0
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.