I am reading "An Introduction to the Theory of Piezoelectricity" 2nd edition by Jiashi Yang

I am trying to understand a derivation for an equation that uses multiple Levi-Civita symbols.

The Levi-Civita symbol is defined as: \begin{equation} \varepsilon _{ijk} = \hat i_i \cdot \left( {\hat i_j \times \hat i_k } \right) = \left\{ {\begin{array}{*{20}c} 1 \hfill & {i,j,k = 1,2,3;} \hfill & {2,3,1;} \hfill & {3,1,2,} \hfill \\ { - 1} \hfill & {i,j,k = 3,2,1;} \hfill & {2,1,3;} \hfill & {1,3,2,} \hfill \\ 0 \hfill & {{otherwise}} \hfill & \hfill & \hfill \\ \end{array}} \right. \end{equation}

and the product of two Levi-Civita symbols is given by

\begin{equation} \varepsilon _{ijk} \varepsilon _{pqr} = \left| {\begin{array}{*{20}c} {\delta _{ip} } & {\delta _{iq} } & {\delta _{ir} } \\ {\delta _{jp} } & {\delta _{jq} } & {\delta _{jr} } \\ {\delta _{kp} } & {\delta _{kq} } & {\delta _{kr} } \\ \end{array}} \right| \end{equation} where $\delta$ is the Kronecker delta.

We have a matrix that defines the deformation in three dimensions of

\begin{equation} \left[ {\begin{array}{*{20}c} {y_{1,1} } & {y_{1,2} } & {y_{1,3} } \\ {y_{2,1} } & {y_{2,2} } & {y_{2,3} } \\ {y_{3,1} } & {y_{3,2} } & {y_{3,3} } \\ \end{array}} \right] \end{equation}

where $y_{k,K} = \frac{{\partial y_k }}{{\partial X_K }}$

The Jacobian of the deformation is given as:

$J = \det \left( {y_{k,K} } \right)$

So if I expand it out manually I get

$ J = y_{1,1} \left( {y_{2,2} y_{3,3} - y_{2,3} y_{3,2} } \right) - y_{1,2} \left( {y_{2,1} y_{3,3} - y_{2,3} y_{3,1} } \right) + y_{1,3} \left( {y_{2,1} y_{3,2} - y_{2,2} y_{3,1} } \right) $

The author writes the results (Equation 1.12) using the Levi-Civita symbols and Einstein summation convention notation as

$ \begin{array}{l} J = \det \left( {y_{k,K} } \right) = \varepsilon _{ijk} y_{i,1} y_{j,2} y_{k,3} = \varepsilon _{KLM} y_{1,K} y_{2,L} y_{3,M} = \frac{1}{6}\varepsilon _{klm} \varepsilon _{KLM} y_{k,K} y_{l,L} y_{m,M} \\ \\ \end{array} $

I can see by rearranging (order of terms and factors) my results manually by hand and using the definition of the Levi-Civita symbol that the three versions he gives are consistent with my results (including the 1/6 coefficient in the last one). However I assume there must be an an algebra that can be applied to his three results directly in their form with the Levi-Civita symbols and Einstein notation that can be used to convert from one to the other two.

The issue comes up when I try to follow his derivation for Equation 1.17 for which he says can be verified that for all L, M and N that the following is true:

$\varepsilon _{ijk} y_{i,L} y_{j,M} y_{k,N} = J\varepsilon _{LMN}$

I try to obtain this result by taking his last result in Equation 1.12

$J = \frac{1}{6}\varepsilon _{klm} \varepsilon _{KLM} y_{k,K} y_{l,L} y_{m,M}$

Changing the subscripts klm to ijk and KLM to LMN giving

$J = \frac{1}{6}\varepsilon _{ijk} \varepsilon _{LMN} y_{i,L} y_{j,M} y_{k,N}$

From here I assume I can treat the Levi-Civita symbol as a scalar (which I'm really not sure is correct) and multiply both sides by $\varepsilon _{LMN}$ and I obtain

$J\varepsilon _{LMN} = \frac{1}{6}\varepsilon _{ijk} \varepsilon _{LMN} \varepsilon _{LMN} y_{i,L} y_{j,M} y_{k,N}$

Using the relationship above for the product of two Levi-Civita symbols and the Kronecker delta, $\varepsilon _{LMN}$ multiplied by itself on the right hand side should equal 1 giving

$J\varepsilon _{LMN} = \frac{1}{6}\varepsilon _{ijk} y_{i,L} y_{j,M} y_{k,N}$

which is close to the result of Equation 1.17 except for the 1/6 coefficient.

So it is unclear to me if my calculations are incorrect or if there is a typo in Equation 1.17.


1 Answer 1


I'm sure there is a more elegant way to do this, but this is what I came up with. I will assume the Einstein summation convention throughout the following.

Let \begin{align} Y_{LMN} = \varepsilon_{ijk}y_{iL}y_{jM}y_{kN} \end{align} Then we have, by Eq. 1-12. \begin{align} \varepsilon_{LMN}Y_{LMN} = 6\det y \end{align} so that \begin{align} \det y = \frac{1}{6}\varepsilon_{ABC}Y_{ABC} \end{align} where I have changed the dummy indices from $L$, $M$, $N$ to $A$, $B$, $C$ to avoid confusion in the next step. Now take \begin{align} \varepsilon_{LMN} \det y = \frac{1}{6}\varepsilon_{LMN}\varepsilon_{ABC}Y_{ABC} \end{align} and use the determinant expansion of the product of epsilons. I will skip writing it out explicitly and just note that it yields \begin{align} \varepsilon_{LMN} \det y = \frac{1}{6}(&Y_{LMN} - Y_{LNM} - Y_{MLN}\\ &+ Y_{NLM} + Y_{MNL} - Y_{NML}) \tag{1} \end{align} Since the indices $i$, $j$, $k$ are dummy indices in the sum \begin{align} Y_{LMN} = \varepsilon_{ijk}y_{iL}y_{jM}y_{kN} \end{align} we can exchange them freely. For example, \begin{align} Y_{LMN} = \varepsilon_{jki}y_{jL}y_{kM}y_{iN}. \end{align} Then using the definition of the Levi-Civita symbol, we can permute the indices to get, for example, \begin{align} Y_{LMN} = \varepsilon_{ijk}y_{iN}y_{jL}y_{kM} = Y_{NLM} \end{align} In general, we find that $Y_{LMN}$ has the same behavior under index permutations as the Levi-Civita symbol. That is, for cyclic permutations, \begin{align} Y_{LMN} = Y_{MNL} = Y_{NLM} \end{align} and for anti-cyclic permutations \begin{align} Y_{LMN} = -Y_{LNM} = -Y_{MLN} = -Y_{NML} \end{align} Applying this property to $(1)$, we find \begin{align} \varepsilon_{LMN} \det y &= \frac{1}{6}(6Y_{LMN})\\ &= Y_{LMN} \end{align} as claimed.

  • $\begingroup$ Thanks for the response. I think my main error was thinking that the Levi-Civita times itself "squared" equaled 1 whereas apparently it is equal to 6. I neglected to apply the Einstein summation convention to its subscripts. It is a little confusing for me since with the symbol isolated there is no summation but as soon as it appears times itself you have to account for the summation. I would like to find a table of properties for the symbol such as one would have for something like the Fourier transform. $\endgroup$
    – MikeM
    Feb 11 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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