Question about cross product and tensor notation

I am a bit rusty on tensor algebra and calculus and may use some wrong terminology, but I know that the cross-product can be expressed in tensor notation with the aid of the Levi-Civita tensor as follows: $$\mathbf{A}\times \mathbf{B}=\varepsilon_{ijk}A_iB_j$$ Can someone tell me the domain of validity of the above equation? I think it is only valid for orthogonal coordinate systems where covariant and contravariant vectors are indistinguishable. If this is the case, how is the cross product defined for the most general coordinate system which may not be orthogonal?

Apart from this fact I heard that there is a cross-product tensor and cross-product is analogous to an operation called exterior-product. I think the cross-product tensor is expressed as below: $$(\mathbf{A}\times \mathbf{B})_{ij}=A_iB_j-A_jB_i$$ I also heard that this tensor is defined in 3 or more dimensions (PDF by Patrick Guio) and it is not possible to contract it in spaces of dimensionality 4 or higher since it has $\frac{1}{2}n(n-1)$ independent components. I have troubles with making the connections between cross-product tensor and vector owing to the fact that I do not know why such a definition for the tensor is used.

Furthermore, in Wikipedia-Cross Product website cross-product vector is defined as the Hodge-dual of the bivector $\mathbf{a} \wedge \mathbf{b}$ as follows: $$\mathbf{a}\times \mathbf{b} = *(\mathbf{a} \wedge \mathbf{b})$$ I have no idea about why this equation holds. Can someone provide a concise explanation or provide a reference where I can learn these concepts if possible?

• From the title I thought I could answer. Now your question becomes my own.; Oct 4, 2014 at 13:26
• @amcalde Thanks, I did not have formal training on tensors I learned them by myself, but there are gaps that need to be filled. Oct 4, 2014 at 13:58

You can define a tensor agrees with the Levi-Civita symbol for orthogonal coordinate systems but that has the correct components for non-orthonormal systems.

This, and other results, can be derived in the setting of clifford algebra.

Clifford algebra deals with a "quotient" of the tensor algebra--an interesting subset, if you will, of tensors that correspond to vectors, planes, and other objects that can often be interpreted geometrically.

To facilitate this, clifford algebra introduces a "geometric product" of vectors, which has the following laws:

1. If two vectors $a, b$ are orthogonal, then $ab = -ba$ under the product.
2. The product of a vector with itself is a scalar, i.e. $aa = |a|^2$.
3. The product is associative: $(ab)c = a(bc)$ for all vectors $a, b, c$.
4. The product is distributive over addition: $a(b+c) = ab + ac$.

From this definition, we can build up various objects that are not vectors but are produced from products of vectors under the geometric product.

With the geometric product in place, consider two vectors $a, b$, and write $b = b_\parallel + b_\perp$, the parts of $b$ parallel and perpendicular to $a$, respectively. Now then, we can write the product $ab$ as

$$ab = a b_\parallel + a b_\perp$$

The first term, $a b_\parallel$ is a scalar: $b\parallel = \alpha a$ for some scalar $\alpha$, and $aa = |a|^2$, a scalar, under rule 2.

The second term cannot be reduced, but we know from rule 1 that it anticommutes: $a b_\perp = -b_\perp a$. This is just like the cross product.

Indeed, if you write out this product with components, you get the following:

$$ab = (a^1 b^1 + a^2 b^2 + a^3 b^2) + (a^1 b^2 - a^2 b^1) e_1 e_2 + \ldots = a \cdot b + \frac{1}{2} a^i b^j e_i e_j$$

(summation implied). The latter term is called a bivector and is traditionally denoted $a \wedge b$.

You might have noticed now that we have at least three different kinds of objects: vectors, scalars, and bivectors. In clifford algebra, we number these objects by the number of vectors needed to form them, and we call this number the grade of the object. Scalars are grade-0, vectors grade-1, and bivectors grade-2. In 3d, you can also form a grade-3 object, a trivector. One choice might be $\epsilon = e_1 e_2 e_3$.

Now, what happens when you multiply a bivector with $\epsilon$?

First, the result must be a vector. Each bivector can be written as a linear combination of $e_1 e_2, e_2 e_3, e_3 e_1$, and $\epsilon$ has all of those in it. You can see that $e_1 e_2 \epsilon = e_1 e_2 e_1 e_2 e_3 = -e_3$ (use rule 1 for anticommuting swaps and rule 2 for same vectors to annihilate). The same holds for all other terms.

By convention, then, we can define a product

$$a \times b = -\epsilon (a \wedge b)$$

This coincides with the usual definition of the cross product. You can verify this term by term of you like; it's not that interesting to do algebraically, but geometrically, one comes to understand that multiplication by $\epsilon$ produces orthogonal complements of subspaces: a vector goes to its complementary plane, a plane to its normal vector, and so on. That is why I called this 3-vector $\epsilon$, as well: its components are those of the correct Levi-Civita tensor (not the symbol) that should have different components in nonorthonormal coordinate systems. And this is exactly what is meant in differential forms parlance when one uses the Hodge star operator.

Outside of 3d, the dual of a bivector is no longer a vector (4d: a bivector has another bivector totally complementary to it), and so the cross product as we typically imagine it no longer makes sense.

• Thanks for the through answer Muphrid. Apart from this can you point me to a source where I can learn all these concepts and which is not too advanced? Oct 4, 2014 at 21:45
• I would recommend Alan Macdonald's two books on geometric algebra and calculus. They are designed for an undergraduate audience and try to highlight how to relate the material to traditional linear algebra (esp. with matrices) and to vector calculus as it is traditionally taught. Oct 4, 2014 at 21:46
• Well thanks, I will check if our library has a copy of it. Oct 4, 2014 at 22:18
• By the way what about the star that is the Hodge dual operetion, does it have an intuitive interpretation? I mean the line $\mathbf{a}\times \mathbf{b} = *(\mathbf{a} \wedge \mathbf{b})$ in my OP. In your case you depicted this a bit differently. Oct 4, 2014 at 22:24
• Yeah, differential forms people use $\star$, and clifford algebra people just multiply by $\epsilon$ (or $-\epsilon$, depending on the particular case--the star is somewhat inconsistent in this respect). They're two different notations for the same thing, but the geometric interpretation is always that, if a $k$-vector corresponds to a subspace, then its dual (found by the star, or by multiplication with $\epsilon$) is that subspace's orthogonal complement--it's perpendicular to the original. Oct 4, 2014 at 23:13

To answer the first part of your question: The first equation you have is incorrect as written for the following reason: the cross product $A \times B$ is a vector independent of any basis. On the right hand side, you have (in einstein summation convention) the components of this cross product in a cartesian basis. To set the equation right, you'll have to introduce the cartesian basis vector on the right hand side, $$A \times B = \epsilon_{ijk} A_j B_k \hat{e}_i$$ where $\hat{e}_i$ is a cartesian basis vector. It is obvious that this expression is valid only for a cartesian basis.

• Thanks for the answer, I think this is due to the implied summation convention and what I wrote is only the component. Oct 4, 2014 at 21:41

Levi-Civita tensor is okay for any coordinate system, including curvilinear ones.

For any three coordinates $$q^i$$ ($$i = 1,2,3$$), the location vector can be uniquely represented as function of these coordinates $$\boldsymbol{r}=\boldsymbol{r}(q^i)$$. Then basis (“tangent”) vectors are $$\boldsymbol{r}_i \equiv \partial_i \boldsymbol{r}$$ (in Leibnitz’s notation $$\partial_i \equiv \frac{\partial}{\partial q^i}$$, $$\boldsymbol{r}_i = \frac{\partial}{\partial q^i} \boldsymbol{r} = \frac{\partial \boldsymbol{r}}{\partial q^i}$$). Then cobasis (dual basis, “cotangent” basis) vectors $$\boldsymbol{r}^i$$ can be found using fundamental property of cobasis $$\boldsymbol{E} = (\sum_i)\, \boldsymbol{r}^i \boldsymbol{r}_i = (\sum_i)\, \boldsymbol{r}_i \boldsymbol{r}^i \,\Leftrightarrow\: \boldsymbol{r}^i \cdot \boldsymbol{r}_j = \delta^i_j$$, where $$\boldsymbol{E}$$ is the bivalent “unit” tensor which is neutral to dot product operation (another names for this the same thing are “metric” tensor and “identity” tensor), and $$\delta^i_j$$ is Kronecker’s delta.

Here comes the trivalent Levi-Civita (“volumetric”, “trimetric”) tensor:

$${^3\!\boldsymbol{\epsilon}} = (\sum_{i,j,k})\, \boldsymbol{r}_i \times \boldsymbol{r}_j \cdot \boldsymbol{r}_k \; \boldsymbol{r}^i \boldsymbol{r}^j \boldsymbol{r}^k = (\sum_{i,j,k})\, \boldsymbol{r}^i \times \boldsymbol{r}^j \cdot \boldsymbol{r}^k \; \boldsymbol{r}_i \boldsymbol{r}_j \boldsymbol{r}_k$$

with its components $$\boldsymbol{r}_i \times \boldsymbol{r}_j \cdot \boldsymbol{r}_k \equiv \epsilon_{ijk}$$ or $$\boldsymbol{r}^i \times \boldsymbol{r}^j \cdot \boldsymbol{r}^k \equiv \epsilon^{ijk}$$.

Then for some two vectors $$\boldsymbol{a} = (\sum_i)\, a_{i} \boldsymbol{r}^i = (\sum_i)\, a^{i} \boldsymbol{r}_i$$ and $$\boldsymbol{b} = (\sum_i)\, b_{i} \boldsymbol{r}^i = (\sum_i)\, b^{i} \boldsymbol{r}_i$$

$$\boldsymbol{a} \times \boldsymbol{b} = (\sum_i)\, a_i \boldsymbol{r}^i \times (\sum_j)\, b_j \boldsymbol{r}^j = (\sum_{i,j})\, a_i b_j \: \boldsymbol{r}^i \times \boldsymbol{r}^j,$$

where cross product of cobasis vectors (as well as cross product of basis vectors, if you prefer to use another decomposition with second set of components) comes from definition of components of Levi-Civita tensor:

$$\boldsymbol{r}^i \times \boldsymbol{r}^j \cdot \boldsymbol{r}^k = \epsilon^{ijk} \,\Leftrightarrow\: (\sum_k)\, \boldsymbol{r}^i \times \boldsymbol{r}^j \cdot \boldsymbol{r}^k \boldsymbol{r}_k = (\sum_k)\, \epsilon^{ijk} \boldsymbol{r}_k \,\Leftrightarrow\: \boldsymbol{r}^i \times \boldsymbol{r}^j \cdot \boldsymbol{E} = (\sum_k)\, \epsilon^{ijk} \boldsymbol{r}_k$$ and finally $$\boldsymbol{r}^i \times \boldsymbol{r}^j = (\sum_k)\, \epsilon^{ijk} \boldsymbol{r}_k$$.

Thus

$$\boldsymbol{a} \times \boldsymbol{b} = (\sum_{i,j,k})\, a_i b_j \, \epsilon^{ijk} \, \boldsymbol{r}_k = \boldsymbol{b} \boldsymbol{a} \cdot \! \cdot \, {^3\!\boldsymbol{\epsilon}} = - \boldsymbol{a} \boldsymbol{b} \cdot \! \cdot \, {^3\!\boldsymbol{\epsilon}}$$