ the pushforward is

$L_{ g \star} X = X' \in T_{gh} G$ and therefore this is a map between different tangent spaces which does not help me, because A Lie algebra homomorphism has to be a map from $T_e G$ to $T_e G$.

Any help or ideas finding the corresponding Lie algebra homomorphism for left translations would be much appreciated.


Left and right translations are not Lie group homomorphisms; they don't even preserve the identity, and the induced map on Lie algebras is obtained by looking at derivatives at the identity. However, conjugation by a fixed element $g \in G$ is, and the induced map on $\mathfrak{g}$ gives a representation $G \to \text{Aut}(\mathfrak{g})$ called the adjoint representation. This is itself a Lie group homomorphism, and differentiating it gives the adjoint representation

$$\mathfrak{g} \ni x \mapsto (y \mapsto [x, y]) \in \text{Der}(\mathfrak{g})$$

of $\mathfrak{g}$ (and this is one way to define the Lie bracket).

  • $\begingroup$ Thanks for you quick answer. I always thought the definition of a homomorphism is that its a map $f: G\rightarrow H$, for which $f(x\star y) = f(x)\circ f(y)$ holds for all $x,y \in G$. Is the requirement that the identity is preserved implicit included in this definition? $\endgroup$ – JakobH Jun 28 '14 at 8:22
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    $\begingroup$ @Jakob: it is implied by that axiom (and if you haven't done this as an exercise yet then you really, really should), but it's good practice to state it as a separate axiom because you don't automatically get preservation of identities in greater generality, e.g. for rings. Also, note that left and right translation don't preserve the group operation either. $\endgroup$ – Qiaochu Yuan Jun 28 '14 at 8:23
  • $\begingroup$ now I feel noobish. My line of thoughts was, since left translation is a diffeomorphism (math.stackexchange.com/questions/668139/…), which is a isomorphisms of smooth manifolds ( en.wikipedia.org/wiki/Diffeomorphism ) , which is a homomorphism plus inverse ( en.wikipedia.org/wiki/Isomorphism ) , left translations are homomorphisms. I can't quite get your argument why they are not, from the definition of a homomorphism. $\endgroup$ – JakobH Jun 28 '14 at 8:30
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    $\begingroup$ @Jakob: you're confusing two uses of of inversion here. On the one hand, diffeomorphisms having inverses means that e.g. the self-diffeomorphisms of a manifold form a group. But this isn't the group that's relevant; the group that's relevant here is $G$. Look, all I'm saying here concretely is that if $a, b, c$ are three elements of a group then $a(bc) \neq a(b) a(c)$ in general. That's what it means for left multiplication not to be a group homomorphism. $\endgroup$ – Qiaochu Yuan Jun 28 '14 at 8:33
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    $\begingroup$ @Jakob: it's also true that left translations are homomorphisms of manifolds, in the appropriate sense, but that doesn't imply that they're homomorphisms of Lie groups. $\endgroup$ – Qiaochu Yuan Jun 28 '14 at 8:35

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