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the reason for my question is the following: the two-dimensional canonical singularities are the ADE-singularities, which all are quotients of either affine space or another ADE-singularity by finite abelian Groups and even more, quotients of affine 2-space by finite solvable Groups but one - namely $E_8$, which is factorial and a quotient of $A_1$ by the perfect Group $A_5$.

Now a generalization of the ADE-singularities in Dimension 3 are the compound du Val singularities, which are canonical and are formally equivalent to $f(x,y,z)+tg(x,y,z,t)$, where f is the polynomial of an ADE-singularity.

Now some of these are again quotients of affine 3-space, while a lot more than the one before ($E_8$) are factorial. For example the one in the heading. Which means in particular, that they cannot be a Quotient of affine space by a solvable Group.

As you may see now, it would be very nice, if these singularities were also some quotients by any Groups. Any idea? I am quite sure that it can't be a finite Group and affine 3-space.

Thank you and looking forward to your suggestions, Lukas

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    $\begingroup$ For the record, at time of writing this question has attracted a vote to close as "unclear what you're asking". I strongly disagree with this reason given $\endgroup$ – Yemon Choi May 8 '17 at 11:17
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Let $B=k[x,y,z,u]$ be a polynomial ring in four variables over a field $k$ of characteristic $0$, and let $X$ be 4-dimensional affine space over $k$. A locally nilpotent derivation $D$ of $B$ induces an algebraic action of the additive group $G=(k,+)$ on $X$ via the exponential mapping ${\rm exp} (tD)$, $t \in k$. The ring of invariants of the group action is equal to the kernel of the derivation, which is a UFD of transcendence degree $3$ over $k$ (It is not known whether the kernel must be affine).

Consider the triangular derivation $D: u \to z \to y \to x^n$ and $x \to 0$, where $n$ is a positive integer. The kernel of $D$ is of the form $k[x,f,g,h]$, where $x^{2n}u+f^3+g^2=0$. The corresponding algebraic variety $X/G$ is therefore of the type you are looking for.

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