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The alternating operator $A$ produces for any $k$-linear map $f$ an alternating $k$-linear map $Af$ (the alternatization of $f$):

$$Af(v_1, \ldots, v_k) = \sum_{\sigma \in S_k} \text{sgn}(\sigma)f\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)$$

I've tried to prove that the definition of an alternating $k$-linear map holds (and most of all, to find an intuitive explanation for this!), but haven't made much progress...

If I think about the definition where having two equal arguments must make the map evaluate to $0$, then having two equal arguments means transposing them doesn't change anything, so two terms cancel out in the sum, but what about the rest?

Proof idea for alternative definition

Not what I was looking for, but I figured out a proof outline for an alternative definition of an alternating map. An alternating map is also defined as a map where transposing two arguments changes the sign of the map (e.g. $f(y, x, z) = - f(x, y, z)$).

Composing any permutation with a transposition changes its sign, so if the sum gets decomposed into a sum of all even permutations and all odd permutations, all even permutations will get mapped to odd permutations and vice-versa, so the two sums change signs. As a consequence, the entire alternatized function also changes sign.

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Fix $\pi \in S_k$.

\begin{align*} Af(v_{\pi(1)},\dots,v_{\pi(k)})&= \sum_{\sigma\in S_k}\operatorname{sgn}(\sigma)f(v_{\sigma(\pi(1))},\dots,v_{\sigma(\pi(n))}) \\ &=\operatorname{sgn}(\pi)\sum_{\sigma\in S_k}\operatorname{sgn}(\pi)\operatorname{sgn}(\sigma)f(v_{\sigma(\pi(1))},\dots,v_{\sigma(\pi(n))}) \\ &=\operatorname{sgn}(\pi)\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma\pi)f(v_{\sigma\pi(1))},\dots,v_{\sigma\pi(n))}) \\ &=\operatorname{sgn}(\pi)Af(v_1,\dots,v_k) \end{align*} where the last equality holds since right multiplication $\sigma\mapsto\sigma\pi$ yields a bijection $S_k\to S_k$.

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  • $\begingroup$ This is what I was looking for! Unfortunately, although I'm sure it's obvious, I'm having trouble understanding how you get from the expression after the first equal sign to that after the second equal sign, where $\text{sgn}(\pi)$ appears twice. I understood the rest. $\endgroup$
    – David Cian
    Feb 24, 2021 at 22:58
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    $\begingroup$ $\operatorname{sgn}(\pi) = \pm 1$, so $\operatorname{sgn}(\pi)^2 = 1$. $\endgroup$
    – Nico
    Feb 24, 2021 at 23:01
  • $\begingroup$ Can't believe I missed that geez... Thanks! $\endgroup$
    – David Cian
    Feb 24, 2021 at 23:03
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Your first definition implies your second definition, but your second does not imply your first for rings in which $1 + 1 = 0$. Moreover, some authors such as Hoffman and Kunze in their text Linear Algebra use your first definition. It therefore seems best to use your first definition.

You found two terms that cancel in the sum. Extend that idea: Pair the permutations into those that have the two equal arguments in the same two positions. The sum of each pair of corresponding terms is zero. Thus, the overall sum is zero which proves the claim. If that reasoning is not obvious, here it is with symbols:

We want to show that $A f(v_1,\dotsc,v_r) = 0$ whenever $v_i =v_j$ with $i\ne j$. The number of terms (permutations) in the sum $Af =\sum_\sigma (\operatorname{sgn}\sigma)f_\sigma$ is even because there are $r!$ terms, and the definition of alternating implicitly assumes $r\ge 2$. Pair each permutation $\sigma$ with the permutation $\sigma\circ(ij)$ by which we denote the composition of the permutation $(ij)$ that interchanges $v_i,v_j$ and the permutation $\sigma$. That pairing is well-defined because $\sigma\circ(ij)$ pairs with $(\sigma\circ(ij))\circ(ij) =\sigma$. Clearly, we can think of the index of summation in the definition of $A$ equivalently as running over all such pairs. The following shows that the pair of terms in that definition associated with the pair $\sigma$, $\sigma\circ(ij)$ add up to zero. We assume without loss of generalization that $i < j$.\begin{align*} & (\operatorname{sgn}(\sigma\circ(ij)))f(v_{(\sigma\circ(ij))1},\dotsc,v_{(\sigma\circ(ij))i},\dotsc,v_{(\sigma\circ(ij))j},\dotsc,v_{(\sigma\circ(ij))r})\\ &\quad = (\operatorname{sgn}\sigma)(\operatorname{sgn}(ij))f(v_{\sigma1},\dotsc,v_{(\sigma\circ(ij))j},\dotsc,v_{(\sigma\circ(ij))i},\dotsc,v_{\sigma r}) && (\sigma\circ(ij))k =\sigma k\text{ if }k\ne i,j;\ v_i = v_j\\ &\quad = -(\operatorname{sgn}\sigma)f(v_{\sigma1},\dotsc,v_{\sigma i},\dotsc,v_{\sigma j},\dotsc,v_{\sigma r}).\end{align*} Thus, because the sum of each such pair of terms is zero, $Af(v_1,\dotsc,v_r) = 0$.

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