1
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.