All 1-tensors are alternating

Question

This statement from page 155 of Guillemin and Pollack's Differential Topology. I would assume because 1-tensors can not alternate because they have nothing to alternate with, so they are alternating...?

Answer 1

The statement is vacuously true: If $\sigma \in S_1$ is a permutation, then we have $$T(v_{\sigma(1)}) = (-1)^{\sigma} T(v_1)$$ for any $1$-tensor $T$ and any $v_1$. This is because $S_1$ only has the trivial permutation, which is even, so that $(-1)^\sigma = 1$.

Note that this is just a rigorous way to say your assumed solution. "Nothing to alternate with" is captured by the fact that $S_1 = \{\mathrm{Id}\}$.

Answer 2

Well you could say this. I think this is a bit more general. When you use a ''for all'' statement, i.e., $A:= \forall x\ ( H[x] ) $, where $H[x]$ is some statement that depends on $x$, and you want to deny it, then logically it means that you can find a $x_0$ such that $H[x_0]$ is false. Since $A \wedge \lnot A$ is true, then if you can't find any element such that $H[x_0]$ doesn't hold, then it must be that $A$ is true. I know it's kind of itchy, but... I think that's the explanation.