Proof that every vector field on a Lie group is left-invariant

I am just starting a course on Lie groups and I'm having some difficulty understanding some of the ideas to do with vector fields on Lie groups. Here is something that I have written out, which I know is wrong, but can't understand why:

Let $X$ be any vector field on a Lie group $G$, so that $X\colon C^\infty(G)\to C^\infty(G)$. Write $X_x$ to mean the tangent vector $X_x\in T_x G$ coming from evaluation at $G$, that is, define $X_x(-)=(X(-))(x)$ for some $-\in C^\infty(G)$. We also write $L_g$ to mean the left-translation diffeomorphism $x\mapsto gx$.

Now \begin{align} X_g(-) = (X(-))(g) &= (X(-))(L_g(e))\\ &= X(-\circ L_g)(e)\\ &= X_e(-\circ L_g) \\ &= ((DL_g)_eX_e)(-). \end{align} Using this we can show that $((L_g)_*X)_{L_g(h)}=X_{L_g(H)}$ for all $h\in G$, and thus $(L_g)_*X=X$, i.e. $X$ is left-invariant.

I'm sure that the mistake must be very obvious, but I'm really not very good at this sort of maths, so a gentle nudge to help improve my understanding would be very much appreciated!

• The passage from the first line to the second is faulty - chain rule. It looks like you are in effect writing something like that if $b = u(a)$, then $v'( b) = ( v\circ u)'(a)$. – peter a g Oct 7 '15 at 15:32
• @peterag That's a very enlightening comment actually, makes a lot more sense when I think of what I've written like that, thanks! – Tim Oct 7 '15 at 15:51
• Right - and $X$ has to be 'chosen correctly' to account for the (missing) chain rule of translation... As an explicit example, try out the multiplicative group $\mathbb R^*$, and $X = x d/dx$, and let $u(x) = ax$, and compare $$X_a (f) = a f'(a)$$ with $$X_1 (f \circ u) = 1 f'(a) u'(1) = f'(a) a.$$ Without the $x$ in $X$, (or scalar multiple thereof) it wouldn't have worked... – peter a g Oct 7 '15 at 16:26
• @peterag So is saying that $X$ is left-invariant sort of saying that $X$ has been 'chosen correctly', as you say? Is this a necessary and sufficient condition, or just sufficient, or neither? – Tim Oct 7 '15 at 17:12
• It's necessary and sufficient... In the previous example, you could have chosen (erroneously) $X=d/dx$, but then $X_a (f)= f'(a)$, while $(DL_aX)_1 (f) = a f'(a)$. So, by just looking at it, the only possible way to correct $X$ to get it to match up (i.e., be invariant under translation) is with the factor $x$... but that's precisely what you get by defining $X_a = (DL_a X)_1$. OK? – peter a g Oct 7 '15 at 17:23

Your problem is with the equality $(X(f))(L_g(e))= X(f\circ L_g)(e).$ Note that in $(X(f))(L_g(e))$ you first get the derivative of $f$ with respect to $X$ evaluated at $g.$ But, in $X(f\circ L_g)(e)$ you modify the function by a left translation. It holds that $(f\circ L_g)(e)=f(g)$ but you cannot say anything at nearby points, which is essential to get $X(f\circ L_g)(e).$
• Ok, so if we define $X_g=(DL_g)_e X_e$, then this doesn't align the idea of $X_g$ given by evaluation at $g$? This is the reason that I asked: the notes I'm using define $X_g$ as above, and I thought that this would be consistent with the idea of writing $X_x$ to mean the tangent vector coming from evaluation at $x$. – Tim Oct 7 '15 at 15:43