How can the dual tensors derivation be achieved using rotation matrices?

[euphoricrhino]

Hello,
I'm reading Group Theory in a nutshell for physicist by A Zee. When he introduced Dual tensors (pp 192), he made a claim with a light hint, and I have had great trouble deriving this claim, any help would be appreciated -

Let $R \in SO(N)$ be an $N$-dimensional rotation, then the following is true
$$
\epsilon^{ijk\cdots n}R^{ip}R^{jq}=\epsilon^{pqr\cdots s}R^{kr}\cdots R^{ns}
$$
(where $\epsilon$ is the antisymmetric symbol and the above uses repeated index summing convention).
The hint was to use the $N\times N$ matrix determinant
$$
\epsilon^{ijk\cdots n}R^{ip}R^{jq}R^{kr}\cdots R^{ns}=\epsilon^{pqr\cdots s} \mbox{det}R=\epsilon^{pqr\cdots s} \quad(\mbox{since }R\mbox{ is special})
$$
and multiply it "by a bunch of $R^T$s carrying appropriate indices".

I have tried to understand the claim with $N=3$ which I think is the cross product relation, but I couldn't see how that could be obtained by involving $R^T$, and how it could be extended to $N$ dimensions.

Thanks for the help!

 

[Paul Colby]

I don't have Zee's book but does he talk about,

$\epsilon^{ijk\cdots n}R^{ip}R^{jq}\cdots R^{ns} = \det(R)\epsilon^{pqr\cdots s}$

which I think this follows from the definition of the determinate. Now, for $SO(N)$ one has $\det(R) = 1$. From this one applies the group relation $R^{ij}R^{ik} = \delta^{jk}$ to both sides twice and you have it. Hope this helps.

 

[euphoricrhino]

Thanks for the reply!

However I must be missing something really obvious, I don't see how to "apply the group relation $R^{ij}R^{ik}=\delta^{jk}$ to both sides" of the determinant equality.

The LHS of the determinant relation is a sum of $N!$ terms, each of which is a product whose factors don't share any index. Can you kindly elaborate for the $N=3$ case here?

From determinant equality
$$
\epsilon^{ijk}R^{ip}R^{jq}R^{kr}=\epsilon^{pqr}
$$
where $(pqr)$ is a given permutation
how to derive (for any given $p,q,k$)
$$
\epsilon^{ijk}R^{ip}R^{jq}=\epsilon^{pqr}R^{kr}
$$

Thank you very much!

 

[Paul Colby]

In matrix form

$R^TR = RR^T = I$

so, one also has the relation,

$R^{ip}R^{kp} = \delta^{ik}$

 

[euphoricrhino]

I finally figured it out, it's actually quite simple, but all the symbols there have been distracting.

The determinant relation
$$
\epsilon^{ijk\cdots n}R^{ip}R^{jq}R^{kr}\cdots R^{ns}=\epsilon^{pqr\cdots s}
$$
can be viewed as an inner product relation
$$
v^nR^{ns}=\epsilon^{pqr\cdots s}
$$
where $v^n$ is defined by
$$
v^n=\epsilon^{ijk\cdots n}R^{ip}R^{jq}R^{kr}\cdots
$$
Since columns $R^{\cdot s}$ form an orthonormal basis of the $N$-dimensional space, the inner product relation above actually gives the decomposition of vector $v$ into this basis, i.e.
$$
v=\epsilon^{pqr\cdots s}R^{\cdot s}
$$
Taking the $n$-th component of this yields
$$
\epsilon^{ijk\cdots n}R^{ip}R^{jq}R^{kr}\cdots=v^n=\epsilon^{pqr\cdots s}R^{ns}
$$

Now we just need to repeat the same argument on all the other $R$s on the left until only $R^{ip}$ and $R^{jq}$ were left