I'm reading about the element of volume:
let $(V,\varphi)$ be an oriented Euclidean vector space; then exists a unique alternating tensor $\sigma\in\mathcal A^d(V)$ s.t. $\sigma(v_1,\dots,v_d)=1$, when $(v_1,\dots,v_d)$ is a positively oriented and orthonormal basis for $V$.
Then I read that, explicitely, $\sigma$ can be written as $d!(v^*_1\wedge\dots\wedge v^*_d)$, where $v_i$ is the dual vector of $v$, and this is the reason:
$v^*_1\wedge\dots\wedge v^*_d(v_1,\dots,v_d)=\frac 1 {d!}\sum_{\tau\in S_d}v^*_{\tau (1)}\oplus\dots\oplus v^*_{\tau (d)}(v_1,\dots,v_d)=\frac 1 {d!}$ where $S_d$ is the group of permutations and $\oplus $ stands for the tensor product.
Now what I don't understand, is why $\sum_{\tau\in S_d}v^*_{\tau (1)}\oplus\dots\oplus v^*_{\tau (d)}(v_1,\dots,v_d)$ is equal to $1$, since the determinant is given by $v^*_1\wedge\dots\wedge v^*_d$ and the determinant of a orthonormal basis is $1$; instead, my lectures put $v^*_1\wedge\dots\wedge v^*_d$ equal to $\frac 1 {d!}$. Thank you in advance
Since $v_i^\star(v_j)$ = 1 iff i = j, the sum on the right only has 1 non-zero term, namely the term indexed by the identity permutation. Otherwise, there exists $i$ such that $\sigma(i) \not= i$. Thus there exists a tensor which will be 0.