# determinant of curvature (notation issue)

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?

• Try reading the nice short paper of Chern, Vector bundles and connections, in the Global Analysis and Geometry book. – Ben McKay Dec 23 '18 at 11:52
• 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. – Ben McKay Dec 23 '18 at 11:55
• @BenMcKay Ok... $\gamma_{k}$ is complex valued $2k$ form on $M$... what does it then mean to say $[\gamma_k]\in H^{2k}(M,\mathbb{R})$? – Praphulla Koushik Dec 23 '18 at 12:34
• 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 – Praphulla Koushik Dec 23 '18 at 13:00
• @BenMcKay Can you see my answer and let me know if that is correct? – 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 $$p^*(\gamma_2)=\Omega_{11}\wedge\Omega_{12}-\Omega_{21}\wedge \Omega_{12}$$

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