I am reading about the Hodge theorem for functions on a compact, connected, oriented Riemannian manifold $M$. My book (Rosenberg - The Laplacian on a Riemannian manifold) uses a heat kernel method to diagonalize the laplacian on $L^2(M)$.
The set up is:
- $\Delta f= \delta d f$ is the laplacian on functions on $M$. The sign convention is that it should match $\Delta f = -\sum \partial_i^2 f$. So the heat equation is $(\partial_t + \Delta)f=0$.
- Using a heat kernel $e(t,x,y)$ we define the heat operator $$e^{-t\Delta}f:=\int_M e(t,x,y)f(y) ~dy$$ which takes a function $f\in L^2(M)$ to its solution to the heat equation at time $t>0$. This operator is self-adjoint.
- We've already diagonalized the heat operator $e^{-t\Delta}$ such that we have an orthonormal basis $\{\omega_i(t)\}\subset L^2(M)$, with eigenvalues $\gamma_i(t)$. So $e^{-t\Delta}\omega_i(t) = \gamma_i(t) \omega_i(t)$.
- Moreover, we've shown that $\omega_i$ is independent of $t$, and that the eigenvalues $\gamma_i(t)$ are (strictly) positive.
Since $\gamma_i>0$, and $$0 = (\partial_t + \Delta )e^{-t\Delta}\omega_i = \dot\gamma_i\omega_i + \gamma_i \Delta\omega_i,$$ we can write $$\Delta\omega_i = -\frac{\dot\gamma_i}{\gamma_i}\omega_i.$$ The left side is independent of $t$, so $$-\frac{\dot\gamma_i}{\gamma_i} = C,$$ Thus $\gamma_i(t) = C'e^{-\lambda_i t}$ for some real $\lambda_i$.
My book states:
As $t\to 0$, $e^{-t\Delta}$ goes to the identity. This forces $\gamma_i(t)\to 1$ as $t\to 0$, and so we must have $C'=1$ and $\lambda_i\geq 0$.
Q: How do we conclude that $\lambda_i\geq 0$ here?
Note that he later remarks that we can also show the eigenvalues of the laplacian are nonnegative by $$\langle \Delta f, f\rangle = \langle \delta d f, f\rangle = \langle df, df\rangle \geq 0,$$ but he brings this up specifically as an alternate method.
As a reference, I am including some relevant pages of the book:
Diagonalization
