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Let $T = \mathbb{S}^1\times \mathbb{S}^1$. I'm trying to find a volume form for the torus, that is, a nowhere-vanishing 2-form. I tried following the idea on this post: $M \times N$ orientable iff both $M$ and $N$ are orientable proof in terms of volume forms, that is, taking a volume form $\omega$ in $\mathbb{S}^1$ and defining $\omega_T = \pi_1^*\omega\wedge\pi_2^*\omega$, where $\pi_1$ and $\pi_2$ are the first and second projections, respectively. I'm just having a hard time proving that this is indeed nowhere-vanishing. I tried using $T_{(p,q)}\mathbb{S}^1\times\mathbb{S}^1\sim T_p\mathbb{S}^1\times T_q\mathbb{S}^1$, but I'm just getting confused with the tangent vectors. Any help would be appreciated.
Trying to write thing explicitely, it gets even more confusing I find: that is, consider $\omega = ydx-xdy$ as a volume form in $\mathbb{S}^1$. If I see $T_p\mathbb{S}^1$ as a subspace of $T_p\mathbb{R}^2$, then I could see the tangent space of $T$ as a subspace of $T_p\mathbb{R}^4$, that is as the span of $y\partial x-x\partial y$ and $z\partial w -w\partial z$. But then what is $\pi_1^*\omega\wedge\pi_2^*\omega(y\partial x-x\partial y,z\partial w -w\partial z)$? Is it simply $\omega(y\partial x-x\partial y)\omega(z\partial w -w\partial z)=0$? I'm sorry for the ignorance. Also, how could I write $\omega_T$ in terms of $dx,dy,dw$ and $dz$?

Edit: If I consider $\pi_1(x,y,z,w) = (x,y)$, $\pi_2(x,y,z,w) = (z,w)$, then $\pi_1^*\omega = ydx-xdy$ and $\pi_2^*\omega=wdz-zdw$. Then $$\pi_1^*\omega\wedge\pi_2^*\omega = ywdx\wedge dz-yzdx\wedge dw-xwdy\wedge dz+xz dy\wedge dw$$ so that $\pi_1^*\omega\wedge\pi_2^*\omega(y\partial x-x\partial y,z\partial w -w\partial z)=(x^2+y^2)(z^2+w^2)=1\neq0$. Is this correct?

MathNewbie
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  • How are you getting confused? Consider $(v,w)\in T_pS^1\times T_qS^1$. Then $\omega_T(v,w) = \omega(v)\omega(w)$. How can this vanish? (It might be less confusing to have different manifolds with different volume forms, but ...) – Ted Shifrin Jul 07 '22 at 17:49
  • @TedShifrin TI think I'm mistaking the definition of nowhere-vanishing: does simply require that there exists $(v,w)\in T_p\mathbb{S}^1\times T_q\mathbb{S}^1$ such that $\omega_T(v,w)\neq 0$? In this case this would make sense. But shouldn't it be $\omega_T(v,w) =\omega(v)\omega(w)-\omega(w)\omega(v)$? Since you must take the tensor product. – MathNewbie Jul 07 '22 at 18:10
  • As i said, it’s more confusing having the same manifold in both slots. What is $\pi_1^*\omega(0,w)$? A $k$-form $\phi$ is nowhere-vanishing if for all points $p$, $\phi(p)$ is not the $0$-form on the tangent space at $p$. – Ted Shifrin Jul 07 '22 at 18:13
  • @TedShifrin Shouldn't $\pi_1^*\omega(0,w) = \omega(0) = 0$? I don't see how this implies the formula you gave though... In the post I reference, they mention that simply multiplying the forms wouldn't result in a differential form since it would fail to be antisymmetric. – MathNewbie Jul 07 '22 at 18:23
  • Write out $\pi_1^\omega\otimes\pi_2^\omega - \pi_2\omega\otimes\pi_1^\omega$ evaluated on the pair of vectors. The anti-symmetry complaint is misleading. Work out the details. – Ted Shifrin Jul 07 '22 at 18:33
  • Re your edit: Write out $\pi_1$ and $\pi_2$ explicitly and compute the pullbacks. – Ted Shifrin Jul 07 '22 at 19:23
  • @TedShifrin Did that, is it ok? I just don't know why I can't figure it out in the general case, I think I don't understand how to compute the wedge product in general. – MathNewbie Jul 07 '22 at 19:37
  • You don’t have to expand, but that’s fine. Why don’t you follow my previous suggestions? Use your correct observation three comments earlier. – Ted Shifrin Jul 07 '22 at 20:06
  • @TedShifrin I tried, but I keep getting zero – MathNewbie Jul 07 '22 at 20:43
  • It can’t be. Only the “missing” term is $0$. I gave you the answer in the first comment. – Ted Shifrin Jul 07 '22 at 20:45

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