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I have been looking for an excplicit example of a birational, divisorial Mori contraction such that the exceptional locus is not equidimensional onto its image.

To agree with the setup I like, the total space should be smooth, and since the exceptional locus $E$ has codimension one the length of this contraction $f$ should be equal to $n-1$, where $n = dim E - dim f(E)$ (this is the case of a Mori contraction of submaximal length, according to the Ionescu-Wisniewski inequality, see for instance Wisniewski's paper: On contractions of extremal rays of Fano manifolds)

In particular, the total space should have dimension $5$ or more. I may imagine an example where $dim E = 4,dim f(E) = 2$ and the length is $1$, with a special fibre of dimension $3$.

This is a very specific question, but who knows? Thanks :-)

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Let $$ Y = \left\{ A = \left(\begin{smallmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{smallmatrix}\right) \right\} \cong \mathbb{A}^6. $$ Let also $$ X = \{ (A,v) \in Y \times \mathbb{P}^2 \mid A \cdot v = 0 \}. $$ Then $X \to \mathbb{P}^2$ is a smooth morphism with fiber $\mathbb{A}^4$, hence $X$ is a smooth sixfold with Picard number 1, and $X \to Y$ is a proper birational morphism, and it is easy to see it is a divisorial Mori contraction. Its fiber over a point $A \in Y$ is a single point, if the rank of $A$ is 2, a line, if the rank of $A$ is 1, and a plane, if $A = 0$. Thus, it is not equidimensional.

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  • $\begingroup$ Thank you, this is an interesting example. The relative dimension of the exceptional locus is 1 though, so the length is maximal. $\endgroup$
    – p0lydactyl
    Commented Feb 19 at 17:12

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