Can all G-connections on a Riemann surface X be induced by maps from X to G - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:08:35Z http://mathoverflow.net/feeds/question/18876 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18876/can-all-g-connections-on-a-riemann-surface-x-be-induced-by-maps-from-x-to-g Can all G-connections on a Riemann surface X be induced by maps from X to G Bo Peng 2010-03-20T20:10:34Z 2010-03-21T07:20:20Z <p>There is the invariant Maurer–Cartan 1-form on a compact semi-simple Lie group G. So if we pull it back using a map from X to G then we get a G-connection on X. The question is, can all G-connections on X ( here just regarded as elements in ${\Omega}^1(X, \mathfrak{g})$, i.e. we are using the trivial G-bundle over X ) arise in this way?</p> <p>For flat connections, I believe the answer is yes, which can be seen when one "cut" the Riemann surface into a disc with boundary conditions. But I have no idea about the general case.</p> <p>Some facts that may be useful ( when I tried to figure out an answer ) :</p> <ol> <li><p>It is well-known that all G-connections on X can arise from a map from X to BG by pulling back a special connection on BG.</p></li> <li><p>If <code>${\pi}_1(G) = {\pi}_2(G) = 0$</code>, then I guess there shall be no homotopic non-trivial map from X to G ( note ${\pi}_n(X) = 0$ when $n > 2$ ), and hence all maps from X to G are "integrable" in the sense that they can be induced by applying the exponential map on maps from X to $\mathfrak{g}$. Not sure about how this will help though.</p></li> <li><p>Actually I am not even sure about the $G = U(1)$ case. On the other hand, if the answer is true, then probably the curvature of the induced connection will have some topological constraints coming from the map <code>${\pi}_1(X) \rightarrow \mathbb{\pi}_1(G)$</code>. </p></li> </ol> <p>I haven't touch math for quite some time due to personal reasons, so I hope I am not making any trivial mistakes in this post ;-)</p> http://mathoverflow.net/questions/18876/can-all-g-connections-on-a-riemann-surface-x-be-induced-by-maps-from-x-to-g/18881#18881 Answer by José Figueroa-O'Farrill for Can all G-connections on a Riemann surface X be induced by maps from X to G José Figueroa-O'Farrill 2010-03-20T21:16:00Z 2010-03-20T21:16:00Z <p>The answer is most certainly negative, since the structure equation for the Maurer-Cartan form will tell you that the pullback connection is flat.</p> <p>In fact, I don't really recognise the initial construction that you mention. The connection one-form on the total space of a principal $G$-bundle on $X$ restricts on each fibre to the pullback of the Maurer-Cartan one-form, but it also has a "gauge field" component which is what defines the horizontal distribution.</p> <p>More precisely, let $\pi : P \to X$ be a principal $G$-bundle and let $U \subset X$ be a trivialising neighbourhood. This means that we have a bundle isomorphism $$\pi \times g : \pi^{-1}U \to U \times G.$$ This trivialisation defines a section $s: U \to \pi^{-1}U$ by $g(s(x)) = e$, the identity of $G$, for all $x \in U$.</p> <p>Let $\mathfrak{g}$ denote the Lie algebra of $G$ and let $\omega \in \Omega^1(P;\mathfrak{g})$ be a connection 1-form on $P$. Then $$A = s^*\omega \in \Omega^1(U,\mathfrak{g})$$ is the corresponding gauge field on $U$.</p> <p>The connection one-form $\omega$, restricted to $\pi^{-1}U$ can be reconstructed in terms of the gauge field as <code>$$\omega = \mathrm{Ad}_{g^{-1}} \circ \pi^* A + g^* \theta$$</code> where $\theta$ is the Maurer-Cartan form. There you can see that the part which is the pull-back by the Maurer-Cartan form is only one component of the connection one-form, corresponding to the choice of zero gauge field. And the gauge field can be chosen to be zero (locally) if and only if the connection is flat.</p> http://mathoverflow.net/questions/18876/can-all-g-connections-on-a-riemann-surface-x-be-induced-by-maps-from-x-to-g/18885#18885 Answer by Tom Church for Can all G-connections on a Riemann surface X be induced by maps from X to G Tom Church 2010-03-20T23:38:49Z 2010-03-20T23:38:49Z <p>José Figueroa-O'Farrill has already pointed out one necessary condition, namely that your connection must be flat. The remaining condition is that the monodromy should be trivial. In what follows $X$ is any connected smooth manifold, not necessarily a surface, and $G$ is any Lie group.</p> <p>Let's first consider the analogous situation when $G$ is replaced by $\mathbb{R}$. You can think of a one-form $\omega\in \Omega^1(X;\mathfrak{g})$ as potentially being the derivative of a map $X\to G$, just as a one-form $\eta\in \Omega^1(X;\mathbb{R})$ is potentially the derivative of a map $X\to \mathbb{R}$. We want to know when these really are the derivative of some map, i.e. when we can integrate these forms. (You mentioned the exponential map, but I think integration is the right metaphor here.)</p> <p>There is a local obstruction, namely that if $\eta$ is to be integrable (meaning $\eta=df$ for some $f$) it must be closed, meaning $d\eta=0$; the Poincaré lemma tells us this is a sufficient condition for $\eta$ to be locally integrable. Then there is also a global condition, that the integral of $\eta$ around every closed loop must be 0 (unlike $d\theta$ on the circle, which has integral $2\pi$); Stokes' theorem tells us this is a necessary condition for $\eta$ to be globally integrable. If we have these conditions, recovering the map $f$ from $\eta$ is easy; just write $f(p)=\int_\ast^p \eta$, which is well-defined by the above two conditions.</p> <p>Now let's try to do the same for $\mathfrak{g}$-valued one-forms. Start with a connection on the trivial $G$-bundle $X\times G$ with connection form $\omega\in \Omega^1(X;\mathfrak{g})$. We've talked about the connection being flat, which means that $d\omega+\frac{1}{2}[\omega,\omega]=0$; but what does that have to do with flatness or integrability? Well, you can show that $d\omega+\frac{1}{2}[\omega,\omega]$ measures the Lie bracket of two horizontal vector fields, or rather measures the vertical part of the Lie bracket. Thus if this vanishes, the bracket of two horizontal vector fields is horizontal. By the Frobenius integrability theorem, this implies that the horizontal distribution of the connection is integrable; another way to say this is that parallel transport is locally well-defined. Now pick a basepoint $\ast$ and restrict your attention to a small neighborhood $U$ of $\ast$. Since parallel transport is well-defined on $U$, we get a function $T\colon U\to G$ by saying that the parallel transport from $\ast$ to $u$ (along any path) is multiplication by $T(u)$.</p> <p><strong>Key point:</strong> if you pull back the tautological form on $G$ by this "parallel transport" map $T$, the form you get is the same as your original $\omega$!</p> <p>What this tells us is that if a flat connection on $X\times G$ comes from a map $f:X\to G$, then you can recover $f$ by looking at the parallel transport of the connection. (The analogue is that if $\eta=f^\ast(dx)$ for some $f\colon X\to \mathbb{R}$, you can recover $f$ by integrating $\eta$, also known as the fundamental theorem of calculus.) Thus flatness, in the form of the Maurer-Cartan equation, is the local obstruction to integrability; here the Frobenius integrability theorem plays the role that the Poincaré lemma does in the real case. To prove the key point is really just a matter of definitions: think about the correspondence between a connection, its connection form, and its parallel transport.</p> <p>In particular, this tells us that parallel transport must be not just locally well-defined, but globally well-defined (meaning independent of the path), since transport along any path from $\ast$ to $p$ is always multiplication by $f(p)\in G$. The monodromy of a flat connection is the map $\pi_1(X,\ast)\to G$ which sends a loop to the parallel transport around that loop, and so another way to say "parallel transport is globally well-defined" is that the monodromy is trivial.</p> <p>This can all be summed up by saying that if $X$ is simply connected, we have an on-the-nose bijection <code>$C^\infty(X,G)\longleftrightarrow \{\omega\in \Omega^1(X;\mathfrak{g})\vert d\omega+\frac{1}{2}[\omega,\omega]=0\}$</code>. (Here on the left we assume the maps take the basepoint $\ast\in X$ to $1\in G$.) If $X$ has fundamental group, we need to add on the right side the additional condition that the monodromy of $\omega$ be 0. This is hard to write down just in terms of $\omega$, but for the corresponding connection it is just that parallel transport is totally path-independent.</p> http://mathoverflow.net/questions/18876/can-all-g-connections-on-a-riemann-surface-x-be-induced-by-maps-from-x-to-g/18891#18891 Answer by Paul for Can all G-connections on a Riemann surface X be induced by maps from X to G Paul 2010-03-21T01:34:57Z 2010-03-21T01:34:57Z <p>Said slightly differently, <em>flat</em> connections can be obtained by pulling back the form via an <em>equivariant</em> map $\tilde X\to G$, where $\tilde X\to X$ denotes the universal cover. Equivariance is measured WRT to the covering action of $\pi=\pi_1(X)$ on $\tilde X$, and a homomorphism (the holonomy) $h:\pi\to G$ acting on $G$ by conjugation. This map is often called the developing map. What you get is a section of $T^*\tilde X \otimes g$ which descends to give you a connection on the principal $G$-bundle $\tilde X\times_\pi G\to X$. </p> <p>To link to what Tom said: if the holonomy (or monodromy) is trivial, then the map $\tilde X\to G$ descends to $X\to G$. </p>