In Calculus on Manifolds – Michael Spivak is proven that If $S \in \mathfrak{J}^k(V)$ and $T \in\mathfrak{J}^k(V)$ and $A l t(S)=0,$ then
$$Alt(S \otimes T)=Alt(T \otimes S)=0$$
where $\mathfrak{J}^m(V)$ is the set of all m-tensors and $Alt(T)$ is a tensor defined by $$Alt(T)(v_1,\dots,v_k)=\dfrac{1}{k!}\sum_{\sigma\in S_k}\operatorname{sgn}\sigma \cdot T\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)$$. On the book the property is proven as follows
$$(k+l)! Alt(S \otimes T)\left(v_{1}, \ldots, v_{k+l}\right)=\sum_{\sigma \in S_{k+l}} \operatorname{sgn} \sigma \cdot S\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right) \cdot T\left(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}\right)$$
If $G \subset S_{k+l}$ consists of all $\sigma$ which leave $k+1, \ldots$, $k+l$ fixed, then
\begin{align*}
&\sum_{\sigma \in G} \operatorname{sgn} \sigma \cdot S\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right) \cdot T\left(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}\right)\\
&=\left[\sum_{\sigma^{\prime} \in S_{k}} \operatorname{sgn} \sigma^{\prime} \cdot S\left(v_{\sigma^{\prime}(1)}, \ldots, v_{\sigma^{\prime}(k)}\right)\right] \cdot T\left(v_{k+1}, \ldots, v_{k+l}\right)\\
&=0
\end{align*}
(*) Suppose now that $\sigma_{0} \notin G .$ Let $G \cdot \sigma_{0}=\left\{\sigma \cdot \sigma_{0}: \sigma \in G\right\}$
and let $v_{\sigma_{0}(1)}, \ldots, v_{\sigma_{0}(k+l)}=w_{1}, \ldots, w_{k+l}$, Then
\begin{align*}
&\sum_{\sigma \in G \cdot \sigma_{0}} \operatorname{sgn} \sigma \cdot S\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right) \cdot T\left(v_{\sigma(k+1)}, \ldots, v_{\sigma(k+l)}\right)\\
&=\left[\operatorname{sgn} \sigma_{0} \cdot \sum_{\sigma^{\prime} \in G} \operatorname{sgn} \sigma^{\prime} \cdot S\left(w_{\sigma^{\prime}(1)}, \ldots, w_{\sigma^{\prime}(k)}\right)\right]\cdot T\left(w_{k+1}, \ldots, w_{k+l}\right)\\
&=0
\end{align*}
Notice that $G \cap G \cdot \sigma_{0}=\varnothing .$ In fact, if $\sigma \in G \cap G \cdot \sigma_{0}$, then $\sigma=\sigma^{\prime} \cdot \sigma_{0}$ for some $\sigma^{\prime} \in G$ and $\sigma_{0}=\sigma \cdot\left(\sigma^{\prime}\right)^{-1} \in G,$
a contradiction. We can then continue in this way, breaking $S_{k+l}$ up into disjoint subsets; the sum over each subset is $0$ , so that the sum over $S_{k+l}$ is $0 .$ The relation $\operatorname{Alt}(T \otimes S)=0$ is proved similarly. $\blacksquare$
I undertood all the proof but from (*). Is there another way to understand these logical steps?
Best Answer
Note that $G$ is a subgroup of $S_{k+l}$ (with $|G| = k!$), and the right cosets $\{G \sigma| \sigma \in S_{k+l} \}$ form a partition of $S_{k+l}$ (hence there are $n={(k+l)! \over k!}$ of them).
Let $\sigma_1,..,\sigma_{n}$ be such that the $\{ G \sigma_k\}$ form the partition, we can assume that $\sigma_1 = e$, the identity.
Then we have $\sum_{\sigma \in S_{k+l}} = \sum_{\sigma \in G} + \sum_{\sigma \in G\sigma_2} + \cdots + \sum_{\sigma \in G\sigma_{n}}$