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I have a fairly specific question. My intuition says the answer is "yes", but there is a natural generalizations in which I take out all the "physics", and then I think the answer is "no".

Edit number 2: the question without all the background

In response to Andrew's comments, here's the question I want to ask without all the infinite-dimensional preamble:

On $\mathbb R^d$ with its usual metric, pick a differential one-form $b$ and a smooth function $c$, and suppose that each has compact support. Consider the following (nondegenerate, nonlinear, second-order) differential equation for a path $\gamma(t)$: $$ \ddot \gamma = db \cdot \dot\gamma + dc $$ This is the Euler-Lagrange equation, and so I will abbreviate it as (EL). In coordinates, it is: $$ \ddot \gamma^i = (\partial\_i b\_j - \partial\_j b\_i) \dot\gamma^j + \partial\_i c $$ Since (EL) is nondegenerate and $b,c$ have compact support, every solution to (EL) extends to have domain all of $\mathbb R$, and the solutions are in bijection with the tangent bundle ${\rm T}\mathbb R^d = \mathbb R^{2d}$ by identifying $\gamma$ with $(\dot\gamma(0),\gamma(0))$.

For each $(v,q) \in {\rm T}\mathbb R^d$, define a second-order linear differential operator $h\_{(v,q)}$, given in coordinates by: $$ h\_{(v,q)}[\eta]^j(t) = \ddot\eta^j(t) + \bigl(\partial\_i b\_j|\_{\gamma(t)} - \partial\_j b\_i|\_{\gamma(t)}\bigr) \dot\eta^i(t) + \bigl( \partial\_i \partial\_k b\_j|\_{\gamma(t)} \dot\gamma^k(t) - \partial\_i\partial\_j b\_k|\_{\gamma(t)} \dot\gamma^k(t) - \partial\_i\partial\_j c|\_{\gamma(t)}\bigr) \eta^j(t) $$ where $\gamma$ is the solution to (EL) with initial conditions $(\dot\gamma(0),\gamma(0)) = (v,q)$.

Let $C = {\rm T}\mathbb R^d \times \mathbb R\_{\>0}$. For $(v,q,T) \in C$, consider the operator $h\_{(v,q)}$ as a map $$ h\_{(v,q,T)} : \bigl\{ \eta: [0,T] \to \mathbb R^d \text{ s.t. } \eta(0) = 0 = \eta(T) \bigr\} \to \bigl\{ \eta: [0,T] \to \mathbb R^d \bigr\}$$ Define $C' \subseteq C$ to be the set $\{ (v,q,T) \in C \text{ s.t. } \ker h\_{(v,q,T)} = 0\}$.

Then I have the following questions:

  1. Is $C'$ open (with the topology induced from $C$)? I asserted that it was, because the coefficients of the second-order operator depend smoothly, and I think that kernels can only jump in dimension at closed regions. But I'm not 100% sure.
  2. Is $C'$ (path) connected? This was my originally-posed question, and it definitely requires that $b,c$ have compact support.
  3. Actually, for my research I need that for each $T > 0$, we have $C' \cap {\rm T}\mathbb R^d \times \{T\}$ is path connected. Since $C'$ includes every $\gamma \in C$ with $\gamma([0,T])$ always outside the support of $b,c$, and since this set is path connected if the supports are compact, 3. implies 2., but perhaps 3. is stronger. Also, perhaps 3. does not require that $b,c$ have compact support?

Bonus question: I used the metric exactly once in (EL) and exactly once in (HJ), to compare the folks with raised indices to the ones with lowered indices. Does anything happen if I change the signature of the metric?

The rest is what I wrote before:

Background and definitions

On $\mathbb R^d$ (with its usual metric), pick a differential one-form $b$ and a smooth function $c$. The tangent bundle $T\mathbb R^d$ is just $\mathbb R^{2d}$; define the Lagrangian $L: T\mathbb R^d \to \mathbb R$ by $L(v,q) = \frac12 |v|^2 + b(q)\cdot v + c(q)$, where $v$ is the fiber coordinate on $T\mathbb R^d$, $q$ is the base coordinate on $\mathbb R^d$, and $\cdot$ is the canonical pairing of a one-form with a vector. A path of length $t$ is a smooth map $\gamma: [0,t] \to \mathbb R^d$; it has a canonical lift $(\dot\gamma,\gamma): [0,t] \to T\mathbb R^d$. The action of a path $\gamma$ of length $t$ is the integral $A[\gamma] = \int_0^t L(\dot\gamma(\tau),\gamma(\tau))d\tau$. By adjusting signs, one can include paths of negative length; a path of length $0$ is a point in $T\mathbb R^d$ and has zero action.

Consider the set $P$ of all paths (of arbitrary length); it is an infinite-dimensional smooth manifold. There are various natural projections from $P$ to finite dimensions. The "initial-value map" $P \to T\mathbb R^d \times \mathbb R$ takes a path $\gamma: [0,t]\to \mathbb R^d$ to the triple $(\dot\gamma(0),\gamma(0),t)$. I will be more interested in the "boundary-value map" $P \to \mathbb R^d \times \mathbb R^d \times \mathbb R$ taking $\gamma \mapsto (\gamma(0),\gamma(t),t)$. The fiber over a point in $\mathbb R^d \times \mathbb R^d \times \mathbb R$ is an affine space modeled on the space of Dirichlet paths $\gamma: [0,t] \to \mathbb R^d$ with $\gamma(0) = 0 = \gamma(t)$.

I like to think of the action $A$ as a Morse function on fibers of the boundary-value map. Let $C \subset P$ be the set of classical paths, i.e. paths $\gamma$ so that $dA|_\gamma \cdot \xi = 0$ if $\xi$ is Dirichlet ($dA|_\gamma$ is the differential of the action at $\gamma$; $\cdot$ is the canonical pairing). Equivalently, $\gamma \in C$ if $\gamma$ satisfies the Euler-Lagrange equations $\frac{\partial L}{\partial q}(\dot\gamma,\gamma) = \frac{d}{d\tau}\bigl[ \frac{\partial L}{\partial v}(\dot\gamma,\gamma) \bigr]$. Since the Euler-Lagrange equations are second-order nondegenerate, the initial-value map restricts to a diffeomorphism of $C$ to an open subset of $T\mathbb R^d \times \mathbb R$ containing $T\mathbb R^d \times \{0\}$.

If I really want to think of $A$ as a Morse function, I should require that its critical points (the classical paths) be nondegenerate. Let $\gamma$ be a (classical) path of length $t$, and $V$ the vector space of Dirichlet paths of length $t$. Then the second derivative or Hessian of $A$ is well-defined as a map $H : V \to V^*$. In fact, the Hessian makes sense as a second-order linear differential operator on the space of all paths of length $t$. Let's say that a classical path is nondegenerate if $\ker H = 0$ (or, rather, does not intersect the space $V$ of Dirichlet paths). The set $C'$ of nondegenerate classical paths is an open (I'm pretty sure) subset of $C$.

My question

Is the space $C'$ (path) connected?

Bonus question: what happens if you change the signature of the metric on $\mathbb R^d$?

Edit

The answer to my original question is "no". Let $d = 1$, $b = 0$, and $c(q) = \frac12 q^2$. Then a classical path of length $t$ is nondegenerate if and only if $t$ is not an integer multiple of $\pi$. This is a very nongeneric Lagrangian (it is the harmonic oscillator, and is exactly solvable). Also, I think with my definitions, paths of length $0$ are always degenerate.

So let me ask a more restricted question. Let's suppose that $b$ and $c$ are only supported in a compact neighborhood. Then classical paths that do not enter this neighborhood are precisely the straight lines, and they are all generic (provided $t \neq 0$). Is it true that the space of classical nondegenerate paths with positive length is connected with the restriction that $b,c$ have compact support?

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Apologies for being picky, but you say that P is an infinite dimensional smooth manifold. How are you patching together the pieces with paths of different lengths? –  Loop Space Nov 6 '09 at 20:38
    
@Andrew: hm, that's a good question. Let t, t' be nearby lengths, with t < t', and \gamma,\gamma' paths of length t, t'. I want \gamma,\gamma' to be close if the restriction of \gamma' to [0,t] is close to \gamma. This is equivalent to extending \gamma by a straight line (we are on R^d, so this makes sense) to a path of length t', and then asking for the extension to be close to \gamma'. This does allow functions like the Action to be smooth: its derivative in a direction that changes the length is its usual functional derivative plus a boundary term. –  Theo Johnson-Freyd Nov 6 '09 at 21:58
    
Your paths are smooth so you can't "extend in a straight line". There is an extension map (old result of R. Seeley) which you could use, I suppose. But I'd like to see an actual definition of open sets rather than just "I want this close to that". Infinite dimensional manifolds are tricky things, and limits/colimits like the one you are taking can really mess things up. –  Loop Space Nov 7 '09 at 21:42
    
Oops! Well, for any formal power series, there is a function with those asymptotics, so I can extend to a smooth function. So I'm not too worried. –  Theo Johnson-Freyd Nov 8 '09 at 5:19
    
Anyway, though, all of Andrew's questions are beside the point, because with whatever the correct definitions are, it's true that $C$ is naturally an open subset of $TR^n \times R$, and I'm asking about a subset of it. –  Theo Johnson-Freyd Nov 8 '09 at 5:21
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1 Answer 1

up vote 2 down vote accepted

If you consider a Riemannian manifold with the Lagrangian $L(\nu,q)=|\nu|^2$, where $q$ is a point in the manifold, $\nu$ is a tangent vector, and $|\nu|$ is defined using the metric at the point $q$, then the first variation gives you the geodesic equation and the second variation gives Jacobi fields. This case has been thoroughly studied (for example, it is one of the main topics of the beautiful book "Morse Theory" by J. Milnore). The space of paths $P$ between two fixed points (or the space of loops, which are Dirichlet paths) with the compact/open topology turns out to be homotopy equivalent to a cell complex with dimensions of cells determined by conjugate points (degenerate paths in your case).

In the case of your question, if we put $b=0$, we will get the equation $$\ddot\eta^i=-R_{ij}\eta^j,$$ where $R_{ij}=\partial_{ij}c$. This is exactly the Jacobi equation along a fixed geodesic if we let $R$ be the curvature tensor with respect to the tangent vector of the geodesic. (It may be that in the case $b=0$ we can interpret our problem in the sense of the first paragraph of this answer for a certain Riemannian metric, but I am not sure. If this is true, then the harmonic oscillator should become the round sphere.) We should expect that the curvature will put certain restrictions on when conjugate points occur or do not occur. For example, if $R$ is positive and large, then by Bonnet-Myers Theorem, there exists certain $T_0$ such that for any $\nu$ and $q$, there exists $T\in (0,T_0)$ such that $(\nu,q,T)\not\in C$; therefore, $C$ cannot be connected. This is exactly what happens in the case of the harmonic oscillator and this would happen if we restricted our attention to solutions with bounded value of $|\nu|$, because in that case our solution will stay for time $T_0$ in a fixed compact region, where we could cook up $c$ so that $R$ is very large. However, if $|\nu|$ is too large, then $\gamma$ will escape from our compact region very fast and conjugate points will not have enough time to develop. (In other words, $(\nu,q,T)\in C$ for all $T>0$.) I think that in the nonnegative curvature case, the complement of $C$ should consist of a set of closed hypersurfaces that go to infinity as $|\nu|$ approaches a certain threshold; therefore, $C$ will still not be connected. On the other hand, if $R$ is always negative, then Hadamard-Cartan Theorem gives that all possible points lie in $C$ (thus it is connected).

Here is a particular outcome of the thoughts in the previous paragraph. For each nonnegative integer $k$, consider the set $$ C_k=\{(\nu,q,T)\in C\mid N(T'\in (0,T)\mid (\nu,q,T')\not\in C)=k\}; $$ namely, we count how many degenerate points we had before time $T$. (We might need to introduce some multiplicities.) Then each $C_k$ is open; since $C_k$ form a nonintersecting cover of $C$, they should represent different connected components. The component $C_0$ is always nonempty; if we are able to prove that, say, $C_1$ is nonempty, then $C$ is not connected.

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