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Setup: Let $(M,g)$ be a (possibly non-compact) Riemannian manifold with volume density $d_gV$. Then one may think of $(M,g)$ as a measure space $(\Omega,\mathcal{A},\mu)$, where $\Omega:=M$, $\mathcal{A}:=\sigma(\tau_M)$ is the $\sigma$-Algebra generated by the topology $\tau_M$ of $M$ and for any $A \in \mathcal{A}$, $\mu(A):=\int_M{\chi_A d_gV}$, where $\chi_A:M \to [0,1]$ is the characteristic function of $A$. We obtain $\int_{M}{f d\mu} = \int_M{f d_gV}$, where the left hand side is understood to be an integral in the measure theoretic sense and the right hand side is an integration of a density. This enables us to define the space $L_p(\mu)$ with norm $\|f\|_{L_p(M)}^p = \int_{M}{|f|d_gV}$ on a manifold and apply all the results from integration theory to it, e.g. that it is a Banach space and so on.

My question is: Does this work in the following more general setup: Extend the Riemannian metric on $M$ to a fibre metric in $\bigwedge^k T^{\;*}M$, $0 \leq k \leq m$, (as described in the paragraph below). Then one may define $L_p$-spaces of differential forms by setting $\|\omega\|_{L_p(M)}^p := \int_{M}{|\omega|^pd_gV}$ and setting $L_p^k(M)$ to be the space of all measurable $k$-forms on $M$ (i.e. with Lebesgue measurable coefficient functions in any chart) such that $\|\omega\|_{L_p(M)}<\infty$. Is it possible to construct a measure space $(M,\mathcal{A},\mu)$ such that $L_p^k(M)$ may be thought of as an $L_p(\mu)$ as well?. The problem obviously is the range of a differential form. Formally it is a map $\omega\colon M \to \bigwedge^k T^\*M$, i.e. it takes values in the vector bundle $\bigwedge^kT^\*M$. Even if integration theory is available for functions on measure spaces with values in Banach spaces, this does not help since the bundle itself is not a vector space. I am interested in this question, because otherwise I see no alternative but to establish all the results about integration theory for $L_p^k(M)$ again, i.e. that it is a Banach space, Lebesgue Dominated Convergence Theorem, Fubini/Tonelli etc. That seems a bit exaggerated since intuitively this space is not so fundamentally different.

Construction of the fibre metric: For any $0 \leq k \leq m$ the Riemannian metric may be extended canonically to differential forms in $\Omega^k(M)$ in the following way: For one forms $\omega,\eta \in \Omega^1(M)$ define $g(\omega,\eta):=g(\omega^\sharp, \eta^\sharp)$, where $\sharp:T^*M \to TM$ is the sharp operator with respect to $g$. Then define $g$ on decomposable forms by $g(\omega^1 \wedge \ldots \wedge \omega^k, \eta^1 \wedge \ldots \wedge \eta^k):= \det(g(\omega^i, \eta^j))$.

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Everything is done locally but locally $M$ looks like an open set of $R^n$ and a differential form just looks like a vector-valued function. So it all works without any additional effort. –  Deane Yang Jan 2 '11 at 15:08
    
What exactly do you mean when you say that “L_p^k(M) may be thought of as an L_p(μ) as well”? This does not look like a rigorous statement. Do you mean that L_p^k(M) is isometrically isomorphic to L_p(μ) as a (quasi-)Banach space? –  Dmitri Pavlov Jan 2 '11 at 15:59
    
(Just edited $\tau_M$ in place of $\mathcal{O}_M$, as the latter is usually deserved for the structure sheaf of the manifold) –  Qfwfq Jan 2 '11 at 17:12
    
I know it is not rigorous. I indeed mean that $L_p^k(M)$ is isometrically isomorphic to $L_p(\mu)$, but if $L_p(\mu)$ is only defined for real-valued functions this cannot be true. So this implicitely contains the question if $L_p(\mu)$ makes sense for functions $f:\Omega \to E$, were $E$ is a vector bundle of finite rank rather than a vector space. –  Meneldur Jan 2 '11 at 17:19
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The answer to your question as stated is negative and the easiest counterexample arises when M is a point and the vector bundle V for which you are trying to find a measurable space M such that L_p(V) is isomorphic to L_p(M) has dimension 2. Indeed, L_p(V) has dimension 2, hence M must consist of two isolated points. But the norm on L_p(V) is different from the norm on L_p(pt⊔pt), because the former space is a Hilbert space and the latter space isn't. If we require that V is the bundle of k-forms, then the easiest example arises for dim M=2. However, see my answer for a possible solution. –  Dmitri Pavlov Jan 2 '11 at 19:23

3 Answers 3

Let E denote a vector bundle over a manifold M equipped with a metric, and L_p(E) the space of measurable sections of E with finite L_p norm. Obviously, in general, one can't identify L_p(E) with an L_p space of vector valued functions.

First assume that M is compact. To understand L_p(E), we use a finite set of trivializations (U_i, h_i) of E which cover M. Each trivialization identifies E|_{U_i} with U_i x R^n (or U_i x C^n). We choose the trivializations such that the bundle norm is equivalent to the euclidean norm, i.e. bounded from above and below. Then an L_p section of E is equivalent to a set of R^n (or C^n) valued measurable functions {f_i} on U_i satisfying the transition law, such that \sum |f_i|_p is finite.

Using this one can easily extend all the basic theorems to L_p(E). In particular, one shows that L_p(E) is equivalent to the completion of C^{\infty}(E) (the space of smooth sections on M) w.r.t. the L_p norm.

If M is noncompact, we write it as the union of locally finite compact subsets A_i, such that the intersections of A_i have zero measure. Then the L_p norm of a section s is given by (\sum \int_{A_i} |s|^p)^{1/p}. Then the arguments for the compact case can easily be carried over. (We argue on each A_i, and then combine.)

----Rugang Ye

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Yes, Lp-spaces can be defined for arbitrary hermitian vector bundles.

For the sake of convenience I denote Lp=L1/p (see this answer for a motivation), in particular L0=L and L1/2=L2. As explained in the link above, p is an arbitrary complex number such that ℜp≥0.

Suppose M is an arbitrary smooth manifold, possibly non-compact, and V is a finite-dimensional hermitian vector bundle over M. Let me stress that we do not need any additional data on M such as a metric, a volume form, a density, or an orientation.

Recall the definition of the line bundle Densp(M) of p-densities on M for an arbitrary complex number p (no restrictions on the real part of p): Every fiber of Densp(M) is the vector space of all set-theoretical maps f: Λtop(TM) \ {0} → C such that for all λ∈C \ {0} and for all x∈Λtop(TM) \ {0} we have f(λx)=|λ|pf(x). In particular, Dens1(M) is the tensor product of the line bundle of top-degree differential forms on M and the line bundle of orientations of M.

Note that for all p∈R the line bundle Densp(M) has a canonical orientation, in particular it is trivaliazable. Moreover, for all p∈C \ R the line bundle Densp(M) is also trivializable, even though it does not possess a canonical orientation. However, only Dens0(M) has a canonical trivialization.

Observe that all bundles Densp combine together in a C-graded unital *-algebra, i.e., we have the unit C→Dens0(M), the multiplication Densp(M)⊗Densq(M)→Densp+q(M), and the involution (Densp)*→Densp* (the first star denotes the conjugation of the complex structure on a vector bundle, the second star denotes the conjugation of complex numbers). All these morphisms are isomorphisms of line bundles.

For any t∈C and any p>0 we also have the power operation Densp+(M)→Denstp(M). This is not a morphism of vector bundles, because it is non-linear for t≠1 and Densp+(M) is a fiber bundle, not a vector bundle. However, the power operation is still a morphism of fiber bundles, in particular we can talk about powers of positive sections of Densp(M).

We have a canonical integration map ∫: C(Dens1(M))→C. We use this map to define norms on spaces C(Densp(M)) for all p∈C such that ℜp>0. First we send f∈C(Densp(M)) to |f|=(f*f)1/2∈C(Densℜp(M)). Observe that f*f and |f| are positive with respect to the canonical orientations on Dens2ℜp(M) and Densℜp(M). Then |f|1/ℜp∈C(Dens1+(M)) and we set ‖f‖:=∫⁠(|f|1/ℜp). This is a norm for ℜp≤1 and a quasi-norm for ℜp>1.

The (quasi-)Banach space Lp(M) is the completion of C(Densp(M)) in this (quasi-)norm.

If ℜp=0, then we complete C(Densp(M)) in the weak topology induced by C(Dens1−p(M)) and obtain the Banach space Lp(M). (The norm of f∈C(Densp(M)) can be defined as the supremum of |f|∈C(Dens0(M))=C(M), however, C(Densp(M)) is not dense in Lp(M) in the norm topology.)

Thus we defined Lp-spaces of the trivial line bundle on M for an arbitrary p∈C such that ℜp≥0. To extend this definition to an arbitrary hermitian vector bundle V we replace f*f by μ(f*f) in the above definition of norm. Here μ denotes the hermitan pairing on V.

All the usual theorems of measure theory like Radon-Nikodym, Riesz, Fubini, Tonelli etc. hold in this more general setting.

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(MathOverflow seems to be eating my at-prefix.) Of course you want $L_p \mathbin{:=} L^{1/p}$, right? –  L Spice Jul 13 at 1:22
    
@LSpice: Yes, of course. Thanks for the correction. –  Dmitri Pavlov Jul 13 at 9:56

Thanks for your posts. I am summing up what we have got so far.

@Dmitri Pavlov: You explained why the answer to my question is negative and give an alternative approach to $L_p$-spaces on hermitian vector bundles in your answer. You claim that all the usual theorems of measure theory hold in this more general setting. Can you give a reference for that?

@Deane Yang: You claim that this can be done locally. That idea seems natural to me, but I'm afraid I cant make this rigorous: Let $E \to M$ be a real vector bundle of rank $k$ with fibre metric $h$ and let $(M,g)$ be a Riemannian $m$-manifold. Assume $U \subset M$ is open and sufficiently small such that there exists a chart $\varphi:U \to R^m$ and a local trivialization $\Psi=(\Psi_1,\Psi_2):\pi^{-1}(U) \to U \times R^k$. Let $\mu_g$ be the Riemannian volume density on $M$, $\tau_g:=\sqrt{\det(g_{ij})}$ and $\left\| \cdot \right\|_h$ be the norm induced by $h$. Then for any section $s \in \Gamma(E)$, we obtain

$$\int_{U}{\left\| s \right\|_h^p \mu_g} = \int_{V}{(\left\| s \right\|_h^p \tau_g \circ \varphi^{-1} )(x) dx}=\int_{V}{\left\| s(\varphi^{-1}(x)) \right\|^p_{h(\varphi^{-1}(x))} \tau_g(\varphi^{-1}(x)) dx }$$

Now of course $\tilde s:= \Psi_2 \circ s\circ \varphi^{-1}:V \to R^k$ is a vector valued function, but you can't choose a norm $\left| \cdot \right|$ on $R^k$ such that for any $x \in V$, we obtain $\left| \tilde s(x) \right| = \left\| s(\varphi^{-1}(x) \right\|_{h(\varphi^{-1}(x))} $, because the fibre metric may change in every point.

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@Meneldur: No, I don't have any reference for this. However, all hermitian vector bundles of dimension n over a measurable space are non-canonically isomorphic to the trivial vector bundle of dimension n. Thus the problem is reduced to the case of trivial vector bundles, provided that you check independence from the non-canonical isomorphism. The situation is similar to the case of differential operators over smooth manifolds: The most common case is the one of a differential operator from the trivial line bundle to itself, but often you need to consider arbitrary vector bundles. –  Dmitri Pavlov Jan 3 '11 at 15:06
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An easy way to transfer all your knowledge about measures to the vector bundle setting is to observe that there is a natural way to map the space of sections of a vector bundle (equipped with a fibre metric) to the space of real-valued functions by simply mapping a section to its pointwise norm. You can then define the $L_p$ norm of a section to be the $L_p$ norm of its pointwise norm function. –  Deane Yang Jan 3 '11 at 17:17

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