Transformation rule in differential geometry

Question

I am reading Walds General Relativity and am looking at Question 8, Chapter 2. In the solutions to this question it states that the metric is determined by the transformation rule

$$g_{\alpha\beta}^{'} = \frac{\partial x^{\mu}}{\partial x^{'\alpha}}\frac{\partial x^{\nu}}{\partial x^{'\beta}}g_{\mu\nu}$$

I really cannot see how this identity works. I cannot see how the partial derivatives act on the tensor $g_{\mu\nu}$. Nor can I see why the 's are on the bottom of the partial derivatives.

Next is says that $\frac{\partial x^{\mu}}{\partial x^{'\alpha}}=({J^{-1}})^{\mu}_{\alpha}$. I cannot see why this should be $J^{-1}$ and not $J$?

Then it says that

$$g_{\alpha\beta}^{'} = ({J^{-1}})^{\mu}_{\alpha}g_{\mu\nu}({J^{-1}})^{\nu}_{\beta}$$ I cannot see here why you are allowed to swap round the partial derivatives and the tensor.

Answer 1

This should be read as an addition to the post of Upax.

Some setup

Let $M$ be the spacetime at hand (whose dimension is $n$). It is to be emphasized, that your equation only holds in local coordinates, i.e in a chart. Let $U,U' \subset M$ with $U \cap U' \neq \emptyset$, and let $$x:U \to \mathbb{R}^n, p \mapsto (x^1(p),...,x^n(p))$$ $$x': U' \to \mathbb{R}^n, p \mapsto (x'^1(p),...,x'^n(p))$$ be the chosen charts, i.e, the coordinates at hand. We denote by $x^{\mu}, x'^{\alpha}$ (where $\mu, \alpha \in \{1,...,n\}$)the mappings $$ x^{\mu}:U \to \mathbb{R}, p \mapsto x^{\mu}(p)$$ $$x'^{\alpha}:U' \to \mathbb{R}, p \mapsto x'^{\alpha}(p).$$ In the same manner, $x^{\nu}$ and $x'^{\beta}$ are defined. Observe that the different choice of indices is merely a matter of convention! Locally, these charts determine frame fields, i.e, they determine a basis for each $p \in U \cap U'$ (we restrict ourself to the intersection since your equation is meaningless elsewhere), which are frequently denoted by $$\{\frac{\partial}{\partial x^1}, ..., \frac{\partial}{\partial x^n}\} = \{\frac{\partial}{\partial x^{\mu }}\}_{\mu=1,...,n}$$ $$\{\frac{\partial}{\partial x'^1}, ..., \frac{\partial}{\partial x'^n}\}= \{\frac{\partial}{\partial x'^{\alpha}}\}_{\alpha=1,..,n},$$ (again, the indices are equivalently denoted by $\nu$ and $\beta$ respectively). See also this Wikipedia article for more information on how exactly this basis is defined. As you (hopefully) know, the metric is given by a (family of) mapping(s) $$g_p: T_pM \times T_pM \to \mathbb{R}, (X_p,Y_p) \mapsto g_p(X_p,Y_p)$$ which varies smoothly with $p$, i.e, the mapping $p \mapsto g_p(X_p,Y_p)$ is smooth for any choice of vector fields $X$ and $Y$. See this article for the case of definite signature (which carries over to the indefinite case).

How this (hopefully) helps

Using the preceeding, your equation $$g'_{\alpha \beta} = \frac{\partial x^{\mu}}{\partial x'^{\alpha}} \frac{\partial x^{\nu}}{\partial x'^{\beta}} g_{\mu \nu}$$ reads as $$g\left(\frac{\partial}{\partial x'^{\alpha}}, \frac{\partial}{\partial x'^{\beta}}\right)= \frac{\partial x^{\mu}}{\partial x'^{\alpha}} \frac{\partial x^{\nu}}{\partial x'^{\beta}}g\left({\frac{\partial}{\partial x^{\mu}}, \frac{\partial}{\partial x^{\nu}}}\right).$$ Now, since $\{\frac{\partial}{\partial x^{\mu}}\}_{\mu=1,...,n}$ and $\{\frac{\partial}{\partial x'^{\alpha}}\}_{\alpha=1,...,n}$ are frame fields on $U \cap U'$, there does exist a (smooth) family $A=(A_{\mu \alpha})_{\mu \alpha =1,...,n}$ of matrices (i.e for $p \in U \cap U'$ we have $A(p): T_pM \to T_pM$) satisfying $\frac{\partial}{\partial x^{\mu}} = A_{\mu \alpha} \frac{\partial}{\partial x'^{\alpha}}$. At this point, it's linear algebra (you only have to verify your equation pointwise). To do this, I suggest you recall the definition of the Gramian Matrix and use it's transformation properties (you should try to understand how the Gramian Matrix is related to the ''metric tensor''-observe that you deal with indefinite signature, so not all properties do hold). Finally, using this and the definition of our frame fields, you should be able to verify that the sought matrix $A$ is indeed the (inverse) of the jacobian corresponding to the coordinates. How the change of coordinates affects the frame fields can be found (or, at least, deduced from) in any book on differential geometry (you don't need the definite signature for that), as in do Carmo - Riemannian Geometry. I do, however, encourage you to take some book on pseudo riemannian geometry and read that too.

A Remark

It did not execute the caluclations myself thus I might have confused the indices: It might very well be that at some point the indices should be written as (e.g) ''$x_{\mu}$'' instead of ''$x^{\mu}$''. Thus, if you find that it does not make sense the way I wrote it, don't worry too much.

Answer 2

Let consider the position vector $r$ of a point P relative to the common origin of two orthonormal bases with basis vectors $e_j$ and $f_j$. The coordinates of P in the rectangular systems defined by these bases are denoted by $x_j$ and $\overline{x}_i$. So we can write: \begin{equation} \mathbb{r}=x^i \mathbb{e_i} \end{equation} and \begin{equation} \mathbb{r} = \overline{x}^j \mathbb{\overline{e}_j} \end{equation} the relationship between the two sets $x_j$ and $\overline{x}_i$ is given by: \begin{equation} \mathbb{r} =x^i \overline{a}_{ij} \mathbb{\overline{e}_j} = {a}_{ji} x^i \mathbb{\overline{e}_j} \end{equation} This is because the basis $\mathbb{e_i}$ can be written as the product $\overline{a}_{ij} \mathbb{\overline{e}_j}$ where $\overline{a}_{ih} {a}_{hj}= \delta_{ij}$. By comparing the above relation with the right hand side of the $\mathbb{r} = \overline{x}^j \mathbb{\overline{e}_j}$ we have that $\overline{x}^j=a_{ji} x^i$. The importance of the above relationships is due to the fact that they allow us to specify the transformation of any given function of the coordinates under a change of orthonormal bases. The metric tensor $g_{\mu \nu}$ establishes a natural direct product on the tangent plane $T_p S$ that ultimately leads to formulas for length of vectors in $T_p S$, arc length, angle between vectors in $T_p S$ and area formulas on S. It is the restriction of the usual dot product in $\mathbb{R}^3$ to the tangent plane $T_p S$. Based on what I wrote, you can consider the product of 2 vectors $A$ and $B$. Their components transform under a change of basis as \begin{equation} \overline{A_j} = a_{jh} A_h \end{equation} and \begin{equation} \overline{B_k} = a_{kl} B_l \end{equation} So that \begin{equation} \overline{A_k} \overline{B_k} = a_{jh} a_{kl} A_h B_l \end{equation} A set og quantities $T_{jk}$ is said to constitute the components of an affine tensor of rank 2, if under the orthogonal coordinate transformation $\overline{x}^j=a_{ji} x^i$, these quantities transform according to the transformation law \begin{equation} \overline{T_{jk}} = a_{jh} a_{kl} T_{hl} \end{equation} Since
\begin{equation} a_{jh} =\frac{\partial \overline{x}^j}{\partial x^h} \end{equation} and \begin{equation} a_{kl} =\frac{\partial \overline{x}^k}{\partial x^l} \end{equation} you can conclude that the first fundamental form $g$ is a tensor of rank (0,2).