Difference between upper and lower indices in Einstein notation

Question

Consider a $(2,0)$ tensor $X^{\mu \nu}$ that can be represented in matrix form by:

$$X^{\mu \nu} = \pmatrix{ a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p}\tag{1}$$

Here $a, b, c, ..., l \in \mathbb{R}$ are scalars.

We can obtain ${{X}_{\mu}}^{\nu}$ and ${{X^{\mu}}_{\nu}}$ by using the Minkowski metric $\eta_{\nu \mu}$: $${X^{\mu}}_{\nu} = \eta_{\nu \sigma}X^{\sigma\mu}$$ and $${{X}_{\mu}}^{\nu} = \eta_{\mu \sigma}X^{\nu\sigma} = X^{\nu\sigma}\eta_{\sigma \mu},$$ where $\eta_{\sigma \mu} = \eta_{\mu \sigma}$, because the Minkowski metric is symmetric.

${X^{\mu}}_{\nu}$ and ${{X}_{\mu}}^{\nu}$ can be represented in matrix form by considering the matrix representations of $X^{\mu \nu}$ and $\eta_{\mu \sigma}$. ${X_{\nu}}^{\mu}$ and ${{X}_{\mu}}^{\nu}$ are transpose matrices of each other. My question is what is the representation of each one and how can we uniquely distinguish the two matrices, so that they cannot be interchanged.

${X_{\nu}}^{\mu} \neq {{X}_{\mu}}^{\nu}$ (unless ${X_{\nu}}^{\mu}$ is symmetric), but what is the representation of each matrix in terms of the elements $a, b, c, d $ etc. of ${X^{\mu}}_{\nu}$?

The context of my question is that textbooks that give the electromagnetic tensor (in the $(-,+,+,+)$ metric signature) as

$$F^{\mu \nu} = \begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & B_z & -B_y \\ -E_y & -B_z & 0 & B_x \\ -E_z & B_y & -B_x & 0 \end{pmatrix}\tag{2}$$

then, using the Minkowski metric, they lower each index to get

$${F^{\mu}}_ \nu = \begin{pmatrix} 0 & E_x & E_y & E_z \\ E_x & 0 & B_z & -B_y \\ E_y & -B_z & 0 & B_x \\ E_z & B_y & -B_x & 0 \end{pmatrix}\tag{3}$$

and

$${F_{\mu}}^ \nu = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ -E_x & 0 & B_z & -B_y \\ -E_y & -B_z & 0 & B_x \\ -E_z & B_y & -B_x & 0 \end{pmatrix}\tag{4}$$

My question is how to distinguish the two. Surely once $F^{\mu \nu}$ is given, the terms of ${F^{\mu}}_ \nu$ and ${F_{\mu}}^ \nu$ can be found in a universally accepted way, such that we know which one's which. Is it that the upper index always numbers the different columns, and the lower index always numbers the different rows?

Answer 1

If $\mathbf X$ is a $(2,0)$-tensor, then that means that $\mathbf X$ is a map which eats two covectors $\boldsymbol \alpha$ and $\boldsymbol \beta$ and spits out a scalar $\mathbf X(\boldsymbol \alpha,\boldsymbol \beta)$. If we choose a basis $\hat e_\mu$ for the vector space and the corresponding basis $\hat \epsilon^\mu$ for the dual space (where $\hat \epsilon^\mu(\hat e_\nu) = \delta^\mu_{\nu}$), then the components $X^{\mu\nu}$ are defined by $$X^{\mu\nu} \equiv \mathbf X(\hat \epsilon^\mu, \hat \epsilon^\nu)$$

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Given any $(0,2)$-tensor $\boldsymbol \eta$, we can define a map from the space of vectors to the space of covectors. A vector $\mathbf v$ is mapped to the covector $\mathbf v^\flat$ which acts on a vector $\mathbf q$ as $$\mathbf v^\flat(\mathbf q) = \boldsymbol \eta(\mathbf q,\mathbf v)$$ The components of $\mathbf v^\flat$ in a particular basis are given by $$(\mathbf v^\flat)_\mu \equiv v^\flat(\hat e_\mu) = \boldsymbol \eta(\hat e_\mu, \mathbf v)$$ $$= \boldsymbol \eta(\hat e_\mu, v^\nu \hat e_\nu) = v^\nu \boldsymbol \eta(\hat e_\mu,\hat e_\nu) \equiv \eta_{\mu\nu} v^\nu$$

If $\boldsymbol \eta$ happens to be non-degenerate, then this mapping is invertible; we define the $(2,0)$-tensor $\boldsymbol \eta^\uparrow$ whose components $(\boldsymbol \eta^\uparrow)^{\mu\nu}$ are the matrix inverse of the components $\eta_{\mu\nu}$, so $$(\boldsymbol \eta^\uparrow)^{\mu\nu} \eta_{\nu \rho} = \delta^\mu_\rho$$

With this structure in place, we can map any vector $\mathbf v$ to a covector $\mathbf v^\flat$, and any covector $\boldsymbol \alpha$ to a vector $\boldsymbol \alpha^\sharp$ which is defined as you might expect: $$\boldsymbol \alpha^\sharp(\boldsymbol \beta) = \boldsymbol \eta^\uparrow(\boldsymbol \beta,\boldsymbol \alpha)$$ $$(\boldsymbol \alpha^\sharp)^\mu = (\boldsymbol \eta^\uparrow)^{\mu\nu} \alpha_\nu$$

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Having defined this so-called musical isomorphism between vectors and covectors, we can address the remainder of your question. Given a $(2,0)$-tensor $\mathbf X$ and a special choice of non-degenerate $(0,2)$-tensor $\eta$, we can define a $(1,1)$-tensor $\tilde {\mathbf X}$ which takes a covector $\boldsymbol \alpha$ in its first slot and a vector $\mathbf v$ in its second slot and returns the value $$\tilde{\mathbf X}(\boldsymbol \alpha,\mathbf v) = \mathbf X(\boldsymbol \alpha, \mathbf v^\flat) \qquad (\tilde {\mathbf X})^\mu_{\ \ \nu} \equiv \tilde {\mathbf X}(\hat \epsilon^\mu,\hat e_\nu) = X^{\mu\rho}\eta_{\nu\rho}$$ We could also define a $(1,1)$-tensor $\tilde {\mathbf X}^T$ which takes a vector $\mathbf v$ in its first slot and a covector $\boldsymbol \alpha$ in its second slot and returns the value $$\tilde {\mathbf X}^T(\mathbf v,\boldsymbol \alpha)= \mathbf X(\mathbf v^\flat,\boldsymbol \alpha) \qquad (\tilde {\mathbf X}^T)_\mu^{\ \ \nu} \equiv \tilde {\mathbf X}^T(\hat e_\mu, \hat \epsilon^\nu) = \eta_{\mu\rho} X^{\rho\nu}$$

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Much of this notation may appear foreign, but that's because it's traditional to drop the symbols $\sharp,\flat,\tilde{\ }, ^T$ and $\uparrow$ and distinguish e.g. $\boldsymbol \eta$ from $\boldsymbol \eta^\uparrow$ or $\mathbf v$ from $\mathbf v^\flat$ by the placement of the indices on their components. This saves a lot of writing and makes the expressions look neater, but comes at the cost of conceptual clarity. That's why I prefer to keep the extra symbols until a student gets so comfortable with them that they feel cumbersome and annoying, and only then drop them to yield expressions like $X_\mu^{\ \ \nu} = \eta_{\mu\rho} X^{\rho\nu}$, which (naively) would seem to suggest that the $X$ on the left is the same as the $X$ on the right when in fact they are different objects.

My question is what is the representation of each one and how can we uniquely distinguish the two matrices, so that they cannot be interchanged.

If you write down a list of numbers, there's no way for me to tell whether those numbers are the components of a covector or the components of a vector. You have to tell me in order for me to be able to interpret what you're saying correctly. Similarly, if you write down a square grid of numbers, I have no idea whether they're the components of a $(2,0)$-tensor, or a $(0,2)$-tensor, or a $(1,1)$-tensor, or of a linear transformation. You must provide this context.

Is it that the upper index always numbers the different columns, and the lower index always numbers the different rows?

Tradition dictates that the first (i.e. left-most) index labels the rows and the second index labels the columns, but that's just a convention. Whether the index is upstairs or downstairs is determined by whether the corresponding slot of the tensor eats a covector or vector, respectively. This in turn has implications for how the components of the tensor transform under a generic change of basis.

Answer 2

The first index always gives the row number, and the second index gives the column number of the matrix elements.

_{image from Matrix (mathematics)}

So, given $$X^{\mu\nu}= \begin{pmatrix} a&b&c&d\\ e&f&g&h\\ i&j&k&l\\ m&n&o&p \end{pmatrix}$$ and the Minkowski metric $\eta$ with $(-+++)$ signature, then you get $$X^\mu{}_\nu=X^{\mu\sigma}\eta_{\sigma\nu}= \begin{pmatrix} -a&b&c&d\\ -e&f&g&h\\ -i&j&k&l\\ -m&n&o&p \end{pmatrix}$$

$$X_\mu{}^\nu=\eta_{\mu\sigma}X^{\sigma\nu}= \begin{pmatrix} -a&-b&-c&-d\\ e&f&g&h\\ i&j&k&l\\ m&n&o&p \end{pmatrix}$$

Answer 3

With Einstein notation you have to distinguish between whether you are trying to use the matrices or just diagram the values.

Let me give you an example. The electromagnetic field tensor $F_{\alpha\beta}$ will, in relativistic electromagnetism expressions, unify the magnetic and electric fields. In Gaussian-type units in the $({+}{-}{-}{-})$ signature we would write, $$ F_{\alpha\beta}\leftrightarrow\begin{pmatrix} 0&E_x&E_y&E_z\\ -E_x&0&-B_z&B_y\\ -E_y&B_z&0&-B_x\\ -E_z&-B_y&B_x&0 \end{pmatrix}$$ Note that this has a very clear antisymmetry, that $F_{\alpha\beta}=-F_{\beta\alpha},$ or in other words its transpose is its negative.

By prioritizing transpose here, our matrix is no longer algebraic. You learned about matrix multiplication, well this is a matrix that was never intended to be multiplied with anything. The problem is that if you take a column vector of something which has raised indices, this matrix will give you back a column vector, but it will be of the values that have lowered indices. Now you have to color every vector that you write, to remind yourself whether it is the upper index color (say, “red”) or the lower index color (say, “blue”), and you have to color this matrix to remind you that it takes red vectors and gives you blue vectors, some nonsense like that.

A similar situation exists if I write $\eta_{\alpha\beta} =\operatorname{diag}(1,-1,-1,-1),$ technically speaking this is nonsense, the expression on the right is a diagonal matrix, a diagonal matrix should always map column vectors to other column vectors, taking $v^\mu$ to be a column vector then we should have $\eta^\alpha_\beta$ which is actually the identity matrix. The given diagonal matrix is, to use the previous paragraph's idea, bi-colored: it maps red vectors to blue vectors.

If I wanted to prioritize the algebra then I would have always, always, always depicted a lowered index as a row vector and a raised index as a column vector, and then I would have written $$ F^\alpha_{\phantom\alpha\beta}=\begin{bmatrix} 0&E_x&E_y&E_z\\ E_x&0&B_z&-B_y\\ E_y&-B_z&0&B_x\\ E_z&B_y&-B_x&0 \end{bmatrix}$$ And now you can see that it takes a second to confirm visually that this is an anti-symmetric tensor underlying this operation: the magnetic field part is fine, that is its own negative transpose, but you have to mentally flip the electric field column at the beginning.

From here you can actually compose this with the 4-velocity $v^\mu$ to find $c\gamma$ times the usual Lorentz force expression, the $\gamma$ gets absorbed into the $\mathrm dt=\gamma~\mathrm d\tau$ of an expression $${\mathrm dp^\alpha\over\mathrm d\tau}=\frac{q}{c} ~F^\alpha_{\phantom\alpha\beta}~v^\beta$$ and so everything works as expected to give you a familiar $\mathrm d\vec p/\mathrm dt.$ (Exercise: do this matrix multiplication explicitly and confirm.)

Now to answer your question, if you are doing this consistently then $F_\alpha^{\phantom\alpha\beta}$ would have had $\beta$ going up-down, and $\alpha$ going left-right. There would have been no ambiguity in what's algebraically correct. But people like to show off things like the antisymmetry of $F_{\alpha\beta}$ or the symmetry of $\eta_{\alpha\beta}$ and they don't like to write it as something long but correct like, $$\eta_{\alpha\beta} = \\\begin{bmatrix} [{-1}~0~0~0]&[0~{1}~0~0]&[0~0~{1}~0]&[0~0~0~1]\end{bmatrix},$$ mapping a column vector to a row vector clearly. And this throws everything off, there is always ambiguity, you just have to read the thing carefully and maybe work out the sign of the (0,3) and (1,2) components of the given matrix in terms of the formulas that you have, see does this match up or is there a sign error from what I think it should be, and then you know what exactly was intended.

One different notation that I like a lot for this is to use the $\dagger$ operator for the column-to-row transpose operator in matrix expressions, since it “lowers the index,” then you can write again, $$\eta_{\alpha\beta} = \dagger_\alpha \begin{bmatrix}{-1}&&&\\&1&&\\&&1&\\&&&1\end{bmatrix}_\beta,$$ Note that this gives you a very nice visual place to tell you what index is going down the rows side, $\alpha$, which would have otherwise been ambiguous. And this serves nicely to “color” a vector that is incorrectly written as a column vector when in fact it is a row vector: those column vectors have a little dagger in front of them.