Let $\omega = \sum_{j=1}^{n+1} x_j dy_j - y_j dx_j $ be a differential form on the sphere $S^{2n +1}$.

Let $G = Z_2$ be the group acting on the sphere.

I want to apply the following proposition to show that $\omega$ is also a differential form on the projective space:

Let $G$ be a discrete group acting properly and smoothly on a manifold $X$, and let $\pi : X → X/G$ be the covering map to the quotient. If $\omega$ is a differential form on $X$ such that $g^\ast \omega=\omega $ for all $g\in G$, then there exists a unique differential form $\alpha$ on $X/G$ such that $\pi^\ast \alpha = \omega$.

Here is what I did:

$$ \begin{align} \omega(v_x^1, \dots, v_x^{2n+1}) &= \sum_{j=1}^{n+1} x_j v_x^{j+n} - y_j v_x^j \\ &= - \sum_{j=1}^{n+1} - x_j v_{j+n}^j + y_j v_x^j \\ &= -\omega(-v_x^1, \dots, -v_x^{2n+1}) \\ \end{align}$$

So basically I deduced $\omega (v) = -g^\ast \omega (v)$ when what I needed was $\omega (v) = g^\ast \omega (v)$.

I don't see where in the calculation I made a mistake and I'm sure that $\omega (v) = g^\ast \omega (v)$. So my question is:

Could someone please point out to me where in the calculation I made a mistake?

  • $\begingroup$ Recall that when you pull-back you form you need also to change the point of evaluation. As you transformation is the antipodal map, the new coordinates of your points are nothing but $-x_i,-y_i$. Thus, when you evaluate your pull-backed form in the same point $+x_i,+y_i$, you need to account for this $\endgroup$ – b00n heT May 31 '16 at 9:29

Expanding my comment: \begin{align*}(g^*\omega)(x)(v_x^1,\dots,v_x^{2n+1}) & =\omega(g(x))(d_xg(v_x^1),\dots,d_xg(v_x^{2n+1})) \\ & =\omega(\color{red}{-x})(-v_x^1,\dots,-v_x^{2n+1})\\ & =\sum_{j=1}^{n+1} (-x_j) (-v_x^{n+j}) - (-y_j) (-v_x^j)\\ &=\sum_{j=1}^{n+1} x_jv_x^{n+j} - y_jv_x^j\\ &=\omega(x)(v_x^1,\dots,v_x^{2n+1}).\end{align*} As claimed


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