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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 $\mathfrak{g}/\mathfrak{h}$

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  • $\begingroup$ Pedro -- what exactly do you want to know about this isomorphism? $\endgroup$ – algori Jun 3 '10 at 22:24
  • $\begingroup$ I want to know how it works... I mean a relation that expresses this isomorphism $\endgroup$ – Pedro Jun 3 '10 at 22:27
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    $\begingroup$ Why is the question titled "a metric on T(G/H)"? The isomorphism you mention has nothing to do with metrics. $\endgroup$ – José Figueroa-O'Farrill Jun 3 '10 at 23:54
  • $\begingroup$ Ah yes! because with this isomorphisme I can construct a metric on T(G/H) using the metric on g/h $\endgroup$ – Pedro Jun 4 '10 at 8:25
  • $\begingroup$ 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? $\endgroup$ – José Figueroa-O'Farrill Jun 4 '10 at 8:29
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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".

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  • $\begingroup$ Interesting convention this backwards arrow, Mariano. $\endgroup$ – Georges Elencwajg Jun 4 '10 at 0:57
  • $\begingroup$ I just did not want to write too much :P I can trace the notation to the additive relations in MacLane's Homology $\endgroup$ – Mariano Suárez-Álvarez Jun 4 '10 at 1:14
  • $\begingroup$ Nice! Now I want to know if this is true $G\times_H\mathfrak g/\mathfrak h \sim G/H \times (\mathfrak g/\mathfrak h)/Ad(H)$ $\endgroup$ – Pedro Jun 4 '10 at 11:57
  • $\begingroup$ 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. $\endgroup$ – José Figueroa-O'Farrill Jun 5 '10 at 19:02

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