Using the First Property to Prove the Second Property of the Exterior Product

[mathmari]

Hey!

Two of the properties of the exterior product are the following:

- Let $\psi_1, \ldots , \psi_k, n_{1}, \ldots , n_{\ell}\in V^{\star}$ then it holds that $$\left (\psi_1\land \ldots \land \psi_k\right )\land \left (n_1\land \ldots \land n_{\ell}\right )=\psi_1\land \ldots \land \psi_k\land n_{1}\land \ldots \land n_{\ell}$$
- Let $\omega \in \land^kV^{\star}$ and $\sigma\in \land^{\ell}V^{\star}$. Then it holds that $$\omega\land \sigma=(-1)^{k\ell}\sigma\land \omega$$

I want to show the second property using besides other properties also the first one.

I have done the following:

Let $\psi_1, \ldots , \psi_k, n_{k+1}, \ldots , n_{k+\ell}\in V^{\star}$. Let $\omega \in \land^kV^{\star}$ and $\sigma\in \land^{\ell}V^{\star}$. Then we have that $\omega= \psi_1\land \ldots \land \psi_k$ and $\sigma=n_{k+1}\land \ldots \land n_{k+\ell}$.

We get the following:
\begin{align*}\omega\land \sigma=\left (\psi_1\land \ldots \land \psi_k\right )\land \left (n_{k+1}\land \ldots \land n_{k+\ell}\right ) \\ \overset{ \text{First property }}{ = } \psi_1\land \ldots \land \psi_k\land n_{k+1}\land \ldots \land n_{k+\ell}\end{align*}

Then we have to take a permutation of the form $\pi: (1, \ldots , k, k+1, \ldots , k+\ell) \mapsto (k+1, \ldots , k+\ell, 1, \ldots , k)$, right? (Wondering)

Which is the sign of that permutation? (Wondering)

Does it holds then that \begin{align*}\psi_1\land \ldots \land \psi_k\land n_1\land \ldots \land n_{\ell}&=n_{\pi(k+1)}\land \ldots \land n_{\pi(k+\ell)}\land \psi_{\pi(1)}\land \ldots \land \psi_{\pi(\ell)} \\ & =\text{sign}(\pi)\cdot n_{k+1}\land \ldots \land n_{k}\land \psi_{k+1}\land \ldots \land \psi_{k+\ell} \\ & = \text{sign}(\pi)\left (n_{k+1}\land \ldots \land n_{k}\right )\land \left (\psi_{k+1}\land \ldots \land \psi_{k+\ell}\right )\\ & = \text{sign}(\pi)\sigma\land \omega \end{align*} ? (Wondering)

 

[I like Serena]

Then we have to take a permutation of the form $\pi: (1, \ldots , k, k+1, \ldots , k+\ell) \mapsto (k+1, \ldots , k+\ell, 1, \ldots , k)$, right? (Wondering)

Which is the sign of that permutation? (Wondering)

Hey mathmari! (Smile)

Yep, that's a good permutation.

The sign of a permutation depends on the number of pairwise transpositions that can compose it.
If it's an odd number of transpositions, the sign is -1, and if it's even, the sign is +1.
For instance for $\pi: (1, 2, 3, 4) \mapsto (3, 4, 1, 2)$, we can write $\pi = (13)(24)$, which is an even permutation, meaning its sign is +1.

We can also use that a cyclic transposition of k elements (a k-cycle) has sign -1 if k is even, and sign +1 if k is odd.
(Note that sign and parity are exactly the other way around here. Can you tell us why? (Wondering))
So for the same example, we can write $\pi=(1234)^2$, which is 2 applications of an odd permutation, yielding an even permutation, that is, sign +1.

Which cycle might we use to build $\pi$?

Does it holds then that \begin{align*}\psi_1\land \ldots \land \psi_k\land n_1\land \ldots \land n_{\ell}&=n_{\pi(k+1)}\land \ldots \land n_{\pi(k+\ell)}\land \psi_{\pi(1)}\land \ldots \land \psi_{\pi(\ell)} \\ & =\text{sign}(\pi)\cdot n_{k+1}\land \ldots \land n_{k}\land \psi_{k+1}\land \ldots \land \psi_{k+\ell} \\ & = \text{sign}(\pi)\left (n_{k+1}\land \ldots \land n_{k}\right )\land \left (\psi_{k+1}\land \ldots \land \psi_{k+\ell}\right )\\ & = \text{sign}(\pi)\sigma\land \omega \end{align*} ?

Yep. (Nod)

 

[mathmari]

Yep, that's a good permutation.

The sign of a permutation depends on the number of pairwise transpositions that can compose it.
If it's an odd number of transpositions, the sign is -1, and if it's even, the sign is +1.
For instance for $\pi: (1, 2, 3, 4) \mapsto (3, 4, 1, 2)$, we can write $\pi = (13)(24)$, which is an even permutation, meaning its sign is +1.

We can also use that a cyclic transposition of k elements (a k-cycle) has sign -1 if k is even, and sign +1 if k is odd.
(Note that sign and parity are exactly the other way around here. Can you tell us why? (Wondering))
So for the same example, we can write $\pi=(1234)^2$, which is 2 applications of an odd permutation, yielding an even permutation, that is, sign +1.

Which cycle might we use to build $\pi$?

What is exactly the difference between sign and parity? I got stuck right now... (Wondering) We can write the permutation as the product of cycles: $(1 \ \ k+1)(2 \ \ k+2) \ldots (k \ \ k+\ell )$, or not? The number of transpositions is $k$, or not? (Wondering)

Yep. (Nod)

So, is the way I wrote the indices, i.e., at $\psi$ I wrote the indices from $1$ to $k$ and at $n$ I wrote the indices from $k+1$ to $k+\ell$, correct? (Wondering)

 

[I like Serena]

What is exactly the difference between sign and parity? I got stuck right now...

Parity is even or odd, while sign is +1 or -1.
A permutation with odd parity is a permutation that has sign -1.
A k-cycle with even k is an odd permutation and therefore the permutation has sign -1.
As an example the permutation given by $(23)$ consists of 1 pairwise transposition (odd), while at the same time the number of elements in the cycle is 2 (even). Either way, the sign of the permutation is -1. (Nerd)

We can write the permutation as the product of cycles: $(1 \ \ k+1)(2 \ \ k+2) \ldots (k \ \ k+\ell )$, or not? The number of transpositions is $k$, or not?

This only works if $\ell = k$ doesn't it? (Wondering)
In that case the number of transitions is indeed $k$, and therefore the sign is $(-1)^k$.

So, is the way I wrote the indices, i.e., at $\psi$ I wrote the indices from $1$ to $k$ and at $n$ I wrote the indices from $k+1$ to $k+\ell$, correct?

I believe so yes.
What do you think might be wrong with it? (Wondering)

 

[mathmari]

Parity is even or odd, while sign is +1 or -1.
A permutation with odd parity is a permutation that has sign -1.
A k-cycle with even k is an odd permutation and therefore the permutation has sign -1.
As an example the permutation given by $(23)$ consists of 1 pairwise transposition (odd), while at the same time the number of elements in the cycle is 2 (even). Either way, the sign of the permutation is -1. (Nerd)

Ah ok! (Nerd)

This only works if $\ell = k$ doesn't it? (Wondering)
In that case the number of transitions is indeed $k$, and therefore the sign is $(-1)^k$.

We have the permutation $\pi: (1, \ldots , k, k+1, \ldots , k+\ell) \mapsto (k+1, \ldots , k+\ell, 1, \ldots , k)$.

We shift the element $(k+1)$ $k$ positions to the left, then we shift the element $(k+2)$ also $k$ positions to the left, etc.

So we shift each of the elements $\{k+1, \ldots k+\ell\}$ $k$ positions to the left.

At a shift of one position the sign is equal to $-1$. At a shift for $k$ positions the sign becomes $(-1)^k$. We shift totally $\ell$ times. Therefore, the sign becomes $((-1)^k)^{\ell}=(-1)^{k\ell}$.

Is everything correct? (Wondering)

Is this a formal proof? (Wondering)

 

[mathmari]

I got stuck right now.. (Wondering)

We have the permutation $\pi: (1, \ldots , k, k+1, \ldots , k+\ell) \mapsto (k+1, \ldots , k+\ell, 1, \ldots , k)$.

Is this correct to say that $\psi_1=n_{\pi (k+1)}$ ? Shouldn't it be that $\psi_{\pi(1)}=n_{k+1}$ ?

Is the following correct? Or should I use in an other way the permutation?
\begin{align*}\psi_1\land \ldots \land \psi_k\land n_1\land \ldots \land n_{\ell}&=n_{\pi(k+1)}\land \ldots \land n_{\pi(k+\ell)}\land \psi_{\pi(1)}\land \ldots \land \psi_{\pi(\ell)} \\ & =\text{sign}(\pi)\cdot n_{k+1}\land \ldots \land n_{k}\land \psi_{1}\land \ldots \land \psi_{\ell} \\ & = \text{sign}(\pi)\left (n_{k+1}\land \ldots \land n_{k}\right )\land \left (\psi_{1}\land \ldots \land \psi_{\ell}\right )\\ & = \text{sign}(\pi)\sigma\land \omega \end{align*}

Shouldn't it be as follows?
\begin{align*}\psi_1\land \ldots \land \psi_k\land n_{k+1}\land \ldots \land n_{k+\ell}&=\text{sign}(\pi)\psi_{\pi(1)}\land \ldots \land \psi_{\pi(k)}\land n_{\pi(k+1)}\land \ldots \land n_{\pi(k+\ell)} \\ & =\text{sign}(\pi)\cdot n_{k+1}\land \ldots \land n_{k+\ell}\land \psi_{1}\land \ldots \land \psi_{k} \\ & = \text{sign}(\pi)\left (n_{k+1}\land \ldots \land n_{k+\ell}\right )\land \left (\psi_{1}\land \ldots \land \psi_{k}\right )\\ & = \text{sign}(\pi)\sigma\land \omega \end{align*}
(Wondering)