Talk:Sphere bundles and spin (Ex)

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	m>4\implies x\in\pi_2(M)\quad &\text{can be embedded}\end{aligned}$$Hence we take an embedded sphere $\Sigma$ on which $w_2(M)$ does not vanish and call this the ''weird sphere''. Then this $\Sigma$ must have twisted normal bundle. Consider that for our untwisted surgery, $S^1$ must initially bound a disk and we can extend the normal framing of $S^1$ to the normal framing on $D^2$ (which is necessarily trivial). For the twisted surgery, $S^1$ must have the ''other'' framing. Take the $D^2$ bounded by $S^1$ to be inside a hemisphere of the weird sphere. Now form an isotopy moving $S^1$ from one hemisphere to the other of the weird sphere. This must necessarily exchange the framing we have on $S^1$ from trivial to twisted as the normal bundle of the weird sphere is twisted. However, the surgery at either end of an isotopy gives a diffeomorphic effect.<br /><br />		m>4\implies x\in\pi_2(M)\quad &\text{can be embedded}\end{aligned}$$Hence we take an embedded sphere $\Sigma$ on which $w_2(M)$ does not vanish and call this the ''weird sphere''. Then this $\Sigma$ must have twisted normal bundle. Consider that for our untwisted surgery, $S^1$ must initially bound a disk and we can extend the normal framing of $S^1$ to the normal framing on $D^2$ (which is necessarily trivial). For the twisted surgery, $S^1$ must have the ''other'' framing. Take the $D^2$ bounded by $S^1$ to be inside a hemisphere of the weird sphere. Now form an isotopy moving $S^1$ from one hemisphere to the other of the weird sphere. This must necessarily exchange the framing we have on $S^1$ 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.<br /><br />
−	To see this last part a different way, we may take a cylinder $M\times [0,1]$. This induces a trivial isotopy of the embedded $S^1$. At some $t\in[0,1]$, take a connect sum with the weird sphere and the isotopy $S^1\times[0,1]$. The $S^1$ now slides along the tube and then over the weird sphere. This is not an isotopy but is now a concordance. However, within codimension $>3$ we can improve a concordance to an isotopy.<br /><br />	+	To see this last part a different way, we may take a cylinder $M\times [0,1]$. This induces a trivial isotopy of the embedded $S^1$. At some $t\in[0,1]$, take a connect sum with the weird sphere and the isotopy $S^1\times[0,1]$. Now make the $S^1$ slide along the tube and then over the weird sphere and down to the bottom of the tube as $t$ increases from 0 to 1. This is not an isotopy but is now a concordance. However, within codimension $>3$ we can improve a concordance to an isotopy.<br /><br />
	This does not cover the case $m=4$, I think it is still possible here but I don't have the solution.		This does not cover the case $m=4$, I think it is still possible here but I don't have the solution.
	</wikitex>		</wikitex>

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Revision as of 23:11, 29 March 2012

Part 1
This is a standard clutching construction. Fix $<wikitex>; '''Part 1'''<br />This is a standard clutching construction. Fix $k\geq 2$ and suppose we have a linear $S^k$ bundle $A$ over the sphere $S^2=D_-^2\cup D_+^2$. A fibre bundle over a contractible space is trivial (up to bundle isomorphism), so without loss of generality $A|_{D_+}\cong A|_{D_-}\cong D^2\times S^k$. We can now glue back the bundles via an automorphism of the fibre at every point on the boundary $\partial D_\pm=S^1$, varying continuously on the base i.e. a continuous map $f:S^1\to \text{Aut}(S^k)$. In fact, a homotopic map $f\simeq g$ will produce an isomorphic bundle so we are interested in a class of $$[S^1,\text{Aut}(S^k)]=[S^1,SO(k+1)]=\Z_2$$ (as we are in the stable range). Hence there are two choices and so two bundles up to isomorphism. One the trivial bundle $S^k\times S^2$ and so the other will be called the ''twisted bundle'' $S^k\tilde{\times} S^2$.<br /><br /> '''Part 2'''<br />The sphere bundle of a 2-plane bundle is an $S^1$-bundle, so the arguments from above carry through here as well. The sphere bundle of $E_k$ is given by the clutching construction above with the clutching map an element of $$[S^1,\text{Aut}(S^1)]=[S^1,SO(2)]=[S^1,S^1]=\Z,$$where the isomorphism is the winding number and this is the same as the Euler number of the resulting bundle. Now embed an $S^1$ in $S^3$ for surgery. We use the standard embedding of a sphere inside a larger sphere: $$S^3=\partial(D^2\times D^2)=S^1\times D^2\cup_{id}D^2\times S^1.$$Form a framed embedding $S^1\times D^2$ in the first factor where the framing is given by twisting the meridian around $k$ times as we pass around the $S^1$ i.e. it is the element $k\in[S^1,\text{Aut}(D^2)]=\Z$ represented by a map $\omega:S^1\to \text{Aut}(D^2)$. Now do surgery. The effect is the gluing $$D^2\times S^1\cup_{f}D^2\times S^1$$where $f(v,x)=(\omega(x)(v),x)$. This is now just the clutching construction as above.<br /><br /> '''Part 3'''<br />As $S^1$ is nullhomotopically embedded, we may consider this inside a contractible disk, or in the second summand of $M\cong M\# S^m$. Moreover we may embed it using the standard embedding $S^m=\partial(D^2\times D^m-1)$ as above. Hence the result will be $M'=M\#N$ where $N$ is either the trivial or twisted linear $(m-2)$-sphere bundle over $S^2$. As $M$ is spin, if $N$ is the trivial bundle then the effect of surgery is also spin. However, if $N$ is twisted then $M'$ cannot be spin as connect sum results in direct sum of second Stiefel-Whitney classes and $w_2(S^k\tilde{\times} S^2)$ is non-vanishing. To see this consider that the bundle over $S^2$ itself is not spin (that it is clutched by the non-trivial element of $\pi_1(SO(m-1))$ is more or less the definition of the obstruction to lifting to the spin group) and that this implies that the total space $S^k\tilde{\times} S^2$ is also spin.<br /><br /> Now the hard part! Assume $M$ is not spin, this means that there is a cocycle on which $w_2(M)$ does not vanish. $$ \begin{aligned}H_2(M;\Z_2)\cong H_2(M;\Z)\otimes\Z_2\qquad &\text{(Univ. coeff. thm.)}\ H_2(M;\Z)\cong\pi_2(M)\qquad &\text{(Hurewicz)}\ m>4\implies x\in\pi_2(M)\quad &\text{can be embedded}\end{aligned}$$Hence we take an embedded sphere $\Sigma$ on which $w_2(M)$ does not vanish and call this the ''weird sphere''. Then this $\Sigma$ must have twisted normal bundle. Consider that for our untwisted surgery, $S^1$ must initially bound a disk and we can extend the normal framing of $S^1$ to the normal framing on $D^2$ (which is necessarily trivial). For the twisted surgery, $S^1$ must have the ''other'' framing. Take the $D^2$ bounded by $S^1$ to be inside a hemisphere of the weird sphere. Now form an isotopy moving $S^1$ from one hemisphere to the other of the weird sphere. This must necessarily exchange the framing we have on $S^1$ 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.<br /><br /> To see this last part a different way, we may take a cylinder $M\times [0,1]$. This induces a trivial isotopy of the embedded $S^1$. At some $t\in[0,1]$, take a connect sum with the weird sphere and the isotopy $S^1\times[0,1]$. Now make the $S^1$ slide along the tube and then over the weird sphere and down to the bottom of the tube as $t$ increases from 0 to 1. This is not an isotopy but is now a concordance. However, within codimension $>3$ we can improve a concordance to an isotopy.<br /><br /> This does not cover the case $m=4$, I think it is still possible here but I don't have the solution. </wikitex>k\geq 2$ and suppose we have a linear $S^k$ bundle $A$ over the sphere $S^2=D_-^2\cup D_+^2$. A fibre bundle over a contractible space is trivial (up to bundle isomorphism), so without loss of generality $A|_{D_+}\cong A|_{D_-}\cong D^2\times S^k$. We can now glue back the bundles via an automorphism of the fibre at every point on the boundary $\partial D_\pm=S^1$, varying continuously on the base i.e. a continuous map $f:S^1\to \text{Aut}(S^k)$. In fact, a homotopic map $f\simeq g$ will produce an isomorphic bundle so we are interested in a class of (as we are in the stable range). Hence there are two choices and so two bundles up to isomorphism. One the trivial bundle $S^k\times S^2$ and so the other will be called the twisted bundle $S^k\tilde{\times} S^2$.

Part 2
The sphere bundle of a 2-plane bundle is an $S^1$-bundle, so the arguments from above carry through here as well. The sphere bundle of $E_k$ is given by the clutching construction above with the clutching map an element of where the isomorphism is the winding number and this is the same as the Euler number of the resulting bundle. Now embed an $S^1$ in $S^3$ for surgery. We use the standard embedding of a sphere inside a larger sphere: Form a framed embedding $S^1\times D^2$ in the first factor where the framing is given by twisting the meridian around $k$ times as we pass around the $S^1$ i.e. it is the element $k\in[S^1,\text{Aut}(D^2)]=\Z$ represented by a map $\omega:S^1\to \text{Aut}(D^2)$. Now do surgery. The effect is the gluing where $f(v,x)=(\omega(x)(v),x)$. This is now just the clutching construction as above.

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

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

Hence we take an embedded sphere $\Sigma$ on which $w_2(M)$ does not vanish and call this the weird sphere. Then this $\Sigma$ must have twisted normal bundle. Consider that for our untwisted surgery, $S^1$ must initially bound a disk and we can extend the normal framing of $S^1$ to the normal framing on $D^2$ (which is necessarily trivial). For the twisted surgery, $S^1$ must have the other framing. Take the $D^2$ bounded by $S^1$ to be inside a hemisphere of the weird sphere. Now form an isotopy moving $S^1$ from one hemisphere to the other of the weird sphere. This must necessarily exchange the framing we have on $S^1$ 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 $M\times [0,1]$. This induces a trivial isotopy of the embedded $S^1$. At some $t\in[0,1]$, take a connect sum with the weird sphere and the isotopy $S^1\times[0,1]$. Now make the $S^1$ slide along the tube and then over the weird sphere and down to the bottom of the tube as $t$ increases from 0 to 1. This is not an isotopy but is now a concordance. However, within codimension $>3$ we can improve a concordance to an isotopy.

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