0

I don't understand the first paragraph of the proof of the Theorem 20.27 from Lee's Introduction to Smooth Manifolds. First, I don't understand why he uses $t=0$ in the derivate, I think that the map $t \mapsto \text{exp}\ tX$ satisfies $t=t_0$ and $X=X_{\text{exp}\ t_0X}$, so shouldn't the derivate be evaluated at $t_0$ instead of $0$? If If I were to carry out to do the computation, I would do this $$\left(\text{Ad}_*X\right)_M=d(L_M)_I\left(d\left(\text{Ad}\right)_eX_e\right)=d(L_M)_I\frac{d}{dt}\Bigg|_{t=0}\left(\text{Ad}\exp t X_e\right),$$ where $M\in GL(\mathfrak{g})$ and $I$ is the identity of $GL(\mathfrak{g})$, but I can't obtain the same expression as in the book.

And second, I don't understand why in the second equality he put the $Y$ inside the derivative:

$$\left(\frac{d}{dt}\Bigg|_{t=0}\text{Ad}\exp t X\right)Y = \frac{d}{dt}\Bigg|_{t=0}\left(\text{Ad}(\exp t X) Y\right)$$

IELaSalvia
  • 1

1 Answers1

1

On question 1: For a homomorphism between Lie groups (like $Ad:G\to GL(G)$) one easily proves (as you suggest) that the derivative in any point is determined by the derivative at the neutral element. So when one talks about the "derivative of a homomorphism" one usually means the derivative in the neutral element (which is a linear map between the Lie algebras of the groups). And this is what is computed in the proof of Lee's book.

For the second question, the abstract reason is that for any vector space $V$ and any $v\in V$ the map $GL(V)\to V$, $f\mapsto f(v)$ is the restriction of a linear map $L(V,V)\to V$. Hence for a smooth curve $f_t$ in $GL(V)$ you get $\frac{d}{dt}(f_t(v))=(\frac{d}{dt}f_t)(v)$. Alternatively, you can view $f_t$ as a matrix $(a_{ij}(t))$ with entries depending smoothly on $t$ and then $f'(t)=(a_{ij}'(t))$, which easily implies the claimed result.

Andreas Cap
  • 22,380