I'm very new to differential geometry and am struggling to get a grasp of left (or right) invariant vector fields.

The cause of my confusion is the defining condition for a left invariant vector field $V$:

$$ L_{g^*} V(h) = V(gh) $$

(I'm guessing it should hold for all functions $h$?) From what I understand, $L_{g^*} V(h)$ is a pushforard, defined as $V(L_{g}^* h) = V(h \circ L_g)$. However, I'm not sure if I'm interpreting this expression correctly. What exactly is $V(h \circ L_g)$? If $L_g(x) = gx$, does this mean that $V(h \circ L_g) = V(h(gx))$, i.e. a map from $x$ being a function to a real number? (Using corresponding notation, the original expression would be $L_{g^*} V(h(x))$, where x is a function.)

Is this correct? In that case, how do I generally confirm whether a vector field is left invariant? In my notes, for a vector field $T_+$ defined on $SU(2)$, given by:

$$ T_+ = e^{2i\theta} \left( \epsilon_z^2 \partial_{\bar{z}} - \frac{iz}{2} \partial_{\theta}\right) $$

where $h$ are elements of $SU(2)$, parametrized by:

$$ h = \frac{1}{\epsilon} \pmatrix{1 & \bar{z} \\ -z & 1} \pmatrix{e^{i \theta} & 0 \\ 0 & e^{-i \theta}} $$

is said to be left invariant. In the notes, there is a proof for this. It is first written that $L_{h^*} T_+ = e^{2i\theta} \left( \epsilon_z^2 \partial_{\bar{z}} - \frac{iz}{2} \partial_{\theta}\right)$. Is this incorrect? Moreover, if my interpretation of the pushforward is correct, wouldn't that expression be different? (I.e, plugging in $gx$ into $h$ for a general function $x$, then differentiating.)

Furthermore, the notes show that:

$$ (L_{h^*} T_+) \circ h = \frac{1}{\epsilon} \pmatrix{1 & \bar{z} \\ -z & 1} \pmatrix{e^{i \theta} & 0 \\ 0 & e^{-i \theta}} \pmatrix{0 & 1 \\ 0 & 0} $$

How does this statement prove that the vector field is invariant? And why do you left multiply by the same element as you put into the vector field? Shouldn't it, for generality, be a different function?

To summarize:

  1. Is my interpretation of $L_{g^*} V(h)$ correct?
  2. In the example of $T_+$, how exactly does the above proof of left-invariance work?
  3. If the proof from the notes happen to be wrong, how would you prove that $T_+$ is left-invariant in this particular case?

Any answer whatsoever is appreciated! I've been spending hours on this.

  • 2
    $\begingroup$ I'm commenting just on the very first thing. Here $g$ and $h$ are elements of $G$. There are no functions involved. We're comparing the values of the vector field at the points $h$ and $gh$ of $G$. $\endgroup$ Mar 11, 2020 at 18:45
  • $\begingroup$ Ted Shifrin Thank you for clarifying this. I'll go through it again and see if it changes anything. $\endgroup$
    – Max
    Mar 11, 2020 at 19:10
  • $\begingroup$ You should read some basics about Lie groups. In your note, $g,h,x$ are elements of a Lie group, $L$ is left multiplication ($L_g(x)=gx$), $V$ is a vector field, assigning a tangent vector on each point (=group element). Matrix groups are primary examples of Lie groups. $\endgroup$
    – Berci
    Mar 11, 2020 at 20:25
  • $\begingroup$ Berci Thank you! I know that there is a way of expressing vector fields as "things" acting on functions, but now that you say this isn't the case in the notes, it changes a lot. In the case that $h \in G$, what would the expression $L_{g_*} V(h)$ yield? Since $L_g$ is a function, isn't pushforward defined such that $L_{g_*} V(h) = V(h \circ L_{g})$. But that expression won't make sense for $h \in G$, will it? (If it was the other way around ($L_g \circ h$), I'd see how that is $gh$, but then the definition of left invariant vector fields I posted at the top of my question would be pointless.) $\endgroup$
    – Max
    Mar 11, 2020 at 21:09
  • $\begingroup$ @Berci I just realized that I used the wrong @-syntax, so I'll just plug it in here $\endgroup$
    – Max
    Mar 12, 2020 at 9:29


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