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...?
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1In the sense that the condition in the definition can't fail. [Why is the empty set a subset of every set? Maybe this is a bad analogy.] – TTS Jun 20 '13 at 20:34
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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 $S_1 = \{\mathrm{Id}\}$.
Henry T. Horton
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Thanks for your patient explanation, especially for reassuring my assertion. – 1LiterTears Jun 20 '13 at 20:54
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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.
Ognan
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Or Henry T. Horton's solution, which is nice. It depends of course on the definition of alternating tensor that you consider. – Ognan Jun 20 '13 at 20:42