Talk:Sphere bundles and spin (Ex)
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− | 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 $f:S^1\to Aut(S^k)$. 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 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 $f:S^1\to Aut(S^k)$. 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 /> |
− | 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 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 /> |
− | 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 /> | + | '''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. | Now the hard part! Assume $M$ is not spin, this means that there is a cocycle on which $w_2(M)$ does not vanish. |
Revision as of 22:06, 29 March 2012
This is a standard clutching construction. Fix and suppose we have a linear bundle over the sphere . A fibre bundle over a contractible space is trivial (up to bundle isomorphism), so without loss of generality . We can now glue back the bundles via an automorphism of the fibre at every point on the boundary , varying continuously on the base i.e. a continuous map . In fact, a homotopic map will produce an isomorphic bundle so we are interested in a class of
Part 2
The sphere bundle of a 2-plane bundle is an -bundle, so the arguments from above carry through here as well. The sphere bundle of is given by the clutching construction above with the clutching map an element of
Part 3
As is nullhomotopically embedded, we may consider this inside a contractible disk or, in the second summand of . Moreover we may embed it using the standard embedding as above. Hence the result will be where is either the trivial or twisted linear -sphere bundle over . As is spin, if is the trivial bundle then the effect of surgery is also spin. However, if is twisted then cannot be spin as connect sum results in direct sum of second Stiefel-Whitney classes and is non-vanishing. To see this consider that the bundle over itself is not spin (that it is clutched by the non-trivial element of is more or less the definition of the obstruction to lifting to the spin group) and that this implies that the total space is also spin.
Now the hard part! Assume is not spin, this means that there is a cocycle on which does not vanish.
To see this last part a different way, we may take a cylinder . This describes a (boring!) isotopy of the embedded . At some , take a connect sum with the weird sphere and the isotopy . The now slides along the tube and then over the weird sphere. This is not an isotopy but is now a concordance. However, within codimension we can improve a concordance to an isotopy.