6
$\begingroup$

I need a bit of clarification about some of the geometry underlying the connection between Lie algebroids and foliations. In case of any confusion I'm using the definition of Lie algebroid from here.

Suppose $\mathcal{F}$ is a foliation on a manifold $M$. For simplicity let's assume it's regular (no singular points). Then the tangent space $T\mathcal{F} \to M$ is a Lie algebroid with anchor map $\rho:~ T\mathcal{F} \to TM$ given by the identity.

If I replace $\rho$ with a different injective mapping $\tilde \rho$, then I get a new foliation which is the image of this map. This is possibly, but not necessarily, isomorphic to the original one.

Question 1: How is this new foliation related to the original one? Is there an interesting non-trivial example?

Let $L$ be a leaf of the foliation determined by a Lie algebroid. When restricted to $L$, the map $\rho$ is a fibrewise surjection. Thus on each leaf it has a right inverse, which is a connection on the bundle $T\mathcal{F} \to M$ (or do we have to restrict to the leaf in question, since leaves may not be isomorphic?).

Question 2: Once we have a connection we can talk about curvature. Am I correct in thinking that the curvature of this connection tells us about how the leaves curve?

Question 3: Suppose the connection is flat. Flat bundles correspond to locally constant sheaves, which correspond to representations of the fundamental groupoid, usually called monodromy representations. The image of the corresponding representation is a groupoid object in some category. If we equip it with an appropriate smooth structure, is this the monodromy groupoid of the foliation? (I know we can get the monodromy groupoid by integrating the algebroid, I was just wondering if this was another approach.) If it is the case, can we do something similar for non-flat algebroids? I don't entirely understand the things I'm talking about here so this may be a stupid question.

Question 4: Whether or not the bundle is flat, we can construct its sheaf of sections, $\Gamma \mathcal{F}$. Does the cohomology of this sheaf tell us anything about the foliation? I'm thinking that non-trivial elements of $H^2(M,~ \Gamma \mathcal{F})$ might have something to do with curvature, but I can't see this concretely.

I've tried to think about question by looking at the foliation on $\mathbb{R}^3$ given by the kernel of the 1-form $dz$. This is a foliation by stacked planes $\mathbb{R}^2 \times \{ z \}$, for each $z$. Obviously these leaves are flat. If I'm correct in my thinking then replacing the anchor map with a new one should produce a foliation whose leaves are still planes, but embedded in $\mathbb{R}^3$ in a different way, perhaps with some curvature. However I have not been able to construct a nice example I can picture.

Can anyone give an example, not necessarily of this form, where we can calculate things like the connection and curvature explicitly? This would really help clarify things.

I know that's a lot of questions, I'd appreciate answers to any of them.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ What do you mean by "the map $\rho$ is surjective on the leaves"? If $\mathcal{L}\subseteq M$ is any leaf, then $\rho:T\mathcal{F}|_{\mathcal{L}}\to TM|_{\mathcal{L}}$ is a map of vector bundles which is still injective on its fibers. $\endgroup$
    – Qfwfq
    Commented Nov 17, 2017 at 0:08
  • $\begingroup$ @Qfwfq A Lie algebroid is called transitive if $\rho$ is fibrewise surjective. The textbook I am using (Mackenzie, General Theory of Lie Groupoids and Lie Algebroids) says that if the image of $\rho$ gives a foliation, then when we restrict the algebroid to a leaf $L$ of the foliation, it is transitive. $\endgroup$
    – Joe Pollard
    Commented Nov 17, 2017 at 10:08
  • $\begingroup$ I don't know, but maybe it's meant that $T\mathcal{F}|_{\mathcal{L}}\to T\mathcal{F}|_{\mathcal{L}}$ is surjective in this case (of course it is), not the analogous map to $TM$ which is certainly not surjective (for a foliation of positive codimension). Am I missing something? $\endgroup$
    – Qfwfq
    Commented Nov 18, 2017 at 23:32
  • $\begingroup$ I have missed a requirement. The Lie algebroid $A \to TM$ is said to be regular if the map $\rho$ (called the anchor) is of locally constant rank. The exact comment in the book is then "If $A$ is regular then the image of the anchor defines a foliation, the characteristic foliation of $A$, and over each leaf of the foliation, the Lie algebroid is transitive." There's no further explanation so I suppose it's meant to be 'obvious', but it's not entirely obvious to me. $\endgroup$
    – Joe Pollard
    Commented Nov 19, 2017 at 20:59
  • 1
    $\begingroup$ I don't think the leaves look the same in general. Consider the distribution generated by $\partial_x$ in the torus $\mathbb{R}^2/\mathbb{Z}^2$, and the foliation $\mathcal{F}_t$ generated by $\partial_{x} + t\cdot \partial_y$ where $t$ is a constant. I haven't thought carefully, but I think the distributions for $\mathcal{F}_0$ and for $\mathcal{F}_{\sqrt 2}$ are isotopic, but the first one has compact leaves while the second has dense leaves. $\endgroup$
    – Qfwfq
    Commented Nov 19, 2017 at 22:32

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .