2
$\begingroup$

Let $G$ be a compact Lie group with Lie algebra $\mathfrak{g}.$ Let $K$ be a closed subgroup of $G$ with Lie algebra $\mathfrak{k}.$ We define the manifold $$\mathcal{E}:= G \times_K \mathfrak{g}$$ to be the quotient of the $K$ action on $G \times \mathfrak{g}$, where $K$ acts on $G$ by right multiplication, and on $\mathfrak{g}$ by the adjoint action.

Let $$ \pi: G \times \mathfrak{g} \rightarrow G \times_K \mathfrak{g}$$ be the projection map; which associates to $(g,X) $ its equivalence class $[g,X] $ in $G \times_K \mathfrak{g}$.

Let $(g,X) \in G \times \mathfrak{g}$, if we identify $T_{(g,X)} (G \times \mathfrak{g})$ with $\mathfrak{g} \times \mathfrak{g}$ and identify $T_{\pi(g,X)}(G \times_K \mathfrak{g})$ with $\mathfrak{g}/\mathfrak{k} \times \mathfrak{g}$, can we give an explicit expression for the derivative of $\pi$ at $(g,X) ?$

What I've tried is to take $Z_1= \alpha'(t)$ , $Z_2= \beta'(t)$ in $\mathfrak{g} \times \mathfrak{g}$ and $$d\pi_{(g,X)}(Z_1,Z_2)= d\pi_{(g,X)}(Z_1,0)+ d\pi_{(g,X)}(0,Z_2)= \frac{d}{dt} \Bigg|_{t=0} \pi(\alpha(t),X) +\frac{d}{dt} \Bigg|_{t=0} \pi(g,\beta(t)),$$ but then I don't how to continue?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ How do you identify $T_{\pi(g, X)}(G \times_K \mathfrak g)$ with $\mathfrak g/\mathfrak t \oplus \mathfrak g$? $\endgroup$
    – LSpice
    Commented Feb 24, 2022 at 2:09
  • 1
    $\begingroup$ @LSpice, I used the fact that $G \times_K \mathfrak{g} \rightarrow G/K$ is locally a trivial fibre bundle with fibre $\mathfrak{g}$ and since the quotient $G \times_K \mathfrak{g}$ is the associated bundle to the principal bundle $G \rightarrow G/K$, then the tangent space will split to be isomorphic to the vector space direct sum $\mathfrak{g}/\mathfrak{k}\oplus \mathfrak{g}$. $\endgroup$
    – Mira
    Commented Feb 24, 2022 at 2:25

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .