Identity used to prove the Chern-Weil theorem

Question

I'm reading the proof of Chern-Weil theorem found in Nakahara's second edition book (page 422) but I got stucked at the very beginning of the proof. It says that from the identity $$\tilde{P}(Ad_{g_t}X_1,\ldots, Ad_{g_t}X_r)= \tilde{P}(X_1,\ldots,X_r)$$ where $g_t=\exp{tX}$, $\tilde{P}$ is an invariant polynomial, and $X, X_i\in\mathfrak{g}$, one can recover the identity $$\sum_{i=1}^r\tilde{P}(X_1,\ldots,[X_i,X],\ldots,X_r)=0$$ by differentiation at $t=0$. I understand that $${\frac{d}{dt}Ad_{g_t}X_i}\vert_{t=0}=[X_i,X]$$ but I still feel there is a chain rule missing when doing the differentiation of the first identity. Can someone please explain me with some degree of detail how is this differentiation done?

Answer 1

Ok, I undestood what is going on here. It actually a triviality, but a very confusing one. Everything is clear if one have in mind that $\tilde{P}$ is actually a multilinear function. Recall that the differential of a linear transformation is actually the same linear transformation. For example $$T:\mathbb{R^n}\rightarrow \mathbb{R}$$ then the differential $D(T):T_x\mathbb{R^n}\rightarrow T_{Tx}\mathbb{R}$ is given by the same transformation, i.e., $D(T)=T$. If there is another function $f:\mathbb{R}\rightarrow\mathbb{R^n}$ then $D(T\circ f)=D(T)\circ D(f)=TD(f)$. All this put together in the previous identity show that $$\begin{align} \frac{d}{dt}\tilde{P}(Ad_{g_t}X_1,\ldots,Ad_{g_t}X_r)&=\sum_{i=1}^rD_i\tilde{P}\left(\frac{d}{dt}Ad_{g_t}X_i\right) \\ &=\sum_{i=1}^r\tilde{P}\left(X_1,\ldots,\frac{d}{dt}Ad_{g_t}X_i,\ldots,X_r\right) \end{align}$$ from which the desired identity follows. Everything gets confusing because of the fact that a polynomial is, in general nonlinear but the associated multilinear function is linear.