The Existence of the Exponential Function - Revision history

Frohlich: /* The "Duflo Homomorphism" Equation */ Remove triple quote which causes strange bold-effects in the table of contents.

2019-11-28T15:36:34Z

The "Duflo Homomorphism" Equation: Remove triple quote which causes strange bold-effects in the table of contents.

← Older revision		Revision as of 11:36, 28 November 2019
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	See {{ref\|Bar-Natan_Le_Thurston_03}}.		See {{ref\|Bar-Natan_Le_Thurston_03}}.
	{{End Side Note}}		{{End Side Note}}
	===The "Duflo Homomorphism" Equation~~'''~~===		===The "Duflo Homomorphism" Equation===
	This is the equation		This is the equation

Drorbn: /* Acknowledgment */

2007-03-08T15:43:47Z

Acknowledgment

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	===Acknowledgment===		===Acknowledgment===

	Significant parts of this paperlet were contributed by '''Omar Antolin Camarena'''. Further thanks to Yael Karshon, to Peter Lee and ~~for~~ the students of [[06-1350\|Math 1350]] (2006) in general.		Significant parts of this paperlet were contributed by '''Omar Antolin Camarena'''. Further thanks to Yael Karshon, to Peter Lee and to the students of [[06-1350\|Math 1350]] (2006) in general.

	==The Scheme==		==The Scheme==

Drorbn: /* Acknowledgment */

2007-03-08T15:42:30Z

Acknowledgment

← Older revision		Revision as of 11:42, 8 March 2007
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	===Acknowledgment===		===Acknowledgment===

	Significant parts of this paperlet were contributed by '''Omar Antolin Camarena'''.		Significant parts of this paperlet were contributed by '''Omar Antolin Camarena'''. Further thanks to Yael Karshon, to Peter Lee and for the students of [[06-1350\|Math 1350]] (2006) in general.

	==The Scheme==		==The Scheme==

Drorbn: /* Finding a Syzygy */

2007-03-08T15:32:46Z

Finding a Syzygy

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	==Finding a Syzygy==		==Finding a Syzygy==

			[[Image:A Syzygy for Exponentiation.png\|320px\|right]]
	So what kind of relations can we get for <math>M</math>? Well, it measures how close <math>e_7</math> is to turning sums into products, so we can look for preservation of properties that both addition and multiplication have. For example, they're both commutative, so we should have <math>M(x,y)=M(y,x)</math>, and indeed this is obvious from the definition. Now let's try associativity, that is, let's compute <math>e_7(x+y+z)</math> associating first as <math>(x+y)+z</math> and then as <math>x+(y+z)</math>. In the first way we get		So what kind of relations can we get for <math>M</math>? Well, it measures how close <math>e_7</math> is to turning sums into products, so we can look for preservation of properties that both addition and multiplication have. For example, they're both commutative, so we should have <math>M(x,y)=M(y,x)</math>, and indeed this is obvious from the definition. Now let's try associativity, that is, let's compute <math>e_7(x+y+z)</math> associating first as <math>(x+y)+z</math> and then as <math>x+(y+z)</math>. In the first way we get

Drorbn: /* Computing the Homology, Hard but Rewarding */

2007-03-06T21:27:04Z

Computing the Homology, Hard but Rewarding

← Older revision		Revision as of 17:27, 6 March 2007
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	Let <math>C_n^k</math> denote the space of degree <math>n</math> polynomials in (commuting) variables <math>x_1,\ldots,x_k</math> (with rational coefficients) and let <math>d^k:C_n^k\to C_n^{k+1}</math> be defined by <math>d^k=\sum_{i=0}^{k+1}(-)^i d^k_i</math>, where <math>(d^k_0f)(x_1,\ldots,x_{k+1}):=f(x_2,\ldots,x_{k+1})</math>, <math>(d^k_i)(x_1,\ldots,x_{k+1}):=f(x_1,\ldots,x_i+x_{i+1},\ldots,x_{k+1})</math> for <math>1\leq i\leq k</math> and <math>(d^k_{k+1}f)(x_1,\ldots,x_{k+1}):=f(x_1,\ldots,x_k)</math>. It is easy to verify that <math>{\mathcal C}_n:=(C_n^\star, d)</math> is a chain complex, and that (for <math>k=1,2,3</math>) it agrees with the degree <math>n</math> piece of the complex in {{EqRef\|Complex}}. We need to show that <math>H^1({\mathcal C}_n)=0</math> for <math>n>1</math> (we don't need the vanishing of <math>H^1</math> for <math>n=0,1</math> as these degrees are covered by the initial condition {{EqRef\|Init}}). This follows from the following theorem.		Let <math>C_n^k</math> denote the space of degree <math>n</math> polynomials in (commuting) variables <math>x_1,\ldots,x_k</math> (with rational coefficients) and let <math>d^k:C_n^k\to C_n^{k+1}</math> be defined by <math>d^k=\sum_{i=0}^{k+1}(-)^i d^k_i</math>, where <math>(d^k_0f)(x_1,\ldots,x_{k+1}):=f(x_2,\ldots,x_{k+1})</math>, <math>(d^k_i)(x_1,\ldots,x_{k+1}):=f(x_1,\ldots,x_i+x_{i+1},\ldots,x_{k+1})</math> for <math>1\leq i\leq k</math> and <math>(d^k_{k+1}f)(x_1,\ldots,x_{k+1}):=f(x_1,\ldots,x_k)</math>. It is easy to verify that <math>{\mathcal C}_n:=(C_n^\star, d)</math> is a chain complex, and that (for <math>k=1,2,3</math>) it agrees with the degree <math>n</math> piece of the complex in {{EqRef\|Complex}}. We need to show that <math>H^1({\mathcal C}_n)=0</math> for <math>n>1</math> (we don't need the vanishing of <math>H^1</math> for <math>n=0,1</math> as these degrees are covered by the initial condition {{EqRef\|Init}}). This follows from the following theorem.

	'''Theorem.''' <math>H^k({\mathcal C}_n)</math> is <math>{\mathbb Q}</math> if <math>k=~~n</math>, and <math>~~0</math> ~~otherwise~~.		'''Theorem.''' <math>H^1({\mathcal C}_1)</math> is <math>{\mathbb Q}</math>; otherwise <math>H^k({\mathcal C}_n)=0</math>.

			'''Proof''' (sketch). It is easy to verify "by hand" that <math>\dim H^k({\mathcal C}_1)=\delta_{k1}</math>. For <math>n>1</math> let <math>{\mathcal C}_1^{\otimes n}</math> be the <math>n</math>th interior power of <math>{\mathcal C}_1</math>, whose <math>k</math>th chain group is <math>(C_1^k)^{\otimes n}</math> and whose differential is defined using the diagonal action of the <math>d^k_i</math>'s. The permutation group <math>S_n</math> acts on <math>{\mathcal C}_1^{\otimes n}</math> by permuting the tensor factors. If <math>R</math> denotes the trivial representation of <math>S_n</math>, then <math>R\otimes_{S_n}{\mathcal C}_1^{\otimes n}={\mathcal C}_n</math>, and so
	'''Proof.'''

			{{Equation*\|<math>H^\star({\mathcal C}_n) = H^\star(R\otimes_{S_n}{\mathcal C}_1^{\otimes n}) = R\otimes_{S_n}H^\star({\mathcal C}_1^{\otimes n}) = R\otimes_{S_n}H^\star({\mathcal C}_1)^{\otimes n}</math>}}

			and the results readily follows. Note that the last equality uses the Eilenberg-Zilber-Künneth formula, which holds because <math>{\mathcal C}_n</math> (and especially <math>{\mathcal C}_1</math>) is a co-simplicial space with the <math>d^k_i</math>'s as co-face maps and with <math>(s^k_i f)(x_1,\ldots,x_{k-1}):=f(x_1,\ldots,x_{i-1},0,x_i,\ldots,x_{k-1})</math> as co-degeneracies.

	==Further Examples==		==Further Examples==

Drorbn: /* The "Formal Quantization" Equation */

2007-03-06T01:11:46Z

The "Formal Quantization" Equation

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	{{Equation*\|<math>(f\star g)\star h = f\star(g\star h),</math>}}		{{Equation*\|<math>(f\star g)\star h = f\star(g\star h),</math>}}

			[[Image:The Pentagon for an Associative Product.png\|right]]
	~~[[Image:The Pentagon for an Associative Product.png\|right]]~~ written for an unknown "<math>\star</math>-product", within a certain complicated space of "potential products" which resembles <math>\operatorname{Hom}(V\otimes V,V)</math> or <math>V^\star\otimes V^\star\otimes V</math> for some vector space <math>V</math>. This is a typical "'''algebraic structure wanted'''" equation, in which the unknown is an "algebraic structure" (an associative product <math>\star</math>, in this case) and the equation is "the structure satisfies a law" (the associative law, in our case). Note that algebraic laws are often non-linear in the structure that they govern (the example relevant here is that the associative law is "quadratic as a function of the product"). Other examples abound, with the "structure" replaced by a "bracket" or anything else, and the "law" replaced by "Jacobi's equation" or whatever you fancy.		written for an unknown "<math>\star</math>-product", within a certain complicated space of "potential products" which resembles <math>\operatorname{Hom}(V\otimes V,V)</math> or <math>V^\star\otimes V^\star\otimes V</math> for some vector space <math>V</math>. This is a typical "'''algebraic structure wanted'''" equation, in which the unknown is an "algebraic structure" (an associative product <math>\star</math>, in this case) and the equation is "the structure satisfies a law" (the associative law, in our case). Note that algebraic laws are often non-linear in the structure that they govern (the example relevant here is that the associative law is "quadratic as a function of the product"). Other examples abound, with the "structure" replaced by a "bracket" or anything else, and the "law" replaced by "Jacobi's equation" or whatever you fancy.

	'''Where from cometh the syzygies?''' From the pentagon shown on the right.		'''Where from cometh the syzygies?''' From the pentagon shown on the right.

Drorbn: /* Computing the Homology, Hard but Rewarding */

2007-03-06T01:10:48Z

Computing the Homology, Hard but Rewarding

← Older revision		Revision as of 21:10, 5 March 2007
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	==Computing the Homology, Hard but Rewarding==		==Computing the Homology, Hard but Rewarding==

	Let <math>C_n^k</math> denote the space of degree n polynomials in (commuting) variables <math>x_1,\ldots,x_k</math> and let <math>d^k:C_n^k\to C_n^{k+1}</math> be defined by <math>d^k=\sum_{i=0}^{k+1}(-)^i d^k_i</math>, where <math>(d^k_0f)(x_1,\ldots,x_{k+1}):=f(x_2,\ldots,x_{k+1})</math>, <math>(d^k_i)(x_1,\ldots,x_{k+1}):=f(x_1,\ldots,x_i+x_{i+1},\ldots,x_{k+1})</math> for <math>1\leq i\leq k</math> and <math>(d^k_{k+1}f)(x_1,\ldots,x_{k+1}):=f(x_1,\ldots,x_k)</math>. It is easy to verify that <math>{\mathcal C}_n:=(C_n^\star, d)</math> is a chain complex, and that (for <math>k=1,2,3</math>) it agrees with the degree <math>n</math> piece of the complex in {{EqRef\|Complex}}. We ~~are hoping~~ to ~~prove~~ that <math>H^1({\mathcal C}_n)=0</math> for <math>n>1</math> (we don't need the vanishing of <math>H^1</math> for <math>n=0,1</math> as these degrees are covered by the initial condition {{EqRef\|Init}}).		Let <math>C_n^k</math> denote the space of degree <math>n</math> polynomials in (commuting) variables <math>x_1,\ldots,x_k</math> (with rational coefficients) and let <math>d^k:C_n^k\to C_n^{k+1}</math> be defined by <math>d^k=\sum_{i=0}^{k+1}(-)^i d^k_i</math>, where <math>(d^k_0f)(x_1,\ldots,x_{k+1}):=f(x_2,\ldots,x_{k+1})</math>, <math>(d^k_i)(x_1,\ldots,x_{k+1}):=f(x_1,\ldots,x_i+x_{i+1},\ldots,x_{k+1})</math> for <math>1\leq i\leq k</math> and <math>(d^k_{k+1}f)(x_1,\ldots,x_{k+1}):=f(x_1,\ldots,x_k)</math>. It is easy to verify that <math>{\mathcal C}_n:=(C_n^\star, d)</math> is a chain complex, and that (for <math>k=1,2,3</math>) it agrees with the degree <math>n</math> piece of the complex in {{EqRef\|Complex}}. We need to show that <math>H^1({\mathcal C}_n)=0</math> for <math>n>1</math> (we don't need the vanishing of <math>H^1</math> for <math>n=0,1</math> as these degrees are covered by the initial condition {{EqRef\|Init}}). This follows from the following theorem.

			'''Theorem.''' <math>H^k({\mathcal C}_n)</math> is <math>{\mathbb Q}</math> if <math>k=n</math>, and <math>0</math> otherwise.

			'''Proof.'''

	==Further Examples==		==Further Examples==

Drorbn: /* The Scheme */

2007-03-06T01:06:24Z

The Scheme

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	As we shall see momentarily by "finding syzygies", <math>\epsilon</math> and <math>M</math> fit within the 1st and 2nd chain groups of a rather short complex		As we shall see momentarily by "finding syzygies", <math>\epsilon</math> and <math>M</math> fit within the 1st and 2nd chain groups of a rather short complex

	{{Equation*\|<math>\left(\epsilon\in C^1={\mathbb Q}[[x]]\right)\longrightarrow\left(M\in C^2={\mathbb Q}[[x,y]]\right)\longrightarrow\left(C^3={\mathbb Q}[[x,y,z]]\right)</math>,}}		{{Equation\|Complex\|<math>\left(\epsilon\in C^1={\mathbb Q}[[x]]\right)\longrightarrow\left(M\in C^2={\mathbb Q}[[x,y]]\right)\longrightarrow\left(C^3={\mathbb Q}[[x,y,z]]\right)</math>,}}

	whose first differential was already written and whose second differential is given by <math>(d^2m)(x,y,z)=m(y,z)-m(x+y,z)+m(x,y+z)-m(x,y)</math> for any <math>m\in{\mathbb Q}[[x,y]]</math>. We shall further see that for "our" <math>M</math>, we have <math>d^2M=0</math>. Therefore in order to show that <math>M</math> is in the image of <math>d^1</math>, it suffices to show that the kernel of <math>d^2</math> is equal to the image of <math>d^1</math>, or simply that <math>H^2=0</math>.		whose first differential was already written and whose second differential is given by <math>(d^2m)(x,y,z)=m(y,z)-m(x+y,z)+m(x,y+z)-m(x,y)</math> for any <math>m\in{\mathbb Q}[[x,y]]</math>. We shall further see that for "our" <math>M</math>, we have <math>d^2M=0</math>. Therefore in order to show that <math>M</math> is in the image of <math>d^1</math>, it suffices to show that the kernel of <math>d^2</math> is equal to the image of <math>d^1</math>, or simply that <math>H^2=0</math>.

Drorbn: /* Computing the Homology, Hard but Rewarding */

2007-03-06T01:05:38Z

Computing the Homology, Hard but Rewarding

← Older revision		Revision as of 21:05, 5 March 2007
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	==Computing the Homology, Hard but Rewarding==		==Computing the Homology, Hard but Rewarding==

	Let <math>C_n^k</math> denote the space of degree n polynomials in (commuting) variables <math>x_1,\ldots,x_k</math> and let <math>d^k:C_n^k\to C_n^{k+1}</math> be defined by <math>d^k=\sum_{i=0}^{k+1}(-)^i d^k_i</math>, where		Let <math>C_n^k</math> denote the space of degree n polynomials in (commuting) variables <math>x_1,\ldots,x_k</math> and let <math>d^k:C_n^k\to C_n^{k+1}</math> be defined by <math>d^k=\sum_{i=0}^{k+1}(-)^i d^k_i</math>, where <math>(d^k_0f)(x_1,\ldots,x_{k+1}):=f(x_2,\ldots,x_{k+1})</math>, <math>(d^k_i)(x_1,\ldots,x_{k+1}):=f(x_1,\ldots,x_i+x_{i+1},\ldots,x_{k+1})</math> for <math>1\leq i\leq k</math> and <math>(d^k_{k+1}f)(x_1,\ldots,x_{k+1}):=f(x_1,\ldots,x_k)</math>. It is easy to verify that <math>{\mathcal C}_n:=(C_n^\star, d)</math> is a chain complex, and that (for <math>k=1,2,3</math>) it agrees with the degree <math>n</math> piece of the complex in {{EqRef\|Complex}}. We are hoping to prove that <math>H^1({\mathcal C}_n)=0</math> for <math>n>1</math> (we don't need the vanishing of <math>H^1</math> for <math>n=0,1</math> as these degrees are covered by the initial condition {{EqRef\|Init}}).

	==Further Examples==		==Further Examples==

Drorbn: /* The Scheme */

2007-03-05T21:46:06Z

The Scheme

← Older revision		Revision as of 17:46, 5 March 2007
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	As we shall see momentarily by "finding syzygies", <math>\epsilon</math> and <math>M</math> fit within the 1st and 2nd chain groups of a rather short complex		As we shall see momentarily by "finding syzygies", <math>\epsilon</math> and <math>M</math> fit within the 1st and 2nd chain groups of a rather short complex

	{{Equation*\|<math>\left(\epsilon\in ~~C_1~~={\mathbb Q}[[x]]\right)\longrightarrow\left(M\in ~~C_2~~={\mathbb Q}[[x,y]]\right)\longrightarrow\left(~~C_3~~={\mathbb Q}[[x,y,z]]\right)</math>,}}		{{Equation*\|<math>\left(\epsilon\in C^1={\mathbb Q}[[x]]\right)\longrightarrow\left(M\in C^2={\mathbb Q}[[x,y]]\right)\longrightarrow\left(C^3={\mathbb Q}[[x,y,z]]\right)</math>,}}

	whose first differential was already written and whose second differential is given by <math>(d^2m)(x,y,z)=m(y,z)-m(x+y,z)+m(x,y+z)-m(x,y)</math> for any <math>m\in{\mathbb Q}[[x,y]]</math>. We shall further see that for "our" <math>M</math>, we have <math>d^2M=0</math>. Therefore in order to show that <math>M</math> is in the image of <math>d^1</math>, it suffices to show that the kernel of <math>d^2</math> is equal to the image of <math>d^1</math>, or simply that <math>H^2=0</math>.		whose first differential was already written and whose second differential is given by <math>(d^2m)(x,y,z)=m(y,z)-m(x+y,z)+m(x,y+z)-m(x,y)</math> for any <math>m\in{\mathbb Q}[[x,y]]</math>. We shall further see that for "our" <math>M</math>, we have <math>d^2M=0</math>. Therefore in order to show that <math>M</math> is in the image of <math>d^1</math>, it suffices to show that the kernel of <math>d^2</math> is equal to the image of <math>d^1</math>, or simply that <math>H^2=0</math>.