I am reading "Calculus on Manifolds" by Michael Spivak.
I think the author's proof of Theorem 4-4(1) is not correct.
Am I right?
The author wrote as follows in the proof of Theorem 4-4(1):
$$\sum_{\sigma\in G\cdot\sigma_0} \text{sgn }\sigma\cdot S(v_{\sigma(1)},\dots,v_{\sigma(k)})\cdot T(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)}) =\left[\text{sgn }\sigma_0\cdot\sum_{\sigma'\in G}\text{sgn }\sigma'\cdot S(w_{\sigma'(1)},\dots,w_{\sigma'(k)})\right]\cdot T(w_{k+1},\dots,w_{k+l})
=0.$$
But I think the following is correct:
$$\sum_{\sigma\in \sigma_0\cdot G} \text{sgn }\sigma\cdot S(v_{\sigma(1)},\dots,v_{\sigma(k)})\cdot T(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)})
\\=\sum_{\sigma'\in G} \text{sgn }(\sigma_0\cdot\sigma')\cdot S(v_{\sigma_0\cdot\sigma'(1)},\dots,v_{\sigma_0\cdot\sigma'(k)})\cdot T(v_{\sigma_0\cdot\sigma'(k+1)},\dots,v_{\sigma_0\cdot\sigma'(k+l)})
\\=\text{sgn }\sigma_0\cdot\sum_{\sigma'\in G} \text{sgn }\sigma'\cdot S(w_{\sigma'(1)},\dots,w_{\sigma'(k)})\cdot T(w_{\sigma'(k+1)},\dots,w_{\sigma'(k+l)})
\\=\text{sgn }\sigma_0\cdot\sum_{\sigma'\in G} \text{sgn }\sigma'\cdot S(w_{\sigma'(1)},\dots,w_{\sigma'(k)})\cdot T(w_{k+1},\dots,w_{k+l})
\\=\left[\text{sgn }\sigma_0\cdot\sum_{\sigma'\in G}\text{sgn }\sigma'\cdot S(w_{\sigma'(1)},\dots,w_{\sigma'(k)})\right]\cdot T(w_{k+1},\dots,w_{k+l})
=0.$$
The author defined $w_1,\dots,w_{k+l}$ as follows:
$v_{\sigma_0(1)},\dots,v_{\sigma_0(k+l)}=w_1,\dots,w_{k+l}$.
So, $v_{\sigma_0\cdot\sigma'(1)},\dots,v_{\sigma_0\cdot\sigma'(k+l)}=w_{\sigma'(1)},\dots,w_{\sigma'(k+l)}$.
Best Answer
I thought I recognized this but didn't remember Spivak being so unclear. So I looked it up and it turns out this is basically the same proof Spivak gives in "A Comprehensive Introduction to Differential Geometry" but it seems your version is from the 1960s and this is the corrected updated 1999 version: