Anti-symmetric tensor of second order from a vector

Question

When given a vector $\overrightarrow V$ = $(x, x+y, x+y+z)$. Find the second order antisymmetric tensor associated with it.

This problem needs to be solved in cartesian coordinate system. The problem I'm facing is that how will I create a tensor of rank 2 with just one vector. To use cross product, i need at least two vectors. I'm have just started studying tensor, and also, if i create a second order tensor, to get the an antisymmetric tensor, I'll just have to subtract the tensor from its transpose and half it. $\frac{W+W^T}{2}$ where W be the second order tensor. Is that correct?

Answer 1

There are two ways of creating a tensor from a given vector: the first is by using the tensor product:

$$ T=V\otimes V . $$

The componentsof this tensor are $T_{ij}=u_iu_j$. It means it is a symmetric tensor, so if you substract it the symmetric part will be zero.

The another way is more subtle. It needs exterior algebra. Roughly speaking, the exterior algebra of a vector space $V$ is got by computing all tensor products ($V$, $V\otimes V$,$\dots$) and then antisymmetrizating them (for example for $u\otimes v$ one obtains $u\otimes v-v\otimes u \equiv u\wedge v$). Since there are $n$ linear-independent vectors, you can't have more than $n$ antisymmetric products $v_1\wedge \dots \wedge v_n$ (in other case there would be two equal vectors and then the product would be zero). If $\{e_i\}$ is a basis of $V$, then

$$ \Lambda^k(V) \equiv \langle \{ e_{i_1} \wedge \dots \wedge e_{i_k} | i_1,\dots,i_k = 1,\dots, n \}\rangle_{\mathbb R}. $$ ($\Lambda^0(V)=\mathbb K$ and $\Lambda^1(V)=V$ is assumed).

If $V$ is also a metric space, then there exists an isomorphism (called the Hodge star operator) between $\Lambda^k(V)$ and $\Lambda^{n-k}(V)$. For $V=\mathbb R^3$, the Hodge star operator gives an isomorphism between vectors and skew-symmetric tensors of rank 2. In coordinates, this isomorphism reads

$$ T_{ij} = \epsilon_{ijk} v_k. $$

If you compute the matrix of $T$ yo get

$$ T= \begin{pmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v1 & 0 \end{pmatrix}. $$

This matrix is just the matrix associated to $v$ when you want to compute the cross product $v\times w$:

$$ v\times w = \begin{pmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v1 & 0 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2\\ w_3 \end{pmatrix} = \cdots. $$

In fact:

$$ (v\times w)_i = \epsilon_{ijk}v_jw_k. $$

Exterior algebra is a powerful thing and it appears in differential geometry for example. If you are interested in these topics you should read about it. Peter Szekeres, Mathematical Physics is (for me) a good reference for start. I hope my answer helps you.

There are aslso related wuestions in this foro.

Answer 2

The other answer by @Dog_69 explained the required theory regarding the Hodge star operator and the Hodge dual in $\mathbb{R}^3$ so I will go ahead and compute the resulting dual antisymmetric second rank tensor for the vector in the question.

For any vector $V \in \mathbb{R}^3$, there exists an isomorphism to an antisymmetric second rank tensor via the Hodge star operator, called its dual tensor or Hodge dual. In index notation (while employing Einstein summation convention), the dual tensor isomorphism is given as,

$$ F_{ij} = \epsilon_{ijk}V_k \tag{1} $$ and its inverse, $$ V_{k} = \frac{1}{2}\epsilon_{ijk}F_{ij} \tag{2} $$ Which can be obtained by multiplying both sides of $(1)$ with $\epsilon_{pqr}$ and contracting via $q = i$, $r = j$. Here, $\epsilon_{ijk}$ is the totally antisymmetric three-indexed Levi-Civita symbol.

Given a vector $V$ as,

$$ V = \begin{pmatrix} x \\ x + y \\ x + y + z \end{pmatrix} $$

We compute its Hodge dual tensor as follows,

$$ F_{ij} = \epsilon_{ijk}V_k $$ Summing over the $k$-index from $k = 1$ to $k = 3$

$$ \begin{align*} F_{ij} &= \epsilon_{ij1}V_1 + \epsilon_{ij2}V_2 + \epsilon_{ij3}V_3 \\ F_{ij} &= \epsilon_{ij1}x + \epsilon_{ij2}(x + y) + \epsilon_{ij3}(x + y + z) \end{align*} $$

Since $F_{ij}$ is antisymmetric, we have that, $$ F_{ij} = -F_{ji} $$ Which implies that all the diagonal elements are $0$, i.e., $$ F_{11} = F_{22} = F_{33} = 0 $$ This reduces the number of independent components in the tensor from the usual $9$ to $3$, with only $F_{12}$, $F_{13}$, and $F_{23}$ being independent as the remaining non-zero components can be deduced by employing the antisymmetric nature of the tensor,

$$ F_{21} = -F_{12} \\ F_{31} = -F_{13} \\ F_{32} = -F_{23} $$

We now compute those three independent components in our case as follows,

$$ F_{12} = \epsilon_{121}x + \epsilon_{122}(x + y) + \epsilon_{123}(x + y + z) $$ Since the Levi-Civita symbol $\epsilon_{ijk}$ is totally antisymmetric in all its indices, any repetition of an index will cause it to vanish so the terms containing $\epsilon_{121}$ and $\epsilon_{122}$ disappear, leaving us with, $$ F_{12} = \epsilon_{123}(x + y + z) $$ Which by evaluation of $\epsilon_{123}$ is, $$ F_{12} = x + y + z $$

Similarly, we can compute $F_{13}$ and $F_{23}$ as,

$$ \begin{align*} F_{13} &= \epsilon_{131}x + \epsilon_{132}(x + y) + \epsilon_{133}(x + y + z) \\ &= \epsilon_{132}(x + y) \\ &= -(x + y) \end{align*} $$

$$ \begin{align*} F_{23} &= \epsilon_{231}x + \epsilon_{232}(x + y) + \epsilon_{233}(x + y + z) \\ &= \epsilon_{231}x \\ &= x \end{align*} $$

The corresponding "mirror" terms can now be obtained via the antisymmetric nature of $F_{ij}$ as,

$$ \begin{align*} F_{21} &= -F_{12} \\ &= -(x + y + z) \\ F_{31} &= -F_{13} \\ &= x + y \\ F_{32} &= -F_{23} \\ &= -x \end{align*} $$

These are all the components of the antisymmetric dual tensor of $V$. We can now write out the dual tensor in full using its matrix representation:

$$ F_{ij} = \begin{pmatrix} 0 && x + y + z && -(x + y) \\ -(x + y + z) && 0 && x \\ x + y && -x && 0 \end{pmatrix}_{ij} $$