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Let $X$ and $Y$ be algebraic varieties over $\mathbb{C}$. I am repeatedly encountering references to crepant morphisms $f:X\rightarrow Y$. I have found several definitions of such a morphism, one of which is the condition that $f^*(K_Y)=K_X$. Is this generally accepted to be the meaning of a crepant morphism, or is this only the definition in some more restrictive context? Also, I would appreciate some perspective on why one naturally considers this class of morphisms.

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Crepant stands for non-discrepant. It's frequently applied to resolutions of singularities or birational maps (but can be applied more generally).

Let's start with the birational case, since that's where the history is. If $f : X \to Y$ is birational, and $K_Y$ is $\mathbb{Q}$-Cartier, then $f^*(K_Y)$ makes sense. In particular, if $nK_Y$ is Cartier, then $f^*(K_Y) = \frac{1}{n} f^*(nK_Y)$ by definition.

Write $K_X - f^* K_Y = \sum a_i E_i$ where we pick $K_X$ and $K_Y$ which agree where $f$ is an isomorphism. The $\sum a_i E_i$ is then independent of choices.

Now, the numbers $a_i$ are called discrepancies. If there are no discrepancies (ie, all the $a_i$ are zero (for example, if $f$ is a small map), then the map is called crepant. Of course, all $a_i = 0$ if and only if $K_X = f^* K_Y$.

Of course, if the pullback of $K_X$ is $K_Y$, then this can be applied to many things. The existence of a crepant resolution of singularities also can be quite useful. Let me give a nonstandard example in characteristic $p > 0$, if $Y$ is Frobenius split and $f : X \to Y$ is crepant, then $X$ is also Frobenius split. Some variant of this appeared in the work of Mehta-van der Kallen and also Mehta-Srinivas.

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  • $\begingroup$ Suppose that the morphism $f:X\rightarrow Y$ is finite, dominant, and birational. Suppose also that $X$ and $Y$ are affine. Is it then automatic that $f$ is crepant? It seems to be the case for the morphisms $\mathbb{C}^*\rightarrow\mathbb{C}^*$, $z\mapsto z^n$. $\endgroup$ May 16 '13 at 23:18
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    $\begingroup$ Dear PDC, I don't think the map you wrote down is birational. A finite dominant birational morphism between normal varieties is automatically an isomorphism, is it not? $\endgroup$ May 17 '13 at 1:19

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