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This question is about how to construct a Fermat pencil of quintics and the mirror family over $\mathbb{A}\setminus {0,1}$ as opposed to over $\mathbb{A}^{1}\setminus {0,\mu_{5}}$, where $\mu_{5}$ is the fifth roots of unity. Most sources say that you can do this, but don't say exactly how.

Let me first recall the the standard things that people say about constructing the family of Calabi-Yaus mirror dual to a quintic 3-fold in $\mathbb{P}^{4}$, and then I'll point out where I have a question (probably pretty simple).

Consider first the pencil of quintics $\mathcal{X}$ in $\mathbb{P}^{4} \times \mathbb{A}^{1}$ with equation $z_{0}^{5} + ... + z_{4}^{5}+5\psi z_{0}\cdots z_{4}=0$. It is easy to check that this is smooth for $\psi^{5} \neq 1$, so away from $5$th roots of unity. Note that the pencil is somewhat redundant, since for any 5th root of unity $\alpha$, $\mathcal{X}_{\psi}$ is isomorphic to $\mathcal{X}_{\alpha\psi}$ via $[a_{0}:...:a_{4}] \mapsto [\alpha^{-1}a_{0}:...:a_{4}]$.

At this point, people usually say it is more natural to take a coordinate $\lambda=\psi^{5}$ for the pencil.

I would like to interpret this in the following way: Consider the action of the group of fifth roots of unity on $\mathbb{P}^{4} \times \mathbb{A}^{1}$ via $([a_{0}:...:a_{4}],b) \mapsto ([\alpha^{-1}a_{0}:...:a_{4}],\alpha b)$, which is free on $\mathbb{P}^{4} \times (\mathbb{A}^{1} \setminus{0})$ and restricts to a free action on $\mathcal{X}$ (minus the zero fibre). We should therefore be able to take a quotient family $\mathcal{X}/\mu_{5} \rightarrow \mathbb{A}^{1} \setminus 0$, which will be smooth away from $1$.

  1. Does this work?

  2. Is there a more concrete description of this family, in terms of an equation for some pencil?

Now to get the mirror family from this point, we take a further quotient of the second family to get a third family, fibrewise. This third family has quotient singularities along the fibres and we want to construct a simultaneous crepant resolution. Surely there is some work involved in showing that you can do this. Usually people do this first, before taking the redundancy out of the family with parameter $\psi$.

  1. Can we construct a simultaneous crepant resolution of the second family with 'coordinate $\lambda=\psi^{5}$'?

  2. Is there a concrete way to write down such a family of mirror quintics, smooth over $\mathbb{A}^{1} \setminus 0,1$? Say as a pencil of hypersurfaces in some ambient toric variety?

Note that I really want the resulting family to be non-redundant. If it were even the restriction of the universal family over the moduli stack of mirror quintics, that would be ideal.

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An explicit quintic parametrized by $\lambda = \psi^5$ is $$ \lambda^{-1} \alpha_0^5 + \alpha_1^5 + \alpha_2^5 + \alpha_3^5 + \alpha_4^5 = 5 \alpha_0 \alpha_1 \alpha_2 \alpha_3 \alpha_4. $$ Indeed if $\lambda = \psi^5$ then the first term is $(\alpha/\psi)^5$, and then replacing $\alpha$ by $\psi\alpha$ recovers the fiber of $\cal X\phantom.$ at $\psi$. – Noam D. Elkies Dec 23 '11 at 18:42

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