Confusion about when an alternating tensor changes sign On page 155 of Guillemin and Pollack's Differential Topology, it says:
A tensor $T$ is alternating if the sign of $T$ is reversed whenever two variables are transposed:
$$T(v_1, \ldots, v_i, \ldots, v_j, \ldots, v_p) = -T(v_1, \ldots, v_j, \ldots, v_i, \ldots, v_p)$$
But as far as I know, from linear algebra, the anticommutative multilinear form got the property only when interchanging neighbors. But from here, it certainly said anticommutativity between any two vectors?
 A: The definition in Guillemin & Pollock (definition $1$ below) is equivalent to the definition you are using (definition $2$ below).

The symmetric group $S_p$ acts on $p$-tensors as follows: given $\sigma \in S_p$ and a $p$-tensor $T$, we obtain a new $p$-tensor $\sigma^*T$ which satisfies $(\sigma^*T)(v_1, \dots, v_p) = T(v_{\sigma(1)}, \dots, v_{\sigma(p)})$.
Definition $1$: A $p$-tensor $T$ is said to be alternating if $\tau^*T = - T$ for any transposition $\tau = (i\ \ j)$.
Definition $2$: A $p$-tensor $T$ is said to be alternating if $\tau_i^*T = - T$ for any 'consecutive' transposition $\tau_i = (i\ \ i+1)$.
Clearly a $p$-tensor which is alternating by the first definition is also alternating by the second definition. 
Now suppose $T$ is a $p$-tensor which is alternating by the second definition. Fix a transposition $\tau = (i\ \ j)$ with $i < j$; we can do this as $(i\ \ j) = (j\ \ i)$. We want to show that $\tau^*T = -T$.
Note that $(i\ \ j)$ can be written as a product of consecutive transpositions. For example
\begin{align*}
(i\ \ j) &= (j-1\ \ j)\cdots(i+1\ \ i+2)(i\ \ i+1)(i+1\ \ i+2)\cdots(j-1\ \ j)\\ 
&= \tau_{j-1}\cdots\tau_{i+1}\tau_i\tau_{i+1}\cdots\tau_{j-1}.
\end{align*}
As $(i\ \ j)$ can be written as a product of an odd number of consecutive transpositions (namely $2(j-i)-1$ of them), we see that
\begin{align*}
\tau^*T &= (\tau_{j-1}\cdots\tau_{i+1}\tau_i\tau_{i+1}\cdots\tau_{j-1})^*T\\ 
&= \tau_{j-1}^*\cdots\tau_{i+1}^*\tau_i^*\tau_{i+1}^*\cdots\tau_{j-1}^*T\\ 
&= (-1)^{2(j-i)-1}T\\ 
&= -T.
\end{align*}
Therefore $T$ is also alternating according to the first definition.

The key here is that the collection of consecutive transpositions $\{(1\ \ 2), \dots, (p-1\ \ p)\}$ generates all transpositions.
Note, there is a third possible definition.
Definition $3$: A $p$-tensor $T$ is said to be alternating if $\sigma^*T = (\operatorname{sign}\sigma)T$ for any permutation $\sigma \in S_p$.
This is equivalent to the other two definitions because transpositions generate the full symmetric group.
