I am following a proof in Hamilton's Mathematical Gauge Theory and getting a bit stuck on a step, although I'll include a bit more in case there are other issues I'm not aware of. The proof is to show the vector space isomorphism between left-invariant vector fields and the tangent space $T_eG$.

The evaluation map is defined as the map from any left-invariant vector field $X\in L(G)$ (i.e. those forming the Lie algebra of the group $G$, with the Lie bracket of vector fields)

\begin{align} \text{ev}:L(G) &\rightarrow T_eG\\ X &\mapsto X_e. \end{align}

We now want to prove that this map is a vector space isomorphism. It's clear to me that this is linear.

Firstly, we want to construct the inverse of a vector $x\in T_eG$ under ev by constructing a vector field $X$ on $G$, which we define point-wise by

\begin{align} X_h = (D_eL_h)x. \end{align} $L_h: G \mapsto G$ is left translation on the group, and so if $x = \dot{\gamma}(0)$ for some curve $\gamma$ on $G$, then the differential is \begin{align} X_h = (D_eL_h)x = (\dot{L_h \circ \gamma})(0). \end{align} $L_h\circ \gamma$ is clearly a curve on $G$ s.t. $L_h \circ \gamma (0) = L_h(e) = h$, and $X_h$ really is a vector at $h\in T_hG$. Lastly, $X_e = (\dot{L_e \circ \gamma})(0) = \dot{\gamma}(0) = x$, and so the evaluation map acts on this vector field to give $x \in T_eG$ as claimed.

The next step is to consider the map

\begin{align} \mu: G \times G &\rightarrow G \\ (h, g)& \mapsto hg. \end{align}

The claim is that this map has differential

\begin{align} (*)\qquad D\mu:TG \times TG &\rightarrow TG\\ ((h, Y), (g, X)) &\mapsto (D_gL_h)(X)+(D_hR_g)(Y). \end{align}

It's not clear to me why this is the case. What I have tried is

\begin{align} (D\mu(h, g))(X, Y) = (D(hg))(X, Y) = D(L_h)(g)(X, Y) + (h)(DR_g)(X, Y), \end{align} but I don't really know what I'm doing here.

Given the above, the map \begin{align} G&\rightarrow TG \\ h& \mapsto D\mu((h, 0), (e, x)) = (D_eL_h)x \end{align} is smooth (since left-translation is smooth), and this is just the vector $X_h$ from before. Since this map is smooth, we have proved the corresponding vector field $X$ is smooth.

Lastly, the vector field is left invariant, because \begin{align} (D_hL_g)X_h = (D_hL_g)(D_eL_h)x = (D_{L_h(e)}L_g)(D_eL_h)x = (D_e(L_g \circ L_h))x = (D_e(L_{gh}))x = X_{gh}. \end{align}

So the map

\begin{align} T_eG &\rightarrow L(G)\\ x&\mapsto X \end{align}

is the inverse of ev. Thus ev is a smooth invertible map from $L(G) \rightarrow T_eG$.

Once I have this, I am happy with the other vector space structure, so the starred $(*)$ equation is really where I'm stuck.

Queries in response to Didier's answer.

I don't understand why \begin{align} (D\mu (X, Y))f = (X, Y)(f \circ \mu), \end{align} by definition.

To verify this, I tried to consider a pointwise (or 'bi-pointwise'?) definition as \begin{align} D_{(g, h)}\mu : T_gG \times T_hG &\rightarrow T_pG \\ (X_g, Y_h) &\mapsto D_{(g, h)}\mu(X_g, Y_h). \end{align} Now the problem for me is that I don't really know what the point $p$ should be, and I don't know which definitions to apply to find the action of this map on a function $f$. Basically, I don't know how to go from this to $(X, Y)(f\circ \mu)$ - or maybe you have taken a different approach?

Additionally, I don't understand what is meant by \begin{align} (X, Y)(f \circ \mu) = X(f \circ \mu) + Y(f \circ \mu). \end{align} As far as I can tell, \begin{align} f \circ \mu: G\times G \rightarrow G \rightarrow \mathbb{R} \end{align} requires two arguments on $G$, and so I don't understand what it means to take e.g. $X(f\circ \mu)$, which pointwise, with $X_p = \dot{\gamma}(0)$ for some curve $\gamma$ (and $\gamma(0) = p$), I would expect to be \begin{align} X(f\circ \mu) = (\dot{f \circ \mu \circ \gamma})(0), \end{align} but this map doesn't make sense since what $\gamma: \mathbb{R} \rightarrow G$, but then $\mu$ is a map from $G\times G$.

Lastly, I don't see where the second line in \begin{align} \left(D_{h,g}\mu (X_h,Y_g)\right)f &= X_h\left((f\circ R_g)(h)\right) + Y_g\left((f\circ L_h)(g) \right)\\ &= D_{hg}f\left(D_hR_g(X_h)\right)+ D_{hg}f\left( D_gL_h(Y_g)\right)\\ &= \left(D_hR_g(X_h) + D_gL_h(Y_g)\right)f \end{align} comes from.

  • $\begingroup$ See the accepted answer of this $\endgroup$
    – Didier
    Nov 10, 2022 at 9:17
  • $\begingroup$ Perhaps that answers my question, but I don't understand exactly how their construction relates to mine. Are you able to explain this to me in some detail? $\endgroup$
    – Bedge
    Nov 10, 2022 at 11:59

1 Answer 1


Consider a vector field $(X,Y)$ on $G\times G$, where $X$ is tangent to the first component and $Y$ to the second. Fix $f$ any smooth function on $G$. By definition, $$ \left(D\mu (X,Y)\right)f = (X,Y)\left(f\circ \mu\right). $$ Now, by linearity, $(X,Y)=(X,0) + (0,Y)$, and $(X,Y)$ acts as the sum $$ (X,Y)(f\circ\mu) = X\left(f\circ\mu\right) + Y \left(f\circ\mu\right). $$ Fix $(h,g)\in G\times G$. Remark that $\mu(h,g)=R_g(h) = L_h(g)$, and that $X_h$ only acts on what depends on $h$ while $Y_g$ only acts on what depends on $g$, so that \begin{align} \left(D_{h,g}\mu (X_h,Y_g)\right)f &= X_h\left((f\circ R_g)(h)\right) + Y_g\left((f\circ L_h)(g) \right)\\ &= D_{hg}f\left(D_hR_g(X_h)\right)+ D_{hg}f\left( D_gL_h(Y_g)\right)\\ &= \left(D_hR_g(X_h) + D_gL_h(Y_g)\right)f. \end{align} Hence, $D_{h,g}\mu(X_h,Y_g) = D_hR_g(X_h)+D_gL_h(Y_g)$.

  • $\begingroup$ Thank you for your answer. I am still a bit confused by some of the definitions you use. I think it's very close to my initial question, so added on to the question to address these specific steps. If you are able to go into some more depth, that would be great. $\endgroup$
    – Bedge
    Nov 10, 2022 at 19:09
  • $\begingroup$ @Bedge Your first and last questions in the edit can be solved by learning about the chain rule in the context of differential geometry. Regarding the one regarding the separate actions of $X$ and $Y$: this is a notation that tells you that $X$ only acts on the first variable, letting the second fixed, and similarly for $Y$. More rigorously, should be $(X,0)$ ans $(0,Y)$. $\endgroup$
    – Didier
    Nov 10, 2022 at 19:36
  • $\begingroup$ Thanks for you answers. I won't accept the answer for now in case someone else is able to help, as I'm still stuck. $\endgroup$
    – Bedge
    Nov 10, 2022 at 21:00

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