Index of elliptic operators III: H-structure invariant under a group G - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:53:38Z http://mathoverflow.net/feeds/question/71560 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71560/index-of-elliptic-operators-iii-h-structure-invariant-under-a-group-g Index of elliptic operators III: H-structure invariant under a group G J. Fabian Meier 2011-07-29T10:28:49Z 2011-11-28T23:22:11Z <p>In the Atiyah-Singer paper mentioned above, they introduced on p.557 a concept called $H$-structure which is used to describe the Chern character of special elements of $K(TX)$. It is roughly the following: Take a principal $H$-bundle $P$, so that $P \times_H V \cong TX$ for some $V$ and use this to construct a map from $K_G(V)$ to $K(TX)$.</p> <p>On p. 559, Remark 2, they describe how this concept can be generalised to $H$-structures with an action of a group $G$. For the special case that $G$ acts trivially on the manifold $X$, they state that </p> <p>"the action of $G$ on the principal $H$-bundle $P$ is given by some homomorphism $\rho:G \to$ Centre($H$)"</p> <p>I actually do not believe this is true (or I misunderstood what kind of actions are considered here). In my opinion we only have fibre-preserving left action of $G$ on $P$, which can be given by any map $G \to H$ (we assume that the base space $X$ is connected). </p> <p>Can somebody familiar with the paper give me argument for (or against) it?</p> <p>PS: Any source where the case of Remark 2 is calculated in special cases would also be helpful.</p> http://mathoverflow.net/questions/71560/index-of-elliptic-operators-iii-h-structure-invariant-under-a-group-g/79656#79656 Answer by Ulrich Pennig for Index of elliptic operators III: H-structure invariant under a group G Ulrich Pennig 2011-10-31T21:39:39Z 2011-10-31T21:39:39Z <p>Maybe I am mistaken, but I think the actions they consider should be such that the projection map $\pi \colon P \to X$ is equivariant with respect to $G$. So, if the action of $G$ on $X$ is trivial (as they state in before the part you quote), this would mean that $p \cdot h \cdot g = p \cdot g \cdot h$ for all $p \in P$, $h \in H$ and $g \in G$. Thus, the action map $G \to H$ would have to have its image in the center.</p>