Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A Euclidean metric on a manifold $X$ defines a function on the symplectic space $T^*X$ whose Hamiltonian flow gives geodesics. Is there a similar interpretation of the Levi-Civita connection?

share|improve this question
    
I am just guessing that what you have in mind is to take a vector field $X$ and a 1-form $\xi$, define a function $H_x(p)=<p,X>$ on $T^* M$, and a function $H_{\xi}(p)=<p,\xi>$ using the metric, and then perhaps $[H_{\xi},H_X]$ is related to $H_{\nabla_X \xi}$. Is that the idea? –  Ben McKay Apr 12 '13 at 7:27
    
@Ben: I'm afraid that the problem with trying to relate $[H_\xi,H_X]$ with $H_{\nabla_X\xi}$ is that the former expression is a little too symmetric with respect to $X$ and $\xi$, involving the same number of derivatives of each one, which the covariant derivative does not. In my answer below, I outline what I think is probably the simplest way to get to the Levi-Civita connection on the cotangent bundle. The key is to pull the metric $g$ up to $T^\ast M$ as a quadratic form as well as to regard it as defining a Hamiltonian function on $T^\ast M$. The interaction of the two does the trick. –  Robert Bryant Apr 14 '13 at 1:27

3 Answers 3

up vote 6 down vote accepted

The intuition is that the Levi-Civita connection corresponds to the linearization of the geodesic flow plus a simple projective-geometric construction:

Let $c(t)$ be an orbit of the geodesic flow (projecting down to a geodesic), consider the vertical subspaces $V(t)$ along $c(t)$ and bring them back to the tangent space of the cotangent bundle over the point c(0) by using the differential of the flow. You get a family of (Lagrangian) subspaces $l(t) := D\phi_{-t}(V(t))$ that is "fanning" or "regular".

Now forget you ever had a geodesic flow: all that you need is the curve of subspaces. A bit of differential projective geometry shows that you also get a second curve $h(t)$ of (Lagrangian subspaces) in $T_{c(0)}(T^*M)$that is transversal to $l(t)$. The subspace h(0) is the horizontal subspace of the connection and $T_{c(0)}(T^*M) = l(0) \oplus h(0)$ is the decomposition into vertical and horizontal subspaces.

I think this is nicely written in this paper ;-)

However, this is classical: it's just the geometry behind the Schwartzian derivative. There are plenty of references in the paper if you're interested in this.

Addendum on the geometry of the Schwartzian derivative.

Since the Schwartzian derivative has many "standard" geometric interpretations, for completeness sake I'll sketch the (new?) one that I'm refering to. I will do it in a way just a bit more conceptual than was done by Duran and myself in the paper. Actually, it's just a "high brow" version of what was done in the paper, but I will do it here just for curves on the projective line. The reader may amuse himself/herself at extending this to curves on the Grassmannian of $n$-planes in $\mathbb{R}^{2n}$ and checking against the paper.

Consider the action of the linear group $GL(2;\mathbb{R})$ on the projective line and lift it to an action on its cotangent bundle. The moment map of this action takes values on the set of nilpotent matrices. In fact, if you take away the zero section, the moment map takes value in the coadjoint orbit formed by $2 \times 2$ nilpotent matrices of rank one. In higher dimensions, one needs to take away a bit more than the zero section and "rank one" is replanced by "the kernel and the image of the matrix (seen as linear transformation) are the same".

We will also need a neat thing about the geometry of the tangent space of the projective line: there is a canonical way to identify non-zero tangent vectors to non-zero covectors. In fact, the tangent space of the projective line at a line $\ell$ is the space of linear maps between $\ell$ and the quotient space $\mathbb{R}^2/\ell$, which is the tensor product $(\mathbb{R}^2/\ell) \otimes \ell^* $. If the map is invertible, its inverse is a map from $\mathbb{R}^2/\ell$ to $\ell$ and, therefore, an element of $ \ell \otimes (\mathbb{R}^2/\ell)^*$, which is the cotangent space at $\ell$.

Now consider a curve $\gamma(t)$ on the projective line whose derivative never vanishes. We lift the curve to the curve $\dot{\gamma}(t)$ on the tangent bundle and use the isomorphism described above to obtain a curve $\Gamma(t)$ on the cotangent of the projective line. Use the moment map to obtain a curve $F(t)$ of nilpotent matrices. Note that everything we have done is projective-equivariant.

Finally we come to the little miracle: the time derivative of $F(t)$ is a curve of reflections $\dot{F}(t)$ (i.e., $\dot{F}(t)^2 = I$) whose -1 eigenspace is the curve of lines $\gamma(t)$ and whose $1$-eigenspace defines a "horizontal curve" $h(t)$ equivariantly attached to $\gamma(t)$. This is the construction that yields the Levi-Civita connection (and what is behind the formalisms of Grifone and Foulon for connections of second order ODE's on manifolds).

Differentiate $F(t)$ a second time to find the Schwartzian derivative. Geometrically, it just describes how the curve $h(t)$ moves with respect to $\gamma(t)$. For comparison, recall that the curvature of a connection is obtained by differentiating (i.e., bracketing) horizonal vector fields and projecting onto the vertical bundle.

share|improve this answer
    
Thank you! I looked at the paper. In terms of its notation, are you saying that the horizontal section is the 1-eigenspace of the derivative of the fundamental endomorphism? –  Dmitry Vaintrob Apr 13 '13 at 4:28
    
Yes. If you look at the papers of Grifone and Foulon this is exactly what they do. Duran and I just provided the geometric meaning that their construction had at the level of the projective geometry of curves in the Grassmannian. –  alvarezpaiva Apr 13 '13 at 15:18

I think that you are asking for a symplectic interpretation of the splitting of the tangent bundle of $T^\ast M$ that is induced by the Levi-Civita connection on a Riemannian $n$-manifold $(M,g)$. I.e., to know how one can see, from the symplectic geometry of the cotangent bundle and the information of the Hamiltonian, the splitting $T(T^\ast M) = D\oplus V$, where $D$ is the $n$-plane field that is transverse to $V$, the tangent bundle to the fibers of $\pi:T^\ast M\to M$, that has the property that parallel transport of covectors with respect to the Levi-Civita connection gives curves in $T^\ast M$ that are tangent to $D$.

There is a simple way to get $D$ in this case: Set $\gamma = \pi^\ast g$, and let $\dot \gamma$ be the Lie derivative of $\gamma$ with respect to the Hamiltonian vector field associated to the Hamiltonian $H$ computed from $g$. Then $\dot\gamma$ is a nondegenerate quadratic form on $T^\ast M$ of split type $(n,n)$, and one can easily show that $D\subset T(T^\ast M)$ is the unique $n$-plane field that is Lagrangian, transverse to $V$, and $\dot\gamma$-null.

Historical Background: This sort of construction was discussed thoroughly by Patrick Foulon in his (mid-1980s) thesis, and, yes, something like it works more generally for pseudo-Riemannian and Finsler geometries (see below). The key extra ingredient, besides knowing the symplectic form $\omega$ and the Hamiltonian $H$, is that one also needs to know the Lagrangian foliation $\mathcal{F}$ that is defined by the fibers of $\pi:T^\ast M\to M$. With just these, one can construct the complementary $n$-plane field $D$ (which, of course, is Lagrangian but not integrable) in a canonical manner. Thus, one knows what it means to have a 'parallel transport' in these more general cases.

É. Cartan did something along these lines in his book on the geometry of Finsler spaces, but it's not written in a modern manner and so it's a bit hard to read for most geometers nowadays.

Here is one way to describe the construction: The starting data are $(X,\omega,\mathcal{F},H)$, where $X$ is a manifold of dimension $2n$, $\omega$ is a symplectic form on $X$, $\mathcal{F}$ is an $\omega$-Lagrangian foliation, and $H$ is a function on $X$. The goal is to construct an $n$-plane field $D\subset TX$ that is $\omega$-Lagrangian and transverse to $\mathcal{F}$ in a canonical way. This will require making a nondegeneracy assumption on $H$; for example, this clearly cannot be done if $H$ is constant.

Now, the data $(X,\omega,\mathcal{F})$ has no local geometry, in the sense that all Lagrangian foliations of a symplectic manifold are locally equivalent. This means that, locally, one can always choose canonical coordinates $(q,p)=(q^i,p_i)$ in which the symplectic form $\omega$ is $dp_i\wedge dq^i$ and the leaves of $\mathcal{F}$ are given by $dq^i=0$. Let's consider the derivatives of $H$ in these coordinates: Define $H^i$ and $H^{ij}=H^{ji}$ by the conditions $$ dH \equiv H^i\ dp_i\ \text{mod}\ dq\qquad\text{and}\qquad dH^i \equiv H^{ij}\ dp_j\ \text{mod}\ dq $$ It is easy to check that, if one were to make a different choice $\bar q = F(q)$ for the $\mathcal{F}$-null coordinates and complete to canonical $\omega$-coordinates with a corresponding new $\bar p$, then the Jacobian change of variables formula yields $$ \bar H^{ij} d\bar p_i\circ d\bar p_j \equiv H^{ij}\ dp_i\circ dp_j\ \text{mod}\ dq $$ so that the quadratic form $\eta_H = H^{ij}\ dp_i\circ dp_j$ is well-defined on the leaves of $\mathcal{F}$. (What is going on here is that, for any Lagrangian foliation $\mathcal{F}$ on a symplectic manifold $M$, there is a canonical, torsion-free flat connection on each leaf of $\mathcal{F}$. The quadratic form $\eta_H$ is merely the leafwise Hessian of $H$ with respect to this canonical connection.)

Let us say that $H$ is $\mathcal{F}$-nondegenerate if $\eta_H$ is a nondegenerate quadratic form (i.e., pseudo-Riemannian metric) on each $\mathcal{F}$-leaf.

From now on, assume that $H$ is $\mathcal{F}$-nondegenerate. Let $G = (G_{ij})$ be the inverse matrix of $(H^{ij})$. The change of variables formula then shows that the quadratic form $$ \gamma_H = G_{ij}\ dq^i\circ dq^j $$ is well-defined on $X$, independent of the choice of coordinates. (What is going on here is that, for each $x\in X$, with $V_x\subset T_xX$ being the tangent to the $\mathcal{F}$-leaf through $x$, one has a canonical isomorphism ${V_x}^\ast \simeq T_xX/V_x$, and $\gamma_H$ at $x$ is simply the canonical quadratic form on $T_x/V_x$ induced by this isomorphism from the canonical dual quadratic form $\eta_H^\ast$ (on ${V_x}^\ast$) of the quadratic form $\eta_H$ on $V_x$.)

Let $\dot\gamma_H$ denote the Lie derivative of $\gamma_H$ with respect to the Hamiltonian vector field $X_H$. One can now easily verify that $\dot\gamma_H$ is a nondegenerate quadratic form of type $(n,n)$ on $X$. Moreover, there is a unique $n$-plane field $D\subset TX$ on $X$ that is transverse to the leaves of $\mathcal{F}$, Lagrangian with respect to $\omega$, and null with respect to $\dot\gamma_H$. This $D$ is the desired splitting.

Example: As stated at the beginning, in the case that $g$ is a pseudo-Riemannian metric on $M^n$, when one defines the associated Hamiltonian $H$ on $X=T^\ast M$, then one finds that $\gamma_H$ is simply ${\pi^\ast}g$, and then $D$ is indeed the plane field on $T^\ast M$ induced by the Levi-Civita connection.

Remark: Note that, in the pseudo-Riemannian case, $\gamma_H$ is expressed in terms of the original metric using no derivatives, so that $D$, which is defined algebraically from $\dot\gamma_H$, uses only one derivative of the metric, which is as it should be, since the Levi-Civita connection depends on only one derivative of the metric. However, in the general case outlined above, $\gamma_H$ depends on two derivatives of $H$ and then $\dot\gamma_H$ depends on three derivatives. Thus, for example, in the general case, the so-called 'nonlinear connection' in Finsler geometry depends on three derivatives of the Finsler structure, and 'curvature' depends on four derivatives. This is one reason that the general Finsler case is more difficult to study than the Riemannian case.

share|improve this answer
    
Thanks! Yes, this is exactly what I need. Do you have a link for Foulon's thesis? –  Dmitry Vaintrob Apr 13 '13 at 4:13
    
@Dmitry: As far as I know, Foulon's thesis is not online. I think you can find what you want, though, in P. Foulon, Géométrie des équations différentielles du second ordre, Ann. Inst. H. Poincaré Phys. Théor. 45 (1986), 1–28. –  Robert Bryant Apr 13 '13 at 13:12

I do not know whether this is the answer you want since in my understanding it is well known: If we have a Riemannian metric $g$ on $M$, i.e. a (nondegenerate) bilinear form on each $T_xM$, then we can canonically identify the tangent and cotangent bundles $T_xM \stackrel{g}{=}T^*_xM$ and thus have an euclidean structure on $T^*_xM$ which we denote by $g^*$. In the coordinates, the matrix of this euclidean structure is the inverse of $g$.

Now consider the Hamiltonian on $T^*M$ , $H:T^*M\to R$, $H(x, p)= g_x^*(p, p)$. In the coordinates, $H= g^{ij}p_ip_j$. Then, the projection of the solutions of the Hamilton system of ODE are geodesics, exactly as in the euclidean case.

One can obtain the above observation by direct calculations. If you have experience with variational theory, the explanation of the above observation is as follows: geodesics are locally shortest curves so they are extremals of the corresponding Largange functional, and therefore correspond to the solutions of a Hamilton equation.

Similar things are true in the pseudo-Riemannian case and in the finsler cathegory.

share|improve this answer
    
@Vladimir: I think that what Dmitry wanted was to understand how to get the 'rest' of the Levi-Civita connection using symplectic geometry, since using the Hamiltonian to define the geodesic flow seems to give one only the covariant derivative of the velocity vector (or, more precisely, the momentum covector) of a curve along itself; it doesn't (apparently) tell one how to parallel transport other (co-)vectors along the curve. In fact, except in the pseudo-Riemannian case, one has to go to higher derivatives to complete the 'nonlinear connection', as the discussion in my answer indicates. –  Robert Bryant Apr 17 '13 at 12:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.