I have a three part question.

  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 $W^{k,2}$.
  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 $H^{1}({{\cal{U}}_{t_{1}}^{+}},\nu_{g})$. 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:


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?

  • $\begingroup$ For the first part, the Sobolev spaces are essentially domains of powers of the Laplace operator. Have you tried approaching the problem that way? No ideas on the second part. $\endgroup$ Apr 11, 2015 at 10:17

1 Answer 1


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.

  • $\begingroup$ Thank you for your answers and comments. I realize that it was very broad question and that actually I am interested in the wave operator rather than the Laplacian. So I change the question accordingly. $\endgroup$
    – yess
    May 25, 2015 at 14:30
  • $\begingroup$ You are welcome. It might be cleaner to post those questions as separate new questions and provide links from this large question to the small ones and vice versa. It can get messy if people answer your question and then you edit it and the answers are no longer valid. But it's up to you, of course. $\endgroup$ May 25, 2015 at 16:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .