# Is a terminal symplectic variety S_4?

For purposes of this question "terminal symplectic variety" means a normal variety which is symplectic (in the usual sense of symplectic singularities) and whose singular locus has codimension $\geq 4$ (the equivalence of this with the usual definition of terminal is a theorem of Namikawa).

Symplectic varieties has lots of nice properties; For example, they are always Cohen-Macaulay, which is a condition about niceness with respect to depth. So, one can hope for others.

Is a terminal symplectic variety necessarily $S_4$?

I should warn any potential answerers that my understanding of the $S_4$ property is very poor (it's an exceptionally tough thing to Google, since it's not even the dominant use of that term in mathematics). I hesitate to even give a definition for fear of messing it up; I believe it means that every ideal sheaf of codimension $\leq 3$ has depth equal to its codimension.

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The condition Cohen-Macaulay is equivalent to $\text{S}_n$ holding for every integer $n > 0$. – Karl Schwede May 6 '11 at 1:46
Maybe I should ellaborate on my comment. A ring $R$ is $S_n$ if the depth of every local ring $R_{\mathfrak p}$ is at least $\text{min}(\dim R_{\mathfrak p}, n)$. Cohen-Macaulay means that the depth at every local ring equals the dimension of that local ring. – Karl Schwede May 6 '11 at 1:55
@Karl: That looks like it should be an answer and not just a comment. – Noah Snyder May 6 '11 at 2:10
The definition of $S_n$ for sheaves is discussed here, in case someone wants to see the definition(s): mathoverflow.net/questions/22228/… – Hailong Dao May 6 '11 at 4:44

Ben,

a terminal singularity is Cohen-Macaulay and Cohen-Macaulay implies $S_n$ for all $n$.

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You're right; I was being silly and somehow missed that. Thanks for the (many) answers. – Ben Webster May 6 '11 at 2:41
No worries. Cheers. :) – Sándor Kovács May 6 '11 at 3:18

I'm posting this as an answer due to encouragement to do so from Noah Snyder in the comments.

As mentioned above in my comment. A ring $R$ is ${\text{S}}_n$ if for every prime ideal $\mathfrak{q} \subseteq R$, $R_q$ has depth at least $\text{min}(\dim R_{\mathfrak q}, n)$.

A ring is Cohen-Macaulay if the depth of each $R_{\mathfrak{q}}$ is equal to $\dim R_{\mathfrak q}$. Thus a ring is Cohen-Macaulay if and only if it is $\text{S}_n$ for every $n$. In particular, any terminal, or even log terminal singularity is Cohen-Macaulay and thus also $\text{S}_4$.

Log canonical singularities don't satisfy this property (they need to be Cohen-Macaulay). However, if you have a family of varieties with log canonical singularities say over a smooth curve, the $\text{S}_n$ property is is closed (in particular, a limit of $\text{S}_n$ varieties is $\text{S}_n$). See the paper of Kollar-Kovacs, HERE

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Thanks to you too for your many answers. Hopefully I can stop being stupid about algebraic geometry soon. – Ben Webster May 6 '11 at 3:09