The $n$-torus is parallelizable.

Question

I'm trying to show that the $n$-torus is parallelizable, meaning there is a diffeomorphism $T(\mathbb{T}^n) \to \mathbb{T}^n \times \mathbb{R}^n$, such that each $T_p(\mathbb{T}^n)$ is carried linearly isomorphically onto $\{p\} \times \mathbb{R}^n$. This is a question in Bredon's Topology and Geometry, and I'm trying to understand this information before my qualifying exam.

I don't really know how to show this. Is it sufficient to show it for $n=2$ and induct? TIA

Answer 1

Can you prove that any Lie group is parallelizable? Well, n-torus is a Lie group since it is the product $S^1 \times S^1 \times \ldots \times S^1 $. If you can't do that I will change the answer to add more details.

Edit: suppose $G$ is a lie group of dimension n. To prove it is parallelizable we need to show that there are b vector fields $e_i: G \to TG$ which are linearly independent at all $g \in G$ which is to say $\langle e_1(g), e_2(g), \ldots, e_n(g) \rangle = T_gG$ for all $g \in G$. But it is easy. Indeed, note that each element $g\ \in G$ defines a diffeomorphism $g: G \to G, h \mapsto gh$. Now take a basis $\{ v_i \}$ in the tangent space at identity $T_eG$. The desired vector field are defined as follows $e_i(g) : = dgv_i, $ where $dg$ is the differential (the tangent map of $g: G \to G$). Since $g$ was a diffeomorphism it takes linearly independent vectors at $e$ to linearly independent vectors at $g$.

Answer 2

$n$-torus is the quotient of $\mathbb{R}^n$ by the group generated by $t_i(x)=x+e_i$ where $e_i=(0,..,1,0..)$, $1$ at the $i$-place.

Consider the vector field $X_i(x)=e_i$ defined on $\mathbb{R}^n$, it is invariant by $t_i$, it defines a vector field $Y_i$ on $\mathbb{T}^n$, for every $x\in \mathbb{T}^n, Y_1(x),...,Y_n(x)$ is a basis of $T_x\mathbb{T}^n$.

Answer 3

Note that an equivalent definition of what it means for an $n$ dimensional manifold to be parallelizable is that there exists $n$ linearly independent sections into the tangent bundle. In other words, can you find $n$ smoothly varifying vector fields on the manifold such that each point on the manifold, those vector fields evaluated at $p$ span the tangent space of the manifold at $p$.

The key to understanding that $S^n$ is parallelizable is to first understand why $S^1$ is parallizable. Since $S^1$ is a dimension 1 manifold, it's tangent space at each point is also a 1 dimensional manifold. Thus the question becomes, Is there a non-vanishing global vector field we can define on $S^1$? Yes there is: indeed, if we parameterize $S^1$ by the global parameterization of $(cos(\theta t),sin(\theta t))$ then the derivative of this parameterization, thought of as the velocity vector of the curve that traces out the circle, is tangent to the manifold at every point, and it never vanishes, and thus spans the (1 dimensional) tangent space of the manifold at that point.

Now, can we define a global vector field on $S^1 \times .... \times S^1$? Sure, just do it component wise!!

Anyway, in the case of $n=1$, we need to find a diffeomorphism $S^1 \cong S^1 \times \mathbb{R}$. By the discussion above, given any point $p$ on the manifold, we have a tangent vector at that point $v_p$, so we have a collection of tuples, $(p,v_p)$. Thus we can diffeomorhpsim $S^1 \rightarrow S^1 \times \mathbb{R}$ by $p \rightarrow (p,span(v_p))$.

And so, for the diffeomorphism $S^1 \times .... \times S^1 \rightarrow S^1 \times ... \times S^1 \times \mathbb{R}^n$, you first pick out a tangent vector for each point of each copy of $S^1$ seperately, so we have a bunch of tuples $(p_i,v_{p_i})$. Then our diffeomorphism will be:

$(p_1,......,p_n) \rightarrow (p_1,...,p_n, span(v_{p_1}) \oplus span(v_{p_2}) \oplus ... \oplus span(v_{p_n}))$

Where $span(v_{p_1}) \oplus span(v_{p_2}) \oplus ... \oplus span(v_{p_n}) \cong \mathbb{R} \oplus .... \oplus \mathbb{R} \cong \mathbb{R}^n$ as all finite vectors spaces of the same dimension are isomorphic