Rasalhague
Nov25-11, 04:06 AM
An Invariant I is of degree p if it is a linear invariant of the p-fold tensor product of the variable with itself, that is,
IA=J(A\otimes ...\otimes A)
- Bishop & Goldberg: Tensor Analysis on Manifolds, Dover 1980, p. 86
...
Problem 2.14.3. Show that (Tr A)p, where A is a tensor of type (1,1), in an invariant of degree p. [The fact that it is an invariant follows from the fact that Tr A is invariant. The question is whether (Tr A)2 is a linear function of the coefficients of A\otimes A etc.]
- ~ p. 87
I think I've misunderstood their definition of an invariant. The pth power of the trace function seems to be homogeneous of degree p (http://en.wikipedia.org/wiki/Homogeneous_function) rather than linear:
I(\lambda A) = J((\lambda A)\otimes (\lambda A)) = (\text{Tr}\, \lambda A)^2
=\lambda\lambda A^i_kA^r_m\delta^k_i\delta^m_r=\lambda\lambda A^k_kA^m_m=\lambda^2 (\text{Tr}\, A)^2= \lambda^2 IA \neq \lambda IA,
where \lambda is a scalar. (Summing over like indices.)
IA=J(A\otimes ...\otimes A)
- Bishop & Goldberg: Tensor Analysis on Manifolds, Dover 1980, p. 86
...
Problem 2.14.3. Show that (Tr A)p, where A is a tensor of type (1,1), in an invariant of degree p. [The fact that it is an invariant follows from the fact that Tr A is invariant. The question is whether (Tr A)2 is a linear function of the coefficients of A\otimes A etc.]
- ~ p. 87
I think I've misunderstood their definition of an invariant. The pth power of the trace function seems to be homogeneous of degree p (http://en.wikipedia.org/wiki/Homogeneous_function) rather than linear:
I(\lambda A) = J((\lambda A)\otimes (\lambda A)) = (\text{Tr}\, \lambda A)^2
=\lambda\lambda A^i_kA^r_m\delta^k_i\delta^m_r=\lambda\lambda A^k_kA^m_m=\lambda^2 (\text{Tr}\, A)^2= \lambda^2 IA \neq \lambda IA,
where \lambda is a scalar. (Summing over like indices.)