I'm trying to understand why on earth the first chern class of a line bundle in K-theory $c_1(L) = 1-L$.

I understand that the first Chern class of the trivial bundle is zero, and that $H-1$ generates the reduced K-theory of $CP^1$ (where H is the canonical line bundle over $CP^1$), but there must be more reasoning behind this!

Additionally, I have seen $c_1(L)$ as both $L-1$ and $1-L$, and I'm curious if this has to do with choosing the universal line bundle to be $\mathcal{O}(1)$ or $\mathcal{O}(-1)$, respectively.


1 Answer 1


This comes from the choice of the $K$-theory Thom class for complex vector bundles.

Firstly, recall that $K$-theory $K^0(X)$ can be described as the group of bounded chain complexes of vector bundles on $X$, modulo the relation of forcing short exact sequences of such chain complexes to split. This is related to the usual description of $K$-theory by sending a chain complex of vector bundles to its Euler characteristic. However, in this model $K^0(X,A)$ is described by the bounded chain complexes of vector bundles on $X$ which are exact over $A$.

If $\pi : E \to B$ is an $n$-dimensional complex vector bundle, then we can pullback the bundle $\pi^*E \to E$, and get chain complex $$0 \to \wedge^0 \pi^* E \to \wedge^{1} \pi^* E \to \wedge^{2} \pi^* E \to \cdots \to \wedge^n \pi^* E \to 0$$ where over a point $v \in E$ the maps are each given by wedge with $v$. By basic linear algebra, for $v \neq 0$ this is exact. The class $\lambda_E \in K^0(E, E-0)$ is then that represented by the linear dual of this complex. This is a possible choice of Thom class.

In particular, if $L \to B$ is a complex line bundle then $\lambda_L$ is the chain complex $$0 \to \pi^*L^* \to \underline{\mathbb{C}} \to 0,$$ where $\underline{\mathbb{C}}$ lies in degree zero. The Euler class of a bundle is by definition the pullback of the Thom class along the zero-section: pulling this back along the zero-section $s : B \to E$ we obtain the complex $$0 \to L^* \overset{0}\to \underline{\mathbb{C}} \to 0$$ which, on taking Euler characteristics, gives $1-L^* \in K^0(B)$. On the other hand, the Euler class and first Chern class of a line bundle are the same thing (by definition), which gives the familiar $$c_1^K(L) = 1-L^* \in K^*(B).$$

It is an important point that the convention for the Thom class is only a convention. You can choose perfectly reasonable other conventions, by dualising E and replacing wedge with contraction, or not dualising the first complex above, leading to any of $L-1$, $1-L$, $L^*-1$, and $1-L^*$ as the first Chern class. Of these $1-L^*$ above and $L-1$ are slightly preferable as then the perverse equation $c_1(c_1^K(L)) = c_1(L)$ holds. I forget which way round they go, but one of these is standard in algebraic topology and the other is standard in algebraic geometry. This makes combining formulas from papers in these two subjects a uniquely frustrating experience.

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    $\begingroup$ $0 \to \land^0\pi^*L \to \land^1\pi^*L \to 0$ is $0 \to \underline{\mathbb{C}} \to \pi^*L \to 0$. I would think that taking the linear dual would then be $0 \to \pi^*L^* \to \underline{\mathbb{C}} \to 0$. Did you mean that we should take the dual of $0 \to \land^0\pi^*L^* \to \land^1\pi^*L^* \to 0$? $\endgroup$ Mar 17, 2015 at 0:24
  • $\begingroup$ You are right, I messed up the dualisation. I think it is right now. $\endgroup$ Mar 17, 2015 at 8:27
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    $\begingroup$ @OscarRandal-Williams: Do you have an idea about the reasons behind these conventions? I've never understood the choice to dualize; I've just seen it in most sources essentially as "now take the dual" without further explanation. Does something work out better with this convention? $\endgroup$
    – jdc
    Oct 20, 2019 at 21:07

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