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Let $G$ be a compact Lie group and $P\rightarrow M$ be a principal $G$ bundle. We want to associate Chern classes for this bundle.

This article says in its second page that "A compact Lie group $G$ may be shown (with some work, using the theory of compact operators) to be none other than a closed subgroup of the Unitary group $U(n)$ of matrices
for some $n$". Thus, for structure group $G$ of $P(M,G)$ there exists an embedding $G\hookrightarrow U(n)$.

See the embedding $G\hookrightarrow U(n)$ as an action of $G$ on $U(n)$. Given a manifold $P'$ and an action of $G$ on $P'$ there is a notion of what is called an associated bundle whose fibres are $P'$. In case $P'$ is a Lie group $H$, we get a principal $H$ bundle. This $G$ action on $U(n)$ gives a principal $U(n)$ bundle $Q\rightarrow M$.

For this principal $U(n)$ bundle $Q\rightarrow M$, we fix a connection $\Gamma$ with curvature $\Omega$. We then get what is called Weil homomorphism $W:I(U(n))\rightarrow H^*(M,\mathbb{R})$.

We choose elements $f_k:\mathfrak{u}(n,\mathbb{C})\rightarrow \mathbb{R}$ from $I(U(n))$ such that $$\text{det}\left(\lambda I-\frac{1}{2\pi\sqrt{-1}}X\right)=\sum_{k=0}^n f_k(X)\lambda^{n-k}$$ The images $W(f_k)\in H^{2k}(M,\mathbb{R})$ are called the $k$-th Chern classes of $Q\rightarrow M$.

How do we define Chern classes of the principal $G$ bundle $P\rightarrow M$ that we have started with?

Is the $k$-th Chern class of $G$ bundle $P(M,G)$ the $k$-th Chern class of the $U(n)$ bundle $Q(M,U(n))$?

Does this depend on the embedding $G\hookrightarrow U(n)$ we have chosen?

In case it is dependent on the choice of $G\hookrightarrow U(n)$, what invariants of $P(M,G)$ does these give?


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