Boundary operator on p-chains

[PsychonautQQ]

Can someone help me to understand what the boundary operator on a p-chain is doing exactly? Or boundary operators in general? I really need to develop a better intuition on the matter.

 

[lavinia]

Can someone help me to understand what the boundary operator on a p-chain is doing exactly? Or boundary operators in general? I really need to develop a better intuition on the matter.

How are you defining chains?

For a p-simplex the boundary is an oriented sum of its p-1 faces.

 

[fresh_42]

Can someone help me to understand what the boundary operator on a p-chain is doing exactly? Or boundary operators in general? I really need to develop a better intuition on the matter.

Maybe this helps a bit for a better intuition: https://www.physicsforums.com/threads/why-the-terms-exterior-closed-exact.871875/#post-5474443
It's a bit of the algebraic rather than the geometric side, where the name comes from, of the coin.

 

[WWGD]

Other than EDIT lowering/raising dimension (depending on whether use Homology or Cohomology) and satisfying $\partial^2=0$ , I don't see how to generalize, like @lavinia .suggested. Also, your operator classifies cycles up to (co)boundaries, by definition of (Co)Homology.

 

[fresh_42]

Other han lowering dimension and satisfying $\partial^2=0$ , I don't see how to generalize, like @lavinia .suggested.

You can equally increase the index and still have $d^2=0$. Chain, cochain, who cares?

 

[WWGD]

You can equally increase the index and still have $d^2=0$. Chain, cochain, who cares?

Yes, but since OP was defined on chains, it suggested Homology. Can you go up in Homology?

 

[WWGD]

Ultimately, you may say that the boundary operator helps classify cycles: closed curves that are piecewise continuous into classes. In , e.g., Simplicial Homology, two cycles are equivalent if their pointset difference is the boundary of a manifold ( of one dimension lower ), e.g., if you take a cylinder, then the top, bottom circles are considered the same, e.g., they are homologous, because their difference is the cylinder they are bounding. Homology attaches an Algebraic object to a Topological space in such a way that If X,Y have non-isomorphic objects attached , then they are not homeomorphic ( nor even Homotopic) , though if X,Y do have the same object attached they are not necessarily ( almost never) homeomorphic.

 

[lavinia]

Maybe an example will help:

A 1 simplex $[a,c]$ is an oriented line segment. One thinks of it as pointing from $a$ to $c$. Geometrically its boundary is the union of its two end points $a$ and $c$. But algebraically, the boundary is the difference of the two 0-simplices $[a]$ and $[c]$. Symbolically one writes $∂[a,c] = [c]-[a]$. This algebraic boundary of the line segment is an element of the free abelian group on the set of all zero simplices, the so called group of 0- chains.

Consider a triangle $Δ(a,b,c)$ with vertices $a$ $b$ and $c$ and suppose moving around it counterclockwise corresponds to moving from $a$ to $b$ then $b$ to $c$ and finally from $c$ back to $a$. One can think of the triangle algebraically as the sum of three 1 simplices. If it is oriented in the counterclockwise direction the sum is $[a,b] + [b,c] + [c,a]$. In the clockwise direction it is $[a,c] + [c,b] + [b,a]$.

Geometrically, the triangle has empty boundary. Algebraically, the boundary operator applied to the 1-chain $[a,b] + [b,c] + [c,a]$ is 0 in the group of 0-chains. This is clear since each vertex occurs twice, once as the end point of a line segment, and once as the beginning point. So the vertices cancel in pairs.

 

[fresh_42]

Yes, but since OP was defined on chains, it suggested Homology. Can you go up in Homology?

I don't know. In case one finds an example, which presumably can be done one way or another on a suitable sequence, it's probably not called boundary operator, chain and homological complex anymore. But the dualism invites to consider both. The cohomology sequences are the more interesting ones anyway. <duck & cover >

 

[WWGD]

Maybe an example will help:Geometrically, the triangle has empty boundary. Algebraically this means that the boundary operator applied to the 1-chain $[a,b] + [b,c] + [c,a]$ is 0 in the group of 0-chains. This is clear since each vertex occurs twice, once as the end point of a line segment, and once as the beginning point of a line segment so the vertices cancel in pairs.

To add a bit and to refresh my homology, formally, after Lavinia set the whole thing up: $\partial ( [a,b]+[b,c]+[c,a] )= \partial [a,b] + \partial [b,c] + \partial [c,a] = (a-b)+(b-c)+(c-a) =(a-a)+(b-b)+(c-c)=0$

 

[lavinia]

To add a bit and to refresh my homology, formally, after Lavinia set the whole thing up: $\partial ( [a,b]+[b,c]+[c,a] )= (a-b)+(b-c)+(c-a) =(a-a)+(b-b)+(c-c)=0$

Yes although usually the convention is $∂[a,c] = [c]-[a]$ with the opposite sign.

There is another way to do this with incidence numbers but it is equivalent.

 

[lavinia]

In general chains are the free abelian group on continuous maps of simplices into a topological space. The boundary operator is defined in the same way.

I like to think of smooth chains in manifolds as the objects over which one integrates differential forms.

 

[mathwonk]

I agree with lavinia. The fundamental fact in this topic is the stokes theorem, that says the integral of a differential form over the boundary of a surface is equal to the integral of the (exterior) derivative of the form over the surface itself. To actually compute this integral one must orient the surface and the boundary; then reversing the orientation gives the negative of the integral. Now we observe that differential forms can be added and multiplied by scalars and hence form a vector space. This allows us to make algebraic calculations with them that are quite useful.

The desire to imitate this computational facility in geometry leads us to ask for some similar structure on surfaces and their boundaries. A first enhancement is the observation that we can just as easily integrate, by pulling back via a parametriation, a parametrized surface, whether or not it is an embedding. The second one, is the fact that we might as well consider sums and scalar multiples of surfaces and curves, so we just define "chains" to be such formal linear combinations of parametrized surfaces. Their integrals are easily defined as the corresponding linear combinations of the integrals of the parametrized objects. Now we have made the set of parametrized surfaces, or curves, or..., into a vector space also.

To extend the stokes theorem to these generalized objects we must define the boundary of a linear combination of surfaces. Of course it should be the corresponding linear combination of the boundaries of the component surfaces in the linear expression. Thus the basic definition of a boundary of a p-chain, is as lavinia has illustrated, the case of the boundary of a p dimensional cube, e.g. an interval in case p=1. The next case is that of a square, where one has to accept that the boundary of a square should be the sum of the 4 edges, but that using algebra, we can orient an edge backwards and include a minus sign to correct this. So just look at the definition of the boundary of a square and see that it amounts to the oriented sum of the 4 edges.

Finally we obtain a beautiful duality between the geometric objects and the differential ones. I.e. the deRham theorem, that says (on an oriented compact manifold) the vector space of p-cycles modulo boundaries is isomorphic to the space of "closed" p-forms (derivative = 0, also called cocycles) modulo "exact" forms (those of form dw, also called coboundaries).

the success of this theory leads us to generalize it into the realm of topological manifolds, and define cohomology by introducing abstract cochains analogous to differential forms, in the sense that they "act" on cycles as linear functionals, i.e. a cochain, like a differential form, is an animal that sees a chain and spits out a number. (paraphrased from remarks of Raoul Bott.)

I hope this does not have too many errors.

 

[WWGD]

I don't know. In case one finds an example, which presumably can be done one way or another on a suitable sequence, it's probably not called boundary operator, chain and homological complex anymore. But the dualism invites to consider both. The cohomology sequences are the more interesting ones anyway. <duck & cover >

I have seen integration viewed as a homological operator while differentiation is a cohomological one. Or do I have them mixed up?

 

[WWGD]

I agree with lavinia. The fundamental fact in this topic is the stokes theorem, that says the integral of a differential form over the boundary of a surface is equal to the integral of the (exterior) derivative of the form over the surface itself. To actually compute this integral one must orient the surface and the boundary; then reversing the orientation gives the negative of the integral. Now we observe that differential forms can be added and multiplied by scalars and hence form a vector space. This allows us to make algebraic calculations with them that are quite useful.

The desire to imitate this computational facility in geometry leads us to ask for some similar structure on surfaces and their boundaries. A first enhancement is the observation that we can just as easily integrate, by pulling back via a parametriation, a parametrized surface, whether or not it is an embedding. The second one, is the fact that we might as well consider sums and scalar multiples of surfaces and curves, so we just define "chains" to be such formal linear combinations of parametrized surfaces. Their integrals are easily defined as the corresponding linear combinations of the integrals of the parametrized objects. Now we have made the set of parametrized surfaces, or curves, or..., into a vector space also.

To extend the stokes theorem to these generalized objects we must define the boundary of a linear combination of surfaces. Of course it should be the corresponding linear combination of the boundaries of the component surfaces in the linear expression. Thus the basic definition of a boundary of a p-chain, is as lavinia has illustrated, the case of the boundary of a p dimensional cube, e.g. an interval in case p=1. The next case is that of a square, where one has to accept that the boundary of a square should be the sum of the 4 edges, but that using algebra, we can orient an edge backwards and include a minus sign to correct this. So just look at the definition of the boundary of a square and see that it amounts to the oriented sum of the 4 edges.

Finally we obtain a beautiful duality between the geometric objects and the differential ones. I.e. the deRham theorem, that says (on an oriented compact manifold) the vector space of p-cycles modulo boundaries is isomorphic to the space of "closed" p-forms (derivative = 0, also called cocycles) modulo "exact" forms (those of form dw, also called coboundaries).

I hope this does not have too many errors.

Can you generalize Stoke's theorem to more abstract theorems like Cech, etc?

 

[mathwonk]

I am sure lavinia will tell us, but i think yes, the stokes theorem should generalize to the topological case, e.g. using cech cohomology, more or less tautologically. but this is not an expert statement, as i have not checked my sources. if need be i can do so later.

 

[fresh_42]

I have seen integration viewed as a homological operator while differentiation is a cohomological one. Or do I have them mixed up?

I tend to confuse the two, too, so don't ask me. I only know I have a damn cochain complex which refuses to reveal its secrets, let alone to be integrated. So in case you remember the source, I would be interested to know.

 

[WWGD]

I tend to confuse the two, too, so don't ask me. I only know I have a damn cochain complex which refuses to reveal its secrets, let alone to be integrated. So in case you remember the source, I would be interested to know.

I'll ask my Korean friend-classmate who told me about it. Hope he remembers.

 

[WWGD]

I tend to confuse the two, too, so don't ask me. I only know I have a damn cochain complex which refuses to reveal its secrets, let alone to be integrated. So in case you remember the source, I would be interested to know.

The confusing thing is that in either case, you can evaluate and get a number in a linear way, so either could be a Cohomological operation. OTOH, when iterating the operation, it seems like you're going down dimension-wise: a triple integral becomes a double integral, going from 3D to 2D. Similar when you differentiate in that, unless you iterate ( as in Sin, Cos) , you end up with a ( 0-D) constant.
Anyway, will look it up.

 

[mathwonk]

maybe i can't help, but integration is the operation that makes a differential form into a cochain, i.e. that is how it spits out a number when it sees a chain. so an integral sign with a diff form under it is a (de Rham) cochain; and thus (exterior) differentiation is analogous to the coboundary operator. But the term "cohomology operation" means something more sophisticated to me, again ask lavinia, but it means to me something like the steenrod squaring operation that is used to define (in milnor's book) certain characteristic classes. so indeed the coboundary operation does act on cohomology classes but i don't call that a cohomology operation.

https://en.wikipedia.org/wiki/Cohomology_operation

by the way WWGD, I think not only stokes but also de rham has a topological version, i.e. cohomology is dual to homology. but this is after dinner conversation not scholarly commentary. look in a good book like dold, lectures in algebraic topology.

 

[WWGD]

maybe i can't help, but integration is the operation that makes a differential form into a cochain, i.e. that is how it spits out a number when it sees a chain. so an integral sign with a diff form under it is a (de Rham) cochain; and thus (exterior) differentiation is analogous to the coboundary operator. But the term "cohomology operation" means something more sophisticated to me, again ask lavinia, but it means to me something like the steenrod squaring operation that is used to define (in milnor's book) certain characteristic classes. so indeed the coboundary operation does act on cohomology classes but i don't call that a cohomology operation.

https://en.wikipedia.org/wiki/Cohomology_operation

by the way WWGD, I think not only stokes but also de rham has a topological version, i.e. cohomology is dual to homology. but this is after dinner conversation not scholarly commentary. look in a good book like dold, lectures in algebraic topology.

Thanks; I always wonder about DeRham homology, one does not hear about it very often.

 

[lavinia]

Thanks; I always wonder about DeRham homology, one does not hear about it very often.

I have not heard of De Rham homology. What is it?

 

[WWGD]

I have not heard of De Rham homology. What is it?

Me neither, but since for each theory there is a dual, there must be a theory of DeRham homology. I just have no clue on what it is, how it works.

 

[WWGD]

Me neither, but since for each theory there is a dual, there must be a theory of DeRham homology. I just have no clue on what it is, how it works.

Seems like a nice exercise , to dualize a theory, or, in this case , to " anti-dualize" , since cohomology is already a dual to homology. We know the dual, we need the " anti-dual". I assume in the finite-dimensional case, it should not be too hard.

 

[lavinia]

A little elaboration on @mathwonk 's explanations.

A cochain complex is a graded collection of modules $C^{i}$ together with a collection of module homomorphisms $δ^{p}$ that map $C^{p}$ into $C^{p+1}$
such that $δ^{p+1}δ^{p} = 0$.

Any cochain complex gives rise to cohomology groups $H^{p}$ of $p$-cocycles modulo $p$-coboundaries. Differential forms on a manifold together with the exterior derivative form a cochain complex of modules over the real(or complex) numbers and therefore determine cohomology groups in each dimension $p$. $H^{p}_{DeRham}$ is the group of closed $p$-forms modulo the exact $p$-forms.

One might ask what the relation of these cohomology groups is to singular cohomology with real coefficients. A priori they could be different. DeRham's Theorem says that they are the same even though singular cohomology is defined entirely differently. As @mathwonk explained, if one integrates a $p$-form over a $p$-chain one gets a homomorphism of the group of smooth singular chains into the real numbers. This interprets the $p$-form as a smooth singular cochain. Stokes Theorem says that a closed $p$- form becomes a closed singular $p$-cocycle and an exact $p$-form becomes a singular coboundary. So one has a map of cochain complexes, that is: a collection of linear maps $C^{p}_{DeRham} → C^{p}_{smooth- singular}$ that commute with the coboundary operators. This gives a collection of homomorphisms of cohomology groups $H^{p}_{DeRham}→H^{p}_{smooth- singular}$. DeRham's Theorem says that these homomorphisms are isomorphisms.

The isomorphism between Czech cohomology and DeRham cohomology that I have seen does not use integration. Rather, it uses a double cochain complex which I am happy to explain if someone wants.

- The map from DeRham cohomology is to smooth singular cohomology i.e. where the simplices are required to be smooth. In general singular theory, simplices need only be continuous. One then proves that smooth singular and continuous singular are isomorphic.

BTW: The wedge product of differential forms determines a product on the DeRham cohomology groups. This product gives more information than the cohomology groups themselves. One might ask whether this product corresponds to another product on singular cohomology groups. The answer is yes. It is called the ""cup product".

One of the advantages of cohomology over homology is this cup product since it gives more information than cohomology alone.

 

[WWGD]

A little elaboration on @mathwonk 's explanations.

A cochain complex is a graded collection of modules $C^{i}$ together with a collection of module homomorphisms $δ^{p}$ that map $C^{p}$ into $C^{p+1}$
such that $δ^{p+1}δ^{p} = 0$.

Any cochain complex gives rise to cohomology groups $H^{p}$ of $p$-cocycles modulo ${P}-coboundaries. Differential forms on a manifold together with the exterior derivative form a cochain complex of modules over the real(or complex) numbers and therefore determine DeRham cohomology groups in each dimension$p$are the closed$p-forms$modulo the exact${p}$-forms. One might ask what the relation of these cohomology groups is to singular cohomology with real coefficients. A priori they could be different. DeRham's Theorem says that they are the same even though singular cohomology is defined entirely differently. As mathwonk explained, if one integrates a${p}$-form over a${p}$-chain one gets a homomorphism of the group of smooth singular chains into the real numbers. This interprets the${p}$-form as a singular cochain. Stokes Theorem says that a closed${p}$- form becomes a closed singular${p}$-cochain and an exact${p}$-form becomes a singular coboundary. So one has a map of cochain complexes, that is: a collection of linear maps$C^{p}_{DeRham → C^{p}_{singular}$that commute with the coboundary operators on each complex. This gives a collection of homomorphisms of cohomology groups$H^{p}_{DeRham}→H^{p}_{singular}##. DeRham's Theorem says that these homomorphisms are isomorphisms.

The isomorphism between Czech cohomology and DeRham cohomology that I have seen does not use integration. Rather, it uses double cochain complex which I am happy to explain if someone wants.

BTW: The wedge product of differential forms determines a product on the DeRham cohomology groups. This product gives more information than the cohomology groups themselves. One might ask whether this product corresponds to another product on singular cohomology groups. The answer is yes. It is called the ""cup product".

Why don't you do an " Insights" on this? and , please, edit your previous; seems interesting, but difficult to read.

 

[lavinia]

Why don't you do an " Insights" on this? and , please, edit your previous; seems interesting, but difficult to read.

I fixed it.

 

[WWGD]

A little elaboration on @mathwonk 's explanations.

A cochain complex is a graded collection of modules $C^{i}$ together with a collection of module homomorphisms $δ^{p}$ that map $C^{p}$ into $C^{p+1}$
such that $δ^{p+1}δ^{p} = 0$.

Any cochain complex gives rise to cohomology groups $H^{p}$ of $p$-cocycles modulo $p$-coboundaries. Differential forms on a manifold together with the exterior derivative form a cochain complex of modules over the real(or complex) numbers and therefore determine cohomology groups in each dimension $p$. $H^{p}_{DeRham}$ is the group of closed $p$-forms modulo the exact $p$-forms.

One might ask what the relation of these cohomology groups is to singular cohomology with real coefficients. A priori they could be different. DeRham's Theorem says that they are the same even though singular cohomology is defined entirely differently. As @mathwonk explained, if one integrates a $p$-form over a $p$-chain one gets a homomorphism of the group of smooth singular chains into the real numbers. This interprets the $p$-form as a smooth singular cochain. Stokes Theorem says that a closed $p$- form becomes a closed singular $p$-cocycle and an exact $p$-form becomes a singular coboundary. So one has a map of cochain complexes, that is: a collection of linear maps $C^{p}_{DeRham} → C^{p}_{smooth- singular}$ that commute with the coboundary operators. This gives a collection of homomorphisms of cohomology groups $H^{p}_{DeRham}→H^{p}_{smooth- singular}$. DeRham's Theorem says that these homomorphisms are isomorphisms.

The isomorphism between Czech cohomology and DeRham cohomology that I have seen does not use integration. Rather, it uses a double cochain complex which I am happy to explain if someone wants.

- The map from DeRham cohomology is to smooth singular cohomology i.e. where the simplices are required to be smooth. In general singular theory, simplices need only be continuous. One then proves that smooth singular and continuous singular are isomorphic.

BTW: The wedge product of differential forms determines a product on the DeRham cohomology groups. This product gives more information than the cohomology groups themselves. One might ask whether this product corresponds to another product on singular cohomology groups. The answer is yes. It is called the ""cup product".

One of the advantages of cohomology over homology is this cup product. It gives more information than cohomology alone.

I vaguely remember using cup products to distinguish between spaces in ways cohomology alone could not, i.e., the spaces had identical
cohomology groups but their cup product structures were non-isomorphic ( or whatever category applies).

 

[lavinia]

I vaguely remember using cup products to distinguish between spaces in ways cohomology alone could not, i.e., the spaces had identical
cohomology groups but their cup product structures were non-isomorphic ( or whatever category applies).

right