I want to know whether any of the following questions are true or not. If they are not true I will appreciate a counter-example:
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If $X$ is a smooth complex non-simply connected projective surface, with a normal crossing divisor $D$ on $X$ then $H^1(X, \mathbb{Q})\rightarrow H^1(X-D, \mathbb{Q})$ is an isomorphism.
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The mixed Hodge structure on the first rational cohomology of smooth complex quasi-projective surfaces is always pure.
Best Answer
It depends on numbers of irreducible components of $D$ as well as the relations of their fundamental classes in $H^2(X)$.
Assume that $X$ is smooth projective, and $D$ is a normal crossing divisor. There is an exact sequence $$H^1(X,X-D,\mathbb Q)\to H^1(X,\mathbb Q)\xrightarrow{f} H^1(X-D,\mathbb Q)\to H^2(X,X-D,\mathbb Q)\xrightarrow{g} H^2(X,\mathbb Q)$$
So to determine if $f$ is isomorphism, we need to compute $H^i(X,X-D,\mathbb Q)$, $i=1,2$ first.
When $D$ is smooth, or it has only one component, Thom isomorphism implies that $H^i(X,X-D)$$\cong H^{i-2}(D)$, which is $\mathbb Q$ when $i=2$ and $0$ when $i=1$. The Gysin homomorphism $g$ sends a generator to the fundamental class of $D$, which is not trivial, so $f$ is an isomorphism.
In general, by Fujiki's paper Duality of Mixed Hodge Structures of Algebraic Varieties, there is a duality between mixed Hodge structures
$$H^{2n-i}(D,\mathbb Q)\times H^i(X,X-D,\mathbb Q)\to \mathbb Q.$$
Set $i=1,2$, we find $h^1(X,X-D,\mathbb Q)=0$ and $h^2(X,X-D,\mathbb Q)=h^2(D,\mathbb Q)=k$, where $k$ is the number of irreducible components of $D$. The exact sequence becomes
$$0\to H^1(X,\mathbb Q)\xrightarrow{f} H^1(X-D,\mathbb Q)\to \mathbb Q^k\xrightarrow{g} H^2(X,\mathbb Q).$$
$g$ sends each component $D_i$ of $D$ to its fundamental class. So the conclusion is
Claim: $H^1(X,\mathbb Q)\to H^1(X-D,\mathbb Q)$ is an isomorphism if and only if the mixed Hodge structure $H^1(X-D,\mathbb Q)$ is pure of weight one if and only if the fundamental classes $[D_i]$ of each component of $D$ are linearly independent in $H^2(X,\mathbb Q)$.
For example, when $X=\mathbb P^2$, $D=L_1\cup L_2$ is union of two lines, the relation $[L_1]=[L_2]$ tells us that the mixed Hodge structure on $H^1(\mathbb P^2\setminus (L_1\cap L_2))$ is concentrated on weight two (and pure).
When $X=Bl_p\mathbb P^2$ is blowup of one point on $\mathbb P^2$ and $D=L\cup E$, where $D$ is the strict transform of a line through $p$, and $E$ is the exceptional divisor, the MHS on $H^1(X\setminus D)$ is pure.
To give an example where $H^1(X\setminus D)$ is not pure, just take any $X=\mathbb P^1\times C$ where $C$ is a curve with genus $g\ge 1$. Take $D$ to be the union of two disjoint fibers of $X\to \mathbb P^1$, then $W_1H^1(X\setminus D)=H^1(X)=\mathbb Q^{2g}$ and $W_2/W_1\cong \mathbb Q$, $H^1(X\setminus D)$ is not pure.