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Prove $\vec i^i = \vec i_i$ where $\vec i_i$ is a basis vector and $\vec i^i$ is a dual basis vector in rectangular cartesian coordinates.

My attempt:

$\vec i^i \cdot \vec i_j = \delta_j^i$
$\delta_j^i = 1$ if $i=j$
$\vec i^i \cdot \vec i_i = 1$

Since the bases in rectangular coordinate systems are orthonormal...

$\vec i_i = 1$

Given that $\vec i^i \cdot \vec i_i = 1$...
$\vec i_i = \vec i^i = 1$

Thanks!

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    $\begingroup$ Letter $i$ to denote tensor, letter $i$ to denote index... now all you need is the letter $i$ to denote the complex scalar and you'll have a truly unholy trinity. $\endgroup$ – anon Aug 26 '17 at 2:09
  • $\begingroup$ Haha yes, the notation is pretty redundant. $\endgroup$ – Resonance Aug 26 '17 at 12:02
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I see this from a geometric perspective, but there may be a more general way to see it!

You see, Euclidean space is actually a Riemannian space with $g=I$ or $g_{ij}=\delta_{ij}$ in Cartesian coordinates. What you seem to be asking is about the musical isomorphism and raising and lowering indices. Here we are talking about vectors and covectors, but more generally one can talk about contravariant and covariant tensors. In any case, when a metric is available, one can lower an index by contracting with the metric.

So, in this case: $$ v_j = g_{ij}{v}^i = \delta_{ij}v^i $$ where notice that the RHS is just the $j$th component of the vector.

Or, in matrix-vector notation: $$ v = Iu = u $$ where $v$ is the dual of $u$.

See also here and here.

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