Chebyshev polynomials and trace of $A in SL_2(mathbb{C})$












1












$begingroup$



Defining $C_n(z) = frac{z^m + z^{-m}}{2}$, the Chebyshev polynomials are defined by



$$T_n(C_1(z)) = C_n(z)$$
and are given by $T_1(z) = z, T_2(z) = 2z^2-1, T_3(z) = 4z^3-3z$, etc. Since for $z=e^{itheta}$ we have $C_1(z) = costheta$, they also satisfy



$$T_n(costheta) = cos(ntheta)$$



thereby generalizing the double angle trig identity.




The notes I'm reading also claim $$T_n(text{tr } A/2) = text{tr } (A^n/2)$$ for $A in SL_2(mathbb{C})$. Why does this follow?



Attempt: if $A= begin{pmatrix} a&b \ c &d end{pmatrix}$ then choosing $z=frac12(sqrt{(a+d)^2-4}- (a+d))$ implies $C_1(z) = text{tr }A/2$, but then $T_n(C_1(z)) = frac{z^n+z^{-n}}{2}$ does not simplify as far as I see to $text{tr} A^n/2$.






complex-analysis functions trigonometry chebyshev-polynomials







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  • 1




    $begingroup$
    Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
    $endgroup$
    – WimC
    Dec 25 '18 at 18:09










  • $begingroup$
    @WimC thanks, that resolves it
    $endgroup$
    – Dwagg
    Dec 25 '18 at 18:57
















1












$begingroup$



Defining $C_n(z) = frac{z^m + z^{-m}}{2}$, the Chebyshev polynomials are defined by



$$T_n(C_1(z)) = C_n(z)$$
and are given by $T_1(z) = z, T_2(z) = 2z^2-1, T_3(z) = 4z^3-3z$, etc. Since for $z=e^{itheta}$ we have $C_1(z) = costheta$, they also satisfy



$$T_n(costheta) = cos(ntheta)$$



thereby generalizing the double angle trig identity.




The notes I'm reading also claim $$T_n(text{tr } A/2) = text{tr } (A^n/2)$$ for $A in SL_2(mathbb{C})$. Why does this follow?



Attempt: if $A= begin{pmatrix} a&b \ c &d end{pmatrix}$ then choosing $z=frac12(sqrt{(a+d)^2-4}- (a+d))$ implies $C_1(z) = text{tr }A/2$, but then $T_n(C_1(z)) = frac{z^n+z^{-n}}{2}$ does not simplify as far as I see to $text{tr} A^n/2$.






complex-analysis functions trigonometry chebyshev-polynomials







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
    $endgroup$
    – WimC
    Dec 25 '18 at 18:09










  • $begingroup$
    @WimC thanks, that resolves it
    $endgroup$
    – Dwagg
    Dec 25 '18 at 18:57














1












1








1





$begingroup$



Defining $C_n(z) = frac{z^m + z^{-m}}{2}$, the Chebyshev polynomials are defined by



$$T_n(C_1(z)) = C_n(z)$$
and are given by $T_1(z) = z, T_2(z) = 2z^2-1, T_3(z) = 4z^3-3z$, etc. Since for $z=e^{itheta}$ we have $C_1(z) = costheta$, they also satisfy



$$T_n(costheta) = cos(ntheta)$$



thereby generalizing the double angle trig identity.




The notes I'm reading also claim $$T_n(text{tr } A/2) = text{tr } (A^n/2)$$ for $A in SL_2(mathbb{C})$. Why does this follow?



Attempt: if $A= begin{pmatrix} a&b \ c &d end{pmatrix}$ then choosing $z=frac12(sqrt{(a+d)^2-4}- (a+d))$ implies $C_1(z) = text{tr }A/2$, but then $T_n(C_1(z)) = frac{z^n+z^{-n}}{2}$ does not simplify as far as I see to $text{tr} A^n/2$.






complex-analysis functions trigonometry chebyshev-polynomials







share|cite|improve this question









$endgroup$





Defining $C_n(z) = frac{z^m + z^{-m}}{2}$, the Chebyshev polynomials are defined by



$$T_n(C_1(z)) = C_n(z)$$
and are given by $T_1(z) = z, T_2(z) = 2z^2-1, T_3(z) = 4z^3-3z$, etc. Since for $z=e^{itheta}$ we have $C_1(z) = costheta$, they also satisfy



$$T_n(costheta) = cos(ntheta)$$



thereby generalizing the double angle trig identity.




The notes I'm reading also claim $$T_n(text{tr } A/2) = text{tr } (A^n/2)$$ for $A in SL_2(mathbb{C})$. Why does this follow?



Attempt: if $A= begin{pmatrix} a&b \ c &d end{pmatrix}$ then choosing $z=frac12(sqrt{(a+d)^2-4}- (a+d))$ implies $C_1(z) = text{tr }A/2$, but then $T_n(C_1(z)) = frac{z^n+z^{-n}}{2}$ does not simplify as far as I see to $text{tr} A^n/2$.






complex-analysis functions trigonometry chebyshev-polynomials




complex-analysis functions trigonometry chebyshev-polynomials






share|cite|improve this question













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asked Dec 25 '18 at 17:44









DwaggDwagg

307111




307111








  • 1




    $begingroup$
    Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
    $endgroup$
    – WimC
    Dec 25 '18 at 18:09










  • $begingroup$
    @WimC thanks, that resolves it
    $endgroup$
    – Dwagg
    Dec 25 '18 at 18:57














  • 1




    $begingroup$
    Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
    $endgroup$
    – WimC
    Dec 25 '18 at 18:09










  • $begingroup$
    @WimC thanks, that resolves it
    $endgroup$
    – Dwagg
    Dec 25 '18 at 18:57








1




1




$begingroup$
Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
$endgroup$
– WimC
Dec 25 '18 at 18:09




$begingroup$
Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
$endgroup$
– WimC
Dec 25 '18 at 18:09












$begingroup$
@WimC thanks, that resolves it
$endgroup$
– Dwagg
Dec 25 '18 at 18:57




$begingroup$
@WimC thanks, that resolves it
$endgroup$
– Dwagg
Dec 25 '18 at 18:57










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$begingroup$

I found several sources with a proof, e.g., the paper by Francis Bonahon, Lemma $8$ on page $9$, using Cayley-Hamilton. Another interesting reference is the paper by Traina on trace polynomials for $SL_2(Bbb{C})$.






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    $begingroup$

    I found several sources with a proof, e.g., the paper by Francis Bonahon, Lemma $8$ on page $9$, using Cayley-Hamilton. Another interesting reference is the paper by Traina on trace polynomials for $SL_2(Bbb{C})$.






    share|cite|improve this answer









    $endgroup$


















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      $begingroup$

      I found several sources with a proof, e.g., the paper by Francis Bonahon, Lemma $8$ on page $9$, using Cayley-Hamilton. Another interesting reference is the paper by Traina on trace polynomials for $SL_2(Bbb{C})$.






      share|cite|improve this answer









      $endgroup$
















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        2








        2





        $begingroup$

        I found several sources with a proof, e.g., the paper by Francis Bonahon, Lemma $8$ on page $9$, using Cayley-Hamilton. Another interesting reference is the paper by Traina on trace polynomials for $SL_2(Bbb{C})$.






        share|cite|improve this answer









        $endgroup$



        I found several sources with a proof, e.g., the paper by Francis Bonahon, Lemma $8$ on page $9$, using Cayley-Hamilton. Another interesting reference is the paper by Traina on trace polynomials for $SL_2(Bbb{C})$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 18:08









        Dietrich BurdeDietrich Burde

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