Consider the following pair of principal bundle descriptions of $\mathbb{CP}^2$: $$ \mathbb{CP}^2 \simeq SU(3)/U(2) \simeq S^5/U(1). $$ If I have a principal $U(2)$-bundle connection for $\mathbb{CP}^2$, will that correspond to a principal $U(1)$-bundle connection? (Where by correspond I suppose I mean give the same covariant derivative.)

I am also interested in how this generalises to the $n$-case $$ \mathbb{CP}^{n} \simeq SU(n+1)/U(n) \simeq S^{2n+1}/U(1). $$

  • $\begingroup$ "Principal bundle" :) $\endgroup$ – Q.Q.J. Feb 17 '10 at 21:39

The answer is no. You can see that for example from the connection one forms which are Lie algebra valued. In the first case they are u(2) valued and in the second case they are u(1) valued. However, in the case of CP2 (which also generalizes to CPn), the U(1) and SU(2) factors of the isotropy group U(2) commute, this means that given a U(2) connection on U(2)-->SU(3)-->CP2, you can project it to the U(1) factor to obtain a U(1) connection on U(1)-->S^5-->CP2, such that both connections will have the same horizontal subspaces.

| cite | improve this answer | |
  • $\begingroup$ I am I right in assuming then that any $U(1)$-connection induces a $U(2)$ connection via the embedding $\mathfrak{u(1)} \to\mathfrak{u(2)}$? $\endgroup$ – Aston Smythe Jan 26 '10 at 23:39
  • $\begingroup$ It is possible. You can induce a connection on an associated bundle, thus if you define an action of U(1) on U(2), then you get an associated U(2) bundle on CP2 and you can compute the induced connection. In our case, the action of U(1) on U(2) is trivial because they commute, thus the induction will be trivial. However, if you want to get any U(2) bundle over CP2 (such that the total space is not necessarily SU(3)), then you may define another action of U(1) on U(2) then induce. $\endgroup$ – David Bar Moshe Jan 27 '10 at 4:29

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.