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The vector $\textbf{a} = a_{i}\textbf{e}^{i}$ in terms of covariant components. In terms of contravariant components, $\textbf{a} = a^{i}\textbf{e}_{i} = a^{j}\textbf{e}_j$. Thus, $a_{i}\textbf{e}^{i} = a^{j}\textbf{e}_j$. Multiplying both sides by the metric tensor, we have

$$g_{ij} a_i \textbf{e}^i = g_{ij} a^j \textbf{e}_j = a_i \textbf{e}_j $$

and

$$g_{ij} \textbf{e}^i = \textbf{e}_j\,.$$

However, the textbook I'm reading (Riley, Hobson and Bence-Mathematical Methods for Physics and Engineering, P959) lists the relation as

$$g_{ij} \textbf{e}^j = \textbf{e}_i \, .$$

Where have I gone wrong? How do I get the correct relationship?

The book defines $\textbf{a}$ in two different ways

$$\textbf{e}_i = \frac{\partial \textbf{r}}{\partial u^i} \quad \text{and}\quad \textbf{e}^i = \nabla u^i \, .$$ From the first equality we see that we may consider a superscript that appears in the denominator of a partial derivative as a subscript.

Given the two bases $\textbf{e}_i$ and $\textbf{e}^i$, we may write a general vector $\textbf{a}$ equally well in terms of either basis as follows: \begin{align} a &= a^1 \textbf{e}_1 + a^2 \textbf{e}_2 + a^3 \textbf{e}_3 = a^i \textbf{e}_i \\ a &= a_1 \textbf{e}^1 + a_2 \textbf{e}^2 + a_3 \textbf{e}^3 = a_i \textbf{e}^i \, . \end{align} The $a^i$ are called the contravariant components of the vector $\textbf{a}$ and the $a_i$ are called the covariant components, the position of the index (either as a subscript or or superscript) serving to distinguish between them. Similarly, we may call the $\textbf{e}_i$ the covariant basis and vectors and the $\textbf{e}^i$ the contravariant ones.

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    $\begingroup$ You went wrong when you wrote expressions with three repeated indices. $\endgroup$ – Diracology Jul 28 '17 at 17:36
  • $\begingroup$ Okay, but I can write $a_{i} = g_{ij}a^{j}$and hence $g_{ij} a^{j} \textbf{e}^{i} = a^{j} \textbf{e}_{j}$, which leads me to $g_{ij}\textbf{e}^{i} = \textbf{e}_j$. None of these expressions have 3 repeated indices. $\endgroup$ – saad Jul 28 '17 at 17:42
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    $\begingroup$ The metric is symmetric (by definition) so $g_{ij}=g_{ji}$. $\endgroup$ – Diracology Jul 28 '17 at 17:48
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    $\begingroup$ Okay got it. This follows from the commutativity of the dot product $\textbf{e}_{i} .\textbf{e}^{j}$. $\endgroup$ – saad Jul 28 '17 at 17:52
  • $\begingroup$ Notice that there is no such thing as covariant components for a vector. Vector components are by definition contravariant: components for one-forms are covariant instead. $\endgroup$ – gented Jul 28 '17 at 20:35
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The indices on both $a^i\,\!\mathbf{e}_i$ and $a_i\,\!\mathbf{e}^i$ are "saturated", meaning that each letter used for an index appears once as a superscript and once as a subscript. This means that you can't "multiply" - or more accurately contract - the indices with the metric tensor.

Generally, if an index appears twice as a superscript, there's something wrong. Similarly if it appears twice as a subscript, there's something wrong. (This is true whenever we're using the Einstein summation convention and the metric has indefinite signature, as we're taking to be the case here - i.e. both time and space dimensions.) Hence your expression $g_{ij} a_i \mathbf{e}^i$ is wrong, as the index $i$ appears twice as a subscript. Similarly, your expression $g_{ij} a^j \mathbf{e}_j$ is wrong, as the index $j$ appears three times, including twice as a subscript.

As you've spotted, what you can do is write $a_i$ as a contraction of $a^i$ with the metric: $a_i = g_{ij} a^j$. The metric therefore relates the components of a (contravariant) vector with the components of the corresponding covariant vector or co-vector. (This is also known as a one-form, as Gennaro Tedesco mentions. The two quantities are said to be "dual" to each other.)

As Diracology has pointed out, your final equation $g_{ij} \mathbf{e}^i = \mathbf{e}_j$ is actually equivalent to the one you're trying to derive. You can see this equivalence as follows. First, for each $i$ and $j$, you can change the letter, as long as you do it consistently. So you can swap over the $i$s and $j$s: $g_{ji} \mathbf{e}^j = \mathbf{e}_i$. Second, the metric is symmetric, i.e. $g_{ji} = g_{ij}$. This gives you $g_{ij} \mathbf{e}^j = \mathbf{e}_i$.

Hope that's some help - hopefully at least some useful tips in manipulating tensors.

To be pedantic, as I understand it, $a^i \mathbf{e}_i$ and $a_i \mathbf{e}^i$ are not actually the same. The first is an element of the tangent space to the manifold, while the second is an element of its dual space. In other language, the first is a vector with components $a^i$, the second is a co-vector with components $a_i$. However, with a lot of the theory on subjects such as this, there are different ways of looking at things, which turn out to be equivalent within a wide range of validity. It looks from the excerpt from the book as though it employs a consistent way of handling the quantities (thanks for adding the excerpt), so I wouldn't worry too much about this matter. (I'm a little surprised about its approach, as most physics texts stick just to the components and books only tend to introduce basis vectors if they're considering the vector spaces the tensors live in, but not something to get hung up about for the problem you've asked about.)

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  • $\begingroup$ Thank you, I did it by setting $a_i = g^{ij} a^j$, something we proved by taking the dot product $\textbf{a}.\textbf{b}$ in four different ways. And then I switched the indices in the last step. $\endgroup$ – saad Jul 29 '17 at 3:52
  • $\begingroup$ I've added an image because I didn't understand what you meant by $a_{i}\textbf{e}^i$ and $a^{i}\textbf{e}_i$ not being the same. $\endgroup$ – saad Jul 29 '17 at 4:36
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$\def\be{{\bf e}} \def\br{{\bf r}} \let\pd=\partial$ T.R Lawrence wrote:

To be pedantic, as I understand it, $a^i\be_i$ and $a_i\be^i$ are not actually the same.

You're not pedantic, you're absolutely right. I find that book very sloppy on this matter. Look at the consequences of $$g_{ij}\,\be^i = \be_j.$$

If we take $g_{ij}=\delta_{ij}$ we find $$\be_i = \be^i$$ or $${\pd\br \over \pd u^i} = \nabla u^i$$ We don't know how $\br$ and $\nabla$ were defined (I guess $u^i$'s are the coordinates), but I'm rather afraid... What does it mean to derive $\br$ wrt a coordinate?

Moreover, I don't like the choice of using the same symbols for bases in vector and co-vector spaces.

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