Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of a vector bundle over P? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T20:45:33Z http://mathoverflow.net/feeds/question/68946 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68946/is-the-lie-algebra-valued-curvature-two-form-on-a-principal-bundle-p-the-curvatur Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of a vector bundle over P? Spencer 2011-06-27T17:33:09Z 2011-12-02T18:32:56Z <p>I am an analyst struggling through some geometry used in physics.</p> <p><strong>Some background:</strong> For some Lie group $G$, let $P$ be a principal $G$-bundle over the smooth manifold $M$. Let $\omega$ be a connection 1-form on $P$ (a "principal connection"). This is a Lie algebra-valued 1-form. </p> <p>As for the curvature two-form, either you see definitions with no explanation at all, e.g. "The curvature is given by $\Omega = d\omega + \frac{1}{2}[\omega,\omega]$". This is obviously less than ideal for improving one's intuitive appreciation. Or one defines something called the exterior covariant derivative $D$ (see <a href="http://en.wikipedia.org/wiki/Exterior_covariant_derivativecurvature?" rel="nofollow">wiki</a>) and then the curvature is simply the exterior covariant derivative of the connection one-form. </p> <p><strong>The issue:</strong> I can't get round the following observation though: From the point of view of the manifold $P$, $\omega$ is just a one-form with values in some vector space which happens to be $\mathfrak{g}$. Usually when you need to covariantly differentiate such an object, you would need a connection in a bundle $E \to P$ with fibre $\mathfrak{g}$, no? Then $\omega$ would be an $E$-valued one-form on $P$, <em>i.e.</em> in $\Gamma(E) \otimes\Omega^1(P)$, and you can differentiate covariantly in the normal way using the connection. Why is this scenario different?</p> <p>The exterior covariant derivative $D$ satisfies $D^2\phi = \Omega\wedge\phi$... So you have some sort of covariant differentiation $D$ which differentiates forms $\eta$ taking values in $\mathfrak{g}$ and for which $D^2$ is some sort of curvature... but of $P \to M$ and not of the bundle in which $\eta$ is taking values. Isn't this strange? Or is this indeed just how things are? This prompts my more precise question: </p> <blockquote> <p>Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of some vector bundle over P with fibre $\mathfrak{g}$?</p> </blockquote> http://mathoverflow.net/questions/68946/is-the-lie-algebra-valued-curvature-two-form-on-a-principal-bundle-p-the-curvatur/68978#68978 Answer by robot for Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of a vector bundle over P? robot 2011-06-27T22:32:02Z 2011-06-27T22:32:02Z <p>A very general treatment of connections is in the (freely accesible) <a href="http://www.emis.de/monographs/KSM/index.html" rel="nofollow">book</a> by Kolář, Michor and Slovák. A connection one-form as the one you have is (I believe) a so called principal connection. Such a thing gives you canonically a connection on any associated bundle to your principal $G$-bundle. I hope this at least gives you a hint. All the details can be found in the mentioned book.</p> http://mathoverflow.net/questions/68946/is-the-lie-algebra-valued-curvature-two-form-on-a-principal-bundle-p-the-curvatur/70168#70168 Answer by Paul Siegel for Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of a vector bundle over P? Paul Siegel 2011-07-12T19:51:41Z 2011-07-12T19:51:41Z <p>To the extent that I can keep all this stuff straight, I find it helpful to inform my intuition with the "path lifting" point of view for connections. </p> <p>Let $\pi: F \to M$ be any smooth fiber bundle. The kernel of the map $\pi_*: TF \to TM$ is the subbundle $VF$ of "vertical vectors"; intuitively, these are the vectors which differentiate only along the fiber directions of $F$. One is led to ask if $VF$ is a direct summand of $TF$; in other words, we ask if there is a global way to write tangent vectors in $TF$ as the sum of a vector which differentiates only along the fiber directions and a vector which differentiates only in directions "parallel to $M$". The answer is yes, but abstractly there is no canonical choice of "horizontal" subbundle $HF$ such that $VF \oplus HF = TF$. Let us think of a connection as a fixed choice of such an $HF$.</p> <p>The advantage of having such an $HF$ is that we can lift paths from $M$ to $F$. Take a smooth path $\gamma(t)$ in $M$, lift $\gamma'(t)$ to a tangent vector $X_t$ in $HF$, and integrate to get a path $\tilde{\gamma}(t)$ which is "parallel" to $\gamma$. Since you're coming from a physics background, you might find it helpful to envision an electron flying around in $\mathbb{R}^3$; if you want to keep track of the phase of its wave function, you need a way to lift paths in $\mathbb{R}^3$ to paths in a $U(1)$-bundle over $\mathbb{R}^3$.</p> <p>Note that the curvature also has a nice geometric interpretation from this point of view: it measures the difference between the Lie bracket of vector fields on $M$ and the Lie bracket of their lifts in $HF$: $\Omega(X,Y) = [X,Y]_H - [X_H, Y_H]$ where $V_H$ refers to the horizontal lift of a vector field $V$ on $M$.</p> <p>Here's how this point of view connects with the language of your question. Instead of viewing a connection as a choice of vector bundle (awkward computationally), express the bundle as the kernel of a bundle map $\omega: TF \to VF$ (also known as a $VF$-valued 1-form) which fixes vertical vectors and view the connection as a choice of such a 1-form. In the case where $F$ is a principal bundle, the fibers are identified with a Lie group $G$ and thus the vertical bundle $VF$ is a bundle whose fibers are identified with the Lie algebra $\mathcal{g}$. It is often implicit when people write "choose a connection on a principal bundle" that the connection should be $G$-invariant in the appropriate sense, so we can think of the connection as a 1-form taking values in just $\mathcal{g}$.</p> <p>So it is not clear to me how one would think of a connection (or its curvature) on a principal bundle $P \to M$ as a differential form with values in a bundle over $P$ - it seems to me that such an object tells us how to lift paths in $P$, not paths in $M$.</p> http://mathoverflow.net/questions/68946/is-the-lie-algebra-valued-curvature-two-form-on-a-principal-bundle-p-the-curvatur/82496#82496 Answer by Robert Bryant for Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of a vector bundle over P? Robert Bryant 2011-12-02T18:11:57Z 2011-12-02T18:32:56Z <p>Your confusion is revealed in this sentence "Or one defines something called the exterior covariant derivative D (see wiki) and then the curvature is simply the exterior covariant derivative of the connection one-form." This is just not true; one does not take the exterior covariant derivative of the connection $1$-form' to get the curvature.</p> <p>Let's be precise: Let $P\to M$ be a principal right $G$-bundle, and let $\omega$ be a $\frak{g}$-valued $1$-form on $P$ that defines a connection on $P$ (I won't repeat the well-known requirements on $\omega$). The curvature $2$-form $\Omega = d\omega +\frac12[\omega,\omega]$ on $P$ is the $2$-form that vanishes if and only if it is possible to find local trivializations $\tau: P_U \to U\times G$ such that $\omega = (\pi_2\circ\tau)^*(\gamma)$ where $\gamma$ is the canonical left-invariant $1$-form on $G$.</p> <p>Now, given a representation $\rho:G\to \text{GL}(V)$, where $V$ is a vector space, one can define an associated vector bundle $E = P\times_\rho V$. Using $\omega$, it is possible to define an 'exterior covariant derivative operator' $D_\omega:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+1})$ (where $A^p\to M$ is the bundle of alternating (i.e., 'exterior') $p$-forms on $M$). The operator ${D_\omega}^2:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+2})$ then turns out to be linear over the $C^\infty$ alternating forms, so it is determined by its value when $p=0$, i.e., by ${D_\omega}^2:\Gamma(E)\to \Gamma(E\otimes A^{2})$, which can be regarded as an section of $\text{End}(E)\otimes A^2$, i.e., a $2$-form with values in $\text{End}(E)$. The formula for this section, when pulled back to $P$, can now be expressed in terms of $\Omega$ in the usual way. In particular, ${D_\omega}^2$ vanishes identically if $\Omega$ does.</p> <p>Note that one does <em>not</em> take the 'exterior covariant derivative' of the $1$-form $\omega$ anywhere. Instead, one takes the exterior covariant derivative of the exterior covariant derivative of a section of $E$.</p> <p>I suspect that what you may be trying to do is interpret $\omega$ as a $1$-form on $P$ with values in the trivial bundle $P\times\frak{g}$ and then say that the curvature is the 'exterior covariant derivative' of $\omega$. However, to make this work, you have to specify a connection on the trivial bundle $P\times\frak{g}$, which amounts to choosing a $1$-form $\eta$ on $P$ that takes values in $\text{End}(\frak{g})$ and setting $D_\eta(s) = ds + \eta\ s$. By setting $\eta = \frac12\text{ad}(\omega)$, one gets $D_\eta\omega = \Omega$ (just by definition), so it is possible to do this, but I don't think that this is that useful an observation, since, after all, you could have taken the connection on the trivial bundle $P\times\frak{g}$ to be $\eta = \frac13\text{ad}(\omega)$ (for example) or even $\eta=0$. What justifies the $\frac12$, other than the desire to get the right' answer?</p>