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My question refers to Piattella's lecture notes on cosmology. On page 15, the Euclidean line element is defined as

$$ ds^2 = \vert d\mathbf{x}\vert^2 = \delta_{ij}dx^idx^j. $$

My first question is simply this: why do we write $\delta_{ij}$ here, when I am used to seeing the Kronecker delta written as $\delta_j^i$, with one raised index? The same question applies to a more general line element

$$ ds^2 = \vert d\mathbf{x}\vert^2 = g_{ij}dx^idx^j. $$ That is, why do we seem to write a metric tensor with two lower indices? In other aspects of my work, involving quantum field theory, the convention is to write the standard Minkowski metric as $\eta^{ij}$, with two raised indices, as opposed to two lowered indices. Why is it different here?

My confusion with raising and lowering indices continues on page 16 of the same document. The components of the spacial Robertson-Walker metric are defined here as

$$g_{ij} = \delta_{ij} + K\frac{x_ix_j}{1-K\vert{\mathbf{x}}\vert^2}$$.

Why, in this formula, have we suddenly gone from writing lengths such as $dx^i$ with raised indices to ones such as $x_i$ with lowered ones? Any help understanding this would be greatly appreciated.

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In general, a $(p,q)$-tensor eats $p$ covectors and $q$ vectors and returns a real number; such an object has $p$ indices upstairs and $q$ indices downstairs. The components of the metric tensor are written $g_{ij}$ because the metric $\mathbf g$ is a $(0,2)$-tensor which eats two vectors (and no covectors) and returns their inner product.

If you see the indices written upstairs, then you're not talking about the components of the metric - you're talking about the components of the dual metric $\tilde{\mathbf g}$, which defines an inner product on the space of covectors. The components of the dual metric are the matrix inverse of the components of the metric, which means $$\tilde g^{ij} g_{jk} = \delta^i_k$$ Conventionally, we drop the tilde and write the components of this object simply as $g^{ij}$, distinguishing it from the metric itself only via placement of the indices.


In Euclidean space and cartesian coordinates, the components of the metric $g_{ij}$ written out in array form are

$$g_{ij} = \pmatrix{1 & 0 &0 &0\\0&1&0&0\\0&0&1&0\\0&0&0&1} \equiv \delta_{ij}$$

My first question is simply this: why do we write $\delta_{ij}$ here, when I am used to seeing the Kronecker delta written as $\delta^i_j$, with one raised index?

Those two symbols have subtly different meanings. They are both equal to $1$ when $i=j$ and $0$ otherwise, but the former transform as the components of a $(0,2)$-tensor (which is to be expected, since it is the metric tensor expressed in a particular choice of coordinates) while the latter transform as the components of a $(1,1)$-tensor (or a linear transformation).

In other aspects of my work, involving quantum field theory, the convention is to write the standard Minkowski metric as $\eta^{ij}$, with two raised indices, as opposed to two lowered indices.

As above, that is the Minkowski dual metric, which is a $(2,0)$-tensor. In cartesian coordinates, the components $\eta^{ij}$ are the same as the components of $\eta_{ij}$, but the same is not true e.g. in spherical coordinates, in which case

$$\eta_{ij} = \pmatrix{-1 &0&0&0\\0&1&0&0\\0&0&r^2&0\\0&0&0&r^2\sin^2(\theta)} \qquad \eta^{ij} = \pmatrix{-1 &0&0&0\\0&1&0&0\\0&0&\frac{1}{r^2}&0\\0&0&0&\frac{1}{r^2\sin^2(\theta)}}$$

Why, in this formula, have we suddenly gone from writing lengths such as $dx^i$ with raised indices to ones such as $x_i$ with lowered ones?

If you go back a step, we have

$$\mathrm ds^2 = |\mathrm d\mathbf x|^2 + K \frac{(\mathbf x\cdot \mathrm d\mathbf x)^2}{a^2-K|\mathbf x|^2}$$ $$ = \delta_{ij} \mathrm dx^i \mathrm dx^j + K \frac{(\delta_{ij} x^i \mathrm dx^j)(\delta_{\ell m} x^\ell \mathrm dx^m)}{a^2 - K|\mathbf x|^2}$$

Relabeling dummy indices on the second term, we have

$$\mathrm dx^2 = \left(\delta_{ij} + K \frac{(\delta_{i\ell} x^\ell \mathrm )(\delta_{jm} x^m \mathrm )}{a^2 - K|\mathbf x|^2}\right) \mathrm dx^i \mathrm dx^j$$

If we use the index-lowering convention $x_i \equiv \delta_{i\ell}x^\ell$, then comparing this expression to the general $\mathrm ds^2 = g_{ij}\mathrm dx^i \mathrm dx^j$ yields the components given in the text. As a side note, it is my personal belief that using the index-lowering convention on the coordinates is not a good idea. It is, however, perfectly well-defined shorthand, so your mileage may vary.

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  • $\begingroup$ Thank you for your answer, that has cleared up most of the issues I had. However, I feel as though I would benefit from a simple worked example to see what the inner product of covectors would look like. Also, why are lengths such as $x^i$ indexed above? Why does multiplying by the metric $\delta_{ij}$ then lower the index (or is this just a form of shorthand)? $\endgroup$
    – wrb98
    Apr 8, 2021 at 22:44
  • $\begingroup$ @wrb98 Operationally, the inner product of two covectors looks the same as the inner product of two vectors, i.e. $\tilde{\mathbf g}(\mathbf p, \mathbf q) = \tilde g^{ij} p_i q_j$. The $x^i$'s are not lengths, they are coordinates, and we write them with indices up by convention. $\endgroup$
    – J. Murray
    Apr 8, 2021 at 22:53
  • $\begingroup$ @wrb98 To each vector $\mathbf X$ with components $X^i$ there corresponds exactly one covector "partner" $\tilde{\mathbf X}$ with components $\tilde X_i = g_{ij}X^j\iff X^i = \tilde g^{ij} \tilde X_j$. In that sense, the metric defines a one-to-one pairing between the space of vectors and the space of covectors. By convention, just as with the dual metric, we simply write $\tilde X_i$ as $X_i$ and distinguish it from its vector partner by the placement of the index. $\endgroup$
    – J. Murray
    Apr 8, 2021 at 22:57
  • $\begingroup$ @wrb98 The coordinates are not the components of a vector, so if you must use this "index-lowering convention* on them, you should understand it purely as a form of notational shorthand. As I mentioned in my answer, I view this as bad practice, but my opinion is not shared by everyone. $\endgroup$
    – J. Murray
    Apr 8, 2021 at 22:57
  • $\begingroup$ Thank you for your comments, they are very helpful. I was reading more on vectors and covectors and came upon this question: math.stackexchange.com/questions/253829/…. In the top answer, 𝑢 in the dual basis appears to be a column vector (the same as 𝑣) when on page 14 of people.math.osu.edu/gerlach.1/math5101/DualOfAVectorSpace.pdf, for example, the basis is expressed as a collection of row vectors. Should it not be a row vector? $\endgroup$
    – wrb98
    Apr 9, 2021 at 11:49

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