Talk:Sphere bundles and spin (Ex)

This is a standard clutching construction. Fix $k\geq 2$$<wikitex>; This is a standard clutching construction. Fix $k\geq 2$ and suppose we have a linear $S^k$ bundle $A$ over the sphere $S^2=D_-^2\cup D_+^2$. A fibre bundle over a contractible space is trivial (up to bundle isomorphism), so without loss of generality $A|_{D_+}\cong A|_{D_-}\cong D^2\times S^k$. We can now glue back the bundles via an automorphism of the fibre at every point on the boundary $\partial D_\pm=S^1$, varying continuously on the base i.e. a continuous map $f:S^1\to Aut(S^k)$. In fact, a homotopic map $f\simeq g$ will produce an isomorphic bundle so we are interested in a class of $$[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$$where $f(v,x)=(\omega(x)(v),x)$. This is now just the clutching construction as above.<br /><br /> As $S^1$ is nullhomotopically embedded, we may consider this inside a contractible disk or, in the second summand of $M\cong M\# S^m$. Moreover we may embed it using the standard embedding $S^m=\partial(D^2\times D^m-1)$ as above. Hence the result will be $M'=M\#N$ where $N$ is either the trivial or twisted linear $(m-2)$-sphere bundle over $S^2$. As $M$ is spin, if $N$ is the trivial bundle then the effect of surgery is also spin. However, if $N$ is twisted then $M'$ cannot be spin as connect sum results in direct sum of second Stiefel-Whitney classes and $w_2(S^k\tilde{\times} S^2)$ is non-vanishing. To see this consider that the bundle over $S^2$ itself is not spin (that it is clutched by the non-trivial element of $\pi_1(SO(m-1))$ is more or less the definition of the obstruction to lifting to the spin group) and that this implies that the total space $S^k\tilde{\times} S^2$ is also spin. </wikitex>k\geq 2$ and suppose we have a linear $S^k$$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 Aut(S^k)$$f:S^1\to Aut(S^k)$. In fact, a homotopic map $f\simeq g$$f\simeq g$ will produce an isomorphic bundle so we are interested in a class of

$\displaystyle [S^1,\text{Aut}(S^k)]=[S^1,SO(k+1)]=\Z_2$ $$\displaystyle [S^1,\text{Aut}(S^k)]=[S^1,SO(k+1)]=\Z_2$$

(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$$S^k\times S^2$ and so the other will be called the twisted bundle $S^k\tilde{\times} S^2$$S^k\tilde{\times} S^2$.

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

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

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$$S^1$ in $S^3$$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$$S^1$ i.e. it is the element $k\in[S^1,\text{Aut}(D^2)]=\Z$$k\in[S^1,\text{Aut}(D^2)]=\Z$ 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.

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.