6
$\begingroup$

Consider the 3-torus $S^1\times S^1\times S^1$ represented by the following fundamental domain

enter image description here

Now remove one point from this space. Then to which well-known space will it be homotopic equivalent?

My guess is that it will be homotopic equivalent to $S^1 \vee S^2$ since the space obtained by removing one point from 2-torus $S^1\times S^1$ is homotopic equivalent to $S^1\vee S^1$ (source).

enter image description here

$\endgroup$
5
  • $\begingroup$ Just a naive question: what is your intuition leading to $S^1\vee S^2$ rather than $S^1\vee S^1 \vee S^1$ or $S^1 \vee RP^2$ for example? $\endgroup$
    – Didier
    Apr 13, 2021 at 7:56
  • $\begingroup$ Not sure what it is, but van Kampen tells you that you'll get a fundamental group of $\mathbb{Z}^3$ $\endgroup$
    – Alvin Jin
    Apr 13, 2021 at 7:59
  • $\begingroup$ @Didier I am imagining that if we remove a point from one of the faces of the cube then that part should deform to $S^1$ and the rest of it should deform to $S^2$ since if we would have removed 2-skeleton from this CW complex then would have obtained a space homeomorphic to $S^3$. I have no good logic to support my guess. $\endgroup$
    – rationalbeing
    Apr 13, 2021 at 8:10
  • 2
    $\begingroup$ I don't think it will be $S^1\vee S^2$. Removing a point should correspond (up to homotopy equivalence) to removing the top cell of its CW-structure. Therefore, it should be $S^1\vee S^1 \vee S^1$ with three $2$-cells attached via the words $aba^{-1}b^{-1}$, $aca^{-1}c^{1}$, and $bcb^{-1}c^{-1}$ where $a$, $b$, and $c$ denotes the three $1$-cells. I don't know which well-known space this is homotopy equivalent to. $\endgroup$
    – Frederik
    Apr 13, 2021 at 8:19
  • 1
    $\begingroup$ I'm not sure to which well-known space is it homotopic equivalent to, but it's like three $T^2$'s glued together along their meridian and longitudes. I don't think it's possible to simplify this anymore. $\endgroup$
    – Kevin.S
    Apr 13, 2021 at 11:26

1 Answer 1

Reset to default
3
$\begingroup$

I don't think that there is a better description of the space than the one i gave in the comments. Let me give slightly more details.

There is a CW-structure for $S^1\times S^1\times S^1$ with a single $0$-cell, three $1$-cells, three $2$-cells, and a single $3$-cell. The picture of this is the one from the question. For more details, see this post. We then get that the $1$-skeleton is $S^1\vee S^1\vee S^1$. The $2$-cells are attached via the words $aba^{-1}b^{-1}$, $aca^{-1}c^{-1}$, and $bcb^{-1}c^{-1}$ where $a$, $b$, and $c$ denotes the three $1$-cells. The $2$-skeleton can also be seen as hollowing out the cube from the question. Note that removing a point from $S^1\times S^1\times S^1$ results in a space that deformation retracts to the $2$-skeleton. This is analogous to the case of the $2$-torus which you have drawn. Let us denote this space with $X$.

Now let me give another geometric description of $X$. Let us attach the $2$-cells one at a time. Attaching the $2$-cell corresponding to the word $aba^{-1}b^{-1}$ yields the space $S^1\vee (S^1\times S^1)$ as seen in the picture:

$S^1\vee (S^1\times S^1)

Then one glues another $2$-cell via the word $aca^{-1}c^{-1}$. This yields a wedge of two tori with two $1$-cells from their $1$-skeleta identified. Attaching the last $2$-cell yields a wedge of three tori such that each torus has the two $1$-cells identified with a $1$-cell from each of the other tori respectively. This yields another description of $X$. Let $(S^1\times S^1)_i$ for $i=1,2,3$ be three tori. Then the space can be described as the quotient $$ X=\frac{(S^1\times S^1)_1\vee (S^1\times S^1)_2 \vee (S^1\times S^1)_3}{(z,*)_1\sim (z,*)_2,\, (*,z)_1\sim(*,z)_3,\, (*,z)_2\sim(z,*)_3} $$ where $*$ denotes the basespoint of the circle. (For example pick $*=(1,0)\in S^1$).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .