What Is the Difference Between M^a_b and M_b^a in Tensor-Index Notation?

[HJ Farnsworth]

Greetings,

There is an aspect of tensor-index notation that I want to know more about. For a simple example, let M be a 1-1-tensor (ie., a matrix), whose elements can be indexed by Latin letters.

What is the difference between the component M^a_{\hspace{2mm}b} and the component M_b^{\hspace{2mm}a}?

The fact that there is a difference between the two components must indicate that the M in the latter case is not the same matrix as the M in the former case. Otherwise, since in both cases M is a rank-2 tensor with one contravariant component, indexed above by a, and one covariant component, indexed above by b, the component M^a_{\hspace{2mm}b} and the component M_b^{\hspace{2mm}a} are both referring to the same matrix element, namely, that in row a and column b. So, if in both cases M is the same matrix, then they both refer to the same matrix element of the same matrix, and therefore must be the same.

So, if M^a_{\hspace{2mm}b} refers to a component of the matrix M, then M_b^{\hspace{2mm}a} must refer to a component in the same row and same column of a different matrix, which I will call M^\#, just as M_a^{\hspace{2mm}b} doesn't refer to a component of M, but instead refers to a component of M^T.

Firstly, could someone please confirm that this is what is going on?

More importantly, is there a name for the transformation that changes the matrix with components M^a_{\hspace{2mm}b} to that with components M_b^{\hspace{2mm}a}? In other words, is there a name for what I used the symbol \# for in the previous paragraph, analogous to the transpose operation T at the end of the paragraph?

Moving beyond the simple case of a 1-1-tensor, is there a general name for operations which take an m-n-tensor indexed by m+n indices, and switches some of the indices horizontally only (ie., switches around the location of contravariant and covariant indices, but doesn't make any contravariant components covariant or vice versa)?

Finally, does anyone have a good interpretation of the difference between the matrices M and M^\#, ie., an intuitive sense of what it mathematically means to switch the horizontal position of two indices of a matrix (and, more generally, for a tensor)? Perhaps a very simple example would be rotations - if M represents a rotation about some axis (not necessarily in Euclidean space, if a different space is needed for the index switch to matter), what exactly does M^\# represent?

Thanks very much for any help that you can give.

-HJ Farnsworth

 

[tiny-tim]

Greetings HJ Farnsworth!

What is the difference between the component M^a_{\hspace{2mm}b} and the component M_b^{\hspace{2mm}a}?

see WannabeNewton's post (#2) in https://www.physicsforums.com/showthread.php?t=695850 …

(i did a tag search )

it makes a difference when you contract

(essentially, it's the same as the difference between M^ab and M^ba)

 

[HJ Farnsworth]

Hi tiny-tim, thanks for the response!

That is definitely a simple way to look at why they are different entities - a nice way to think about it using 0-2-tensors (or 2-0-tensors), since its a lot easier to see how those tensor components become different if the tensor is not symmetrical, and so immediately makes it clear that the horizontal switch does, indeed, have to make a difference. So, thanks very much for the link!

Does anyone know of the answers to my other questions? I think what I am most curious about is what the operation that I called \# in my previous post would be called (ie., is their a name for the matrix M^\# in my previous post, analogous to how M^T is called M-transpose?

Thanks again.

-HJ Farnsworth

 

[WannabeNewton]

$M_{b}{}{}^{a}$ is the transpose of $M^{a}{}{}_{b}$; you're just swapping the indices which is exactly what the transpose is. $M_{a}{}{}^{b}$ is a different object which is given by $M_{a}{}{}^{b} = g_{ac}g^{bd}M^{c}{}{}_{d}$ so it's related to the original tensor $M^{a}{}{}_{b}$ by means of the index raising and lowering operations given by the metric tensor $g_{ab}$; in other words the metric tensor and its inverse define an isomorphism between contravariant indices and covariant indices of a tensor. This is called a musical isomorphism: http://en.wikipedia.org/wiki/Musical_isomorphism

 

[HJ Farnsworth]

Thanks for replying, WannabeNewton, as well as for your post that I was referred to earlier in this thread!

This is called a musical isomorphism: http://en.wikipedia.org/wiki/Musical_isomorphism

That may be just the thing I'm looking for, I'll look it up and learn about it.

M_{b}{}{}^{a} is the transpose of M^{a}{}{}_{b}

This is a minor point, and could just come down to a notational difference, but wouldn't M^{b}{}{}_{a}, rather than M_{b}{}{}^{a}, be the transpose element for M^{a}{}{}_{b}? This seems correct to me just because, if we write the matrix M and a vector v as

M=<br /> \left( \begin{array}{ccc}<br /> M^{1}{}{}_{1} &amp; M^{1}{}{}_{2} \\<br /> M^{2}{}{}_{1} &amp; M^{2}{}{}_{2} \end{array} \right) and v=<br /> \left( \begin{array}{ccc}<br /> v^{1} \\<br /> v^{2} \end{array} \right),

then we have

Mv=<br /> \left( \begin{array}{ccc}<br /> M^{1}{}{}_{1}v^{1}+M^{1}{}{}_{2}v^{2} \\<br /> M^{2}{}{}_{1}v^{1}+M^{2}{}{}_{2}v^{2} \end{array} \right)<br /> =M^{i}{}{}_{j}v^{j},

ie., this convention for M does give us the right result using ESC. Then, taking the transpose of M, we get

M^{T}=<br /> \left( \begin{array}{ccc}<br /> M^{1}{}{}_{1} &amp; M^{2}{}{}_{1} \\<br /> M^{1}{}{}_{2} &amp; M^{2}{}{}_{2} \end{array} \right),

so that it seems like we write the transpose of M as M^{a}{}{}_{b}\rightarrow M^{b}{}{}_{a}.

Thanks again, I appreciate the help!