4
$\begingroup$

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?

$$0 \to \mathcal{O}_{\mathbb{P}^n} \to \mathcal{O}_{\mathbb{P}^n}(1)^{n+1} \to \mathcal{T}_{\mathbb{P}^n} \to 0$$

$\endgroup$
  • 1
    $\begingroup$ If you are working in the setting of algebraic geometry, you can prove that isomorphism via "faithfully flat descent". If you pullback your exact sequence to $G$, then it is the tangent bundle sequence. Next you check that the induced isomorphism between the pullback to $G$ of $T(G/H)$ and $G\times^H (\mathfrak{g}/\mathfrak{h})$ satisfies a "cocycle condition" after pullback to $G \times_{G/H} G \cong H \times G$. Thus the isomorphism of bundles on $G$ is the pullback of an isomorphism on $G/H$. $\endgroup$ – Jason Starr Mar 11 '16 at 13:42
  • 1
    $\begingroup$ See 18.16 (p 225) of mat.univie.ac.at/~michor/dgbook.pdf $\endgroup$ – Peter Michor Mar 11 '16 at 13:44
3
$\begingroup$

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.

$\endgroup$

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.