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Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, and let $H$ be a Lie subgroup of $G$ with Lie algebra $\mathfrak{h}$. I want to prove that for $g \in G$,

$T_{[g]} (G/H ) $ is isomorphic to $\mathfrak{g}/\mathfrak{h}.$

To do so, I considered the map $a_{g^{-1}}: G/H \rightarrow G/H$, which associates to every class $[k] $, the class $ [g^{-1}k].$

By differentiating this map at point $[g]$, we get the required isomorphism

$$da_{g^{-1}}|_{[g]} : T_{[g]} (G/H ) \rightarrow \mathfrak{g}/\mathfrak{h}.$$

Is this true ?

Asma
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    https://math.stackexchange.com/questions/1377112/tangent-space-of-quotient-space might be helpful. – kzkzkzz Mar 07 '22 at 18:02
  • Thanks @kzkzkzz for the link. Could you please comment what do you think of my answer is it true or false ? – Asma Mar 07 '22 at 18:07
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    How would show your map $da_{g^{-1}}$ is both surjective and injective? And you also need to show $T_{[e]} (G/H ) \cong\mathfrak{g}/\mathfrak{h}$. – kzkzkzz Mar 07 '22 at 18:16
  • @kzkzkzz, $da_{g^{-1}}$ is an isomorphism since $a_{g^{-1}}$ is an isomorphism ; it is surjective since every $[k] \in G/H$ is equal to the image $a_{g^-1}[gk]$ and it is injective since if $a_{g^-1}[k]=a_{g^-1}[k']$ then $[k]= [k']$. Is this true? – Asma Mar 07 '22 at 18:57
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    Looks correct. But be careful that $H$ might not be a normal subgroup, and thus $G/H$ is only the space of left cosets. Hence this isomorphism is only a vector space isomorphism. For Lie algebra isomorphism, we need more assumptions: https://math.stackexchange.com/questions/29390/lie-algebra-of-a-quotient-of-lie-groups. – kzkzkzz Mar 07 '22 at 19:29
  • @kzkzkzz thank you very much for your help! – Asma Mar 07 '22 at 19:36
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    You are welcome. – kzkzkzz Mar 07 '22 at 19:40

1 Answers1

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It is true that the map you define is an isomorphism. However, this isomorphism is not canonical. It is true that $T_{[e]}(G/H)$ can be canonically identified with $\mathfrak g/\mathfrak h$, but to define your map, you have chosen a representative $g$ of the coset $[g]$, and choosing a different representative gives a different isomorphism. What can be done canonically here is the identification of $TG/H$ as an associated bundle. Take $G\times(\mathfrak g/\mathfrak h)$ and endow it with a right $H$-action defined by $(g,X+\mathfrak h)\cdot h:=(gh,Ad_{h^{-1}}(X))$. Then $TG/H$ can be canonically identified with the space of $H$-orbits $(G\times(\mathfrak g/\mathfrak h))/H$. Edit: Explicitly, this is induced by the map that sends $(g,X+\mathfrak h)$ to $d_gp(d_e\ell_g(X))\in T_{[g]}G/H$, where $p:G\to G/H$ is the canonical map and $\ell_g$ denotes left multiplication by $g$. This can also be written as $d_[e]a_g(d_ep(X))$, so in a way it encodes the inverses of the isomorphism in the family that you constructed.

Andreas Cap
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