What is the adjoint of the wave operator $\square_{g}$ in Sobolev Spaces? I have a three part question.


*

*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 $W^{k,2}$. 

*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.  

*Below I am trying to calculate explicitly the adjoint for $\square_{g}$ in $H^{1}({{\cal{U}}_{t_{1}}^{+}},\nu_{g})$. 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?
 A: You have many questions in one1), but let me only calculate the formal adjoint.
In the Sobolev space $H^k=W^{k,2}(\mathbb R^n)$, $k\geq1$, one can use the inner product
$$
\langle u,v\rangle_{H^k}
=
\sum_{|\alpha|\leq k}\langle\partial^\alpha u,\partial^\alpha v\rangle_{L^2}.
$$
Compactly supported smooth functions are dense, so let us work with $u,v\in C^\infty_0\subset H^k$.
Since $\Delta^*=\Delta$ in $L^2$ and partial derivatives commute, we have
$$
\langle\partial^\alpha u,\partial^\alpha\Delta v\rangle_{L^2}
=
\langle\partial^\alpha\Delta u,\partial^\alpha v\rangle_{L^2}.
$$
Summing this over the multi-index $\alpha$ gives
$$
\langle u,\Delta v\rangle_{H^k}
=
\langle\Delta u,v\rangle_{H^k}.
$$
That is, $\Delta$ is a symmetric operator also in $H^k$.
Of course, to make $\Delta$ continuous and everywhere defined, you should consider it as a mapping $H^k\to H^{k-2}$ instead of $H^k\to H^k$.
(The adjoint will then be a map $H^{2-k}\to H^{-k}$, given by the distributional Laplacian.)

1) I suggest splitting your question in more parts, since it really contains several different questions.
