# Mechanics of a $(0,2)$-tensor and a bi-vector acting on 2 vectors

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

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)?

• 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 $2$. Mar 1, 2019 at 2:42

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 $$[\alpha \wedge \beta] = \pmatrix{0 & \alpha_1 \beta_2 - \alpha_2 \beta_1 \\ -(\alpha_1 \beta_2 - \alpha_2 \beta_1) & 0} .$$

• 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. Mar 1, 2019 at 2:33
• 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. Mar 1, 2019 at 2:39
• If you have a change, please take a quick look at the matrix expressions in this post. Mar 1, 2019 at 14:58
• 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. Mar 1, 2019 at 17:34
• 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. Mar 4, 2019 at 22:24

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 $$\Lambda^3(\mathbb R^6)^\star:$$

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 $$dx_{65}\in\Lambda^2(\mathbb R^6)^\star:$$

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.

• Here is a great presentation on the topic. Mar 15, 2019 at 14:32