Help w/ Tensor Calc Homework: Bijk Properties Under Rotations

[c299792458]

Homework Statement

if BijkAjk is a vector for all symmetric tensors Ajk, (but Bijk is not necessarily a tensor),
what are the properties of Bijk under rotations of the basis/coordinate axes?

Homework Equations

The Attempt at a Solution

I am not sure what the question is looking for... though I can say that
BijkAjk=BijkAkj(by symmetry of Ajk)=BikjAjk(by relabeling dummy suffices)
Since Ajk is an arbitrary symmetric matrix, they cancel, giving Bijk=Bikj
So Bijk is symmetric wrt the last 2 suffices...

Thanks in advance!

 

[grey_earl]

No, B_{ijk} is not symmetric on the last 2 indices, but only the symmetric part survives. That you have showed right.
Now, we can write B_{ijk} = B1_{ijk} + B2_{ijk} where B1 is symmetric on the last 2 indices and B2 is antisymmetric on the last 2 indices. Then,
B_{ijk} A^{jk} = B1_{ijk} A^{jk} (because the antisymmetric part doesn't survive), and your problem said that this product transforms as a vector, so B1_{ijk} transforms on its first index as a vector. Now you can apply the reduction theorem (don't know if it's called this way) to conclude that B1_{ijk} transforms as a tensor under coordinate transformations, since its contraction with any tensor transforms as a vector. About the transformation properties of B2 you can not say anything.

 

[c299792458]

Thanks, grey_earl.

 

[grey_earl]

The reduction theorem basically says that if you contract an object with unknown transformation properties with an arbitrary tensor (which means multiplying and summing over equal indices, as you did), and this transforms as a scalar (or if you have one index free, as a vector, etc.), then your original object also transforms as a tensor. I think Schouten's tensor calculus has a proof of this.