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.
Crepant stands for nondiscrepant. 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 Mehtavan der Kallen and also MehtaSrinivas.

$\begingroup$ 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

1$\begingroup$ 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