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 $g(\omega^1 \wedge \ldots \wedge \omega^k, \eta^1 \wedge \ldots \wedge \eta^k):= \det(g(\omega^i, \eta^j))$.

  • 10
    $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Jan 2, 2011 at 15:08
  • $\begingroup$ 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? $\endgroup$ Jan 2, 2011 at 15:59
  • $\begingroup$ (Just edited $\tau_M$ in place of $\mathcal{O}_M$, as the latter is usually deserved for the structure sheaf of the manifold) $\endgroup$
    – Qfwfq
    Jan 2, 2011 at 17:12
  • $\begingroup$ 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. $\endgroup$
    – Meneldur
    Jan 2, 2011 at 17:19
  • 2
    $\begingroup$ 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. $\endgroup$ Jan 2, 2011 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 $U_i \times\mathbf R^n$ (or $U_i \times\mathbf C^n$). 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 $\mathbf R^n$ (or $\mathbf C^n$)-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.)


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.

  • $\begingroup$ Editing (a great answer!) after all these years, but no TeX? $\endgroup$
    – LSpice
    May 13, 2020 at 16:57
  • $\begingroup$ @LSpice: Copy-paste is severely broken when using MathJax formulas. It also takes forever to render on Chromium because currently it does not support MathML. That's why for simple formulas that only need subscripts and superscripts it's better not to use MathJax. I think my objection to MathJax will be removed if they can fix copy-paster and Chromium implements MathML. $\endgroup$ May 13, 2020 at 20:10
  • $\begingroup$ OK, fair enough. So I assume that you would also prefer not to have the answer edited to include MathJax? $\endgroup$
    – LSpice
    May 15, 2020 at 16:13
  • $\begingroup$ @LSpice: Yes. Adding dollar signs around formulas is a trivial affair, and if this hasn't been done, then one must operate under presumption that it is for a good reason. I often see people editing a recent answer only to add a dollar sign for some superscript. I honestly do not understand why people are doing this. $\endgroup$ May 15, 2020 at 19:27

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.

  • $\begingroup$ @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. $\endgroup$ Jan 3, 2011 at 15:06
  • 3
    $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Jan 3, 2011 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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