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The map $G \times G \rightarrow G$ defined as $(x,y) \mapsto xy$ is differentiable. The left translation is $L_x(y) = xy$. To show that it is a diffeomorphism, we need it to be a bijection, that is both differentiable and has a differentiable inverse.

It is injective because if $xy_1 = xy_2$ then $y_1=y_2$.

It is surjective because for some arbitrary element $z \in G$ we can multiply on the left by $x^{-1}$ to get $x^{-1}z=y$. Then $L_x(y)=z$, meaning that $z\in \operatorname{im}(L_x)$.

But why is $L_x$ differentiable, and why is its inverse?

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    $\begingroup$ $L_x$ is the restriction of $G\times G\to G$ to $\{x\}\times G$. Its inverse is $L_{x^{-1}}$. $\endgroup$ – Gil Bor Feb 8 '14 at 6:01
  • $\begingroup$ If you restrict a differentiable map of smooth manifolds to a submanifold, the result is differentiable. This can be checked right from the definition of beeing a submanifold. In a product, each factor is a submanifold. $\endgroup$ – archipelago Feb 8 '14 at 9:21
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The multiplication map $(x,y)\mapsto xy$ from $G\times G$ to $G$ is smooth by the definition of a Lie group. Fixing $x$ here results in a smooth map $L_x:G\to G$. Its inverse is $L_{x^{-1}}$; thus the map is a diffeomorphism.

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Proposition. Let $G$ be a Lie group, and let $a\in G$. Then the map $L_a\colon G \to G$ with $L_a(x)=ax$ is a diffeomorphism.

Proof. We have already shown that $L_a$ is bijective. (A quicker way to see it might be to notice that $L_{a^{-1}}$ is the inverse of $L_a$.) Left to show is that $L_a$ and $L_a^{-1}$ are both smooth.

To show that $L_a$ is smooth, notice that $L_a=\mu\circ \iota_a$, where $\mu\colon G\times G\to G$ is defined by $\mu(x,y)=xy$, and $\iota_a\colon G\to G\times G$ is defined by $\iota_a(x)=(a,x)$. It follows from the definition of a Lie group that $\mu$ is smooth, and it's easy to show (how?) that $\iota_a$ is also smooth. The desired result now follows directly from the fact that the composition of two smooth maps is smooth itself.

The smoothness of $L_a^{-1}=L_{a^{-1}}$ follows by replacing $a$ by $a^{-1}$ in the previous paragraph.

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