Manifolds / Lie Groups - confusing notation

[GSpeight]

Hi there,

I'm reading over my Lie groups notes and in them, in the introductory section on manifolds, I've written that F_{\star} is a commonly used notation for d_{x}F and so the chain rule d_{x}{G \circ F}=d_{F(x)}G \circ d_{x}F can be written (G\circ F)_{\star}=G_{\star}\circ F_{\star}

Is what I've written correct? To me this seems horribly confusing since it neglects to mention where you are taking the differential. Should it instead be that F_{\star} is the map from M to d_{x}F. On second thoughts this doesn't make total sense either...

He's gone on to make definitions like:

A vector field X on a Lie group G is called left-invariant if, for all g,h in G, (L_{g})_{\star}X_{h}=X_{gh}=X_{L_{g}(h)} where L_{g} is the left multiplication map by g ,which I'm finding difficult to understand with my current definition of F_{\star}.

Thanks for any replies :)

 

[mrandersdk]

okay so you are asking why we are not writing F_*_x, so I could as well expain the notation

dF

where i have left out the x. So I try to explain that.

dF_x is a map between the tangent space at some point x in a manifold M to the tangent space at some point F(x) in a manifold N, (so F: M \rightarrow N). Now we could as well drop the x and look at dF as a map between the tangent bundles of M and N, that is

dF_x : T_xM \rightarrow T_{F(x)M}

and

dF: TM \rightarrow TN

so now we can think of dF to work on sections of TM (and more common vector fields). Of cause you are right that you have to be carefull about where each X(x) goes.

So dF(X(x)) = dF_x(X_x) \in T_{F(x)}N where X_x=X(x) \in T_xM and dF(X) \in TN

so there is a difference. This is a bit like, if f: R \rightarrow R smooth then

\frac{\partial f}{\partial x} \in C^{\infty}(R) and \frac{\partial f}{\partial x}|_{x=t} \in R. That is the derivative of f is a smooth function again, but if you evauate at t then it is a real number.

It can seem confusing because if we take a simple example f(x) = x^2 then

\frac{\partial f}{\partial x} = 2 x and

[/tex] and \frac{\partial f}{\partial x}|_{x=t} = 2 t

So what is the difference? Remember that t is just a real number, so if you differentiate again with respect to x you see that there is a big difference,

\frac{\partial^2 f}{\partial x^2} = 2 and

[/tex] and \frac{\partial}{\partial x}\frac{\partial f}{\partial x}|_{x=t} = \frac{\partial}{\partial x} 2 t = 0

we often don't make this distinction, but that is because when we are doing a calculation it is clear what is ment.

So i would say that F_* is a notaion for dF and not for dF_x, but often the distiction is not necesary, because it is clear how we could identify dF with dF_x, that is just restrict dF from TM to T_xM, and so it is clear how to identify dF_x with F_*.

You will experience often in differential geometry that in the notation there are a lot of hidden things, that is because it is a very notation heavy subject, so just be carefull and don't worry it will become easier with time.

 

[mrandersdk]

so you see that for your last question:

(L_{g})_{\star}X_{h}

makes more sense because dL_g=(L_{g})_{\star} maps from a vector field to another vectorfield where as d_xL_g only is a map from T_xG to T_L_g(x)G.

 

[GSpeight]

Thank you so much for your response, I think I understand it much better after reading your post!

Am I correct in thinking that, explicitly, (L_{g})_{\star}X_{h}=(d_{h}L_{g})\circ X_{h} since that is the only place we can be taking the differential, since X_{h} \in T_{h}M

This makes the rest of the definitions I have easier to understand too, e.g. if F:M\rightarrow N is a smooth map of manifolds and X, Y are smooth vector fields on M and N respectively then X and Y are F-related if F_{\star}(X_{x})=Y_{F(x)} for all x in M... would just mean d_{x}F \circ X_{x} = Y_{F(x)} for all x in M, since X_{x} \in T_{x}M

 

[mrandersdk]

think you are kind a right but you shouldn't write

(L_{g})_{\star}X_{h}=(d_{h}L_{g})\circ X_{h}

because (L_g)_* acts on a vector field, and should not be put togheter with one that is

(L_{g})_{\star}X_{h}=d_{h}L_{g}(X_{h}).

But maybe try to see it like this: Let X be a vector field. That is X\in TM and X(h) \in T_hM, then (L_g)_*X is a new vector field, call it Y such that Y \in TM and Y(h) = (L_g)_*X(h) \in T_{L_g(h)}M (remember that L_g goes from M into M). Then you have

(L_g)_*X = dL_g(X) = Y \in TM
and

d_hL_g(X_h) = Y_h \in T_{L_g(h)}M where X_h is just shorthand for X(h)

but you could write

(L_g)_*X_h = d_hL_g(X_h) = Y_h \in T_{L_g(h)}M

but this notation is a bit sloppy (I think) because (L_g)_* actually acts on vectorfields not tangent vectors, but because X_h is actually a vector field evaluated at h, we can define

(L_g)_*X_h = (L_g)_*X|_h = d_hL_g(X_h) = Y_h \in T_{L_g(h)}M

so it makes sence. I guess that if you have a tangent vector t \in T_pM you could write

(L_g)_*t

but for this to make sense you need to show that you can extend t to a vector field X in TM with X(p) = t and show that (L_g)_*X(p) only depends on the value of X at p, then you can use the above, and thus make sense to (L_g)_* working only on a tangent vector (I believe that this can be shown, but don't have my books with me), so this is why you can be a bit sloppy the definitions extends and restricts to each others.

 

[GSpeight]

Thanks so much for your help again. So to summarise...

If F:M \rightarrow N is a smooth map of manifolds then F_{\star} maps smooth vector fields on M to smooth vector fields on N according to the rule (F_{\star}X)_{g}=d_{g}F(X_{g})

Alternatively F_{\star} is a map of tangent bundles according to the rule F_{\star}(x,v)=d_{x}F(v)

If it is clear that v \in T_{x}M we will sometimes write F_{\star}(v) instead of d_{x}F(v)

Does this seem like a reasonable working understanding?

 

[mrandersdk]

Alternatively F_{\star} is a map of tangent bundles according to the rule F_{\star}(x,v)=d_{x}F(v)

this is not correct, the only reason i talk about tangent bundles is because a vector field lives in the tangent bundle, so when you correct say "maps smooth vector fields on M to smooth vector fields on N" then it is a map from the tangent bundle of M to the tangent bundle of N.

Else I think what you are saying are correct, maybe someone should read it through to check it. But I'm pretty sure that F_*X defines a vector field on N by
(F_*X)(g)=d_gF(X(g)) as you say.

 

[GSpeight]

I meant to say F_{\star} is a map between tangent bundles, not sure if that makes any difference, but, in any case, will forget about that bit.

I'm sure it will become clearer as I work through my notes and rewrite pieces that confuse me, just as long as I have a working understanding for now.

Thanks again!

 

[mrandersdk]

no problem, using the definition is the only way to go. Maybe just a quick comment:

I guess the notation dF and d_xF is a bit like this:

d_xF could be written just as well as dF(x) just like if you have a normal function f, and you could write f(x) and f_x (not so commen, but is used). Let's take the example with f which is just a familiar function.

f is the object calld a function, and f(x) is actually that object called a function evaluated at x. We often say that f(x) is the function, but this is actually not correct, it is the function evaluated at x. You see this more clear if I write

\frac{\partial f(x)}{\partial x} that you wouldn't write you would write
\frac{\partial f}{\partial x} and clearly \frac{\partial f(x)}{\partial x} should be \frac{\partial f}{\partial x}|_x, but here you would have know what i meant by \frac{\partial f(x)}{\partial x}.

Just like here dF is the "operator" and d_xF is the operator dF evaluated at x.

The thing that may be difficult here is that dF is defined as what it does on a vector field at a point in the manifold x. So dF is kinda defined through d_xF but it is an object in it self.

But if you think about it this not stranger than a regular function f, this is also defined by what it is at some point x. Because if i tell you that you have a function f you wouldn't know much before i tell you what f(x) is for all x, but still the object f is interesting in it self, that is why we sometimes write f and sometimes f(x) it depends on what we are doing (and how lazy we are, authors often are :-)).

I guess you would never be confused here because you are so familiar with functions.

You could also look at the ordinary derivetive \frac{\partial}{\partial x} this is an operator from let's say smooth functions on R to smooth functions on R, but it is defined by what it does to f at every point x. If you look in you ordinary analysis book you will find that you actually only have a definition for

\frac{\partial f}{\partial x}|_x not for \frac{\partial f}{\partial x}

 

[Hurkyl]

If we're writing points in the tangent bundle as point-tangent vector pairs, then we do have

F_*(p, v) = (F(p), (d_p F)(v))

 

[Hurkyl]

no problem, using the definition is the only way to go. Maybe just a quick comment:

If I may expand...

A more sophisticated way of thinking here is to forget about points, and think about functions. If I let * denote the manifold with a single point, then the notion of a point P of M is equivalent to the notion of a map P : * \to M. We also have coordinate charts x : \mathbb{R}^n \to M, and the "generic element" z which is the identity function z : M \to M.


Now, the trick is to no longer think of f(P) as evaluating a function (in the classical sense) -- instead think of it as function composition: f(P) = f \circ P.


Let F : M \to N be a map of manifolds. If you've ever taken a coordinate chart x for M, and written F(x), then this is exactly what you're doing. Intuitively, we think of "x" denoting some generic point of M described by our coordinate chart, and let F(x) denote the image of that point after applying F. If you plug in an actual coordinate vector a, then by F(x) we mean F(x)(a) := F(x(a)). Note if we think of the coordinate vector a as being a map * \to \mathbb{R}^n, then everything is as before, because: F(x)(a) = (F \circ x) \circ a = F \circ (x \circ a) = F(x(a)).


The identity function z is even cooler -- it exactly represents the intuitive notion of a "indeterminate variable". In fact, f(z) = f, and this gives us a fully rigorous way to slip indeterminate variables into and out of an expression without changing its meaning. e.g. if f is a function M \to N, then f(z) is also a function M \to N. But if a denotes an 'ordinary' element * \to M, then f(a) would be an 'ordinary' element of N.

 

[GSpeight]

Sorry for the late reply - I was reading through some Functional Analysis notes which I found much more easy-going than Lie Groups

Thank you a lot to both of you for the idea of the intuition behind this. I understand it now and it's made the next few pages of my notes much clearer.

I'm just reading about how vector fields can be viewed as derivations of the space of smooth real valued functions on the manifold - it seems a common characteristic here that some objects can be viewed in several different ways (as with tangent spaces).

I don't want to spam this forum with threads so, if I have any other questions on Lie Groups, I'll possibly post them here in case anyone has time to share some insight.

Thanks again.

 

[GSpeight]

I've just read the proof of the result that the exponential (from the lie algebra to the lie group) restricts to a diffeomorphism U -> V for a nbhd U of 0 and nbhd V of e (e the identity in G).

Shortly after this is a corollary that every element of a connected lie group is a finite product of exponentials. There's then a one line proof of this simply stating that V=exp(U) is open in G, so generates G.

Can anyone elaborate on this? I don't quite see why an open set should generate the lie group (clearly we need to use connectedness somewhere here).

I expect it's something obvious, since otherwise I wouldn't have a one line proof, so sorry if this is a stupid question.

Thanks again.

 

[Hurkyl]

Can anyone elaborate on this? I don't quite see why an open set should generate the lie group (clearly we need to use connectedness somewhere here).

A connected space has exactly two subsets that are both open and closed. The subgroup generated by V is clearly open, and not the empty set...

 

[mrandersdk]

Here is the proof of what you need:

The theorem states:

Let G be a Lie group and U a nbh. of e. Then U generates G, i.e. G = \bigcup_{n=1}^{\infty} U^n where U^n consists of all n-fold products of elements of U.

proof:

We may assume U is open without loss of generality. Define V=U \cap U^{-1} \subset U where U^{-1} is the set of all invers elements of u. This is an open set because inverse in continuous, so V is open. Define H = \bigcup_{n=1}^{\infty} V^n, by construction H is an open subgroup containing e. For a g \in G write gH = \{gh | h \in H\}. The set gH contains g , thus G=\bigcup_{g\in G}gH. gH is open because multiplication from the left with g^{-1} is continuous. We now pick a representative g_{\alpha}H for each coset in G/H then G \bigcup_{\alpha}g_{\alpha}H, but all g_{\alphaH} is disjoint, so because G is connected there can only be one coset in G/H thus H = eH = G. That is V generates G, but V \subset U so U generates G q.e.d.

connectedness:

http://mathworld.wolfram.com/ConnectedSet.html

 

[Hurkyl]

My idea was somewhat different.

Let H be the subgroup of G generated by V.

(1) H is open

This is clear, because for any point x of H, the entire (open) set xV is contained in H.

(2) H is closed

This is clear, because lie groups are nice topological spaces, so any point in the closure of H is the limit of a sequence of points.

Suppose that the sequence x_1, x_2, x_3, ... of points in H converged to x. Then the open set xV contains, say, x_n. It follows that x must be an element of H.

(3) H = G

H contains the identity, and G is the only nonempty set that is open and closed.

 

[mrandersdk]

I fail to see something, maybe you can elabrorate. You want to show that if you have a sequence in H that is convergent then the limit is in H?

how can you say: "Then the open set xV contains, say, x_n. It follows that x must be an element of H"

xV is open i give you that but, i can't see that you have xV is a subset of H, because x need not to be in H.

And where do you use the that G is connected?

 

[Hurkyl]

I fail to see something, maybe you can elabrorate. You want to show that if you have a sequence in H that is convergent then the limit is in H?

how can you say: "Then the open set xV contains, say, x_n. It follows that x must be an element of H"

Because that means there is a v in V such that x_n = xv. And since both x_n and v are in H...

And where do you use the that G is connected?

That's why G is the only nonempty subset of G that is both open and closed. If G were not connected, then there would be other nonempty clopen subsets.

 

[mrandersdk]

hey thanks. Nive proof, always good to see other ways to prove something.

 

[GSpeight]

Thanks so much to both of you for your help - I've rewritten that bit of my notes so that I now have a complete proof.

I really appreciate it - thanks again!

 

[GSpeight]

Hi there,

I have my Lie groups exam on Monday and am currently working on a past paper. At the moment I'm stuck on the following question:

Define the terms homomorphism of Lie groups and homomorphism of Lie algebras (done).
If \phi:G->H is a homomorphism of Lie groups show that its differential at the identity gives a homomorphism of Lie algebras (done).
Show \phi (\exp \zeta)=exp \phi_{\star} \zeta (done).
Hence (or otherwise) find all homomorphisms of Lie groups \phi:S^{1}->SL(2,\mathbb{R} )

If we consider the circle S^1 as the interval [0,2pi) with addition modular 2pi then the exponential map is (I think) just the identity modulo 2pi.

Then if \phi is a homomorphism, \phi (\zeta)=\exp(\phi_{\star}(\zeta))=I_{2}+\phi_{\star}(\zeta)+\frac{1}{2}\phi_{\star}(\zeta)^2+\ldots

Am I even on the right track? Any ideas where I go from here?

 

[GSpeight]

Just a quick message to say thanks to mrandersdk and Hurkyl for the help you've both given. I really appreciate it. I had my Lie Groups exam today and it went very well :) Now to go revise some topology!

 

[mrandersdk]

Congratulations, sorry that i couldn't help you with the last question.