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I think this will be an easy question for numerous people.

Let $\pi:P\rightarrow M$ be a principal bundle and $\rho:G\rightarrow GL(V)$ a representation.

The space of $k$ forms on $M$ with values in $P\times_G V$ (denote as $\Omega^k(M;P\times_G V)$ can be identified with with the space of horizontal, right invariant $k$-forms on $M$ (denote as $\Omega^k_G(P;V))$.

Ie, there is an isomorphism:

$\Omega^k_G(P;V)\cong \Omega^k(M;P\times_G V)$.

I am reading through some lecture notes which say

Let $\overline{\zeta}\in \Omega^k_G(P;V)$. Define $\zeta_{\alpha}=s_{\alpha}^*\overline{\zeta}\in \Omega^k(U_{\alpha};V)$. ($s_{\alpha}$ is the local section $s_{\alpha}:U_{\alpha}\rightarrow P$).

It then asks to show that $\{\zeta_{\alpha}\}$ define a form in $\Omega^k(M;P\times_G V)$ by showing that the 'gluing' equation is satisfied

$\zeta_{\alpha}=\rho(g_{\alpha\beta})\circ \zeta_{\beta}$.

Here $g_{\alpha\beta}$ is the transition functions related to the local trivialisations which satisfy $s_{\beta}(m)=s_{\alpha}(m)g_{\alpha\beta}(m)$.

I have managed to show that the required equation holds. My question is - why do the constructed $\zeta_{\alpha}\in \Omega^k(U_{\alpha};V)$ define forms in $\Omega^k(M;P\times_G V)$? I understand that $P\times_G V$ has the structure of a fibre bundle with typical fibre $V$, but I am not sure why the gluing equation is important. I am guessing it has something to do with the fact that because the equation holds, one is able to extend the local definition go a global one. I'm not sure. If someone can help that would be great.

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    $\begingroup$ In general, a vector bundle can be specified by the data of a vector space $V$, a cover $U_\alpha$ of your manifold, and functions $h_{\alpha\beta} : U_\alpha\cap U_\beta \to GL(V)$ satisfying the cocycle condition. Under this definition of a vector bundle, a section is a collection of functions $\zeta_\alpha : U_\alpha \to V$ such that $\zeta_\alpha(x) = h_{\alpha\beta}(x) \zeta_\beta(x)$. The associated vector bundle has the same transition functions as the principal bundle, but now acting via the rep, so the transition functions are $\rho(g_{\alpha\beta})$. $\endgroup$ – Eric O. Korman Mar 16 '15 at 6:19
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    $\begingroup$ So on $U_\alpha$ you just have a trivial vector bundle with fiber $V$ but then the transition functions tell you how to glue (i.e. identify) these trivial vector bundles together on overlaps. To get a global section you then need the sections on each piece to be compatible via the transition functions. $\endgroup$ – Eric O. Korman Mar 16 '15 at 6:22
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This is a situation where generalizing makes things clearer. Suppose that $X$ is a manifold and $G$ is a lie group acting freely and properly discontinuously on $X$. It follows that the quotient $X/G$ exists in the category of manifolds and we have a smooth map $ \rho : X \to X/G$. Assume that $ \pi : E \to X $ is a $G$-equivariant vector bundle. That is, $E$ is a vector bundle with a $G$ action such that $ \pi $ is $G$-equivariant and $ g : E_x \to E_{gx}$ is a linear isomorphism. Note that if $s$ is a section of $E$, then $g s g^{-1}$ is a section of $E$. Therefore $G$ acts on the sections of $E$.

The quotient $E/G$ is a vector bundle on $X/G$. In the quotient $E/G$, $E_x$ and $E_{gx}$ get identified via multiplication by $g$. Moreover, if $ U \subseteq X/G$ is open, then $$ H^0(E/G,U) = H^0(E,\pi^{-1}(U))^G $$ If this is not clear, try and draw some cartoons in the case when $G$ is a finite group. This is a special case of the following more general fact: suppose that $E$ and $F$ are $G$-equivariant vector bundles on $X$. Then $$ {\rm Hom} \, (E/G,F/G) = {\rm Hom}(E,F)^G $$

Now consider the following exact sequence of $G$-equivariant vector bundles on $X$: $$ 0 \to V \to TX \xrightarrow{\rho'} \rho^* T(X/g) \to 0 $$ We call $V$ the vertical subbundle: It consists of vectors which are tangent to the orbits. The isomorphism $ \rho' : TX / V \to \rho^* T(X/G)$ induces an isomorphism $(TX/V)/G \cong T(X/G)$. If $E$ is a $G$-equivariant vector bundle on $X$ then $$ {\rm Hom} \, (T(X/G),E/G) = {\rm Hom} \, ((TX/V)/G,E/G) = {\rm Hom} \, (TX/V,E)^G $$ This says that $E/G$-valued 1-forms on $X/G$ are the same as $E$-valued 1-forms $\eta$ on $X$ such that

  • $\eta(v) = 0 $ for tangent vectors $v \in V$
  • $L_g^* \eta = g \cdot \eta $

extending this to $p$-forms is easy: the first condition becomes $\eta(v_1,\dots,v_p) = 0 $ whenever one of the $v_i$ is tangent to a orbit and the second does not change.

Now let us apply this framework to your problem. Assume that $M$ is a manifold, $G$ is a Lie group, $ \pi : P \to G$ is a principal $G$-bundle and $V$ is a $G$-representation. Then $ P / G = M $. The vector bundle $P \times V$ is $G$-equivariant with $G$-action given by $$ (p,v) \cdot g = (pg,g^{-1}v) $$ The quotient bundle $P \times V / G = P \times_G V $ is a vector bundle on $M$. A $P \times_G V$-valued $p$-form on $M$ is the same as a $V$-valued $p$-form $\eta$ on $P$ such that

  • $\eta(v_1,\dots,v_p) = 0 $ when one of the $v_i$ is tangent to the fiber
  • $R_g^* \eta = g^{-1} \cdot \eta $

The formulas look a little different because $G$ acts on $P$ from the right instead of the left. This is exactly this isomorphism $\Omega^p_G(P,V) \cong \Omega^p( M, P\times_G V) $ which you were asking about.

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Same thing as your other question, here is a note I took: https://www.evernote.com/shard/s318/sh/e4354637-e8a0-4a89-9043-4507303f7006/af3513ed77440494665094c5f4270cc3

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  • $\begingroup$ Thanks for the notes... yes they are good lecture notes but the author's notes have lots of different notation and there are a lot of things I think he takes for granted which I don't understand. Still. they're very good $\endgroup$ – beedge89 Apr 1 '15 at 10:51
  • $\begingroup$ I think this is the price to pay to be more explicit, diff. geo. for gauge theories works with objects that can have several "lives" in different structures $\endgroup$ – ubugnu Apr 1 '15 at 11:44

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