From: Terry Pilling Subject: Help with Steenrod Date: Thu, 14 Sep 2000 08:53:30 -0400 Newsgroups: sci.math Summary: Classifying fibre bundles up to isomorphism I would like someone to help me with the proof of a lemma given in Steenrod's "The topology of fibre bundles". The lemma is on page 10 in the 1999 paperback edition: ----- Lemma: Let B and B' be coordinate bundles having the same fibre Y and group G, and let \overline{h}: X --> X' be a map of one base space into the other. Finally, let _ _ g_{kj}: V_j \cap h^{-1}(V_k') --> G be a set of continuous maps satisfying the conditions _ _ _ g_{kj}(x) g_{ji}(x) = g_{ki}(x), x \in V_i \cap V_j \cap h^{-1}(V'_k) _ _ _ _ g'_{lk}(h(x)) g_{kj}(x) = g_{lj}(x), x \in V_j \cap h^{-1}(V'_k \cap V'_j) _ Then there exists one and only one map h: B --> B' inducing h and _ having { g_{jk} } as its mapping transformations. ------ _ But g_{jk} is defined on the previous page as g_{kj}(x) = (\phi'_{k, x'})^{-1} h_x \phi_{j,x} where h_x: Y_x --> Y_{x'} from the fibre over x in X to the fibre over _ x' in X' is the map INDUCED by h. (note also that x' = h(x)). My problem with this is that he proves the lemma by constructing h locally in the bundle and in the construction his h_{kj} actually _ _ depends on g_{kj}(x). We see by the definition above that g_{kj}(x) has h_x in its definition, which is itself induced by h. So I am confused, it seems like he is defining his unique h but using h in its own definition! It seems circular. Help! -Ter ============================================================================== From: "glenn" Subject: Re: Help with Steenrod Date: Thu, 14 Sep 2000 18:49:02 +0300 Newsgroups: sci.math [Previous article quoted in toto --djr] No, the map h_x is induced by h(x^prime=h^bar(x)). There is a BIG difference. ============================================================================== From: Terry Pilling Subject: Re: Help with Steenrod Date: Thu, 14 Sep 2000 11:41:42 -0400 Newsgroups: sci.math On Thu, 14 Sep 2000, glenn wrote: > > No, the map h_x is induced by h(x^prime=h^bar(x)). There is a BIG difference. > I don't see it. Please dummy it up for me a bit. I hope that you don't think that when he writes h(x' = h^bar(x)) he actually means h(x')... that would be nonsense since h is a map from B to B' and x' lives in B'. He is just using (x' = h^bar(x) ) as a parenthetical note in definition of x'. I may be asking a stupid question but please don't assume that it is more stupid than it is. -Ter ============================================================================== From: "glenn" Subject: Re: Help with Steenrod Date: Thu, 14 Sep 2000 21:07:59 +0300 Newsgroups: sci.math [Previous article quoted in toto --djr] When he writes h(x' = h^bar(x)) he means "h OF all those x for which the equation x'=h^bar(x) is valid". It's a bit indirect but you have to used to it. ============================================================================== From: Terry Pilling Subject: Re: Help with Steenrod Date: Thu, 14 Sep 2000 13:44:27 -0400 Newsgroups: sci.math On Thu, 14 Sep 2000, glenn wrote: [Previous article quoted in toto --djr] uhm... but x' = h^bar(x) is the DEFINITION of x'.... ANY x is an x for which the equation is valid. This is sci.math right? not sci.monkeys_banging_on_keyboards? Listen closely, Okay: I have a pair of fibre bundles B and B'. They have the same fibre and the same group. I also have a map h^bar from the base space of one to the base space of the other (i.e. h^bar is _given_ it is not a restriction of h to the base space it is simply a continuous map between two spaces). Finally I have a set of maps, one for each neighboorhood in the base space of B which return a group element (basically the group element that tells you how the fibre over x' = h^bar(x) has to be twisted to match up with the fibre over x.) I define h_{kj}(b) where b is in the bundle B over the neighboorhood _ V_j intersect with h^{-1}(V'_k) by _ _ h_{kj}(b) := \phi'_k( h(x), g_{kj}(x) \cdot p_j(b) ) Let me draw what this means... B B' ------------ --------------- | | h_x | | | . b | ----------> | . h_{kj}(b)| | | | | <--- Bundles | | | | ------------ X ====> --------------- X' _ | ^ h(x) ^ | | | p_j | | | | | \phi_j | \phi'_k V | | ------------ ------------- | | | | | | _ | | | . p_j(b)| g_{kj} | . b' | <--- Local | | | | Trivializations | | ------> | | | | | | | | _ | | | x | h(x) | x' | ---.-------- X ====> ----.-------- X' V_j \cap h^{-1}(V'_k) V'_k Now if you examine the diagram everything seems fine _ until you notice that the definition of g_{kj} in Steenrod actually goes back the other way as: _ g_{kj} := \phi'_k^{-1} h_x \phi_j so in the definition of h_{kj} you use h_x but h_x is the restriction of h to the fibre over x... unless I am totally misreading Steenrod. So we are defining h as a map from B to B' but implicitly using h in the definition. Look, it is already clear to me intuitively that the map is unique. I am given the map of the base space, and I am given the group element associated to each neighboorhood in the bundle which will map it to the other bundle. It is clear that it is unique since those are the only two items needed to specify every point in B' from one in B. My problem is with his proof... Actually it is not even with his proof, which is totally valid if I am given the g_{jk}'s on each neighboorhood. The only thing that invalidates it is that he says the g_{jk} are the mapping transformations, which have already been defined in (17) with h incorporated in the definition. Perhaps the solution is as simple as this: 1) g_{jk} GIVEN 2) DEFINE g_{jk} = (\phi'_k)^{-1} h \phi_j 3) find h from 1) and 2) argh. -Ter ============================================================================== From: "glenn" Subject: Re: Help with Steenrod Date: Thu, 14 Sep 2000 23:46:21 +0300 Newsgroups: sci.math -- glenn "Terry Pilling" wrote in message news:Pine.LNX.4.10.10009141237230.1447-100000@offshell.phys.ndsu.nodak.edu... > On Thu, 14 Sep 2000, glenn wrote: > > > "Terry Pilling" wrote in message > > news:Pine.LNX.4.10.10009141135100.1132-100000@offshell.phys.ndsu.nodak.edu... > > > On Thu, 14 Sep 2000, glenn wrote: > > > > > > > > > > > No, the map h_x is induced by h(x^prime=h^bar(x)). There is a BIG > > difference. > > > > > > > > > > I don't see it. Please dummy it up for me a bit. > > > > > > I hope that you don't think that when he writes h(x' = h^bar(x)) > > > he actually means h(x')... that would be nonsense since h is > > > a map from B to B' and x' lives in B'. > > > He is just using (x' = h^bar(x) ) as a parenthetical note in > > > definition of x'. > > > > > > I may be asking a stupid question but please don't assume that > > > it is more stupid than it is. > > > > > > > When he writes h(x' = h^bar(x)) he means > > "h OF all those x for which the equation x'=h^bar(x) is valid". > > It's a bit indirect but you have to used to it. > > > > uhm... but x' = h^bar(x) is the DEFINITION of x'.... > ANY x is an x for which the equation is valid. Really? If you look closer to the page, you will notice that x and x' enter to the text, long before the relation x' = h^bar(x). And maybe, maybe I say, you should also put ahead of (16) an existential quantifier (you know), like: For every x and x' of the base spaces of X and X' (these being the total spaces) etc etc. Also ahead of (17) a similar quantifier (you know) like the previous one, with the restriction that x belongs to blah blah. And then maybe, I say maybe (and this is a rare posibility as I can see) you will understand that the index at h, is calculated by x' via h^bar, as ITS BEEN DONE in the top of page 11, in another ocassion. [deletia -- djr] ============================================================================== From: hook@nas.nasa.gov (Ed Hook) Subject: Re: Help with Steenrod Date: 15 Sep 2000 13:51:29 GMT Newsgroups: sci.math In , Terry Pilling writes: |> I would like someone to help me with the proof of a lemma given |> in Steenrod's "The topology of fibre bundles". |> The lemma is on page 10 in the 1999 paperback edition: Fortunately, it's in the same place in the ancient hardcover edition that I own ... :-) |> ----- |> Lemma: Let B and B' be coordinate bundles having the same fibre Y |> and group G, and let \overline{h}: X --> X' be a map of one base |> space into the other. |> Finally, let |> _ _ |> g_{kj}: V_j \cap h^{-1}(V_k') --> G |> be a set of continuous maps satisfying the conditions |> _ _ _ |> g_{kj}(x) g_{ji}(x) = g_{ki}(x), x \in V_i \cap V_j \cap h^{-1}(V'_k) |> _ _ _ _ |> g'_{lk}(h(x)) g_{kj}(x) = g_{lj}(x), x \in V_j \cap h^{-1}(V'_k \cap V'_j) |> _ |> Then there exists one and only one map h: B --> B' inducing h and |> _ |> having { g_{jk} } as its mapping transformations. |> ------ |> _ |> But g_{jk} is defined on the previous page as |> g_{kj}(x) = (\phi'_{k, x'})^{-1} h_x \phi_{j,x} |> where h_x: Y_x --> Y_{x'} from the fibre over x in X to the fibre over |> _ |> x' in X' is the map INDUCED by h. (note also that x' = h(x)). |> My problem with this is that he proves the lemma by constructing h |> locally in the bundle and in the construction his h_{kj} actually |> _ _ |> depends on g_{kj}(x). We see by the definition above that g_{kj}(x) |> has h_x in its definition, which is itself induced by h. |> So I am confused, it seems like he is defining his unique h but using |> h in its own definition! It seems circular. |> Help! OK ... you're misunderstanding the "big picture" here. In the discussion prior to this Lemma, Steenrod defines what a bundle map is ... and, as part of that definition, he introduces the maps g_{kj}(x): Y_x --> Y_{h#(x)} (where each g_{kj} is defined on a certain open subset of the base space). (Here, 'h#' is what Steenrod denotes by 'h' with an overbar -- too hard to do that here :-) Because these maps g_{kj} arise from a bundle map, they satisfy some conditions that are listed in (19). The _purpose_ of the Lemma is to show that you can "go backwards" -- given a map of the base spaces and a collection of maps g_{kj} defined on the right sorts of open sets in X and satisfying the conditions (19) [which, you'll notice, do _not_ involve the original bundle map h], there is a unique bundle map h which lifts the given map of the base spaces _and_ has the given collection of g_{kj}'s as its "mapping transformations". The proof (as you observe) proceeds by defining h locally -- a definition that's pretty much forced -- and then using the conditions on the g_{kj} to show that the various local definitions agree wherever they overlap. Having done that, one can paste the local definitions together to get the global map h: B --> B'. That's the hard part -- the facts that h _is_ a bundle map and that the g_{kj}'s give the corresponding mapping transformations just drop out from the way in which h was defined. Did that help ?? -- Ed Hook | Copula eam, se non posit Computer Sciences Corporation | acceptera jocularum. NAS, NASA Ames Research Center | All opinions herein expressed are Internet: hook@nas.nasa.gov | mine alone ============================================================================== From: Terry Pilling Subject: Re: Help with Steenrod Date: Fri, 15 Sep 2000 10:09:22 -0400 Newsgroups: sci.math On 15 Sep 2000, Ed Hook wrote: > OK ... you're misunderstanding the > "big picture" here. In the discussion > prior to this Lemma, Steenrod defines > what a bundle map is ... and, as part > of that definition, he introduces the > maps g_{kj}(x): Y_x --> Y_{h#(x)} (where > each g_{kj} is defined on a certain open > subset of the base space). (Here, 'h#' > is what Steenrod denotes by 'h' with an > overbar -- too hard to do that here :-) > Because these maps g_{kj} arise from a > bundle map, they satisfy some conditions > that are listed in (19). Alright! So I was confused because I was thinking that the g_{kj} had to be defined in the same way as he had done earlier (i.e. with the h). > > The _purpose_ of the Lemma is to show > that you can "go backwards" -- given > a map of the base spaces and a collection > of maps g_{kj} defined on the right sorts > of open sets in X and satisfying the > conditions (19) [which, you'll notice, > do _not_ involve the original bundle map h], > there is a unique bundle map h which > lifts the given map of the base spaces > _and_ has the given collection of g_{kj}'s > as its "mapping transformations". > So he is giving me the collection of maps g_{ik} to the group. Which are defined only by the fact that they obey the transitivity-type conditions, agreeing on the overlapping sets. This is actually what I thought the answer to my problem was in the first place, but I didn't want to jump to that conclusion. I wanted to be sure I knew the proof clearly. In fact in one of my posts I said that perhaps it was: given g_{jk} and h# find h. I guess that that was correct. > The proof (as you observe) proceeds by > defining h locally -- a definition that's > pretty much forced -- and then using the > conditions on the g_{kj} to show that the > various local definitions agree wherever > they overlap. Having done that, one can > paste the local definitions together to > get the global map h: B --> B'. That's the > hard part -- the facts that h _is_ a bundle > map and that the g_{kj}'s give the > corresponding mapping transformations just > drop out from the way in which h was defined. > > Did that help ?? That totally helped. It is clear as an azure sky of deepest summer. :) This is the kind of reply that I wanted. Thank-you. One more little thing: I noticed, as was also mentioned in Steenrod, that in his definition of a bundle (at the beginning of the book anyway) does not distinguish between the Klein bottle and the twisted torus. Defining a bundle by: generic bundle = {X, Y, p(x,y), \phi(x,y), H,G} X = base space Y = Typical fibre p = projection \phi = local trivialization H = Bundle group G = Isometry group of fibre The bundles for each are Klein = {S^1, S^1, p(x,y)=x, \phi(x,y)=(x,y), Z_2, SO(2)} T-Torus = {S^1, S^1, p(x,y)=x, \phi(x,y)=(x,y), Z_2, SO(2)} which are identical. So the definition does not distinguish between these two different bundles. I was thinking that it _would_ distinguish them if we added one thing to our definition, namely generic bundle = {X, Y, p(x,y), \phi(x,y), H, G, H/G} and since H/G = Z_2 for the Klein bottle and H/G = 1 for the twisted torus they would be distinguished. I think Steenrod suggests that one state whether H is a subgroup of G or not to distinguish them, this works too, but isn't there some cases where H/G would distinguish two bundles that H < G would not? Like for example when H and G are different, but they both have a common subgroup, and similarly for another bundle but the common subgroup is different? -Ter ============================================================================== From: hook@nas.nasa.gov (Ed Hook) Subject: Re: Help with Steenrod Date: 17 Sep 2000 18:41:12 GMT Newsgroups: sci.math In , Terry Pilling writes: |> One more little thing: |> I noticed, as was also mentioned in Steenrod, that in |> his definition of a bundle (at the beginning of the book |> anyway) does not distinguish between the Klein bottle and |> the twisted torus. Actually, that's not true -- the Klein bottle and the twisted torus _are_ distinguished by the the definitions that Steenrod provides. Just for starters, if the Klein bottle and the twisted torus _were_ equivalent as bundles, the "bundle spaces" would have to be homeomorphic. And the Klein bottle and the twisted torus are not homeomorphic. In fact, the twisted torus is (as a space) just the torus ... |> Defining a bundle by: |> generic bundle = {X, Y, p(x,y), \phi(x,y), H,G} |> X = base space |> Y = Typical fibre |> p = projection |> \phi = local trivialization |> H = Bundle group |> G = Isometry group of fibre I don't think that Steenrod's definition has this G -- it certainly has H and (at least implicitly) an action of H on the fibers (which you can think of as a homomorphism H --> Homeo(Y)). |> The bundles for each are |> Klein = {S^1, S^1, p(x,y)=x, \phi(x,y)=(x,y), Z_2, SO(2)} ... I'm not sure that I buy this. Actually, I'm troubled by the fact that, in this case, the group G should probably be (at least) O(2). That's because the nontrivial element of the bundle group is _not_ an element of SO(2). |> T-Torus = {S^1, S^1, p(x,y)=x, \phi(x,y)=(x,y), Z_2, SO(2)} |> which are identical. So the definition does not distinguish between |> these two different bundles. No -- the definition involves more than just the various objects that you list (and, as noted above, doesn't involve *all* of those :-). But, more to the point, there's a separate definition of just what it means for two bundles to be equivalent = isomorphic = indistinguishable. And, according to that definition, the Klein bottle and the twisted torus are _not_ the same as Z_2-bundles with fibre S^1 ... |> I was thinking that it _would_ distinguish them if we added |> one thing to our definition, namely |> generic bundle = {X, Y, p(x,y), \phi(x,y), H, G, H/G} |> and since H/G = Z_2 for the Klein bottle and H/G = 1 for |> the twisted torus they would be distinguished. I don't understand your notation. But I'm pretty sure that the Steenrod definitions really _do_ handle this situation ... |> I think Steenrod suggests that one state whether H is a |> subgroup of G or not to distinguish them, this works |> too, but isn't there some cases where H/G would distinguish |> two bundles that H < G would not? Like for example when |> H and G are different, but they both have a common subgroup, |> and similarly for another bundle but the common subgroup |> is different? Where does he say this stuff ?? If you give me a hint, I'll take a look and see if I have anything useful to contribute to further discussion ... -- Ed Hook | Copula eam, se non posit Computer Sciences Corporation | acceptera jocularum. NAS, NASA Ames Research Center | All opinions herein expressed are Internet: hook@nas.nasa.gov | mine alone ============================================================================== From: Terry Pilling Subject: Re: Help with Steenrod Date: Sun, 17 Sep 2000 14:52:25 -0400 Newsgroups: sci.math On 17 Sep 2000, Ed Hook wrote: > In , > Terry Pilling writes: > > |> One more little thing: > > |> I noticed, as was also mentioned in Steenrod, that in > |> his definition of a bundle (at the beginning of the book > |> anyway) does not distinguish between the Klein bottle and > |> the twisted torus. > > Actually, that's not true -- the Klein bottle > and the twisted torus _are_ distinguished by > the the definitions that Steenrod provides. No they aren't. They both have the same bundle group which is Z_2 (integers modulo 2) the _difference_ between them is that the Z_2 for the twisted torus is a subgroup of SO(2) of the fibre. The Z_2 for the Klein bottle is also a reflection but it is _not_ a subgroup of SO(2) since it is reflection through the diameter of S^1 rather than the center (so a subgroup of O(2)). Perhaps later in the book he refines his definition. But at the beginning of the book at least he only _says_ that they are distinguished by the fact that the group of the twisted torus is contained in the full group of rotations of the fibre (i.e. SO(2)) and thus is homeomorphic to the regular torus. This is just _stated_ but is (so far) not built in to the definition of the bundle. Look at the definition that he gives for a bundle (first page?) and you will see that within that definition they are not distinguished. > > Just for starters, if the Klein bottle and the > twisted torus _were_ equivalent as bundles, the > "bundle spaces" would have to be homeomorphic. I am _not_ saying that they are equivalent. I know that they are not. What I am saying is that his definition of a fibre bundle at the beginning of the book is not sufficient to distinguish the two. > |> Klein = {S^1, S^1, p(x,y)=x, \phi(x,y)=(x,y), Z_2, SO(2)} > > ... I'm not sure that I buy this. Actually, > I'm troubled by the fact that, in this case, > the group G should probably be (at least) > O(2). That's because the nontrivial element > of the bundle group is _not_ an element of > SO(2). Yeah. hmmm.... maybe I am just screwed up here. My reasoning is this. The group of the fibres are both Z_2, the difference between them is that the Z_2 for the twisted torus is equivalent to rotation by pi and is therefore a subgroup of SO(2) of the fibre. The Z_2 OTOH is not equivalent to an element of SO(2) and, as you mentioned, it is equivalent to a reflection through the diameter or the parity operation and is thus an element of O(2). So I guess my whole idea is gonzo now. The fibres have group O(2). My problem is that you have to tell me _which_ Z_2 you are referring to in order that I know which bundle you are talking about. i.e. For the torus R = rotation by \pi, R^2 = e => Z_2 For the KB R = reflection through x axis, R^2 = e => Z_2 The bundle groups are isomorphic as groups, the difference is that they are not isomorphic as subgroups of O(2). > > Where does he say this stuff ?? If you > give me a hint, I'll take a look and see > if I have anything useful to contribute > to further discussion ... He discusses the whole twisted torus - Klein bottle problem at the beginning of the book. He also explains why they are different. But I don't think that his definition handles it without at least extra words. He says subgroup of the rotation group of the fibre which I take to mean SO(2) not O(2). -Ter ============================================================================== From: hook@nas.nasa.gov (Ed Hook) Subject: Re: Help with Steenrod Date: 18 Sep 2000 17:19:38 GMT Newsgroups: sci.math In , Terry Pilling writes: |> On 17 Sep 2000, Ed Hook wrote: |> > In , |> > Terry Pilling writes: |> > |> One more little thing: |> > |> I noticed, as was also mentioned in Steenrod, that in |> > |> his definition of a bundle (at the beginning of the book |> > |> anyway) does not distinguish between the Klein bottle and |> > |> the twisted torus. |> > Actually, that's not true -- the Klein bottle |> > and the twisted torus _are_ distinguished by |> > the the definitions that Steenrod provides. |> No they aren't. They both have the same bundle group |> which is Z_2 (integers modulo 2) the _difference_ |> between them is that the Z_2 for the twisted torus |> is a subgroup of SO(2) of the fibre. |> The Z_2 for the Klein bottle is also a reflection |> but it is _not_ a subgroup of SO(2) since it is |> reflection through the diameter of S^1 rather than |> the center (so a subgroup of O(2)). I think that I wasn't sufficiently clear ... There are (at least) two definitions involved here -- there's (first) the definition of a fibre bundle and then there's the definition of just exactly what it means for two such to be equivalent/isomorphic/indistinguishable. And, using those definitions (which actually appear in the book _after_ the initial discussion about the "intuitive" notion that the definitions are trying to capture), the Klein bottle and the twisted torus are _not_ equivalent as bundles over S^1 with group Z_2. To be explicit, the *definition* of 'fibre bundle' is found in 2.4 (but requires the definition of 'coordinate bundle' that Steenrod gives in 2.3). And the notion of equivalence of two bundles is first _defined_ in the discussion following Lemma 2.7. And, using *that* definition, the Klein bottle and the twisted torus are easily seen to be distinct -- that's because "equivalence as bundles" implies that the total spaces (sorry, "bundle spaces") have to be homeomorphic (_via_ a homeomorphism with some additional properties relating to bundlehood). Since the Klein bottle and the torus are _not_ homeomorphic, the corresponding bundles can't be equivalent. |> Perhaps later in the book he refines his definition. |> But at the beginning of the book at least he only |> _says_ that they are distinguished by the fact that |> the group of the twisted torus is contained in the |> full group of rotations of the fibre (i.e. SO(2)) and |> thus is homeomorphic to the regular torus. This is just |> _stated_ but is (so far) not built in to the |> definition of the bundle. Look at the definition |> that he gives for a bundle (first page?) and you |> will see that within that definition they are not |> distinguished. Again, that's *not* the _definition_. Read carefully the last paragraph in 1.1 -- it's basically saying that the purpose of the remainder of Section 1 is to give a "brief and intuitive" discussion of some examples that will motivate the *actual* definitions (that will be coming along a little later) -- the tentative definitions with which Steenrod begins Section 1 are just that -- a first stab at the thing that (he knows) will need to be refined if they're to be useful ... Admittedly, this can be a bit confusing - but that's how it goes. Also (as an aside) you're slightly misreading the stuff about the twisted torus. What Steenrod is actually alluding to is the fact that (since the bundle group is a Z_2 that "naturally" lives in SO(2)), the twisted torus can be viewed as a bundle over S^1 with bundle group SO(2) -- that's kinda squishy at the point where this discussion is taking place, but it's crystal clear once Steenrod has given the actual definition. And, if you enlarge the group in this way, then it's (relatively) easy to show that the twisted torus is equivalent _as_SO(2)_bundle_ to the product bundle S^1xS^1 --> S^1. In the same vein, if you try to play the same game with the Klein bottle = regarding it as an O(2)-bundle over S^1, it does _not_ become equivalent to a product bundle. If you persevere, you'll find these results proved in 4.4 ... |> > Just for starters, if the Klein bottle and the |> > twisted torus _were_ equivalent as bundles, the |> > "bundle spaces" would have to be homeomorphic. |> I am _not_ saying that they are equivalent. I know that |> they are not. What I am saying is that his definition |> of a fibre bundle at the beginning of the book is not |> sufficient to distinguish the two. Hopefully, I've convinced you that his definitions actually _do_ suffice ... ?? |> > |> Klein = {S^1, S^1, p(x,y)=x, \phi(x,y)=(x,y), Z_2, SO(2)} |> > ... I'm not sure that I buy this. Actually, |> > I'm troubled by the fact that, in this case, |> > the group G should probably be (at least) |> > O(2). That's because the nontrivial element |> > of the bundle group is _not_ an element of |> > SO(2). |> Yeah. hmmm.... maybe I am just screwed up here. |> My reasoning is this. The group of the fibres are both |> Z_2, the difference between them is that the Z_2 |> for the twisted torus is equivalent to rotation by |> pi and is therefore a subgroup of SO(2) of the fibre. |> The Z_2 OTOH is not equivalent to an element of SO(2) and, |> as you mentioned, it is equivalent to a reflection through |> the diameter or the parity operation and is thus an |> element of O(2). It's probably better (more "intrinsic", somehow ?) to simply realize that there's more than one effective action of Z_2 on the fibre S^1 -- if you let the nontrivial element act by reflection across the x-axis, you get one action; if you let it act _via_ rotation through a straight angle, you get another. The difference between the actions is reflected in the difference between the corresponding bundles over S^1 ... Later, you can get into the question of just how you can enlarge the group (or, for that matter, shrink it -- which is frequently more useful). |> So I guess my whole idea is gonzo now. The fibres have |> group O(2). No -- the group is Z_2, as noted. Initially, it's best to just work with the definitions. And they say that you need to fix a group -- in some sense, the most natural such groups are Homeo(Y) (or Diffeo(Y), if appropriate), but the most useful choices are much _smaller_ groups. So it's very important that the definition allow you to restrict yourself to a particular group ... |> My problem is that you have to tell me _which_ Z_2 you |> are referring to in order that I know which bundle you |> are talking about. Of course -- for these particular examples, that's really just about all that distinguishes them. But the definition incorporates that observation, because it requires that there be _given_ an effective action of the bundle group on the fibre ... really ... check it out. In order to define the Klein bottle as a Z_2-bundle over S^1 with fibre S^1 (following the definition that you'll find in 2.3/2.4, you *will* have to say just how the nontrivial element of Z_2 is supposed to act on the fibre -- at that point, you'll find yourself saying, "It's by reflection across the x-axis" (or something equivalent). If you then carry out the same exercise for the twisted torus, then (when you reach that same point) you'll be saying, "It's by rotation through 180 degrees (or \pi radians)". At that point, you should find yourself having a "V8 moment" -- _of_course_, they're different bundles -- they involve different actions on the fibre -- *sheesh* :-) |> i.e. |> For the torus R = rotation by \pi, R^2 = e => Z_2 |> For the KB R = reflection through x axis, R^2 = e => Z_2 |> The bundle groups are isomorphic as groups, the difference |> is that they are not isomorphic as subgroups of O(2). No, I think you really want to concentrate here on the "basics" -- the difference is that you have two distinct actions of the group on the fibre ... |> > Where does he say this stuff ?? If you |> > give me a hint, I'll take a look and see |> > if I have anything useful to contribute |> > to further discussion ... |> He discusses the whole twisted torus - Klein bottle problem |> at the beginning of the book. He also explains why they are |> different. But I don't think that his definition handles it |> without at least extra words. |> He says subgroup of the rotation group of the fibre which |> I take to mean SO(2) not O(2). Actually ... if you read the discussion in 1.5 a bit more carefully, it's really _all_ about the twisted torus (except for the parenthetical remark that the Klein bottle is _not_ S^1 x S^1). And the rotation group of the fibre _is_ SO(2) -- you're right about that. Again, check out what's said later (in 4.4) -- I think/hope that it'll clear up any remaining confusion ... -- Ed Hook | Copula eam, se non posit Computer Sciences Corporation | acceptera jocularum. NAS, NASA Ames Research Center | All opinions herein expressed are Internet: hook@nas.nasa.gov | mine alone ============================================================================== From: Terry Pilling Subject: Re: Help with Steenrod Date: Mon, 18 Sep 2000 13:43:29 -0400 Newsgroups: sci.math On 18 Sep 2000, Ed Hook wrote: > Also (as an aside) you're slightly misreading > the stuff about the twisted torus. What Steenrod > is actually alluding to is the fact that (since > the bundle group is a Z_2 that "naturally" lives > in SO(2)), the twisted torus can be viewed as > a bundle over S^1 with bundle group SO(2) -- that's > kinda squishy at the point where this discussion is > taking place, but it's crystal clear once Steenrod > has given the actual definition. And, if you enlarge > the group in this way, then it's (relatively) > easy to show that the twisted torus is equivalent > _as_SO(2)_bundle_ to the product bundle S^1xS^1 --> S^1. > In the same vein, if you try to play the same game > with the Klein bottle = regarding it as an O(2)-bundle > over S^1, it does _not_ become equivalent to a > product bundle. If you persevere, you'll find these > results proved in 4.4 ... > I get it now, I don't think I had read quite far enough. Let's see... For one bundle to be G-equivalent to another, he says that the coordinate transformations must satisfy (Lemma 2.10): g_{ji}' = \lambda_j(x)^{-1} g_{ji}(x) \lambda_i(x) where \lambda_j: V_j --> G. This means that if the transition functions for the two bundles are in the same conjugacy class in G, the bundles are G-equivalent. In particular, the trivial bundle has g_{ji}' = 1 so that if a bundle is G-equivalent to the trivial bundle then it must have g_{ji}(x) = \lambda_j(x) \lambda_i(x)^{-1} which is given in section 4.3. For the circle take the base space S^1, cover it with 2 open neighborhoods, V_1 and V_2 i.e. semi-circles, the intersection of the two neighborhoods is the union of two small arcs U and W. Define g_{12} = 1 in U and define g_{12} = -1 in W, *** which is the same as rotation by 180 degrees in G = SO(2) ***. Take \lambda_2(V_2) = e and take \lambda_1(x) be a function over V_1 which forms a continuous path in SO(2) from 1 to -1. Therefore the twisted torus is SO(2) equivalent to the product bundle. For the Klein bottle we can't take SO(2) since the group of the bundle is not even in SO(2) and it can't be SO(2)-equivalent to the product. If we think for a second that it might be O(2) equivalent to the product bundle we immediately see that it isn't. To see it clearly we try the same construction that we did for the twisted torus, defining g_{12} = 1 in U and g_{12} = -1 in W, *** which is the same as the reflection operation in O(2) ***. Now in order for the Klein bottle to be O(2) equivalent to the trivial bundle we need to find a continuous path \lambda_1(x) leading from 1 to -1 in O(2). This is impossible since 1 and -1 live in different connected components of O(2) and there can be no continuous path connecting them. So the twisted torus and the Klein bottle are not equivalent in any sense. Thanks for the help Ed, I feel I have a much better understanding of the first few sections of Steenrod now. -Ter ============================================================================== From: hook@nas.nasa.gov (Edward C. Hook) Subject: Re: Help with Steenrod Date: 18 Sep 2000 22:20:32 GMT Newsgroups: sci.math In , Terry Pilling writes: |> I get it now, I don't think I had read quite far enough. Let's see... |> For one bundle to be G-equivalent to another, he says that the |> coordinate transformations must satisfy (Lemma 2.10): |> g_{ji}' = \lambda_j(x)^{-1} g_{ji}(x) \lambda_i(x) |> where \lambda_j: V_j --> G. |> This means that if the transition functions for the two |> bundles are in the same conjugacy class in G, the bundles are |> G-equivalent. Almost ... the transition functions have to be conjugate in G "in a continuous manner", which is (almost surely) a stronger condition. The point is that the 'lambda's have to be continuous maps of the coordinate neighborhoods into the group G. But you've certainly got the idea correctly scoped out ... |> In particular, the trivial bundle has g_{ji}' = 1 so that |> if a bundle is G-equivalent to the trivial bundle then it |> must have |> g_{ji}(x) = \lambda_j(x) \lambda_i(x)^{-1} |> which is given in section 4.3. |> For the circle take the base space S^1, cover it with 2 open |> neighborhoods, V_1 and V_2 i.e. semi-circles, the intersection |> of the two neighborhoods is the union of two small arcs U and W. |> Define g_{12} = 1 in U and define g_{12} = -1 in W, |> *** which is the same as rotation by 180 degrees in G = SO(2) ***. |> Take \lambda_2(V_2) = e and take \lambda_1(x) be a function |> over V_1 which forms a continuous path in SO(2) from 1 to -1. |> Therefore the twisted torus is SO(2) equivalent to the product |> bundle. |> For the Klein bottle we can't take SO(2) since the group of the |> bundle is not even in SO(2) and it can't be SO(2)-equivalent to |> the product. If we think for a second that it might be O(2) |> equivalent to the product bundle we immediately see that it isn't. |> To see it clearly we try the same construction that we did for |> the twisted torus, defining g_{12} = 1 in U and g_{12} = -1 in W, |> *** which is the same as the reflection operation in O(2) ***. Nope -- yet another nit to pick :-) The antipodal map (aka '-1') is _not_ the same as the reflection operation that you want here. Just for starters, the involution that occurs in the definition of the Klein bottle has 2 fixed points on S^1, whereas the antipodal map never has _any_ fixed points. But the argument doesn't depend on this point -- the crux of the matter is that this involution and the identity map lie in different components of O(2) -- and any trivialization of the Klein bottle would be equivalent to a path in O(2) joining those two maps ... |> Now in order for the Klein bottle to be O(2) equivalent to the |> trivial bundle we need to find a continuous path \lambda_1(x) leading |> from 1 to -1 in O(2). This is impossible since 1 and -1 live in |> different connected components of O(2) and there can be no continuous |> path connecting them. |> So the twisted torus and the Klein bottle are not equivalent in |> any sense. |> Thanks for the help Ed, I feel I have a much better understanding |> of the first few sections of Steenrod now. Good -- keep plugging away and you'll probably learn a *lot* of interesting/useful stuff ... -- Ed Hook | Copula eam, se non posit Computer Sciences Corporation | acceptera jocularum. NAS, NASA Ames Research Center | All opinions herein expressed are Internet: hook@nas.nasa.gov | mine alone