Computing the pull-back of $i: S^3 \longrightarrow \mathbb{R}^4$

differential-formsdifferential-geometrypullback

Let $\alpha$ be the differential 1-form defined by $$\alpha = xdy-ydx+zdt-tdz$$ and let $$i: S^3 = \{(x,y,z,t) \in \mathbb{R}^4 | x^2+y^2+z^2+t^2=1\} \longrightarrow \mathbb{R}^4$$ be the inclusion map. I need to show that $i^*(\alpha)$does not vanish in any point.

I tried two ways of doing that, but I can't conclude in any of them.

  1. For the first method I defined the spherical coordinates system in $\mathbb{R}^4$ $$\cases{x = r\cos\theta\\y= r\sin\theta \cos\phi\\ z= r\sin\theta\sin\phi\cos\gamma \\ t=r\sin\theta\sin\phi\sin\gamma}$$ Then, $i^*(\alpha) = \sum (\alpha_k \circ i) d(x^k \circ i)$, where $\alpha = \sum \alpha_k dx^k $, so $i^*(\alpha) = \cos\phi d\theta+sin^2\phi\sin^2\theta d\gamma-\cos\theta\sin\theta\sin\phi d\phi$. And, if I'm not mistaken, this does vanish in some point.

  2. The second method is trying to use the fact that $i^*(\alpha)(p)(v) = \alpha (p)(i_*|_p(v)) $, $\forall p\in S^3, \forall v \in T_pS^3$. So, I have to find and expression for $i_*|_p(v)$. $$v \in T_pS^3 \iff v = (\lambda_1 \frac{\partial}{\partial{x}}+\lambda_2 \frac{\partial}{\partial{y}}+\lambda_3 \frac{\partial}{\partial{z}}+\lambda_4 \frac{\partial}{\partial{t}}), (\lambda_1,…,\lambda_4)\bot p$$ but then I can't find a coordinates expression for $i_*|_p(v)$. How the differential can be calculated?

Is there another way of doing this?

Best Answer

For any $p=(x, y, z, t)$ in $\mathbb S^3$, the vector

$$ v =(y, -x, t, -z)= y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y} + t\frac{\partial}{\partial z}-z\frac{\partial}{\partial t}$$

is in $T_p\mathbb S^3$ since it's perpendicular to the normal vector $(x, y, z, t)$. Another way to see that: consider the curve

$$\gamma (t) = \sin t (y, -x, t, -z) + \cos t (x, y, z, t).$$ Since $ (y, -x, t, -z)$ is perpendicular to $ (x, y, z, t)$, $|\gamma (t)|^2 =1$. Thus $\gamma(t)$ is a curve in $\mathbb S^3$ and $\gamma(0) = p$. Then $\gamma'(0) =(y, -x, t, -z)$ is an element in $T_p\mathbb S^3$.

Then $$\alpha (v) = -y^2 -x^2 - t^2 - z^2 = -1\neq 0$$ Thus $\alpha$ does not vanish at any point.

Related Question