Timeline for Number of elements of "$\mathrm{SL}_n(\mathbb{F}_p^\times)$" mod $p$
Current License: CC BY-SA 3.0
10 events
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| Dec 25, 2014 at 21:11 | history | edited | Krishanu Sankar | CC BY-SA 3.0 |
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| Dec 25, 2014 at 20:03 | comment | added | alpoge | Yeah --- we can gchat or whatever's easiest. | |
| Dec 25, 2014 at 19:31 | comment | added | Krishanu Sankar | (continued) ... by picking any $k$ rows which are linearly independent, freely adding on a nonzero entry to the end, and then noting that the remaining entries of that last column are determined. If this number is $(p-1)^k$, then the number of ways to add another column so as to INCREASE the rank to $k+1$ must be $(p-1)^m-(p-1)^k$. Use this computation twice, considering the three cases given above (the rank $k-1$ cases splits into two subcases) to get the recurrence. I can explain this more in depth in a PM if you're interested. | |
| Dec 25, 2014 at 19:26 | comment | added | Krishanu Sankar | Of those 256 3x3 matrices with 1s in the first row and column and entries between 1 and 4, $\frac{46080}{4^5}=45$ should have determinant nonzero, unless I've messed up my computation. Does that sound reasonable? To explain where the recurrence comes from: if you have an $n \times n$ matrix whose rank is $k$, then its upper left $(n-1) \times (n-1)$ minor has rank $k-2, k-1,$ or $k$. Then, given an $m \times n$ matrix $M$ of a particular rank $k$, the number of ways to add a column of nonzero entries to the right side of the matrix so that the result has rank $k$ is $(p-1)^k$... | |
| Dec 25, 2014 at 19:02 | comment | added | alpoge | (By the way, these small numbers are the (alleged) counts once you fix the first row and column, which is why they're small --- the 4^5 was factored out. Can you explain the recurrence?) | |
| Dec 25, 2014 at 18:13 | comment | added | alpoge | Whoops! I made a computational mistake, as expected (I counted those with nonzero determinant, rather than determinant not zero mod p!). It seems the congruence mod p stabilizes in the 2x2 case to 2, and in the 3x3 case to -6? Or maybe I mixed up a sign. Anyway now I get 201 matrices, though we'll see how long that lasts... | |
| Dec 25, 2014 at 18:03 | comment | added | alpoge | But there are only 256 3x3 matrices with 1s in the first row and column and entries between 1 and 4! | |
| Dec 25, 2014 at 17:36 | comment | added | Krishanu Sankar | You can scale columns so the first row is all 1's, and then scale rows $2$ to $n$ so the first column is all 1's. Then you can subtract the first row from all of the other rows, so that you see $f(n, n)$ is $(p-1)^{2n-1}$ times the number of invertible $(n-1)\times (n-1)$ matrices that don't contain a $-1$. For $n=3$ and $p=5$, this gives that your result is divisible by $4^5$ ($4^4$ in the determinant $1$ case). $212$ isn't (in fact it's waaay too small!), so I think you made a computation error. My computation shows that $f(3, 3)=p(p-1)^6(p-2)^2$, so your answer should be $46080$... | |
| Dec 25, 2014 at 16:15 | comment | added | alpoge | Now I'm confused --- here's why I thought being nonzero was plausible. I think it suffices to count matrices with first row and column all 1s, say. But in the 3x3 case I think I get 212 such matrices over $\mathbb{F}_5$? (I just brute forced it but I probably just made a mistake in a reduction step.) Same sort of deal for 7, 11, 17, 19, 23 and then I stopped. (Not 13, though.) | |
| Dec 25, 2014 at 15:24 | history | answered | Krishanu Sankar | CC BY-SA 3.0 |