# Vector proofs using index notation

Index notation provides a very powerful tool for proving many identities in vector calculus, or for manipulating formulae for multi-dimensional calculus. The power of index notation is usually first revealed when you're forced to prove identities that involve the (three-dimensional) cross product. We know that there are symmetries in such proofs, but it's difficult to see what these symmetries are, exactly. Index notation allows us to take advantage of these symmetries.

There are only a few basic rules. First, a sum is represented by a repeated index. So for instance, for a vector with three components. $\bm{u} \cdot \bm{u} = \sum_{i=1}^3 u_i u_i = u_i u_i.$

Second, we introduce the Kronecker delta symbol, $\delta_{ij} = \begin{cases} 0 & i \neq j \\ 1 & i = j \end{cases}$

So for instance, the dot product can be alternatively written as $\bm{u} \cdot \bm{u} = u_i u_j \delta_{ij}.$

which is a double sum over all values of $i, j = 1, 2, 3$.

Finally, the cross product is represented using the Levi-Civita symbol, $\epsilon_{ijk} = \begin{cases} 1 & \textrm{if cyclic} \\ -1 & \textrm{if acyclic} \\ 0 & \textrm{otherwise} \end{cases}.$

To determine which case it is, start counting from the number one. A cyclic sequence is one in which the $ijk$ occurs in cyclic order (count to the right), so 123, 231, 312. Acyclic order occurs for 321, 132, 213 (count to the left). finally, the symbol is zero if there is a repeated index (i.e. otherwise).

The key rule is the following definition for the cross product $\bm{a} \times \bm{b} = \epsilon_{ijk} \bm{a}_i \bm{b}_j.$

Once you have learned this, then generally, the only rule you must memorize in order to do most vector proofs is the unintuitive rule that allows you to manipulate cross products of cross products. This rule is given by the bolded identity in the table below.

### Basics from vector calculus

 Dot product $\mb{u} \cdot \mb{v}$ $u_i v_i$ Cross product $\mb{u} \times \mb{v}$ $\epsilon_{ijk}u_j v_k$ Gradient $\nabla f$ $\partial_i f$ Divergence $\nabla \cdot \mb{u}$ $\partial_i u_i$ Curl $\nabla \times \mb{u}$ $\epsilon_{ijk} \partial_k u_k$ Directional derivative $\mb{u} \cdot \nabla f$ $u_i \partial_i v_j$ Vector Laplacian $\nabla^2 \mb{u}$ $\partial_i \partial_i u_j$

### Identities

 Antisymmetry of permutation $\epsilon_{ijk} = -\epsilon_{jik} = \epsilon_{jki} = \ldots$ Derivatives of orthonormal basis $\partial_i e_j = 0$ Derivatives of coordinates $\partial_i x_j = \delta_{ij}$ Permutations differing by 1 index $\epsilon_{jmn}\epsilon_{imn} = 2\delta_{ij}$ Permutations with identical indices $\epsilon_{ijk}\epsilon_{ijk} = 6$ Permutations differing by 2 indices $\epsilon_{ijk} \epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$

### Examples

#### 1. Triple scalar product

Show $\mb{a} \cdot (\mb{b} \times \mb{c}) = \mb{b} \cdot (\mb{c} \times \mb{a}) = \mb{c} \cdot (\mb{a} \times \mb{b})$ $a_i (\epsilon_{ijk}b_j c_k) = b_j \epsilon_{ijk}a_i c_k = b_j (\epsilon_{jki} c_k a_i)$

The last equality follows from $\epsilon_{ijk} = \epsilon_{jki}$. Others are done similarly.

#### 2. Triple vector product

Show $\mb{a} \times (\mb{b} \times \mb{c}) = \mb{b}(\mb{a} \cdot \mb{c}) - \mb{c}(\mb{a}\cdot\mb{b})$ $\epsilon_{ijk}a_k(\epsilon_{klm}b_l c_m) = \epsilon_{ijk}\epsilon_{klm} a_k b_l c_m = (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})a_j b_l c_m = a_j b_i c_j - a_j b_j c_i = b_i(a_j c_j) - c_i(a_j b_j)$

Third equality follows from $\delta_{ij} = 1$ if $i = j$ and $= 0$ otherwise.

#### 3. Curl grad equals zero

Show $\nabla \times (\nabla f) = 0$ $\epsilon_{ijk} \partial_j \partial_k f = \epsilon_{ijk} \partial_k \partial_j f = -\epsilon_{ikj} \partial_k \partial_j f = -\epsilon_{ijk} \partial_j \partial_k f = 0.$

First equality follows from equality of mixed partials.

#### 4. Div curl equals zero

Show $\nabla \cdot (\nabla \times \mb{u}) = 0$ $\partial_i \epsilon_{ijk} \partial_j u_k = \epsilon_{ijk} \partial_i \partial_j u_k = \epsilon_{ijk} \partial_j \partial_i u_k = -\epsilon_{jik} \partial_j \partial_i u_k = -\epsilon_{ijk} \partial_i \partial_j u_k = 0.$

#### 5. Curl of the curl

Show $\nabla \times (\nabla \times \mb{u}) = \nabla(\nabla \cdot \mb{u}) - \nabla^2 \mb{u}$ $\epsilon_{ijk}\partial_j(\epsilon_{klm}\partial_l u_m) = \epsilon_{ijk}\epsilon_{klm} \partial_j \partial_l u_m = (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_j \partial_l u_m = \partial_j \partial_i u_j - \partial_j \partial_j u_i = \partial_i(\partial_j u_j) - \partial_j \partial_j u_i$ 