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One definition of Mixed Hodge Structure could be found at Wikipedia

https://en.wikipedia.org/wiki/Hodge_structure#Mixed_Hodge_structures

The definition assumes an abelian group $H_{\mathbb{Z}}$, but the Hodge filtration $F^p$ and weight filtration $W_j$ are associated with $H_{\mathbb{C}}=H_{\mathbb{Z}} \otimes \mathbb{C}$ and $H_{\mathbb{Q}}=H_{\mathbb{Z}} \otimes \mathbb{Q}$. In the process of tensorring with $\mathbb{Q}$ or $\mathbb{C}$, the torsion part of $H_{\mathbb{Z}}$ is killed. So my questions are

  1. Does the torsion of $H_{\mathbb{Z}}$ matter? What is its importance?

  2. Why use a lattice instead of the $\mathbb{Q}$ vector space $H_{\mathbb{Q}}$ in the defition?

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The cohomology of an algebraic variety is not a mere abelian group. It has way more structure. In particular, it has a mixed Hodge structure on it. This structure is indeed the data of some filtrations on $H_{\mathbb{Q}}$ and $H_{\mathbb{C}}$. But the cohomology of an algebraic variety in not only a $\mathbb{Q}$-vector space with a weight filtration and Hodge filtration after tensoring by $\mathbb{C}$. It also has an integral structure $H_{\mathbb{Z}}$.

Of course, we can forget about the Hodge structures. There are nice theorems which do not need them. Or we can forget about integral coefficients for other purposes. We may also want a structure which forgets none of them. This is why there is a definition of integral mixed Hodge stucture.

Here is an important example where it plays a role : consider a proper flat map $f:X\rightarrow D$ where $D$ is a small disk centered at 0, whose fibers are projective varieties, smooth except above 0 where it is a normal crossing divisor. Then the groups $H^*(X_t)$ carry a Hodge structure depending on $t$. If we "fix" the Hodge filtration, then the weight filtration is also fixed, but the lattices vary. (We can also "fix" the lattices, then the weight filtration is also fixed, and this is the Hodge filtration that varies).

However, it is true that the Hodge structure does not give anything on the torsion in cohomology. But it does not mean torsion is not interesting. It is just unrelated to the Hodge structure.

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    Thank you. In your example, if the Hodge filtration is fixed, how are the lattices varied? – 2017-02-24
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    Think about the elliptic curves $y^2=x^3-x^2+t$ and the $H^1$. If $t\neq 0$, it is smooth, so the $H^1$ is pure of weight 1. The weight filtration is not interesting here. The Hodge filtration is the subspace of the de Rham cohomology spanned by $\frac{dx}{y}$. So it is fixed (does not depends on $t$). However, the lattice comes from singular cohomology, this is the abelian group spanned by two differential forms $\omega_1,\omega_2$ such that $\int_{\eta_i}\omega_j=\delta_{ij}$ where $\eta_1,\eta_2$ is a basis of $H_1$. These forms $\omega_i$ do depend on $t$. – 2017-02-24
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    Instead, we could fix the lattice, that is we can look at $H^1_{\mathbb{Z}}\subset H^1_{sing}(X_t,\mathbb{C})$. But then, the differential form $\frac{dx}{y}$ should be seen as an element of $Hom(H_1,\mathbb{C})$. With this point of view, it depends on $t$, so the Hodge filtration varies. – 2017-02-24
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    I see, something like Einstein's relativity. – 2017-02-24
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    Sorry to bother you again. There are definitions of rational and real mixed Hodge structures. The definitions need a rational vector space or a real vector space instead of an abelian group $H_{\mathbb{Z}}$ in integral mixed Hodge structures. The morphisms in integral mixed hodge structures maps $H_{\mathbb{Z}}$ to $H'_{\mathbb{Z}}$. I guess in the category of rational or real mixed hodge structures, you just need to map the $\mathbb{Q}$ or $\mathbb{R}$ vector spaces to $\mathbb{Q}$ or $\mathbb{R}$ vector spaces! Is this true? – 2017-02-27
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    @Wenzhe Yes, a morphism of $\mathbb{Q}$-Hodge structures is a $\mathbb{Q}$-linear map $H_\mathbb{Q}\rightarrow H'_\mathbb{Q}$ compatible with the filtrations (after proper scalar extension). Similarly for $\mathbb{R}$-Hodge structures. – 2017-02-27