This is probably a very simple question, and I think I know the answer, but I cannot find a place to solidly confirm this.

So if I want to write a vector $\mathbf{V}$ in terms of its contravariant (or ordinary) components, I would write it as: \begin{equation} \mathbf{V} = V^{\mu}\hat{e}_{(\mu)} \end{equation} where $V^{\mu}$ are the components of the vector and $\hat{e}_{(\mu)}$ are the unit vectors and the Einstein convention is implied. Similarly, for the covector components: \begin{equation} \mathbf{V} = V_{\mu}\hat{\theta}^{(\mu)} \end{equation} The general rule is, for covector components, the index is downstairs, and for ordinary vector components, the index is upstairs. In order to make dot products and things like that work out, you would write ordinary vectors as column vectors and covectors as row vectors.

But why do the the basis vectors have the index in the opposite position as the index? Is it merely to let the math work out with this convention of "row and columns"? For example: \begin{equation} \mathbf{V} = \left(\begin{array}{c} V^0 \\ V^1 \\ V^2 \\ V^3 \end{array}\right) \left( \begin{array}{cccc} \hat{e}_{(0)} & \hat{e}_{(1)} & \hat{e}_{(2)} & \hat{e}_{(3)} \end{array} \right) = V^0\hat{e}_{(0)}+V^1\hat{e}_{(1)}+V^2\hat{e}_{(2)}+V^3\hat{e}_{(3)} = \sum_{\mu = 0}^{3} V^{\mu} \hat{e}_{(\mu)} \rightarrow V^{\mu} \hat{e}_{(\mu)} \end{equation}

So it would be incorrect to say that the basis vectors $\hat{e}_{(\mu)}$ are covectors correct? They are just basis vectors, end of story. $\mathbf{V}$ is the object that is either a vector or a covector.

Is this right?If not, where am I going wrong?

EDIT: So this answer I came up with about rows and columns doesn't hold when you want to calculate something like $V^{\mu}V^{\mu}$, which you can certainly do I would think. What gives?

  • $\begingroup$ Related: physics.stackexchange.com/q/144089/2451 , physics.stackexchange.com/q/79013/2451 and links therein. $\endgroup$ – Qmechanic Oct 20 '16 at 9:10
  • $\begingroup$ I think there are probably already good answers to this: @Qmechanic's pointers are likely to be worth following. But one thing to think about is what $v^\mu v^\mu = \sum_\mu v^\mu v^\mu$ is. It should be a scalar, right, as it has no indices and can not be the component of anything? Well, then if you change basis it should be unchanged: is it? OK, so does it in fact make any sense at all, and did the notation give you a clue? $\endgroup$ – tfb Oct 20 '16 at 12:48

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