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!

  • $\begingroup$ 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)$. $\endgroup$ – peter a g Oct 7 '15 at 15:32
  • $\begingroup$ @peterag That's a very enlightening comment actually, makes a lot more sense when I think of what I've written like that, thanks! $\endgroup$ – Tim Oct 7 '15 at 15:51
  • $\begingroup$ 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... $\endgroup$ – peter a g Oct 7 '15 at 16:26
  • $\begingroup$ @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? $\endgroup$ – Tim Oct 7 '15 at 17:12
  • $\begingroup$ 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? $\endgroup$ – 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 $X(f\circ L_g)(e).$

  • $\begingroup$ 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 $x$. $\endgroup$ – Tim Oct 7 '15 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.