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Let $P =M\times G\to M$ be a principal $G$-bundle on $M$(first coordinate projection)

What is $ad(P)$?

Here $ad(E) = E\times_{Ad}g$ is a vector bundle on $M$ [where $g$ = Lie$(G)$, $E$ is any principal $G$-bundle on $M$] . My expectation is that it is the trivial bundle on $M$.

Any element in $ad(P)$ is $[(m,g),v] \sim [(m,e).g,v] \sim [(m,e), Ad_{g}v]$. I need an isomorphism from $ad(P) \to M\times g$. Here $e$ is the identity element in $G$.

All spaces are smooth manifolds and groups are Lie groups.

Edit : I am adding the wiki link for adjoint bundle to avoid any confusion. https://en.wikipedia.org/wiki/Adjoint_bundle

Angry_Math_Person
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  • what is $(m,e) $ ? I don't get your definition. Can you write down which variables holds in which space ? – InfiniteLooper Feb 25 '22 at 17:33
  • e is the identity of G. m is any element of M. Sorry if I wasn't too clear with my notation. – Angry_Math_Person Feb 25 '22 at 17:34
  • G acts on $M \times G$ as right multiplication on second factor. – Angry_Math_Person Feb 25 '22 at 17:35
  • I mean edit your question. In the end you want a vector bundle on $G \times M$ ? Sorry but it's really unclear. – InfiniteLooper Feb 25 '22 at 17:40
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    Why on $G \times M$? The adjoint bundle is a vector bundle on the base space which is $M$. The principal bundle is the trivial one namely $M\times G$. I'm asking what is the adjoint bundle explicitly? – Angry_Math_Person Feb 25 '22 at 17:44
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    The adjoint bundle corresponding to a principal G bundle is a new vector bundle on the base space. Not on the starting bundle. – Angry_Math_Person Feb 25 '22 at 17:46
  • For a general principal bundle on $M$ you don't have a map $M \times G \to M$. It's still rather unclear, can you edit your question please ? – InfiniteLooper Feb 25 '22 at 17:50
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    I don't have a general principal bundle. I have the trivial bundle $M \times G$ on $M$ which is the first coordinate projection. This has a corresponding adjoint bundle which is a vector bundle on $M$. I am asking what is that vector bundle? I have edited the question but I'm not sure what exactly you are finding unclear. – Angry_Math_Person Feb 25 '22 at 17:54
  • @InfiniteLooper, if $G$ acts on a vector space, then there is an induced vector bundle. Details can be found, for example, here: https://math.stackexchange.com/questions/649142/how-a-principal-bundle-and-the-associated-vector-bundle-determine-each-other – Deane Feb 25 '22 at 18:04
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    Hint: Try to show that the vector bundle is trivial, no matter what representation of $G$ you use (i.e., it's not special to $Ad$). As a further hint, given a representation $V$ of $G$, pick a basis of $V$ and use that to create independent sections, trivializing the bundle. – Jason DeVito - on hiatus Feb 25 '22 at 18:04
  • Correct me if I'm wrong. Here's the map from the associated bundle for any representation to the trivial vector bundle. Send $[(x,g),v]$ to $(x,g.v)$ Here $g.v$ is the representation action of $g$ on $v$ – Angry_Math_Person Feb 25 '22 at 18:13
  • I thought by "a pincipal bundle" $P$ was any pincipal bundle was intended. My english may be bad, sorry. – InfiniteLooper Feb 25 '22 at 19:04

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Ok, you got everything right so far and your guess is spot on. We need an isomorphism between the vector bundles ${\rm Ad}(M\times G)\to M$ and $M\times\mathfrak{g}\to M$. Let's take $\Phi \colon M\times\mathfrak{g} \to {\rm Ad}(M\times G)$ given by $$\Phi(x,v) = (x,[(x,e),v]).$$The inverse will be $\Phi^{-1}\colon {\rm Ad}(M\times G) \to M\times \mathfrak{g}$ given by $$\Phi^{-1}(x,[(x,g),v]) = (x, {\rm Ad}(g)v).$$

Clearly $\Phi^{-1}$ is well defined, since by replacing $g$ with $gh$ and $v$ with ${\rm Ad}(h^{-1})v$ we have that ${\rm Ad}(gh){\rm Ad}(h^{-1})v = {\rm Ad}(g)v$. It should not be hard to see that they both act linearly on fibers.

Now, let's check that these maps are indeed inverses: $$\begin{align} \Phi\Phi^{-1}(x,[(x,g),v]) &= \Phi(x, {\rm Ad}(g)v) \\ &= (x,[(x,e), {\rm Ad}(g)v]) \\ &= (x, [(x,g),v]),\end{align}$$and $$\begin{align} \Phi^{-1}\Phi(x,v) &= \Phi^{-1}(x, [(x,e),v]) \\ &= (x, {\rm Ad}(e)v) \\ &= (x,v). \end{align}$$

In fact, there's nothing special about ${\rm Ad}$ and $\mathfrak{g}$ here, the same argument shows that if $M\times G \to M$ is the trivial principal $G$-bundle and $\rho\colon G \to {\rm GL}(V)$ is a representation of $G$ on a vector space $V$, then $\rho(M\times G) \cong M\times V$.

Ivo Terek
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    Yeah. I worked it out just now. And as mentioned in the comments, we don't need it to be adjoint representation. Any representation will work. Thank you nonetheless. – Angry_Math_Person Feb 25 '22 at 18:16
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    Ok, great! I ended up answering without looking at the comments, but it's good that more people are confirming what happens here. – Ivo Terek Feb 25 '22 at 18:16