# Manifolds / Lie Groups - confusing notation

## Main Question or Discussion Point

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 :)

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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 alot 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.

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so you see that for your last question:

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

makes more sence 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 $$T_L_g(x)G$$.

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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 $X_{x} \in T_{x}M$

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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 sence 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 sence 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.

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?

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.

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!

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). Lets 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 lets 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
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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
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Gold Member
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.

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.

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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
Staff Emeritus
Gold Member
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....

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
Staff Emeritus
Gold Member
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.

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?

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Hurkyl
Staff Emeritus
Gold Member
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.

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

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!

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?

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!