# Trace of a tensor product

1. Nov 9, 2008

### JasonJo

Hey guys,

How exactly do you take the trace of a tensor product? Do I take the trace of each tensor individually and multiply their traces?

For example, how would I take the trace of this tensor product:

$$-B^{c}_b B_{ac}$$

Last edited by a moderator: Nov 10, 2008
2. Nov 10, 2008

### tiny-tim

Hi JasonJo!

Isn't that just -gabBbcBac = BacBac ?

3. Nov 10, 2008

### astros

No! it's like matrices... the trace of a product is not the product of traces.
Now if you have a product of tensors which is an 1-1 tensor the trace is the contraction of these two indices, if you have more than 2 indices you must specify over which indices the trace must be done, and then you contract...

4. Nov 10, 2008

### locality

Yes, you can take the trace of each tensor individually and multiply their traces.
That is, Tr(A \Otimes B) = Tr(A)Tr(B)

5. Nov 11, 2008

### tiny-tim

No … it's as astros says:
For example,
Code (Text):
0 1
1 0
has trace 0, but its square has trace 2.

6. Nov 11, 2008

### Vid

If you mean the tensor product defined as how it's defined at the beginning of Ch 4 of Spivak's Calculus on Manifolds, then yes what locality said is correct.
Let V be a Vector space over R
A k tensor on V is a multilinear map from V^k into R.
The tensor product $\otimes$ of S:V^k -> R and T:V^l -> R is the map S$\otimes$T: V^(k+l) -> R defined by
S$\otimes$T(v_1,....v_k,....,v_(k+l)) = S(v_1,..v_k)*T(v_k+1,...,v_l).

In the special case S and T are square matrices,
tr(S$\otimes$T) = tr(S)tr(T)

Last edited: Nov 11, 2008
7. Nov 12, 2008

### tiny-tim

ah, but the OP's question was clearly about tensor product in the sense of contraction (analogously to matrix product ), not in the ⊗ sense of Spivak.

8. Nov 20, 2008

### JasonJo

That makes sense. But then what do I do this the B^ac B_ac term? What exactly is this term? Am I contracting here? When do we know something has a trace?

And when we are taking traces of tensors, what does it mean to be tracefree? Are we referring to the antisymmetric part?

If anyone is wondering, I am trying to derive Raychaudhuri's Equation.

9. Nov 21, 2008

### tiny-tim

BacBac is the "quadratic invariant" of the tensor.

It's a double contraction.

(It's a sort-of "trace of a trace", but I really don't think thinking of it like that helps at all)

It may help to look at http://en.wikipedia.org/wiki/Raychaudhuri's_equation#Mathematical_statement … are you trying to derive ω2 and σ2?
Tracefree simply means that the trace is zero.

An antisymmetric tensor must be tracefree, but not vice versa.

For example, the LHS of Einstein's field equations is Rij - (1/2)R gij, where Rij is the Ricci curvature tensor, which is symmetric, with 10 independent parameters, and R is its trace (a scalar, obviously only 1 parameter) … the tracefree part of Rij is Rij - (1/4)R gij, with 9 independent parameters, and the EFE can be written:

Rij - (1/4)R gij = 8π(Tij - (1/4)T gij)

and R = -8πT,

where the first line equates the tracefree parts, with 9 independent equations, and the second line (anti-)equates the traces, with just 1.