Convert 2x2 Matrix to 1x1 Tensor

[Vitani1]

TL;DR Summary

Trying to figure out how to transform tensors appropriately.

If I have a matrix representing a 2nd order tensor (2 2) and I want to convert this matrix from M$$\textsuperscript{ab}$$ to $$M\textsubscript{b}\textsuperscript{a}$$ what do I do? I'm given the matrix elements for the 2x2 tensor. When applying the metric tensor to this matrix I understand symbolically how to get this 1st tank tensor but I don't understand how to write it out.

 

[Vitani1]

Sorry - these should be proper super/subscripts.

 

[Ibix]

To write subscripts and superscripts use _ and ^. To get $M^{ab}$ use M^{ab}. To get $M^a{}_b$ do M^a{}_b (the empty {} produce a zero width blank which the subscript hangs off - missing it out gives $M^a_b$).

Applying the metric tensor to a rank-2 tensor doesn't give you a rank-1 tensor - it gives you a (1,1) tensor, which is also a rank-2 tensor. If you are representing the components in a matrix, it's still a 4×4 matrix. If you know how to write the expression, do you understand the Einstein summation convention? If so, can you write out explicitly what the $\mu\nu$ component of the result is in terms of a sum of products of the components of $M$ and the metric?

 

[PeterDonis]

@Vitani1 I'm confused about what you are trying to do. Are you trying to raise an index on the tensor $M$? Or are you trying to compute its trace, which is a scalar?

Raising an index, which is what is implied by your saying you want to convert the matrix from $M_{ab}$ to $M_a{}^b$, would look like this:

$$
M_{ab} \ g^{bc} = M_a{}^c
$$

Taking the trace, which is what is implied by your saying you want to convert the matrix to a 1 x 1 tensor, which is a scalar, would look like this:

$$
M_{ab} \ g^{ab}
$$

 

[Vitani1]

The first thing you stated is correct. Say I have a tensor Mab (superscripts) and I apply this metric tensor to this matrix to lower one of its indices - how would I multiply this result out?

 

[PeterDonis]

Say I have a tensor Mab (superscripts) and I apply this metric tensor to this matrix to lower one of its indices - how would I multiply this result out?

So you have $M^{ab} \ g_{bc} = M^a{}_c$ [Edit: fixed] in the standard notation using the Einstein summation convention. Do you understand how that convention works?

 

[Vitani1]

I take this to mean I multiply this matrix by a scalar given by the metric tensor g?

 

[Ibix]

No.

Do you understand the Einstein summation convention?

 

[Vitani1]

I guess not. I'll look it up. Thank you.

 

[Ibix]

I'll look it up.

See the link in @PeterDonis' last post.

 

[weirdoguy]

So you have

Shouldn't it be $M^{a}{}_{c}$?

 

[vanhees71]

So you have $M^{ab} \ g_{bc} = M_{ac}$ in the standard notation using the Einstein summation convention. Do you understand how that convention works?

It's $M^{ab} g_{bc}={M^a}_c$, and whenever two indices are the same (where one must be an upper and the other necessarily a lower index) you sum over $a$ (in GR from 0 to 3).

The metric components $g_{ab}$ are used to convert an upper index (contravariant) to a lower index (covariant).

 

[Vitani1]

Solved. Thanks! By the way there is a nice problem about this in A First Course In General Relativity by Schutz (Chapter 3, problem 24)

 

[PeterDonis]

Shouldn't it be $M^{a}{}_{c}$?

Yes, you're right, I'll fix the post.