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I have a three part question, which I could only received an answer for the first part here.

  1. The Laplace-Beltrami operator is an operator which is the typical example of a self-adjoint operator in $L^{2}$. I am wondering if this is also true for other Hilbert spaces $H^{k}$.
  2. The second part of the question and in which I am more interested is the following: How much the results are true for metrics with are not positive definitive. For example, the D'Alambertian(wave operator) in a general Pseudo-riemannian space.
  3. Below I am trying to calculate explicitly the adjoint for $\square_{g}$ in $H^{1}({{\cal{U}}_{t_{1}}^{+}},\nu_{g}=\sqrt{-g}d^{4}x)$. Any comments will be appreciated.

The wave operator is self-adjoint in $L^{2}({{\cal{U}}_{t_{1}}^{+}},\nu_{g})$ which means that for all $\psi,\omega \in C_{c}^{\infty}({{\cal{U}}_{t_{1}}^{+}})$ it is true that: $$(\psi,\square\omega)_{L^2}=(\square\psi,\omega)_{L^2}$$.

which allow us to write:

$$\int_{{{\cal{U}}_{t_{1}}^{+}}} \square_{g}\psi\omega \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi\square_{g}\omega \nu_{g}$$

Now taking into account the contracted Ricci identity:

$$\square(\psi_{,i})=(\square\psi)_{,i}+R^{j}_{i}\psi_{,j}$$

we have the following equality:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}\psi)_{,i}\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} (\square_{g}(\psi_{,i})-R^{j}_{i}\psi_{,j})\omega_{,i} \nu_{g} $$

Now the first term using the self adjointness of $\square_{g}$ can be rewritten as: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square_{g}(\psi_{,i})\omega_{,i} \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} $$ which again using the contracted Ricci identities gives: $$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{,_{i}}\square_{g}(\omega_{,i}) \nu_{g} = \int_{{{\cal{U}}_{t_{1}}^{+}}} \psi_{,_{i}}((\square_{g}\omega)_{,i}+R^{j}_{i}\omega_{,j}) \nu_{g} $$

Now putting together the above equalities we have that:

$$ \int_{{{\cal{U}}_{t_{1}}^{+}}}\square\psi\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}(\square\psi)_{i}\omega_{i} \nu_{g}=\int_{{{\cal{U}}_{t_{1}}^{+}}}\psi\square\omega \nu_{g}+ \int_{{{\cal{U}}_{t_{1}}^{+}}}\psi_{i}(\square\omega)_{i} \nu_{g}+\int_{{{\cal{U}}_{t_{1}}^{+}}}(R^{i}_{j}-R^{j}_{i})\psi_{,j}\omega_{,i} \nu_{g} $$

It seems that in the case $R^{i}_{j}=0$ then the $\square_{g}$ is self adjoint. Is that correct?

So to sum up.

My main question is:

How can I calculate the adjoint of the wave operator $\square_{g}$ in Sobolev Spaces?

Improve this question
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  • $\begingroup$ Just a few remarks: the Laplace-Beltrami operator $\Box_g$ on a Riemannian manifold $(M,g)$ is essentially self-adjoint with domain of smooth $p$-forms with compact support when $(M,g)$ is complete (particularly when $M$ is compact), as shown by Chernoff, Strichartz and other people. In complete generality, $\Box_g$ has at least one self-adjoint extension - namely, the Friedrichs extension, since the quadratic form associated to $\Box_g$ is bounded from below. The latter property is lost when $g$ is not positive definite. All we can say then is that $\Box_g$ is symmetric. $\endgroup$
    – Pedro Lauridsen Ribeiro
    May 31, 2015 at 15:33
  • $\begingroup$ Computing the adjoint of an unbounded operator such as $\Box_g$ involves not only finding its formula, but also its domain from that of $\Box_g$. It's not clear from your reasoning which is the domain you assume for $\Box_g$ - notice that it cannot be the whole Hilbert space, be it $L^2$ of $H^k$. If you assume, as in my former comment, its domain to be smooth forms of compact support, then the symmetry of $\Box_g$ follows immediately from the definition, since $\Box_g=d\delta+\delta d$, where the co-differential $\delta$ is the formal adjoint of $d$. $\endgroup$
    – Pedro Lauridsen Ribeiro
    May 31, 2015 at 15:50
  • $\begingroup$ In other words, the formal adjoint of $\Box_g$ (i.e. the restriction of the adjoint of $\Box_g$ to the latter's domain) is equal to $\Box_g$ itself. $\endgroup$
    – Pedro Lauridsen Ribeiro
    May 31, 2015 at 15:53
  • $\begingroup$ By the way, what is your definition of $\mathcal{U}^+_t$? $\endgroup$
    – Pedro Lauridsen Ribeiro
    May 31, 2015 at 15:59
  • $\begingroup$ Thank you for you comments. ${{\cal{U}}_{t_{1}}^{+}}$ is just any open set with compact closure on a manifold $M$. Is the comment you make about symmetry of $\square_{g}$ that follows from the definition using the differential and the codifferential applies in $H^{k}$ or only in L2? $\endgroup$
    – yess
    Jun 1, 2015 at 11:16

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