# A metric on T(G/H)

I want to know which isomorphism of vector bundles is the following?

alt text http://2.bp.blogspot.com/_uGcgLiQvkI8/TAgpIivl6oI/AAAAAAAAAKk/DCg9iK2W3rA/s1600/Capture-52.png

where $G/H$ is the quotient of group $G$ by its subgroup $H$ and $T(G/H)$ is the tangent bundle, $G\times_H \mathfrak{g}/\mathfrak{h}$ is the bundle associated to the principal bundle $G\to G/H$ via the adjointe representation on $\mathfrak{g}/\mathfrak{h}$

• Pedro -- what exactly do you want to know about this isomorphism? – algori Jun 3 '10 at 22:24
• I want to know how it works... I mean a relation that expresses this isomorphism – Pedro Jun 3 '10 at 22:27
• Why is the question titled "a metric on T(G/H)"? The isomorphism you mention has nothing to do with metrics. – José Figueroa-O'Farrill Jun 3 '10 at 23:54
• Ah yes! because with this isomorphisme I can construct a metric on T(G/H) using the metric on g/h – Pedro Jun 4 '10 at 8:25
• By this do you mean perhaps that if you have an inner product on g/h which is invariant under the adjoint action of H, then it extends to a G-invariant metric on G/H? – José Figueroa-O'Farrill Jun 4 '10 at 8:29

There is an obvious map $G\times\mathfrak g\to G\times_H\mathfrak g/\mathfrak h$, and an isomorphism $TG\to G\times\mathfrak g$. On the other hand, the projection $G\to G/H$ gives a map $TG\to T(G/H)$. Now you can construct a map $T(G/H)\to G\times_H\mathfrak g/\mathfrak h$ as the composition $$T(G/H)\leftarrow TG\to G\times\mathfrak g\to G\times_H\mathfrak g/\mathfrak h.$$ Here the backwards arrow means "pick any preimage".
• Nice! Now I want to know if this is true $G\times_H\mathfrak g/\mathfrak h \sim G/H \times (\mathfrak g/\mathfrak h)/Ad(H)$ – Pedro Jun 4 '10 at 11:57
• That cannot be true for a number of reasons, the most basic perhaps is the following. The left-hand side is a homogeneous vector bundle over $G/H$, whereas the right-hand side (assuming you mean what I understand by your notation) is not even a manifold in general, since the orbit structure of the vector space $\mathfrak{g}/\mathfrak{h}$ under the adjoint action of $H$ can be quite complicated. – José Figueroa-O'Farrill Jun 5 '10 at 19:02