Let $T=D^2 \times S^1$ be a solid torus, where $D^2$ is a 2-dimensional disk and $S^1$ is a circle.

Suppose we have another solid torus $T'$ and we have a homeomorphism $f$ sending a meridian $\partial D^2\times \{*\} \subset T$ to a longitude $\{*\}\times S^1\subset T'$.

Then gluing $T, T'$ along the boundary torus via the homeomorphism $f$ results in a 3-sphere.

Instead of thinking another torus $T'$, we can think about the complement of $T'$ in $S^3$, which is also a solid torus.

(Now I am getting confused.)

Then, the above homeomorphism corresponds to a homeomorphism sending a meridian to a meridian, or orientation reversed meridian?

So my questions are:

  1. Once we choose a meridian and longitude of a torus, is there a canonical choice for a meridian and longitude of the compliment torus?
  2. Let $(a, b)\in \mathbb{Z} \times \mathbb{Z}$ represent a curve in a torus winding $a$ times in the meridian direction and $b$ times in the longitude direction.

    I read somewhere that if we turn a torus inside out, the homeomorphism sending a meridian to $(m,l)$ corresponds to the homeomorphism sending a meridian to $(-l, -m)$.Is this right? Or, for this to be true, what choice should I make? (orientation, meridian, longitude.)

I appreciate any help.


I am sorry that I cannot add a comment due to lack of reputation.

First when you write down $(m,l)$ that means you have already choose a basis for $H_1(T)$ and $(m,l)$ is the "coordinate" of your element in the homology. In this case it may be convenient to say the element $(1,0)$ is meridian, $(0,1)$ is longitude and the orientation is $(1,0)$ followed by $(0,1)$ .

Second, when you think of another torus as the complement of $T'$ in $S^3$ and this $S^3$ is obtained by your previous construction(i.e. identifying meridian to longitude), then in this perspective the homeomorphism you want on the boundary is the identity map. It cannot be anything else because you already identified the boundary.

If you want to "turn a torus inside-out", I think you might want to consider a map from $S^3 \subset \mathbb{C}^2$ to itself by sending $(z_1,z_2)$ to $(\bar{z}_2,\bar{z}_1)$. Now consider a subspace in $S^3$ which contains points $ (r_1e^{i\theta_1},r_2e^{i\theta_2})$ where $r_1 = r_2 = \sqrt{2}/2$. Notice that this is a torus and it has a coordinate $(\theta_1,\theta_2)$ and the above map on $S^3$ induce a homeomorphism on the torus by sending $(\theta_1,\theta_2)$ to $(-\theta_2,-\theta_1)$. Thus it sends $(m,l)$ to $(-l,-m)$. If you want to see why it turns the torus inside-out, you can simply check the map changes the places of two solid torus it bounds(For instance $(0,1)$ goes to $(1,0)$ and a path connecting these two points must pass through our torus). If you imagine $S^3$ and the torus in the way Hatcher drawing Hopf fibration in his book, this map will send the unit circle in xy plane to the z-axis with infinity and so on.

It is a little too detailed but I wish it helps.

  • 2
    $\begingroup$ I just mention an old trick. Take a pair of men's pyjamas, preferable striped, and sew or pin the two legs two legs together at their ends. Now turn the result inside out through the top! $\endgroup$ – Ronnie Brown Feb 24 '14 at 10:32

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