Raising and lowering indices and tensor contraction

I'm really confused by the notation of raising and lower indices in tensors when mixed with einstein summation notation and referencing the metric tensor. I need help separating several conflicting concepts and notations into a single framework that I can work with for reading literature.

Here Suppose $$A$$ is a first order "tensor" that is a vector, then abstractly $$A$$ can be viewed as a list of numbers we shall call this list $$\alpha_i$$ indexed by $$i$$. Now the tensor can many in forms:

My understanding is as follows:

First Model: Index Location Transpose

$$A^{i} = \begin{bmatrix} \alpha_{0} \\ \alpha_1 \\ \vdots \\ \alpha_n\end{bmatrix}$$

$$A_{i} = \begin{bmatrix} \alpha_{0} & \alpha_1 & \dots & \alpha_n\end{bmatrix}$$

Naturally then

We have the dot product or inner product:

$$A_iA^i = \alpha_{0}^2 +\alpha_1^2 + \dots + \alpha_n^2$$

And we have the tensor product or "outer product":

$$A^iA_i = \begin{bmatrix} \alpha_0^2 & \alpha_0 \alpha _1 & \dots & \alpha_0 \alpha_n \\\ \alpha_0 \alpha_1 & \alpha_1^2 & \dots & \alpha_1 \alpha_n \\\ \vdots & \vdots & \ddots & \vdots \\\ \alpha_0\alpha_n & \alpha_1 \alpha_n & \dots & \alpha_n^2 \end{bmatrix}$$ So far all is well.

In this model $$A^i A^i$$ is undefined and same with $$A_i A_i$$. Like these expressions cannot be well defined.

Second Model: Einstein Summation, ALL Repeated Indices are Summed Over

Here $$A^{i}$$ cannot be intrinsically viewed as a column or row vector, and therefore $$A^{i} = \alpha_i$$ with no "matrix like behavior". What can be said though is that in an expression if an index is repeated then we sum over the index, and moreover once we want to concretely evaluate $$A^{j}$$ versus $$A_{j}$$ then we just access the $$j^{\text{th}}$$ element of the list as per the answer here

Thus:

$$A_{i} A_{i} = A_{i}A^{i} = A^{i}A^{i} = A^{i}A_{i} = \alpha_0^2 + \alpha_1^2 + \dots + \alpha_n^2$$

This is weird to me because now its not clear at ALL what the difference between $$A_{i}$$ and $$A^{i}$$ is and this is in DIRECT conflict with the first model that can't even evaluate two of the sums listed above.

Third Model: The metric tensor model + Einstein convention

I still don't have rigorous definition of what $$A^{i}$$ or $$A_{i}$$ mean much like the einstein summation model BUT we do know the following that given a metric tensor (as a rank 2 tensor) $$g_{ij}$$ that we have

$$g_{ij}A^{i} = A_{j}$$ $$g^{ij}A_{j} = A^{i}$$

This third model is the most interesting, we have literally that:

$$g_{ij}A^{i} = \sum_{i=0}^{n}g_{ij}A^{i} = g_{0j}\alpha_0 + g_{1j}\alpha_1 \dots g_{nj}\alpha_n$$

But it's clear this is a different model as well, because of the following:

1. if we interpret $$A^{i}$$ to be a column vector then by definition $$A^{j}$$ is a column vector.

2. If we interpret $$A^{i}$$ to be a row vector, then by definition $$A^{j}$$ is a row vector.

3. If we assume NO matrix structure on $$A^{i}$$ that is, it is just a one-dimensional array of numbers and nothing more, then we have that $$A_{j}$$ is also JUST a one dimensional array of numbers and nothing more.

Qualitatively this is in DIRECT conflict with the first model, although it may not be in direct conflict with the second model, DEPENDING on your choice of metric tensor $$g_{ij}$$, of we let our metric tensor stray from being the euclidean metric (all 1's on the diagonal and zeros elsewhere) then this does not lead to the same results as the einstein summation convention I think.

• possible dup?: Lorentz invariance of the Minkowski metric. Aug 28 '19 at 22:13
• Not a full answer, just some advice: don't get hung up on row vs column vectors. In index notation, the components of a vector are just a list of numbers: the operations are defined by the index expressions you write down, not by the rules of matrix multiplication. Rows and columns are something to help you be faster once you're comfortable with indices. Aug 28 '19 at 22:18
• Don't think about what's a row or a column vector or how matrices are laid out. These are all ways of representing the components of tensors which can be used to speed up calculations, which depend entirely on the conventions you're using. The tensors themselves don't know or care. Aug 28 '19 at 23:56
• You referenced my answer to your previous, similar question. I didn’t say that $A^\mu$ and $A_\mu$ are the same. In general, they are not. Aug 29 '19 at 1:44
• @G.Smith yes, in fact looking back at it, I didn’t realize how literally to take your advice on “avoiding thinking in terms of a matrix”, this question I think clarifies what you were warning me could happen if I didn’t let go of the old algebraic notions Aug 29 '19 at 1:46

Let $$\mathbf A$$ be a vector. Given a choice of basis $$\{\hat e_i\}$$, we can write

$$\mathbf A= A^i \hat e_i = A^1 \hat e_1 + A^2 \hat e_2 +\ldots$$

where we've adopted the Einstein summation convention. The numbers $$A^1,A^2,\ldots$$ are called the components of $$\mathbf A$$ in the basis $$\{\hat e_i\}$$. In particular, $$A^i$$ (read "the $$i$$-th component of $$\mathbf A$$") does not have any particular structure - it's just a number.

If you wish, you can think of $$\mathbf A$$ as a column vector. As you note in your question, each column vector can be uniquely associated to a row vector; however, you are mistaken when you claim that the row vector is obtained simply by turning the column vector on its side.

Let $$\mathbf A$$ be a column vector and let $$\mathbf A^\flat$$ be its corresponding row vector. In general, the components $$A^\flat_i$$ of the row vector are related to the components $$A^i$$ of the column vector through the metric tensor as follows:

$$A^\flat_i = g_{ij} A^j$$

Explicitly, $$A^\flat_0 = g_{00}A^0 + g_{01}A^1 + g_{02}A^2 + g_{03}A^3$$ $$A^\flat_1 = g_{10}A^0 + g_{11}A^1 + g_{12}A^2 + g_{13}A^3$$

so on and so forth. From this point of view, obtaining a row vector by simply turning the corresponding column vector on its side is the special case where the components $$g_{ij}$$ of the metric are given by

$$g_{ij}=\begin{cases}1 & i=j \\ 0 & \text{otherwise}\end{cases}$$

It is standard practice to drop the $$\flat$$ notation with the understanding that we're talking about the column vector when the index is upstairs and the corresponding row vector when the index is downstairs. In my mind, it's worth adding a little extra notation to be extra clear until you're comfortable with the basics.

Additionally, though I worded this answer in the language of row vectors and column vectors, you should try to transition away from those ideas as soon as possible. Arranging components into rows and columns can be a useful mnemonic device, but thinking of a vector as a list of its components is deceiving (and incorrect). The language of differential geometry, in which vectors are understood as differential operators, can be difficult to get into at first, but is the language in which this subject really starts to become clear.