Consistent matrix index notation when dealing with change of basis

[Shirish]

Until now in my studies - matrices were indexed like ##M_{ij}##, where ##i## represents row number and ##j## is the column number. But now I'm studying vectors, dual vectors, contra- and co-variance, change of basis matrices, tensors, etc. - and things are a bit trickier.

Let's say I choose to index matrices like: superscript denotes row number, subscript denotes column number. Let's try to apply this in change of basis equations for vectors and covectors. Suppose for a vector space ##V##, we switch from basis ##H## to ##\tilde H##. The corresponding dual bases are ##\Theta## and ##\tilde\Theta##. If ##b=[id_V]^H_{\tilde H}##, then $$\tilde h_j=b^i_jh_i$$ and indeed ##b^i_j## is the ##i,j##-th element of ##[id_V]^H_{\tilde H}##, since the ##j##-th column of this matrix is the representation of ##id_V(\tilde h_j)## in the ##H## basis. Given some vector ##v##, multiplying both sides by ##\tilde v^j##, we get $$v=b^i_j\tilde v^jh_i\implies v^i=b^i_j\tilde v^j\implies v(\theta^i)=b^i_jv(\tilde\theta^j)=v(b^i_j\tilde\theta^j)\\
\implies \theta^i=b^i_j\tilde\theta^j$$ The issue is that in THIS case, I cannot interpret ##b^i_j## as the ##i,j##-th element of ##[id_V]^H_{\tilde H}##, since the coefficients in the RHS are supposed to represent the column elements of the CoB matrix. In other words ##j## represents the row number this time, and ##i## represents the column number!

Does any consistent notation exist for tagging row/column indices in matrices in this whole treatment? Something that holds true for both the CoB equations for ##\tilde h_j## and ##\theta^i##? I have a vague idea about some staggered notation, but even if I were to stagger the subscript a bit to the right in all equations above, it would still give inconsistent results.

 

[PeroK]

Let's say I choose to index matrices like: superscript denotes row number, subscript denotes column number.

That might be the problem. The superscript denotes whether you have a covariant or contravariant component (if I can phrase it like that). If you are putting the coeffients in a matrix, then you can have a convention like the first index is the row and the second index is the olumn. In any case, there is a difference between ##b^i_{\ \ j}## and ##b_j^{\ \ i}##

 

[Shirish]

That might be the problem. The superscript denotes whether you have a covariant or contravariant component (if I can phrase it like that). If you are putting the coeffients in a matrix, then you can have a convention like the first index is the row and the second index is the olumn. In any case, there is a difference between ##b^i_{\ \ j}## and ##b_j^{\ \ i}##

So ##b^i_{\ \ j}## and ##b_j^{\ \ i}## represent transpose matrices?

 

[PeroK]

So ##b^i_{\ \ j}## and ##b_j^{\ \ i}## represent transpose matrices?

Not necessarily. In general, we have that ##b_i^{\ \ j}## and ##c^i_{\ \ j}## are inverse matrices. Only when the transformation is orthogonal is the inverse also the transpose.

In general we have (I'll use my notes as it's easier for me). The notation should be obvious with ' instead of tilde):
$$e'_i = \frac{\partial x^j}{\partial x'^i}e_j = b_i^{ \ \ j}e_j \ \ \text{(basis vectors)}$$$$\theta'^i = \frac{\partial x'^i}{\partial x^j}\theta^j = c^i_{ \ \ j}\theta^j \ \ \text{(basis dual vectors)}$$$$v'^i = \frac{\partial x'^i}{\partial x^j}\theta^j = c^i_{ \ \ j}v^j \ \ \text{(vector components)}$$$$w'_i = \frac{\partial x^j}{\partial x'^i}w_j = b_i^{ \ \ j}w_j \ \ \text{(dual vector components)}$$If we take an example:
$$x^1 = x'^1 + x'^2, \ x^2 = 2x'^1 - 2x'^2$$$$e'_1 = e_1 + 2e_2, \ e'_2 = e_1 - 2e_2$$This gives the first transformation matrix:
$$b_i^{\ \ j} =
\begin{bmatrix}
1& 2\\
1&-2
\end{bmatrix}
$$And:
$$x'^1 = \frac 1 2 x^1 + \frac 1 4 x^2, \ x'^2 = \frac 1 2 x^1 - \frac 1 4 x^2$$$$\theta'^1 = \frac 1 2 \theta^1 + \frac 1 2 \theta^2, \ \theta'^2 = \frac 1 4 \theta^1 - \frac 1 4 \theta^2$$This gives the second transformation matrix:
$$c^i_{\ \ j} =
\begin{bmatrix}
\frac 1 2 & \frac 1 2\\
\frac 1 4 & -\frac 1 4
\end{bmatrix}
$$

 

[Maarten Havinga]

So ##b^i_{\ \ j}## and ##b_j^{\ \ i}## represent transpose matrices?

These indicate two tensors that in one chosen basis could be represented as indeed transpose matrices, right?

 

[PeroK]

These indicate two tensors that in one chosen basis could be represented as indeed transpose matrices, right?

The components of a transformation matrix cannot be those of a tensor. A tensor transforms in a prescribed way under a change of basis. The transformation matrix described that change of basis in a particular case.

 

[Maarten Havinga]

You talk only about transformation matrices. But I thought any tensor with 1 covariant, 1 contravariant dimension can, in a given basis, be represented as a matrix. Since matrices transform the same way under basis transformations.

 

[PeroK]

You talk only about transformation matrices. But I thought any tensor with 1 covariant, 1 contravariant dimension can, in a given basis, be represented as a matrix. Since matrices transform the same way under basis transformations.

Nevertheless, a coordinate transformation or basis transformation is not a tensor.

 

[Maarten Havinga]

Well I'm also a mathematician that first studied matrices and later the contra and covariant tensor index notation. I guessed that the question of

So ##b^i_{\ \ j}## and ##b_j^{\ \ i}## represent transpose matrices?

was actually meant for any tensor ##b_i^j##.

If basis transformation matrices are no tensors, as a mathematician I consider the notation ##b_i^j v_j## to be an abuse of notation.

 

[PeroK]

If basis transformation matrices are no tensors, as a mathematician I consider the notation ##b_i^j v_j## to be an abuse of notation.

Coordinates, coordinate derivatives and Christoffel symbols are examples of things that are not tensors, but have raised and/or lower indices. Tensors are not the only thing that have indexed components.

 

[Shirish]

...
In general we have (I'll use my notes as it's easier for me). The notation should be obvious with ' instead of tilde):
$$e'_i = \frac{\partial x^j}{\partial x'^i}e_j = b_i^{ \ \ j}e_j \ \ \text{(basis vectors)}$$$$\theta'^i = \frac{\partial x'^i}{\partial x^j}\theta^j = c^i_{ \ \ j}\theta^j \ \ \text{(basis dual vectors)}$$$$v'^i = \frac{\partial x'^i}{\partial x^j}\theta^j = c^i_{ \ \ j}v^j \ \ \text{(vector components)}$$$$w'_i = \frac{\partial x^j}{\partial x'^i}w_j = b_i^{ \ \ j}w_j \ \ \text{(dual vector components)}$$

So from your notes, ##b_i^{\ \ j}=([id_V]_{H'}^H)_{ij}## and ##c^i_{\ \ j}=([id_{V*}]_{\Theta'}^{\Theta})_{ji}##. I was able to get the latter relation only because I've memorized the fact that columns of the CoB matrix represent the components of ##\theta'^i## in the ##\Theta## basis.

I guess I can't rely on some standardized correspondence between super/subscript (staggered) indices and matrix row/column numbers.

 

[PeroK]

So from your notes, ##b_i^{\ \ j}=([id_V]_{H'}^H)_{ij}## and ##c^i_{\ \ j}=([id_{V*}]_{\Theta'}^{\Theta})_{ji}##.

I don't understand why you have transposed ##i## and ##j## here.

I was able to get the latter relation only because I've memorized the fact that columns of the CoB matrix represent the components of ##\theta'^i## in the ##\Theta## basis.

All these relations can be proved using the definition of dual vectors and components.

I guess I can't rely on some standardized correspondence between super/subscript (staggered) indices and matrix row/column numbers.

It depends what you mean by standardised. Those four equations tell you all you need to know. They give you a defined way of writing the transformation coefficients in matrix form.

 

[Shirish]

I don't understand why you have transposed ##i## and ##j## here.

You mean for the second relation? Sorry I should've explained better. From what I've learned:

If ##A:V\to W## is a linear map (where ##H,F## are resp. bases of ##V,W##), let ##A(h_j)=a^i_jf_i##. Then we can collect all these coefficients into the matrix.

The representation of this linear map w.r.t. bases ##H,F## is ##[A]_H^F##. By convention, its columns denote the representation of the images of ##H## basis vectors in terms of the ##F## basis vectors. i.e., since ##a^1_1, a^2_1, a^3_1, \ldots, a^m_1## represent components of ##A(h_1)## in the ##F## basis, therefore they form the first column of ##[A]_H^F##.

Following the above definition/convention, we can replace ##V## and ##W## by ##V^*##, ##A## by ##id_{V^*}##, ##H## by ##\Theta'## and ##F## by ##\Theta##. Then ##A(h_j)=a^i_jf_i## becomes ##id_{V^*}(\theta'^j)=c^j_i\theta^i##. Same thing after interchanging ##i,j## is ##\theta'^i=c^i_j\theta^j##.

Now again, ##c^1_1, c^1_2,\ldots, c^1_n## represent components of ##\theta'^1## in the ##\Theta## basis, therefore they form the first column of ##[id_{V^*}]_{\Theta'}^{\Theta}##.

I hope this justifies why I switched ##i## and ##j## in ##c^i_{\ \ j}=([id_{V^*}]_{\Theta'}^{\Theta})_{ji}##