Differential of the multiplication and inverse maps on a Lie group

Solution 1:

One of Tu's previous exercises, ex. $8.7^*$ shows that for $M,N$ manifolds, by defining $\pi_1:M\times N\to M$ and $\pi_2: M\times N \to N$, one can show for $(p,q)\in M\times N$ that $$\pi_{1*}\times\pi_{2*}:T_{(p,q)}(M\times N)\to T_pM\times T_qN$$ is an isomorphism. Also $T_pM\times T_qN\cong T_pM \oplus T_q N$, which shows up in other texts like Lee.

The proof of this is actually in the back of Tu's book, which I assume you have or had, but I can supply it if requested.

Anyway the key is to use this and his proposition $8.17$ which shows how to compute differentials with curves. The usefulness of the above exercise is that we can write a tangent vector at $(a,b)$ as $(X_a,Y_b)$ rather than just something like $v$, as $T_{(a,b)}(G\times G)\cong T_aG\times T_bG$

For $Q1$ we use first use that $\mu_*$ is a linear map from $T_a G\times T_b G \to T_{\mu(a,b)}G$ so that $$\mu_{*,(a,b)}(X_a,Y_b)=\mu_{*,(a,b)}(X_a,0)+\mu_{*,(a,b)}(0,Y_b)$$

Then by proposition $8.17$ we can compute these with curves. For $\mu_{*,(a,b)}(X_a,0)$ consider a curve $c(t)$ in $G$ such that $c(0)=a$ and $c'(0)=X_a$, then we define a curve $\gamma(t)$ in $G\times G$ by $\gamma(t)=(c(t),b)$. Thus $$\mu_{*,(a,b)}(X_a,0)=\frac{d}{dt}\bigg|_0 \mu\circ\gamma=\frac{d}{dt}\bigg|_0 \mu(c(t),b)$$

On the other hand consider the right multiplication map by $r_b:G\to G$ by $a\mapsto \mu(a,b)$. Then $(r_{b})_*:T_a(G)\to T_{\mu(a,b)}G$. Again we can compute this by the same curve $c(t)$ so that $(r_{b})_*(X_a)=\dfrac{d}{dt}\bigg|_0 r_b\circ c$. We can then identify this with the above, since

$$(r_{b})_*(X_a)=\dfrac{d}{dt}\bigg|_0 r_b\circ c= \dfrac{d}{dt}\bigg|_0 \mu(c(t),b) $$

A similar argument with left multiplication shows that:

$$\mu_{*,(a,b)}(X_a,Y_b)=\mu_{*,(a,b)}(X_a,0)+\mu_{*,(a,b)}(0,Y_b)=(r_{b})_*(X_a)+(l_{a})_*(Y_b)$$

I will update when I have question $2$ worked out nicely.

Update(long overdue - apologies)

Given $\iota:G\to G$, $\iota(a)=a^{-1}$, and $\iota_{*,a}:T_a G\to T_{a^{-1}}G$, we are asked to show that $$\iota_{*,a}(Y_a)=-(r_{a^{-1}})(l_{a^{-1}})_*(Y_a)$$ where $(l_{a^{-1}})_*=(l_{a^{-1}})_{*,a}$ and $(r_{a^{-1}})_*=(r_{a^{-1}})_{*,e}$. Since we know what the answer is supposed to look like and we're reminded that in problem $8.8(b)^\&$ we proved that $\iota_{*,e}(X_e)=-X_e$, the trick is to notice that $\iota=r_{a^{-1}}\circ\iota\circ l_{a^{-1}}$ via checking that (using juxtaposition instead of writing the $\mu$'s) $$r_{a^{-1}}\circ\iota\circ l_{a^{-1}}(x)=(a^{-1}x)^{-1}a^{-1}=x^{-1}aa^{-1}=x^{-1}=\iota(x)$$

So using this and the chain rule we have that

$$\iota_{*,a}(Y_a)=(r_{a^{-1}})_{*,e}\circ\iota_{*,e}\circ (l_{a^{-1}})_{*,a}(Y_a)$$

Calling for convenience $X_e:=(l_{a^{-1}})_{*,a}(Y_a)$, and using $\iota_{*,e}(X_e)=-X_e$ and linearity of the differential/pushforward maps we have

$$\iota_{*,a}(Y_a)=(r_{a^{-1}})_{*,e}(-X_e)=-(r_{a^{-1}})_{*,e}(l_{a^{-1}})_{*,a}(Y_a)$$

$^\&$Proof of $8.8(b)$ for completeness sake:

The pushforward $\iota_{*,e}:T_e G\to T_e G$ of the inverse map can be computed as in the first question by choosing a curve. If we want to calculate $\iota_{*,e}(X_e)$, choose a curve $c(t)$ in $G$ such that $c(0)=e$ and $c'(0)=X_e$.

We know that $\iota_{*,e}(X_e)=\frac{d}{dt}\big|_0 \iota\circ c$ which we call $Y_e$ for convenience, that is set $Y_e=\iota_{*,e}(X_e)$. The hint given says to use that $\mu(c(t),\iota\circ c(t))=e$. This can be seen as a composition of the maps $\mu \circ \left(c\times (\iota\circ c)\right)$, which is a constant map from some (say) open interval $(a,b)\times (a,b)\to G\times G$.

Using that the differential of a constant map is the zero map, then $$\mu_{*,(e,e)}(X_e,Y_e)=\mu_{*,(e,e)}(X_e,0)+\mu_{*,(e,e)}(0,Y_e)=0$$

We just need that $\mu_{*,(e,e)}(X_e,0)=X_e$ and $\mu_{*,(e,e)}(0,Y_e)=Y_e$, from $8.8(a)$, which essentially follows similarly to problem 1, defining curves $\gamma(t)=(c(t),e)$ and $\tilde{\gamma}(t)=(e,\iota\circ c(t))$ $$\mu_{*,(e,e)}(X_e,0)=\frac{d}{dt}\bigg|_0 \mu\circ \gamma=\frac{d}{dt}\bigg|_0 \mu(c(t),e)=\frac{d}{dt}\bigg|_0 c=X_e$$

and

$$\mu_{*,(e,e)}(0,Y_e)=\frac{d}{dt}\bigg|_0 \mu\circ \tilde{\gamma}=\frac{d}{dt}\bigg|_0 \mu(e,\iota\circ c(t))=\frac{d}{dt}\bigg|_0 \iota\circ c=Y_e$$

so we have that $X_e+Y_e=0$ and so that

$$\iota_{*,e}(X_e)=Y_e=-X_e$$

Solution 2:

Here $\exp(tX)$ is a curve at identity $e$ whose tangent is $X\in T_eG$

(1) $$\frac{d}{dt} \mu (a,\exp(tY)b)= \frac{d}{dt} a(\exp (tY)b)=dl_a\ Y_b$$

$$\frac{d}{dt} \mu (a\exp(tX), b)= \frac{d}{dt} (a \exp (tX)) b=dr_b\ X_a$$ Hence $$ dl_a\ Y_b + dr_b\ X_a= d\mu\ \{ \frac{d}{dt} (a,\exp(tY)b) + \frac{d}{dt} (a\exp (tX),b) \} =d\mu \{ (0,Y_b) + (X_a,0) \} $$

(2) $$ 0= \frac{d}{dt}\mu (a\exp (tY), [\exp (tY)]^{-1} a^{-1} ) =\frac{d}{dt}\mu (a\exp (tY), i (a \exp (tY) ) ) = dr_{a^{-1}} Y_a + dl_a \ di_a\ Y_a $$ so that $$ di_a \ Y_a= - (dl_a)^{-1} dr_{a^{-1}}\ Y_a $$

( We will show that $(dl_a)^{-1}=dl_{a^{-1}} $ : $ dl_{a^{-1}} dl_a\ X = \frac{d}{dt} \exp (tX) =X $)