I Embedding homeomorphic manifolds

[mathwonk]

yes, in my terminology, [0,1) is called a "manifold with boundary". i suppose you can generalize it to that case by saying the map restricts to a bijection between the two respective boundaries.

this phenomenon does fail for differentiable real manifolds however, (project the graph of y = x^3 onto the y axis), but it holds again for complex analytic manifolds, e.g. riemann surfaces.

 

[mathwonk]

to add a little detail to lavinia's beautiful post #28, recall that poincare duality for manifolds is via intersection theory, which we thus want to relate to winding numbers. To specify a one diml cohomology class on M we want to see how it acts on a 1 diml cycle. If we have a map f from M to the circle, and a one cycle c on M, we can define the action of the corresponding cohomology class [f] on c by mapping c to the circle by f and asking how many times the image winds around the circle, i.e. we integrate dtheta/2π over the 1-cycle, which is the action of the pullback via f of the fundamental cohomology generator of the circle, on c.

Now as lavinia observed, we could alternatively look at the inverse image of a point p on the circle, via f, choosing p as a regular value, and obtain a sub manifold N of dimension n-1 in M. Now this sub manifold is poincare dual to a 1-cocycle [N], and the correspondence is by intersection of N with a one cycle. So [N] acts on c by intersecting N with c. Now almost by definition of N, this (oriented) intersection number equals the number of oriented preimages of the point p under the map f restricted to c, i.e. it is the number of times the path f(c) passes through the point p on the circle. (The orientation on the intersection cycle is sensitive to which direction the path passes through p, hence turning around and passing in the opposite direction subtracts an intersection, as it should).

Now by the basic principle of degree theory that the number of times you go around a monopoly board equals the number of times you pass GO, this oriented number of preimages equals the winding number of the path f(c) around the circle. Hence as lavinia said, the two actions, that of the class [N], and that of f*(dtheta) on c are equal.

Presumably this generalizes to higher dimensions, but it is a little harder for me to visualize it since the target spaces in higher dimensions tend to be infinite dimensional, i.e. to construct a K(Z,2) space, you start from a simply connected space with π2=Z, but then you have to add cells in all dimensions where your space has higher homotopy. I.e. you have to kill off all higher homotopy in all higher dimensions; the circle is nice since its universal cover is contractible, hence has no higher homotopy.

I suppose when you map a finite dimensional manifold into an infinite diml classifying space, you can prove the image can be considered as only lying in a finite amount of the skeleton, but to make sense of the theory of inverse images, we need the map representing a 2-cocycle say, to map into a 2 dimensional space, in order to claim the inverse image is of codimension 2. so it wold be nice if the 2-sphere were the classifying space, but as mentioned before, that is not true since the 2-sphere has homotopy in dimension 3 for instance, so you have attach a 4-cell to kill that element off...nonetheless Brown's theorem is true, but I have forgotten the arguments these 50 years later.

Lets see, maybe if we approximate our map by a simplicial map, hmmmmm... I wonder if even when our source space is higher diml, maybe we can homotopy the image of our map into the 2-skeleton using the vanishing of higher homotopy...?

 

[mathwonk]

Ok here is a speculative suggestion of how to generalize lavinia's discussion to higher dimensions. Given a 2-cohomology class on an n manifold M, find the sub manifold N of codimension two which is poincare dual to it, then use some tubular nbhd argument (?) to realize N as the fiber over a regular value of a smooth map from M to the 2-sphere, i.e. CP^1. (Or do we need a triviality resultful the normal 2-plane bundle?) I.e. thus we would get a surjection from homotopy classes of maps M-->CP^1 to H^2(M), but then we presumably need the higher dimensional cells in CP^infinity, to kill the kernel of this surjection, i.e. two maps from M to CP^1 might give the same fiber but not be homotopic. ?

no that seems unlikely to work. e.g. let M = CP^2 and the the 2nd cohomology is poincare dual to the subspace CP^1, which seems unlikely to be represented by the fiber of a map to S^2 = CP^1. and the normal bundle is not trivial, since the zero locus of a section is represented by the self intersection of CP^1 i.e. degree one, not zero.

the map representing this cohomology class would seem to be the inclusion of CP^2 in CP^infinity. ?

 

[lavinia]

Presumably this generalizes to higher dimensions, but it is a little harder for me to visualize it since the target spaces in higher dimensions tend to be infinite dimensional, i.e. to construct a K(Z,2) space, you start from a simply connected space with π2=Z, but then you have to add cells in all dimensions where your space has higher homotopy. I.e. you have to kill off all higher homotopy in all higher dimensions; the circle is nice since its universal cover is contractible, hence has no higher homotopy.

The arguments involving regular values generalize to any sphere but one can not in general obtain the entire second integer cohomology group from regular values of maps into the 2 sphere. For instance, if a cohomology class has a non-trivial cup product with itself it can not be the pull back of the fundamental class of the 2 sphere. In order to get 2 dimensional cohomology classes with non zero self cup products the classifying map must be into a higher dimensional complex projective space.

For a given manifold, I think that a single finite dimensional complex projective space will classify all of $H^2(M;Z)$. But if its dimension is greater that the dimension of $M$ it is not immediately clear to me how to use the Implicit Function Theorem.

Also, no torsion class can be Poincare dual to an oriented smooth submanifold this because the Poincare dual can be represented by a smooth differential form. This form is cohomologous to the Thom class of the normal bundle to the submanifold.

 

[mathwonk]

We can get some insight about the relation between second cohomology of a manifold M and subspaces of codimension two in M, from the situation in algebraic geometry. As lavinia has explained, not all 2-cohomology classes occur as poincare dual to embedded oriented sub manifolds, but we could ask which ones do. We could also ask which ones are poincare dual to the fiber of a map to a 2-sphere. But we know that all classes are pullbacks via maps to infinite dimensional projective space, although for a given class on a given manifold, presumably the map can be chosen as going into a specific finite dimensional projective space, although possibly of high dimension, i.e. not only greater than 2, but greater than the dimension of the given manifold M.Thus it seems the right way to get a subspace of M from a map on M, is to look at the pullback, not of a point, but of a subspace of codimension 2 in the target space, e.g. of a hyperplane in projective space, in case our map goes into a projective space of any dimension. Thus given a map f:M—>P^r, for any r, we choose a hyperplane (this mathematically naive spell checker prefers hydroplane) of P^r that does not contain the full image f(M), and we pull back that hyperplane H to its inverse image f^-1(H) in M.In algebraic geometry, an algebraic map, or in complex geometry an analytic map, from M to P^r is defined by sections of a line bundle L on M, which has a chern class c, in H^2(M;Z), and under some hypotheses, which I am embarrassed to say I cannot state explicitly at the moment, (perhaps that the space of global sections of L has no "base divisor"), this chern class equals the pullback of the unique distinguished generator of H^2(P^r;Z) = Z, i.e. of the cohomology class of a hyperplane.Now it is a famous theorem of Lefschetz that a cohomology class c in H^2(M;Z), where M is a compact complex manifold, is the chern class of some line bundle if and only if it has “type (1,1)”, i.e. if under the Hodge decomposition of H^2(M;C) = H^(0,2) + H^(1,1) + H^(0,2), the class c maps into the component of type (1,1), by the map H^2(M;Z)—>H^2(M;C).In sheaf cohomology language, in which line bundles are classified up to isomorphism by classes in the group H^1(M;O*), this says that c goes to zero under the map H^2(Z)—>H^2(O), induced by the sheaf sequence 0—>Z—>O—>O*—>0 and the following map from the corresponding long exact sequence H^2(Z)—>H^2(O). I.e. we have

H^1(O*)—>H^2(Z)—>H^2(O), where the first map is the chern class map.Now for a “very ample” line bundle L, the corresponding map to projective space f:M—>P^r given by a basis of the space of global sections of L, is an embedding, and the standard generator of P^r does pull back to the chern class of L. But it seems that the corresponding maps to projective spaces of lower dimensions g:M—>P^s, s ≤ r, given by subspaces of the space of global sections of L, may also pull back the standard generator to the chern class of L, at least under some conditions.In particular, if we project down to P^1, that is CP^1 ≈ S^2 the usual 2-sphere, we may get a map with fiber poincare dual to the chern class. The catch is that projection tends not to be well defined when the center of projection meets the embedded manifold. Still it becomes well defined on a “blowup” of that manifold. In particular every pair of independent global sections of L defines a map M’—>P^1, where M’ is the blowup of M along the (complex) codimension 2 intersection in m of the zero loci of the 2 sections.
So we are getting, at least for certain 2 diml cohomology classes, some geometric realization via a map to both P^r for large r, and in some sense also via a map to P^1 ≈ S^2.

In the category of smooth topology, given a smooth map M-->P^r, presumably we want to choose our hyperplane to satisfy some transversality condition wrt the map f. No doubt a generalization of Sard guarantees existence of such, perhaps see Guillemin and Pollack.

 

[WWGD]

We can get some insight about the relation between second cohomology of a manifold M and subspaces of codimension two in M, from the situation in algebraic geometry. As lavinia has explained, not all 2-cohomology classes occur as poincare dual to embedded oriented sub manifolds, but we could ask which ones do. We could also ask which ones are poincare dual to the fiber of a map to a 2-sphere. But we know that all classes are pullbacks via maps to infinite dimensional projective space, although for a given class on a given manifold, presumably the map can be chosen as going into a specific finite dimensional projective space, although possibly of high dimension, i.e. not only greater than 2, but greater than the dimension of the given manifold M.Thus it seems the right way to get a subspace of M from a map on M, is to look at the pullback, not of a point, but of a subspace of codimension 2 in the target space, e.g. of a hyperplane in projective space, in case our map goes into a projective space of any dimension. Thus given a map f:M—>P^r, for any r, we choose a hyperplane (this mathematically naive spell checker prefers hydroplane) of P^r that does not contain the full image f(M), and we pull back that hyperplane H to its inverse image f^-1(H) in M.In algebraic geometry, an algebraic map, or in complex geometry an analytic map, from M to P^r is defined by sections of a line bundle L on M, which has a chern class c, in H^2(M;Z), and under some hypotheses, which I am embarrassed to say I cannot state explicitly at the moment, (perhaps that the space of global sections of L has no "base divisor"), this chern class equals the pullback of the unique distinguished generator of H^2(P^r;Z) = Z, i.e. of the cohomology class of a hyperplane.Now it is a famous theorem of Lefschetz that a cohomology class c in H^2(M;Z), where M is a compact complex manifold, is the chern class of some line bundle if and only if it has “type (1,1)”, i.e. if under the Hodge decomposition of H^2(M;C) = H^(0,2) + H^(1,1) + H^(0,2), the class c maps into the component of type (1,1), by the map H^2(M;Z)—>H^2(M;C).In sheaf cohomology language, in which line bundles are classified up to isomorphism by classes in the group H^1(M;O*), this says that c goes to zero under the map H^2(Z)—>H^2(O), induced by the sheaf sequence 0—>Z—>O—>O*—>0 and the following map from the corresponding long exact sequence H^2(Z)—>H^2(O). I.e. we have

H^1(O*)—>H^2(Z)—>H^2(O), where the first map is the chern class map.Now for a “very ample” line bundle L, the corresponding map to projective space f:M—>P^r given by a basis of the space of global sections of L, is an embedding, and the standard generator of P^r does pull back to the chern class of L. But it seems that the corresponding maps to projective spaces of lower dimensions g:M—>P^s, s ≤ r, given by subspaces of the space of global sections of L, may also pull back the standard generator to the chern class of L, at least under some conditions.In particular, if we project down to P^1, that is CP^1 ≈ S^2 the usual 2-sphere, we may get a map with fiber poincare dual to the chern class. The catch is that projection tends not to be well defined when the center of projection meets the embedded manifold. Still it becomes well defined on a “blowup” of that manifold. In particular every pair of independent global sections of L defines a map M’—>P^1, where M’ is the blowup of M along the (complex) codimension 2 intersection in m of the zero loci of the 2 sections.
So we are getting, at least for certain 2 diml cohomology classes, some geometric realization via a map to both P^r for large r, and in some sense also via a map to P^1 ≈ S^2.

In the category of smooth topology, given a smooth map M-->P^r, presumably we want to choose our hyperplane to satisfy some transversality condition wrt the map f. No doubt a generalization of Sard guarantees existence of such, perhaps see Guillemin and Pollack.

Is the Algebraic Geometric map you refer to the Lefschetz Hyperplane Theorem?

 

[mathwonk]

no that is a much deeper theorem. Given a complex line bundle L on a compact complex manifold M, there is a finite dimensional vector space H^0(L) of global sections of that bundle. If s0,s1,...,sr is a basis of that space of sections, we define a map on M as sending a point p to [so(p),...,sr(p)]. Now the values sj(p) all lie in the same one dimensional complex vector space, the "line" over p. Thus their ratios determine a unique point of projective space P^r. This defines a map M-->P^r which is defined at those points p at which at least one section is non zero. "L has no base points" means it is defined at all points. Then we get a map M-->P^r defined everywhere and we can look at the intersection of the image f(M) with a hyperplane. That hyperplane pulls back to a (complex) codimension one subspace N of M that in good cases is poincare dual to the pullback of the standard generator of H^2(P^r;Z).

The Lefschetz hyperplane theorem gives a strong relationship between the low degree homology, cohomology, and homotopy groups of M and N. I.e. if (complex) dim(M) = n, then any of these groups is the same for M as for N, in degrees < n-1. The hypothesis is that M is embedded in P^r and that if M is singular, then N contains all the singularities of M.

If we apply this to the case of M = CP^3, and N a non singular surface embedded in P^3, then the case of π1 implies that N, i.e. any smooth complex surface embedded in CP^3, is simply connected, just because CP^3 is simply connected.

 

[mathwonk]

The theorem of Lefchetz was first proved by him with the aid of a decomposition of the embedded algebraic manifold by means of a "Lefschetz pencil". This concept is just an example of the procedure we defined of projecting a manifold that is embedded in a higher dimensional projective, and projecting it down to P^1. E.g. consider a surface M embedded in P^3 and a line L (not a line bundle), meeting M at a finite set of points. Then consider the pencil of planes containing L. This pencil is parametrized by a P^1, so we get a map from M to this P^1 sending each point p of M to the plane determined by L and p. This is undefined precisely at the points where L meets M, but can be extended to the blowup of M at this points, a space obtained from M by replacing each of those points by a copy of P^1.

This blowup space is then fibered over P^1 ≈ S^2, with fibers isomorphic to the intersections of M with the planes through L. These intersections have been made disjoint by blowing up M. I.e. each point of MmeetL has been replaced by a whole P^1 of points, one for each plane. Thus the map from the blowup to P^1 is now isomorphic on each copy of P^1 replacing a point where L met M. I.e. those plane sections of M which used to meet each other, are now disjoint fibers of the map from the blowup M' to P^1 ≈ S^2. I suppose one can use the fairly simple relation between the cohomology of M and its blowup to say something of value. Note that in the new blowup space, the normal bundle of one of these plane sections has apparently been rendered trivial. I am not at all expert on this but it is such fun to see things becoming somewhat more clear and more overlapping with the concepts above of Brown's theorem and lavinia's geometric realization in degree one.

It appears to me that one can always project an embedded manifold M down into a space of the same dimension as M and still pullback a hyperplane to give a copy of a hyperplane section of the original embedded manifold M. This may help with lavinia's concern. Thus we may be able to realize a chunk of H^2, i.e. the (1,1) part, by a map to CP^n, where n = (complex) dim(M), in particular CP^n and M have also the same real dimension 2n.

This is super fun, not necessarily proving anything, but beginning to understand some things I had studied and memorized partially. Thank you!

 

[lavinia]

Thanks to @mathwonk for suggesting inverse images of higher dimensional submanifolds. Here is a thought on how to approach his idea.

There is a generalization of the Implicit Function Theorem due to Thom which says that if a map is transverse to a submanifold then the inverse image of the submanifold is a submanifold of the same codimension. If $f:M→N$ is transverse to $S⊂N$ where $S$ is a closed smooth submanifold of codimesnion k, then $f^{-1}(S)$ is a closed submanifold of $M$ of codimension k.

Transverse means that the dimension of the span of the tangent space to the submanifold together with the image of $df$ equals the dimension of the ambient manifold. This generalizes the idea of a regular value. A regular value is a zero dimensional submanifold on which $f$ is transverse.

The key theorem is that in any homotopy class of smooth maps, there is a map that is transverse to a given manifold. So in the case of complex projective spaces one can choose a nearby function that is transverse to the complex projective subspace of 1 less complex dimension. For instance, in the case of the 2 sphere this would be a point. In the case pf $CP^2$ it would be $CP^1$.

If $f$ is transverse to the complex hypersurface $CP^{m-1}$ then $f^{-1}(CP^{m-1})$ is a codimension 2 submanifold of $M$.

The Poincare dual of $CP^{m-1}$ given the canonical orientation is the generator of the integer cohomology group $H^2(CP^{m},Z)$ and this is just the fundamental class $ω$ of the 2 sphere embedded and oriented in $CP^{m}$ as $CP^1$. Its pull back is Poincare dual to the codimension 2 hypersurface $f^{-1}(CP^{m-1})$.

As mathwonk pointed out, $f^{*}(ω)$ is the Chern class of the pull back bundle $f^{*}(γ^{m})$ of the canonical(also called tautological I think) line bundle $γ^{m}$

It remains to show that this map is bijective if $CP^{m}$ has sufficiently high dimension..

Surjective I think that one could start with an orientable codimension 2 hypesurface $H$ and consider the classifying map for its normal 2 plane bundle $E$ (which is a complex line bundle) into some complex projective space $CP^{m}$. One gets a bundle morphism from $E$ into the canonical line bundle over $CP^{m}$. Since $H$ and $E$ are orientable there is a tubular neighborhood of $H$ in $M$ which is homeomorphic to $E$. This tubular neighborhood maps onto a tubular neighborhood $T_{m}$ of $CP^{m}$ in the canonical line bundle. Note that the classifying map is transverse to $CP^{m}$ in $T_{m}$.

The normal bundle to $CP^{m}$ in $CP^{m+1}$ is also isomorphic to the canonical line bundle so $T_{m}$ may be taken to be a tubular neighborhood of $CP^{m}$ in $CP^{m+1}$.

Identify the boundary of $T_{m}$ to a point $p$ and map the rest of $M$ to $p$ . I think $T_{m}$ modulo its boundary is homeomorphic to $CP^{m+1}$ because $CP^ {m+1}$ is homeomorphic to $CP^{m}$ with a $2(m+1)$ dimensional ball attached via the projection of $S^{2m+1}$ onto $CP^{m}$. I think this makes it homeomorphic to the Thom space of the canonical (tautological) line bundle over $CP^{m}$.

 

[WWGD]

I don't know if this has been mentioned, but (transversal) intersection is Poincare dual to the cup product of homology classes. I think this is true for manifolds. @mathwonk : are there related notions of cup product, Poincare Duality for non-smooth spaces, like some of these Algebraic-Geometric objects with kinks/cusps?

 

[mathwonk]

@WWGD, brilliantly insightful question! answer: YES! It is called intersection (co)homology I believe, and involves "perverse sheaves", but i never learned this theory. I.e. in the case of singular spaces, one restricts which cycles to include in order to still have poincare duality. I believe the original approach, by Clint McCrory in his 1972 Brown PhD thesis, involved stratifications and cycles transverse to certain strata. The theory was later developed by Mark Goresky and Robert Macpherson, and also (fields medalist) Pierre Deligne.

https://www.genealogy.math.ndsu.nodak.edu/id.php?id=5577

https://en.wikipedia.org/wiki/Intersection_homology@lavinia, your interesting post is food for thought...I need time to digest it, and a little research on such things as thom spaces.

 

[lavinia]

@lavinia, your interesting post is food for thought...I need time to digest it, and a little research on such things as thom spaces.

@mathwonk This is the perhaps erroneous thought.

The Thom space of a vector bundle is its 1 point compactification.

With respect to a Riemannian metric on the bundle this is the same as identifying the boundary of the unit ball bundle to a point. Hence the tubular neighborhood idea.

In the case of the Mobius band this is the same as attaching a disk to its boundary circle since the disk is contractible radially to a point. This gives the real projective plane. Similarly one gets $RP^3$ by attaching a 3 ball to the canonical real line bundle over the projective plane. And so on for arbitrary $n$. Each real projective space looks like the Thom space of the canonical real line bundle over the real projective space of 1 less dimension.

I guess one would map the tubular neighborhood $T_{m}$ in $CP^{m+1}$ onto all of $CP^{m+1}$ by sending its boundary to the center of the attached $2(m+1)$ ball and its interior to all of $CP^{m+1}$ minus this central point - sort of like mapping a closed interval onto the 1 point compactification of the real line..

 

[WWGD]

Please forgive any confusion, I am not well acquainted with topological analysis and differential geometry, and I'm a novice with regards to this topic.

According to this theorem (I don't know the name for it), we cannot embed an n-dimensional space in an m-dimensional space, where n>m, without the former losing some of its structure.

So, does that mean that you can't render a higher n-dimensional space (for example, 4D), into a lower dimensional m-dimensional space (3D)? Is information lost, when trying to do so?

Edit: Does this relate or is applicable to holographic renderings?

Not quite true : an n- ball can be embedded in $\mathbb R^n$ without any loss.