Sheaves and Differential Equations - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T07:08:49Z http://mathoverflow.net/feeds/question/9764 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9764/sheaves-and-differential-equations Sheaves and Differential Equations John Mangual 2009-12-26T01:34:58Z 2010-01-23T05:57:30Z <p>How do sheaves arise in studying solutions to ordinary differential equations?</p> <p><strong>EDIT:</strong> Is it possible to construct non-isomorphic sheaves on a domain $D \subset \mathbb{R}^n$ using solution sets to differential equations?</p> <p><strong>EDIT:</strong> Is the sheaf of vector spaces arising from the solution set of a linear ODE necessarily a vector bundle?</p> http://mathoverflow.net/questions/9764/sheaves-and-differential-equations/9767#9767 Answer by Steve Huntsman for Sheaves and Differential Equations Steve Huntsman 2009-12-26T02:02:41Z 2009-12-26T02:02:41Z <p><a href="http://en.wikipedia.org/wiki/Jet%5Fbundle" rel="nofollow">Jet bundles.</a></p> http://mathoverflow.net/questions/9764/sheaves-and-differential-equations/9769#9769 Answer by Vinoth for Sheaves and Differential Equations Vinoth 2009-12-26T02:06:55Z 2009-12-26T02:06:55Z <p>One way is through $D$-modules, perverse sheaves, and the Riemann-Hilbert correspondence. A good reference is: "D-Modules, Perverse Sheaves, and Representation Theory", by Hotta, Takeuchi &amp; Tanisaki. </p> http://mathoverflow.net/questions/9764/sheaves-and-differential-equations/9771#9771 Answer by Ben Webster for Sheaves and Differential Equations Ben Webster 2009-12-26T02:16:25Z 2009-12-26T02:16:25Z <p>Being a solution to a differential equation is a local condition, so solutions to a differential equation are naturally a sheaf.</p> http://mathoverflow.net/questions/9764/sheaves-and-differential-equations/9775#9775 Answer by Mariano Suárez-Alvarez for Sheaves and Differential Equations Mariano Suárez-Alvarez 2009-12-26T03:43:33Z 2009-12-26T04:07:40Z <p>Let $U$ be an open subset of $\mathbb R^n$, and let $X$ be a vector field on $U$. You can construct a sheaf $\mathcal F$ of solutions of the ODE $Xf=0$ by letting $\mathcal F(U)$, for each open subset $V\subseteq U$, be the vector space of all $C^\infty$ functions $f$ on $V$ such that $Xf=0$.</p> <p>By changing the field $X$ you can certainly change the isomorphism clas of $\mathcal F$.</p> <p>Let $U=\mathbb R^2\setminus\{(0,0)\}$, define fields $X_1(x,y)=\Bigl((\frac1r-1)\frac xr-y,(\frac1r-1)\frac yr+x\Bigr)$ and $X_2(x,y)=(y,-x)$ and consider the corresponding sheaves $\mathcal F_1$ and $\mathcal F_2$. It is not difficult show show that $\mathcal F_1(U)$ is one-dimensional as a real vector space, while $\mathcal F_2(U)$ is infinite dimensional. It follows that $\mathcal F_1\not\cong\mathcal F_2$.</p> <p><img src="http://img109.imageshack.us/img109/8818/plots.png" alt="alt text" /></p> <p>Notice that $\mathcal F_1$ and $\mathcal F_2$ are locally isomorphic. This follows easily from the fact that the fields $X_1$ and $X_2$ are non-zero on their domain.</p> http://mathoverflow.net/questions/9764/sheaves-and-differential-equations/11498#11498 Answer by jvp for Sheaves and Differential Equations jvp 2010-01-12T03:35:12Z 2010-01-12T13:34:47Z <p>I will start commenting on Mariano's answer. I believe it is a perfect answer for the question </p> <blockquote> <p>How do sheaves arise in studying solutions of differential equations ?</p> </blockquote> <p>but not for the question </p> <blockquote> <p>How do sheaves arise in studying solutions to <strong>ordinary</strong> differential equations ? </p> </blockquote> <p>According to the current terminology a function $f$ satisfying $X(f)=0$ is not a solution of the vector field $X$ but a first integral. Moreover, if $X = a(x,y) \partial_x + b(x,y) \partial_y$ then $$X(f) = a \partial_x f + b \partial_y f .$$ Thus $X(f)=0$ is a PDE and not an ODE. Indeed t3suji made the same point at a comment on Mariano's answer. I understand the solutions of (the ODE determined by) $X$ as functions $\gamma : V \subset \mathbb R \to U$ satisfying $X(\gamma(t))=\gamma'(t)$ for every $t \in V$. Notice that here indeed we have a system of ODEs. </p> <p>A vector field can be thought as autonomous differential equation and I do not see clearly how to consider the sheaf of its solutions. </p> <p>On the other hand when we have a non-autonomous ordinary differential equation then there is its sheaf of solutions. This sheaf is a sheaf over the time variable only and not the whole space. ( At this point it is natural to talk about connections and/or jet bundles but I will try to keep things as elementary as possible. ) </p> <p>Note that in general the sheaf of solutions will not be a sheaf of vector spaces: the sum of two solutions, or the multiplication of a solution by a constant need not to be a solution. This will occur only when the differential equation is linear.</p> <p>The differential equations $y'(t) = y$ and $y'(t) = y^2$, both defined over the whole real line, are examples of differential equations with non-isomorphic sheaves of solutions. The solutions of the first ODE are the multiples of $\exp t$ and define a sheaf of $\mathbb R$-modules. The solutions of the second ODE are zero and $\frac{1}{\lambda - t}$ with $\lambda \in \mathbb R$. They do define a sheaf of sets, but not a sheaf of $\mathbb R$-modules.</p> <p>To obtain examples of linear differential equations with non-isomorphic sheaves, one has to have nontrivial fundamental group on the time-variable of the differential equation. Thus it is natural to consider complex differential equations over $\mathbb C^{\ast}$. </p> <p>The equations $y'(z) = \frac{ \lambda y(z)}{z}$ parametrized by $\lambda \in \mathbb C$ have non-isomorphic sheaves of solutions. More precisely,</p> <ul> <li>if $\lambda \in \mathbb Z$ then the solution sheaf is the free $\mathbb C$-sheaf of rank one (solutions of the ODE are complex multiples of $z^{ \lambda }$); </li> <li>if $\lambda \in \mathbb Q - \mathbb Z$ then the solution sheaf has no global sections but some tensor power of it does;</li> <li>if $\lambda \in \mathbb C - \mathbb Q$ then the solution sheaf has no global sections nor any of its powers does.</li> </ul>