# Justifications for the index upper lower labels in tensor component transformations

A (1,0)-type tensor may be written as $$V = V^{\mu} e_{\mu}$$ The component transforms as $$V^{\nu} = A^{\nu}{}_{\mu^\prime} V^{\mu^\prime}$$ (the basis can transform similarly) My question is, what is the justification that the label of $$A$$ is one upper script and one lower script? In Wald's General relativity p26, the author showed

$$V^{\mu^\prime} = \sum_{\mu=1}^{n} V^{\mu} \frac{ \partial x^{\prime\mu^\prime} }{\partial x^{\mu}} \qquad\qquad(2.3.6)$$

The term $$\frac{ \partial x^{\prime\mu^\prime} }{\partial x^{\mu}}$$ can justify the label should be one upper and one lower. But, the approach in Wald's book is based on the geometry aspect of tensor.

A tensor may be defined in more general content, just an element in vector space products. Vector space is something that satisfies a couple of axioms. By this means, how to justify the term in basis transformation has the structure of upper/lower indices? Should I regard basis transformation as a tensor contraction between (1,1) and (1,0) to (1,0) something like that? Or the index positions of transormation matrix are not always strict?

Where you place the indices of a transformation matrix depends on what kind of object it transforms into what other kind of object. For example, let $$g^{\mu\nu} = g_{\mu\nu}{}^1$$ be the components of the metric tensor $$g=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}~, \tag{1}$$ which has the property that it maps the covariant position vector $$x_\mu = (ct,-\vec x)_\mu$$ onto the contravariant one, $$x^\mu = g^{\mu\nu} x_\nu =(ct,\vec x)^\mu~,$$ and thus can be used to "pull indices up or down". Multiplying the above equation with $$g_{\rho\mu}$$ yields $$x_\rho=g_{\rho\mu} x^\mu = g_{\rho \mu}g^{\mu\nu} x_\nu~,$$ so $$g_{\rho}^{~~~\mu} = g_{\rho\mu} g^{\mu\nu}$$ is the $$\rho$$-$$\nu$$-component of the unit matrix $$g^2$$.
Likewise, your example $$V'^\mu = A^\mu_{~~~\nu} V^\nu$$ can be multiplied with $$g_{\rho \mu}$$ to obtain $$V'_{\rho} = g_{\rho\mu} V'^\mu = g_{\rho\mu} A^\mu_{~~~\nu} V^\nu = A_{\rho\nu} V^\nu~.$$ This means, $$A^\mu_{~~~\nu}$$ maps a contravariant vector to another contravariant vector, while $$A_{\mu\nu}$$ maps a contravariant vector to a covariant vector. Which one you use depends on what vectors you have or need and you can easily convert between the variants by multipliying equations with $$g^{\mu\nu}$$ or $$g_{\mu\nu}$$.
$${}^1$$ Strictly speaking, this is wrong, because free indices on the left- and righthand side of an equation have to match. What I wanted to express is simply, that the co- and contravariant forms of the metric tensor are identical.
As you say, the components of the vector $$V$$ transform as $$\tag{1} V^\mu={{A^\mu}_{\mu'}}V^{\mu'}.$$ Likewise, the basis vectors transform as $$\tag{2} \boldsymbol{e}_\mu={{A_\mu}^{\mu'}}\boldsymbol{e}_{\mu'}\,.$$ where $${A_\mu}^{\mu'}$$ is the inverse of $${{A^\mu}_{\mu'}}\,.$$ That's the reason why it is common to call the transformaton (1) contravariant and (2) covariant. The position of the index (upper or lower) is a convenient way of underpinning this in the notation.