This is when studying about Chern classes from Kobayashi and Nomizu.


Let $\pi:E\rightarrow M$ be a complex vector bundle with fibre $\mathbb{C}^r$ and Group $G=GL(r,\mathbb{C})$.


Let $p:P\rightarrow M$ be associated principal $G$ bundle. Let $\mathfrak{g}=\mathfrak{gl}(r,\mathbb{C})$ denote the Lie algebra of $G$.


Given $B\in \mathfrak{g}$, the determinant $\text{det}\left(\lambda I\_r-\frac{1}{2\pi\sqrt{-1}} B\right)$ is $\sum\_{k=0}^r a\_k\lambda^{r-k}$ for some $a\_k\in \mathbb{C}$. 


Given $B\in \mathfrak{g}$, we have $r$ elements in $\mathbb{C}$. Varying $B$ over $\mathfrak{g}$, gives $r$ functions $f\_k:\mathfrak{g}\rightarrow \mathbb{C}$.


We have 
$$\text{det}\left(\lambda I\_r-\frac{1}{2\pi\sqrt{-1}} X\right)=\sum\_{k=0}^r f\_k(X) \lambda ^{r-k}$$ 
These $f\_k:\mathfrak{gl}(r,\mathbb{C})\rightarrow \mathbb{C}$ are homogeneous, degree $k$ polynomial functions on $\mathfrak{gl}(r,\mathbb{C})$. I can recall what are polynomial functions on a vector space if some one needs it. 


These are $GL(r,\mathbb{C})$ invariant i.e., $f\_k(X)=f\_k(DXD^{-1})$ for all $D\in Gl(r,\mathbb{C})$. These $f\_k$ gives a symmetric, multilinear, $Gl(r,\mathbb{C})$ invariant $\mathbb{C}$ valued mappings $f\_k\in I\_{\mathbb{C}}^k(G)$. 


Let $\Gamma$ be a connection on $P(M,G)$ and $\Omega$ be its curvature form. This $f\_k$ gives a $\mathbb{C}$ valued $2k$ form $f\_k(\Omega)$ on $P$. Let $\gamma\_k$ be the unique $\mathbb{C}$ valued closed $2k$-form on $M$ such that $p^\*(\gamma\_k)=f\_k(\Omega)$. 


This gives $[\gamma\_k]\in H^{2k}(M,\mathbb{C})$. But $k$-th Chern class are supposed to take values in $H^{2k}(M,\mathbb{R})$. What is the obvious map $H^{2k}(M,\mathbb{C})\rightarrow H^{2k}(M,\mathbb{R})$ to look for to get an element in $H^{2k}(M,\mathbb{R})$?


We then have
$$\sum\_{k=0}^r f\_k(\Omega)=\sum\_{k=0}^rp^\*(\gamma\_k)=p^\*(1+\gamma\_1+\cdots+\gamma\_r)$$ 
Then,they (Kobayashi and Nomizu,page no $307$) write
$$\text{det}\left(I\_r-\frac{1}{2\pi \sqrt{-1}} \Omega\right)= p^\*(1+\gamma\_1+\cdots+\gamma\_r)$$
I see that they are just replacing $X$ with $\Omega$. But what does it mean to say determinant of $\Omega$? 


We have $\gamma\_k\in H^{2k}(M,\mathbb{R})$. So, $1+\gamma\_1+\cdots+\gamma\_r\in H^\*(M,\mathbb{R})$ which then imply that $p^\*(1+\gamma\_1+\cdots+\gamma\_r)\in H^\*(P,\mathbb{R})$. So, $\text{det}\left(I\_r-\frac{1}{2\pi i} \Omega\right)\in H^\*(P,\mathbb{R})$. How is it defined? 



determinant of curvature (notation issue)
Curvature $\Omega$ is a $\mathfrak{g}$ valued $2$-form on $P$ i.e., for each $p\in P$, we have $\Omega(p):T\_pP\times T\_pP\rightarrow \mathfrak{g}$. 


As $\mathfrak{g}$ is $\mathfrak{gl}(r,\mathbb{C})$, given $(v\_1,v\_2)\in T\_pP\times T\_pP$, we get a $r\times r$ matrix $\Omega(p)(v\_1,v\_2)=[a\_{ij}]$. 


Once we fix $(v\_1,v\_2)$ we get $a\_{ij}\in \mathbb{C}$, these depend on $(v\_1,v\_2)$. So, we have $$\Omega(p)(v\_1,v\_2)=[a\_{ij}(v\_1,v\_2)]$$ 
Here $a\_{ij}$ are functions $a\_{ij}:T\_pP\times T\_pP\rightarrow \mathbb{C}$ with $(v\_1,v\_2)\mapsto a\_{ij}(v\_1,v\_2)\in \mathbb{C}$. 


We can write $\Omega(p)(v\_1,v\_2)=[a\_{ij}(v\_1,v\_2)]$ as $\Omega(p)=[a\_{ij}]$. 


So, for each $p\in P$, $\Omega(p)$ is a matrix of functions $a\_{ij}:T\_pP\times T\_pP\rightarrow \mathbb{C}$. 


Thus, we can write $\Omega$ as an $r\times r $ matrix $[\Omega\_{ij}]$ where $\Omega\_{ij}$ is a $\mathbb{C}$ valued $2$ form on $P$ given by $\Omega\_{ij}(p):T\_pP\times T\_pP\rightarrow\mathbb{C}$ is the map $a\_{ij}:T\_pP\times T\_pP\rightarrow \mathbb{C}$. 


Thus, $I\_r-\frac{1}{2\pi\sqrt{-1}}\Omega$ is an $r\times r$ matrix whose $i,j$-th entry is $\delta\_{ij}-\frac{1}{2\pi \sqrt{-1}}\Omega\_{ij}$. 


Let $\omega,\tau$ be $p$-form and $q$-form respectively. Then, $\omega\wedge \tau=(-1)^{pq}\tau\wedge \omega$. We are only dealing with $2$-forms here. So, $pq=4$ and $\omega\wedge \tau=\tau\wedge \omega$ for each $\omega,\tau$. 


We have a matrix $I\_r-\frac{1}{2\pi\sqrt{-1}}\Omega$ whose entries are coming from a commutative ring (of $2$-forms). 


We can then talk about determinant of that matrix and $\text{det}(I\_r-\frac{1}{2\pi\sqrt{-1}}\Omega)$ is precisely that.


For simplicity, take $r=2$. Then,
$$\text{det}(I-\Omega)=\text{det}\begin{bmatrix}1-\Omega\_{11}&-\Omega\_{12}\\
-\Omega\_{21}&1-\Omega\_{22}\end{bmatrix}=1-\underbrace{(\Omega\_{11}+\Omega\_{22})}\_{2-form}+\underbrace{\Omega\_{11}\wedge\Omega\_{12}-\Omega\_{21}\wedge \Omega\_{12}}\_{4-form}$$
We then have 
$$p^\*(1+\gamma\_1+\gamma\_2)=1+\underbrace{p^\*(\gamma\_1)}\_{2-form}++\underbrace{p^\*(\gamma\_2)}\_{4-form}$$
So, it makes sense to say $\text{det}(I-\Omega)=p^\*(1+\gamma\_1+\gamma\_2)$ (I ignored $\frac{1}{2\pi \sqrt{-1}}$ to keep it simple).


We have $p^\*(\gamma\_1)=-\Omega\_{11}-\Omega\_{22}$ and $p^\*(\gamma\_2)=\Omega\_{11}\wedge\Omega\_{12}-\Omega\_{21}\wedge \Omega\_{12}$


