0
$\begingroup$

Let $\boldsymbol{a},\boldsymbol b$ be vectors in $\mathbb{R}^3$ with components $a^i,b^i$ given in some arbitrary (not necessarily orthogonal!) normalized curvilinear coordinate system. Wikipedia says

The cross product can alternatively be defined in terms of the Levi-Civita symbol $\varepsilon_{ijk}$ and a dot product $\eta^{mi}$ (=$\delta^{mi}$ for an orthonormal basis) which are useful in converting vector notation for tensor applications: $$\boldsymbol a\times\boldsymbol b=\boldsymbol c\iff c^m=\eta^{mi}\varepsilon_{ijk}a^jb^k$$

My question is, what is $\boldsymbol \eta$ ? What happens when our basis is not orthogonal? I have a feeling that it should be something like

$$\eta^{ij}=\frac{g^{ij}}{\sqrt{g^{ii}g^{jj}}}$$ But I'm not sure. Can someone please clarify for me?

$\endgroup$
1
  • $\begingroup$ I understand it as the contravariant that contra-vary with the basis change. I must admit that I don't know right now what is the consequence when the basis is not orthogonal. I remember that we need "only" ensure that the basis need to be linear independent: en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors $\endgroup$
    – user736865
    Sep 17, 2021 at 17:54

2 Answers 2

1
+200
$\begingroup$

The general setting for defining a cross product of two vectors is as follows. Consider an $3$-dimensional, oriented, pseudo inner-product space $(V,\text{Or}, g)$, and let $\mu$ be the volume form on $V$ defined with respect to the given orientation and $g$ (see this question and answer of mine for more about the definition of this volume form).

Now, given $v,w\in V$, we define \begin{align} v\times w&:= g^{\sharp}\left(\mu(v,w,\cdot)\right) \end{align} Here, $g^{\flat}:V\to V^*$, $x\mapsto g(x,\cdot)$ is the flat-mapping, which one can easily show is an isomorphism, and $g^{\sharp}:V^*\to V$ is the inverse isomorphism. Just so we're clear: $\mu$ is the volume form on $V$, which means it is an alternating multilinear mapping $V\times V\times V\to\Bbb{R}$, so by feeding in $v,w$, we have $\mu(v,w,\cdot)\in V^*$, hence by applying $g^{\sharp}$ to it, we get an element of $V$. (and in the case of $\Bbb{R}^3$ with the standard inner product, we have $\mu=\det$ being the usual determinant of a matrix thought of as an alternating $(0,3)$-tensor on $\Bbb{R}^3$... and it coincides with the usual definition).

Now, extracting the components is basic linear algebra. Let $\{f_1,f_2,f_3\}$ be a positively-oriented (otherwise, there will be an extra minus sign at the end) basis (we make no assumptions about orthogonality) of $V$, and let $\{\phi^1,\phi^2,\phi^3\}$ be the dual basis for $V^*$. Then, as shown in my linked answer, one has $\mu= \sqrt{|g|_F}\,\phi^1\wedge \phi^2\wedge \phi^3$, where the square root is short for the following $3\times 3$ determinant: \begin{align} \sqrt{|g|_F}:= \sqrt{|\det [g(f_i,f_j)]_{i,j=1}^3|} \end{align} Also, I shall use the notation $g_{ab}=g(f_a,f_b)$ and $g^{ab}$ shall denote the $(a,b)$-entry of the matrix inverse to $[g_{ij}]_{i,j=1}^3$. I hope you remember how elements of $V$ and $V^*$ can be expanded in terms of their bases. In what follows, all indices range over $1,2,3$, regardless of whether I use Latin/Greek indices. We have \begin{align} v\times w&:=g^{\sharp}\bigg(\mu(v,w,\cdot)\bigg)\\ &=g^{\sharp}\bigg(\mu(v,w,f_{\alpha})\phi^{\alpha}\bigg)\\ &=\mu(v,w,f_{\alpha})\cdot g^{\sharp}(\phi^{\alpha})\tag{$g^{\sharp}$ is linear}\\ &= \mu(v^{\beta}f_{\beta},w^{\gamma}f_{\gamma},f_{\alpha})\cdot g^{i\alpha}f_i\\ &= \mu(f_{\beta},f_{\gamma},f_{\alpha})\cdot g^{i\alpha}v^{\beta}w^{\gamma}\,f_i\\ &=\left(\sqrt{|g|_F}\,\phi^1\wedge \phi^2\wedge \phi^3\right)(f_{\beta},f_{\gamma},f_{\alpha})\cdot g^{i\alpha}v^{\beta}w^{\gamma}\,f_i\\ &= \sqrt{|g|_F}\,\, \varepsilon_{\beta\gamma\alpha}\cdot g^{i\alpha}v^{\beta}w^{\gamma}\,f_i \end{align} In the last line, I used the fact that $\phi^1\wedge\phi^2\wedge\phi^3(f_{\beta},f_{\gamma},f_{\alpha})=\det(e_{\beta},e_{\gamma},e_{\alpha})=\varepsilon_{\beta\gamma\alpha}$, where $e_1,e_2,e_3$ are the standard basis vectors of $\Bbb{R}^3$ (the reason why this holds, you just unwind the definition of the wedge product, and use the fact that the $\phi$'s are dual to the $f$'s, so eventually, determinants start popping up).

In other words, if we take a look at the $i^{th}$ component only, we have

\begin{align} (v\times w)^i&=\sqrt{|g|_F}\,\, \varepsilon_{\beta\gamma\alpha}\cdot g^{i\alpha}v^{\beta}w^{\gamma}\\ &=\sqrt{|g|_F}\,\,g^{i\alpha}\varepsilon_{\alpha\beta\gamma}v^{\beta}w^{\gamma}, \end{align} where in the last line, I permuted the indices in the Levi-Civita symbol appropriately to make things "look nice". Note that if $g$ is a proper inner-product and the vectors $\{f_1,f_2,f_3\}$ are orthonormal, then the determinant equals $1$ and we have a Kronecker delta, so \begin{align} (v\times w)^i&=\sum_{\beta,\gamma=1}^3\varepsilon_{i\beta\gamma}v^{\beta}w^{\gamma}, \end{align} which is exactly what we expect

$\endgroup$
6
  • $\begingroup$ I think this answer goes a little above my pay grade, but I will do my best to understand it! I presume $\{f_i\}$ is supposed to represent the basis of $V$? Also the notation for that square root determinant is a little complicated... is that just the usual $\sqrt{\left|\det\mathbf{g}\right|}$ that we see in e.g the Voss-Weyl formula? $\endgroup$
    – K.defaoite
    Sep 17, 2021 at 23:21
  • $\begingroup$ @K.defaoite yes $\{f_1,f_2,f_3\}$ is a basis for $V$, and indeed it's the 'usual' square root determinant of the metric like in Voss-Weyl. I used the subscript $F$ merely to emphasize that the matrix whose determinant we're calculating is $(g_{ij})$ where the components are with respect to the F basis: i.e $g_{ij}=g(f_i,f_j)$. The reason I emphasize this is that the determinant of a bilinear form $g:V\times V\to\Bbb{R}$ is not well-defined (unlike an endomorphism $T:V\to V$), so we can only speak of the determinant of a matrix representation, with respect to a given basis, of a bilinear form. $\endgroup$
    – peek-a-boo
    Sep 18, 2021 at 0:22
  • $\begingroup$ Ok, I'm going to accept this answer, mainly because I can understand the final result unlike md2perpe 's answer. As to the working that precedes it, I don't think I've gotten far enough in differential geometry yet to actually understand it. But I will be sure to come back to this answer as soon as I have the appropriate amount of rigor. Much appreciated. $\endgroup$
    – K.defaoite
    Sep 18, 2021 at 11:27
  • $\begingroup$ Btw, this has zero differential geometry, meaning no manifolds at all; it's all (multi)linear algebra :) (knowing definition of (pseudo) inner product, orientation of vector spaces, and how an inner product sets up an isomorphism with the vector space and its dual, how vectors can be expanded in terms of a basis, and how elements of the dual can be expanded in terms of the dual basis etc. The rest is an 'straightforward' calculation unwinding these definitions) $\endgroup$
    – peek-a-boo
    Sep 18, 2021 at 16:55
  • 1
    $\begingroup$ Well, good, because I am taking a course in advanced linear algebra this semester so perhaps that will give me a way to make sense of all this. $\endgroup$
    – K.defaoite
    Sep 18, 2021 at 16:58
0
$\begingroup$

A few lines up in the Wikipedia article it says "a dot product $\eta_{mi}$ ($= \delta_{mi}$ for an orthonormal basis)".

Below, lowercase Latin indices $i,j,k$ will refer to an orthonormal basis and lowercase Greek indices $\alpha,\beta,\sigma$ will refer to a general basis.

In an orthonormal basis $\mathbf{e}_i$ one has $[\mathbf{a}\times\mathbf{b}]^i=\epsilon^{i}{}_{jk}a^j b^k.$ In a general basis $\mathbf{e}_\sigma=T^i_\sigma \mathbf{e}_i$ (and reversely, $\mathbf{e}_i = T_i^\sigma \mathbf{e}_\sigma$) we get $$ [\mathbf{a}\times\mathbf{b}]^\sigma = T^\sigma_i [\mathbf{a}\times\mathbf{b}]^i = T^\sigma_i \epsilon^i{}_{jk} a^j b^k = T^\sigma_i \epsilon^i{}_{jk} (T^j_\alpha a^\alpha) (T^k_\beta b^\beta) = \epsilon^i{}_{jk} T^\sigma_i T^j_\alpha T^k_\beta a^\alpha b^\beta =: \epsilon^\sigma{}_{\alpha\beta} a^\alpha b^\beta, $$ where in the last step I have introduced $\epsilon^\sigma{}_{\alpha\beta}=\epsilon^i{}_{jk} T^\sigma_i T^j_\alpha T^k_\beta.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .