# In the proof of showing $\operatorname{Ad}_{*} =\operatorname{ad}$ (adjoint representations of Lie Group(Algebra).

I'am reading the Lee's Introduction to smooth manifolds, 2nd edition, p.534, Theorem 20.27 and stuck at understanding some equality :

Why the underlined equality is true? Here, $$\operatorname{ad} : \mathfrak{g} \to \mathfrak{gl(g)}$$ is the adjoint representation of $$\mathfrak{g}$$ and $$\operatorname{Ad}_{*} :\mathfrak{g} \to \mathfrak{gl(g)}$$ is the induced Lie algebra representation which is defined as follows (His book p.195) :

Here, the 'L' in the above underlined definition of $$Y$$ is defined as follows (His book, p.191) : Let $$G$$ be a Lie group. Let $$v\in T_{e}G$$ be arbitrary. Then define a (left invariant) vector field $$v^{L}$$ on $$G$$ by

$$v^{L}|_g = d(L_g)_{e}(v)$$.

Note that by the proof of his book, Theorem 8.37 (p.191), the correspondence $$L$$ is an isomorphism $$T_eG \to \operatorname{Lie}(G) =: \mathfrak{g}$$ with inverse $$\epsilon : \operatorname{Lie}(G) \to T_eG$$, given by $$\epsilon(X) := X_e$$.

It seems that our question is somewhat basic but I don't know how to establish the equality rigorously. What should I notice?

First attempt : Let $$H:=GL(\mathfrak{g})$$ and let $$\mathfrak{h}:=\mathfrak{gl(g)}$$ be its Lie algebra. Let $$\gamma : t \mapsto \operatorname{exp}tX$$.

Then $$\frac{d} {dt} |_{t=0} (\operatorname{Ad}(\operatorname{exp}tX)) = (\operatorname{Ad} \circ \gamma)^{'}(0) = d(\operatorname{Ad})_{\gamma(0)}(\gamma^{'}(0)) = d(\operatorname{Ad})_{e}(X_e) \in T_eH$$.

And, $$\operatorname{Ad}_{*}(X) = (d(\operatorname{Ad})_{e}(X_e))^{L} \in \operatorname{Lie}H =: \mathfrak{h} = \mathfrak{gl(g)}$$

And why the difference appear? I think that for the first equality, more correct form is,

$$(\operatorname{Ad}_{*}X)Y = (\frac{d} {dt} |_{t=0} (\operatorname{Ad}(\operatorname{exp}tX)))^{L}(Y)$$

Did I make point out well?

If so, it remains(the second equality) to show that

$$(\frac{d} {dt} |_{t=0} (\operatorname{Ad}(\operatorname{exp}tX)))^{L}(Y) = \frac{d} {dt} |_{t=0}(\operatorname{Ad}(\operatorname{exp}tX)Y)$$

How can we show this? How the passage of $$Y$$ into the derivative $$\frac{d}{dt}|_{t=0}$$ ( ; i.e., the second equality) possible? If we can use the Chain rule, how can we apply the chain rule more rigorously?

Can anyone helps?

• The first equals sign is basically the definition of the induced homomorphism and the second is just chain rule. Aug 30, 2022 at 14:15
• My question is, "the 'L' really apears in the $\operatorname{Ad}_{*}(X)$, as formula above? If so, on the other side, why L doesn't apears in for formula for $\frac{d} {dt} |_{t=0} \operatorname{Ad}(\operatorname{exp}tX))$." And for the second equality, what chain rule you mentioned exactly means? I know a version of chain rule for multivariable calculus. And in our case, what does it means? Where can I find assoicated data/reference? Sep 1, 2022 at 3:06
• So for the first point $X$ and $X^L$ are being identified here. Sep 1, 2022 at 21:02
• @Callum : Can you explain how to apply the Chain rule? Jan 1, 2023 at 0:55

Here are my thoughts, which are too long to write down in the comment section. I believe what follows is substantially correct, with "holes"perhaps. I would appreciate any corrections/feedback.

Ad maps $$G$$ into $$h=GL(\frak g)$$ so Ad$$_*$$ maps $$\frak g$$ into $$\frak h$$ which is just $$GL(\frak g)$$ itself with the Lie bracket, and is written $$\frak gl(g).$$

Now, $$X\in \frak g$$ is identified with $$X_e$$ because of the $$X\to X_e$$ isomorphism of $$\frak g$$ with $$T_eG.$$ Similarly, Ad$$_*X$$ as an element of $$\frak gl(g)$$ is identified with an element of the tangent space at the identity in the tangent bundle of $$\frak gl(g),$$ which is just an element of $$\frak gl(g)$$ itself because the latter is a vector space. And so the formula Ad$$_*X(Y)$$ makes sense.

We also know that $$X$$ is a left-invariant vector field on $$M$$ so there is a one-parameter subgroup $$\gamma$$ which satisfies $$X_{\gamma(t)}=\gamma'(t).$$ And of course, $$\gamma(t)=\exp tX.$$

So, by definition of the pushforward, with which the induced Lie algebra representation is identified (I have not verified the validity of the identification), and with test function $$f$$, we get

Ad$$_*X_ef=$$Ad$$_*X_{\gamma(0)}f=$$Ad$$_*\gamma'(0)f=\frac{d}{dt}|_{t=0}f($$Ad$$\circ \gamma(t))=\frac{d}{dt}|_{t=0}f($$Ad$$(\exp tX)).$$

Hence, the first equality in the formula is correct.

Edit: here is a more careful proof:

$$c^g:G\to G$$ is defined by $$c^g(x)=g^{-1}xg.$$ Henceforth, we simply write this as $$c.$$

$$\operatorname{Ad}:G\to \operatorname{GL}(\frak g)$$ is defined by $$\operatorname{Ad}(g)=c_*:\frak g \to \frak g,$$ the induced Lie Algebra homomorphism.

$$\operatorname{ad}:\frak g\to \operatorname{Hom}(\frak g, \frak g)=\frak gl(g)$$ is defined by $$\operatorname{ad}(X)=\operatorname{Ad_*}(X).$$

$$\operatorname{Hom}(\frak g, \frak g)$$ is the Lie algebra of the Lie group $$\operatorname{GL}(\frak g)$$ in the same way as the Lie algebra $$gl_n(\mathbb{C})$$ of the general linear group $$\operatorname{GL}_n(\mathbb C)$$ can be identified with the space $$M_n(\mathbb C)$$ of $$n×n$$ complex matrices, with the Lie bracket $$[A,B]=AB−BA$$.

The key identification is in what follows. By definition,

$$\operatorname{Ad_*}(X)=(d\operatorname{Ad}_e(X_e))^L: \frak g\to \frak gl(g)$$

which is the same as

$$d\operatorname{Ad}_e(X):T_eG\to T_I(\operatorname{GL}(\frak g)).$$

Hence,

$$\operatorname{Ad_*}(X)=d\operatorname{Ad}_e(X)=\frac{d}{dt}|_{t=0}\operatorname{Ad}(\exp tX)dt.$$

For the second part, it's just a matter of unwinding the definitions:

Fix $$X\in \frak g.$$ We have $$\operatorname{Ad}\exp 0X=\operatorname{Ad}e=I$$ (because $$\operatorname{Ad}$$ is a homomorphism.) Write $$F:=\operatorname{Ad}\exp,$$ take a $$Y\in \frak g$$ and compute, at $$t=0$$, the difference quotient:

$$\tag1 \left(n\left(\left(F(t+\frac{1}{n}\right)X-I\right)\right)Y=n\left(\left(F(t+\frac{1}{n}\right)X)(Y)-Y\right).$$

Now, define $$F(tX)=\tilde F_X(t)$$ and notice that $$\tilde F_X(t)\in \operatorname{GL}(\frak g).$$ Using these facts, rewrite $$(1):$$

$$\tag2 n\left(\tilde F_X\left(t+\frac{1}{n}\right)-I\right)Y=n\left(\tilde F_X\left(t+\frac{1}{n}\right)Y-Y\right).$$

The limit on the left-hand side is

$$\tag3 \left(\frac{d}{dt}|_{t=0}\tilde F_X(t)\right)Y=\left(\frac{d}{dt}|_{t=0}\operatorname{Ad}\exp tX\right)Y$$

so the limit on the right-hand side exists and is equal to it. Now notice that $$\tilde F_X(0)Y=Y$$ so the limit on the right-hand side of $$(2)$$ is

$$\tag4 \frac{d}{dt}|_{t=0}\tilde F_X(t)Y=\frac{d}{dt}|_{t=0}\left(\operatorname{Ad}\exp X\right)Y$$

• O.K. Thank you. For the first equality, I've edited my question after some thought. Can you see? And For the second equality : in your answer, what is $h$? And how can we apply your comment to our situation? In our situation, what will be $F$, $G$, and $L$? Dec 31, 2022 at 6:52
• C.f. Precisely, symbols $(\operatorname{Ad}(\operatorname{exp}tX))_e$ and $\operatorname{Ad}(\operatorname{exp}tX)^{L}|_g$ are nonsense. :) Because $\operatorname{Ad}(\operatorname{exp}tX)$ is neither included in $\mathfrak{g}$ nor $T_eG$. Jan 1, 2023 at 1:06
• O.K. For the second part, regarding the chain rule, I think it needs some more thought. A priori, it seems possible ( since $Y$ does not depends on $t$) but I couldn't make a formal proof. Now I am trying to apply the Loring Tu's p.88's version of Chain rule (Theorem 8.5). Can we apply the theorem successfully ? If so, how? Jan 1, 2023 at 1:12
• I have added a more careful proof to my answer. I will delete all my previous comments. Jan 1, 2023 at 5:32
• O.K. I think I should point out few things. First, you defined $\operatorname{ad}$ as $\operatorname{Ad}_{*}$. But in john lee's book, $\operatorname{ad}$ is defined differently, and $\operatorname{ad} = \operatorname{Ad}_{*}$ is objective/goal of the his book Theorem 20.27, not a definition. Second, following his book (e.g. Theorem 8.44 as above image), $\operatorname{Ad}_{*}(X)$ is an element of $\mathfrak{gl}(\mathfrak{g})$. I don' know how can you view $\operatorname{Ad}_{*}(X)$ as $\mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$. Jan 1, 2023 at 5:58