# Relative Characteristic classes

A pair of vector bundles over a base space $X$ is a pair $(E,F)$ where $E$ is a vector bundle over $X$ and $F$ is a sub-bundle of $E$. Two pairs $(E_{1},F_{1})$ and $(E_{2}, F_{2})$ are isomorphic if there is an isomorphism from $E_{1}$ to $E_{2}$ which send $F_{1}$ onto $F_{2}$. The direct sum of two pairs has a natural definition.

Are there some type of characteristic classes with values in the cohomology of base space with the following properties?:

It is trivial $("1")$ on the trivial pairs. Isomorphism pairs have the same characteristic classes. They commute with the pull back operation. They restrict (or have a reasonable relation)to the usual characteristic classes on pairs $(E,0)$ and $(\epsilon_{n}, F)$ where $\epsilon_{n}$ is the trivial $n$ dimensional bundle.

## 2 Answers

You obviously have all characteristic classes of $E$, of $F$, and of $E/F$, with some relations because $E\cong F\oplus E/F$. To see if there are more, consider the classifying space for your pairs. If you are working over $\Bbbk$, $\mathrm{rk}(F)=k$, $\mathrm{rk}(E)=\ell$, it is the colimit over $n$ of the space of partial flags of length $(k,\ell)$ in $\Bbbk^n$. For a fixed $n$, this is homotopy equivalent to $$G_{k,\ell,n}(\Bbbk)=U(n,\Bbbk)/(U(k,\Bbbk)\times U(\ell-k,\Bbbk)\times U(n-\ell,\Bbbk))\;.$$ This space is the total space of (at least) three fibre bundles $G_{k,\ell}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{\ell,n}(\Bbbk)$ and $G_{\ell-k,n-k}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{k,n}(\Bbbk)$ and $G_{k,n+k-\ell}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{\ell-k,n}(\Bbbk)$. Hence, you can try the Leray-Serre spectral sequence to find characteristic classes not generated by those mentioned above.

In the holomorphic setting, you see Bott-Chern forms. If $E$ and $F$ are flat, you see the forms described in the appendix of Bismut-Lott. In both settings, you can sometimes extract cohomological information.

I am not sure there are similar classes in a purely topological setting, but I would personally go for $\Bbbk=\mathbb R$ first and look at $\mathbb Z/2$-valued cohomology.

EDIT: It seems there are no additional characteristic classes that do not come from $F$ or $E/F$. For consider the fibre bundle $G_{\ell-k,n-k}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{k,n}(\Bbbk)$. For sufficiently large $n$, all classes of $G_{\ell-k,n-k}(\Bbbk)$ of small degree come from the vector bundle $E/F$, and so are pulled back from $G_{k,\ell,n}(\Bbbk)$. Hence by the Leray-Hirsch theorem, in small degrees the cohomology ring of $G_{k,\ell,n}(\Bbbk)$ is generated by the characteristic classes of the vector bundles $F$ and $E/F$.

• Thanks for your answer.as I understand from your answer, the isomorphic type of a pair $(E,F)$ is completely determined by $E,F,E/F$, since every short exact sequence of vector bundles split automatically(for compact base space). this motivates us to ask the following question: Let $A$ be a noncommutative $C^{*}$ algebra. Is it true to say that every short exact sequence of finite protective $A$- modules split automatically? – Ali Taghavi Dec 18 '15 at 20:31
• If they are projective, that should follow from the definition. But you are right - your argument is way shorter than the one I gave. – Sebastian Goette Dec 18 '15 at 20:57

The characteristic classes of the virtual bundle $E-F$ seem to have all the properties you asked for.

If $c$ is a characteristic class of vector bundles satisfying a sum formula $$c(E\oplus F) = c(E)c(F),$$ then $c(E-F)$ can be defined via the formula $$c(E-F)c(F) = c(E),$$ or if the base $X$ is compact, $$c(E-F) = \frac{c(E)}{c(F)}.$$