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Nov 14, 2019 at 3:01 comment added Student Combining both of your comments, it seems to me that larger eigenvalues tell us more about global properties, while the smaller eigenvalues tell us more about the local. Is that a bad way to look that it as I do? (Also, I would love to see a reference for a proof of the fourth-order-jet statement.. thank you!)
Nov 11, 2019 at 11:02 comment added Liviu Nicolaescu On a compact Riemann manifold, the metric and its curvature are determined by the behavior as $t\searrow 0$ of the fourth order jets of the heat kernel along the diagonal.
Nov 11, 2019 at 7:16 comment added Jochen Glueck [continued] (see Can one hear the shape of a drum?). On the other hand, as mentioned by @LiviuNicolaescu, it is known that the heat flow determines $\Omega$ in the sense that if there is a lattice isomorphism $L^2(\Omega_1) \to L^2(\Omega_2)$ that intertwines the Dirichlet Laplace operators, then $\Omega_1$ and $\Omega_2$ are congruent.
Nov 11, 2019 at 7:16 comment added Jochen Glueck A few more comments about spectral properties and their relation to the geometry of the underlying domain (for the sake of simplicity, I'll stick to bounded domains $\Omega \subseteq \mathbb{R}^d$ here): simple geometric properties of $\Omega$ are encoded in the large eigenvalues of the Laplace operator rather than in the small ones; see Weyl's law. Moreover, it is a major open problem whether two smooth domains $\Omega_1$ and $\Omega_2$ for which all eigenvalues of the Dirichlet Laplace operator coincide are always congruent.
Nov 11, 2019 at 7:06 comment added Jochen Glueck @Student: Yes, I mean that the eigenvalues of $T$ should be located "on the right" rather than "on the left" if $-T$ is supposed to generate a $C_0$-semigroup. Details can be found in the Hille-Yosida theorem already mentioned by Liviu Nicolaescu.
Nov 10, 2019 at 9:27 comment added Liviu Nicolaescu Ona compact manifold, the heat kernel determines the metric.
Nov 9, 2019 at 12:38 comment added Student Another aspect that interests me: Under some settings, the lower eigenvalues tell us more about the topology of the space. And as the parameter $t$ flows to infinity, the higher eigenvalues' contributions decrease much faster than the lowers'. Through this process we might be able to capture some topological data of the original underlying space. I'd hope to see a more rigorous treatment about this, and perhaps even arguments that tell us exactly how much topological details can we see from that process.
Nov 9, 2019 at 12:35 comment added Student @LiviuNicolaescu that's a useful pointer! Suppose everything is good can I safely say: we can then use Hille-Yosida theorem to study equations by studying equations with one more degree -- e.g. we can use our understanding of heat equations to attack Poisson equations, and we can use our understanding of wave equation to deal with heat equations? (I might be wildly wrong..)
Nov 9, 2019 at 12:11 comment added Student @JochenGlueck as I am aware of there's also no standard convention for what $\Delta$ means ;) Do you mean I need the eigenvalues of $T$ to be nonnegative?
Nov 9, 2019 at 10:12 comment added Liviu Nicolaescu There is such an approach and $T$ need not be linear. For $T$ linear look at Hille-Yosida theorem. This was generalized to $T$ nonlinear and maximal monotone. For example the theory works when $T$ is the (sub)gradient of a convex function.
Nov 9, 2019 at 9:34 comment added Jochen Glueck Now, concerning your question itself: what you are probably looking for is the theory of $C_0$-semigroups. If $-T$ generates a $C_0$-semigroup, it is not uncommon to denote this $C_0$-semigroup by $(e^{-tT})_{t \ge 0}$, and the integral $\int_0^s e^{tT}\,dt$ then exists (in the strong sense) for every $s \ge 0$. If the growth bound of the $C_0$-semigroup is strictly smaller than $0$, then $-T$ is bijective from its domain onto the entire Banach space, and $\int_0^s e^{-tT}\,dt$ does indeed converge (with respect to the operator norm) to $T^{-1}$ as $s \to \infty$.
Nov 9, 2019 at 9:29 comment added Jochen Glueck Small remark concerning signs: do you mean the operator $\Delta$ or the operator $-\Delta$ when you write Laplacian? (I'm asking because your approach only works for $T = -\Delta$.)
Nov 9, 2019 at 1:57 history asked Student CC BY-SA 4.0