Exercise on computing the pullback via the antipodal map I'm trying to do the following exercise. Let 
$$ V =\{ (x,y,z,t) \in \mathbb{R}^4 \ | \ x^3+y^3+z^3+t^3=0\ , \  x^2+y^2+z^2+t^2=1 \}.$$
V is a smooth manifold of dimension $2$ by the preimage theorem. Let $a: V \rightarrow V $ be the antipodal map, given by $a(x,y,z,t)=(-x,-y,-z,-t)$. For $p \in V$, compute the induced map $a^*:\Omega^i(V)_{a(p)}\rightarrow \Omega^i(V)_p$ for $i=1,2$. 
I just tried to apply the definition:  given $\omega \in \Omega^i(V)_{a(p)}$ and $v_1,...,v_i \in T_pV$
$$a^*(w)(v_1,...,v_i)=w(da_p(v_1),...,da_p(v_i))$$
but I'm not really seeing how I can proceed more explicitly and how to use the expression of $V$.
 A: Define $F(x,y,z,t) = (x^3 + y^3 + z^3 + t^3, x^2 + y^2 + z^2 + t^2 - 1)$. The map $a$ leaves $V = F^{-1}(0)$ invariant. The tangent plane to $V$ at $(x,y,z,t)$ can be naturally identified with
$$ \ker(Df|_{(x,y,z,t)}) = \ker \begin{pmatrix} 3x^2 & 3y^2 & 3z^2 & 3t^2 \\
2x & 2y & 2z & 2t \end{pmatrix} = \operatorname{span} \{ v_1 = 2\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} , v_2 = 3\begin{pmatrix} x^2 \\ y^2 \\ z^2 \\ t^2 \end{pmatrix} \}^{\perp} \subseteq \mathbb{R}^4. $$
Note that $\ker(Df|_{(x,y,z,t)}) = \ker(Df|_{a(x,y,z,t)})$ and so $da|_{(x,y,z,t)}$ leaves the tangent plane $T_{(x,y,z,t)}V$ invariant and acts on it as the differential $da|_{(x,y,z,t)} = -I$. Hence
$$ a^{*}(\omega)|_{p}(v_1, \dots, v_k) = \omega|_{a(p)}(-v_1, \dots, -v_k) = (-1)^k \omega_{a(p)}(v_1, \dots, v_k). $$
You can probably be even more explicit and find a formula for $\ker(Df|_{(x,y,z,t)}$) by computing the orthogonal projection (note that $v_1$ has length $2$ and $v_2$ is orthogonal to $v_1$ so this is not so bad).
