What is the generalization of the tensor product definition for a three form?

[JonnyMaddox]

Hello,
the tensor product definition of a two form is
$\alpha^{1} \wedge \beta^{1} := \alpha \otimes \beta - \beta \otimes \alpha$
$\alpha \wedge \beta(v,w) = \beta(v)\alpha(w)-\beta(w)\alpha(v)$
But what is the definition in this sense for a three form?

 

[Matterwave]

That...looks like a 2-form to me?

I'm not understanding what your question is.

 

[JonnyMaddox]

That...looks like a 2-form to me?

I'm not understanding what your question is.

Ohhps sry, of course it is a two form.

 

[Ben Niehoff]

You can work out the definition by carefully using the definition of the interior product $\iota$. For a p-form $\alpha$ and q-form $\beta$, the interior product distributes over the wedge product just like the exterior derivative:

$$\iota_X (\alpha \wedge \beta) = (\iota_X \alpha) \wedge \beta + (-1)^p \, \alpha \wedge (\iota_X \beta).$$
Therefore, for 1-forms $\alpha, \beta, \gamma$ and vector fields $X, Y, Z$, we have

$$(\alpha \wedge \beta \wedge \gamma)(X,Y,Z) = \iota_Z (\iota_Y (\iota_X (\alpha \wedge \beta \wedge \gamma))),$$
which can then be expanded into

$$\begin{align*}(\alpha \wedge \beta \wedge \gamma)(X,Y,Z) &= \alpha(X) \beta(Y) \gamma(Z) + \alpha(Y) \beta(Z) \gamma(X) + \alpha(Z) \beta(X) \gamma(Y) \\ &\qquad - \alpha(X) \beta(Z) \gamma(Y) - \alpha(Y) \beta(X) \gamma(Z) - \alpha(Z) \beta(Y) \gamma(X).\end{align*}$$
Then you can see that you can write

$$\begin{align*}\alpha \wedge \beta \wedge \gamma &= \alpha \otimes \beta \otimes \gamma + \beta \otimes \gamma \otimes \alpha + \gamma \otimes \alpha \otimes \beta \\ & \qquad - \alpha \otimes \gamma \otimes \beta - \beta \otimes \alpha \otimes \gamma - \gamma \otimes \beta \otimes \alpha. \end{align*}$$
In general, the wedge product of 1-forms is the sum of tensor products of those 1-forms in every permutation, weighted by the sign of the permutation. I would say it is the "anti-symmetrized" tensor product, but authors have various conventions as to whether the antisymmetrized product should be weighted by $1/n!$.

 

[Terandol]

[edit] Looks like I type too slowly...This post doesn't really add anything to the previous one except a few explicit formulas

The general definition for the wedge product of $\alpha\in \Omega^m(M)$ and $\beta\in \Omega^n (M)$ is
$$\alpha\wedge \beta<br /> = \frac{(m+n)!}{m!n!}\mathrm{Alt }(\alpha\otimes\beta)$$
where here we are regarding the tensors as multilinear maps and the alternation of a multilinear map is defined for $\gamma\in \Omega^p(M)$ by
$$\mathrm{Alt }(\gamma)(v_1,\cdots, v_p)<br /> =\frac{1}{p!}\sum_{\sigma\in S_p} \mathrm{sgn}(\sigma) \gamma(v_{\sigma(1)}, \cdots, v_{\sigma(p)} )$$

So, evaluating the multilinear maps we get the formula
$$\alpha\wedge\beta (v_1,\cdots, v_{m+n})<br /> =\frac{1}{m!n!}\sum_{\sigma\in S_{m+n}} \mathrm{sgn}(\sigma) <br /> \alpha(v_{\sigma(1)},\cdots v_{\sigma(m)})<br /> \beta(v_{\sigma(m+1)},\cdots, v_{\sigma(m+n)})$$
which you can easily work out for the special case of two three forms or whatever else you would like.

You should be a little bit careful about the algebra structure here though. There is a canonical choice of multiplication for $\bigwedge V^*$, however to transfer this multiplication over to alternating multilinear maps can be done using the isomorphisms $\bigwedge V^* \cong \left(\bigwedge V\right)^*\cong A(V)$ where $A(V)$ is the alternating multilinear maps on $V$. Since there are different ways to pair $\bigwedge V$ with $\bigwedge V^*$ the first isomorphism is not unique and this results in different numerical factors sometimes showing up in the definition of the wedge product. So you always have to check the definitions of whatever source you are using to make sure you get the right ones..

 

[JonnyMaddox]

Hi, thank you guys. I will certainly look that up when I'm more advanced at this stuff (if ever). I needed the three form because I wanted to calculate the cross product like this $\iota_{A \times B}vol^{3} \rightarrow A \times B$ if someone is interested. But I think I should do some more calculations with the exterior and interior product now. Thx

 

[JonnyMaddox]

Hi, I just want to say that I found an easier and more intuitive way to figure this out. I just think of forms as multilinear and alternating functions like the determinant !

$dx\wedge dy\wedge dz(X,V,W) = \begin{vmatrix} dx(X) & dx(V) & dx(W) \\ dy(X) & dy(V) & dy(W) \\ dz(X) & dz(V) & dz(W) \end{vmatrix}$

Now I just write it as a typical determinant expression and I get the same as Ben gets in his post :) and I don't have to struggle with this index and summation mambo jumbo and it is even easier to think about the interior product.

 

[WWGD]

Hi, I just want to say that I found an easier and more intuitive way to figure this out. I just think of forms as multilinear and alternating functions like the determinant !

$dx\wedge dy\wedge dz(X,V,W) = \begin{vmatrix} dx(X) & dx(V) & dx(W) \\ dy(X) & dy(V) & dy(W) \\ dz(X) & dz(V) & dz(W) \end{vmatrix}$

Now I just write it as a typical determinant expression and I get the same as Ben gets in his post :) and I don't have to struggle with this index and summation mambo jumbo and it is even easier to think about the interior product.

But you may need to understand it if you want to, e.g., evaluate a wedge $dx \wedge dy$ when, say, dx is a 2-form and dy is a 1-form.

 

[JonnyMaddox]

@WWGD: But when dx is a two form and dy is a one from then $dx \wedge dy$ is just a three form or not?

Greets

 

[WWGD]

@WWGD: But when dx is a two form and dy is a one from then $dx \wedge dy$ is just a three form or not?

Greets

Yes, but how do you evaluate $dx \wedge dy (X,Y,Z)$ , or $dy \wedge dx (X,Y,Z)$? for dx a 2-form and dy a 1-form. Of course, if you know one you know the other, but can you see how to do the evaluation? What if you had a p-form wedged with a q-form: how do you evaluate the wedge on a (p+q)-ple of vector fields without using more general results?

 

[JonnyMaddox]

I'm not sure this is appropriate but in Frankel's book there is a generalization of this.

Let $\tau^{1},...,\tau^{n}$ be any n-tuple of 1-forms, and expand each in terms of a basis (not assuming any scalar product) $\tau^{i}=T^{i}_{j}\sigma^{j}$ Then $\tau^{1} \wedge ...\wedge \tau^{n}= \sum\limits_{J} T^{1}_{j_{1}...}T^{n}_{j_n}\sigma^{j_{1}}\wedge ...\wedge \sigma^{j_{n}}= \sum\limits_{J} T^{1}_{j_{1}...}T^{n}_{j_n} \delta_{12...n}^{j_{1}...j_{n}}\sigma^{1}\wedge...\wedge\sigma^{n}$ that is $\tau^{1}\wedge...\wedge \tau^{n}= (det T)\sigma^{1} \wedge...\wedge \sigma^{n}$ Exterior product yield a coordinate-free expression for the determinant! ...

If you have a $\alpha^{1} \wedge \beta^{2}=(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})dx^{1} \wedge dx^{2} \wedge dx^{3}$ Which you can express with a determinant.