# Definition of Hodge tensor

The following is the definition of Hodge structures as given in [Milne]:

Let $$R$$ be one of $$\mathbb{R}, \mathbb{Q}$$ or $$\mathbb{Z}$$. And, let $$(V,h)$$ be an $$R$$-Hodge structure of weight $$n$$. Then, multilinear form $$V^r\rightarrow R$$ is a Hodge tensor is the corresponding map $$V^{\otimes r}\rightarrow R(-nr/2)$$ is a morphism of Hodge structures.

Probably the most famous example is when $$r=2$$, you get a polarization. This in tern helps define an equivalence between the category of Abelian varieties over $$\mathbb{C}$$ and polarizable $$\mathbb{Z}$$-Hodge structures of type $$(-1,0),(0,-1)$$.

My question is kind of stupid. What is $$R(m/2)$$ when $$m$$ is odd? If my Hodge structure is of odd weight (as it is in the above example), can we not have any Hodge tensors where $$r$$ is also odd? My best guess is that $$R(m/2)$$ when $$m$$ is odd is just zero. But, I am not entirely confident to continue reading with a shakey guess.

References:

Milne, J. S., Introduction to Shimura varieties, Arthur, James (ed.) et al., Harmonic analysis, the trace formula, and Shimura varieties. Proceedings of the Clay Mathematics Institute 2003 summer school, Toronto, Canada, June 2–27, 2003. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3844-X/pbk). Clay Mathematics Proceedings 4, 265-378 (2005). ZBL1148.14011.

$$R(m/2)$$ is not defined if $$m$$ is odd, so, in fact, there are no Hodge tensors unless $$nr$$ is even. This reflects the fact that there are no Hodge classes in $$H^i$$ unless $$i$$ is even.