Talk:Sphere bundles and spin (Ex)

Part 1
This is a standard clutching construction. Fix $k\geq 2$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}k\geq 2$ and suppose we have a linear $S^k$$S^k$ bundle $A$$A$ over the sphere $S^2=D_-^2\cup D_+^2$$S^2=D_-^2\cup D_+^2$. A fibre bundle over a contractible space is trivial (up to bundle isomorphism), so without loss of generality $A|_{D_+}\cong A|_{D_-}\cong D^2\times S^k$$A|_{D_+}\cong A|_{D_-}\cong D^2\times S^k$. We can now glue back the bundles via an automorphism of the fibre at every point on the boundary $\partial D_\pm=S^1$$\partial D_\pm=S^1$, varying continuously on the base i.e. a continuous map $f:S^1\to \text{Aut}(S^k)$$f:S^1\to \text{Aut}(S^k)$. In fact, a homotopic map $f\simeq g$$f\simeq g$ will produce an isomorphic bundle so we are interested in a class of

$\displaystyle [S^1,\text{Aut}(S^k)]=[S^1,SO(k+1)]=\Zz_2$ $$\displaystyle [S^1,\text{Aut}(S^k)]=[S^1,SO(k+1)]=\Zz_2$$

(as we are in the stable range). Hence there are two choices and so two bundles up to isomorphism. One the trivial bundle $S^k\times S^2$$S^k\times S^2$ and so the other will be called the twisted bundle $S^k\tilde{\times} S^2$$S^k\tilde{\times} S^2$.

Part 2
The sphere bundle of a 2-plane bundle is an $S^1$$S^1$-bundle, so the arguments from above carry through here as well. The sphere bundle of $E_k$$E_k$ is given by the clutching construction above with the clutching map an element of

$\displaystyle [S^1,\text{Aut}(S^1)]=[S^1,SO(2)]=[S^1,S^1]=\Zz,$ $$\displaystyle [S^1,\text{Aut}(S^1)]=[S^1,SO(2)]=[S^1,S^1]=\Zz,$$

where the isomorphism is the winding number and this is the same as the Euler number of the resulting bundle. Now embed an $S^1$$S^1$ in $S^3$$S^3$ for surgery. We use the standard embedding of a sphere inside a larger sphere:

$\displaystyle S^3=\partial(D^2\times D^2)=S^1\times D^2\cup_{id}D^2\times S^1.$ $$\displaystyle S^3=\partial(D^2\times D^2)=S^1\times D^2\cup_{id}D^2\times S^1.$$

Form a framed embedding $S^1\times D^2$$S^1\times D^2$ in the first factor where the framing is given by twisting the meridian around $k$$k$ times as we pass around the $S^1$$S^1$ i.e. it is the element $k\in[S^1,\text{Aut}(D^2)]=\Zz$$k\in[S^1,\text{Aut}(D^2)]=\Zz$ represented by a map $\omega:S^1\to \text{Aut}(D^2)$$\omega:S^1\to \text{Aut}(D^2)$. Now do surgery. The effect is the gluing

$\displaystyle D^2\times S^1\cup_{f}D^2\times S^1$ $$\displaystyle D^2\times S^1\cup_{f}D^2\times S^1$$

where $f(v,x)=(\omega(x)(v),x)$$f(v,x)=(\omega(x)(v),x)$. This is now just the clutching construction as above.

Part 3
As $S^1$$S^1$ is nullhomotopically embedded, we may consider this inside a contractible disk, or in the second summand of $M\cong M\# S^m$$M\cong M\# S^m$. Moreover we may embed it using the standard embedding $S^m=\partial(D^2\times D^{m-1})$$S^m=\partial(D^2\times D^{m-1})$ as above. Hence the result will be $M'=M\#N$$M'=M\#N$ where $N$$N$ is either the trivial or twisted linear $(m-2)$$(m-2)$-sphere bundle over $S^2$$S^2$. As $M$$M$ is spin, if $N$$N$ is the trivial bundle then the effect of surgery is also spin. However, if $N$$N$ is twisted then $M'$$M'$ cannot be spin as connect sum results in direct sum of second Stiefel-Whitney classes and $w_2(S^k\tilde{\times} S^2)$$w_2(S^k\tilde{\times} S^2)$ is non-vanishing. To see this consider that the bundle over $S^2$$S^2$ itself is not spin (that it is clutched by the non-trivial element of $\pi_1(SO(m-1))$$\pi_1(SO(m-1))$ is more or less the definition of the obstruction to lifting to the spin group) and that this implies that the total space $S^k\tilde{\times} S^2$$S^k\tilde{\times} S^2$ is also spin.

Now the hard part! Assume $M$$M$ is not spin, this means that there is a cocycle on which $w_2(M)$$w_2(M)$ does not vanish.

$\displaystyle \begin{aligned}H_2(M;\Zz_2)\cong H_2(M;\Zz)\otimes\Zz_2\qquad &\text{(Univ. coeff. thm.)}\\ H_2(M;\Zz)\cong\pi_2(M)\qquad &\text{(Hurewicz)}\\ m>4\implies x\in\pi_2(M)\quad &\text{can be embedded}\end{aligned}$ $$\displaystyle \begin{aligned}H_2(M;\Zz_2)\cong H_2(M;\Zz)\otimes\Zz_2\qquad &\text{(Univ. coeff. thm.)}\\ H_2(M;\Zz)\cong\pi_2(M)\qquad &\text{(Hurewicz)}\\ m>4\implies x\in\pi_2(M)\quad &\text{can be embedded}\end{aligned}$$

Hence we take an embedded sphere $\Sigma$$\Sigma$ on which $w_2(M)$$w_2(M)$ does not vanish and call this the weird sphere. Then this $\Sigma$$\Sigma$ must have twisted normal bundle. Consider that for our untwisted surgery, $S^1$$S^1$ must initially bound a disk and we can extend the normal framing of $S^1$$S^1$ to the normal framing on $D^2$$D^2$ (which is necessarily trivial). For the twisted surgery, $S^1$$S^1$ must have the other framing. Take the $D^2$$D^2$ bounded by $S^1$$S^1$ to be inside a hemisphere of the weird sphere. Now form an isotopy moving $S^1$$S^1$ from one hemisphere to the other of the weird sphere. This must necessarily exchange the framing we have on $S^1$$S^1$ from trivial to twisted as the normal bundle of the weird sphere is twisted. However, the surgery at either end of an isotopy gives a diffeomorphic effect.

To see this last part a different way, we may take a cylinder $M\times [0,1]$$M\times [0,1]$. This induces a trivial isotopy of the embedded $S^1$$S^1$. At some $t\in[0,1]$$t\in[0,1]$, take a connect sum with the weird sphere and the isotopy $S^1\times[0,1]$$S^1\times[0,1]$. Now make the $S^1$$S^1$ slide along the tube and then over the weird sphere and down to the bottom of the tube as $t$$t$ increases from 0 to 1. This is not an isotopy but is now a concordance. However, within codimension $>3$$>3$ we can improve a concordance to an isotopy.

This does not cover the case $m=4$$m=4$, I think it is still possible here but I don't have the solution.