Does Antisymmetrization in Wedge Product Skip Basis Effects?

[PhyAmateur]

If I want to take the wedge product of $$\alpha = a_i\theta^i $$ and $$\beta = b_j\theta^j$$ I get after applying antisymmetrization,$$ \alpha \Lambda \beta = \frac{1}{2}(a_ib_j - a_jb_i)\theta^i\theta^j$$

My question is it seems to me that antisymmetrization technique doesn't apply to the basis $$\theta^i , \theta^j$$ right? Is it that the wedge product antisymmetrization jumps over those basis only affecting the components? Or is there something I am missing?

Thanks![PLAIN]http://physics.stackexchange...rization-and-antisymmetrization-combinatorics[/PLAIN]

 

[WannabeNewton]

$\alpha \wedge \beta = \alpha_i \beta_j \theta^i \wedge \theta^j = \frac{1}{2}\alpha_i \beta_j (\theta^i \theta^j - \theta^j \theta^i) = \frac{1}{2}(\alpha_i \beta_j - \alpha_j \beta_i)\theta^i \theta^j$ is how the wedge product works on the basis 1-forms.

 

[PhyAmateur]

Thank you for the reply, I guess you meant $$a_i , b_j $$ instead of $$\alpha_i , \beta_j$$?

And so now, what I said in my question was totally wrong, it means according to this that it works on the basis rather than on the components. And then we flip indices accordingly, right?

 

[vanhees71]

Shouldn't it be (concerning the factors 1/2):
$$\alpha \wedge \beta=\alpha_i \beta_j \Theta^{i} \wedge \Theta^{j}=\frac{1}{2} (\alpha_i \beta_j - \alpha_j \beta_i) \Theta^{i} \wedge \Theta^{j} = (\alpha_i \beta_j-\alpha_j \beta_i) \Theta^{i} \otimes \Theta^{j}.$$

 

[pervect]

If I want to take the wedge product of $$\alpha = a_i\theta^i $$ and $$\beta = b_j\theta^j$$ I get after applying antisymmetrization,$$ \alpha \Lambda \beta = \frac{1}{2}(a_ib_j - a_jb_i)\theta^i\theta^j$$

My question is it seems to me that antisymmetrization technique doesn't apply to the basis $$\theta^i , \theta^j$$ right? Is it that the wedge product antisymmetrization jumps over those basis only affecting the components? Or is there something I am missing?

Thanks!

I'm not sure what the question is. I'd interpret what you wrote as the wedge product of a vector with a multiple of itself, which would be zero.

 

[vanhees71]

No, $\alpha$ and $\beta$ are different vectors. Note that the Einstein summation convention is used here, and $\Theta^j$ is a basis of the dual space (space of linear forms).

 

[PhyAmateur]

@vanhees71 thanks for your reply. But where are the components$$ a_i , b_j $$ And is there a general rule to this wedge product?

 

[vanhees71]

I called them $\alpha_i$ and $\beta_j$. Why should I introduce more symbols?

The wedge product is defined as follows. If you have two antisymmetric tensors $\omega_1$ and $\omega_2$, i.e., antisymmetric linear mappings $\omega_1:V^p \rightarrow \mathbb{R}$ and $\omega_2:V^q \rightarrow \mathbb{R}$ (also called $p$- and $q$-forms respectively), the wedge product is a $(p+q)$ form which is defined by
$$(\omega_1 \wedge \omega_2)(\vec{v}_1,\ldots,\vec{v}_p,\vec{v}_{p+1},\ldots,\vec{v}_{p+q})=\sum_{\tau \in S_{p+q}} \frac{\text{sign}(\tau)}{p! q!} \omega_1((\vec{v}_1,\ldots,\vec{v}_p) \omega_2(\vec{v}_{p+1},\ldots,\vec{v}_{p+q}).$$
Here $S_{p+q}$ denotes the permutation of $(p+q)$ elements.

 

[Matterwave]

I called them $\alpha_i$ and $\beta_j$. Why should I introduce more symbols?

The wedge product is defined as follows. If you have two antisymmetric tensors $\omega_1$ and $\omega_2$, i.e., antisymmetric linear mappings $\omega_1:V^p \rightarrow \mathbb{R}$ and $\omega_2:V^q \rightarrow \mathbb{R}$ (also called $p$- and $q$-forms respectively), the wedge product is a $(p+q)$ form which is defined by
$$(\omega_1 \wedge \omega_2)(\vec{v}_1,\ldots,\vec{v}_p,\vec{v}_{p+1},\ldots,\vec{v}_{p+q})=\sum_{\tau \in S_{p+q}} \frac{\text{sign}(\tau)}{p! q!} \omega_1((\vec{v}_1,\ldots,\vec{v}_p) \omega_2(\vec{v}_{p+1},\ldots,\vec{v}_{p+q}).$$
Here $S_{p+q}$ denotes the permutation of $(p+q)$ elements.

There's a mismatched parenthesis in the last expression here.