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Both bivectors and $(0,2)$-tensors are mathematical structures that take in $2$ vectors and produce a scalar. Similar as in this prior post I wrote, I would like to dumb down the mechanics of these operations with a simple example.

The goal is to understand at a very immediate level why

$C_{ij} \,dx^i \wedge dx^j = C_{ij}\, \frac 1 2 \left(dx^i \otimes dx^j - dx^j \otimes dx^i \right)\tag 1$

Let's say that we have two covectors: $\beta =\begin{bmatrix}\sqrt \pi &\sqrt[3]\pi\end{bmatrix}$ and $\gamma=\begin{bmatrix}\frac 1 3&\frac 15\end{bmatrix},$ which will form the coefficients for the wedge and tensor products.

The wedge product is $\beta \wedge \gamma$

$$\begin{align} \beta \wedge \gamma&= \sqrt \pi \frac 13 \;e^1\wedge e^1 + \sqrt\pi \frac 15 \;e^1 \wedge e^2 + \sqrt[3]\pi\frac 13\; e^2 \wedge e^1 + \sqrt[3]\pi \frac 15 e^2\wedge e^2 \\[2ex] &= 0 + \sqrt\pi \frac 15 \;e^1 \wedge e^2 + \sqrt[3]\pi\frac 13\; e^2 \wedge e^1 + 0 \\[2ex] &=\left( \sqrt \pi \frac 15 - \sqrt[3]\pi \frac 13 \right) \; e^1\wedge e^2\tag 2 \end{align}$$

If we feed two vectors to this form, say $\vec v=\begin{bmatrix} 2&3\end{bmatrix}^\top$ and $\vec w=\begin{bmatrix}4&5 \end{bmatrix}^\top$ we end up with

$$\begin{align} \left(\left( \sqrt \pi \frac 15 - \sqrt[3]\pi \frac 13 \right) \; e^1\wedge e^2\right)[\vec v, \vec w] &=\left( \sqrt \pi \frac 15 - \sqrt[3]\pi \frac 13 \right) \; \det\begin{bmatrix}2&4\\3&5 \end{bmatrix}\\[2ex] &=\left( \sqrt \pi \frac 15 - \sqrt[3]\pi \frac 13 \right)\left(2\cdot 5-3\cdot 4\right) \end{align}\tag 3$$


Now coparing to the tensor product $\beta\otimes\gamma$:

$$\beta \otimes \gamma =\begin{bmatrix} \sqrt \pi \frac 13 \; e^1\otimes e^1 & \sqrt \pi \frac 1 5 \;e^1\otimes e^2\\ \sqrt[3]\pi\frac 1 3 \, e^2\otimes e^1 & \sqrt[3]\pi \frac 1 5\,e^2\otimes e^2 \end{bmatrix}$$

Feeding the vectors $\vec v$ and $\vec w$ first as $\left(\beta \otimes \gamma\right)[\vec v, \vec w],$ followed by $\left(\beta \otimes \gamma\right)[\vec w, \vec v]:$

$$\begin{align} \left(\beta \otimes \gamma\right)[\vec v, \vec w]&= \begin{bmatrix}2&3\end{bmatrix} \begin{bmatrix} \sqrt \pi \frac 13 \; e^1\otimes e^1 & \sqrt \pi \frac 1 5 \;e^1\otimes e^2\\ \sqrt[3]\pi\frac 1 3 \, e^2\otimes e^1 & \sqrt[3]\pi \frac 1 5\,e^2\otimes e^2 \end{bmatrix} \begin{bmatrix}4\\5\end{bmatrix}\\[2ex] &=\sqrt \pi \frac 13 4\cdot 2 + \sqrt \pi \frac 15 5\cdot 2+\sqrt[3]\pi \frac 13 4\cdot 3+\sqrt[3]\pi\frac 15 5 \cdot 3 \end{align}$$

And

$$\begin{align} \left(\beta \otimes \gamma\right)[\vec w, \vec v]&= \begin{bmatrix}4&5\end{bmatrix} \begin{bmatrix} \sqrt \pi \frac 13 \; e^1\otimes e^1 & \sqrt \pi \frac 1 5 \;e^1\otimes e^2\\ \sqrt[3]\pi\frac 1 3 \, e^2\otimes e^1 & \sqrt[3]\pi \frac 1 5\,e^2\otimes e^2 \end{bmatrix} \begin{bmatrix}2\\3\end{bmatrix}\\[2ex] &=\sqrt \pi \frac 13 2\cdot 4 + \sqrt \pi \frac 15 3\cdot 4+\sqrt[3]\pi \frac 13 2\cdot 5+\sqrt[3]\pi\frac 15 3 \cdot 5 \end{align}$$

The difference is therefore:

$$\left(\beta \otimes \gamma\right)[\vec v, \vec w]-\left(\beta \otimes \gamma\right)[\vec w, \vec v]=\left(\sqrt \pi \frac 15-\sqrt[3]\pi\frac 13\right)\left(2\cdot 5 - 3 \cdot 4 \right)\tag 4$$

Now, (4) is identical to (3). The question is, then, about the $\frac 12$ factor to fulfill equation (1)? Why is there no need to divide (4) by $2$ to fulfill equation (1)?

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  • $\begingroup$ It's pretty standard to use the term $2$-form for an alternating covariant tensor of rank $2$ on a given vector space and reserve bivector for an alternative contravariant tensor of rank $2$. $\endgroup$ Commented Mar 1, 2019 at 2:42

2 Answers 2

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It's unusual---and possibly a source of confusion here---to write tensor products as matrices with tensor entries.

Denote the underlying vector space by $\Bbb V$. For a covariant $2$-tensor $T$ we pick a basis, say, $(E^a)$ of $\Bbb V$, denote the dual frame by $(e^a)$, and form the matrix $[T]$ whose $(a, b)$ entry respectively is the component $T_{ab} = T(E^a, T^b)$, that is, the coefficient $T_{ab}$ in the decomposition $T = \sum_{a, b} T_{ab} e^a \otimes e^b$. If $T$ is antisymmetric, then $T_{ab} = T(E_a, E_b) = -T(E_b, E_a) = T_{ba}$, in which case $[T]$ itself is antisymmetric.

In the case in the question, in which $\Bbb V = \Bbb F^2$ and $T$ is a wedge product $\alpha \wedge \beta$ of $1$ forms, $[\alpha \wedge \beta]$ has a single independent component, namely, $$[\alpha \wedge \beta]_{12} = (\alpha \wedge \beta)_{12} = (\alpha \wedge \beta)(E_1, E_2) = \alpha(E_1) \beta(E_2) - \beta(E_1) \alpha(E_2) = \alpha_1 \beta_2 - \alpha_2 \beta_1 ,$$ so that $$[\alpha \wedge \beta] = \pmatrix{0 & \alpha_1 \beta_2 - \alpha_2 \beta_1 \\ -(\alpha_1 \beta_2 - \alpha_2 \beta_1) & 0} .$$

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  • $\begingroup$ I totally see that I went down the rabbit's hole with the matrix expression, but I would like, nonetheless, to understand at which point there is a mistake in the OP. $\endgroup$ Commented Mar 1, 2019 at 2:33
  • $\begingroup$ I would say on in first display equations where the $2 \times 2$ matrices occur, since the notation with tensor entries doesn't have any predefined meaning. $\endgroup$ Commented Mar 1, 2019 at 2:39
  • $\begingroup$ If you have a change, please take a quick look at the matrix expressions in this post. $\endgroup$ Commented Mar 1, 2019 at 14:58
  • $\begingroup$ Anything in particular about them? That last display equation looks to be the same as the last display equation in my answer, except that it's generalized to an $n$-dimensional vector space. $\endgroup$ Commented Mar 1, 2019 at 17:34
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    $\begingroup$ I still don't quite follow all of your calculations, but if the only outstanding issue is the factor of $\frac{1}{2}$ then your calculation must be correct for some choice of convention. $\endgroup$ Commented Mar 4, 2019 at 22:24
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Just as a pictorial illustration of the workings of alternating linear algebra and wedge products - kind of a footnote.


Alternating multilinear functions:

The following is an example of $\Lambda^3(\mathbb R^6)^\star:$

enter image description here

a function $dx$ taking in $3$ vectors in $\mathbb R^6,$ i.e. $dx_{134}(v_ {1},v_{2},v_{k=3}),$ and returning a determinant of a matrix of the rows $1,$$3$ and $4$ - the $k\times k$ matrix composed of the corresponding elements $1,$ $3$ and $4$ of the input vectors: $$\mathbb R^6 \times \mathbb R^6 \times \mathbb R^6 \to\mathbb R.$$ All these possible such operations form a vector space with basis

$$\begin{align}\{dx_{123},dx_{124},dx_{125},dx_{126},\\dx_{134},dx_{135},dx_{136},\\dx_{145},dx_{146},\\dx_{156},\\dx_{234},dx_{235},dx_{236},\\dx_{245},dx_{246},\\dx_{256},\\dx_{345},dx_{346},dx_{356},\\dx_{456}\}\end{align} $$ of dimension ${6\choose 3}=20 .$


Wedge product:

The wedge product of, for example, an element $dx_{134}\in\Lambda^3(\mathbb R^6)^\star$ and an element $dx_{65}\in\Lambda^2(\mathbb R^6)^\star:$

enter image description here

will be an element $dx_{134}\wedge dx_{65} =dx_{13465}=-dx_{13456}\in\Lambda^5(\mathbb R^6)^\star.$

Or, as a different example, $dx_i \wedge dx_j = dx_{ij} = - dx_j \wedge dx_i = d_{ji}.$

These operators ($dx_i,$ $dx_i\wedge dx_j,$ etc) can have coefficients, and these coefficients can be functions. In fact, both the coefficients of these forms and the vectors fed into them can be functions, as in $z dx\wedge dy.$ If $g(u,v)=\begin{bmatrix}v \cos u & v\sin v & 3u \end{bmatrix}^\top,$ the pullback of $z dx\wedge dy$ on $g$ will be the determinant of the matrix of partial derivatives:

$$Dg =\begin{bmatrix} -v\sin u & \cos u \\ v \cos u & \sin u \\ 3 & 0 \end{bmatrix}$$

$$g^\star (z dx\wedge dy) = 3u\left(g^\star dx \wedge g^\star dy \right)=3u\left( -v \sin^2u \,du\wedge dv + v \cos^2u \,dv\wedge du\right)$$

as explained here.

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  • $\begingroup$ Here is a great presentation on the topic. $\endgroup$ Commented Mar 15, 2019 at 14:32

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