# Coordinate charts and change of basis

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1. Jun 3, 2015

### "Don't panic!"

So I know that this involves using the chain rule, but is the following attempt at a proof correct.

Let $M$ be an $n$-dimensional manifold and let $(U,\phi)$ and $(V,\psi)$ be two overlapping coordinate charts (i.e. $U\cap V\neq\emptyset$), with $U,V\subset M$, covering a neighbourhood of $p\in M$, such that $p\in U\cap V$. Consider a function $f:M\rightarrow\mathbb{R}$, and let $x=\phi(p)$, $y=\psi(p)$. It follows then that $$\frac{\partial f}{\partial x^{\mu}}(p)=\frac{\partial}{\partial x^{\mu}}\left((f\circ\phi^{-1})(\phi(p))\right)=\frac{\partial}{\partial x^{\mu}}\left[(f\circ\psi^{-1})\left((\psi\circ\phi^{-1})(\phi(p))\right)\right]\\ \qquad \quad\; =\frac{\partial}{\partial y^{\nu}}\left[(f\circ\psi^{-1})\left((\psi\circ\phi^{-1})(\phi(p))\right)\right]\frac{\partial}{\partial x^{\mu}}\left[\left((\psi\circ\phi^{-1})^{\nu}(\phi(p))\right)\right]\\ =\frac{\partial f}{\partial y^{\nu}}(p)\frac{\partial y^{\nu}}{\partial x^{\mu}}(p)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\,$$ where $y(x)=(\psi\circ\phi^{-1})(\phi(p))$.

Hence, as $f$ is an arbitrary differentiable function, we conclude that $$\frac{\partial }{\partial x^{\mu}}=\frac{\partial y^{\nu}}{\partial x^{\mu}}\frac{\partial }{\partial y^{\nu}}$$ From this, we note that as $\lbrace\frac{\partial }{\partial x^{\mu}}\rbrace$ and $\lbrace\frac{\partial }{\partial y^{\nu}}\rbrace$ are two coordinate bases for the tangent space $T_{p}M$ at the point $p$, the two bases must be related by the formula above.

2. Jun 8, 2015

### Orodruin

Staff Emeritus
Looks fine to me. Note that $(\psi \circ \phi^{-1})(\phi(p)) = \psi(\phi^{-1}(\phi(p))) = \psi(p)$. It is a bit shorter to write ...

3. Jun 8, 2015

### "Don't panic!"

Cheers for taking a look. Yeah, apologies for the explicitness, just wrote it out in full to make sure that I was understanding it correctly and show that the transition functions behave as the "new" coordinates as functions of the "old" coordinates...

4. Jun 8, 2015

### "Don't panic!"

So, is my intuition correct in the following analysis.

In the intersection of two coordinate charts (with coordinates $x$ and $y$ for simplicity) we can equally well describe functions (0-forms) in terms of the $x$-coordinates or the $y$-coordinates. Similarly, the tangent space to a point $p$ in the intersection can equally be described in terms of the coordinate basis $\frac{\partial}{\partial x^{\mu}}$ induced by the $x$ coordinate chart, or in terms of the coordinate basis $\frac{\partial}{\partial y^{\nu}}$ induced by the $y$ coordinate chart. As such, we have a situation in which a given function $f$ could be described in terms of the $x$ or the $y$ coordinate chart at a given point in the intersection, thus we must express one as a function of the other in order for the function to be coordinate independent. Similarly, either of the coordinate bases $\frac{\partial}{\partial x^{\mu}}$ and $\frac{\partial}{\partial y^{\nu}}$ could be used to describe a tangent vector $X$ in the tangent space at that point. Again this tangent vector must be coordinate independent, and so, in this intersection we must have that $$X[f]=X^{\mu}\frac{\partial f(y)}{\partial x^{\mu}}=X^{\mu}\frac{\partial f(y(x))}{\partial x^{\mu}}=X^{\mu}\frac{\partial y^{\nu}(x)}{\partial x^{\mu}}\frac{\partial f(y)}{\partial y^{\nu}}=\tilde{X}^{\nu}\frac{\partial f(y)}{\partial y^{\nu}}$$ where $X^{\mu}$ and $\tilde{X}^{\mu}$ are the components of $X$ with respect to the two bases $\frac{\partial}{\partial x^{\mu}}$ and $\frac{\partial}{\partial y^{\nu}}$ respectively. Hence, this implies that the components of the vector transform as $$\tilde{X}^{\nu}=\frac{\partial y^{\nu}(x)}{\partial x^{\mu}}X^{\nu}.$$ Would this be a correct understanding at all?

Last edited: Jun 8, 2015
5. Jun 8, 2015

### Orodruin

Staff Emeritus
Just one detail, the $X^\mu$ are the components of the vector (in the appropriate basis), not coordinates.

6. Jun 8, 2015

### "Don't panic!"

Sorry, that's what I'd meant (typed coordinates by mistake). Have corrected the post now.

so otherwise, have I understood the notion correctly?

7. Jun 8, 2015

### Fredrik

Staff Emeritus
The notation in post #1 is ambiguous and very confusing.

Here $\frac{\partial}{\partial x^\mu}$ apparently denotes the $\mu$th partial derivative with respect to the coordinate system $\phi$. Why a notation that involves $x$ instead of one that involves $\phi$? I don't think the fact that you chose to denote $\phi(p)$ by $x$ automatically makes this notation appropriate. Consider e.g. the situation where $\phi(p)=\psi(p)$ (for the specific p that we're considering). Then your notation suggests that $\frac{\partial}{\partial x^\mu}=\frac{\partial}{\partial y^\mu}$.

I can't tell what the intention is here. Are you using the definition of the previous expression, or are you just inserting $\phi^{-1}\circ\phi$ in the middle? If it's the latter, then this step doesn't take you closer to where you want to go. If it's the former, then the notation doesn't make sense. The notational issues only get bigger as get further into the calculation.

I would use a notation like this:

\begin{align*}
&\frac{\partial}{\partial\phi^\mu}\bigg|_p f =(f\circ\phi^{-1})_{,\mu}(\phi(p)) = (f\circ\psi^{-1}\circ\psi\circ \phi^{-1})_{,\mu}(\phi(p)) =(f\circ\psi^{-1})_{,\nu}\big((\psi\circ\phi^{-1})(\phi(p))\big) (\psi\circ\phi^{-1})^\nu{}_{,\mu}(\phi(p))\\
&=(f\circ\psi^{-1})_{,\nu}(\psi(p)) (\psi^\nu\circ\phi^{-1})_{,\mu}(\phi(p)) = \bigg(\frac{\partial}{\partial\phi^\mu}\bigg|_p \psi^\nu\bigg) \frac{\partial}{\partial\psi^\nu}\bigg|_p f.
\end{align*}
Edit: Actually, as you have seen in some of my posts before, I like to call the coordinate systems x and y instead of $\phi$ and $\psi$, and I like to use Latin indices instead of Greek, to save myself some typing. So I'd write this result as
$$\frac{\partial}{\partial y^i}\bigg|_p f =\bigg(\frac{\partial}{\partial y^i}\bigg|_p x^j\bigg) \frac{\partial}{\partial x^j}\bigg|_p f.$$

Last edited: Jun 9, 2015
8. Jun 8, 2015

### "Don't panic!"

Ok, thanks Fredrik, sorry for the notational issues.

Would the intuitive description behind why this is so be correct at all? (Quoted below from one of my previous posts)

9. Jun 9, 2015

### Fredrik

Staff Emeritus
The sentence that involves the words "one as a function of the other" is a bit odd, and the notation y(x) is too. But you found the correct formula for how the components of a vector transforms under the change of coordinates $x\to y$.

Last edited: Jun 9, 2015
10. Jun 9, 2015

### "Don't panic!"

So is the point that in the coordinate chart overlap one represent a function in terms of either set of coordinates and then transition between these two descriptions via transition functions. Then if we use the coordinate basis $\lbrace\frac{\partial}{\partial x^{\mu}}\rbrace$ induced by one coordinate system $\phi$ to act on a function $f\circ\psi^{-1}$ that is represented in the other coordinate system $\psi$, then we must use transition functions $\psi\circ\phi^{-1}$ so that we can relate this coordinate basis to the coordinate basis $\lbrace\frac{\partial}{\partial y^{\nu}}\rbrace$ induced by the coordinate system $\psi$ that the function is represented in?

11. Jun 9, 2015

### Fredrik

Staff Emeritus
$\frac{\partial}{\partial x^\mu}\big|_p$ acts on $f$, not on $f\circ\psi^{-1}$.

12. Jun 9, 2015

### "Don't panic!"

I thought one needed to introduce a coordinate chart before doing calculus though? That is, don't we need to have $f\circ\psi^{-1}:\mathbb{R}^{n}\rightarrow\mathbb{R}$?

13. Jun 9, 2015

### Fredrik

Staff Emeritus
That's correct. That's why we define $\frac{\partial}{\partial x^\mu}\big|_p$ by
$$\frac{\partial}{\partial x^\mu}\bigg|_p f=(f\circ\phi^{-1})_{,\mu}(\phi(p))$$ for all smooth $f:M\to\mathbb R$.

Since the $\frac{\partial}{\partial x^\mu}\big|_p$ defined this way is the partial derivative functional associated with the point p and the coordinate system $\phi$, I would prefer to denote it by $\frac{\partial}{\partial\phi^\mu}\big|_p$. Another option is to denote the cordinate system by $x$ instead of $\phi$.

Last edited: Jun 9, 2015
14. Jun 9, 2015

### "Don't panic!"

Ah ok. I think I was really trying to justify why one needs to do the whole procedure in the first place? Is the point that in the coordinate chart overlap we have two different way of representing the same object and so we require a way of relating these to representations (achieved via the appropriate change of basis between the two coordinate bases)?

15. Jun 9, 2015

### Fredrik

Staff Emeritus
I guess you could say that (if the last "to" is a typo, supposed to be "two"), since the n-tuple of components of a vector v with respect to the ordered basis associated with a particular coordinate system can be thought of as a representation of the vector v. Change the coordinate system, and you change the ordered basis, which changes the components.

16. Jun 9, 2015

### "Don't panic!"

Yes sorry it was meant to be "two" and not "to".

Could one also derive the change of coordinate basis by considering to coordinate systems $x$ and $y$ and noting that as $\lbrace\frac{\partial}{\partial x^{i}}\rbrace$ and $\lbrace\frac{\partial}{\partial y^{j}}\rbrace$ are bases for the same tangent space we can express each basis vector $\frac{\partial}{\partial x^{i}}$ as a linear combination of the basis vectors $\lbrace\frac{\partial}{\partial y^{j}}\rbrace$ such that $$\frac{\partial}{\partial x^{i}}=A^{i}_{j}\frac{\partial}{\partial y^{j}}$$ Then if we act on the $j$th coordinate function $y^{j}$ (of the $y$ coordinate system) we find that $$\frac{\partial y^{j}}{\partial x^{i}}=A^{i}_{j}$$ as required. Alternatively, we can equally express each basis vector $\frac{\partial}{\partial y^{j}}$ as a linear combination of the basis vectors $\lbrace\frac{\partial}{\partial x^{i}}\rbrace$ such that $$\frac{\partial}{\partial y^{j}}=\tilde{A}^{i}_{j}\frac{\partial}{\partial x^{i}}$$ Then, acting on the $i$th coordinate function $x^{i}$ (of the $x$ coordinate system) we have that $$\frac{\partial x^{i}}{\partial y^{j}}=\tilde{A}^{i}_{j}.$$ We also note that this implies that $$A^{i}_{j}\tilde{A}^{j}_{k}=\delta^{i}_{k}$$ and so $\tilde{A}^{i}_{j}=(A^{-1})^{i}_{j}$

17. Jun 9, 2015

### Fredrik

Staff Emeritus
Yes, that's accurate, and is proved by using the definition of the $\frac{\partial}{\partial x^\mu}\big|_p$ notation and the chain rule, as discussed above.

18. Jun 9, 2015

### "Don't panic!"

ok great, thanks.