# Homogenous vector bundles

In the definition of homogenous vector bundles, an equivalence class is defined.

Briefly:

G is a lie group and H a (lie) subgroup. Define

$$\rho : H \rightarrow GL(V)$$

where V is a vector space.

The equivalence class is defined as

$$(g_1, v_1) = (g_1h, \rho(h^{-1})v_1)$$

The question is why the above definition and not

$$(g_1, v_1) = (g_1h, \rho(h)v_1) ?$$

-

The first definition $$(g, v).h ~:=~ (gh, \rho(h^{-1})v)$$ defines a right group action $G \times V \times H \to G \times V$, $$((g, v).h).k~=~(g, v).(h.k),$$ while the second definition $$(g, v) ~\mapsto~ (gh, \rho(h)v)$$ is neither a left nor a right group action, cf. Chris Gerig's comment.

-
This completes the picture. The second definition wouldn't work in an obvious way –  Tbh Dec 20 '11 at 18:21

Following up on Qmechanic's response, here is a a step by step demonstration of why one definition works and the other does not:

$$(g,v)hk = ((g,v)h)k = ((gh, \rho(h)v)k = ((ghk, \rho(k)\rho(h)v) = ((ghk, \rho(kh)v) \neq (g,v)hk$$

From the above the second definition does not work.

On the other hand

$$(g,v)hk = ((g,v)h)k = ((gh, \rho(h^{-1})v)k = ((ghk, \rho(k^{-1})\rho(h^{-1})v) = ((ghk, \rho(k^{-1}h^{-1})v) =$$

$$((ghk, \rho((hk)^{-1})v) = (g,v)hk$$

-