Is the space of nondegenerate classical paths connected? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T21:35:01Z http://mathoverflow.net/feeds/question/4424 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4424/is-the-space-of-nondegenerate-classical-paths-connected Is the space of nondegenerate classical paths connected? Theo Johnson-Freyd 2009-11-06T19:59:18Z 2009-11-12T17:44:25Z <p>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".</p> <h3>Edit number 2: the question without all the background</h3> <p>In response to Andrew's comments, here's the question I want to ask without all the infinite-dimensional preamble:</p> <p>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 <strong>Euler-Lagrange equation</strong>, 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 $(\dot\gamma(0),\gamma(0))$.</p> <p>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 $(\dot\gamma(0),\gamma(0)) = (v,q)$.</p> <p>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 $\{ (v,q,T) \in C \text{ s.t. } \ker h_{(v,q,T)} = 0\}$.</p> <p>Then I have the following questions:</p> <ol> <li>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.</li> <li>Is $C'$ (path) connected? This was my originally-posed question, and it definitely requires that $b,c$ have compact support.</li> <li>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?</li> </ol> <p>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?</p> <p>The rest is what I wrote before:</p> <h3>Background and definitions</h3> <p>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 <strong>Lagrangian</strong> $L: T\mathbb R^d \to \mathbb R$ 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 <strong>path of length $t$</strong> 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 <strong>action</strong> 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.</p> <p>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 <strong>Dirichlet</strong> paths $\gamma: [0,t] \to \mathbb R^d$ with $\gamma(0) = 0 = \gamma(t)$.</p> <p>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 <strong>classical</strong> paths, i.e. paths $\gamma$ so that <img src="http://latex.mathoverflow.net/png?%24dA%7C%5F%20%5Cgamma%20%5Ccdot%20%5Cxi%20%3D%200%24" alt="$dA|\sb \gamma \cdot \xi = 0$" title="" /> if $\xi$ is Dirichlet (<img src="http://latex.mathoverflow.net/png?%24dA%7C%5F%5Cgamma%24" alt="$dA|\sb \gamma$" title="" /> is the differential of the action at $\gamma$; $\cdot$ is the canonical pairing). Equivalently, $\gamma \in C$ if $\gamma$ satisfies the <strong>Euler-Lagrange equations</strong> $\frac{\partial L}{\partial q}(\dot\gamma,\gamma) = \frac{d}{d\tau}\bigl[ \frac{\partial L}{\partial v}(\dot\gamma,\gamma) \bigr]$. 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 $T\mathbb R^d \times {0}$.</p> <p>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 <strong>nondegenerate</strong> 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 $C$.</p> <h3>My question</h3> <p>Is the space $C'$ (path) connected?</p> <p>Bonus question: what happens if you change the signature of the metric on $\mathbb R^d$?</p> <h3>Edit</h3> <p>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 <strong>harmonic oscillator</strong>, and is exactly solvable). Also, I think with my definitions, paths of length $0$ are always degenerate.</p> <p>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?</p> http://mathoverflow.net/questions/4424/is-the-space-of-nondegenerate-classical-paths-connected/5210#5210 Answer by Semyon Dyatlov for Is the space of nondegenerate classical paths connected? Semyon Dyatlov 2009-11-12T17:44:25Z 2009-11-12T17:44:25Z <p>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).</p> <p>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).</p> <p>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.</p>