Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The adjoint of the exterior derivarive is defined by

$\delta:=(-1)^k\ast^{-1}d\ast$,

but I need a way which avoids the Hodge $\ast$ operator.

Is there another definition?

For example, for positive definite metric there is the alternative of defining it by

$\int_M \langle d\alpha,\beta\rangle_{k+1} d_{vol} = \int_M\langle\alpha,\delta\beta\rangle_k d_{vol}$

but I don't know if we can apply the Riesz representation theorem to differential forms on semi-Riemannian manifolds, since they do not form a Hilbert space, because the metric is not positive definite.

Can you provide a link or a reference?

share|improve this question
    
Slightly tangential: it is bad form to ask essentially the same question twice: mathoverflow.net/questions/42370/… –  Willie Wong Oct 16 '10 at 12:59
add comment

3 Answers 3

up vote 5 down vote accepted

On any Riemannian manifold, any differential operator has a formal adjoint. This has nothing to do with functional analysis, it is a pure calculus fact that uses little more than Stokes theorem.

This is because taking adjoints is linear, preserves (or rather reverses) composition and commutes with multiplication by real-valued functions. It is also clear how the adjoint of an order 0 operator (i.e. vector bundle homomorphisms) has to be formed. The last and most important step is to construct the adjoint of a vector field on functions. Once this is done, the existence of an adjoint is proven, because the above operators generate the algebra of differential operator. This last argument also generalizes to differential operators on vector bundles. The formula below shows you also - for free - that the adjoint is a differential operator, which is something you have to work on if you define adjoints on the (pre) Hilbert-space level.

Let $X$ be a vector field, viewed as a differential operator on compactly supported function. To compute the adjoint of $X$, start with the formula $\int_M Lie_X \alpha = \int_M d \iota_X \alpha + \int_M \iota_x d \alpha =0$ for any $n$-form $\alpha$ and any vector field $X$. The first equality is the Cartan infinitesimal homotopy formula. The second summand is zero for degree reasons, the first by Stokes. Now let $\omega$ the volume form and $f,g$ two functions. Therefore $0= \int_M Lie (fg\omega)= \int_M (Xf) g \omega + \int_M f f (Xg) \omega+ \int_M fg \cdot div (X) \omega$ by the definition of the divergence. Hence the adjoint of $X$ is $-X - div (X)$.

So: Once you know how define the volume form on a semi-Riemannian manifold, you know that formal adjoints exists. So you can say that taking adjoints is a canonical operation on the algebra of differential operators, depending only on the volume form.

The argument above can be easily made into a formula for the adjoint once your operator is given in local coordinates. Whether this formula is useful or enlightening is a matter of taste.

For the exterior derivative, that formula can be written in terms of the Hodge star. This is a (rather simple) theorem. I do not know whether there is any other formula for $d^*$.

Pseudodifferntial operators also have adjoint, which is proven in a different fashion.

It is more difficult to show that the formal adjoint agrees with the Hilbert space adjoint (and I am not really qualified to say much on it and refer you to a book or a real expert)

share|improve this answer
1  
By the way, I should say where I learnt all this stuff: Bernhelm Booss, "Topologie und Analysis". There is an english translation and it is a wonderful book. –  Johannes Ebert Oct 16 '10 at 12:57
    
The way I like to think about it is this: give a smooth manifold with a volume form $dvol$, then you can identify a k-vector $v$ with the linear functional $\omega\in \Omega^k \to \int \omega(v) dvol$. So the exterior derivative has a formal adjoint that leads to the divergence operator for k-vectors. Then the metric structure allows you to identify k-vectors and k-forms. It is the same thing Johannes wrote, but I like this way because it makes the role of the metric clear. –  Willie Wong Oct 16 '10 at 13:29
    
@Johannes: What you are writing is not correct! Firstly, taking the formal adjoint of a differential operator does not only depend on the volume form if the differential operator acts on sections of a vector bundle! Obviously the metric on the fibers plays an important role! Secondly, the formal adjoint does not agree with the Hilbert space adjoint as I pointed out in my comment/answer. –  Orbicular Oct 16 '10 at 14:28
    
@orbicular: "Firstly, taking the formal adjoint of a differential operator does not only depend on the volume form if the differential operator acts on sections of a vector bundle! Obviously the metric on the fibers plays an important role" Sure –  Johannes Ebert Oct 16 '10 at 17:51
    
@orbicular: Here is what I meant with the Hilbert spaces. I am cautious and assume that the manifold is closed. If $D: \Gamma (E) \to \Gamma(F)$ is an operator of order $k$, then it induces $ W^{s,2} (E) \to W^{s-k,2}(F)$ for all real (!) $s$. The spaces $W^{s,2} (E)$ and $W^{-s,2} (E)$ are naturally dual (if $E$ has a metric and $M$ a volume form). The Hilbert space adjoint of $D:W^{s,2} (E) \to W^{s-k,2}(F)$ is a map $W^{k-s,2}(F) \to W^{-s,2}(E)$, and this is the Sobolev completion of the formal adjoint $D^*$. –  Johannes Ebert Oct 16 '10 at 18:03
show 1 more comment

An alternative definition of the codifferential is given as formal adjoint of $d$ when this one is expressed using the covariant derivative:

$\delta\omega=-\frac 1 {(k-1)!}\nabla^i\omega_{i i_2 \ldots i_k}d x^{i_2}\wedge\ldots\wedge d x^{i_k}$

I found this formula in Pit-Mann Wong - Einstein Manifolds.

Eventually it turned out that this definition fits my needs.

I wish tot thank to Orbicular, Johannes Ebert and Willie Wong for their help.

I would like to keep the question open for a while, because I am interested in other alternative answers.

share|improve this answer
1  
You can find this version of the formula in many Riemannian geometry textbooks. I am pretty sure it is in Jurgen Jost's book and also in Gallot-Hulin-Lafontaine. For semi-Riemannian geometry, what I do is just examine the proofs in the Riemannian case and see if and when positive definiteness is required. In very many cases (but not all!) it is not required so the same results hold in the semi-Riemannian case. –  Spiro Karigiannis Nov 29 '10 at 13:23
add comment

This is no answer, just a too long comment expressing my confusion. I am concerned your claim that one can define the "adjoint" of $d$ in terms of the Riesz representation theorem on Riemannian manifolds. I don't understand that.

Take your favorite compact Riemannian manifold $(M,g)$. Then the exterior derivative is an operator $d:\Omega(M)\rightarrow \Omega(M),$ taking smooth forms to smooth forms. Thus, after suitable Sobolev completions, one can expect $d$ to be a bounded linear map between two Hilbert spaces.

On the other hand, the Riesz representation theorem concerns bounded linear functionals on a Hilbert space $H$, i.e. bounded linear maps $H\rightarrow \mathbb{R}.$ It says that any such map is of the form $x\mapsto (x,y)$ for some $y\in H$ (where $(.,.)$ denotes the inner product on $H$). Consequently it is not applicaple to your problem.

I could rather image you are concerned with the theorem on the Hilbert space adjoint. Here one is considering a bounded linear map $A:H_1\rightarrow H_2$ between two different Hilbert spaces. It says there is a bounded linear map $A^*:H_2\rightarrow H_1$ characterized uniquely by the relation $(Ax,y)_2=(x,A^*y)_1$ for any $x\in H_1$ and $y\in H_2$.

In your example you want to apply it to the Hilbert space $H_1=H_2=L^2(\Lambda M),$ the space of square-integrable differential forms. Unfortunately, $d$ is not defined on all of $L^2(\Lambda M)$, but only on the space $W^{1,2}(\Lambda M),$ the space of square-integrable differentiable forms whose first derivative is also in $L^2$. Hence $d$ descends to an operator $d:W^{1,2}(\Lambda M)\rightarrow L^2(\Lambda M)$ and its adjoint is an operator $d^*:L^2(\Lambda M)\rightarrow W^{1,2}(\Lambda M)$. But this adjoint is NOT the operator you are looking for!

share|improve this answer
    
Yes, you are right, it is the Hilbert space adjoint. On the other hand, if we fix $\alpha$ we have a functional, and that's why I mentioned the Riesz representation theorem. I avoided to mention "Hilbert space" because I am interested in the case of indefinite semi-Riemannian metric. What kind of space is that replacing $L^2(\Lambda M)$ and $W^{1.2}(\Lambda M)$ for manifolds with indefinite metrics? The indefinite metric is not nice from this viewpoint, because 1. the square of the differential forms can be negative, 2. we don't have a CBS inequality etc. –  Cristi Stoica Oct 16 '10 at 10:33
    
I am a bit confused, because I was convinced that $\delta$ is the Hilbert space adjoint of $d$: [en.wikipedia.org/wiki/Hodge_theory][1]. Anyway, I am interested in any alternative definition of $\delta$ which avoids the use of the Hodge star. Thanks for your help. [1]: en.wikipedia.org/wiki/Hodge_theory –  Cristi Stoica Oct 16 '10 at 10:34
1  
I am confused now, as well. Concerning the wikipedia definition: Fix $\beta.$ You want to consider the functional $L^2\ni \alpha\mapsto (d\alpha,\beta)$ in order to apply Riesz. But this functional is not continous. So it seems to me there is a problem in the definition. The phrase I am familiar with is that $\delta$ is the formal adjoint of $d$ wrt. the $L^2$ inner product. –  Orbicular Oct 16 '10 at 11:14
2  
For adjoints of unbounded operators, you may want to consult Reed-Simon, Vol 1. Basically, for any densely defined operator, you can define the notion of an adjoint. But the domain of the adjoint operator may be arbitrarily small (an example is given in the book where the adjoint is only defined on the set {0}). That the adjoint is also densely defined is equivalent to the operator being closable. And $d$ is closable over $C^1\subset L^2$ by an approximation to the identity argument. But you are right, it doesn't just "follow" from Riesz. –  Willie Wong Oct 16 '10 at 13:13
1  
@Willie: To be precise: Taking the formal adjoint of $d$ has nothing whatsoever to do with Riesz. And talking about densely defined operators is also a bit complicated. You don't even need Hilbert spaces at all. Just compute the formal adjoint in local coordinates (as Johannes points out) and check that it is consistent! –  Orbicular Oct 16 '10 at 14:31
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.