The following is according to Lee's introduction to smooth manifolds chapters 12 and 14.
Let V be a vector space, $k\in \Bbb Z_{\ge 0}, \alpha $ a covariant k-tensor, i.e. $\alpha \in T^k(V^*)$ a multilinear map from $V\times...\times V\to \Bbb R$ .
$\Lambda^k (V^*)$ is the subspace of alternating covariant rank-k tensors
I proved the following property for $k \ge 2$.
Alt($\alpha)=\alpha \iff \alpha \in \Lambda^k (V^*)$
where Alt($\alpha)$ is a projection called alternation defined as
But the book states it is true for $k \ge 0$. My question if about the $k=0$ case. I know that the right side of the implication is vacuosly true, that is every covariant rank-0 tensor is alternating. (here $\Lambda^k (V^*) = \Bbb R $ ) so for the statement to be true, I would need the left side to be true as well, wouldn't I? so Alt($\alpha)=\alpha$ should be true, but then I would have an empty sum, wouldn't I? (https://en.wikipedia.org/wiki/Empty_sum) And that is defined to be 0, so I would get $\alpha=0$ and therefore $0=\alpha \iff \alpha \in \Lambda^0 V^*$
Which is not true as $\alpha$ can be any real number. How do I resolve this situation?
If we think of $S_n$ as the set (in fact, group) of bijections $\{1, \ldots, n\} \to \{1, \ldots, n\}$, $S_0$ contains precisely the empty map $e : \varnothing \to \varnothing$, which is consistent with the convention $0! = 1$. The empty map is the (empty) product of zero transpositions we should take its sign to be $+1$, giving $$\operatorname{Alt} \alpha = \sum_{\sigma \in S_0} (\operatorname{sgn} \sigma) ({}^\sigma\alpha) = (\operatorname{sgn} e) ({}^e\alpha) = \alpha$$ as desired.