I'm trying to solve the following two problems:

Let $G$ be a Lie group with multiplication map $\mu\colon G\times G\to G$, and let $\ell_a\colon G\to G$ and $r_b\colon G\to G$ be left and right multiplication by $a$ and $b$ in $G$, respectively. Show that the differential of $\mu$ at $(a,b)\in G\times G$ is $$ \mu_{*, (a,b)}(X_a,Y_b) = (r_b)_*(X_a) + (\ell_a)_*(Y_b) \quad \text{for } X_a\in T_aG,\ Y_b\in T_bG. $$


Let $G$ be a Lie group with multiplication map $\mu\colon G\times G\to G$, inverse map $\iota\colon G\to G$, and identity element $e$. Show that the differential of the inverse map at $a\in G$, $$ \iota_{*,a} \colon T_aG \to T_{a^{-1}}G, $$ is given by $$ \iota_{*,a}(Y_a) = -(r_{a^{-1}})_* (\ell_{a^{-1}})_* Y_a, $$ where $(r_{a^{-1}})_* = (r_{a^{-1}})_{*,e}$ and $(\ell_{a^{-1}})_* = (\ell_{a^{-1}})_{*,a}$.

I am wondering if I should try to find curves through these points, or just apply either side to functions in $C^\infty(G)$? It looks like there are some questions on here about the differential of multiplication and inversion at the identity, but I'm trying to evaluate these at arbitrary points... the author (Tu) talks about the matrix exponential in this section but should I use that here? These Lie groups don't necessarily contain matrices.


4 Answers 4


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:


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


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


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

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


  • $\begingroup$ thanks very useful. I am trying to find the same result for exercize 2 using a slightly different approach but got stuck. I want to use a curve $c(t)$ in G whose value at $t_0$ is $a$ and whose differential is $X_a$ and use again the fact that $\mu(c(t),\iota\circ c(t))$ is the constant map. I then apply the result from exercise 1, getting $(r_{a^-1})_{*,a}X_a+(l_{a})_{*,a}\iota_{*,a}(X_a)=0$ and then just $\iota_{*,a}(X_a)=((l_a)_{*,a})^{-1}(r_{a^-1})_{*,a}X_a$. Where am I wrong? thanks $\endgroup$ Dec 25, 2020 at 11:25
  • $\begingroup$ actually $\iota_{*,a}(X_a)=-((l_a)_{*,a})^{-1}(r_{a^-1})_{*,a}X_a$. also, what is the point of the subscript $e$ opposed to $a$ for the differential of the right and left multiplication? thanks $\endgroup$ Dec 25, 2020 at 11:33

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 $)


I know this is an old question. Here is my elementary solution, which does not make use of curves.

We start with a theorem:

Theorem: Let $M$ and $N$ be smooth manifolds. Let $p\in M$ and $q\in N$. Let $$\pi_1: M\times N \to M, \quad \pi_2: M\times N \to N$$ be the projection maps. Let $$\eta_1: M\to M\times N: m\mapsto (m,q), \quad \eta_2: N \to M\times N: n \mapsto (p,n)$$ be the inclusion maps. We have canonical maps $$\varphi: T_{(p,q)}(M\times N) \to T_p M \times T_q N: z \mapsto ((\pi_1)_{*, (p,q)}(z), (\pi_2)_{*, (p,q)}(z))$$ $$\psi: T_p M\times T_q N\to T_{(p,q)}(M\times N): (x,y) \mapsto (\eta_1)_{*,p}(x)+ (\eta_2)_{*,q}(y).$$ Then $\varphi$ and $\psi$ are inverse to each other.

Proof: Just calculate a composition using the chain rule.

Corollary: Let $G$ be a Lie group. If $p,q\in G$, and $l_p: G \to G$ is left multiplication and $r_q: G \to G$ is right multiplication in $G$ and $\kappa = \mu_{*,(p,q)}\circ \psi$ is the composition $$T_p G \times T_q G \cong T_{(p,q)}(G\times G)\to T_{pq}(G)$$ then $$\kappa(X_p, X_q) = (r_q)_{*,p}(X_p) + (l_p)_{*,q}(Y_p).$$ In particular, taking $p=q=e$ yields $\kappa(X_e, Y_e) = X_e + Y_e.$

Proof: Note that $\mu \circ \eta_1 = r_q, \mu \circ \eta_2 = l_p$. Hence, $$\kappa(X_p, X_q) = \mu_{*,(p,q)}((\eta_1)_{*,p}(X_p) + (\eta_2)_{*,q}(X_q))= (r_q)_{*,p}(X_p) + (l_p)_{*,q}(X_q)$$ by the chain rule.


Here is a proof for question (b):
Define $K=id\times \ell:G\rightarrow G\times G$, where "$id$" is the identity. Then $F=\mu\circ K$ is a constant map towards unit. Hence, $\forall X_{e}\in T_{e}G$: \begin{align} (\mathrm{d}F)_{e}(X_{e})&=\mathrm{d}\mu_{(e,e)}\circ (\mathrm{d}K)_{e}(X_{e})\\ &=\mathrm{d}\mu_{(e,e)}((X_{e},(\mathrm{d}\ell)_{e}(X_{e})))\\ &=X_{e}+(\mathrm{d}\ell)_{e}(X_{e})=0 \end{align} The result follows.

  • $\begingroup$ What is $i\text{d}_e(X)$? And why redefine all the notation from the original problem? $\endgroup$
    – semper-lux
    Mar 14, 2022 at 23:10
  • 1
    $\begingroup$ Sorry, it is a typo. Actually, I mean $(di)_{e}(X_{e})$. And I have adapt my notations to the orginal problem now. $\endgroup$ Mar 15, 2022 at 2:37

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