Let $H \subset G$ be a closed subgroup of a lie group and $G/H$ the homogeneous coset space. There's an exact sequence of adjoint representations of $H$:
$$0 \to \mathfrak{h} \to \mathfrak{g} \to \mathfrak{g/h}\to 0 $$
The canonical principal $H$-bundle $G \to G/H$ gives an exact functor from representations of $H$ to vector bundles over $G/H$. The corresponding sequence of vector bundles is:
$$0 \to G \times_H \mathfrak{h} \to G \times_H \mathfrak{g} \to G \times_H \mathfrak{g/h}\to 0 $$
I'm pretty convinced that $G\times_H \mathfrak{g/h}$ is canonically isomorphic to $T(G/H)$ but a simple proof alludes me so help here would be welcome.
Taking the projective space as an example we have $\mathbb{CP}^n \cong \frac{U(n+1)}{U(n)\times U(1)}$ and the corresponding exact sequence:
$$0 \to U(n+1) \times_{U(n)\times U(1)} \mathfrak{u(n)\times u(1)} \to U(n+1) \times_{U(n)\times U(1)} \mathfrak{u(n+1)} $$
$$\to U(n+1) \times_{U(n)\times U(1)} \mathfrak{\frac{u(n+1)}{u(n)\times u(1)}}\to 0 $$
Where the last vector bundle is the tangent bundle $T\mathbb{CP}^n$. How does this relate to the euler sequence if at all?
$$0 \to \mathcal{O}_{\mathbb{P}^n} \to \mathcal{O}_{\mathbb{P}^n}(1)^{n+1} \to \mathcal{T}_{\mathbb{P}^n} \to 0$$
Best Answer
The isomorphism $G\times_H(\mathfrak g/\mathfrak h)\to T(G/H)$ is induced by the map $G\times (\mathfrak g/\mathfrak h)$ mapping $(g,X+\mathfrak h)$ to $T_gp\cdot L_X(g)\in T_{gH}(G/H)$.
The Euler sequence corresponds to an exact sequence for the restriction of the standard representation of $G$ to $H$. This is better seen when viewing $\mathbb CP^n$ as a homogeneous space of $G=SL(n+1,\mathbb C)$. Then $H\subset G$ is the group of block-upper-triangular matrices with blocks of sizes $1$ and $n$, so this is a semi-direct product of $S(GL(1,\mathbb C)\times GL(n,\mathbb C))\cong GL(n,\mathbb C)$ and $\mathbb C^{n*}$. In particular, there are natural completely reducible representations for $H$ on $\mathbb C$ and $\mathbb C^n$, say $V$ and $W$. Now consider the standard representation $\mathbb C^{n+1}$ of $G$ and restrict it to $H$. The result is indecomposable but not irreducible, since is contains an $H$-invaraint line but no invariant complement. It fits into an exact sequence $0\to V\to\mathbb C^{n+1}\to W\to 0$ and the Euler-sequence is the short exact sequence of homogeneous vector bundles over $\mathbb CP^n$ corresponding to the tensor product of this sequence with $V^*$. In the picture of unitary groups, the picture is not as clear, since $H$ itself is semisimple, so the restriction of $\mathbb C^{n+1}$ is completely reducible, and it is not obvious, which "direction" of the resulting exact sequence to use. This corresponds to the fact that the Euler sequence admits a split which is $U(n+1)$-equivariant, but not one, which is $SL(n+1,\mathbb C)$-equivariant.