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Let $L$ be an indefinite {\it non-unimodular} integral lattice. I am particularly interested in unimodular cases, such as $U(2)\oplus A_4, U\oplus D_4$. Are there any general method to determine whether or not the orthogonal group $O(L)$ is a finite group?

I am aware of this similar question Automorphism groups of indefinite non-unimodular integer lattices The difference is, I am interested only in whether or not $O(L)$ is finite.

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  • $\begingroup$ I have no idea how any of these could be finite. Indefinite binary forms (nonsquare discriminant) have infinite groups, here you have at least total dimension 4. If you have any finite examples, please let me know. $\endgroup$
    – Will Jagy
    Commented Aug 5, 2013 at 23:29
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    $\begingroup$ @Will: In all dimensions $\ge 3$ it has to be infinite: the special orthogonal and spin groups are always semisimple (even if not absolutely simple for dimension 4), so one can apply strong approximation relative to the infinite place for the simply connected spin group and open subset of its finite-adelic points corresponding to the $\widehat{\mathbf{Z}}$-points relative to the $\mathbf{Z}$-structure provided by the lattice (which gives a $\mathbf{Z}$-structure to the Clifford algebra and hence to the spin group, thereby making sense of the $\widehat{\mathbf{Z}}$-points just mentioned). $\endgroup$
    – user36938
    Commented Aug 6, 2013 at 4:38
  • $\begingroup$ @user36938, thanks. There are times that I wish I had actually studied the full theory. Lots of stuff I like is written in adelic language, probably starting with Kneser 1961. $\endgroup$
    – Will Jagy
    Commented Aug 6, 2013 at 4:51
  • $\begingroup$ @user36938, you might like this question on the other site, math.stackexchange.com/questions/459064/adelic-lattices The kid asking is a graduate student at Harvard and might well start using MO... $\endgroup$
    – Will Jagy
    Commented Aug 6, 2013 at 14:23
  • $\begingroup$ @Will: Someone fortunate enough to be a graduate student at Harvard has many people nearby to speak with in person about things like that. (For a slick proof, he should make more effective use of the fact that any two adelic lattices are commensurable with each other, and that tensoring against $\widehat{\mathbf{Z}}$ has no effect on the "lattice" of finite-index subgroups.) $\endgroup$
    – user36938
    Commented Aug 6, 2013 at 15:04

1 Answer 1

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Maybe it is too late. A lattice has a finite orthogonal group if and only if it is definite or it is on the hyperbolic plane. This can be deduced from Satz (30.4) in Kneser's book Quadratische Formen (taking L = M there).

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  • $\begingroup$ Do you mean $U(k)$ by the hyperbolic plane? $\endgroup$
    – user2013
    Commented Oct 31, 2013 at 20:14
  • $\begingroup$ Hyperbolic plane is the rank 2 lattice with Gram matrix $\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$. This is probably the $U$ in the original question. $\endgroup$
    – WKC
    Commented Nov 1, 2013 at 2:16
  • $\begingroup$ Isn't $O(U(K)))$ finite? $\endgroup$
    – user2013
    Commented Nov 2, 2013 at 22:23
  • $\begingroup$ I assume that $U(k)$ is the orthogonal sum of $k$ copies of the hyperbolic plane. When $k \geq 2$, $O(U(k))$ is infinite. Suppose $k = 2$, and $e,f$ and $x,y$ are two orthogonal pairs of hyperbolic pairs. Then for any integer $a$, the linear map such that $e\mapsto e + ax$, $f \mapsto f + ax$, $x \mapsto -x$, and $y \mapsto ae + af + a^2x - y$, is an isometry of $U(2)$. $\endgroup$
    – WKC
    Commented Nov 3, 2013 at 0:36
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    $\begingroup$ I see what you mean (isn't your conditions on $e$ and $f$ should be then $e^2 = f^2 = (e,f) - k = 0$?). It is my fault--Kneser's statement should be ``...it is definite or it is on the hyerbolic plane..." $\endgroup$
    – WKC
    Commented Nov 3, 2013 at 1:53

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