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Are there known examples of a non-multiplicative Euler-Poincaré characteristic on varieties?

Let $\mathbf{Var}/k$ be the category of varieties over a filed $k$, i.e. the category of reduced separated schemes of finite type over $k$, and let $\mathop{\mathrm{K}}_0(\mathbf{Var}/k)$ be its Grothendieck ring.

A (generalised) Euler-Poincaré characteristic is a group homomorphism from $\mathop{\mathrm{K}}_0(\mathbf{Var}/k)$. All examples of Euler-Poincaré characteristics that I am aware of happen to be multiplicative, i.e. each of them is given by a ring homomorphism $ \mathop{\mathrm{K}}_0(\mathbf{Var}/k) \to R$, for some ring $R$. Ramachandran and Tabuada's Exponentiation of Motivic Measures provides examples of multiplicative Euler-Poincaré characteristics, e.g. counting rational point over a finite field, the Hodge characteristic, the Poincaré characteristic, the Larsen-Lunts measure, the Albanese measure, and the Gillet-Soulé measure.

I am mainly interested in 'non-artificial' characteristics (ones that people actually use); Nevertheless, I would like to know if any 'artificial' or 'non-artificial' example is known.

Edit: I would like to know an example of non-multiplicative characteristic that is not arising from a multiplicative one.

Edit 2: Apparently, I am only interested in 'non-artificial' non-multiplicative characteristics that people actually use, did not arise by forgetting some of the data a multiplicative characteristic give, or by composing with an arbitrary group homomorphism.

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    $\begingroup$ Just compose any of the characteristics you know with an abelian group homomorphism. For example, you can take your favorite Hodge number, or your favorite linear combination of Hodge numbers. $\endgroup$
    – Qiaochu Yuan
    Commented Jun 1, 2016 at 18:51
  • $\begingroup$ @QiaochuYuan: Thanks, yea sure, my formulation for the question was rather silly. What is relevant to me is characteristics that are not coming from multiplicative ones. $\endgroup$
    – user337830
    Commented Jun 1, 2016 at 19:47
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    $\begingroup$ I'm not sure the question makes sense as edited. By “arising from a multiplicative one”, I understand that an additive invariant $\alpha$ is induced by a multiplicative invariant $\chi$ composed with a morphism of abelian groups. But every additive invariant has this form, with $\chi$ the identity... $\endgroup$
    – ACL
    Commented Jun 1, 2016 at 22:29

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