October 10, 2004

Where have I been?

Given smooth manifolds $M$ and $N$, $T(M\times N)\cong TM\times TN$.

Proof. Let $p : T(M\times N)\to TM\times TN$ given by $p(x,X) =
((\pi_M(x),X_M),(\pi_N(x),X_N))$, where $\pi_M$ is the projection
$M\times N\to M$, $X_m(f)=X(f\circ \pi_M)$ for $f:M\to \R$, and
$\pi_N, X_N$ are defined analogously. $p^{-1}$ is given as
follows: Choose coordinates $x_1,\ldots,x_m$ on $M$, and
$y_1,\ldots,y_n$ on $N$; note $x_1,\ldots,x_m,y_1,\ldots,y_n$ are
coordinates on $M\times N$. Then $p^{-1}$ takes $((x,X),(y,Y))$ to
\[\left((x,y),\sum_{i=1}^m X(x_i) \frac{\partial}{\partial x_i} +
\sum_{j=1}^n Y(y_j) \frac{\partial}{\partial y_j}\right)\] by
Lemma 2.3. Since $p$ and $p^{-1}$ are both clearly continuous, $p$
is the desired diffeomorphism.\qed

(Plus 5 other problems like this, and 120 pages of John Locke to read.)


MosBen said...

Come on, you just mashed your hands on the keyboard to make that. That's gibberish.

LameAim said...

My head hurts.

LameAim said...

Plus, you forgot to close your italics tag.