Differential of Multiplication Map of Lie Group

Question

Checking some basic facts about $G$, a Lie Group. Proposition $3.14$ provides an isomorphism $\alpha: T_e(G\times G)\to T_eG\oplus T_eG:v\mapsto (d\pi_1(v),d\pi_2(v))$, where $\pi_i;\ i=1,2$ are the projections to the first and second factor, respectively.

Here is the exercise:

I thought this would be routine, but I have a problem justifying a step in $(1)$. Using the identification as a hint, write $dm_{(e,e)}(X,Y ) = dm_{(e,e)}(X, 0) + dm_{(e,e)}(0,Y ).$ Each term on the right hand side is the restriction of the differential $dm_{(e,e)}$ to vector spaces that are isomorphic to $T_eG$ so it is intuitively obvious that $dm_{(e,e)}(X,0)=dm'_e(X)$ where $m'(x)=m(x,e)$ and $dm_{(e,e)}(0,Y)=dm''_e(Y)$ where $m''(y)=m(e,y).$ Now the result follows because $m'$ and $m''$ are the identity on $G$. How do I prove rigorously what I think is "intuitively obvious"?

Remark: for part $(2)$ I fiddled around with maps until I got one that has zero differential: $x\overset{\Delta}{\mapsto}(x,x)\overset{f}{\mapsto}(x,i(x))\overset{m}{\mapsto}e.$ The result now follows by the chain rule, using $(1)$ at the end.

Answer 1

First note that the inverse of the isomorphism $\alpha$ is the map $\tau :T_pM \oplus T_qN \longrightarrow T_{(p,q)}(M\times N)$ defined as $$ \tau(v,w)=d\iota_1(v)+d\iota_2(w), $$ where $\iota_1 : M \hookrightarrow M\times N$ defined as $\iota_1(x)=(x,q)$ and $\iota_2 : N \hookrightarrow M\times N$ defined as $\iota_2(x) = (p,x)$.

When working with identification of tangent space $T_{(p,q)}(M\times N)$ as $T_pM \oplus T_qN$, i always keep track the elements in $T_{(p,q)}(M\times N)$ as the image of $\tau$. For me it's easier to work this way. You’re right about the equality $dm_{(e,e)}(X,0)=dm'_e(X)$, but it’s not directly clear since we treat $dm_{(e,e)}$ as a map from $T_eG\oplus T_eG$.

Now, in our case the map $ dm_{(e,e)} : T_eG \oplus T_eG \longrightarrow T_eG$ in the text is actually the composition $$ T_eG \oplus T_eG \xrightarrow{\tau}T_{(e,e)}(G\times G) \xrightarrow{dm_{(e,e)}} T_eG. $$ Define $\iota_1,\iota_2 : G \hookrightarrow G\times G$ as $\iota_1(x)=(x,e)$ and $\iota_2(x)=(e,x)$, we have \begin{align} dm_{(e,e)}(\tau(X,Y))&= dm_{(e,e)} (d\iota_1(X)+d\iota_2(Y)) \\ &= dm_{(e,e)} (d\iota_1(X)) + dm_{(e,e)} (d\iota_2(Y)) \\ &= d(m \circ \iota_1)(X) + d(m \circ \iota_2)(Y) \\ &= X + Y, \end{align} since $d(m \circ \iota_i) = \text{Id}_{T_eG}$, because $m \circ \iota_i = \text{Id}_{G}$.

For (2), the equation $di_e(X) = -X$ or (equally) $di_e(X) + d\text{Id}_G(X) = 0$ probably give some clue that we need to consider the trivial map $T : g \mapsto e$ and break it as the composition $$ g \mapsto (g,g^{-1}) \mapsto e=gg^{-1}. $$ That is $$ T : G \xrightarrow{E} G \times G \xrightarrow{m} G, $$ where $E(g):=(g,g^{-1})$. Since $T$ is a constant map, $dT_e= 0$. Use (1) and Proposition 3.14 again to show that $di_e(X) = -X$.

Answer 2

Oftentimes the way you prove what a differential looks like is through taking the derivative of the map composed with a smooth curve. Let $\gamma:(-\epsilon, \epsilon) \to G$ be a curve such that $\gamma(0) = e$ and $\dot{\gamma}(0) = X$, and similarly we can define some curve $\eta$ where $\eta(0) = e$ and $\dot{\eta}(0) = Y$. Then we can see that

\begin{eqnarray*} dm_{(e,e)}(X,Y) & = & \left . \frac{d}{dt} m\circ (\gamma(t), e) \right |_{t=0} + \left . \frac{d}{ds} m\circ (e, \eta(s)) \right |_{s=0} \\ & = & \left . \frac{d}{dt} \gamma(t) \right |_{t=0} + \left . \frac{d}{ds} \eta(s) \right |_{s=0} \\ & = & X + Y. \end{eqnarray*}