1

I'm learning about manifolds and Lie groups, and have come across the following definition of a topological group:

A topological group or continuous group consists of

  1. An underlying $\eta$-dimensional manifold $\mathscr{M}$.
  2. An operation $\phi$ mapping each pair of points ($\beta, \alpha$) in the manifold into another point $\gamma$ in the manifold.
  3. In terms of the coordinate systems around the points $\gamma, \beta, \alpha$, we write

$$\gamma^\mu = \phi^\mu(\beta^1, ..., \beta^\eta; \alpha^1, ..., \alpha^\eta); \mu=1, 2, ..., \eta$$

("Lie groups, Lie algebras, and some of their applications," Robert Gilmore, 1974, p.63)

I don't understand what's going on in the third definition. He says "in terms of the coordinate systems around the points," but doesn't specify which charts are to be used. $\beta$ and $\alpha$ will have different coordinates under different charts. And we can't even assume they're being mapped by the same chart.

The video series I'm following glosses over this by pretending that the space is globally Euclidean (or at least that it can be given uniform global coordinates -- I'm not sure if these are the same thing).

How should I think about this? Why does it make sense to specify a function of 2$\eta$ real variables without indexing it by the relevant charts?

Edit: Since my question apparently wasn't clear, let me try rephrasing it. It appears as though he's claiming that there exists a function $\phi^\mu$ that can take the coordinates of any two points whatsoever, using any (applicable) coordinate charts whatsoever and produce a result. This clearly can't be what he's saying.

A_P
  • 1,007

2 Answers2

2

Since $M$ is a manifold, there is an open set $U\subseteq M$ such that $\alpha\in U$ and a smooth chart (i.e. a function) $\varphi:U\to \mathbb R^{\eta}$. So, $\alpha$ may be identified with its image $\varphi (\alpha)$ in $\mathbb R^{\eta}.$ That is, with the tuple $(\varphi^1(\alpha),\cdots ,\varphi^{\eta}(\alpha))$. These are the "coordinates" $\alpha^i$ of $\alpha$ in $M$. Similarly, there is an open set $V\subseteq M$ such that $\beta\in V$ and a smooth chart $\psi:V\to \mathbb R^{\eta}.\ \beta$ is then identified with the tuple $(\varphi^1(\beta),\cdots ,\varphi^{\eta}(\beta))$, which are the coordinates $\beta^i.$

$\phi$ is then a map that sends these coordinates to the coordinates, defined in the same way as for those of $\alpha$ and $\beta$, of $\gamma.$ The way to see this is to note that in order to analyze an abstract manifold, one develops the machinery that allows us to "calculate locally" in the easy to understand Euclidean space $\mathbb R^{\eta}$ and then transfer the results back to $M$ using the charts.

Matematleta
  • 29,139
  • Thanks. What I'm not understanding, though, is this: suppose I have $\alpha_1$ and (a different) $\alpha_2$ that both happen to be mapped, by different charts, to the same coordinates. $\phi$, seeing only these coordinates, is forced to map them the same way. Is this expected? – A_P Jul 23 '19 at 22:09
  • $\phi$ is just the group operation in coordinates. That is, $m:M\times M$ is the smooth group operation on $M: \alpha\times \beta\mapsto \alpha\cdot \beta=\gamma.$ So, $m(\alpha\times \beta)=\gamma.$ Now, in coordinates (on an open $W$ in $M$), $\gamma$ is represented by the tuple $(\gamma^{\mu})$ and $\phi$ is the map $\varphi (\alpha)\times \psi(\beta)\mapsto \alpha\times \beta\mapsto m(\alpha\times \beta)=\gamma \mapsto (\gamma^{\mu}.)$ It is smooth, being a composition of smooth maps. – Matematleta Jul 23 '19 at 22:34
  • I apologize, I'm not understanding the notation. In particular, I don't understand what the operation $\times$ is on either points ($\alpha \times \beta$) or coordinates ($\phi(\alpha) \times \psi(\beta))$.

    We could have charts that map $\alpha$, $\beta$, $\eta$, etc. all to (0, 0, ...). Then the product of any two of these is forced to always have the same coordinates, given by $\phi(0, 0, ...; 0, 0, ...)$, because $\phi$ was never defined to depend on $\alpha$ or $\beta$. I suppose that's probably where my error is: $\phi$ must obviously depend on $\alpha$ and $\beta$... right?

     – A_P Jul 24 '19 at 03:47
  • Since $M$ is a group, there is a map $m: G\times G\to G:(p,q)\mapsto pq.$ (multiplication). Now, $\varphi (\alpha)\times \psi(\beta)\in \mathbb R^{\eta}\times \mathbb R^{\eta}$ and the map I wrote is the composition $\gamma\circ m\circ (\varphi^{-1}\times \psi^{-1}).$ – Matematleta Jul 24 '19 at 20:38
  • Okay, so $\alpha \times \beta$ appears to be another way to write $(\alpha, \beta)$? My key confusion remains the same though: $\phi$ is specific to a choice of $\alpha$ and $\beta$ (and their charts) right? I.e., if we pick different points and charts, we get a different $\phi$? If not, I'm still totally lost. – A_P Jul 24 '19 at 20:44
  • Yes, correct. $\phi$ is defined on $\varphi (U)\times \psi (V)$ for suitably chosen opens $U\ni \alpha$ and $V\ni \beta$. For different charts, you get (possibly) different domains. – Matematleta Jul 24 '19 at 22:22
  • Thanks! Sorry to make you write so much. That was my only question. So, is this all obvious from the notation alone, or only from the fact that it wouldn't otherwise make sense? – A_P Jul 24 '19 at 22:43
1

Why does it make sense to specify a function of 2 real variables without indexing it by the relevant charts?

My question was phrased poorly, but the answer I was looking for was that $\phi$ is "indexed by" (i.e., specific to) the particular elements. There's not one global $\phi$ as I thought the text suggested.

A_P
  • 1,007
  • Right. Happy to upvote this. Notice in our comments, I wrote that $\phi$ is defined locally: that is, on $\varphi (U)\times \psi (V).$ – Matematleta Jul 25 '19 at 00:08
  • Thanks. Somehow I didn't understand that until your last comment. Since () $\in \mathbb{R}^\eta$, when you wrote " is the map ()×() ↦ ..." I kept thinking that $\phi$ was meant to be defined on the whole of the (unique) $\mathbb{R}^\eta \times \mathbb{R}^\eta$.

    Don't know if you saw my last comment on your answer. What in the text gives away the fact that $\phi$ is local, other than the fact that it makes no sense otherwise?

     – A_P Jul 25 '19 at 00:33
  • "In terms of the coordinate systems around the points γ,β,α" is the giveaway. It is usually not possible to cover a manifold with one (a global) chart. Try to do it for $S^1$ and see what happens. – Matematleta Jul 25 '19 at 00:52
  • Yes, it's clear to me that this isn't generally possible; hence my confusion. For some reason the "in terms of the coordinate systems around the points" didn't suggest to me anything about $\phi$ itself being local, but only that the points were being expressed in local coordinates. I thought there may be some other trickery going on that I didn't understand. – A_P Jul 25 '19 at 02:07