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?

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    $\begingroup$ Yes. I'm surprised your linear algebra experience imposed that restriction. Note that to switch $1$ and $3$, you can switch $1$ and $2$, then $1$ and $3$, then $3$ and $2$. And so on. $\endgroup$ – Ted Shifrin Jun 21 '13 at 15:47
  • $\begingroup$ @TedShifrin Oh no, my linear algebra gets really rusty! In my impression, Det[v1,v2,v3] = -Det[v2,v1,v3] = Det[v2,v3,v1]. Oh that's why I drew the wrong conclusion because I've been swapping neighbors but non-neighbors can be swapped as well. @.@ Thanks! $\endgroup$ – 1LiterTears Jun 21 '13 at 15:53
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    $\begingroup$ Do one more step: Swap $v_2$ and $v_3$, and with one additional minus sign, you have $$\det(v_3,v_2,v_1)=-\det(v_1,v_2,v_3).$$ $\endgroup$ – Ted Shifrin Jun 21 '13 at 16:02
  • $\begingroup$ Got it, thanks @TedShifrin $\endgroup$ – 1LiterTears Jun 21 '13 at 16:04

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.


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