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Let $(0\in X)$ be the germ of a normal surface singularity and let $f: Y \to X$ be the minimal resolution.

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(1) How can I define a map $f_*\mathcal{O}_Y(K_Y)\hookrightarrow \mathcal{O}_X(K_X)$?

(2) If $f_*\mathcal{O}_Y(K_Y)=\mathcal{O}_X(K_X)$ and $K_X$ is Catier, then $f^*\mathcal{O}_X(K_X)=\mathcal{O}_Y(K_Y)$. Why?

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  • $\begingroup$ (1) is the trace map in Grothendieck duality. There are various references for this. (2) is false. The first condition of (2) is the condition for $X$ to have rational singularities. It does not imply that $X$ is Gorenstein ($K_X$ Cartier). $\endgroup$ May 28, 2015 at 9:30
  • $\begingroup$ @Donu Arapura- Sorry, I corrected (2). And thank you for the answer. $\endgroup$
    – Donghoa
    May 28, 2015 at 10:08

1 Answer 1

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Concerning (1), the map in question can be described in more down-to-earth terms as follows. Let $E = \sum E_i \subset Y$ be the exceptional divisor of $f$. A section of $f_* \mathcal O_Y(K_Y)$ is by definition nothing but a section of $\mathcal O_Y(K_Y)$, which you can restrict to $Y \setminus E$. As $f$ restricts to an isomorphism $Y \setminus E \to X \setminus \{ 0 \}$, you get a section $\sigma$ of $\mathcal O_X(K_X)$ on $X \setminus \{ 0 \}$. Since $\mathcal O_X(K_X)$ is reflexive (as is any Weil divisorial sheaf on a normal variety), you can extend $\sigma$ to all of $X$. This defines a map $f_* \mathcal O_Y(K_Y) \to \mathcal O_X(K_X)$. (This actually works for normal varieties of arbitrary dimension.)

Regarding (2), as Donu pointed out, the condition $f_* \mathcal O_Y(K_Y) = \mathcal O_X(K_X)$ is equivalent to $X$ having rational singularities. (Beware, this only holds in dimension two. In general you additionally need $X$ to be Cohen-Macaulay, which is automatic for normal surface singularities.)

If $f_* \mathcal O_Y(K_Y) = \mathcal O_X(K_X)$ and $K_X$ is Cartier, then essentially by definition $X$ has canonical singularities, also known as the Du Val or ADE singularities. Now if you write down the ramification formula $$ K_Y = f^* K_X + \sum a_i E_i $$ (which makes sense because $K_X$ is Cartier), then having canonical singularities exactly means that all $a_i \ge 0$. On the other hand, since $f$ is the minimal resolution, $K_Y$ is $f$-nef and then so is $\sum a_i E_i$. By the Negativity Lemma, all $a_i \le 0$. Hence all $a_i = 0$, and $K_Y = f^* K_X$. (Obviously this fails if we blow up $Y$ further.)

A general reference for these things is Kollár-Mori, "Birational geometry of Algebraic Varieties", Chapters 4 and 5.

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