# Very specific question about index notation

1. Jul 31, 2013

### mindarson

I am reading through this text

http://www.ita.uni-heidelberg.de/~dullemond/lectures/tensor/tensor.pdf

and am having a bit of trouble with one of the arguments that is put in index notation. Specifically, equation (3.3). I was wondering if anyone could have a look at it and clear up a confusion for me.

I understand the argument, i.e. that the 'old definition' (eqn (3.2) in the text) of the inner product is not invariant under coordinate transformation in general, which is why we need covectors, covariant components, etc.

My specific question is about how the index notation is used in eqn (3.3). The authors write that

s' = <a',b'> = Aμ αaαAμ βbβ = (AT)μ αAμ βaαbβ (3.3)

They then argue that this shows that only if A-1 = AT (so the 2 matrices together equal δβα) (i.e. only if the transformation is orthonormal) will the inner product actually come out to the same value that it had in the untransformed coordinate system.

My question is how to express the inverse and transpose of a matrix in index notation. Where did the transpose come from in the 3rd equality, and why did the indices on it not change position at all? How is the relationship between a matrix, its transpose, and its inverse expressed in index notation? How, exactly, do the authors read off from (3.3) the fact that A-1 must equal AT?

I do understand that, to complete the argument, we ultimately need α = β, but how does one get, in practice, from 2 matrices to the Kronecker delta? What would the multiplication of a matrix by its inverse to get the Kronecker delta actually look like when written out?

I understand the argument, but I need clarification on how the argument is being expressed specifically using index notation.

2. Jul 31, 2013

### Bill_K

This author is being very careless. Eq 3.3 is invalid, since it has two μ's upstairs. It should be written

Aμα aα Aμβ bβ

The transpose of a tensor is obtained the same way as the transpose of a matrix - by interchanging rows and columns. So Aμα = (AT)αμ. Thus we have

(AT)αμ Aμβ aα bβ

The inverse of Aμβ is defined as the tensor (A-1)αμ such that

(A-1)αμ Aμβ = δβα

and thus comparing the last two eqs we have (AT)αμ = (A-1)αμ

3. Jul 31, 2013

### dextercioby

Hi Bill, it's valid as he didn't assume that <mu> upstairs differs from <mu> downstairs. Actually he uses the metric as the unit matrix, so he's free to place the indices wherever he wants. It's like special relativity with x4=ict.

4. Jul 31, 2013

### Bill_K

Are you sure? He does draw a distinction between co- and contravariant indices. In fact he says earlier

5. Jul 31, 2013

### dextercioby

Then you're right and the notes are badly written.