4
$\begingroup$

Assume that $P\to M$ is a principal $G$-bundle where $G$ is some (compact) Matrix group. Let $\rho\colon G \to \operatorname{Gl}(\mathbb{R}^n)$ be the tautological representation and $\rho^\prime\colon G\to \operatorname{Gl}(V)$ some other representation.

Let $$ E = P \times_{\rho}\mathbb{R}^n, \qquad \text{and} \qquad F = P\times_{\rho^{\prime}}V $$ be the associated vector bundles.

Using the Chern-Weyl theory and the usual definitions can associate characteristic classes to $P$, $E$ and $F$.

Are they equal?

To make this less vague, assume $G$ is a complex Matrix group and let`s look at $c_1(P)$, $c_1(E)$ and $c_1(F)$. I assume that $c_1(P) = c_1(E)$. Is this also true for $c_1(F)$?

To make it even less vague, consider the Hopf bundle $\mathbb{S}^3 \to \mathbb{S}^2$, and let $\rho_n\colon U(1)\to \operatorname{Gl}(\mathbb{C})$ be the usual irreducible representations for $n\in\mathbb{Z}$. Then $c_1$ of these vector bundles are a complete invariant. Assuming $c_1(\operatorname{Hopf}) = -1$, what is $$c_1\left(\operatorname{Hopf}\times_{\rho_n}\mathbb{C}\right)? $$

I apologize if this is too trivial.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Why should they coincide if $\rho'$ is altogether different? Say, if $V$ is of different dimension... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 29, 2016 at 11:52
  • 1
    $\begingroup$ Fair point, but nevertheless the characteristic classes of $P$ and $\rho^\prime$ should uniquely determine the characteristic lasses of the associated bundle. I can guess a formula for my last example but I have no idea how this can be stated generally. $\endgroup$
    – Montecristo
    Commented Jun 29, 2016 at 12:00

2 Answers 2

Reset to default
4
$\begingroup$

The first Chern class of the dual $-L$ of a line bundle $L$ is the negative of the first Chern class of $L$. In general Chern classes of different associated vector bundles are unrelated, and when they are related the story is complicated. In your example, when you tensor line bundles, the first Chern class scales.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you for the answer. Is it true that the characteristic classes of the tautological asociated bundle agrees with the ones from the principal bundle? $\endgroup$
    – Montecristo
    Commented Jun 29, 2016 at 12:03
  • 3
    $\begingroup$ The Chern classes of the principal bundle arise from the $G$-invariant polynomials on the Lie algebra, while those on each associated vector bundle of rank $N$ arise from $SO(N)$-invariant polynomials on $so(N)$. There might be fewer from the vector bundle, because $G$ sits inside $SO(N)$, so $SO(N)$-invariance is stronger. But there might be fewer on the principal bundle, because $G$ might have a small Lie algebra, as in your example. There is no general rule here. $\endgroup$
    – Ben McKay
    Commented Jun 29, 2016 at 12:30
3
$\begingroup$

Where you seem to be going wrong here is assuming you can talk about the Chern classes of $P$. That's not a well-defined concept (unless $G$ is $GL_n(\mathbb{C})$); at best, you're just fixing $\rho$ and defining the Chern classes of $P$ to be those of $E$.

A principal $G$-bundle is always the pullback of the tautological bundle on $BG$. The Chern classes of the associated bundles $E$ and $F$ are just the pullbacks of the associated bundles to $\rho$ and $\rho'$ on $BG$. So, the Chern classes will always be the same if and only if the associated bundles on $\rho$ and $\rho'$ have the same Chern classes on $BG$. As Ben McKay notes, there are some cases where you can write a formula for one in the terms of the other, but if you don't have a lot of control over the representation theory of $G$, there's no hope.

Improve this answer
$\endgroup$

You must log in to answer this question.

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