MHB Reciprocal Vectors: Representing $\mathbf{u}_i$ in Indicial Notation

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The discussion centers on representing vectors in indicial notation, specifically addressing the representation of the reciprocal basis vectors $\mathbf{u}^1$, $\mathbf{u}^2$, and $\mathbf{u}^3$ derived from the base vectors $\mathbf{u}_1$, $\mathbf{u}_2$, and $\mathbf{u}_3$. It clarifies the correct formulation of the reciprocal vectors, correcting initial typographical errors in their definitions. The importance of the dot product and cross product in establishing relationships between these vectors is emphasized, particularly in showing that certain dot products yield zero due to orthogonality. Additionally, the volume of the parallelepiped formed by the base vectors is noted as significant in the calculations. The conversation concludes with a demonstration of the anti-symmetry of the Levi-Civita symbol, reinforcing the zero result in specific dot products.
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How do I even represent the u's in indicial notation since the u's are labeled 1,2,3? I can't have components 1,2,3 and say u is $u_i$.

With respect to the triad of base vectors $\mathbf{u}_1$, $\mathbf{u}_2$, and $\mathbf{u}_3$ (not necessary unit vectors), the triad $\mathbf{u}^1$, $\mathbf{u}^2$, and $\mathbf{u}^3$ is said to be the reciprocal basis if $\mathbf{u}_i\cdot\mathbf{u}^j = \delta_{ij}$.
Show that to satisfy these conditions
$$
\mathbf{u}^1 = \frac{\mathbf{u}_3\times\mathbf{u}_1}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]};\quad
\mathbf{u}^2 = \frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]};\quad
\mathbf{u}^3 = \frac{\mathbf{u}_1\times\mathbf{u}_2}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]}
$$
and determine the reciprocal basis for the specific base vectors
\begin{alignat*}{3}
\mathbf{u}_1 & = & 2\hat{\mathbf{e}}_1 + \hat{\mathbf{e}}_2,\\
\mathbf{u}_2 & = & 2\hat{\mathbf{e}}_2 - \hat{\mathbf{e}}_3,\\
\mathbf{u}_3 & = & \hat{\mathbf{e}}_1 + \hat{\mathbf{e}}_2 + \hat{\mathbf{e}}_3
\end{alignat*}
 
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You can write $\mathbf u_i$ for instance as $u_{ij}$.
A basis can be seen as just a matrix containing the basis vectors.

You appear to have a couple of typos in your problem.
The numerators of the first two reciprocal vectors have the wrong indices.

To verify, consider that the two vectors in a cross product are both perpendicular to the resulting vector.
That should make it easy to calculate each of the dot products.
 
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ILikeSerena said:
You can write $\mathbf u_i$ for instance as $u_{ij}$.
A basis can be seen as just a matrix containing the basis vectors.

You appear to have a couple of typos in your problem.
The numerators of the first two reciprocal vectors have the wrong indices.

To verify, consider that the two vectors in a cross product are both perpendicular to the resulting vector.
That should make it easy to calculate each of the dot products.
$$
\mathbf{u}^1 = \frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]};\quad
\mathbf{u}^2 = \frac{\mathbf{u}_3\times\mathbf{u}_1}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]};\quad
\mathbf{u}^3 = \frac{\mathbf{u}_1\times\mathbf{u}_2}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]}
$$
 
All I have to do is show that $\mathbf{u}_1\cdot \mathbf{u}^j = \delta_{1j}$ where j = 1,2,3.
How do I do vector division?
\begin{alignat}{3}
\frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]}\cdot \mathbf{u}_1 & = & \frac{\mathbf{u}_2\times\mathbf{u}_3}{\mathbf{u}_1 \cdot(\mathbf{u}_2 \times\mathbf{u}_3)}\cdot \mathbf{u}_1\\
& = & \frac{\mathbf{u}_1}{\mathbf{u_1}}\\
& = & 1\\
\frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]}\cdot \mathbf{u}_2 & = & \frac{\mathbf{u}_2\times\mathbf{u}_3}{\mathbf{u}_1 \cdot(\mathbf{u}_2 \times\mathbf{u}_3)}\cdot \mathbf{u}_2\\
& = & \frac{\mathbf{u}_2}{\mathbf{u_1}}\\
& = & ?
\end{alignat}
 
The cross product in the numerator is a vector.
The dot product in the denominator is a scalar.

You are supposed to do a dot product of $\mathbf u_1$ with the nominator, and afterward divide by the scalar denominator.

Btw, the dot product in the denominator represent the volume of the parallellepipedum between the 3 basis vectors.
It is symmetric in each cyclic combination of the basis vectors.
 
ILikeSerena said:
The cross product in the numerator is a vector.
The dot product in the denominator is a scalar.

You are supposed to do a dot product of $\mathbf u_1$ with the nominator, and afterward divide by the scalar denominator.

Btw, the dot product in the denominator represent the volume of the parallellepipedum between the 3 basis vectors.
It is symmetric in each cyclic combination of the basis vectors.
\begin{alignat}{3}
\frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]} \cdot\mathbf{u}_2 & = & \frac{\mathbf{u}_2\times\mathbf{u}_3}{\mathbf{u}_1 \cdot(\mathbf{u}_2 \times\mathbf{u}_3)} \cdot\mathbf{u}_2\\
& = & \frac{\mathbf{u}_2\cdot(\mathbf{u}_2\times\mathbf{u}_3)}{\mathbf{u}_1 \cdot(\mathbf{u}_2 \times\mathbf{u}_3)}\\
& = & \frac{\varepsilon_{ijk}u_ju_k\hat{\mathbf{e}}_i \cdot \mathbf{u_2}}{\mathbf{u}_1 \cdot(\mathbf{u}_2 \times\mathbf{u}_3)}
\end{alignat}
How do I show the numerator is zero now?

Is it because the two left over are always the reverse?
(123) and (132) = 1 - 1 = 0
 
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The cross product of $\mathbf u_2$ and $\mathbf u_3$ is perpendicular to $\mathbf u_2$, therefore their dot product is zero $\square$.

Or if you want to do it with indices (your indices of the basis vectors are not quite right):
$\mathbf u_2 \cdot (\mathbf u_2 \times \mathbf u_3) $

$\qquad = u_{2i}\mathbf{\hat e}_i \cdot \varepsilon_{ijk}u_{2j}u_{3k}\mathbf{\hat e}_i$

$\qquad = \varepsilon_{ijk}u_{2i}u_{2j}u_{3k}$​

Since $\varepsilon_{ijk}$ is anti-symmetric in i and j, the result is zero $\square$.
 
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