# Question from Jones-Groups,Reps and Physics

1. Jul 23, 2010

### LAHLH

Hi,

I'm reading ch5 of jones about the conductivity tensor (p80 in my edition). He explains that the conductivity tensor is symmetric, and in general we can decompose any two index tensor into its antisymmetric and symmetric parts $$T_{ij}=T_{\{ij\}}+T_{[ij]}$$, where the symmetric part $$T_{[ij]}=\tfrac{1}{2}(T_{ij}-T_{ji})$$ and antisym part is $$T_{\{ij\}}=\tfrac{1}{2}(T_{ij}+T_{ji})$$. Which is stuff I already know and am happy with.

He then goes on to give the transformation law for such a tensor of rank 2, as $$T^{'}_{ij}=D^{V}_{ik}D^{V}_{jl}T_{kl}$$, again I'm happy with this. However he then talks about how the antisym part and sym part of a tensor lie in invariant subspaces, transforming independently under rotations, and the transformation matrices (presumably of the antisym part and sym part) can be found by subbing the decomposition of the tensor into its antisym/sym parts into $$T^{'}_{ij}=D^{V}_{ik}D^{V}_{jl}T_{kl}$$.

So I try to do this:

The LHS:

$$T^{'}_{ij}=\tfrac{1}{2}(T^{'}_{ij}-T^{'}_{ji})+\tfrac{1}{2}(T^{'}_{ij}+T^{'}_{ji})$$
$$=\tfrac{1}{2}(D^{V}_{ik}D^{V}_{jl}T_{kl}-D^{V}_{jk}D^{V}_{il}T_{kl})+\tfrac{1}{2}(D^{V}_{ik}D^{V}_{jl}T_{kl}+D^{V}_{jk}D^{V}_{il}T_{kl})$$

He states that the matrices are:

$$D^{(V\otimes V)\pm}_{(ij),(kl)}=\tfrac{1}{2}\left[D^{V}_{ik}D^{V}_{jl} \pm D^{V}_{il}D^{V}_{jk} \right]$$

If I just take line 2 of my LHS:

$$\tfrac{1}{2}(D^{V}_{ik}D^{V}_{jl}T_{kl}-D^{V}_{jk}D^{V}_{il}T_{kl})+\tfrac{1}{2}(D^{V}_{ik}D^{V}_{jl}T_{kl}+D^{V}_{jk}D^{V}_{il}T_{kl})$$

and equate this with my RHS in the form $$D^{V}_{ik}D^{V}_{jl}T_{kl}$$ I find:

$$\left[\tfrac{1}{2}(D^{V}_{ik}D^{V}_{jl}-D^{V}_{jk}D^{V}_{il})+\tfrac{1}{2}(D^{V}_{ik}D^{V}_{jl}+D^{V}_{jk}D^{V}_{il})\right]T_{kl}=D^{V}_{ik}D^{V}_{jl}T_{kl}$$

This does give the obvious $$D^{V}_{ik}D^{V}_{jl}=\left[\tfrac{1}{2}(D^{V}_{ik}D^{V}_{jl}-D^{V}_{jk}D^{V}_{il})+\tfrac{1}{2}(D^{V}_{ik}D^{V}_{jl}+D^{V}_{jk}D^{V}_{il})\right]$$

But this is just the decomposition of product rep into its sym and antisymmetric parts (wrt to i and j ), it's not as if the symmetric part of T transforms by $$D^{(V\otimes V)+}_{(ij),(kl)}$$ and the antisymmetric part transforms only by $$D^{(V\otimes V)-}_{(ij),(kl)}$$. Which is what I thought the point was going to be...

I've checked how say the symmetric part transforms $$T^{'}_{[ij]}=\tfrac{1}{2}(T^{'}_{ij}-T^{'}_{ji})=\tfrac{1}{2}(D^{V}_{ik}D^{V}_{jl}T_{kl}-D^{V}_{jk}D^{V}_{il}T_{kl})=\tfrac{1}{2}(D^{V}_{ik}D^{V}_{jl}T_{kl}-D^{V}_{jl}D^{V}_{ik}T_{lk})=D^{V}_{ik}D^{V}_{jl}\tfrac{1}{2}\left[T_{kl}-T_{lk}\right]=D^{V}_{ik}D^{V}_{jl}\tfrac{1}{2}T_{[kl]}\right]$$. This isn't therefore according to $$D^{(V\otimes V)-}_{(ij),(kl)}$$ then.

I'm rather lost at why the symmetric conductivity tensor should be invariant only under $$D^{(V\otimes V)+}_{(ij),(kl)}$$

Thanks for any help

Last edited: Jul 23, 2010
2. Jul 24, 2010

### LAHLH

Sorry if the above is a bit long winded, perhaps my question is lost in there somewhere. All I'm really wondering is why the conductivty tensor (or any symmetric tensor) would transform under $$D^{(V\otimes V)+}_{(ij),(kl)}$$. It seems to me that it doesn't, but rather just transforms under the whole tensor product rep, and that $$D^{(V\otimes V)+}_{(ij),(kl)}$$ is of no significance except being the symmetric part of the product rep.

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