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In Road to Reality, section 14.6 on Lie derivative Penrose writes:

Now $\epsilon^2 [j,h]$ corresponds to an $O(\epsilon^2)$ gap in the ‘parallelogram’ whose initial sides are $e_j$ and $e_h$ at the origin I. The relevant notion of ‘parallelism’ comes from the group action, supplying the needed notion of ‘parallel transport’, which actually gives a connection with torsion but no curvature.[14.17]

and poses the the exercise:

[14.17] Try to explain why there is torsion but no curvature.

I find it surprising that one could get a connection just from the Lie bracket, since connections depend on additional structure like the metric tensor, while Lie brackets do not.

One forum has a proposed answer giving the Lie bracket itself as the connection $\nabla_{L}M = [L, M]$ if I understood it correctly. The proof proposed there convincingly shows that there would be torsion & no curvature, but the proposed answer does not seem to be a connection in the first place since multiplying the the vector fields $L$ and $M$ by a scalar field $\phi$ does not satisfy a condition required given earlier in the book:

  • linearity in $L$ : $\nabla_{\phi L}M = \phi\nabla_{L}M$

Treating veactors as directional derivative operators on scalar field shows how linearity in $L$ fails

$$\begin{align} [\phi L, M](\psi) &= \phi L(M(\psi)) - M(\phi L(\psi)) \\ &= \phi L(M(\psi)) - (M(\phi)L(\psi) + \phi M(L(\psi))) \\ &= \phi[L, M](\psi) - M(\phi)L(\psi) \end{align}$$

because of the additional $M(\phi)L(\psi)$ term.

Also if it is possible to define a connection just from the Lie bracket, why is it usually said that you need additional structure to define parallel transport? Is it just that the all torsion no curvature property stops this connection from being useful, so people disregard it? So what is Penrose getting at? Am I missing something?

EDIT

From the answers it seems that the Lie bracket really is a connection, just not a linear or affine one. The book does not formally define what a connection is in general or specify the type of connection is intended when the term is used without qualification. It only gives the derivative style algebraic laws connections a connection must satisfy, including linearity wrt $L$ above. So the Lie bracket does seem to be not a connection in the sense the term used in the rest of the book. That is what I was trying to figure out

I am still curious why it is so frequently claimed that connections, parallel transport and covariant derivatives (which I in my understanding are equivalent concepts) require extra structure on a manifold while Lie derivatives do not, if the Lie bracket is a connection and this connection is hardly ever mentioned.

Also is there a name for the Lie derivative as a connection? Lie connection??

Daniel Mahler
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    Presumably Penrose allows for connections to be arbitrary Ehresmann connections, which do not need to be linear. You appear to be thinking of linear/affine connections. If that was the case, then yes you are correct, the Lie bracket is not such a connection by your argument. – Ryan Budney Sep 20 '13 at 06:56
  • Most of my understanding of differential geometry comes from R2R. The only specific types of connections mentioned in that capter are torsion free & the Levi-Civita. Other wise it just says connection, but linearity linearity & Leibniz law are stated for those. What do people usually mean if they just say "connection". – Daniel Mahler Sep 20 '13 at 07:06
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    A Levi-Cevita connection is a linear/affine connection that is compatible with the metric. One way of stating the metric compatibility is that if two vector fields are parallel along a curve, their inner product does not change (along the curve). You have not stated this property in your definition above. What you've stated above is what people would call just a linear/affine connection. An Ehresmann connection is even weaker -- it depends on more than just the infinitesimal change in the vector fields at that point. Feel free to look up the definition. My webpage has course notes on it. – Ryan Budney Sep 20 '13 at 08:06
  • I meant that when the book states the conditions I gave in the question above as general properties of connections, without qualifying the kind connection, and then talks about torsion free and Levi-Civita connections as special cases with the additional properties you mentioned. Thanks for the pointer to your notes. – Daniel Mahler Sep 20 '13 at 22:14
  • It's not clear to me his comment that the "Lie bracket is a connection" is of much value, so if you can't find a reason to see value in it you might prefer to just move on. You have already seen that it's not a traditional connection, so perhaps you'd enjoy learning about other kinds of connections and how they're somewhat similar to linear connections, but also how they can fail to be similar. – Ryan Budney Sep 20 '13 at 23:50
  • Thanks. That is my plan. I just wanted to make sure I was not missing anything important and that my initial concern was reasonable and I think I got that now. – Daniel Mahler Sep 21 '13 at 02:30
  • @Daniel Mahler My understanding is that a connection is an extra structure on a manifold. It does not require anything extra in general. You don't need a metric to define a connection, and even if a metric is available there are plenty of linear connections incompatible with it. On the other hand, both parallel transport and covariant derivative require a connection. They just cannot be computed if a connection is not given, but a Lie derivative can be computed without a connection. The fact that it can be used to define some non-linear connection is beside the point. – Andrey Sokolov Sep 21 '13 at 06:22
  • @RyanBudney I know it has been a while, but can you point me to your lecture notes that you referred to. Thanks. – Daniel Mahler Jun 11 '15 at 06:44
  • This is the stable webpage: http://rybu.org/DGNotes If you click on the March 25th, 2015 edition of the notes, it's around page 74, near the start of Chapter 3. Basic properties of connections continue until around page 92. – Ryan Budney Jun 11 '15 at 15:03
  • @RyanBudney I found those notes useful. Thank you. – ziggurism Dec 02 '17 at 19:17

2 Answers2

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Te connection is indeed $\nabla_LM=[L,M]$ if $L$ is left-invariant; notice that if you have a vector at some point of your Lie group, it can be uniquely extended to a left invariant vector field, so the formula makes sense and defines a connection (btw $\phi L$ is not left-invariant (unless $\phi$ is constant)).

The torsion is (for left-inv. vect. fields) $T(K,L)=[K,L]-[L,K]-[K,L]=[K,L]$, so it's not $0$.

The curvature, on the other hand, is $R(K,L,M)=[K,[L,M]]+\dots=0$ by Jacobi identity. Another way to see that $R=0$ is that any right-invariant vect. field $M$ satisfies $\nabla M=0$, as left and right-inv. fields commute.

user8268
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  • The argument about torsion & curvature is clear to me, but I am not familiar with the concept of invariance of vector fields. I tought that to be a connection, $\nabla$ had to satisfy linearity & Leibniz for all vector fields. – Daniel Mahler Sep 20 '13 at 07:12
  • A vector field $V$ is left-inavriant if $D_gL_h V_{g} = V_{hg}$ where $L_h$ is the map given by $g \mapsto hg$. – Alexander Thumm Sep 20 '13 at 07:19
  • Daniel, it appears that both user8268 and Alexander Thumm are assuming you are interested primarily in Lie Groups, as that's the language they are using in their answers. – Ryan Budney Sep 20 '13 at 21:53
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The parallel transport Penrose is talking about is given by translation in the group. More precisely, the parallel transport $\Gamma(\gamma)^t_s\colon T_{\gamma(s)}G \to T_{\gamma(t)}G$ alongside a curve $\gamma$ is simply the differential of the left translation by $\gamma(t)\cdot \gamma(s)^{-1}$.

We can retrieve the connection from this as follows: Let $\gamma$ be a smooth curve, $X = \dot\gamma(0)$, and $V$ a vector field over $\gamma$ then $$\nabla_X V = \frac{\mathrm d}{\mathrm d t} \Gamma(\gamma)_t^0 V_{\gamma(t)} \bigg\vert_{t=0}.$$

It is now easy to see that $\nabla_XV = 0$ for a left-invariant vector field $V$ and $\nabla_XW = [X,W]$ for a right-invariant vector field $W$. Moreover, as this parallel transport is independent of the curve connecting two points of $G$, the curvature of its associated connection must vanish.

Of course we can define parallel transport by right translation as well and obtain a second connection on $TG$. These two connections agree if $G$ is commutative.

Alexander Thumm
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