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I came up with a pair of strange 3-folds, I was wondering if there is any way to show such thing should not happen just by looking some of the properties they have.

Let $V_1, V_2$ be smooth, projective, 3-folds over complex numbers. I know $V_1, V_2$ both have fiberations over $\mathbb{P}^1$, and for any closed point $t \in \mathbb{P}^1$, the fibres ${(V_1)}_t, (V_2)_t$ are isomorphism. However, I also know $V_1, V_2$ are NOT birational. The other information I know is: The generic fibre are both abelian surfaces.

I really could not imagine such varieties, the only thing might be an analogy is the Möbius strip. I was wondering if there is an argumement to show there must be something wrong in the construction as follows: because the fibres are isomorphism, it is also isomorphism nearby the fibres, and hence they two should be birational.

Edit: Here is how I construct $V_1, V_2$. First, I have a smooth 3-fold $X$ fibreing over $\mathbb{P}^1$, and I have two finite groups $G_1 ,G_2 \subseteq {\rm Aut}(X)$ (one is abelian, the other is not abelian) acting on $X$ without fixed point.

Then $V_1 := X/G_1, V_2 :=X/G_2$. Because I also know the fundamental group of $X$ is trivial, the fundamental groups of $V_1, V_2$ are $G_1, G_2$ respectively. If $V_1, V_2$ are birational, I guess they should have the same fundamental groups. Hence, they are not birational.

As for they have the isomorphic fibres, I did this in a simply-mined, non-rigours way, and that is the part I worry most:

As I know how does the group $G_i$ acting on coordinates of the generic points of $X$, and I know how does the generic points maps to $\mathbb{P}^1$. I just pick a general point $x$, and see what is the images under $G_i$. For $G_i\cdot x$ (i.e. the orbits) are in the fibre of the exactly same four points of $\mathbb{P}^1$ for both $G_1, G_2$. Then "module out" this action on different fibres (i.e. looking at the subgroup of $H_i \subseteq G_i$ such that the orbits are in the same fibre), I get the same group for both $G_1, G_2$. Hence, if I further quotient out $H_i$ for the fibres, I have the same varieties on the fibre. That is how I get the isomrophism on generic fibres. But this procedure is quite intuitive...

Li Yutong
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  • Can you describe how you come up with your 3-folds, and how you know they're not birational? –  Feb 18 '14 at 16:36
  • Please take a look at the edit above. – Li Yutong Feb 18 '14 at 19:41
  • I'm not quite sure if I understand your edits last paragraph. Don't the acting automorphisms respect the fibres? If so, what are the "same four points of $\mathbb{P}^1$" and if not, how are the quotients fibred over $\mathbb{P}^1$? – Ben Feb 18 '14 at 20:51
  • No, both the actions does not respect the fibres. For $G_1$, there is $H_1 \subset G_1,$ with $[G_1 : H_1] =4$. The $H_1$ respect the fiber (i.e. $H_1 \cdot x$ are on the same fibre of $x$). However, the $G_1 \cdot x$ are actually in 4 fibres. Similarly for $G_2$. And the interesting part is $H_2 = H_1$. Feel free to let me know if this is still unclear. – Li Yutong Feb 19 '14 at 02:37

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