Throughout the question, please keep in mind that I know very little differential geometry. I.e., just the intrinsic definitions of differentiable/Riemannian manifolds and the metric tensor, etc. I am trying to understand the definition of the cross product given by Wikipedia here:


The article says that we can define the cross product $c$ of two vectors $u$,$v$ given a suitable "dot product" $\eta^{mi}$ as follows

$c^m := \sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\eta^{mi}\epsilon_{ijk}u^jv^k$

To demonstrate my current understanding of this definition, I will introduce some notation and terminology. Then I will show where my confusion arises with an example. I do apologize in advance for the length of this post.

Let $M$ be a smooth Riemannian manifold on $\mathbb{R}^3$ with the metric tensor $g$. Pick a coordinate chart $(U,\phi)$ with $\phi$ a diffeomorphism. We define a collection $\beta = \{b_i:U \to TM | i\in\{1,2,3\}\}$ of vector fields, called coordinate vectors, as follows

$b_i(x) := \Big(x,\big(\delta_x \circ \frac{\partial{\phi^{-1}}}{\partial{q_i}} \circ \phi\big)(x)\Big)$

where $\delta_x:\mathbb{R}^3 \to T_xM$ denotes the canonical bijection. The coordinate vectors induce a natural basis $\gamma_x$ at each point $x \in U$ for the tangent space $T_xM$. Let $[g_x]_S$ denote the matrix representation of the metric tensor at the point $x$ in the standard basis for $T_xM$ and let $[g_x]_{\gamma_x}$ denote the matrix representation in the basis $\gamma_x$.

My understanding of the above definition of the cross product now follows. Let $u,v \in T_xM$ be tangent vectors and let

$[u]_{\gamma_x}=\begin{bmatrix} u_1\\ u_2\\ u_3 \end{bmatrix}$ $\space \space \space \space \space \space [v]_{\gamma_x}=\begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix}$

denote the coordinates of $u,v$ in the basis $\gamma_x$. Then we define the $m$th coordinate of the cross product $u \times v \in T_xM$ in the basis $\gamma_x$ as

$\big([u \times v]_{\gamma_x}\big)_m := \sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\big([g_x]_{\gamma_x}\big)_{mi}\epsilon_{ijk}u_jv_k$

Now I will demonstrate my apparent misunderstanding with an example. Let the manifold $M$ be the usual Riemannian manifold on $\mathbb{R}^3$ and let $\phi$ be given by

$\phi(x_1,x_2,x_3) = (x_1,x_2,x_3-x_1^2-x_2^2)$


The Jacobian matrix $J$ of $\phi^{-1}$ is

$J=\begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 2q_1 & 2q_2 & 1 \end{bmatrix}$ $\space \space \space \space \space \space J^{-1}=\begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ -2q_1 & -2q_2 & 1 \end{bmatrix}$

And the matrix representation of the metric tensor in the basis $\gamma_x$ is

$[g_x]_{\gamma_x} = J^T[g_x]_SJ = \begin{bmatrix} 1+4q_1^2 & 4q_1q_2 & 2q_1 \\\ 4q_1q_2 & 1+4q_2^2 & 2q_2 \\\ 2q_1 & 2q_2 & 1 \end{bmatrix}$

Now choose $x=(1,1,-1)$. The coordinates of $x$ are evidently $\phi(x) = (1,1,1)$ and the three matrices above become

$J=\begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 2 & 2 & 1 \end{bmatrix}$ $\space \space \space \space \space \space J^{-1}=\begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ -2 & -2 & 1 \end{bmatrix}$ $\space \space \space \space \space \space [g_x]_{\gamma_x} = \begin{bmatrix} 5 & 4 & 2 \\\ 4 & 5 & 2 \\\ 2 & 2 & 1 \end{bmatrix}$

Now we compute the cross product in the basis $\gamma_x$. Using my understanding of the definition as outlined above, I get

$[u \times v]_{\gamma_x} = \begin{bmatrix} 36 \\\ 35 \\\ 16 \end{bmatrix}$

If we instead compute the cross product in the standard basis, then using my understanding of the definition, I get

$[u \times v]_S = \begin{bmatrix} 0 \\\ -1 \\\ 2 \end{bmatrix}$

Naturally, these results ought to agree if we perform a change of basis on $[u \times v]_{\gamma_x}$. Doing just that, I get

$[u \times v]_S = J[u \times v]_{\gamma_x} = \begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 36 \\\ 35 \\\ 16 \end{bmatrix} = \begin{bmatrix} 36 \\\ 35 \\\ 158 \end{bmatrix}$

Clearly, these do not agree. I can think of several reasons for this. Perhaps the definition given on Wikipedia is erroneous or only works for orthogonal coordinates. Perhaps I am misinterpreting the definition given on Wikipedia. Or maybe I have made an error somewhere in my calculation. My question is then as follows. How should I interpret the definition given on Wikipedia, and how should one express that definition using the notation provided here?


The core of the issue is whether the definition of the cross-product is invariant. The definition that Wikipedia provides uses coordinates, and is hence manifestly chart dependent. This is mostly fine for ordinary uses of the cross product, where we are mainly sticking to $\mathbb{R}^3$, but the context here is a bit different. You have a vector space, in your case $T_xM$, and you want to define a cross-product on this vector space in a coordinate invariant manner. It is not at all clear that Wikipedia's definition achieves this.

The question then is really this: given a $3$-dimensional inner product space $V$. How do we define a cross product on $V$ that does not depend on a particular choice of isomorphism with $\mathbb{R}^3$? There is actually a rather natural way of defining an invariant cross product, which is to use the Hodge star.

Definition: Let $(V,g)$ be a $3$-dimensional oriented inner product space. Given $\mathbf{u},\mathbf{v}\in V$, we define their cross-product to be $$\mathbf{u} \times \mathbf{v} \equiv \star(\mathbf{u}\wedge \mathbf{v}),$$ where $\star$ is the Hodge star. This is a manifestly coordinate independent definition, which for $V = \mathbb{R}^3$ reduces to the ordinary cross-product.

Let $(\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3)$ be a basis for $V$. Then in components, we have $$\mathbf{u} \wedge \mathbf{v} = \frac{1}{2}(u^iv^j-u^jv^i)\,\mathbf{v}_i\wedge \mathbf{v}_j,$$ where we use Einstein summation notation for repeated indices. Avoiding a rather lengthy calculation, you can show that the Hodge star is given in coordinates by $$(\star (\mathbf{u} \wedge \mathbf{v}))_k = \sqrt{g}\,u^iv^j\epsilon_{ijk},$$ where $\sqrt{g}$ is the square root of the determinant of $g$ in whatever coordinate system you're in. These are components of a covector, so if we raise the index (using the inverse metric, not the metric itself, as you did in your OP) we get $$(\mathbf{u} \times \mathbf{v})^\ell = \sqrt{g}\,u^iv^jg^{k\ell}\epsilon_{ijk},$$ which is very similar to the expression from Wikipedia, albeit with the inclusion of the determinant factor.

Ultimately, that determinant is the source of all your troubles. When you write down an expression in coordinate form, you need to make sure that it is tensorial so that they define genuinely coordinate invariant quantities. The easiest way to make sure your expressions are tensorial (besides remaining manifestly coordinate invariant) is to make sure the constituent objects you use are tensors themselves. This is fine for the (inverse) metric and your vectors, but the Levi-Civita symbol is not a tensor.

As the name suggests, the Levi-Civita symbol $\epsilon_{ijk}$ is a symbol whose components are defined to be completely alternating. The Levi-Civita symbol exists independently of any choice of coordinate system and is not designed to be a coordinate invariant object that transforms appropriately (for example, it doesn't really make sense to raise and lower indices of the Levi-Civita symbol). However, a closely related concept is the Levi-Civita tensor (the Levi-Civita symbol is distinct from the Levi-Civita tensor, although people are unfortunately quite sloppy about the distinction), which I will denote with a tilde as $\tilde{\epsilon}$. This is the object defined as $$\tilde{\epsilon}_{ijk} = \sqrt{g}\,\epsilon_{ijk}.$$ This is a bonafide tensor (in fact, the Riemannian volume form, which underlies the fact that cross-products give volume) which transforms appropriately. You can see then that the cross-product is given by $$(\mathbf{u}\times \mathbf{v})^\ell = u^iv^jg^{k\ell}\tilde{\epsilon}_{ijk},$$ and everything works out correctly if the $\epsilon$ in the Wikipedia definition was intended to be the Levi-Civita tensor and not symbol (again, people are unfortunately quite sloppy about this). Now you can see that all components of the formula above are genuine tensors, and so the expression is guaranteed to be well-defined in any coordinate system. Try your calculation again, but this time include the determinant factor (and use the inverse metric), and hopefully you'll find that everything works out.

  • $\begingroup$ This was very helpful in understanding my mistakes. Introducing the definition in terms of the Hodge star and exterior product is also good for future reference! $\endgroup$
    – J_Psi
    Jul 8 '19 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.