To define characteristic classes in smooth vector bundles $E\longrightarrow M$ there is a more or less standard procedure: to choose a connection $\nabla$ and to derive the curvature $\Omega$, which is an $End(E)$-valued 2-form. In each chart $U_\alpha$, $\Omega$ may be described by a $r\times r$ matrix ($r$ rank of $E$) whose entries are 2-forms. The matrices change when the charts changes, but due to the tensoriality and the nature of $\Omega$, some quantities such as the trace or the determinant do not change for overlaping charts. (See, for instance, the first chapter of Lecture Notes on Seiberg-Witten Invariants)

Now to take full advantage of these invariant quantities (= to define Chern classes) one considers powers of the curvature matrix and their traces:


The procedure is ok, but if one studies it closer, one realizes that we are indeed defining some 'pseudo-wedge' map


by simply taking the product of matrices whose entries are forms. But the question is

Is there any way to define this pseudo-wedge product intrinsically, that is, by using only the classical wedge product $\Omega^p(M)\times\Omega^q(M)\longrightarrow\Omega^{p+q}(M)$ together with some linear algebra? Perhaps there is already some book making an explicit definition; in this case it would be most helpful for me to have good references.

Any idea or suggestion is welcome.

EDIT: See the discution below about the definition of Wikipedia and the relationship with the curvature of connections in vector bundles.

  • 1
    $\begingroup$ Does en.wikipedia.org/wiki/… help? The wedge product of two vector-valued forms is naturally a form taking values in the tensor product of the bundles. But you are working with End(E) which happens to be an algebra. // For a reference I think this may have been explained in Morita's Differential Forms book, but Google at least yields Greg Naber's Topology, Geometry and Gauge fields: Foundations in section 5.11 around page 323. $\endgroup$ – Willie Wong Mar 17 '15 at 12:15
  • $\begingroup$ @WillieWong Yes, I think it is just the thing I needed. I will work on it more carefully, but I think the reference is the key point to have in mind. By the way, is there anything like the Hodge star for $End(E)$-valued forms? $\endgroup$ – Jjm Mar 17 '15 at 12:30
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    $\begingroup$ Yes, just do it on the base. You should be able to find something in any exposition about Yang-Mills theories (on curved backgrounds) (there instead of End(E) your form takes value in a Lie algebra, but the principles are the same). $\endgroup$ – Willie Wong Mar 17 '15 at 12:47
  • $\begingroup$ @WillieWong According to the definition in Wikipedia, it seems that if $\alpha\in\Omega^p$, $\beta\in\Omega^q$ and $g\in E_1$, $h\in E_2$, then $(\alpha\otimes g)\wedge(\beta\otimes h)=(\alpha\wedge\beta)g\otimes h$. Is that true? In this case, the formula $F_{A+a}=F_A+d_A a+a\wedge a$ for $A$ connection and $a$ $\mathfrak{g}$-valued 1-form makes no sense! $\endgroup$ – Jjm Sep 24 '15 at 8:20
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    $\begingroup$ I think the natural point here is not to view $End(E)$ as an associative algebra but as a Lie algebra (under the commutator of endomorphism). Hence you use the commutator to map $End(E)\otimes End(E)$ to $End(E)$, and there is a corresponding calculus for Lie algebra valued forms. This is consistent with the last comment of @Jjm in which $a\wedge a$ maps $(\xi,\eta)$ to $[a(\xi),a(\eta)]$. It also fits to the fact that the polynomials you use are invariant under conjugation, which infinitesimally means that they are invariant under the adjoint action in the Lie algebra. $\endgroup$ – Andreas Cap Sep 24 '15 at 9:10

Not sure if this answers your question exactly (this is a calculation in cohomology), but, having done this sort of calculation, recently I formalised it as follows.

  • For sheaves $\mathscr{F},\mathscr{G}$ of $\mathcal{O}_X$-modules we have the cup product $$\smile\colon H^m(X,\mathscr{F})\otimes H^n(X,\mathscr{G})\to H^{m+n}(X,\mathscr{F}\otimes\mathscr{G})$$
  • In Čech cohomology, this is give by the tensor product, i.e. $$(a\smile b)_{ijk}=(a)_{ij}\otimes(b)_{jk}$$
  • Given the cocycle $$\omega_{ij}\in\check{\mathscr{C}}_\mathscr{U}^1\big(\mathcal{Hom}(E,E\otimes\Omega_X^1)\big)\cong\check{\mathscr{C}}_\mathscr{U}^1\big(\Omega_X^1\otimes\mathcal{End}(E)\big)$$ representing the Atiyah (i.e. Chern) class $\mathrm{at}_E$ of $E$, we can define $\mathrm{at}_E^2$ as follows:

    1. Take $(\mathrm{at}_E\otimes\mathrm{id}_{\Omega_X^1})\smile\mathrm{at}_E$ in $ H^2\big(X,\mathcal{Hom}(E\otimes\Omega_X^1,E\otimes\Omega_X^1\otimes\Omega_X^1)\otimes\mathcal{Hom}(E,E\otimes\Omega_X^1)\big)$
    2. Apply the composition map $H^m\big(X,\mathcal{Hom}(\mathscr{G},\mathscr{H})\otimes\mathcal{Hom}(\mathscr{F},\mathscr{G})\big)\to H^m\big(X,\mathcal{Hom}(\mathscr{F},\mathscr{H})\big)$
    3. Apply the wedge product $\wedge\colon\Omega_X^i\otimes\Omega_X^j\to\Omega_X^{i+j}$ to obtain $$\mathrm{at}_E^2\in H^2\big(X,\mathcal{End}(E)\otimes\Omega_X^2\big).$$

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