Question about manifolds and coordinate transformations

Question

I know about the definition of a differentiable manifold and that the transition functions:

$$\psi_{a} \circ \psi_{b}^{-1}$$

$$\psi_{b} \circ \psi_{a}^{-1}$$

are the way to construct the notion of coordinate transformation (change of charts).

But even after reading Wald's , Sean Carroll's and Nightingale's books unfortunately I didn't grasp why we perform coordinates transformations like:

$$\displaystyle V'^{a} = \frac{\partial x'^{a}}{\partial x^{b}}V^{b}$$

I mean, I didn't "connect" the abstract notion of coordinate transformation by the transition functions with the notion of coordinate transformation by partial derivatives. Furthermore, I know that the functions $\psi$ are differentiable, but why, a priori, we want to differentiate then?

Answer 1

Assume that $\varphi:U\rightarrow \mathbb{R}^n$ is a chart function. If $p\in M$ is a point, then we write $$ \varphi(p)=(x^1(p),...,x^n(p)), $$ so $\varphi$ as a local $\mathbb{R}^n$-valued function is equal to $n$ local $\mathbb{R}$-valued functions, which are the coordinate functions of the chart.

Consequently (and because $\varphi$ is invertible), the inverse function is given by $\varphi^{-1}:\mathbb{R}^n\rightarrow M$ (I am abusing notation, because usually it doesn't map from all of $\mathbb{R}^n$, interpret this as a partial function). It's value on a given $n$-tuple is described as $$ \varphi^{-1}(x^1,...,x^n). $$ Here I am once again abusing notation, because the $x^\mu$s are now variables in $\mathbb{R}^n$. I am not sure what is the source and topic of your confusion exactly, but I assume it has something to do with this. We use $x^\mu$ both as a (local) function from $M$ to $\mathbb{R}$, and as a coordinate/variable within $\mathbb{R}^n$.

We can also say that $$ p=\varphi^{-1}(x^1(p),...,x^n(p)) $$ and here we didn't abuse notation.

Now let $\psi:V\rightarrow \mathbb{R}^n$ also be a chart function, and let us assume that $U\cap V\neq\emptyset$. To make notation easier, I'll reduce $U$ and $V$ both so that they'll coincide, and I'll just use $U$ for both coordinate domains.

We can write $$ \psi(p)=(y^1(p),...,y^n(p)) $$, so the coordinate functions of $\psi$ are now denoted with $y$. The inverse statement is $$ p=\psi^{-1}(y^1(p),...,y^n(p)), $$ so we once again abuse notation and think of the inverse function $\psi^{-1}$ as being the function of the variables $y^1,...,y^n$.

With these notations, the transition function $\psi\circ\varphi^{-1}$ is an $\mathbb{R}^n\rightarrow\mathbb{R}^n$ function, whose value on a given element of its domain can be written as $$ (\psi\circ\varphi^{-1})(x^1,...,x^n)=(y^1(\varphi^{-1}(x^1,...,x^n)),...,y^1(\varphi^{-1}(x^1,...,x^n)))=(y^1(x^1,...,x^n),...,y^n(x^1,...,x^n)). $$

Here in the last equation, we committed a heinous abuse of notation, and "forgot about" $\varphi^{-1}$ - we simply viewed the transition function $\psi\circ\varphi^{-1}$ as a functional relationship between the dependent variables $y^\mu$ and the independent varibles $x^\mu$.

This abuse of notation is very common in differential geometry - even among mathematicians. Because even simple things would be more or less untractable, if we used a very pedantic notation.

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About the actual question: The optimal answer depends on how you like to think about tangent vectors. Usually it either involves point-derivations on the ring of smooth functions, eg. maps of the form $ f\mapsto v(f)\in\mathbb{R} $ such that this map is $\mathbb{R}$-linear and satisfies $$ v(fg)=v(f)g(p)+f(p)v(g), $$ or as tangent vectors to curves, in which case there is an equivalence relation at play between smooth curves passing through $p$.

The connection between the two can be given by the following: If $\gamma$ is a smooth curve on $M$, passing through $p$ at $t_0$, and $f$ is a smooth function defined in an open neighborhood containing $p$, then the tangent vector of the curve $\gamma$ at $p$ is given by the derivation (at $p$) described as $$ v(f)=\left.\frac{d}{dt}(f\circ\gamma)\right|_{t=t_0}. $$ Furthermore, it can be shown that all derivations arise this way.

I'll use this as an example, because it is very easy to examine the behaviour of vector components this way.

Because $\varphi^{-1}\circ\varphi=\text{Id}$ the identity function, we can write $$ \left.\frac{d}{dt}(f\circ\gamma)\right|_{t=t_0}=\left.\frac{d}{dt}(f\circ\varphi^{-1}\circ\varphi\circ\gamma)\right|_{t=t_0}. $$

But what's $f\circ\varphi^{-1}$? It is the multivariable function that maps the $x$-coordinates to numbers instead of abstract points $p$. And what is $\varphi\circ\gamma$? It is the $\mathbb{R}^n$-valued curve $(\varphi\circ\gamma)(t)=(x^1(t),...,x^n(t))$ (warning!! Heavy abuse of notation here!) that describes a one-parameter family of $x$-coordinates instead of abstract $p$-points!

In particular, we can use the usual chain-rule of ordinary calculus to evaluate this derivative, and we get $$ \left.\frac{d}{dt}(f\circ\varphi^{-1}\circ\varphi\circ\gamma)\right|_{t=t_0}=\frac{\partial(f\circ\varphi^{-1})}{\partial x^\mu}\frac{d (x^\mu\circ\gamma)}{d t}=\frac{\partial f}{\partial x^\mu}\frac{d x^\mu}{dt}, $$ where 1) all derivatives are evalued where needed, 2) in the last equation we did a massive abuse of notation once again, 3) the summation convention is in effect.

But this is of cource $v(f)$, so we can "decouple" $f$ from this, and write $v$ as $$ v=\frac{d x^\mu}{dt}\frac{\partial}{\partial x^\mu}. $$ Once again, the $t$-derivative is evaluated at the correct place, and we note that rigorously, $\partial/\partial x^\mu$ is not a partial derivative, but a derivation that acts by taking the partial derivative of the function's $x$-coordinate representation (!!!) (so $\partial/\partial x^\mu$ acts on $f$, but the actual partial derivatives act on $f\circ\varphi^{-1}$). Here we can write $$ v=v^\mu\frac{\partial}{\partial x^\mu}, $$ where $v^\mu=dx^\mu/dt|_{t=t_0}$ and we call the $v^\mu$ the components of $v$ in the chart $\varphi$.

We can also check that $$ \frac{\partial}{\partial x^\mu}(x^\nu)=\frac{\partial (x^\nu\circ\varphi^{-1})}{\partial x^\mu}=\delta^\nu_\mu, $$ so we have $$ v^\nu=v(x^\nu). $$

We can then ask what are the components of $v$ with respect to the coordinates $y$? We evaluate $$ v(y^\nu)=v^\mu\frac{\partial}{\partial x^\mu}(y^\nu)=v^\mu\frac{\partial(y^\nu\circ\varphi^{-1})}{\partial x^\mu}=v^\mu\frac{\partial y^\nu}{\partial x^\mu}, $$ where the last equation is - essentially an abuse of notation.

Answer 2

(This answer assumes you know the differential geometry and just want to know how the physicist gets that expression.)

Let $V,W\subset\mathbb{R}^n$ and $\psi : V\to U$ and $\phi : W\to U'$ be charts for $U,U'\subset M$ on some manifold $M$. Then the "change of coordinates" on $U\cap U'$ is the transition function $$ \phi^{-1}\circ \psi : V\to W.$$ This function induces a natural map between tangent vectors, the pushforward $$ \mathrm{d}(\phi^{-1}\circ \psi) : TV\to TW,$$ which is the Jacobian of the transformation, i.e. at every point $x\in V$, we have $$ \mathrm{d}(\phi^{-1}\circ \psi)_x : T_xV\to T_xW, v\mapsto J(\phi^{-1}\circ\psi)(x)\cdot v.$$ Written in the standard coordinates of the $\mathbb{R}^n$ both $V$ and $W$ are subsets of, the Jacobian is precisely the matrix with components $\frac{\partial x'^a}{\partial x^b}$ you ask about, where $x' = \phi^{-1}\circ \psi$.

We want the functions $\psi$ to be differentiable precisely because we want them to yield this map between tangent vectors. If the maps are not differentiable, there is no natural map induced on the tangent vectors.

Answer 3

Recall that if $X$ is a vector space with basis $(e_1,\ldots,e_n)$ and corresponding dual basis $(e^1,\ldots,e^n)$, then $w = e^i(w)e_i$, for all $w\in X$.

We apply this for every tangent space of the manifold. The basis is $(\partial/\partial x^1,\ldots,\partial/\partial x^n)$ and the dual is $(dx^1,\ldots,dx^n)$. Meaning that $V^a=dx^a(V)$. Similarly, $V^{'a}=dx^{'a}(V) $. Now, the chain rule says that $$dx^{'a} =\frac{\partial x^{'a}}{\partial x^b} dx^b , $$whence applying it all in $V$ gives$$V^{'a} =\frac{\partial x^{'a}}{\partial x^b} V^b, $$as wanted.