Chebyshev polynomials and trace of $A in SL_2(mathbb{C})$
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
asked 2 days ago
Dwagg
305111
add a comment |
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
asked 2 days ago
Dwagg
305111
1
Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
– WimC
2 days ago
@WimC thanks, that resolves it
– Dwagg
2 days ago
add a comment |
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
asked 2 days ago
Dwagg
305111
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
asked 2 days ago
Dwagg
305111
asked 2 days ago
Dwagg
305111
asked 2 days ago
Dwagg
305111
asked 2 days ago
Dwagg
305111
asked 2 days ago
Dwagg
305111
305111
1
Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
– WimC
2 days ago
@WimC thanks, that resolves it
– Dwagg
2 days ago
add a comment |
1
Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
– WimC
2 days ago
@WimC thanks, that resolves it
– Dwagg
2 days ago
1
1
Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
– WimC
2 days ago
Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
– WimC
2 days ago
@WimC thanks, that resolves it
– Dwagg
2 days ago
@WimC thanks, that resolves it
– Dwagg
2 days ago
add a comment |
1 Answer
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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 $SL_2(Bbb{C})$.
answered 2 days ago
Dietrich Burde
77.4k64386
add a comment |
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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 $SL_2(Bbb{C})$.
answered 2 days ago
Dietrich Burde
77.4k64386
add a comment |
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 $SL_2(Bbb{C})$.
answered 2 days ago
Dietrich Burde
77.4k64386
add a comment |
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 $SL_2(Bbb{C})$.
answered 2 days ago
Dietrich Burde
77.4k64386
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 $SL_2(Bbb{C})$.
answered 2 days ago
Dietrich Burde
77.4k64386
answered 2 days ago
Dietrich Burde
77.4k64386
answered 2 days ago
Dietrich Burde
77.4k64386
answered 2 days ago
Dietrich Burde
77.4k64386
77.4k64386
add a comment |
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Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$.
– WimC
2 days ago
@WimC thanks, that resolves it
– Dwagg
2 days ago