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For oriented manifolds $M$ and $N$, I want to give $M\times N$ an orientation. (*) Additionally, I want to show that this orientation makes positively oriented bases $(v_1, \dots , v_m)$ and $(w_1, \dots , w_n)$ for $T_pM$ and $T_qN$ give a positively oriented basis $(v_1 \times 0, \dots , v_m \times 0, 0 \times w_1, \dots , 0 \times w_n)$ for $T_{(p,q)}(M \times N) = T_pM \times T_qN$.

Let $\{(U_i,\varphi_i)\}_{i\in I}$ be an oriented atlas for $M$ and $\{(V_j,\psi_j)\}_{j\in J}$ an orientable atlas for $N$. Then $M\times N$ has product atlas generated by the charts $\{(U_i\times V_j,\varphi_i\times \psi_j)\}_{i\in I,j\in J}$. These charts cover $M\times N$. Since the charts on $M$ and $N$ are orientable, it follows that for each $p\in U_i\cap U_j$ and $q\in V_i\cap V_j$ that the Jacobian matrices $d(\varphi_j\circ\varphi_i^{-1})_p$ and $d(\psi_j\circ \psi_i^{-1})_q$ have positive determinant.

But then $$\det(d((\varphi_j\times\psi_j)\circ (\varphi_i\times \psi_i)^{-1})_{(p,q)})=\det(d(\varphi_j\circ\varphi_i^{-1})_p\times d(\psi_j\circ\psi_i^{-1})_{q})=\det(d(\varphi_j\circ\varphi_i^{-1})_p)\det(d(\psi_j\circ\psi_i^{-1})_q)>0$$

What I don't know is how an oriented manifold (defined in terms of an oriented atlas, that is an atlas generated by a choice of covering charts, where all chart transitions have positive determinant jacobian matrices) induces an orientation on each tangent space.

Do we just fix an oriented basis $\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in I}$ for the $n$-dimensional manifold $M$, and for each $p\in M$ where $U_\alpha\ni p$ take $\varphi_\alpha^*\omega$ for $\omega$ the standard orientation form on $\Bbb R^n$, which is well defined since chart transitions about $p$ preserve orientation.

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  • $\begingroup$ Given oriented charts, you can find a positive basis at a point by pulling back the standard basis from $\Bbb{R}^n$ via the chart. Since you've defined your charts on the product to be $\phi_i\times \psi_j$, this makes it immediately clear that $(*)$ holds. $\endgroup$ – jgon Jun 14 at 23:15
  • $\begingroup$ @jgon Sorry, I don't follow. You mean for an $m$-dimensional manifold $M$ with a chosen oriented atlas (in the sense that the derivative of the chart transitions all have positive determinant), how do I give $T_pM$ an orientation? What's meant by pulling back the standard basis from $\Bbb R^n$? Do you mean we let $(U,\varphi_U)$ be our oriented chart, then let $det$ be the standard orientation form on $\Bbb R^m$ and just pull this back over $\varphi_U$? $\endgroup$ – F.White Jun 19 at 15:59
  • $\begingroup$ @jgon Maybe I did follow by the end of posting that :P. Is my last paragraph in the edit what you mean? $\endgroup$ – F.White Jun 19 at 16:09
  • $\begingroup$ So your question is not really about the product of manifolds, but you want to understand the connection between an oriented atlas and orientations of the tangent spaces? $\endgroup$ – Paul Frost Jul 4 at 16:13

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