1
$\begingroup$

I'm reading Atiyah-Bott's paper "The Yang-Mills equations over Riemann surfaces" and have a couple of questions on page 547. They define a connection $A$ as a $G$-invariant splitting of the exact sequence $ 0 \rightarrow T_FP \rightarrow TP \rightarrow \pi^{-1}TM \rightarrow 0 $. I'm not sure how the group $G$ acts on $\pi^{-1}TM$.

My guess is that if $h_p = (p,\psi_{\pi(p)})$ is in this space (where $p \in P$, and $\psi_{\pi(p)} \in T_{\pi(p)}M$) then $g \cdot h_p = (pg, \psi_{\pi(pg)=\pi(p)})$ . And I guess $G$-invariance means that if $X_p \in T_pP$ is uniquely written as $V_p + H_p$ (for $V_p \in T_FP_p$ and $H_p$ the image by $\gamma$, the right inverse of $\pi_*$, of an element in $\pi^{-1}TM$, say $h_p$ ) then $G$-invariance means that $g v_p$ is uniquely written as $gX_{p} + g H_p = X_{pg} + \gamma(g h_p) $.

On the same page they want to see the above exact sequence as a pull-back sequence of bundles over $M$. The middle term comes from a bundle $E(P)$ whose fibre over a point $\pi(p)$ should consist of the fibre $T_pP$, but instead they say it is given as, $E(P)_m = \Gamma \{ TP|_{\pi^{-1}(m)} \}^G$. Why are they adding the $G$-invariant?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

The bundle $\pi^{-1}TM$ is the set of tuples $(p,m,v)$ so that $m \in M$, $p \in \pi^{-1}\{m\}$, and $v \in T_m M$. The $G$-action is $g(p,m,v)=(gp,m,v)$, only acting on $p$. So you are correct. They are correct (of course) that $E(P)_m$ should consist of the $G$-equivariant sections of $TP$ along the fiber $P_m = \pi^{-1}\{m\}$. In order to make bundles descend to $M$, not living on $P$, we have to work with $G$-equivariant objects along fibers over each point of $M$. That way we can work on $M$, not on $P$. (Not how I think about principal bundles and connections, but certainly something Atiyah always does.)

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks Ben. I'm not sure I understood the last bit, I understand that there is a 1-1 correspondence between sections of the vector bundle and equivariant functions on $P$. But here we are not making things descend to $M$ but to pullback bundles from $M$ to $P$. After a long though I came to the conclusion that we need the $G$-invariant vector fields because the requirement of the $G$-invariant splitting implies that the vector fields of $TP$ must be $G$-invariant (this follows directly from what I wrote above $gv_p = X_{pg} + H_{pg} = v_{pg}$). Please let me know if you agree with this $\endgroup$
    – Raul
    Commented Sep 18, 2016 at 8:10
  • $\begingroup$ No. We ARE making things descend. They even emphasize in italics that this is an exact sequence on $M$. You start with the sequence on $P$: $0 \to T_F P \to TP \to \pi^{-1}TM$. You then look at the $G$-invariant sections of those bundles over each point of $M$, to construct bundles on $M$: $0 \to ad(P) \to E(P) \to TM \to 0$. So $ad(P)$ means the bundle on $M$ whose local sections over an open subset $U_M \subset M$ are the $G$-equivariant local sections of $T_F P$ defined on the set $U_P=\pi^{-1}U_M$. $\endgroup$
    – Ben McKay
    Commented Sep 18, 2016 at 11:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .