I have to say sorry about my English which is not my native language. So you can ask me for the ambiguous concepts of what I wrote.

Notation: $(G,\cdot)$is a Lie group, $\mathscr{L}$ is left invariant vector field on $G$ and $T_eG$ is tangent space at $e\in G$.

I read a theorem that said $\mathscr{L}$ and $T_eG$ are linearly isomorphic to each other. That is

$$\exists \ \text{an isomorphism}\ \eta:T_eG \rightarrow\mathscr{L} \ \text{s.t.}\ A\mapsto\eta(A)\equiv \bar{A} $$ Where $A\in T_eG$ is a vector and $\bar{A} \in \mathscr{L}$ is a vector field.

Question1: Elements of $T_eG$ are vectors and elements of $\mathscr{L}$ are a vector fields. Why can we map a vector to a vector field?(Is this allowed?)

I mean in the proof we define that $$\forall A\in T_eG, \text{Define} \ \bar{A}|_g:=L_{g*}A, \ \forall g \in G$$ Where $L_{g*}$ is a push forward of left translation and $\bar{A}|_g$ is a vector and $\bar{A}|_g\notin\mathscr{L}$ (because $\bar{A}|_g $is a vector not vector field.) So it is not reasonable for me to say $A\mapsto\eta(A)\equiv \bar{A}$, and $A\mapsto\eta(A)\equiv \bar{A}|_g, \forall g \in G$ is more reasonable for me.

Question2:So I collection all of $\bar{A}|_g$ be a vector field $\mathscr{L}’$(also a vector space)(I don’t know which one is better of I described.) $$\mathscr{L}’:=\{\bar{A}\in\mathscr{L}|\bar{A}|_g=L_{g*}A,g\in G, \ A \in T_eG \}$$ $$\mathscr{L}':=\{\bar{A}_g \ |\ \bar{A}|_g=L_{g*}A, g\in G , A\in T_eG \}$$ Therefore, I may say $T_eG$ and $\mathscr{L}’$ are linear isomorphic to each other (use $\mathscr{L}’$ to instead of $\mathscr{L}$ ) is reasonable for me.

Question3: Are $\mathscr{L}’$ and $\mathscr{L}$ equivalent or not?

Conclusion: For me

Vector($T_eG$) maps to another vector ($\mathscr{L}’$) by $\eta$ is reasonable for me.

Vector($T_eG$) maps to a vector field($\mathscr{L}$) by $\eta$ is not reasonable for me.


Thanks for reading and answering my question. Sorry for wasting your time and my poor English .

Hopefully I have explained enough to let you understand my question as possible as I can.


1 Answer 1


(For technical reasons let's assume $G$ is connected.) To answer your question, let me first explain a little more about left invariant fields. Let's start with two things.

  • Choose some element $g \in G$. Since $G$ is a Lie group, then I get a left translation map $\ell_g: G \to G$ given by $h \mapsto gh$. (You can think of this intuitively as "pushing" $h$ to $gh$.)
  • A vector field $X$ on $G$ is a function $X: G \to \mathsf{T}G$ that gives you a tangent vector $X_h \in \mathsf{T}_hG$ above each point $h \in G$. (You can think of this as a bunch of arrows placed over $G$.)

Now I can push $X$ forward by $\ell_g$, to obtain a new vector field $(\ell_g)_*X$ as follows. Given a point $h \in G$, I need to define a vector $((\ell_g)_*X)_h \in \mathsf{T}_hX$ in the tangent space above $h$. To do this, let me start at the point $g^{-1}h \in G$. The translation map $\ell_g$ sends $g^{-1}h \mapsto h$, so its derivative $\mathsf{d}(\ell_g)_{g^{-1}h}$ at $g^{-1}h$ is a function $\mathsf{T}_{g^{-1}h}G \to \mathsf{T}_hG$. So if I take the tangent vector $X_{g^{-1}h} \in \mathsf{T}_{g^{-1}h}G$, then I can push it forward by this derivative to get a vector in $\mathsf{T}_hX$, which is what I wanted. In other words, I define $$ ((\ell_g)_*X)_h := \mathsf{d}(\ell_g)_{g^{-1}h}(X_{g^{-1}h}). $$ So I've started at the point $g^{-1}h$ and "pushed" the vector $X_{g^{-1}h}$ over to the tangent space $\mathsf{T}_hG$ to get a vector $((\ell_g)_*X)_h$ above $h$. However, there's no reason that this new vector $((\ell_g)_*X)_h$ has to be the equal to the vector $X_h$ that was already there. The original vector field could be changing in totally different ways over $G$. But if $X$ is left invariant, then by the definition of a left invariant vector field, we have exactly $$ X_h = ((\ell_g)_*X)_h. $$ In other words, if $X$ is left invariant, then as long as I know the value of my vector field $X$ at some point of $G$, then I can find the value of $X$ anywhere else just by pushing it forward by the correct translation map. So all of the information you need about the left invariant vector field can be recovered by its value at a single point.

So, if you give me a single tangent vector $A \in \mathsf{T}_eG$, then I can just create an entire vector field $\bar{A}$ on $G$ by defining its value at each point $g \in G$ to be $\bar{A}_g := d(\ell_g)_e(A)$. That is, I'm taking the vector $A$ and pushing it over to $g$ using the map $d(\ell_g)_e: \mathsf{T}_eG \to \mathsf{T}_gG$. By construction, this vector field is left invariant. So I've just defined a function $$ \eta: \mathsf{T}_eG \to \mathscr{L}, \quad A \mapsto \eta(A) := \bar{A}, $$ assigning a left invariant vector field to every vector $A \in \mathsf{T}_eG$. This is definitely allowed! Defining the inverse of $\eta$ is easy. Given a left invariant vector field $X \in \mathscr{L}$, just send it to its value at $e$. That is, define $\eta^{-1}: \mathscr{L} \mapsto \mathsf{T}_eG$ by $X \mapsto X_e$.

To answer your second question, firstly I think you may be confusing the collection $\{\bar{A}_g\}_{g \in G}$ (the values of a single vector field, at each point of $G$) with the collection $\mathscr{L}$ of all left invariant vector fields $A$ on $G$. The first one, you can think of as a vector field (though it's better to think of a vector field as a function $G \to \mathsf{T}G$). The second one is a vector space. Next, yes, $\mathscr{L}'$ is isomorphic to $\mathsf{T}_eG$ as a vector space; that's because we've defined an isomorphism $\eta$ between them.

For your last question, you've essentially defined the space $\mathscr{L}' = \{ \bar{A} = \eta(A) \mid A \in \mathsf{T}_eG \}$ as the image of $\eta$. That is, you took all of the vectors in $\mathsf{T}_eG$ and collected the vector fields that $\eta$ sends them to into a set. Again, every one of these vector fields is left invariant, so what you've defined is a subset $\mathscr{L}'\subset\mathscr{L}$. But actually, $\mathscr{L}'=\mathscr{L}$ because, as we've seen above, every left invariant field $X$ is in the image of $\eta$, since $X = \eta(X_e)$.

Hope this helps!

  • 1
    $\begingroup$ Very thanks for answering my question and I will read you answer carefully. If I have any more question I will leave the question in here. Very thank you! $\endgroup$
    – A student
    Commented Mar 22, 2020 at 18:10

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