Timeline for Matrices in $SL(2,\mathbb{C})$ with characteristic polynomial defined over a subring
Current License: CC BY-SA 3.0
19 events
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| Oct 29, 2016 at 18:27 | comment | added | YCor | @PeterSamuelson it does not distinguish closure of orbits. For instance the set of pairs of matrices of determinant 1 and with all traces of $A,B,AB$ equal to 2 is not the closure of a single orbit. | |
| Oct 29, 2016 at 18:17 | comment | added | Peter Samuelson | @YCor Right, when I wrote "in that sense" I was being lazy/imprecise, and should have said they distinguish closures of orbits. | |
| Oct 29, 2016 at 17:34 | comment | added | YCor | @PeterSamuelson that some polynomials generate the ring of invariant functions does not mean that they separate orbits (a trivial counterexample is: if $A$ is nilpotent, then this triple of traces does not separate $(A,A)$ to $(A,tA)$ although these pairs are not conjugate). | |
| Oct 29, 2016 at 9:28 | comment | added | Peter Samuelson | The functions $tr(A)$, $tr(B)$, and $tr(AB)$ generate the ring of $GL_2$-invariant functions on $Hom(F_2, SL_2)$, so in that sense they completely determine the pair $(A,B)$ up to conjugacy. A good reference is the book by Brumfiel and Hilden, "SL_2 representations of finitely presented groups." They work over $\mathbb Z$ or $\mathbb Z[1/2]$ for the whole book. | |
| Oct 29, 2016 at 2:55 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Oct 29, 2016 at 2:18 | history | edited | YCor | CC BY-SA 3.0 |
I separated some easily solvable part of the question to emphasize the real one
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| Oct 29, 2016 at 1:52 | comment | added | stupid_question_bot | @YCor Ah great! Would you happen to know if the trace of $A,B,AB$ completely determines the pair $(A,B)$ up to conjugacy? Would you happen to know of a good reference for this? | |
| Oct 29, 2016 at 1:40 | comment | added | YCor | Yes assuming that the trace of $AB$ is in $R$ changes the answer, then there's a positive result. | |
| Oct 29, 2016 at 1:37 | comment | added | YCor | $e_{ij}(t)=I_2+tE_{ij}$, where $(E_{ij})$ is the canonical basis of matrices. | |
| Oct 29, 2016 at 1:33 | comment | added | stupid_question_bot | @YCor (my motivation is to see if the character variety of homorphisms $F_2\rightarrow SL(2,\mathbb{C})$ of commutator trace $k$ given by the equation $x^2 + y^2 + z^2 - xyz-2 = k$, where $x,y,z$ are the traces of $A,B,AB$ respectively, is functorial in the ring of coefficients of $SL_2$) | |
| Oct 29, 2016 at 1:25 | comment | added | stupid_question_bot | @YCor Hmm, so in my case I also know that $AB$ also has trace in $R$. Would that change the answer to (2)? | |
| Oct 29, 2016 at 1:23 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Oct 29, 2016 at 0:50 | comment | added | stupid_question_bot | @YCor Sorry, what is $e_{12}(\cdots)$? | |
| Oct 29, 2016 at 0:43 | review | Close votes | |||
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| Oct 29, 2016 at 0:26 | history | edited | YCor |
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| Oct 29, 2016 at 0:25 | comment | added | YCor | 1: always: if $A$ is a scalar matrix (scalar $t$), its trace is $2t$ so $t$ has to be in the ring. If $A$ is not scalar, it's conjugated to its companion matrix, whose entries are $0,1$ or coefficients of the characteristic polynomial. | |
| Oct 29, 2016 at 0:20 | comment | added | YCor | 2: never (unless $R=\mathbf{C}$): pick $t\notin R$, define $A=e_{12}(t)$ and $B=e_{21}(1)$; then $AB$ has trace $t+2$. So $A,B$ have characteristic polynomial $(x-1)^2$ which is in $R[x]$, but $AB$ has trace $t+2$ so $A,B$ are not simultaneously conjugate into $SL_2(R)$. | |
| Oct 29, 2016 at 0:02 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Oct 28, 2016 at 23:50 | history | asked | stupid_question_bot | CC BY-SA 3.0 |