Is this tensor identity correct?

Homework Statement

I do not know if the following is correct;if it is,I will be able to save some calculation while doing a problem.Can you please let me know if it is true:

$$\epsilon_{ijk}\*\epsilon_{lmn} = \left(\begin{array}{ccc}\ g_\ {11}&\ g_\ {21}&\ g_\ {31}\\ g_\ {12}&\ g_\ {22}&\ g_\ {32}\\ g_\ {13}&\ g_\ {23}&\ g_\ {33}\end{array}\right)$$

Homework Equations

definition of $$\epsilon_\ {ijk}$$

definition of the "metric" tensor $$\ g_{ij}$$

The Attempt at a Solution

let me see first if the latex output has come OK.

Last edited by a moderator:

I am sorry.Last time there was a problem and the Latex output failed.

I am to prove $$\ det[g_{ij}]=\ V^2$$ where $$\[g_{ij}]$$ is metric tensor and $$\ V=\epsilon_i\cdot\epsilon_j\times\epsilon_k$$

Now,I was wondering if the identity

$$\epsilon_{ijk}\epsilon_{lmn}= \begin{vmatrix} \ g_{11}&\ g_{21}&\ g_{31}\\ g_{12}&\ g_{22}&\ g_{32}\\ g_{13}&\ g_{23}&\ g_{33} \end{vmatrix}$$

is correct---in which case the problem is solved easily.Note that the antisymmetric tensors are both co-variant.

Surely,the best way to know whether it is correct or not is to prove it...but somehow I am not getting it.If it is correct can anyone give me some hint?

HallsofIvy
Homework Helper
I fixed your Latex. You had {cc} where you wanted {ccc} for three columns. Also you had "\{" where you only wanted "{".

You give as a "relevant equation" "the definition of $\epsilon_{ijk}$. Okay, what is that definition?

We know $$\ g_{ij}=\ e_i\cdot\ e_j$$

Now, $$\ V^2=\ V\ V=\epsilon_{ijk}\epsilon_{lmn}$$

My idea was if we use the defiiniton of $$\epsilon_{ijk}=\ e_i\cdot\ e_j\times\ e_k$$

then after some vector manipulation we would get $$\ g_{ij}=\ e_i\cdot\ e_j$$,but that did not help much...

I believe the equation is correct for it reduces to a known result when one of the epsilons are covariant and the other is contravariant.

gabbagabbahey
Homework Helper
Gold Member
I am sorry.Last time there was a problem and the Latex output failed.

I am to prove $$\ det[g_{ij}]=\ V^2$$ where $$\[g_{ij}]$$ is metric tensor and $$\ V=\epsilon_i\cdot\epsilon_j\times\epsilon_k$$

Your definition of $V$ makes very little sense to me. Are $\epsilon_i$ orthogonal unit vectors sush that $\epsilon_i\epsilon^j=\delta_{ij}$? How are you defining the dot and cross products between covariant vectors?

Now,I was wondering if the identity

$$\epsilon_{ijk}\epsilon_{lmn}= \begin{vmatrix} \ g_{11}&\ g_{21}&\ g_{31}\\ g_{12}&\ g_{22}&\ g_{32}\\ g_{13}&\ g_{23}&\ g_{33} \end{vmatrix}$$

I can tell you just from glancing at your proposed identity that there is no way it is correct. On the LHS of your "identity" you have a 6th rank covariant tensor (Assuming the definition you are using for $\epsilon_{ijk}$ makes it a tensor and not a tensor density----there are at least two different definitions for $\epsilon_{ijk}$, so you need to tell us which one your course/textbook uses!) while on the RHS you have the determinant of a matrix; which is a scalar....how can a 6th rank tensor possibly be equal to a scalar?!

We know $$\ g_{ij}=\ e_i\cdot\ e_j$$

Now, $$\ V^2=\ V\ V=\epsilon_{ijk}\epsilon_{lmn}$$

No, assuming $V$ is a vector (again, your definition is not at all clear to me), then $V^2=V_{a}V^{a}$ is a scalar, why do you think it is equal to $\epsilon_{ijk}\epsilon_{lmn}$???

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Let me tell you step by step:

A.Neither $$\ V$$ is a vector and $$\epsilon_i$$ are orthogonal unit vectors. My proposition was V is the volume of the parallelopiped whose sides are three three linearly independent vectors: $$\ e_i,\ e_j,\ e_k$$.In the first post there was a typo and I meant the unit vectors are $$\ e_i$$ and not $$\epsilon_i$$.And I do not propose $\epsilon_i\epsilon^j=\delta_{ij}$

Regarding definiing dot and cross products between covariant vectors:

With this do you agree: $$\epsilon_{ijk}=(\ e_j\times\ e_k)\cdot\ e_i$$?

Now,we also know:

$$\epsilon_{ijk} =\ +\ V ,[\ i,\ j,\ k \ is\ an\ even\ permutation\ of (\ 1,\ 2,\ 3)]\\ =\ -\ V ,[\ i,\ j,\ k \ is\ an\ odd\ permutation\ of (\ 1,\ 2,\ 3)]\\ =0 ,[\ two\ or\ more\ indices\ are\ equal].$$

Hence, for orthonormal system, V=1 as $$\epsilon_{ijk}=\ 1$$ for even permutation.

I can tell you just from glancing at your proposed identity that there is no way it is correct. On the LHS of your "identity" you have a 6th rank covariant tensor.Assuming the definition you are using for $\epsilon_{ijk}$ makes it a tensor and not a tensor density----there are at least two different definitions for $\epsilon_{ijk}$, so you need to tell us which one your course/textbook uses!) while on the RHS you have the determinant of a matrix; which is a scalar....how can a 6th rank tensor possibly be equal to a scalar?!

Yes,I can clearly see it's wrrong...I am new to tensors.

However, then why the definition that $\epsilon_{ijk}=\pm\ V$ etc. for even or odd permutations of 1,2,3 is correct---$\epsilon_{ijk}$ is also a rank three tensor...Is not it?

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OK...I got it...$$\epsilon_{ijk}=\ +\ V$$ means ijk-th component of epsilon antisymmatric matrix is equal to V.

So,can we write ...$$\epsilon_{ijk}\epsilon_{lmn}=\ V\ V=\ V^2$$ to mean that as we multiply ijk-th element of $$\epsilon$$ tensor and lmn-th element of $$\epsilon$$ tensor, we get the squared volume of the parallelopiped spanned by the vectors $$\ e_i,\ e_j,\ e_k$$?

Let us refer to the link provided by robphy.Refer to the section:1.1 and 1.2---relation to Kronecker Delta and generalisation to n dimensions.

I think there is a problem with covariant and contravariant scheme written out there.$$\epsilon_{ijk}\epsilon^{lmn}$$ and not $$\epsilon_{ijk}\epsilon_{lmn}$$ is equal to the det [Kronecker delta] as long as we are not using orthonormal basis.by [Kronecker delta] I mean the matrix written there.

I thought $$\epsilon_{ijk}\epsilon_{lmn}$$ is equal to det[g_ij] as

$$\ g_{ij}=\ e_i\cdot\ e_j$$ and

$$\ g_{ij}=\ e_i\cdot\ e^j=\delta_i^j$$

Could I convey my thought?

Let $$\ e_i,\ e_j,\ e_k$$ form a basis,not necessarily orthogonal.Then,
$$\ e_i\cdot(\ e_j\times\ e_k)$$ is the volume |V| spanned by the bases.
In the deleted post I intended to prove
$$\ e_i\cdot(\ e_j\times\ e_k)=\epsilon_{ijk}$$

Thus,it follows that for a non-orthogonal basis,
$$\ e_i\cdot(\ e_j\times\ e_k)=\epsilon_{ijk}=+V$$
So,for even permutation of ijk,ijk-th element of $$\epsilon$$ tensor=+V

$$\ e_i\cdot(\ e_j\times\ e_k)=\epsilon_{ijk}=-V$$
So,for odd permutation of ijk,ijk-th element of $$\epsilon$$ tensor=-V

etc.

However,if the basis is orthonormal,we would have |V|=1.

Thus,$$\ V^2=\epsilon_{ijk}\epsilon_{lmn}=[\ e_i\cdot(\ e_j\times\ e_k)][\ e_l\cdot(\ e_m\times\ e_n)]$$ for even permutation of {i,j,k} and {l,m,n} where {i,j,k},{l,m,n}={1,2,3}

Let $$\ e_i=i,\ e_j=j,\ e_k=k,\ e_l=l,\ e_m=m,\ e_n=n$$

I was failing to evaluate the following determinant:

$$\begin{vmatrix}\ i_1&\ i_2&\ i_3\\ j_1&\ j_2&\ j_3\\ k_1&\ k_2&\ k_3 \end{vmatrix} \begin{vmatrix}\ l_1&\ l_2&\ l_3\\ m_1&\ m_2&\ m_3\\ n_1&\ n_2&\ n_3 \end{vmatrix} = \begin{vmatrix}\ i_1\ l_1\ +\ i_2\ l_2\ + \ i_3\ l_3 &\ i_1\ m_1\ + \ i_2\ m_2\ + \ i_3\ m_3 & \ i_1\ n_1\ +\ i_2\ n_2\ + \ i_3\ n_3\\ \ j_1\ l_1\ +\ j_2\ l_2\ + \ j_3\ l_3 &\ j_1\ m_1\ + \ j_2\ m_2\ + \ j_3\ m_3 &\ j_1\ n_1\ +\ j_2\ n_2\ + \ j_3\ n_3\\ \ k_1\ l_1\ +\ k_2\ l_2\ + \ k_3\ l_3 &\ k_1\ m_1\ + \ k_2\ m_2\ + \ k_3\ m_3 & \ k_1\ n_1\ +\ k_2\ n_2\ + \ k_3\ n_3 \end{vmatrix} =\begin{vmatrix} \ i\cdot\ l &\ i\cdot\ m & \ i\cdot\ n\\ \ j\cdot\ l &\ j\cdot\ m & \ j\cdot\ n\\ \ k\cdot\ l &\ k\cdot\ m & \ k\cdot\ n \end{vmatrix} =\begin{vmatrix} \ e_i\cdot\ e_l &\ e_i\cdot\ e_m & \ e_i\cdot\ e_n\\ \ e_j\cdot\ e_l &\ e_j\cdot\ e_m & \ e_j\cdot\ e_n\\ \ e_k\cdot\ e_l &\ e_k\cdot\ e_m & \ e_k\cdot\ e_n \end{vmatrix} =\begin{vmatrix} \ g_{il} &\ g_{im} & \ g_{in}\\ \ g_{jl} &\ g_{jm} & \ g_{jn}\\ \ g_{kl} &\ g_{km} & \ g_{kn} \end{vmatrix} =det[g]$$

Can the moderators please fix the Latex?