Contents

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.

Splitting principle and Chern roots

Under the splitting principle all Chern classes are determnined 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 (Kochmann 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. (Kochmann 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.

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 include

• A. Grothendieck, La théorie des classes de Chern, Bulletin de la Société Mathématique de France 86 (1958), p. 137–154, numdam

Textbook accounts include

A brief introduction is in chapter 23, section 7