# Chern class cohomology coefficients complex/real/integral?

I am reading Chern classes from Kobayashi and Nomizu.

Given a vector bundle $$\pi:E\rightarrow M$$ with fibre $$\mathbb{C}^r$$ and Group $$GL(r,\mathbb{C})$$ they associate for each $$k\leq r$$ a cohomology class of $$M$$ and call it the $$k$$-th Chern class. It looks like cohomology is $$H^*(M,\mathbb{C})$$ and not $$H^*(M,\mathbb{R})$$. Can some one clarify what is happening here?

Given $$\pi:E\rightarrow M$$, let $$p:P\rightarrow M$$ be the associated $$GL(r,\mathbb{C})$$ bundle. We have $$\text{det}\left(\lambda I-\frac{1}{2\pi\sqrt{-1}}X\right)=\sum_{k=0}^rf_k(X)\lambda^{r-k}$$ for $$X\in \mathfrak{gl}(r,\mathbb{C})$$. Here $$f_k:\mathfrak{gl}(r,\mathbb{C})\rightarrow \mathbb{C}$$ are $$GL(r,\mathbb{C})$$ invariant degree $$k$$ homogeneous poly. on $$\mathfrak{gl}(r,\mathbb{C})$$. These $$f_k$$ can be seen as $$GL(r,\mathbb{C})$$ invariant symmetric multilinear map $$\underbrace{\mathfrak{gl}(r,\mathbb{C})\times\cdots\times\mathfrak{gl}(r,\mathbb{C})}_{k-\text{times}}\rightarrow \mathbb{C}$$ giving an element of $$I_{\mathbb{C}}(G)$$.

After fixing a connection on the principal bundle $$P(M,G)$$, there is a complex valued version of Weil homomorphism $$I_{\mathbb{C}}(G)\rightarrow H^*(M,\mathbb{C})$$. These $$f_k\in I^k_{\mathbb{C}}(G)$$ gives an element $$c_k$$ in $$H^{2k}(M,\mathbb{C})$$. But, they write $$c_k\in H^{2k}(M,\mathbb{R})$$.

What am I missing here?

Does it mean $$c_k\in H^{2k}(M,\mathbb{C})$$ is image of some element in $$H^{2k}(M,\mathbb{R})$$ under some map $$H^{2k}(M,\mathbb{R})\rightarrow H^{2k}(M,\mathbb{C})$$? One which induce from $$\mathbb{R}\rightarrow \mathbb{C}$$ defined as $$a\mapsto a+i 0$$?

EDIT : The book Calculus to cohomology by Ib Madsen and Jxrgen Tornehave says in remark $$18.12$$ (page $$189$$) that

Definition $$18.3$$ (of Chern class) gives cohomology classes in $$H^*(M,\mathbb{C})$$, but actually all classes lies in Real cohomology. This follows from (some result before).

There was no clear explanation (for me) for that comment.

EDIT : User Jessica L (last seen 7 years ago) said

Chern classes can be defined by topological means (see Milnor's book on characteristic classes), which yields elements $$c_k(V) \in H^{2k}(M;\mathbb{Z})$$. The normalization in the Chern-Weil theory is chosen so that the associated elements of de Rham cohomology groups $$H^{2k}(M;\mathbb{R})$$ agree with the integral elements, and thus integrate to give integers.

I think this answer and my question are related. So, any reference (which contains more details) for this are also welcome.

EDIT : Kobayashi and Nomizu (Foundations of Differential geometry) in page $$59$$ says the following.

Let $$P(M,G)$$ be a principal fibre bundle over a paracompact manifold $$M$$ with group $$G$$ which is a connected Lie group. It is known that $$G$$ is diffeomorphic with a direct product of any of its maximal compact subgroups $$H$$ and a Euclidean space (Iwasawa). By the same reasoning as above the structure group cann be reduced to $$H$$.

See that $$GL(r,\mathbb{C})$$ is a connected Lie group, $$U(r,\mathbb{C})$$ is a maximal compact group and this says that $$GL_r(\Bbb C) \cong U(r) \times \Bbb R^{r^2}$$. Using above result, we see that structure group $$GL(r,\mathbb{C})$$ for $$P(M,G)$$ can be reduced to $$U(r,\mathbb{C})$$.

Edit : As structure group $$GL(n,\mathbb{C})$$ of $$P\rightarrow M$$ can be reduced to $$U(n)$$, we get a principal $$U(n)$$ bundle $$Q\rightarrow M$$. Now, Lie algebra of $$U(n)$$ is $$\mathfrak{u}(n)$$ of skew Hermitian matrices. For $$X\in \mathfrak{u}(n)$$, I believe (I checked it for some examples) that $$\frac{1}{2\pi \sqrt{-1}}X$$ has characteristic polynomial with real coefficients. So, we have $$\text{det}\left(\lambda I-\frac{1}{2\pi\sqrt{-1}}X\right)=\sum_{k=0}^rf_k(X)\lambda^{r-k}$$ for $$X\in \mathfrak{u}(r,\mathbb{C})$$. Here $$f_k:\mathfrak{u}(r,\mathbb{C})\rightarrow \mathbb{R}$$. I now consider Chern Weil Homomorphism for $$Q\rightarrow M$$ and these $$f_k$$ give real cohomology classes $$H^{2k}(M,\mathbb{R})$$. Thus we get deRham cohomology classes with real coefficients and not just complex coefficients.

• @PhilTosteson Can you give a reference where this is done or can you give an outline as an answer... How do I know that the structure group can be reduced to $U(n)$?
– user537667
Dec 23, 2018 at 16:42
• The reason that every complex vector bundle admits a reduction to $\mathrm U_n$ is because of the Gram-Schmidt process in linear algebra. Working in a trivializing open cover, local bases define transition functions into $\mathrm{GL}_n(\mathbb C)$; then, apply Gram-Schmidt to the local bases to obtain unitary bases. Therefore the transition functions, which are given by change-of-basis matrices, are valued in $\mathrm U_n$.
– Arun Debray
Dec 23, 2018 at 17:03
• @ArunDebray For principal $GL(n,\mathbb{C})$ bundle $P\rightarrow M$, we have an open cover $\{U_\alpha\}$ of $M$ and transition functions $g_{\alpha\beta}:U_\alpha\bigcap U_\beta\rightarrow Gl(n,\mathbb{C})$. I do not completely understand "apply Gram-Schmidt to the local bases to obtain unitary bases"... Can you explain what this is... We want functions $U_\alpha\bigcap U_\beta\rightarrow U(n)$...
– user537667
Dec 23, 2018 at 17:18
• One way to say this is that the Gram-Schmidt process describes a continuous map $\mathit{GS}\colon \mathrm{GL}_n(\mathbb C)\to\mathrm U_n$, given by the explicit formulas in its Wikipedia article: an element of $\mathrm{GL}_n(\mathbb C)$ is a matrix whose columns are a basis. Applying the formulas to those basis vectors, one obtains a unitary basis, so putting those vectors into a matrix, one obtains a unitary matrix. The new transition functions are $\mathit{GS}\circ g_{\alpha\beta}\colon U_\alpha\cap U_\beta\to\mathrm U_n$. (1/2)
– Arun Debray
Dec 23, 2018 at 17:28
• you need it to be smooth so that the composition with transitin functions is smooth so you get a principal bundle... Am I thinking wrong?
– user537667
Dec 23, 2018 at 17:34

Let $$E\rightarrow M$$ be a complex vector bundle with fibre $$\mathbb{C}^r$$ and structure group $$Gl(r,\mathbb{C})$$. For this we want to associate Chern classes. We want it to be elements in $$H^*(M,\mathbb{R})$$.

Chern-Weil theory (as in Kobayashi and Nomizu's Foundations of Differential geometry) does that in following way.

• Given $$E\rightarrow M$$ as above, consider associated Principal $$GL(r,\mathbb{C})$$ bundle $$P\rightarrow M$$.
• See that the structure group $$GL(r,\mathbb{C})$$ of $$P\rightarrow M$$ can be reduced to $$U(n)$$ i.e., there exists a principal $$U(n)$$ bundle $$Q\rightarrow M$$ that is reduced from $$P\rightarrow M$$.
• Consider Weil-homomorphism $$I(U(n))\rightarrow H^*(M,\mathbb{R})$$ for $$Q\rightarrow M$$.

See that $$\mathfrak{u}(r,\mathbb{C})$$ is the Lie algebra of $$U(r,\mathbb{C})$$.

Let $$X\in \mathfrak{u}(r,\mathbb{C})$$ and consider its characteristic polynomial of $$X$$ i.e., $$\text{det}(\lambda I-X)\in \mathbb{C}[\lambda]$$ (polynomial in $$\lambda$$ with coefficients from $$\mathbb{C}$$).

Let $$X\in \mathfrak{u}(r,\mathbb{C})$$. As $$X$$ is skew-Hermitian, $$-iX=\frac{1}{i}X$$ is Hermitian which then imply $$\frac{1}{2\pi\sqrt{-1}}X$$ is Hermitian i.e., characteristic polynomial of $$\frac{1}{2\pi\sqrt{-1}}X$$ is with real coefficients. Thus, we have $$\text{det}\left(\lambda I-\frac{1}{2\pi\sqrt{-1}}X\right)=\sum_{k=0}^ra_k \lambda^{r-k}$$ such that $$a_k\in \mathbb{R}$$. Thus, for $$X\in \mathfrak{u}(r,\mathbb{C})$$ we have $$r+1$$ real numbers. Varying $$X$$ over $$\mathfrak{u}(r,\mathbb{C})$$ we get $$r+1$$ real valued functions $$f_k:\mathfrak{u}(r,\mathbb{C})\rightarrow \mathbb{R}$$ with $$\text{det}\left(\lambda I-\frac{1}{2\pi\sqrt{-1}}X\right)=\sum_{k=0}^rf_k(X) \lambda^{r-k}$$ for $$X\in \mathfrak{u}(r,\mathbb{C})$$. It is to be noted that, these $$f_k$$ are homogeneous $$GL(r,\mathbb{C})$$ polynomials.

I think the main reason to consider $$\frac{1}{2\pi i}X$$ and not $$X$$ to get $$f_k:\mathfrak{u}(r,\mathbb{C})\rightarrow \mathbb{R}$$ and not just $$f_k:\mathfrak{u}(r,\mathbb{C})\rightarrow \mathbb{C}$$. These $$f_k:\mathfrak{u}(r,\mathbb{C})\rightarrow \mathbb{R}$$ under Weil homomorphism gives cohomology classes with real coefficients i.e., in $$H^*(M,\mathbb{R})$$ where as $$f_k:\mathfrak{u}(r,\mathbb{C})\rightarrow \mathbb{C}$$ under Weil homomrphism gives cohomology classes with complex coefficients i.e., in $$H^*(M,\mathbb{C})$$. So, we need $$f_k:\mathfrak{u}(r,\mathbb{C})\rightarrow \mathbb{R}$$, for that we consider characteristic polynomial of $$\frac{1}{i}X$$ and not just $$X$$.

It is another discussion why we condiser $$\frac{1}{2\pi}$$ multiple for $$\frac{1}{\sqrt{-1}}X$$ and not just $$\frac{1}{\sqrt{-1}}X$$ if all we want is maps $$f_k:\mathfrak{u}(r,\mathbb{C})\rightarrow \mathbb{R}$$. I read it is to normalize but I do not understand completely. That is not important for now.

So, by above process, we get Chern classes as elements of $$H^*(M,\mathbb{R})$$.

I do not know how does $$Q\rightarrow M$$ affect the assignment of Chern classes? There might be another reduction $$Q'\rightarrow M$$ of $$P(M,G)$$ with structure group $$U(r)$$. Suppose we get Chern classes $$c'_k\in H^{2k}(M,\mathbb{R})$$ for this reduction $$Q'\rightarrow M$$, Is it the same thing as Chern classes $$c_k\in H^{2k}(M,\mathbb{R})$$ that came from $$Q\rightarrow M$$?

• Any comments are welcome for this.
– user537667
Dec 25, 2018 at 19:01
• There is a reason for the factor of $2\pi$. You want $\int_{\mathbb{C}P^1} c_1(\mathbb{C}P^1) = 1$. One finds that a 2-form representative of $c_1(\mathbb{C}P^1)$ is the Fubini-Study metric. Integrating it over $\mathbb{C}P^1$ requires a factor of $2\pi$ to make sure it comes out to $1$. See Huybrechts' book Complex Geometry, chapters 3.1 and 4.4 for a bit more info. Dec 25, 2018 at 19:29
• @Andrew Thank you. I will see that. Do you have any suggestion on what I have written..
– user537667
Dec 26, 2018 at 1:57