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

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Product Property
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

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)
 
Physics news on Phys.org
mathmari said:
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 $\pi$?

mathmari said:
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)
 
I like Serena said:
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)
I like Serena said:
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)
 
Last edited by a moderator:
mathmari said:
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)

mathmari said:
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 $(-1)^k$.

mathmari said:
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)
 
I like Serena said:
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)
I like Serena said:
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)
 
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)
 
Last edited by a moderator:
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...

Similar threads

Replies
2
Views
2K
Replies
3
Views
2K
Replies
4
Views
1K
Replies
10
Views
2K
Replies
2
Views
1K
Replies
4
Views
2K
Replies
5
Views
2K
Back
Top