Dehn surgery on $RP^2 \times S^1$

Question

A standard example of Dehn surgery is obtaining $S^3$ from $S^2 \times S^1$. Consider a unknot $L$ wrapping the non-trivial cycle $S^1$ in $S^2 \times S^1$. We drill out a tubular neighborhood $T_{L} $ along $L$ which is a solid torus, $T_{L} = D^2 \times S^1$. It is obvious the knot complement $M_{L} = M - T_{L}$ is also a solid torus $M_{L} = D^2 \times S^1$. Now, if we glue back $M_{L}$ and $T_{L}$ along their boundaries with a $S$-transformation of the boundary of $T_{L}$, we will get $S^3$.

I wish to understand the same procedure when $\widetilde{M} = \mathbb{RP}^2 \times S^1$. The first homology group is $H_{1}(\widetilde{M},\mathbb{Z}) \cong \mathbb{Z}_2 \oplus \mathbb{Z} = \langle a,b\,|\, a^2 = 1\rangle$. Consider an unknot $L$ wrapping the non-trivial cycle $b$ (i.e along $S^1$) and drill out a tubular neighborhood, $T_{L} = D^2 \times S^1$. The knot complement in this case is $\widetilde{M}_{L} = \widetilde{M}- T_L = \mathbb{MB}\times S^1$ where $\mathbb{MB}$ is the Mobius band.

My question is, What is the 3-manifold obtained when we glue back $\widetilde{M}_{L}$ and $T_{L}$ with a $S$-transformation on the boundary of $T_{L}$? How do I deduce this 3-manifold?

I am having trouble visualizing the resulting 3-manifold. From this answer by Antonio Alfieri, I believe this manifold is $\mathbb{RP}^3$. Is this correct?

Edit:- To clarify what I mean by an S-transformation, I mean an element of Mapping class group ($SL(2;\mathbb{Z}$) that exchanges the two cycles of the boundary torus. This agrees with the assumption of Sam Nead's answer.

Answer 1

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.