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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.

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  • $\begingroup$ @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 $U(n)$? $\endgroup$
    – user537667
    Dec 23, 2018 at 16:42
  • $\begingroup$ 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 $\mathrm U_n$. $\endgroup$
    – Arun Debray
    Dec 23, 2018 at 17:03
  • $\begingroup$ @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 $U_\alpha\bigcap U_\beta\rightarrow U(n)$... $\endgroup$
    – user537667
    Dec 23, 2018 at 17:18
  • $\begingroup$ 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) $\endgroup$
    – Arun Debray
    Dec 23, 2018 at 17:28
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    $\begingroup$ 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? $\endgroup$
    – user537667
    Dec 23, 2018 at 17:34

1 Answer 1

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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 $Q\rightarrow M$?

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  • $\begingroup$ Any comments are welcome for this. $\endgroup$
    – user537667
    Dec 25, 2018 at 19:01
  • $\begingroup$ 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. $\endgroup$
    – Andrew
    Dec 25, 2018 at 19:29
  • $\begingroup$ @Andrew Thank you. I will see that. Do you have any suggestion on what I have written.. $\endgroup$
    – user537667
    Dec 26, 2018 at 1:57

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