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Sam Nead
Sam Nead
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The boundary of your solid torus $T_L \cong D^2 \times S^1$ is equipped with a pair of foliations by circles (at "right angles"). The first one is the foliation by circles of the form $\partial D^2 \times \{\mathrm{pt}\}$. The second one is the foliation by circles of the form $\{\mathrm{pt}\} \times S^1$. I assume that by "$S$-transformation" you mean "cut $T_L$ out and glue it back in so that the foliations swap palces"places”. Essentially, you are "rotating the gluing map by 90 degrees". Assuming this is correct: we have the following.

The manifold $\tilde{M}$, obtained by cutting out $T_L$ and gluing it back in with an "$S$-transformation", is homeomorphic to $S^2 \, \tilde{\times}\, S^1$: the $S^2$ bundle over $S^1$ with monodromy the antipodal map.

To see this: After removing $T_L$, what remains is $M^2 \times S^1$. Note that $M^2$ is an interval bundle over the circle. Each interval, crossed with a circle, gives an annulus properly embedded in $M^2 \times S^1$. The $S$-transformation glues a pair of disks to the two boundaries of each such annulus. This is enough to show that $\tilde{M}$ is an $S^2$ bundle over the circle. There are only two of these - the product and the twisted bundle. Since $\tilde{M}$ is not orientable, it is homeomorphic to the twisted bundle.

The boundary of your solid torus $T_L \cong D^2 \times S^1$ is equipped with a pair of foliations by circles (at "right angles"). The first one is the foliation by circles of the form $\partial D^2 \times \{\mathrm{pt}\}$. The second one is the foliation by circles of the form $\{\mathrm{pt}\} \times S^1$. I assume that by "$S$-transformation" you mean "cut $T_L$ out and glue it back in so that foliations swap palces". Essentially, you are "rotating the gluing map by 90 degrees". Assuming this is correct:

The manifold $\tilde{M}$, obtained by cutting out $T_L$ and gluing it back in with an "$S$-transformation", is homeomorphic to $S^2 \, \tilde{\times}\, S^1$: the $S^2$ bundle over $S^1$ with monodromy the antipodal map.

To see this: After removing $T_L$, what remains is $M^2 \times S^1$. Note that $M^2$ is an interval bundle over the circle. Each interval, crossed with a circle, gives an annulus properly embedded in $M^2 \times S^1$. The $S$-transformation glues a pair of disks to the two boundaries of each such annulus. This is enough to show that $\tilde{M}$ is an $S^2$ bundle over the circle. There are only two of these - the product and the twisted bundle. Since $\tilde{M}$ is not orientable, it is homeomorphic to the twisted bundle.

The boundary of your solid torus $T_L \cong D^2 \times S^1$ is equipped with a pair of foliations by circles (at "right angles"). The first one is the foliation by circles of the form $\partial D^2 \times \{\mathrm{pt}\}$. The second one is the foliation by circles of the form $\{\mathrm{pt}\} \times S^1$. I assume that by "$S$-transformation" you mean "cut $T_L$ out and glue it back in so that the foliations swap places”. Essentially, you are "rotating the gluing map by 90 degrees". Assuming this is correct we have the following.

The manifold $\tilde{M}$, obtained by cutting out $T_L$ and gluing it back in with an "$S$-transformation", is homeomorphic to $S^2 \, \tilde{\times}\, S^1$: the $S^2$ bundle over $S^1$ with monodromy the antipodal map.

To see this: After removing $T_L$, what remains is $M^2 \times S^1$. Note that $M^2$ is an interval bundle over the circle. Each interval, crossed with a circle, gives an annulus properly embedded in $M^2 \times S^1$. The $S$-transformation glues a pair of disks to the two boundaries of each such annulus. This is enough to show that $\tilde{M}$ is an $S^2$ bundle over the circle. There are only two of these - the product and the twisted bundle. Since $\tilde{M}$ is not orientable, it is homeomorphic to the twisted bundle.

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Sam Nead
Sam Nead
  • 28.1k
  • 5
  • 72
  • 131

The boundary of your solid torus $T_L \cong D^2 \times S^1$ is equipped with a pair of foliations by circles (at "right angles"). The first one is the foliation by circles of the form $\partial D^2 \times \{\mathrm{pt}\}$. The second one is the foliation by circles of the form $\{\mathrm{pt}\} \times S^1$. I assume that by "$S$-transformation" you mean "cut $T_L$ out and glue it back in so that foliations swap palces". Essentially, you are "rotating the gluing map by 90 degrees". Assuming this is correct:

The manifold $\tilde{M}$, obtained by cutting out $T_L$ and gluing it back in with an "$S$-transformation", is homeomorphic to $S^2 \, \tilde{\times}\, S^1$: the $S^2$ bundle over $S^1$ with monodromy the antipodal map.

To see this: After removing $T_L$, what remains is $M^2 \times S^1$. Note that $M^2$ is an interval bundle over the circle. Each interval, crossed with a circle, gives an annulus properly embedded in $M^2 \times S^1$. The $S$-transformation glues a pair of disks to the two boundaries of each such annulus. This is enough to show that $\tilde{M}$ is an $S^2$ bundle over the circle. There are only two of these - the product and the twisted bundle. Since $\tilde{M}$ is not orientable, it is homeomorphic to the twisted bundle.