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Lebesgue measure theory vs differential forms?
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I am currently reading various differential geometry books. From what I understand differential forms allow us to generalize calculus to manifolds and thus perform integration on manifolds. I gather that it is, in general, completely distinct from Lebesgue measure theory and is more like a generalization of Riemann integration.
Ok so here's the problem. I have always viewed Lebesgue measure theory as 'solving the issues with Riemann integration'. For example, a big problem with Riemann integration is that the space of Riemann integral functions is not complete. The fact that $L^p$ spaces in the Lebesgue theory are complete seems like a huge improvement on the Riemann situation, and is vital for so many concepts in functional analysis, PDEs, operator theory, and numerical analysis.
So if we then consider differential geometry and integration via differential forms, unless I am misunderstanding something, we lose all the benefits of Lebesgue theory?
It seems like if do lose all those benefits we are in a very bad situation. For example, how are we supposed to rigorously define solution spaces for PDEs if we can't use $L^p$ spaces and thus can't use Sobolev spaces? How can we obtain acceptable convergence of some sequence that may arise during our work if we are operating in this generalized Riemann setting where we lack completeness?
In summary, if differential forms are a generalization of Riemann integration how are we supposed to perform analysis when we no longer have the power and utility of Lebesgue measure theory?
functional-analysis measure-theory differential-geometry pde differential-forms
asked Jul 20 at 7:18
sonicboom
3,44182550
add a comment |Â
up vote
30
down vote
favorite
I am currently reading various differential geometry books. From what I understand differential forms allow us to generalize calculus to manifolds and thus perform integration on manifolds. I gather that it is, in general, completely distinct from Lebesgue measure theory and is more like a generalization of Riemann integration.
Ok so here's the problem. I have always viewed Lebesgue measure theory as 'solving the issues with Riemann integration'. For example, a big problem with Riemann integration is that the space of Riemann integral functions is not complete. The fact that $L^p$ spaces in the Lebesgue theory are complete seems like a huge improvement on the Riemann situation, and is vital for so many concepts in functional analysis, PDEs, operator theory, and numerical analysis.
So if we then consider differential geometry and integration via differential forms, unless I am misunderstanding something, we lose all the benefits of Lebesgue theory?
It seems like if do lose all those benefits we are in a very bad situation. For example, how are we supposed to rigorously define solution spaces for PDEs if we can't use $L^p$ spaces and thus can't use Sobolev spaces? How can we obtain acceptable convergence of some sequence that may arise during our work if we are operating in this generalized Riemann setting where we lack completeness?
In summary, if differential forms are a generalization of Riemann integration how are we supposed to perform analysis when we no longer have the power and utility of Lebesgue measure theory?
functional-analysis measure-theory differential-geometry pde differential-forms
asked Jul 20 at 7:18
sonicboom
3,44182550
2
What makes you think that differential forms are incompatible with Lebesgue measure and integration theory?
– md2perpe
Jul 20 at 7:53
add a comment |Â
up vote
30
down vote
favorite
up vote
30
down vote
favorite
I am currently reading various differential geometry books. From what I understand differential forms allow us to generalize calculus to manifolds and thus perform integration on manifolds. I gather that it is, in general, completely distinct from Lebesgue measure theory and is more like a generalization of Riemann integration.
Ok so here's the problem. I have always viewed Lebesgue measure theory as 'solving the issues with Riemann integration'. For example, a big problem with Riemann integration is that the space of Riemann integral functions is not complete. The fact that $L^p$ spaces in the Lebesgue theory are complete seems like a huge improvement on the Riemann situation, and is vital for so many concepts in functional analysis, PDEs, operator theory, and numerical analysis.
So if we then consider differential geometry and integration via differential forms, unless I am misunderstanding something, we lose all the benefits of Lebesgue theory?
It seems like if do lose all those benefits we are in a very bad situation. For example, how are we supposed to rigorously define solution spaces for PDEs if we can't use $L^p$ spaces and thus can't use Sobolev spaces? How can we obtain acceptable convergence of some sequence that may arise during our work if we are operating in this generalized Riemann setting where we lack completeness?
In summary, if differential forms are a generalization of Riemann integration how are we supposed to perform analysis when we no longer have the power and utility of Lebesgue measure theory?
functional-analysis measure-theory differential-geometry pde differential-forms
asked Jul 20 at 7:18
sonicboom
3,44182550
I am currently reading various differential geometry books. From what I understand differential forms allow us to generalize calculus to manifolds and thus perform integration on manifolds. I gather that it is, in general, completely distinct from Lebesgue measure theory and is more like a generalization of Riemann integration.
Ok so here's the problem. I have always viewed Lebesgue measure theory as 'solving the issues with Riemann integration'. For example, a big problem with Riemann integration is that the space of Riemann integral functions is not complete. The fact that $L^p$ spaces in the Lebesgue theory are complete seems like a huge improvement on the Riemann situation, and is vital for so many concepts in functional analysis, PDEs, operator theory, and numerical analysis.
So if we then consider differential geometry and integration via differential forms, unless I am misunderstanding something, we lose all the benefits of Lebesgue theory?
It seems like if do lose all those benefits we are in a very bad situation. For example, how are we supposed to rigorously define solution spaces for PDEs if we can't use $L^p$ spaces and thus can't use Sobolev spaces? How can we obtain acceptable convergence of some sequence that may arise during our work if we are operating in this generalized Riemann setting where we lack completeness?
In summary, if differential forms are a generalization of Riemann integration how are we supposed to perform analysis when we no longer have the power and utility of Lebesgue measure theory?
functional-analysis measure-theory differential-geometry pde differential-forms
asked Jul 20 at 7:18
sonicboom
3,44182550
asked Jul 20 at 7:18
sonicboom
3,44182550
asked Jul 20 at 7:18
sonicboom
3,44182550
asked Jul 20 at 7:18
sonicboom
3,44182550
3,44182550
2
What makes you think that differential forms are incompatible with Lebesgue measure and integration theory?
– md2perpe
Jul 20 at 7:53
add a comment |Â
2
What makes you think that differential forms are incompatible with Lebesgue measure and integration theory?
– md2perpe
Jul 20 at 7:53
2
2
What makes you think that differential forms are incompatible with Lebesgue measure and integration theory?
– md2perpe
Jul 20 at 7:53
What makes you think that differential forms are incompatible with Lebesgue measure and integration theory?
– md2perpe
Jul 20 at 7:53
add a comment |Â
1 Answer
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active
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up vote
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People use measure theory in tandem with differential forms all the time—there's no contradiction whatsoever between the formalisms. Be aware, though, that the adjective “Riemannian†in the context of differential geometry refers to constructions depending on Riemannian metrics (which are “Riemannian†in the sense of originating in the work of Bernhard Riemann), not to Riemann integration.
Suppose that $M$ a smooth $n$-manifold. By definition, it's locally homeomorphic to $mathbbR^n$, so that you can define a set $S subset M$ to be measurable if and only if $x(S cap U) subset mathbbR^n$ is Lebesgue measurable for every local coordinate chart $x: U to x(U) subset mathbbR^n$. This gives you a $sigma$-algebra of Lebesgue measurable sets on $M$ that correctly completes the Borel $sigma$-algebra generated by the open sets on $M$ as a topological space. At this point, you have everything you need to define measurable functions, vector fields, differential forms, tensor fields, etc., in a manner compatible with calculations in local coordinates.
Now, suppose that $M$ is a Riemannian manifold, so that it comes equipped with a Riemannian metric $g$—again, the “Riemannian†here does not refer to Riemann integration, but to Riemann himself and his work on differential geometry. On any local coordinate chart $x : U to x(U) subset mathbbR^n$, you can define a measure $lambda_g,x$ on $U$ by setting
$$
lambda_g,x(S cap U) := int_x(S cap U) sqrtdetleft(gleft(tfracpartialpartial x^i,tfracpartialpartial x^jright)right) ,dlambda
$$
for any Lebesgue measurable $S subset M$, where $lambda$ denotes Lebesgue measure on $mathbbR^n$. By paracompactness of the manifold $M$, one can cover $M$ by a locally finite open cover of such local coordinate charts, and hence use a smooth partition of unity subordinate to this cover to patch these local scaled pullbacks of Lebesgue measure together into a single measure $lambda_g$, the Riemannian measure [!] on $M$ with respect to $g$, which is a complete $sigma$-finite measure on the $sigma$-algebra of Lebesgue measurable sets in $M$.
Let me now describe the basic properties of $lambda_g$.
The measure $lambda_g$ is compatible with calculations in local coordinates, in the precise sense that $lambda_g(S cap U) = lambda_g,x(S cap U)$ for any Lebesgue measurable $S$ and any local coordinate chart $x : U to x(U) subset mathbbR^n$.
If $g^prime$ is any another Riemannian metric, then the Riemannian measures $lambda_g$ and $lambda_g^prime$ will be mutually absolutely continuous $sigma$-finite measures with smooth Radon–Nikodym derivative computable directly in terms of $g$ and $g^prime$.
Suppose that $M$ is orientable, and let $mathrmvol_g in Omega^n(M)$ be the Riemannian volume form defined by $g$. Then for any Riemann integrable $f$ on $M$,
$$
int_M f , mathrmvol_g = int_M f ,dlambda_g,
$$
so that $lambda_g$ really is the (completed) Radon measure on $M$ corresponding to the positive functional $C_c(M) ni f mapsto int_M f , mathrmvol_g$ via the Riesz representation theorem. In other words, integration with respect to $lambda_g$ really is the “Lebesgue-ification†of integration against the top-degree form $mathrmvol_g$.
Once you've constructed the Riemannian measure on your Riemannian manifold $(M,g)$, the sky is now the limit—you can construct $L^p$ and Sobolev spaces of functions, vector fields, differential forms, tensor fields, etc., and in particular, you can use them to study, for instance, the geometric partial differential operators (e.g., generalisations of the Laplacian and the Dirac operator) and their associated partial differential equations (e.g., heat equations) to great mathematical effect. As a mathematical researcher, I'm personally most familiar with the mathematical ecosystem centred around the Atiyah–Singer index theorem, which relates quantities from algebraic topology to functional-analytic computations on Riemannian manifolds, but you should be aware, for instance, that Perelman's proof of the Poincaré conjecture involved the detailed analysis of a certain highly non-linear PDE for the Riemannian metric itself [!]. Perhaps the most accessible example of these methods in action is Hodge theory, which basically computes the cohomology of a compact manifold in terms of solutions of the Laplace equation (with respect to some Riemannian metric) on differential forms of various degrees.
P.S. People tend to take the extension of Lebesgue theory from $mathbbR^n$ to manifolds more or less for granted, so precise accounts of this can be oddly hard to find. However, a precise if terse account of Lebesgue theory on manifolds can be found in Dieudonné's Treatise of Analysis, Volume 3, Section 16.22 (especially Theorem 16.22.2 and the following discussion). Dieudonné doesn't require a Riemannian metric, but the point is that Riemannian metric gives a canonical choice of Lebesgue measure in the sense of Dieudonné, in exactly the same way that it gives a canonical volume form in the orientable case. In fact, Lebesgue measures in the sense of Dieudonné can be identified with nowhere vanishing $1$-densities, and the construction of the Riemannian measure $lambda_g$ is really the construction of the canonical $1$-density $lvert mathrmvol_g rvert$ associated to $g$.
ADDENDUM
One can define a measurable $k$-form on $M$ to be a map $omega : M to wedge^k T^ast M$, such that the following hold.
- For every $x in X$, $omega(m) in wedge^k T^ast M_m$ (i.e., $omega$ is a set-theoretic section of $wedge^k T^ast M$).
- For every local coordinate chart $x : U to x(U) subset mathbbR^n$, the pullback $(x^-1)^ast omega : x(U) to wedge^k mathbbR^n$ defined by
$$
(x^-1)^astomega := sum_i_1 < cdots < i_k omegaleft(tfracpartialpartial x^i_1,dotsc,tfracpartialpartial x^i_kright) dx^i_1 wedge cdots wedge dx^i_k
$$
(with the usual abuses of notation) is measurable; this turns out to be equivalent to requiring that $omega(X_1,dotsc,X_k) : M to mathbbR$ be measurable (in the above sense) for any smooth vector fields $X_1,dots,X_k in mathfrakX(M)$.
Now, suppose that $N$ is an oriented $k$-dimensional submanifold of $M$ (compact and without boundary, for simplicity), and let $x : U to x(U) subset mathbbR^n$ be a local coordinate chart of $M$, such that $x(N cap U) = V_x,N times 0$ for some open $V_x,N subset mathbbR^k$, and such that restriction of $x$ to a diffeomorphism $N cap U to V_x,N$ is orientation-preserving. Then we can define
$$
int_N cap U omega := int_V_x,N omegaleft(tfracpartialpartial x^1,dotsc,tfracpartialpartial x^kright) lambda_mathbbR^k
$$
whenever the Lebesgue integral on the right-hand side exists (with $lambda_mathbbR^k$ the Lebesgue measure on $mathbbR^k$). We can then define $omega$ to be integrable on $N$ whenever it's integrable in this way on $N cap U$ for any suitable local coordinate chart $x : U to mathbbR^n$, and then, by exactly the same arguments as in the Riemann integral case, patch these local integrals into a global Lebesgue integral $int_N omega$, which turns out to be independent of all the choices of local coordinate chart and partition of unity made along the way.
answered Jul 20 at 8:52
Branimir Ćaćić
9,69221846
There is something that it's not clear to me. Am I remember wrongly that the integral with respect to the volume form is a oriented one? If so, without any restriction in picking the orientation, how could be possible that $int_M f , mathrmvol_g = int_M f ,dlambda_g$, where LHS could be negative for positive $f$? Can you please explain what am I missing or if I am plainly wrong? (don't know why, but I'm unable to tag you...)
– Bob
Jul 20 at 9:13
The Riemannian volume form on an orientable Riemannian manifold is canonical; in particular, it's constructed so that $int_M f mathrmvol_g geq 0$ for $f geq 0$, and in particular, so that $mathrmVol_g(M) := int_M 1 ,dlambda_g = int_M 1 mathrmvol_g > 0$ if $M$ has finite volume (i.e., if the constant function $1$ is integrable with respect to $lambda_g$).
– Branimir Ćaćić
Jul 20 at 9:21
1
I see, so the point is that you pick a canonical orientation... thanks :)
– Bob
Jul 20 at 9:25
1
The problem here, IMO, is that this is basically only touches upon "lebesgue integration of zero-forms" (together with the observation you can dualize a top forms if you have a metric) -- it doesn't address at all the topic of blending the theory of Lebesgue integration with the theory of differential forms (and their associated path/surface/n-space integrals).
– Hurkyl
Jul 20 at 13:18
1
Excellent answer thanks! It would be great if someone can give me a reference for a book that rigorously deals with extending Lebesgue measure theory to differential forms?
– sonicboom
Jul 20 at 14:19
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
37
down vote
accepted
People use measure theory in tandem with differential forms all the time—there's no contradiction whatsoever between the formalisms. Be aware, though, that the adjective “Riemannian†in the context of differential geometry refers to constructions depending on Riemannian metrics (which are “Riemannian†in the sense of originating in the work of Bernhard Riemann), not to Riemann integration.
Suppose that $M$ a smooth $n$-manifold. By definition, it's locally homeomorphic to $mathbbR^n$, so that you can define a set $S subset M$ to be measurable if and only if $x(S cap U) subset mathbbR^n$ is Lebesgue measurable for every local coordinate chart $x: U to x(U) subset mathbbR^n$. This gives you a $sigma$-algebra of Lebesgue measurable sets on $M$ that correctly completes the Borel $sigma$-algebra generated by the open sets on $M$ as a topological space. At this point, you have everything you need to define measurable functions, vector fields, differential forms, tensor fields, etc., in a manner compatible with calculations in local coordinates.
Now, suppose that $M$ is a Riemannian manifold, so that it comes equipped with a Riemannian metric $g$—again, the “Riemannian†here does not refer to Riemann integration, but to Riemann himself and his work on differential geometry. On any local coordinate chart $x : U to x(U) subset mathbbR^n$, you can define a measure $lambda_g,x$ on $U$ by setting
$$
lambda_g,x(S cap U) := int_x(S cap U) sqrtdetleft(gleft(tfracpartialpartial x^i,tfracpartialpartial x^jright)right) ,dlambda
$$
for any Lebesgue measurable $S subset M$, where $lambda$ denotes Lebesgue measure on $mathbbR^n$. By paracompactness of the manifold $M$, one can cover $M$ by a locally finite open cover of such local coordinate charts, and hence use a smooth partition of unity subordinate to this cover to patch these local scaled pullbacks of Lebesgue measure together into a single measure $lambda_g$, the Riemannian measure [!] on $M$ with respect to $g$, which is a complete $sigma$-finite measure on the $sigma$-algebra of Lebesgue measurable sets in $M$.
Let me now describe the basic properties of $lambda_g$.
The measure $lambda_g$ is compatible with calculations in local coordinates, in the precise sense that $lambda_g(S cap U) = lambda_g,x(S cap U)$ for any Lebesgue measurable $S$ and any local coordinate chart $x : U to x(U) subset mathbbR^n$.
If $g^prime$ is any another Riemannian metric, then the Riemannian measures $lambda_g$ and $lambda_g^prime$ will be mutually absolutely continuous $sigma$-finite measures with smooth Radon–Nikodym derivative computable directly in terms of $g$ and $g^prime$.
Suppose that $M$ is orientable, and let $mathrmvol_g in Omega^n(M)$ be the Riemannian volume form defined by $g$. Then for any Riemann integrable $f$ on $M$,
$$
int_M f , mathrmvol_g = int_M f ,dlambda_g,
$$
so that $lambda_g$ really is the (completed) Radon measure on $M$ corresponding to the positive functional $C_c(M) ni f mapsto int_M f , mathrmvol_g$ via the Riesz representation theorem. In other words, integration with respect to $lambda_g$ really is the “Lebesgue-ification†of integration against the top-degree form $mathrmvol_g$.
Once you've constructed the Riemannian measure on your Riemannian manifold $(M,g)$, the sky is now the limit—you can construct $L^p$ and Sobolev spaces of functions, vector fields, differential forms, tensor fields, etc., and in particular, you can use them to study, for instance, the geometric partial differential operators (e.g., generalisations of the Laplacian and the Dirac operator) and their associated partial differential equations (e.g., heat equations) to great mathematical effect. As a mathematical researcher, I'm personally most familiar with the mathematical ecosystem centred around the Atiyah–Singer index theorem, which relates quantities from algebraic topology to functional-analytic computations on Riemannian manifolds, but you should be aware, for instance, that Perelman's proof of the Poincaré conjecture involved the detailed analysis of a certain highly non-linear PDE for the Riemannian metric itself [!]. Perhaps the most accessible example of these methods in action is Hodge theory, which basically computes the cohomology of a compact manifold in terms of solutions of the Laplace equation (with respect to some Riemannian metric) on differential forms of various degrees.
P.S. People tend to take the extension of Lebesgue theory from $mathbbR^n$ to manifolds more or less for granted, so precise accounts of this can be oddly hard to find. However, a precise if terse account of Lebesgue theory on manifolds can be found in Dieudonné's Treatise of Analysis, Volume 3, Section 16.22 (especially Theorem 16.22.2 and the following discussion). Dieudonné doesn't require a Riemannian metric, but the point is that Riemannian metric gives a canonical choice of Lebesgue measure in the sense of Dieudonné, in exactly the same way that it gives a canonical volume form in the orientable case. In fact, Lebesgue measures in the sense of Dieudonné can be identified with nowhere vanishing $1$-densities, and the construction of the Riemannian measure $lambda_g$ is really the construction of the canonical $1$-density $lvert mathrmvol_g rvert$ associated to $g$.
ADDENDUM
One can define a measurable $k$-form on $M$ to be a map $omega : M to wedge^k T^ast M$, such that the following hold.
- For every $x in X$, $omega(m) in wedge^k T^ast M_m$ (i.e., $omega$ is a set-theoretic section of $wedge^k T^ast M$).
- For every local coordinate chart $x : U to x(U) subset mathbbR^n$, the pullback $(x^-1)^ast omega : x(U) to wedge^k mathbbR^n$ defined by
$$
(x^-1)^astomega := sum_i_1 < cdots < i_k omegaleft(tfracpartialpartial x^i_1,dotsc,tfracpartialpartial x^i_kright) dx^i_1 wedge cdots wedge dx^i_k
$$
(with the usual abuses of notation) is measurable; this turns out to be equivalent to requiring that $omega(X_1,dotsc,X_k) : M to mathbbR$ be measurable (in the above sense) for any smooth vector fields $X_1,dots,X_k in mathfrakX(M)$.
Now, suppose that $N$ is an oriented $k$-dimensional submanifold of $M$ (compact and without boundary, for simplicity), and let $x : U to x(U) subset mathbbR^n$ be a local coordinate chart of $M$, such that $x(N cap U) = V_x,N times 0$ for some open $V_x,N subset mathbbR^k$, and such that restriction of $x$ to a diffeomorphism $N cap U to V_x,N$ is orientation-preserving. Then we can define
$$
int_N cap U omega := int_V_x,N omegaleft(tfracpartialpartial x^1,dotsc,tfracpartialpartial x^kright) lambda_mathbbR^k
$$
whenever the Lebesgue integral on the right-hand side exists (with $lambda_mathbbR^k$ the Lebesgue measure on $mathbbR^k$). We can then define $omega$ to be integrable on $N$ whenever it's integrable in this way on $N cap U$ for any suitable local coordinate chart $x : U to mathbbR^n$, and then, by exactly the same arguments as in the Riemann integral case, patch these local integrals into a global Lebesgue integral $int_N omega$, which turns out to be independent of all the choices of local coordinate chart and partition of unity made along the way.
answered Jul 20 at 8:52
Branimir Ćaćić
9,69221846
There is something that it's not clear to me. Am I remember wrongly that the integral with respect to the volume form is a oriented one? If so, without any restriction in picking the orientation, how could be possible that $int_M f , mathrmvol_g = int_M f ,dlambda_g$, where LHS could be negative for positive $f$? Can you please explain what am I missing or if I am plainly wrong? (don't know why, but I'm unable to tag you...)
– Bob
Jul 20 at 9:13
The Riemannian volume form on an orientable Riemannian manifold is canonical; in particular, it's constructed so that $int_M f mathrmvol_g geq 0$ for $f geq 0$, and in particular, so that $mathrmVol_g(M) := int_M 1 ,dlambda_g = int_M 1 mathrmvol_g > 0$ if $M$ has finite volume (i.e., if the constant function $1$ is integrable with respect to $lambda_g$).
– Branimir Ćaćić
Jul 20 at 9:21
1
I see, so the point is that you pick a canonical orientation... thanks :)
– Bob
Jul 20 at 9:25
1
The problem here, IMO, is that this is basically only touches upon "lebesgue integration of zero-forms" (together with the observation you can dualize a top forms if you have a metric) -- it doesn't address at all the topic of blending the theory of Lebesgue integration with the theory of differential forms (and their associated path/surface/n-space integrals).
– Hurkyl
Jul 20 at 13:18
1
Excellent answer thanks! It would be great if someone can give me a reference for a book that rigorously deals with extending Lebesgue measure theory to differential forms?
– sonicboom
Jul 20 at 14:19
 |Â
show 4 more comments
up vote
37
down vote
accepted
People use measure theory in tandem with differential forms all the time—there's no contradiction whatsoever between the formalisms. Be aware, though, that the adjective “Riemannian†in the context of differential geometry refers to constructions depending on Riemannian metrics (which are “Riemannian†in the sense of originating in the work of Bernhard Riemann), not to Riemann integration.
Suppose that $M$ a smooth $n$-manifold. By definition, it's locally homeomorphic to $mathbbR^n$, so that you can define a set $S subset M$ to be measurable if and only if $x(S cap U) subset mathbbR^n$ is Lebesgue measurable for every local coordinate chart $x: U to x(U) subset mathbbR^n$. This gives you a $sigma$-algebra of Lebesgue measurable sets on $M$ that correctly completes the Borel $sigma$-algebra generated by the open sets on $M$ as a topological space. At this point, you have everything you need to define measurable functions, vector fields, differential forms, tensor fields, etc., in a manner compatible with calculations in local coordinates.
Now, suppose that $M$ is a Riemannian manifold, so that it comes equipped with a Riemannian metric $g$—again, the “Riemannian†here does not refer to Riemann integration, but to Riemann himself and his work on differential geometry. On any local coordinate chart $x : U to x(U) subset mathbbR^n$, you can define a measure $lambda_g,x$ on $U$ by setting
$$
lambda_g,x(S cap U) := int_x(S cap U) sqrtdetleft(gleft(tfracpartialpartial x^i,tfracpartialpartial x^jright)right) ,dlambda
$$
for any Lebesgue measurable $S subset M$, where $lambda$ denotes Lebesgue measure on $mathbbR^n$. By paracompactness of the manifold $M$, one can cover $M$ by a locally finite open cover of such local coordinate charts, and hence use a smooth partition of unity subordinate to this cover to patch these local scaled pullbacks of Lebesgue measure together into a single measure $lambda_g$, the Riemannian measure [!] on $M$ with respect to $g$, which is a complete $sigma$-finite measure on the $sigma$-algebra of Lebesgue measurable sets in $M$.
Let me now describe the basic properties of $lambda_g$.
The measure $lambda_g$ is compatible with calculations in local coordinates, in the precise sense that $lambda_g(S cap U) = lambda_g,x(S cap U)$ for any Lebesgue measurable $S$ and any local coordinate chart $x : U to x(U) subset mathbbR^n$.
If $g^prime$ is any another Riemannian metric, then the Riemannian measures $lambda_g$ and $lambda_g^prime$ will be mutually absolutely continuous $sigma$-finite measures with smooth Radon–Nikodym derivative computable directly in terms of $g$ and $g^prime$.
Suppose that $M$ is orientable, and let $mathrmvol_g in Omega^n(M)$ be the Riemannian volume form defined by $g$. Then for any Riemann integrable $f$ on $M$,
$$
int_M f , mathrmvol_g = int_M f ,dlambda_g,
$$
so that $lambda_g$ really is the (completed) Radon measure on $M$ corresponding to the positive functional $C_c(M) ni f mapsto int_M f , mathrmvol_g$ via the Riesz representation theorem. In other words, integration with respect to $lambda_g$ really is the “Lebesgue-ification†of integration against the top-degree form $mathrmvol_g$.
Once you've constructed the Riemannian measure on your Riemannian manifold $(M,g)$, the sky is now the limit—you can construct $L^p$ and Sobolev spaces of functions, vector fields, differential forms, tensor fields, etc., and in particular, you can use them to study, for instance, the geometric partial differential operators (e.g., generalisations of the Laplacian and the Dirac operator) and their associated partial differential equations (e.g., heat equations) to great mathematical effect. As a mathematical researcher, I'm personally most familiar with the mathematical ecosystem centred around the Atiyah–Singer index theorem, which relates quantities from algebraic topology to functional-analytic computations on Riemannian manifolds, but you should be aware, for instance, that Perelman's proof of the Poincaré conjecture involved the detailed analysis of a certain highly non-linear PDE for the Riemannian metric itself [!]. Perhaps the most accessible example of these methods in action is Hodge theory, which basically computes the cohomology of a compact manifold in terms of solutions of the Laplace equation (with respect to some Riemannian metric) on differential forms of various degrees.
P.S. People tend to take the extension of Lebesgue theory from $mathbbR^n$ to manifolds more or less for granted, so precise accounts of this can be oddly hard to find. However, a precise if terse account of Lebesgue theory on manifolds can be found in Dieudonné's Treatise of Analysis, Volume 3, Section 16.22 (especially Theorem 16.22.2 and the following discussion). Dieudonné doesn't require a Riemannian metric, but the point is that Riemannian metric gives a canonical choice of Lebesgue measure in the sense of Dieudonné, in exactly the same way that it gives a canonical volume form in the orientable case. In fact, Lebesgue measures in the sense of Dieudonné can be identified with nowhere vanishing $1$-densities, and the construction of the Riemannian measure $lambda_g$ is really the construction of the canonical $1$-density $lvert mathrmvol_g rvert$ associated to $g$.
ADDENDUM
One can define a measurable $k$-form on $M$ to be a map $omega : M to wedge^k T^ast M$, such that the following hold.
- For every $x in X$, $omega(m) in wedge^k T^ast M_m$ (i.e., $omega$ is a set-theoretic section of $wedge^k T^ast M$).
- For every local coordinate chart $x : U to x(U) subset mathbbR^n$, the pullback $(x^-1)^ast omega : x(U) to wedge^k mathbbR^n$ defined by
$$
(x^-1)^astomega := sum_i_1 < cdots < i_k omegaleft(tfracpartialpartial x^i_1,dotsc,tfracpartialpartial x^i_kright) dx^i_1 wedge cdots wedge dx^i_k
$$
(with the usual abuses of notation) is measurable; this turns out to be equivalent to requiring that $omega(X_1,dotsc,X_k) : M to mathbbR$ be measurable (in the above sense) for any smooth vector fields $X_1,dots,X_k in mathfrakX(M)$.
Now, suppose that $N$ is an oriented $k$-dimensional submanifold of $M$ (compact and without boundary, for simplicity), and let $x : U to x(U) subset mathbbR^n$ be a local coordinate chart of $M$, such that $x(N cap U) = V_x,N times 0$ for some open $V_x,N subset mathbbR^k$, and such that restriction of $x$ to a diffeomorphism $N cap U to V_x,N$ is orientation-preserving. Then we can define
$$
int_N cap U omega := int_V_x,N omegaleft(tfracpartialpartial x^1,dotsc,tfracpartialpartial x^kright) lambda_mathbbR^k
$$
whenever the Lebesgue integral on the right-hand side exists (with $lambda_mathbbR^k$ the Lebesgue measure on $mathbbR^k$). We can then define $omega$ to be integrable on $N$ whenever it's integrable in this way on $N cap U$ for any suitable local coordinate chart $x : U to mathbbR^n$, and then, by exactly the same arguments as in the Riemann integral case, patch these local integrals into a global Lebesgue integral $int_N omega$, which turns out to be independent of all the choices of local coordinate chart and partition of unity made along the way.
answered Jul 20 at 8:52
Branimir Ćaćić
9,69221846
There is something that it's not clear to me. Am I remember wrongly that the integral with respect to the volume form is a oriented one? If so, without any restriction in picking the orientation, how could be possible that $int_M f , mathrmvol_g = int_M f ,dlambda_g$, where LHS could be negative for positive $f$? Can you please explain what am I missing or if I am plainly wrong? (don't know why, but I'm unable to tag you...)
– Bob
Jul 20 at 9:13
The Riemannian volume form on an orientable Riemannian manifold is canonical; in particular, it's constructed so that $int_M f mathrmvol_g geq 0$ for $f geq 0$, and in particular, so that $mathrmVol_g(M) := int_M 1 ,dlambda_g = int_M 1 mathrmvol_g > 0$ if $M$ has finite volume (i.e., if the constant function $1$ is integrable with respect to $lambda_g$).
– Branimir Ćaćić
Jul 20 at 9:21
1
I see, so the point is that you pick a canonical orientation... thanks :)
– Bob
Jul 20 at 9:25
1
The problem here, IMO, is that this is basically only touches upon "lebesgue integration of zero-forms" (together with the observation you can dualize a top forms if you have a metric) -- it doesn't address at all the topic of blending the theory of Lebesgue integration with the theory of differential forms (and their associated path/surface/n-space integrals).
– Hurkyl
Jul 20 at 13:18
1
Excellent answer thanks! It would be great if someone can give me a reference for a book that rigorously deals with extending Lebesgue measure theory to differential forms?
– sonicboom
Jul 20 at 14:19
 |Â
show 4 more comments
up vote
37
down vote
accepted
up vote
37
down vote
accepted
People use measure theory in tandem with differential forms all the time—there's no contradiction whatsoever between the formalisms. Be aware, though, that the adjective “Riemannian†in the context of differential geometry refers to constructions depending on Riemannian metrics (which are “Riemannian†in the sense of originating in the work of Bernhard Riemann), not to Riemann integration.
Suppose that $M$ a smooth $n$-manifold. By definition, it's locally homeomorphic to $mathbbR^n$, so that you can define a set $S subset M$ to be measurable if and only if $x(S cap U) subset mathbbR^n$ is Lebesgue measurable for every local coordinate chart $x: U to x(U) subset mathbbR^n$. This gives you a $sigma$-algebra of Lebesgue measurable sets on $M$ that correctly completes the Borel $sigma$-algebra generated by the open sets on $M$ as a topological space. At this point, you have everything you need to define measurable functions, vector fields, differential forms, tensor fields, etc., in a manner compatible with calculations in local coordinates.
Now, suppose that $M$ is a Riemannian manifold, so that it comes equipped with a Riemannian metric $g$—again, the “Riemannian†here does not refer to Riemann integration, but to Riemann himself and his work on differential geometry. On any local coordinate chart $x : U to x(U) subset mathbbR^n$, you can define a measure $lambda_g,x$ on $U$ by setting
$$
lambda_g,x(S cap U) := int_x(S cap U) sqrtdetleft(gleft(tfracpartialpartial x^i,tfracpartialpartial x^jright)right) ,dlambda
$$
for any Lebesgue measurable $S subset M$, where $lambda$ denotes Lebesgue measure on $mathbbR^n$. By paracompactness of the manifold $M$, one can cover $M$ by a locally finite open cover of such local coordinate charts, and hence use a smooth partition of unity subordinate to this cover to patch these local scaled pullbacks of Lebesgue measure together into a single measure $lambda_g$, the Riemannian measure [!] on $M$ with respect to $g$, which is a complete $sigma$-finite measure on the $sigma$-algebra of Lebesgue measurable sets in $M$.
Let me now describe the basic properties of $lambda_g$.
The measure $lambda_g$ is compatible with calculations in local coordinates, in the precise sense that $lambda_g(S cap U) = lambda_g,x(S cap U)$ for any Lebesgue measurable $S$ and any local coordinate chart $x : U to x(U) subset mathbbR^n$.
If $g^prime$ is any another Riemannian metric, then the Riemannian measures $lambda_g$ and $lambda_g^prime$ will be mutually absolutely continuous $sigma$-finite measures with smooth Radon–Nikodym derivative computable directly in terms of $g$ and $g^prime$.
Suppose that $M$ is orientable, and let $mathrmvol_g in Omega^n(M)$ be the Riemannian volume form defined by $g$. Then for any Riemann integrable $f$ on $M$,
$$
int_M f , mathrmvol_g = int_M f ,dlambda_g,
$$
so that $lambda_g$ really is the (completed) Radon measure on $M$ corresponding to the positive functional $C_c(M) ni f mapsto int_M f , mathrmvol_g$ via the Riesz representation theorem. In other words, integration with respect to $lambda_g$ really is the “Lebesgue-ification†of integration against the top-degree form $mathrmvol_g$.
Once you've constructed the Riemannian measure on your Riemannian manifold $(M,g)$, the sky is now the limit—you can construct $L^p$ and Sobolev spaces of functions, vector fields, differential forms, tensor fields, etc., and in particular, you can use them to study, for instance, the geometric partial differential operators (e.g., generalisations of the Laplacian and the Dirac operator) and their associated partial differential equations (e.g., heat equations) to great mathematical effect. As a mathematical researcher, I'm personally most familiar with the mathematical ecosystem centred around the Atiyah–Singer index theorem, which relates quantities from algebraic topology to functional-analytic computations on Riemannian manifolds, but you should be aware, for instance, that Perelman's proof of the Poincaré conjecture involved the detailed analysis of a certain highly non-linear PDE for the Riemannian metric itself [!]. Perhaps the most accessible example of these methods in action is Hodge theory, which basically computes the cohomology of a compact manifold in terms of solutions of the Laplace equation (with respect to some Riemannian metric) on differential forms of various degrees.
P.S. People tend to take the extension of Lebesgue theory from $mathbbR^n$ to manifolds more or less for granted, so precise accounts of this can be oddly hard to find. However, a precise if terse account of Lebesgue theory on manifolds can be found in Dieudonné's Treatise of Analysis, Volume 3, Section 16.22 (especially Theorem 16.22.2 and the following discussion). Dieudonné doesn't require a Riemannian metric, but the point is that Riemannian metric gives a canonical choice of Lebesgue measure in the sense of Dieudonné, in exactly the same way that it gives a canonical volume form in the orientable case. In fact, Lebesgue measures in the sense of Dieudonné can be identified with nowhere vanishing $1$-densities, and the construction of the Riemannian measure $lambda_g$ is really the construction of the canonical $1$-density $lvert mathrmvol_g rvert$ associated to $g$.
ADDENDUM
One can define a measurable $k$-form on $M$ to be a map $omega : M to wedge^k T^ast M$, such that the following hold.
- For every $x in X$, $omega(m) in wedge^k T^ast M_m$ (i.e., $omega$ is a set-theoretic section of $wedge^k T^ast M$).
- For every local coordinate chart $x : U to x(U) subset mathbbR^n$, the pullback $(x^-1)^ast omega : x(U) to wedge^k mathbbR^n$ defined by
$$
(x^-1)^astomega := sum_i_1 < cdots < i_k omegaleft(tfracpartialpartial x^i_1,dotsc,tfracpartialpartial x^i_kright) dx^i_1 wedge cdots wedge dx^i_k
$$
(with the usual abuses of notation) is measurable; this turns out to be equivalent to requiring that $omega(X_1,dotsc,X_k) : M to mathbbR$ be measurable (in the above sense) for any smooth vector fields $X_1,dots,X_k in mathfrakX(M)$.
Now, suppose that $N$ is an oriented $k$-dimensional submanifold of $M$ (compact and without boundary, for simplicity), and let $x : U to x(U) subset mathbbR^n$ be a local coordinate chart of $M$, such that $x(N cap U) = V_x,N times 0$ for some open $V_x,N subset mathbbR^k$, and such that restriction of $x$ to a diffeomorphism $N cap U to V_x,N$ is orientation-preserving. Then we can define
$$
int_N cap U omega := int_V_x,N omegaleft(tfracpartialpartial x^1,dotsc,tfracpartialpartial x^kright) lambda_mathbbR^k
$$
whenever the Lebesgue integral on the right-hand side exists (with $lambda_mathbbR^k$ the Lebesgue measure on $mathbbR^k$). We can then define $omega$ to be integrable on $N$ whenever it's integrable in this way on $N cap U$ for any suitable local coordinate chart $x : U to mathbbR^n$, and then, by exactly the same arguments as in the Riemann integral case, patch these local integrals into a global Lebesgue integral $int_N omega$, which turns out to be independent of all the choices of local coordinate chart and partition of unity made along the way.
answered Jul 20 at 8:52
Branimir Ćaćić
9,69221846
People use measure theory in tandem with differential forms all the time—there's no contradiction whatsoever between the formalisms. Be aware, though, that the adjective “Riemannian†in the context of differential geometry refers to constructions depending on Riemannian metrics (which are “Riemannian†in the sense of originating in the work of Bernhard Riemann), not to Riemann integration.
Suppose that $M$ a smooth $n$-manifold. By definition, it's locally homeomorphic to $mathbbR^n$, so that you can define a set $S subset M$ to be measurable if and only if $x(S cap U) subset mathbbR^n$ is Lebesgue measurable for every local coordinate chart $x: U to x(U) subset mathbbR^n$. This gives you a $sigma$-algebra of Lebesgue measurable sets on $M$ that correctly completes the Borel $sigma$-algebra generated by the open sets on $M$ as a topological space. At this point, you have everything you need to define measurable functions, vector fields, differential forms, tensor fields, etc., in a manner compatible with calculations in local coordinates.
Now, suppose that $M$ is a Riemannian manifold, so that it comes equipped with a Riemannian metric $g$—again, the “Riemannian†here does not refer to Riemann integration, but to Riemann himself and his work on differential geometry. On any local coordinate chart $x : U to x(U) subset mathbbR^n$, you can define a measure $lambda_g,x$ on $U$ by setting
$$
lambda_g,x(S cap U) := int_x(S cap U) sqrtdetleft(gleft(tfracpartialpartial x^i,tfracpartialpartial x^jright)right) ,dlambda
$$
for any Lebesgue measurable $S subset M$, where $lambda$ denotes Lebesgue measure on $mathbbR^n$. By paracompactness of the manifold $M$, one can cover $M$ by a locally finite open cover of such local coordinate charts, and hence use a smooth partition of unity subordinate to this cover to patch these local scaled pullbacks of Lebesgue measure together into a single measure $lambda_g$, the Riemannian measure [!] on $M$ with respect to $g$, which is a complete $sigma$-finite measure on the $sigma$-algebra of Lebesgue measurable sets in $M$.
Let me now describe the basic properties of $lambda_g$.
The measure $lambda_g$ is compatible with calculations in local coordinates, in the precise sense that $lambda_g(S cap U) = lambda_g,x(S cap U)$ for any Lebesgue measurable $S$ and any local coordinate chart $x : U to x(U) subset mathbbR^n$.
If $g^prime$ is any another Riemannian metric, then the Riemannian measures $lambda_g$ and $lambda_g^prime$ will be mutually absolutely continuous $sigma$-finite measures with smooth Radon–Nikodym derivative computable directly in terms of $g$ and $g^prime$.
Suppose that $M$ is orientable, and let $mathrmvol_g in Omega^n(M)$ be the Riemannian volume form defined by $g$. Then for any Riemann integrable $f$ on $M$,
$$
int_M f , mathrmvol_g = int_M f ,dlambda_g,
$$
so that $lambda_g$ really is the (completed) Radon measure on $M$ corresponding to the positive functional $C_c(M) ni f mapsto int_M f , mathrmvol_g$ via the Riesz representation theorem. In other words, integration with respect to $lambda_g$ really is the “Lebesgue-ification†of integration against the top-degree form $mathrmvol_g$.
Once you've constructed the Riemannian measure on your Riemannian manifold $(M,g)$, the sky is now the limit—you can construct $L^p$ and Sobolev spaces of functions, vector fields, differential forms, tensor fields, etc., and in particular, you can use them to study, for instance, the geometric partial differential operators (e.g., generalisations of the Laplacian and the Dirac operator) and their associated partial differential equations (e.g., heat equations) to great mathematical effect. As a mathematical researcher, I'm personally most familiar with the mathematical ecosystem centred around the Atiyah–Singer index theorem, which relates quantities from algebraic topology to functional-analytic computations on Riemannian manifolds, but you should be aware, for instance, that Perelman's proof of the Poincaré conjecture involved the detailed analysis of a certain highly non-linear PDE for the Riemannian metric itself [!]. Perhaps the most accessible example of these methods in action is Hodge theory, which basically computes the cohomology of a compact manifold in terms of solutions of the Laplace equation (with respect to some Riemannian metric) on differential forms of various degrees.
P.S. People tend to take the extension of Lebesgue theory from $mathbbR^n$ to manifolds more or less for granted, so precise accounts of this can be oddly hard to find. However, a precise if terse account of Lebesgue theory on manifolds can be found in Dieudonné's Treatise of Analysis, Volume 3, Section 16.22 (especially Theorem 16.22.2 and the following discussion). Dieudonné doesn't require a Riemannian metric, but the point is that Riemannian metric gives a canonical choice of Lebesgue measure in the sense of Dieudonné, in exactly the same way that it gives a canonical volume form in the orientable case. In fact, Lebesgue measures in the sense of Dieudonné can be identified with nowhere vanishing $1$-densities, and the construction of the Riemannian measure $lambda_g$ is really the construction of the canonical $1$-density $lvert mathrmvol_g rvert$ associated to $g$.
ADDENDUM
One can define a measurable $k$-form on $M$ to be a map $omega : M to wedge^k T^ast M$, such that the following hold.
- For every $x in X$, $omega(m) in wedge^k T^ast M_m$ (i.e., $omega$ is a set-theoretic section of $wedge^k T^ast M$).
- For every local coordinate chart $x : U to x(U) subset mathbbR^n$, the pullback $(x^-1)^ast omega : x(U) to wedge^k mathbbR^n$ defined by
$$
(x^-1)^astomega := sum_i_1 < cdots < i_k omegaleft(tfracpartialpartial x^i_1,dotsc,tfracpartialpartial x^i_kright) dx^i_1 wedge cdots wedge dx^i_k
$$
(with the usual abuses of notation) is measurable; this turns out to be equivalent to requiring that $omega(X_1,dotsc,X_k) : M to mathbbR$ be measurable (in the above sense) for any smooth vector fields $X_1,dots,X_k in mathfrakX(M)$.
Now, suppose that $N$ is an oriented $k$-dimensional submanifold of $M$ (compact and without boundary, for simplicity), and let $x : U to x(U) subset mathbbR^n$ be a local coordinate chart of $M$, such that $x(N cap U) = V_x,N times 0$ for some open $V_x,N subset mathbbR^k$, and such that restriction of $x$ to a diffeomorphism $N cap U to V_x,N$ is orientation-preserving. Then we can define
$$
int_N cap U omega := int_V_x,N omegaleft(tfracpartialpartial x^1,dotsc,tfracpartialpartial x^kright) lambda_mathbbR^k
$$
whenever the Lebesgue integral on the right-hand side exists (with $lambda_mathbbR^k$ the Lebesgue measure on $mathbbR^k$). We can then define $omega$ to be integrable on $N$ whenever it's integrable in this way on $N cap U$ for any suitable local coordinate chart $x : U to mathbbR^n$, and then, by exactly the same arguments as in the Riemann integral case, patch these local integrals into a global Lebesgue integral $int_N omega$, which turns out to be independent of all the choices of local coordinate chart and partition of unity made along the way.
answered Jul 20 at 8:52
Branimir Ćaćić
9,69221846
edited Jul 20 at 14:46
answered Jul 20 at 8:52
Branimir Ćaćić
9,69221846
answered Jul 20 at 8:52
Branimir Ćaćić
9,69221846
answered Jul 20 at 8:52
Branimir Ćaćić
9,69221846
9,69221846
There is something that it's not clear to me. Am I remember wrongly that the integral with respect to the volume form is a oriented one? If so, without any restriction in picking the orientation, how could be possible that $int_M f , mathrmvol_g = int_M f ,dlambda_g$, where LHS could be negative for positive $f$? Can you please explain what am I missing or if I am plainly wrong? (don't know why, but I'm unable to tag you...)
– Bob
Jul 20 at 9:13
The Riemannian volume form on an orientable Riemannian manifold is canonical; in particular, it's constructed so that $int_M f mathrmvol_g geq 0$ for $f geq 0$, and in particular, so that $mathrmVol_g(M) := int_M 1 ,dlambda_g = int_M 1 mathrmvol_g > 0$ if $M$ has finite volume (i.e., if the constant function $1$ is integrable with respect to $lambda_g$).
– Branimir Ćaćić
Jul 20 at 9:21
1
I see, so the point is that you pick a canonical orientation... thanks :)
– Bob
Jul 20 at 9:25
1
The problem here, IMO, is that this is basically only touches upon "lebesgue integration of zero-forms" (together with the observation you can dualize a top forms if you have a metric) -- it doesn't address at all the topic of blending the theory of Lebesgue integration with the theory of differential forms (and their associated path/surface/n-space integrals).
– Hurkyl
Jul 20 at 13:18
1
Excellent answer thanks! It would be great if someone can give me a reference for a book that rigorously deals with extending Lebesgue measure theory to differential forms?
– sonicboom
Jul 20 at 14:19
 |Â
show 4 more comments
There is something that it's not clear to me. Am I remember wrongly that the integral with respect to the volume form is a oriented one? If so, without any restriction in picking the orientation, how could be possible that $int_M f , mathrmvol_g = int_M f ,dlambda_g$, where LHS could be negative for positive $f$? Can you please explain what am I missing or if I am plainly wrong? (don't know why, but I'm unable to tag you...)
– Bob
Jul 20 at 9:13
The Riemannian volume form on an orientable Riemannian manifold is canonical; in particular, it's constructed so that $int_M f mathrmvol_g geq 0$ for $f geq 0$, and in particular, so that $mathrmVol_g(M) := int_M 1 ,dlambda_g = int_M 1 mathrmvol_g > 0$ if $M$ has finite volume (i.e., if the constant function $1$ is integrable with respect to $lambda_g$).
– Branimir Ćaćić
Jul 20 at 9:21
1
I see, so the point is that you pick a canonical orientation... thanks :)
– Bob
Jul 20 at 9:25
1
The problem here, IMO, is that this is basically only touches upon "lebesgue integration of zero-forms" (together with the observation you can dualize a top forms if you have a metric) -- it doesn't address at all the topic of blending the theory of Lebesgue integration with the theory of differential forms (and their associated path/surface/n-space integrals).
– Hurkyl
Jul 20 at 13:18
1
Excellent answer thanks! It would be great if someone can give me a reference for a book that rigorously deals with extending Lebesgue measure theory to differential forms?
– sonicboom
Jul 20 at 14:19
There is something that it's not clear to me. Am I remember wrongly that the integral with respect to the volume form is a oriented one? If so, without any restriction in picking the orientation, how could be possible that $int_M f , mathrmvol_g = int_M f ,dlambda_g$, where LHS could be negative for positive $f$? Can you please explain what am I missing or if I am plainly wrong? (don't know why, but I'm unable to tag you...)
– Bob
Jul 20 at 9:13
There is something that it's not clear to me. Am I remember wrongly that the integral with respect to the volume form is a oriented one? If so, without any restriction in picking the orientation, how could be possible that $int_M f , mathrmvol_g = int_M f ,dlambda_g$, where LHS could be negative for positive $f$? Can you please explain what am I missing or if I am plainly wrong? (don't know why, but I'm unable to tag you...)
– Bob
Jul 20 at 9:13
The Riemannian volume form on an orientable Riemannian manifold is canonical; in particular, it's constructed so that $int_M f mathrmvol_g geq 0$ for $f geq 0$, and in particular, so that $mathrmVol_g(M) := int_M 1 ,dlambda_g = int_M 1 mathrmvol_g > 0$ if $M$ has finite volume (i.e., if the constant function $1$ is integrable with respect to $lambda_g$).
– Branimir Ćaćić
Jul 20 at 9:21
The Riemannian volume form on an orientable Riemannian manifold is canonical; in particular, it's constructed so that $int_M f mathrmvol_g geq 0$ for $f geq 0$, and in particular, so that $mathrmVol_g(M) := int_M 1 ,dlambda_g = int_M 1 mathrmvol_g > 0$ if $M$ has finite volume (i.e., if the constant function $1$ is integrable with respect to $lambda_g$).
– Branimir Ćaćić
Jul 20 at 9:21
1
1
I see, so the point is that you pick a canonical orientation... thanks :)
– Bob
Jul 20 at 9:25
I see, so the point is that you pick a canonical orientation... thanks :)
– Bob
Jul 20 at 9:25
1
1
The problem here, IMO, is that this is basically only touches upon "lebesgue integration of zero-forms" (together with the observation you can dualize a top forms if you have a metric) -- it doesn't address at all the topic of blending the theory of Lebesgue integration with the theory of differential forms (and their associated path/surface/n-space integrals).
– Hurkyl
Jul 20 at 13:18
The problem here, IMO, is that this is basically only touches upon "lebesgue integration of zero-forms" (together with the observation you can dualize a top forms if you have a metric) -- it doesn't address at all the topic of blending the theory of Lebesgue integration with the theory of differential forms (and their associated path/surface/n-space integrals).
– Hurkyl
Jul 20 at 13:18
1
1
Excellent answer thanks! It would be great if someone can give me a reference for a book that rigorously deals with extending Lebesgue measure theory to differential forms?
– sonicboom
Jul 20 at 14:19
Excellent answer thanks! It would be great if someone can give me a reference for a book that rigorously deals with extending Lebesgue measure theory to differential forms?
– sonicboom
Jul 20 at 14:19
 |Â
show 4 more comments
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2
What makes you think that differential forms are incompatible with Lebesgue measure and integration theory?
– md2perpe
Jul 20 at 7:53