How would I show if $N_1,\ldots, N_k$ are parallelizable manifolds, then so is $N_1\times \cdots \times N_k$?

Question

A smooth $n$-manifold $N$ is called parallelizable if it admits $n$ smooth vector fields $Y_1;\ :\ :\ :\ ;\ Y_n$ that are linearly independent at every point $p$ in $M$.

How would I show if if $N_1, \ldots, N_k$ are parallelizable manifolds, then so is $N_1\times\cdots\times N_k$?

Use that $T(N_1 \times \cdots \times N_k) = TN_1 \times \cdots \times TN_k$. Where do you see a problem? — t.b., Oct 27 '11 at 18:56
Do it for two factors and use induction, to keep things minimally annoying. — Mariano Suárez-Álvarez, Oct 27 '11 at 23:03

score 1 · Answer 1 · answered Jun 28 '13 at 07:36

1

Following the comments:

It suffices to prove the statement for $k=2$; induction does the rest.
Since $T(N_1\times N_2) = TN_1 \times TN_2$, the union of parallelizations of $TN_1$ and $TN_2$ is a parallelization of $T(N_1)\times N_2)$.

answered Jun 28 '13 at 07:36

ˈjuː.zɚ79365

3,004

How would I show if $N_1,\ldots, N_k$ are parallelizable manifolds, then so is $N_1\times \cdots \times N_k$?

1 Answers1