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?

  • $\begingroup$ Try reading the nice short paper of Chern, Vector bundles and connections, in the Global Analysis and Geometry book. $\endgroup$ – Ben McKay Dec 23 '18 at 11:52
  • 1
    $\begingroup$ Since $\Omega$ is a matrix of 2-forms, and 2-forms commute with one another, they sit in a commutative algebra of even degree forms, so use your favourite explicit formula for determinant in any commutative ring. Reality comes from the fact that you could have picked a reduction of structure group to the unitary group, and computed there instead, giving the same cohomology classes. This is in a lot of reference works, like Griffiths and Harris for example. $\endgroup$ – Ben McKay Dec 23 '18 at 11:55
  • $\begingroup$ @BenMcKay Ok... $\gamma_{k}$ is complex valued $2k$ form on $M$... what does it then mean to say $[\gamma_k]\in H^{2k}(M,\mathbb{R})$? $\endgroup$ – Praphulla Koushik Dec 23 '18 at 12:34
  • $\begingroup$ My problem is I am comfortable with connections on principal bundles and those two references uses connections on vector bundles... Can you think of some book/paper where this is done using principal bundles.. ? @BenMcKay $\endgroup$ – Praphulla Koushik Dec 23 '18 at 13:00
  • $\begingroup$ @BenMcKay Can you see my answer and let me know if that is correct? $\endgroup$ – Praphulla Koushik Dec 23 '18 at 13:41

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 $p^*(\gamma_2)=\Omega_{11}\wedge\Omega_{12}-\Omega_{21}\wedge \Omega_{12}$

  • $\begingroup$ This is done on suggestion of Ben (in comments) $\endgroup$ – Praphulla Koushik Dec 23 '18 at 13:31

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