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I have a past qual question here: let $T^3 = S^1 \times S^1 \times S^1$ be the 3-torus, and let $\Delta = \{ (x,x) \in T^3 \times T^3 \colon x \in T^3 \}$ be the diagonal subspace. Compute $\pi_1(T^3 \times T^3 - \Delta)$.

If one wanted to understand the space $T^1 \times T^1 - \Delta$, one can realized it as $[0,1]^2$ (with opposite sides identified) minus the diagonal; then, one can shear along one of the two axes to get that $T^1 \times T^1 - \Delta$ is just a cylinder. Can this process be generalized to the higher-dimensional case?

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up vote 9 down vote accepted

Removing a subcomplex of codimension 3 or higher from a manifold does not change the fundamental group. Therefore, in this example the fundamental group is $Z^6$.

Edit: Let $M$ be an $n$-manifold and $K\subset M$ be submanifold of codimension $\ge 3$. (I will do this in smooth category but it works in topological setting too: It suffices to assume that $K$ is a closed subset of covering dimension $\le n-3$.) Pick a base-point $m\in N=M-K$ and consider the homomorphism $$ f: \pi_1(N,m)\to \pi_1(M,m) $$ induced by the inclusion $N\to M$. I claim that $f$ is an isomorphism. To show surjectivity, let $\alpha$ be a (smooth) loop in $M$ based at $m$. Then, by the generic transversality theorem, perturbing $\alpha$ a bit, we get a new loop $\beta$ based at $m$ which is transverse to $K$. By dimension count, it follows that $\beta$ is disjoint from $K$ and hence, is a loop in $N$. Nearby loops in a manifold are always homotopic (rel. base-point). Surjectivity follows.

To prove injectivity, you use the same argument: Let $\gamma: S^1\to N$ be a smooth loop and $F: D^2\to M$ be a smooth map of the unit disk extending $\gamma$. Then, by the same argument as above we perturb $F$ to a smooth map $G: D^2\to M$ transverse to $K$. Again, by the dimension count, transversality implies that $G(D^2)$ is disjoint from $K$. Hence, $f$ is an epimorphism. qed

The standard references for transversality are books by Guillemin and Pollack and by Hirsch. Your qual is likely to contain at least some differential topology and transversality material.

Same argument works if you work with a triangulated manifold $M$ where $K$ is a subcomplex of codimension $\ge 3$. (You use the fact that a 2-dimensional affine subspace of $R^n$ is generically disjoint from a subspace of codimension $\ge 3$.)

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Could you give a brief explanation why that is? Or perhaps give a reference? Thanks! – msteve Jul 20 '14 at 2:33
@msteve: See the edit. – studiosus Jul 20 '14 at 17:40

Let $p:\mathbb{R}^6\to T^3\times T^3$ be the universal covering map, which I consider as reducing each coordinate mod $1$. Consider the pre-image $p^{-1}(\Delta)\subset\mathbb{R}^6$. This is the set of $(t_1,\ldots,t_6)\in\mathbb{R}^6$ such that $(t_4,t_5,t_6)-(t_1,t_2,t_3)\in\mathbb{Z}^3$. In other words, $$ p^{-1}(\Delta) \cong \mathbb{R}^3\times\mathbb{Z}^3, $$ where the homeomorphism is given by $(t_1,t_2,t_3,t_1+n_1,t_2+n_2,t_3+n_3)\mapsto (t_1,t_2,t_3,n_1,n_2,n_3)$ ($t_i\in\mathbb{R}, n_i\in\mathbb{Z}$). Using this, we can retract $\mathbb{R}^6$ "diagonally" onto $\mathbb{R}^3$ in a way that retracts $p^{-1}(\Delta)$ onto $\mathbb{Z}^3\subset\mathbb{R}^3$. Then as $\mathbb{R}^3 - \mathbb{Z}^3$ is simply connected, we see that $\mathbb{R}^6-p^{-1}(\Delta)$ is also simply connected, so $\mathbb{R}^6-p^{-1}(\Delta)$ is the universal cover of $T^3\times T^3-\Delta$. The deck transformations of the two covers are the same (as the preimages of points in $T^3\times T^3-\Delta$ are the same), so $\pi_1(T^3\times T^3-\Delta)=\pi_1(T_3\times T_3)=\mathbb{Z}^6$.

It is worth noting why this approach falls apart for the $T^1$ case you describe: the preimage of $T^1\times T^1-\Delta$ is not connected. Also, in terms of your specific question:

Can this process be generalized to the higher-dimensional case?

I tried this as well and didn't get all the way there. My idea was to think of the cylinder you found as a fiber bundle, where the base is one of the $T^1$, and the fiber is $T^1-\{point\}$. In that case, the fact that $T^1-\{point\}$ is contractible makes it easier, but $T^3-\{point\}$ isn't (see Presentation of the fundamental group of a manifold minus some points ), so the same process only got so far. Specifically:

(using notation $\mathbf{t}=(t_1,t_2,t_3) \in [0,1)^3$ as coordinates for $T^3$)

Pick an arbitrary $\mathbf{x}\in T^3$ with $\mathbf{x}\neq (0,0,0)$, and define $Y=\{(\mathbf{t},\mathbf{t}+\mathbf{x})\in T^3\times T^3-\Delta : \mathbf{t}\in T^3 \}$ (addition is mod $1$). Define $\pi : T^3\times T^3-\Delta\to Y$ by $\pi(\mathbf{t},\mathbf{s})=(\mathbf{t},\mathbf{t}+\mathbf{x})$. With a little work, you can see that $\pi$ makes $T^3\times T^3-\Delta$ into a fiber bundle over $Y \cong T^3$ with fiber $T^3-\{point\}$. The resulting exact sequence yields $$ \cdots\to\underset{=0}{\pi_2(Y)}\to\underset{=\mathbb{Z}^3}{\pi_1(T^3-\{pt\})}\to \pi_1(T^3\times T^3-\Delta)\to\underset{=\mathbb{Z}^3}{\pi_1(Y)}\to\underset{=0}{\pi_0(T^3-\{pt\})}\to\cdots $$ This is consistent with the answer given above, but I don't see how it is enough to finish it off.

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Isn't $\pi_0(T^3-\{ pt \}) = \mathbb{Z}$, as $T^3-\{ pt \}$ is still path-connected? – msteve Jul 20 '14 at 3:22
It has one path component, so $\pi_0$ is the $1$-element group. – cws Jul 20 '14 at 3:37
Sorry, $\pi_0$ is the $1$-element set, not group. – cws Jul 20 '14 at 3:48

In terms of coordinates $\pmb{x} = (x_1, ..., x_n) \in (\mathbb{R}/ \mathbb{Z})^n$ for $T^n$, consider the map $f: T^n \times (T^n \setminus \lbrace \pmb{0} \rbrace) \rightarrow (T^n \times T^n) \setminus \Delta$ given by $f(\pmb{x}, \pmb{y}) = (\pmb{x}, \pmb{x}+\pmb{y})$. The map $f$ is continuous with continuous inverse $f^{-1}(\pmb{x}, \pmb{y}) = (\pmb{x}, \pmb{y}-\pmb{x})$. Hence, $T^n \times (T^n \setminus \lbrace pt \rbrace)$ is homeomorphic to $(T^n \times T^n) \setminus \Delta$ and so $\pi_1((T^n \times T^n) \setminus \Delta) \cong \pi_1(T^n) \times \pi_1(T^n \setminus \lbrace pt \rbrace)$. Note that $T^1 \setminus \lbrace pt \rbrace$ is contractible, $T^2 \setminus \lbrace pt \rbrace$ deformation retracts onto $S^1 \vee S^1$, and $\pi_1(T^n \setminus \lbrace pt \rbrace) \cong \pi_1(T^n)$ for $n \ge 3$. Therefore, $$ \pi_1((T^n \times T^n) \setminus \Delta) = \left\{ \begin{array}{lr} \mathbb{Z} & n=1\\ \mathbb{Z}^2 \times (\mathbb{Z}{*}\mathbb{Z}) & n=2\\ \mathbb{Z}^{2n} & n\ge 3 \end{array} \right. $$

Note: The construction of the homeomorphism $f$ above was inspired by a variant of the fiber bundle argument suggested by cws. Specifically, we can realize $(T^n \times T^n) \setminus \Delta$ as a principal $T^n$-bundle over $T^n \setminus \lbrace \pmb{0} \rbrace$ with projection map $\pi(\pmb{x}, \pmb{y}) = \pmb{y}-\pmb{x}$ and group action $\pmb{z} \cdot (\pmb{x}, \pmb{y}) = (\pmb{z}+\pmb{x}, \pmb{z}+\pmb{y})$. This principal $T^n$-bundle admits a global section $\sigma(\pmb{x}) = (\pmb{0}, \pmb{x})$ and is therefore isomorphic as a $T^n$-bundle to the trivial bundle $T^n \times (T^n \setminus \lbrace pt \rbrace)$. The bundle isomorphism is explicitly given by $f$ above.

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$T^3$ is a Lie group,so your space $T^3\times T^3-\Delta$ is a trivial bundle over $T^3$ with fiber $T^3-\{\ast\}$ (google"configuration space" for more details).So $$\pi_1(T^3\times T^3-\Delta)\cong \pi_1(T^3\times (T^3-\{\ast\}))\cong Z^3\times Z^3\cong Z^6$$

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