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Let $M$ be a compact Riemannian manifold and $\Delta$ be the Laplace-Beltrami operator. It is well-known that the solution operator to the heat equation $e^{t \Delta}$ is smoothing for $t>0$ and has a smooth integral kernel $k_t(x, y) \in C^\infty(M \times M)$. Furthermore, $k_t$ has an asymptotic expansion $$ k_t(x, y) \sim \underbrace{(4 \pi t)^{-n/2} \exp \left( -\frac{1}{4t} \mathrm{dist}(x, y)^2 \right)}_{:= e_t(x, y)} \sum_{j=0}^\infty t^j \Phi_j(x, y) $$ meaning that $$ \left| k_t(x, y) - e_t(x, y) \sum_{j=0}^N t^j \Phi_j(x, y) \right| \leq C t^{N+1}$$ uniformly in $x$ and $y$ in a neighborhood of the diagonal.

Now my question is about the Schrödinger equation. The solution operator $e^{it\Delta}$ is not smoothing anymore in this case (as it is unitary), so it cannot have a smooth integral kernel, can it? However, by formally substituting $t \rightarrow it$, one gets the formal asymptotic series $$ e_{it}(x, y) \sum_{j=0}^\infty (it)^j \Phi_j(x, y),$$ but apparently this does not have anything to do with $e^{it\Delta}$?

Honestly, I do not have a precise question, rather a catalogue of questions about this situation:

  • Does this "formal Schrödinger kernel" make any sense?
  • Why is the Schrödinger kernel smooth on $\mathbb{R}^n$ (it is given by $e_{it}(x, y)$), but not an a compact manifold (or am I completely mistaken here)?
  • What can generally be said here to make this situation more clear?
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Why do you think unitary would prevent smoothing? –  Willie Wong Jan 11 '13 at 14:25
    
Because a unitary operator maps $L^2$ to $L^2$ isometrically... –  Kofi Jan 11 '13 at 14:39
    
So? The Schrodinger operator on $\mathbb{R}^n$ is unitary and is smoothing. And for that matter, the heat operator maps $L^1$ to itself preserving norms. Yet it is also smoothing even on manifolds. –  Willie Wong Jan 11 '13 at 14:41
    
I don't see what you mean. The Schrödinger solution operator $U = e^{it\Delta}$ does not map $L^2(\mathbb{R}^n)$ to $C^\infty$, as taking $u := U^{-1}f$ for any non-smooth funtion $f$ gives a contradiction! On the other hand, the heat operator preserves the $L^1$-Norm, but not the $L^2$ norm and is not unitary! –  Kofi Jan 14 '13 at 20:15
    
I meant smoothing in the sense of "having a smooth integral kernel", since that is ultimately what you are interested in. The point is that the smoothing effect as seen from convolving against a smooth integral kernel does apply to a family of functions; it just happens that this family of functions is not the whole of $L^2$. –  Willie Wong Jan 16 '13 at 9:16
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Unitarity has rather little to do with it, as the Schrodinger operator on $\mathbb{R}^n$ is unitary, and for any rapidly decreasing initial data (no regularity assumptions here! just decay ones) we have in fact that the solution is smooth for all positive times.

Compactness, however, of the manifold has quite a lot to do with it. This is because compactness implies that every geodesic is trapped, so we cannot have dispersion to infinity. More precisely:

Consider first the linear wave equation. We know that this equation has propagation of singularities along null geodesics. Roughly speaking we have that all frequencies are transported at the same speed and so if a collection of plane waves add to produce a singularity at time $t$, it will continue to do so at later times.

For the linear Schrodinger equation, the situation is different, the frequencies are not all traveling at the same speed. So if you have a high frequency wave packet and a low frequency one, some time later their spatial support will separate and won't constructively add to a singularity. This is why Schrodinger equation is smoothing for rapidly decaying initial data: if the data is decaying fast, all the action starts out near the origin, and so after some small time the wave packets, which were all originally located near the origin, now burst all over the place and cannot add up to a singularity anymore.

However, if you now try to do Schrodinger's equation on a manifold for which the geodesic flow no longer guarantees that wave packets be transported by distance $\approx |\xi|t$, where $\xi$ is the frequency of the wave packet and $t$ is the elapsed time, then the above smoothing heuristic will no longer work. And in fact, this argument can be made rigorously in the case of non compact, asymptotically flat manifolds. See

Craig, Walter, On the microlocal regularity of the Schrödinger kernel. Partial differential equations and their applications (Toronto, ON, 1995), 71--90, CRM Proc. Lecture Notes, 12, Amer. Math. Soc., Providence, RI, 1997

In the case of a compact manifold, no geodesic can "escape" to infinity, so all wave packets will remain in finite distance of each other. By a covering argument, there will necessarily be points where a infinite number of the wave packets can accumulate and potentially cause the solution to be singular. This intuition has been carried out in special cases. For example, it is known that the Schrodinger kernel on the sphere $\mathbb{S}^d$ is a distribution with singular support in all of $\mathbb{R}\times \mathbb{S}^d$.

You can find more references in this MathOverflow post of Mazzeo.


Edit Let me expand a bit further on my comment, which may give you an answer to your second question.

The main issue is the following: when we think of a "convolution kernel" as a solution to an evolutionary partial differential equation we generally expect the kernel $E_t$ to be in $(C^\infty_c(X))'$, where $X$ is the background manifold. That is to say, we expect $E_t$ to be a distribution for each time $t$. By convolution we can guarantee that for any $v$ a distribution with compact support (in notations, $v\in \mathcal{E}'(X)$) that $E_t*v$ is a distribution, and we have a distributional solution to the Cauchy problem. Now, if for $t > 0$ we have that the singular support of $E_t$ is the empty set, then by the properties of the convolution we have that $E_t*v \in C^\infty(X)$.

This is what I think of as "smoothing", and we see that it is immediately tied to the singular support of the convolution kernel.

Where compactness enters is the following trivial fact:

If $X$ is compact, then $C^\infty(X) = C^\infty_c(X)$, and the space of distributions and the space of distributions with compact support are the same.

We know that $L^2(X) \subset (C^\infty_c(X))'$, that is, $L^2$ functions are locally integrable and can be interpreted as distributions. In general, however, $L^2$ functions do not have compact support. But by the above trivial fact, we have that if $X$ is compact manifold, $L^2(X) \subset \mathcal{E}'(X)$. This implies that a smoothing kernel on a compact manifold will smooth any $L^2(X)$ function. This is what justifies your reasoning that on a compact manifold the Schrodinger kernel cannot be smooth.

On the other hand, this argument breaks whenever $X$ is non-compact. As $L^2(\mathbb{R}^n) \setminus \mathcal{E}'(\mathbb{R}^n)$ is non-empty, the originally defined convolution kernel cannot necessarily be applied to all $L^2$ functions (the convolution of two distributions of non-compact support may fail to be a distribution). For the case of the Schrodinger operators, as it turns out, we can take the convolution and still end up with a distribution, but the uniform estimates required for "smooth kernel implies smooth solution" is no longer true on the whole of $L^2(\mathbb{R}^n)$. Hence in general on a non-compact manifold $X$ one cannot conclude "unitary on $L^2(X) \implies $ lack of smoothing on $\mathcal{E}'(X)$".

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