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An antisymmetric map $\varphi:E\times\ldots\times E\to F$ is a map such that

$$\varphi(v_1,\ldots, v_i\ldots,v_j,\ldots, v_r)=-\varphi(v_1,\ldots, v_j\ldots,v_i,\ldots, v_r)$$

For every $i,j=1,\ldots,r$.

I would like to prove this definition is equivalent to the map has this property:

For every permutation $\sigma$ of the set of the integers $\{1,\ldots,r\}$

$$\varphi(v_{\sigma(1)},\ldots, v_{\sigma(r)})=\epsilon_{\sigma}\cdot\varphi(v_1,\ldots, v_r)$$

Where $\epsilon_{\sigma}=1$ if the permutation is even and $\epsilon_{\sigma}=-1$ if the permutation is odd.

For me this equivalence is extremely obvious, but I don't know how to prove it formally. Any help would be very appreciated.

My attempt

We know if $\sigma$ is a permutation, we know $\sigma$ is a composition of transpositions $\sigma=\pi_1\circ\ldots\circ\pi_k$. If $k$ is even we say $\sigma$ is even, if If $k$ is odd we say $\sigma$ is odd. I'm trying to prove by induction:

$\varphi(v_{\sigma(1)},\ldots v_{\sigma(r)})=(-1)^k\varphi(v_1,\ldots,v_r)$.

For $r=1$, is obviously true.

Suppose it's true for $r$ and let's prove for $r+1$.

  • If $\sigma(r+1)=r+1$, we have:

    $\varphi(v_{\sigma(1)},\ldots, v_{\sigma(r)},v_{\sigma(r+1)})=\varphi(v_{\sigma(1)},\ldots, v_{\sigma(r)},v_{r+1})=(-1)^k\varphi(v_1,\ldots,v_r)$

  • If $\sigma(r+1)\neq r+1$

    Suppose $\sigma (j)=r+1$ for some $j=1,\ldots, r$. Define the permutation $\sigma'$ by $\sigma'(i)=\sigma(i)$ for $i=1\ldots r$ and $i\neq j$ and $\sigma'(j)=\sigma(r+1),\sigma'(r+1)=r+1$. Therefore $\varphi(v_{\sigma(1)},\ldots, v_{\sigma(r)},v_{\sigma(r+1)})=(-1)\varphi(v_{\sigma'(1)},\ldots, v_{\sigma'(r)},v_{\sigma'(r+1)})=(-1)\varphi(v_{\sigma'(1)},\ldots, v_{\sigma'(r)},v_{r+1})=(-1)(-1)^{k-1}\varphi'(v_1,\ldots,v_r)=(-1)^k\varphi'(v_1,\ldots,v_r)$

Am I right? Is there something missing?

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    What is your definition of an "even" and "odd" permutation? – 2017-02-05
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    @HenningMakholm https://en.wikipedia.org/wiki/Parity_of_a_permutation – 2017-02-05
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    Every permutation can be expressed as a product of transpositions (not necessarily uniquely). Even permutations can be expressed as an even number of permutations & odd permutations can be expressed as an odd number of permutations. – 2017-02-05
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    x @user: There are several ways to define the parity in that article. Which of them is _yours_ -- i.e. the one you want your proof here to be based on? – 2017-02-05
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    So you have trouble proving it formally. Why don't you tell us where you started? Where'd you get stuck? What do you actually know to be true? Once you show some effort (and stop being dismissive of people who ask reasonable questions), you might get some real help. – 2017-02-05
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    @JohnHughes I'm editing right now. I don't remember when I have being dismissive of people who ask reasonable questions. Could you give me some example? Thank you – 2017-02-05
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    Your answer to Henning. – 2017-02-05
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    @JohnHughes thank you for the remarks – 2017-02-05
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    @HenningMakholm if $\sigma$ is a permutation, $\sigma$ is a composition of transpositions $\sigma=\pi_1\circ\ldots\circ\pi_k$ (a transposition is when you permute only two elements while the rest remains unchanged). If $k$ is even we say $\sigma$ is even, if $k$ is odd we say $\sigma$ – 2017-02-05
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    @JohnHughes I edited the question, is it good enough? It was the best I could do. – 2017-02-05
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    In this problem, $r$ is fixed; the thing on which you need to do induction would be the number $k$ or transpositions applied. If you define $\sigma_j = \pi_1 \circ \pi_2 \circ \ldots \circ \pi_j$, then you could do induction on $j$: assume that the theorem is true for $\sigma_j$; show it's then true for $\sigma_{j+1}$. Start by proving it's true for $\sigma_0$, which is the identity permutation. Of course, it gets a little tricky for $j > k$; you have to define $\sigma_j$ for that case, and ... well, you'll see what happens. – 2017-02-06
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    I didn't understand why I couldn't use induction over $r$, but I did understand the induction over $k$ you have just said. I will try it, thank you. – 2017-02-06

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