AKT-09/HW1: Difference between revisions
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{{AKT-09/Navigation}} |
{{AKT-09/Navigation}} |
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{{In Preparation}} |
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'''Solve the following problems''' and submit them in class by October 13, |
'''Solve the following problems''' and submit them in class by October 13, 2009: |
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'''Problem 1.''' If <math>f \in {\mathcal V}_n</math> and <math>g \in {\mathcal V}_m</math> then <math>f \cdot g \in {\mathcal V}_{n+m}</math> (as what one would expect by looking at degrees of polynomials) and <math>W_{f \cdot g} = m_\mathbb{Q} \circ (W_f \otimes W_g) \circ \ |
'''Problem 1.''' If <math>f \in {\mathcal V}_n</math> and <math>g \in {\mathcal V}_m</math> then <math>f \cdot g \in {\mathcal V}_{n+m}</math> (as what one would expect by looking at degrees of polynomials) and <math>W_{f \cdot g} = m_\mathbb{Q} \circ (W_f \otimes W_g) \circ \Box</math> where <math>(W_f \otimes W_g) \circ \Box: {\mathcal A} \rightarrow \mathbb{Q} \otimes \mathbb{Q}</math> and <math>m_\mathbb{Q}</math> is the multiplication of rationals. (See {{AKT-09/vps|0924-2}}, minute 36:01). |
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'''Problem 2.''' Let <math>\Theta:{\mathcal A}\to{\mathcal A}</math> be the multiplication operator by the 1-chord diagram <math>\theta</math>, and let <math>\partial_\theta=\frac{d}{d\theta}</math> be the adjoint of multiplication by <math>W_\theta</math> on <math>{\mathcal A}^\star</math>, where <math>W_\theta</math> is the obvious dual of <math>\theta</math> in <math>{\mathcal A}^\star</math>. Let <math>P:{\mathcal A}\to{\mathcal A}</math> be defined by |
'''Problem 2.''' Let <math>\Theta:{\mathcal A}\to{\mathcal A}</math> be the multiplication operator by the 1-chord diagram <math>\theta</math>, and let <math>\partial_\theta=\frac{d}{d\theta}</math> be the adjoint of multiplication by <math>W_\theta</math> on <math>{\mathcal A}^\star</math>, where <math>W_\theta</math> is the obvious dual of <math>\theta</math> in <math>{\mathcal A}^\star</math>. Let <math>P:{\mathcal A}\to{\mathcal A}</math> be defined by |
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{{Equation*|<math>P = \sum_{n=0}^\infty \frac{(-\Theta)^n}{n!}\partial_\theta^n.</math>}} |
{{Equation*|<math>P = \sum_{n=0}^\infty \frac{(-\Theta)^n}{n!}\partial_\theta^n.</math>}} |
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Verify the following assertions, but submit only your work on assertions 4,5,7,11: |
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# <math>\left[\partial_\theta,\Theta\right]=1</math>, where <math>1:{\mathcal A}\to{\mathcal A}</math> is the identity map and where <math>[A,B]:=AB-BA</math> for any two operators. |
# <math>\left[\partial_\theta,\Theta\right]=1</math>, where <math>1:{\mathcal A}\to{\mathcal A}</math> is the identity map and where <math>[A,B]:=AB-BA</math> for any two operators. |
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# <math>\partial_\theta</math> satisfies Leibnitz' law: <math>\partial_\theta(ab)=(\partial_\theta a)b+a(\partial_\theta b)</math> for any <math>a,b\in{\mathcal A}</math>. |
# <math>\partial_\theta</math> satisfies Leibnitz' law: <math>\partial_\theta(ab)=(\partial_\theta a)b+a(\partial_\theta b)</math> for any <math>a,b\in{\mathcal A}</math>. |
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# <math>P</math> is an algebra morphism: <math>P1=1</math> and <math>P(ab)=(Pa)(Pb)</math>. |
# <math>P</math> is an algebra morphism: <math>P1=1</math> and <math>P(ab)=(Pa)(Pb)</math>. |
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# <math>\Theta</math> satisfies the co-Leibnitz law: <math>\Box\circ\Theta=(\Theta\otimes 1+1\otimes\Theta)\circ\Box</math> (why does this deserve the name |
# <math>\Theta</math> satisfies the co-Leibnitz law: <math>\Box\circ\Theta=(\Theta\otimes 1+1\otimes\Theta)\circ\Box</math> (why does this deserve the name "the co-Leibnitz law"?). |
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# <math>P</math> is a co-algebra morphism: <math>\eta\circ P=\eta</math> (where <math>\eta</math> is the co-unit of <math>{\mathcal A}</math>) and <math>\Box\circ P=(P\otimes P)\circ\Box</math>. |
# <math>P</math> is a co-algebra morphism: <math>\eta\circ P=\eta</math> (where <math>\eta</math> is the co-unit of <math>{\mathcal A}</math>) and <math>\Box\circ P=(P\otimes P)\circ\Box</math>. |
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# <math>P\theta=0</math> and hence <math>P\langle\theta\rangle=0</math>, where <math>\langle\theta\rangle</math> is the ideal generated by <math>\theta</math> in the algebra <math>{\mathcal A}</math>. |
# <math>P\theta=0</math> and hence <math>P\langle\theta\rangle=0</math>, where <math>\langle\theta\rangle</math> is the ideal generated by <math>\theta</math> in the algebra <math>{\mathcal A}</math>. |
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# If <math>Q:{\mathcal A}\to{\mathcal A}</math> is defined by {{Equation*|<math>Q = \sum_{n=0}^\infty \frac{(-\Theta)^n}{(n+1)!}\partial_\theta^{(n+1)}</math>}} then <math>a=\theta Qa+Pa</math> for all <math>a\in{\mathcal A}</math>. |
# If <math>Q:{\mathcal A}\to{\mathcal A}</math> is defined by {{Equation*|<math>Q = \sum_{n=0}^\infty \frac{(-\Theta)^n}{(n+1)!}\partial_\theta^{(n+1)}</math>}} then <math>a=\theta Qa+Pa</math> for all <math>a\in{\mathcal A}</math>. |
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# <math>\ker P=\langle\theta\rangle</math>. |
# <math>\ker P=\langle\theta\rangle</math>. |
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# <math>P</math> descends to a Hopf algebra morphism <math>{\mathcal A}^r\to{\mathcal A}</math>, and if <math>\pi:{\mathcal A}\to{\mathcal A}^r</math> is the obvious projection, then <math>\pi\circ P</math> is the identity of <math>{\mathcal A}^r</math>. (Recall that <math>{\mathcal A}^r={\mathcal A}/\langle\theta\rangle</math> |
# <math>P</math> descends to a Hopf algebra morphism <math>{\mathcal A}^r\to{\mathcal A}</math>, and if <math>\pi:{\mathcal A}\to{\mathcal A}^r</math> is the obvious projection, then <math>\pi\circ P</math> is the identity of <math>{\mathcal A}^r</math>. (Recall that <math>{\mathcal A}^r={\mathcal A}/\langle\theta\rangle</math>). |
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# <math>P^2=P</math>. |
# <math>P^2=P</math>. |
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'''Idea for a good deed.''' Later than October 13, prepare a [[AKT-09/Sol1|beautiful TeX writeup]] (including the motivation and all the details) of the solution of this assignment for publication on the web. For all I know this information in this form is not available elsewhere. |
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'''Mandatory but unenforced.''' Find yourself in the class photo and identify yourself as explained in the [[AKT-09/Class Photo|photo page]]. |
'''Mandatory but unenforced.''' Find yourself in the class photo and identify yourself as explained in the [[AKT-09/Class Photo|photo page]]. |
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Latest revision as of 12:00, 19 October 2009
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Solve the following problems and submit them in class by October 13, 2009:
Problem 1. If [math]\displaystyle{ f \in {\mathcal V}_n }[/math] and [math]\displaystyle{ g \in {\mathcal V}_m }[/math] then [math]\displaystyle{ f \cdot g \in {\mathcal V}_{n+m} }[/math] (as what one would expect by looking at degrees of polynomials) and [math]\displaystyle{ W_{f \cdot g} = m_\mathbb{Q} \circ (W_f \otimes W_g) \circ \Box }[/math] where [math]\displaystyle{ (W_f \otimes W_g) \circ \Box: {\mathcal A} \rightarrow \mathbb{Q} \otimes \mathbb{Q} }[/math] and [math]\displaystyle{ m_\mathbb{Q} }[/math] is the multiplication of rationals. (See 090924-2, minute 36:01).
Problem 2. Let [math]\displaystyle{ \Theta:{\mathcal A}\to{\mathcal A} }[/math] be the multiplication operator by the 1-chord diagram [math]\displaystyle{ \theta }[/math], and let [math]\displaystyle{ \partial_\theta=\frac{d}{d\theta} }[/math] be the adjoint of multiplication by [math]\displaystyle{ W_\theta }[/math] on [math]\displaystyle{ {\mathcal A}^\star }[/math], where [math]\displaystyle{ W_\theta }[/math] is the obvious dual of [math]\displaystyle{ \theta }[/math] in [math]\displaystyle{ {\mathcal A}^\star }[/math]. Let [math]\displaystyle{ P:{\mathcal A}\to{\mathcal A} }[/math] be defined by
Verify the following assertions, but submit only your work on assertions 4,5,7,11:
- [math]\displaystyle{ \left[\partial_\theta,\Theta\right]=1 }[/math], where [math]\displaystyle{ 1:{\mathcal A}\to{\mathcal A} }[/math] is the identity map and where [math]\displaystyle{ [A,B]:=AB-BA }[/math] for any two operators.
- [math]\displaystyle{ P }[/math] is a degree [math]\displaystyle{ 0 }[/math] operator; that is, [math]\displaystyle{ \deg Pa=\deg a }[/math] for all [math]\displaystyle{ a\in{\mathcal A} }[/math].
- [math]\displaystyle{ \partial_\theta }[/math] satisfies Leibnitz' law: [math]\displaystyle{ \partial_\theta(ab)=(\partial_\theta a)b+a(\partial_\theta b) }[/math] for any [math]\displaystyle{ a,b\in{\mathcal A} }[/math].
- [math]\displaystyle{ P }[/math] is an algebra morphism: [math]\displaystyle{ P1=1 }[/math] and [math]\displaystyle{ P(ab)=(Pa)(Pb) }[/math].
- [math]\displaystyle{ \Theta }[/math] satisfies the co-Leibnitz law: [math]\displaystyle{ \Box\circ\Theta=(\Theta\otimes 1+1\otimes\Theta)\circ\Box }[/math] (why does this deserve the name "the co-Leibnitz law"?).
- [math]\displaystyle{ P }[/math] is a co-algebra morphism: [math]\displaystyle{ \eta\circ P=\eta }[/math] (where [math]\displaystyle{ \eta }[/math] is the co-unit of [math]\displaystyle{ {\mathcal A} }[/math]) and [math]\displaystyle{ \Box\circ P=(P\otimes P)\circ\Box }[/math].
- [math]\displaystyle{ P\theta=0 }[/math] and hence [math]\displaystyle{ P\langle\theta\rangle=0 }[/math], where [math]\displaystyle{ \langle\theta\rangle }[/math] is the ideal generated by [math]\displaystyle{ \theta }[/math] in the algebra [math]\displaystyle{ {\mathcal A} }[/math].
- If [math]\displaystyle{ Q:{\mathcal A}\to{\mathcal A} }[/math] is defined by
[math]\displaystyle{ Q = \sum_{n=0}^\infty \frac{(-\Theta)^n}{(n+1)!}\partial_\theta^{(n+1)} }[/math] then [math]\displaystyle{ a=\theta Qa+Pa }[/math] for all [math]\displaystyle{ a\in{\mathcal A} }[/math]. - [math]\displaystyle{ \ker P=\langle\theta\rangle }[/math].
- [math]\displaystyle{ P }[/math] descends to a Hopf algebra morphism [math]\displaystyle{ {\mathcal A}^r\to{\mathcal A} }[/math], and if [math]\displaystyle{ \pi:{\mathcal A}\to{\mathcal A}^r }[/math] is the obvious projection, then [math]\displaystyle{ \pi\circ P }[/math] is the identity of [math]\displaystyle{ {\mathcal A}^r }[/math]. (Recall that [math]\displaystyle{ {\mathcal A}^r={\mathcal A}/\langle\theta\rangle }[/math]).
- [math]\displaystyle{ P^2=P }[/math].
Idea for a good deed. Later than October 13, prepare a beautiful TeX writeup (including the motivation and all the details) of the solution of this assignment for publication on the web. For all I know this information in this form is not available elsewhere.
Mandatory but unenforced. Find yourself in the class photo and identify yourself as explained in the photo page.