Suppose I have a Smith normal form $S,$ and I want to have an $M \in SL(n, \mathbb{Z}),$ such that $M - I$ has SNF $S.$ Is this always possible? For a (potentially) somewhat harder question, what if I require that $M \in Sp(n, \mathbb{Z})$ (and again, I want to realize a God-given SNF).
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asked Jan 11, 2014 at 2:05
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$\begingroup$ No idea. There is a book called integral Matrices I've always wanted to own, I think by Newman. amazon.com/Integral-Matrices-Monographs-Textbooks-Mathematics/… Some of this material becomes important for quadratic forms $\endgroup$– Will JagyJan 11, 2014 at 2:14
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$\begingroup$ Take a look at this ams.org/journals/proc/1985-094-01/S0002-9939-1985-0781052-8 could be good for something $\endgroup$– Will JagyJan 11, 2014 at 2:50
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2 Answers
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I suppose you mean $n>1$. For $n=2$, the matrix $A=\left[\begin{array}{cc} 1-r(1+rs) & r\\ -r(1+s+rs) & 1+r \end{array}\right]$ has determinant 1, and $A-I$ has SNF with diagonal $(r,rs)$.
answered Jan 11, 2014 at 3:17
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$\begingroup$ So, presumably that resolves the $SL(n)$ question for any even $n,$ by using this construction in blocks. Does this matrix come from somewhere, or did it just occur to you? $\endgroup$ Jan 11, 2014 at 3:23
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$\begingroup$ And it does the $Sp(2n, \mathbb{Z}) case, since $SL(2) = Sp(2).$ Truly you are wise in the ways of science. $\endgroup$ Jan 11, 2014 at 4:41
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1$\begingroup$ The matrix came from solving the equations $ad-bc=1$, gcd$(a-1,b,c,d-1)=r$, $(a-1)(d-1)-bc=r^2s$. I made the simplifying assumption $b=d=r$, $a-1=\alpha r$, $d-1=\delta r$ and solved. $\endgroup$ Jan 11, 2014 at 15:02
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I think the answer for the $Sl(n,Z)$ is positive if $n>1$. Let $D$ be a diagonal matrix, we ask if there's a $M$ such that $M-I$ is equivalent to $D$.($A$ is equivalent to $B$ if there are invertible $P,Q $such that $PAQ=B$.) This is same as asking if there's an $R$ in $SL(n,Z)$ s.t. $D+R$ is in $Sl(n,Z)$.(Reasoning:$ P(M-I)Q=PMQ-PQ$, and $PMQ$ is in $Sl(n,Z)$, $PQ$ is in $SL(n,Z)$, (neglecting the $\pm1$, which is not important)).
For any $P,Q$ in $Sl(n,Z)$, $D+R$ in $SL(n,Z)$ is same as $P(D+R)Q$ is in $SL(n,Z)$, thus we can replace $D$ with $PDQ$, so we can consider $PDQ$ to be the the follwoing matrix: the entries right above the diagonal are $d_1,d_2,\cdots,d_{n-1}$, the $(n,1)$ entry is $d_n$, all other entries are $0$. Now let $PRQ$ be the identity matrix but only the $(n,1)$ entry being $-d_n$. This matrix $R$ is the one we are looking for.
In this argument, $D$ is not necessarily a diagonal matrix, actually $D$ can be any integer matrix, because we can use SNF to transform it to a diagonal matrix.
Conclusion: if $n>1$, any $n$ by $n$ integer matrix is the difference of a pair matrices in $Sl(n,Z)$.
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$\begingroup$ I think the conclusion must have been a well-known fact: $SL(n,Z)-SL(n,z)=Mat(n,Z)$, for $n>1$. $\endgroup$ Jan 11, 2014 at 16:02