Contents

cohomology

# Contents

## Idea

The ordinary Chern classes are the integral characteristic classes

$c_i : B U \to B^{2 i} \mathbb{Z}$

of the classifying space $B U$ of the unitary group.

Accordingly these are characteristic classes in ordinary cohomology of U-principal bundles and hence of complex vector bundle

The first Chern class is the unique characteristic class of circle group-principal bundles.

The analogous classes for the orthogonal group are the Pontryagin classes.

More generally, there are generalized Chern classes for any complex oriented cohomology theory (Adams 74, Lurie 10).

## Definition

###### Definition

For $n \geq 1$ the universal Chern classes

$c_i \;\in\; H^{2i} \big( B U(n), \mathbb{Z} \big)$

of the classifying space $B U(n)$ of the unitary group are the cohomology classes of $B U(n)$ in integral cohomology that are characterized as follows:

1. $c_0 = 1$ and $c_i = 0$ if $i \gt n$;

2. for $n = 1$, $c_1$ is the canonical generator of $H^2(B U(1), \mathbb{Z})\simeq \mathbb{Z}$;

3. under pullback along the inclusion $i : B U(n) \to B U(n+1)$ we have $i^* c_i^{(n+1)} = c_i^{(n)}$;

4. under the inclusion $B U(k) \times B U(l) \to B U(k+l)$ we have $i^* c_i = \sum_{j = 0}^i c_i \cup c_{j-i}$.

The corresponding total Chern class is the formal sum

$c \;\coloneqq\; 1 + c_1 + c_2 + \cdots \;\in\; \underset{k}{\prod} H^{2k} \big( B U(n) \big)$

## Properties

### Existence

###### Proposition

The cohomology ring of the classifying space $B U(n)$ (for the unitary group $U(n)$) is the polynomial ring on generators $\{c_k\}_{k = 1}^{n}$ of degree 2, called the Chern classes

$H^\bullet(B U(n), \mathbb{Z}) \simeq \mathbb{Z}[c_1, \cdots, c_n] \,.$

Moreover, for $B i \colon B U(n_1) \longrightarrow BU(n_2)$ the canonical inclusion for $n_1 \leq n_2 \in \mathbb{N}$, then the induced pullback map on cohomology

$(B i)^\ast \;\colon\; H^\bullet(B U(n_2)) \longrightarrow H^\bullet(B U(n_1))$

is given by

$(B i)^\ast(c_k) \;=\; \left\{ \array{ c_k & for \; 1 \leq k \leq n_1 \\ 0 & otherwise } \right. \,.$
###### Proof

For $n = 1$, in which case $B U(1) \simeq \mathbb{C}P^\infty$ is the infinite complex projective space, we have (prop)

$H^\bullet(B U(1)) \simeq \mathbb{Z}[ c_1 ] \,,$

where $c_1$ is the first Chern class. From here we proceed by induction. So assume that the statement has been shown for $n-1$.

Observe that the canonical map $B U(n-1) \to B U(n)$ has as homotopy fiber the (2n-1)sphere (prop.) hence there is a homotopy fiber sequence of the form

$S^{2n-1} \longrightarrow B U(n-1) \longrightarrow B U(n) \,.$

Consider the induced Thom-Gysin sequence.

In odd degrees $2k+1 \lt 2n$ it gives the exact sequence

$\cdots \to H^{2k}(B U(n-1)) \longrightarrow \underset{\simeq 0}{\underbrace{H^{2k+1-2n}(B U(n))}} \longrightarrow H^{2k+1}(B U(n)) \overset{(B i)^\ast}{\longrightarrow} \underset{\simeq 0}{\underbrace{H^{2k+1}(B U(n-1))}} \to \cdots \,,$

where the right term vanishes by induction assumption, and the middle term since ordinary cohomology vanishes in negative degrees. Hence

$H^{2k+1}(B U(n)) \simeq 0 \;\;\; for \; 2k+1 \lt 2n$

Then for $2k+1 \gt 2n$ the Thom-Gysin sequence gives

$\cdots \to H^{2k+1-2n}(B U(n)) \longrightarrow H^{2k+1}(B U(n)) \overset{(B i)^\ast}{\longrightarrow} \underset{\simeq 0}{\underbrace{H^{2k+1}(B U(n-1))}} \to \cdots \,,$

where again the right term vanishes by the induction assumption. Hence exactness now gives that

$H^{2k+1-2n}(B U(n)) \overset{}{\longrightarrow} H^{2k+1}(B U(n))$

is an epimorphism, and so with the previous statement it follows that

$H^{2k+1}(B U(n)) \simeq 0$

for all $k$.

Next consider the Thom Gysin sequence in degrees $2k$

$\cdots \to \underset{\simeq 0}{\underbrace{H^{2k-1}(B U(n-1))}} \longrightarrow H^{2k-2n}(B U(n)) \longrightarrow H^{2k}(B U(n)) \overset{(B i)^\ast}{\longrightarrow} H^{2k}(B U(n-1)) \longrightarrow \underset{\simeq 0}{\underbrace{H^{2k +1 - 2n}(B U(n))}} \to \cdots \,.$

Here the left term vanishes by the induction assumption, while the right term vanishes by the previous statement. Hence we have a short exact sequence

$0 \to H^{2k-2n}(B U(n)) \longrightarrow H^{2k}(B U(n)) \overset{(B i)^\ast}{\longrightarrow} H^{2k}(B U(n-1)) \to 0$

for all $k$. In degrees $\bullet\leq 2n$ this says

$0 \to \mathbb{Z} \overset{c_n \cup (-)}{\longrightarrow} H^{\bullet \leq 2n}(B U(n)) \overset{(B i)^\ast}{\longrightarrow} (\mathbb{Z}[c_1, \cdots, c_{n-1}])_{\bullet \leq 2n} \to 0$

for some Thom class $c_n \in H^{2n}(B U(n))$, which we identify with the next Chern class.

Since free abelian groups are projective objects in Ab, their extensions are all split (the Ext-group out of them vanishes), hence the above gives a direct sum decomposition

\begin{aligned} H^{\bullet \leq 2n}(B U(n)) & \simeq (\mathbb{Z}[c_1, \cdots, c_{n-1}])_{\bullet \leq 2n} \oplus \mathbb{Z}\langle 2n\rangle \\ & \simeq (\mathbb{Z}[c_1, \cdots, c_{n}])_{\bullet \leq 2n} \end{aligned} \,.

Now by another induction over these short exact sequences, the claim follows.

### Top Chern class

For $\mathcal{V}_X$ a complex vector bundle of complex rank $n$, the highest degree Chern class that may generally be non-vanishing is $c_n$. This is hence often called the top Chern class of the vector bundle.

###### Proposition

The top Chern class of a complex vector bundle $\mathcal{V}_X$ equals the Euler class $e$ of the underlying real vector bundle $\mathcal{V}^{\mathbb{R}}_X$:

$\mathcal{V}_X \; \text{has complex rank}\;n \;\;\;\;\; \Rightarrow \;\;\;\;\; c_n \big( \mathcal{V}_X \big) \;\; = \;\; e \big( \mathcal{V}^{\mathbb{R}}_X \big) \;\;\;\; \in H^{2n} \big( X; \, \mathbb{Z} \big) \,.$

(e.g. Bott-Tu 82 (20.10.6))

###### Proposition

The top Chern class of a complex vector bundle $\mathcal{V}_X$ equals the pullback of any Thom class $th \;\in\; H^{2n}\big( \mathcal{V}_X; \mathbb{Z} \big)$ on $\mathcal{V}_X$ along the zero-section:

$\mathcal{V}_X \; \text{has complex rank}\;n \;\;\;\;\; \Rightarrow \;\;\;\;\; c_n \big( \mathcal{V}_X \big) \;\;\; = \;\;\; (0_X)^\ast (th) \;\; \in \; H^{2n} \big( X ; \, \mathbb{Z} \big)$

(e.g. Bott-Tu 82, Prop. 12.4)

### Splitting principle and Chern roots

Under the splitting principle all Chern classes are determined by first Chern classes:

Write $i \colon T \simeq U(1)^n \hookrightarrow U(n)$ for the maximal torus inside the unitary group, which is the subgroup of diagonal unitary matrices. Then

$H^\bullet(B T, \mathbb{Z}) \simeq H^\bullet(B U(1)^n, \mathbb{Z})$

is the polynomial ring in $n$ generators (to be thought of as the universal first Chern classes $c_i$ of each copy of $B U(1)$; equivalently as the weights of the group characters of $U(n)$) which are traditionally written $x_i$:

$H^\bullet(B U(1)^n, \mathbb{Z}) \simeq \mathbb{Z}[x_1, \cdots, x_n] \,.$

Write

$B i \;\colon\; B U(1)^n \to B U(n)$

for the induced map of deloopings/classifying spaces, then the $k$-universal Chern class $c_k \in H^{2k}(B U(n), \mathbb{Z})$ is uniquely characterized by the fact that its pullback to $B U(1)^n$ is the $k$th elementary symmetric polynomial $\sigma_k$ applied to these first Chern classes:

$(B i)^\ast (c_k) = \sigma_k(x_1, \cdots, x_n) \,.$

Equivalently, for $c = \sum_{i = 1}^n c_k$ the formal sum of all the Chern classes, and using the fact that the elementary symmetric polynomials $\sigma_k(x_1, \cdots, k_n)$ are the degree-$k$ piece in $(1+x_1) \cdots (1+x_n)$, this means that

$(B i)^\ast (c) = (1+x_1) (1+ x_2) \cdots (1+ x_n) \,.$

Since here on the right the first Chern classes $x_i$ appear as the roots of the Chern polynomial, they are also called Chern roots.

###### Lemma

For $n \in \mathbb{N}$ let $B \iota_n \;\colon\; B (U(1)^n) \longrightarrow B U(n)$ be the canonical map. Then the induced pullback operation on ordinary cohomology

$\left( B \iota_n \right) \;\colon\; H^\bullet( B U(n); \mathbb{Z} ) \longrightarrow H^\bullet( B U(1)^n; \mathbb{Z} )$

is a monomorphism.

A proof of lemma , via analysis of the Serre spectral sequence of $U(n)/U(1)^n \to B U(1)^n \to B U(n)$ is indicated in (Kochman 96, p. 40). A proof via transfer of the Euler class of $U(n)/U(1)^n$, following (Dupont 78, (8.28)), is indicated at splitting principle (here).

###### Proposition

For $k \leq n \in \mathbb{N}$ let $B \iota_n \;\colon\; B (U(1)^n) \longrightarrow B U(n)$ be the canonical map. Then the induced pullback operation on ordinary cohomology is of the form

$(B i_n)^\ast \;\colon\; \mathbb{Z}[c_1, \cdots, c_k] \longrightarrow \mathbb{Z}[(c_1)_1,\cdots (c_1)_n]$

and sends the $k$th Chern class $c_k$ (def. ) to the $k$th elementary symmetric polynomial in the $n$ copies of the first Chern class:

$(B i_n)^\ast \;\colon\; c_k \mapsto \sigma_k( (c_1)_1, \cdots, (c_1)_n ) \,.$
###### Proof

First consider the case $n = 1$.

The classifying space $B U(1)$ is equivalently the infinite complex projective space $\mathbb{C}P^\infty$. Its ordinary cohomology is the polynomial ring on a single generator $c_1$, the first Chern class (prop.)

$H^\bullet(B U(1)) \simeq \mathbb{Z}[ c_1 ] \,.$

Moreover, $B i_1$ is the identity and the statement follows.

Now by the Künneth theorem for ordinary cohomology (prop.) the cohomology of the Cartesian product of $n$ copies of $B U(1)$ is the polynomial ring in $n$ generators

$H^\bullet(B U(1)^n) \simeq \mathbb{Z}[(c_1)_1, \cdots, (c_1)_n] \,.$

By prop. the domain of $(B i_n)^\ast$ is the polynomial ring in the Chern classes $\{c_i\}$, and by the previous statement the codomain is the polynomial ring on $n$ copies of the first Chern class

$(B i_n)^\ast \;\colon\; \mathbb{Z}[ c_1, \cdots, c_n ] \longrightarrow \mathbb{Z}[ (c_1)_1, \cdots, (c_1)_n ] \,.$

This allows to compute $(B i_n)^\ast(c_k)$ by induction:

Consider $n \geq 2$ and assume that $(B i_{n-1})^\ast_{n-1}(c_k) = \sigma_k((c_1)_1, \cdots, (c_1)_{(n-1)})$. We need to show that then also $(B i_n)^\ast(c_k) = \sigma_k((c_1)_1,\cdots, (c_1)_n)$.

Consider then the commuting diagram

$\array{ B U(1)^{n-1} &\overset{ B i_{n-1} }{\longrightarrow}& B U(n-1) \\ {}^{\mathllap{B j_{\hat t}}}\downarrow && \downarrow^{\mathrlap{B i_{\hat t}}} \\ B U(1)^n &\underset{B i_n}{\longrightarrow}& B U(n) }$

where both vertical morphisms are induced from the inclusion

$\mathbb{C}^{n-1} \hookrightarrow \mathbb{C}^n$

which omits the $t$th coordinate.

Since two embeddings $i_{\hat t_1}, i_{\hat t_2} \colon U(n-1) \hookrightarrow U(n)$ differ by conjugation with an element in $U(n)$, hence by an inner automorphism, the maps $B i_{\hat t_1}$ and $B_{\hat i_{t_2}}$ are homotopic, and hence $(B i_{\hat t})^\ast = (B i_{\hat n})^\ast$, which is the morphism from prop. .

By that proposition, $(B i_{\hat t})^\ast$ is the identity on $c_{k \lt n}$ and hence by induction assumption

\begin{aligned} (B i_{n-1})^\ast (B i_{\hat t})^\ast c_{k \lt n} &= (B i_{n-1})^\ast c_{k \lt n} \\ = \sigma_k( (c_1)_1, \cdots, \widehat{(c_1)_t}, \cdots, (c_1)_n ) \end{aligned} \,.

Since pullback along the left vertical morphism sends $(c_1)_t$ to zero and is the identity on the other generators, this shows that

$(B i_n)^\ast(c_{k \lt n}) \simeq \sigma_{k\lt n}((c_1)_1, \cdots, \widehat{(c_1)_t}, \cdots, (c_1)_n) \;\; mod (c_1)_t \,.$

This implies the claim for $k \lt n$.

For the case $k = n$ the commutativity of the diagram and the fact that the right map is zero on $c_n$ by prop. shows that the element $(B j_{\hat t})^\ast (B i_n)^\ast c_n = 0$ for all $1 \leq t \leq n$. But by lemma the morphism $(B i_n)^\ast$, is injective, and hence $(B i_n)^\ast(c_n)$ is non-zero. Therefore for this to be annihilated by the morphisms that send $(c_1)_t$ to zero, for all $t$, the element must be proportional to all the $(c_1)_t$. By degree reasons this means that it has to be the product of all of them

\begin{aligned} (B i_n)^{\ast}(c_n) & = (c_1)_1 \otimes (c_1)_2 \otimes \cdots \otimes (c_1)_n \\ & = \sigma_n( (c_1)_1, \cdots, (c_1)_n ) \end{aligned} \,.

This completes the induction step.

### Whitney sum formula

###### Proposition

For $k\leq n \in \mathbb{N}$, consider the canonical map

$\mu_{k,n-k} \;\colon\; B U(k) \times B U(n-k) \longrightarrow B U(n)$

(which classifies the Whitney sum of complex vector bundles of rank $k$ with those of rank $n-k$). Under pullback along this map the universal Chern classes (prop. ) are given by

$(\mu_{k,n-k})^\ast(c_t) \;=\; \underoverset{i = 0}{t}{\sum} c_i \otimes c_{t-i} \,,$

where we take $c_0 = 1$ and $c_j = 0 \in H^\bullet(B U(r))$ if $j \gt r$.

So in particular

$(\mu_{k,n-k})^\ast(c_n) \;=\; c_k \otimes c_{n-k} \,.$

e.g. (Kochman 96, corollary 2.3.4)

###### Proof

Consider the commuting diagram

$\array{ H^\bullet( B U(n) ) &\overset{\mu_{k,n-k}^\ast}{\longrightarrow}& H^\bullet( B U(k) ) \otimes H^\bullet( B U(n-k) ) \\ {}^{\mathllap{\mu_k^\ast}}\downarrow && \downarrow^{\mathrlap{ \mu_{k}^\ast \otimes \mu_{n-k}^\ast }} \\ H^\bullet( B U(1)^n ) &\simeq& H^\bullet( B U(1)^k ) \otimes H^\bullet( B U(1)^{n-k} ) } \,.$

This says that for all $t$ then

\begin{aligned} (\mu_k^\ast \otimes \mu_{n-k}^\ast) \mu_{k,n-k}^\ast(c_t) & = \mu^\ast_n(c_t) \\ & = \sigma_t((c_1)_1, \cdots, (c_1)_n) \end{aligned} \,,

where the last equation is by prop. .

Now the elementary symmetric polynomial on the right decomposes as required by the left hand side of this equation as follows:

$\sigma_t((c_1)_1, \cdots, (c_1)_n) \;=\; \underoverset{r = 0}{t}{\sum} \sigma_r((c_1)_1, \cdots, (c_1)_{n-k}) \cdot \sigma_{t-r}( (c_1)_{n-k+1}, \cdots, (c_1)_n ) \,,$

where we agree with $\sigma_q((c_1)_1, \cdots, (c_1)_p) = 0$ if $q \gt p$. It follows that

$(\mu_k^\ast \otimes \mu_{n-k}^\ast) \mu_{k,n-k}^\ast(c_t) = (\mu_k^\ast \otimes \mu_{n-k}^\ast) \left( \underoverset{r=0}{t}{\sum} c_r \otimes c_{t-r} \right) \,.$

Since $(\mu_k^\ast \otimes \mu_{n-k}^\ast)$ is a monomorphism by lemma , this implies the claim.

### Chern-, Pontrjagin-, and Euler- characteristic forms

We spell out the formulas for the images under the Chern-Weil homomorphism of the Chern classes, Pontrjagin classes and Euler classes as characteristic forms over smooth manifolds.

#### Preliminaries

Let $X$ be a smooth manifold.

Write

(1)$\Omega^{2\bullet}(X) \;\; \in \; CAlg_{\mathbb{R}}$

for the commutative algebra over the real numbers of even-degree differential forms on $X$, under the wedge product of differential forms. This is naturally a graded commutative algebra, graded by form degree, but since we consider only forms in even degree it is actually a plain commutative algebra, too, after forgetting the grading.

Let $\mathfrak{g}$ be a semisimple Lie algebra (such as $\mathfrak{su}(d)$ or $\mathfrak{so}(d)$) with Lie algebra representation $V \,\in\, Rep_{\mathbb{C}}(\mathfrak{g})$ over the complex numbers of finite dimension $dim_{\mathbb{C}}(V) \,=\, n \,\in\, \mathbb{N}$ (for instance the adjoint representation or the fundamental representation), hence a homomorphism of Lie algebras

$\mathfrak{g} \xrightarrow{\;\;\rho\;\;} End_{\mathbb{C}}(V)$

to the linear endomorphism ring $End_{\mathbb{C}}(V)$, regarded here through its commutator as the endomorphism Lie algebra of $V$.

When regarded as an associative ring this is isomorphic to the matrix algebra of $n \times n$ square matrices

(2)$End_{\mathbb{C}}(V) \;\; \simeq \;\; Mat_{n \times n}(\mathbb{C}) \,.$

The tensor product of the $\mathbb{C}$-algebras (1) and (2)

is equivalently the $n \times n$ matrix algebra with coefficients in the complexification of even-degree differential forms:

$\Omega^{2\bullet} \big(X\big) \otimes_{\mathbb{R}} End_{\mathbb{C}}(V) \;\simeq\; \Omega^{2\bullet}(X) \otimes_{\mathbb{R}} \big( Mat_{n \times n}( \mathbb{R} ) \big) \;\; \simeq \;\; Mat_{n \times n} \big( \Omega^{2\bullet}(X) \otimes_{\mathbb{R}} \mathbb{C} \big) \,.$

The multiplicative unit

(3)$I \;\in\; Mat_{n \times n} \big( \Omega^{2\bullet}(X) \otimes_{\mathbb{R}} \mathbb{C} \big)$

in this algebra is the smooth function (differential 0-forms) which is constant on the $n \times n$ identity matrix and independent of $t$.

Given a connection on a $G$-principal bundle, we regard its $\mathfrak{g}$-valued curvature form as an element of this algebra

(4)$F_\nabla \,\in\, \Omega^2(X) \otimes_{\mathbb{R}} \mathfrak{g} \xrightarrow{\; \rho \;} \Omega^2(X) \otimes_{\mathbb{R}} End_{\mathbb{C}}(V) \xhookrightarrow{\;\;\;} \Omega^{2\bullet}(X) \otimes_{\mathbb{R}} End_{\mathbb{C}}(V)[t] \;\simeq\; Mat_{n \times n} \Big( \mathbb{C} \otimes_{\mathbb{R}} \Omega^{2}(X) \Big) \,.$

#### The formulas

##### Chern forms

The total Chern form $c(\nabla)$ is the determinant of the sum of the unit (3) with the curvature form (4), and its component in degree $2k$, for $k \in \mathbb{N}$, is the $k$th Chern form $c_k(\nabla)$:

$c(\nabla) \;\; \coloneqq \;\; \sum_k \underset{ \mathclap{ deg = 2k } }{ \underbrace{ c_k(\nabla) } } \;\; \coloneqq \;\; det \left( I + t \frac{i F_\nabla}{2\pi} \right) \,.$

By the relation between determinant and trace, this is equal to the exponential of the trace of the logarithm of $I + \frac{i F_\nabla}{2\pi}$, this being the exponential series in the trace of the Mercator series in $\frac{i F_\nabla}{2\pi}$:

(5)\begin{aligned} c(\nabla) & \;=\; det \left( I + t \frac{i F_\nabla}{2\pi} \right) \\ & \;=\; \exp \circ tr \circ ln \left( I + \frac{i F_\nabla}{2\pi} \right) \\ & \;=\; \exp \circ tr \left( - \underset {k \in \mathbb{N}_+} {\sum} \tfrac{1}{k} \left( \frac{F_\nabla}{2\pi i} \right)^k \right) \\ & \;=\; \exp \left( \underset {k \in \mathbb{N}_+} {\sum} \tfrac{1}{k} \left( \frac { - (-i)^k } {(2\pi)^k} tr\big( F_\nabla^{\wedge_k} \big) \right) \right) \\ & \;=\; 1 \\ & \phantom{\;=\;} + \phantom{\frac{1}{1}} \left( i \tfrac{ tr\big(F_\nabla\big) }{2 \pi} + \tfrac{1}{2} \tfrac{ tr\big( (F_\nabla)^{2} \big)}{(2 \pi)^2} -i \tfrac{1}{3} \tfrac{ tr\big( (F_\nabla)^{3} \big)}{(2 \pi)^3} - \tfrac{1}{4} \tfrac{ tr\big( (F_\nabla)^{4} \big)}{(2 \pi)^4} + \cdots \right) \\ & \phantom{\;=\;} + \frac{1}{2} \left( i \tfrac{ tr\big(F_\nabla\big) }{2 \pi} + \tfrac{1}{2} \tfrac{ tr\big( (F_\nabla)^{2} \big)}{(2 \pi)^2} -i \tfrac{1}{3} \tfrac{ tr\big( (F_\nabla)^{3} \big)}{(2 \pi)^3} - \tfrac{1}{4} \tfrac{ tr\big( (F_\nabla)^{4} \big)}{(2 \pi)^4} + \cdots \right)^2 \\ & \phantom{\;=\;} + \frac{1}{6} \left( i \tfrac{ tr\big(F_\nabla\big) }{2 \pi} + \tfrac{1}{2} \tfrac{ tr\big( (F_\nabla)^{2} \big)}{(2 \pi)^2} -i \tfrac{1}{3} \tfrac{ tr\big( (F_\nabla)^{3} \big)}{(2 \pi)^3} - \tfrac{1}{4} \tfrac{ tr\big( (F_\nabla)^{4} \big)}{(2 \pi)^4} + \cdots \right)^3 \\ & \phantom{\;=\;} + \frac{1}{24} \left( i \tfrac{ tr\big(F_\nabla\big) }{2 \pi} + \tfrac{1}{2} \tfrac{ tr\big( (F_\nabla)^{2} \big)}{(2 \pi)^2} -i \tfrac{1}{3} \tfrac{ tr\big( (F_\nabla)^{3} \big)}{(2 \pi)^3} - \tfrac{1}{4} \tfrac{ tr\big( (F_\nabla)^{4} \big)}{(2 \pi)^4} + \cdots \right)^4 \\ & \phantom{\;=\;} + \cdots \\ & \;=\; 1 \\ & \phantom{\;=\;} + i \frac { tr\big(F_\nabla\big) } { 2 \pi } \\ & \phantom{\;=\;} + \tfrac{1}{2} \frac { tr\big( (F_\nabla)^2 \big) } { (2 \pi)^2 } + \frac{1}{2} \left( i \frac { tr\big( F_\nabla \big) } { 2\pi } \right)^2 \\ & \phantom{\;=\;} - i \tfrac{1}{3} \frac { tr\big( (F_\nabla)^3 \big) } { (2 \pi)^3 } + \frac{1}{2} \left( 2 \left( i \frac { tr\big( F_\nabla \big) } { 2 \pi } \right) \left( \tfrac{1}{2} \frac { tr\big( (F_\nabla)^2 \big) } { (2 \pi)^2 } \right) \right) + \frac{1}{6} \left( \left( i \frac { tr\big(F_\nabla\big) } { 2\pi } \right)^3 \right) \\ & \phantom{\;=\;} - \tfrac{1}{4} \frac {tr\big( (F_\nabla)^4 \big)} { (2 \pi)^4 } + \frac{1}{2} \left( \tfrac{1}{2} \frac {tr\big( (F_\nabla)^2 \big)} { (2 \pi)^2 } \right)^2 + \frac{1}{24} \left( i \frac {tr\big( F_\nabla \big)} { 2\pi } \right)^4 \\ & \phantom{\;=\;} + \cdots \\ & \;=\; 1 \\ & \phantom{\;=\;} + \underset{ \color{blue} = c_1(\nabla) }{ \underbrace{ i \frac { tr\big(F_\nabla\big) } { 2 \pi } }} \\ & \phantom{\;=\;} + \underset{ \color{blue} = c_2(\nabla) }{ \underbrace{ \frac {\tr\big( (F_\nabla)^2 \big) - \big( tr(F_\nabla) \big)^2 } { 8 \pi^2 } }} \\ & \phantom{\;=\;} + \underset{ \color{blue} = c_3(\nabla) }{ \underbrace{ i \frac { - 2 \cdot tr\big( (F_\nabla)^3 \big) + 3 \cdot tr(F_\nabla) \cdot tr\big( (F_\nabla)^2 \big) - \big( tr(F_\nabla ) \big)^3 } {48 \pi^3} }} \\ & \phantom{\;=\;} + \underset{ \color{blue} = c_4(\nabla) }{ \underbrace{ \frac { -6 \cdot tr\big( (F_\nabla)^4 \big) + 3 \cdot tr\big( (F_\nabla)^2 \big)^2 + \big( tr(F_\nabla) \big)^4 } {384 \pi^4} }} \\ & \phantom{\;=\;} + \cdots \end{aligned}
##### Pontrjagin forms

Setting $tr(F_\nabla) = 0$ in these expressions (5) yields the total Pontrjagin form $p(\nabla)$ with degree=$4k$-components the Pontrjagin forms $p_{k}(\nabla)$:

\begin{aligned} p(\nabla) & \;\coloneqq\; \underset{k \in \mathbb{N}}{\sum} \underset{ deg = 4k }{ \underbrace{ (-1)^{k} p_{k}(\nabla) } } \\ & \;=\; \underset{k \in \mathbb{N}}{\sum} \underset{ deg = 4k }{ \underbrace{ c_{2k}(\nabla) } } \\ & \;=\; 1 \\ & \phantom{\;=\;} + \underset{ \color{blue} = - p_1(\nabla) }{ \underbrace{ \frac {\tr\big( (F_\nabla)^2 \big) } { 8 \pi^2 } }} \\ & \phantom{\;=\;} + \underset{ \color{blue} = p_2(\nabla) }{ \underbrace{ \frac { - 2 \cdot tr\big( (F_\nabla)^4 \big) + tr\big( (F_\nabla)^2 \big)^2 } {128 \pi^4} }} \\ \phantom{\;=\;} + \cdots \end{aligned}

Hence the first couple of Pontrjagin forms are

\begin{aligned} p_1(\nabla) & \;=\; - \frac {\tr\big( (F_\nabla)^2 \big) } { 8 \pi^2 } \\ p_2(\nabla) & \;=\; \frac { tr\big( (F_\nabla)^2 \big)^2 - 2 \cdot tr\big( (F_\nabla)^4 \big) } {128 \pi^4} \,. \end{aligned}

(See also, e.g., Nakahara 2003, Exp. 11.5)

##### Euler forms

For $n = 2k$ and with the curvature form again regarded as a 2-form valued $(2k) \times (2k)$-square matrix

$F_{\nabla} \;=\; \big( (F_{\nabla})^a{}_b \big)_{1 \leq a,b, \leq 2k}$

the Euler form is its Pfaffian of this matrix, hence the following sum over permutations $\sigma \in Sym(2k)$ with summands signed by the the signature $sgn(\sigma) \in \{\pm 1\}$:

$\chi_{2k}(\nabla) \;=\; \frac {(-1)^k} { (4 \pi)^k \cdot k! } \underset{\sigma}{\sum} sgn(\sigma) \cdot (F_{\nabla})_{\sigma(1)\sigma(2)} \wedge (F_{\nabla})_{\sigma(3)\sigma(4)} \wedge \cdots \wedge (F_{\nabla})_{\sigma(2k-1)\sigma(2k)} \,.$

The first of these is, using the Einstein summation convention and the Levi-Civita symbol:

$\chi_4(\nabla) \;=\; \frac { \epsilon^{ a b c d} (F_{\nabla})_{a b} \wedge (F_\nabla)_{c d} } {32 \pi^2}$

(See also, e.g., Nakahara 2003, Exp. 11.7)

## Examples

### Chern classes of linear representations

Under the Atiyah-Segal completion map linear representations of a group $G$ induce K-theory classes on the classifying space $B G$. Their Chern classes are hence invariants of the linear representations themselves.

See at characteristic class of a linear representation for more.

In Yang-Mills theory field configurations with non-vanishing second Chern class (and minimal energy) are called instantons. The second Chern class is the instanton number . For more on this see at SU(2)-instantons from the correct maths to the traditional physics story.

## References

Original articles:

Textbook accounts:

With an eye towards mathematical physics:

A brief introduction is in chapter 23, section 7