Representative Surfaces (For H_2(M) )Always Exist. If M is a4-manifold.
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From: Bacle on 11 Apr 2010 17:58 Hi, I am trying to show that for M a 4-manifold, and [a]_2 a class in H_2(M,Z) , there is always a surface that represents [a]_2 , i.e., there exists a surface S , and an embedding i of S into M , with [ioS]_2 =[a]_2. ** What I have ** If M is simply-connected, so that Pi_1(M)=0 (Notation: Pi_1:=Fund. Grp.) Then, by the Hurewicz Theorem (Hip, Hip Hurewicz!) Pi_2(M) is actually Isomorphic to H_2(M;Z) , so that every class in H_2(M;Z) can be represented as an embedded sphere S^2 (possibly with self-intersections, which can be smoothed away ). **BUT** I can't think of what can be done if M is not simply-connected. Any Ideas.? Thanks.
From: W. Dale Hall on 12 Apr 2010 01:01 Bacle wrote: > Hi, I am trying to show that for M a 4-manifold, > and [a]_2 a class in H_2(M,Z) , there is always > a surface that represents [a]_2 , i.e., there > exists a surface S , and an embedding i of S into > M , with [ioS]_2 =[a]_2. > > ** What I have ** > > If M is simply-connected, so that Pi_1(M)=0 > (Notation: Pi_1:=Fund. Grp.) > Then, by the Hurewicz Theorem (Hip, Hip Hurewicz!) > Pi_2(M) is actually Isomorphic to H_2(M;Z) , so that > every class in H_2(M;Z) can be represented as an > embedded sphere S^2 (possibly with self-intersections, > which can be smoothed away ). Sorry, you don't smoothe away self-intersections. Removing self-intersections of a k-manifold immersed in a 2k-manifold requires the so-called Whitney trick, which doesn't work in dimension 2k = 4. > > **BUT** I can't think of what can be done if > > M is not simply-connected. > > Any Ideas.? > > Thanks. Here's something that works for any smooth manifold, any dimension: Every element x of H^2(M; Z) is represented by a unique homotopy class of maps f_x : M ---> CP^oo (infinite-dimensional complex projective space). If M is of dimension m, then the cellular approximation theorem shows that f_x can be assumed to map into the m-skeleton of CP^oo, and that map is unique up to homotopy into that m-skeleton (since the m- skeleton is equal to the (m+1)-skeleton if m is even, and if m is odd, it's equal to the (m-1)-skeleton. In either case, maps from an m- dimensional CW-complex can always be assumed to map into an even- dimensional skeleton of CP^oo, and homotopies among such maps can be deformed (modulo the ends) into the same-dimensional skeleton. So, f_x : M^m ---> CP^k for k = [m/2] ([...] the greatest integer function). Make f_x transverse to your favorite CP^(k-1) and consider its preimage under f_x, to get a codimension 2 submanifold F_x, whose fundamental class [F_x] in H_(m-2)(M) is Poincare-dual to x. If M is orientable, and m=4, you find H_2(M) ~ H^2(M), and you're done. Dale
From: W. Dale Hall on 12 Apr 2010 01:16 W. Dale Hall wrote: > Bacle wrote: >> Hi, I am trying to show that for M a 4-manifold, and [a]_2 a class >> in H_2(M,Z) , there is always a surface that represents [a]_2 , >> i.e., there exists a surface S , and an embedding i of S into M , >> with [ioS]_2 =[a]_2. >> >> ** What I have ** >> >> If M is simply-connected, so that Pi_1(M)=0 (Notation: Pi_1:=Fund. >> Grp.) Then, by the Hurewicz Theorem (Hip, Hip Hurewicz!) Pi_2(M) is >> actually Isomorphic to H_2(M;Z) , so that every class in H_2(M;Z) >> can be represented as an embedded sphere S^2 (possibly with >> self-intersections, which can be smoothed away ). > > Sorry, you don't smoothe away self-intersections. Removing > self-intersections of a k-manifold immersed in a 2k-manifold requires > the so-called Whitney trick, which doesn't work in dimension 2k = 4. > >> >> **BUT** I can't think of what can be done if >> >> M is not simply-connected. >> >> Any Ideas.? >> >> Thanks. > > Here's something that works for any smooth manifold, any dimension: > Every element x of H^2(M; Z) is represented by a unique homotopy > class of maps f_x : M ---> CP^oo (infinite-dimensional complex > projective space). If M is of dimension m, then the cellular > approximation theorem shows that f_x can be assumed to map into the > m-skeleton of CP^oo, and that map is unique up to homotopy into that > m-skeleton (since the m- skeleton is equal to the (m+1)-skeleton if m > is even, and if m is odd, it's equal to the (m-1)-skeleton. In either > case, maps from an m- dimensional CW-complex can always be assumed to > map into an even- dimensional skeleton of CP^oo, and homotopies among > such maps can be deformed (modulo the ends) into the same-dimensional > skeleton. > > So, f_x : M^m ---> CP^k for k = [m/2] ([...] the greatest integer > function). Make f_x transverse to your favorite CP^(k-1) and > consider its preimage under f_x, to get a codimension 2 submanifold > F_x, whose fundamental class [F_x] in H_(m-2)(M) is Poincare-dual to > x. I note that I'm a bit rusty, so the issue of [F_x] being Poincare-dual to x may not be as trivial as I'm thinking, if M is not orientable. After all, there is no fundamental class [M] in H_m(M; Z) = 0 in that case. You do get such a class in H_m(M; Z_2), and F_x should be Poincare-dual (mod 2) to x. Dale
From: Bacle on 12 Apr 2010 08:43 W.Dale Hall Wrote, in Part: " Here's something that works for any smooth manifold, any dimension: Every element x of H^2(M; Z) is represented by a unique homotopy class of maps f_x : M ---> CP^oo (infinite-dimensional complex projective space)." Is this because CP^oo is a K(Z,2) space ( i.e. an Eilenberg-McLane construction with Pi_2(S)=Z, and Pi_m(S)=0 for m=/2.?. And I guess then you are using the classifying space,? Thanks.
From: W. Dale Hall on 14 Apr 2010 00:56
Bacle wrote: > W.Dale Hall Wrote, in Part: > > " Here's something that works for any smooth manifold, any > dimension: > > Every element x of H^2(M; Z) is represented by a unique homotopy > class of maps f_x : M ---> CP^oo (infinite-dimensional complex > projective space)." > > Is this because CP^oo is a K(Z,2) space ( i.e. an > Eilenberg-McLane construction with Pi_2(S)=Z, and > Pi_m(S)=0 for m=/2.?. And I guess then you are using > the classifying space,? > > Thanks. That's it exactly. Dale |