Notes for AKT-140310/0:35:45 - Revision history

Cameron.martin at 19:20, 8 August 2018

2018-08-08T19:20:52Z

← Older revision		Revision as of 15:20, 8 August 2018
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	[[File:so(N)_1.jpg\|600px]]		[[File:so(N)_1.jpg\|600px]]

	[[File:notafile.jpg]]

	The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$		The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$

Cameron.martin at 19:20, 8 August 2018

2018-08-08T19:20:10Z

← Older revision		Revision as of 15:20, 8 August 2018
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	[[File:so(N)_1.jpg\|600px]]		[[File:so(N)_1.jpg\|600px]]

	[[File:~~Example~~.jpg]]		[[File:notafile.jpg]]

	The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$		The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$

Cameron.martin at 19:19, 8 August 2018

2018-08-08T19:19:50Z

← Older revision		Revision as of 15:19, 8 August 2018
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	[[File:so(N)_1.jpg\|600px]]		[[File:so(N)_1.jpg\|600px]]

			[[File:Example.jpg]]

	The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$		The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$

Cameron.martin at 23:00, 31 July 2018

2018-07-31T23:00:46Z

← Older revision		Revision as of 19:00, 31 July 2018
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	In this note, we compute and interpret the structure constants $f_{abc}$ of $so(N)$, as well as the two-index tensors $t_{ab}$ encoding the information from the metric. In other words, we follow a similar process to the lecture, while disregarding the representation/skeleton edges.		In this note, we compute and interpret the structure constants $f_{abc}$ of $so(N)$, as well as the two-index tensors $t_{ab}$ encoding the information from the metric. In other words, we follow a similar process to the lecture, while disregarding the representation/skeleton edges.

	Let $so(N) = \{Q \in gl(N) \| Q^TQ = QQ^T = I,~~$and~~ ~~det$Q~~ = 1\}$, with the commutator as its bracket (i.e. $[A,B] = AB - BA$), and the metric $\langle A, B \rangle = tr(AB)$.		Let $so(N) = \{Q \in gl(N) \| Q^TQ = QQ^T = I, detQ = 1\}$, with the commutator as its bracket (i.e. $[A,B] = AB - BA$), and the metric $\langle A, B \rangle = tr(AB)$.

	Let $\{\pm M_{ij}\}_{i < j}$ be a basis for $so(N)$, where $(M_{ij})_{kl} = \delta_{ij}\delta_{jl} - \delta_{il}\delta_{jk}$.		Let $\{\pm M_{ij}\}_{i < j}$ be a basis for $so(N)$, where $(M_{ij})_{kl} = \delta_{ij}\delta_{jl} - \delta_{il}\delta_{jk}$.

Cameron.martin at 22:59, 31 July 2018

2018-07-31T22:59:32Z

← Older revision		Revision as of 18:59, 31 July 2018
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	As in the $gl(N)$ case, we can represent the result $t^{(ij)(kl)} = \delta^{ik}\delta^{jl}$ as a splitting of two lines in the diagram, as in the image below. In addition, we can represent the result $f_{(ij)(kl)(mn)} = \epsilon_{(ij)(kl)(mn)}$ as a trivalent vertex becoming a sum of diagrams, over transpositions of certain lines.		As in the $gl(N)$ case, we can represent the result $t^{(ij)(kl)} = \delta^{ik}\delta^{jl}$ as a splitting of two lines in the diagram, as in the image below. In addition, we can represent the result $f_{(ij)(kl)(mn)} = \epsilon_{(ij)(kl)(mn)}$ as a trivalent vertex becoming a sum of diagrams, over transpositions of certain lines.

	[[File:so(N)_1.jpg\|~~100px~~]]		[[File:so(N)_1.jpg\|600px]]

	The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$		The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$
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	The last line exactly corresponds with the last illustration.		The last line exactly corresponds with the last illustration.

	[[File:so(N)_2.jpg]]		[[File:so(N)_2.jpg\|600px]]

Cameron.martin at 22:58, 31 July 2018

2018-07-31T22:58:41Z

← Older revision		Revision as of 18:58, 31 July 2018
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	As in the $gl(N)$ case, we can represent the result $t^{(ij)(kl)} = \delta^{ik}\delta^{jl}$ as a splitting of two lines in the diagram, as in the image below. In addition, we can represent the result $f_{(ij)(kl)(mn)} = \epsilon_{(ij)(kl)(mn)}$ as a trivalent vertex becoming a sum of diagrams, over transpositions of certain lines.		As in the $gl(N)$ case, we can represent the result $t^{(ij)(kl)} = \delta^{ik}\delta^{jl}$ as a splitting of two lines in the diagram, as in the image below. In addition, we can represent the result $f_{(ij)(kl)(mn)} = \epsilon_{(ij)(kl)(mn)}$ as a trivalent vertex becoming a sum of diagrams, over transpositions of certain lines.

	[[File:so(N)_1.jpg]]		[[File:so(N)_1.jpg\|100px]]

	The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$		The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$

Cameron.martin at 22:57, 31 July 2018

2018-07-31T22:57:15Z

← Older revision		Revision as of 18:57, 31 July 2018
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	Note that this also gives us the inverses $t^{(ij)(kl)} = const\cdot\delta^{ik}\delta^{jl}$.		Note that this also gives us the inverses $t^{(ij)(kl)} = const\cdot\delta^{ik}\delta^{jl}$.

	Now the structure constants: $$f_{(ij)(kl)(mn)} = \langle[M_{ij}, M_{kl}], M_{mn} \rangle = \langle M_{ij}M_{kl}, M_{mn} \rangle - \langle M_{kl}M_{ij}, M_{mn} \rangle~~$$ \\ $$f_{(ij)(kl(mn)} = tr(M_{ij}M_{kl}M_{mn}) - tr(M_{kl}M_{ij}M_{mn}) = const\cdot\epsilon_{(ij)(kl)(mn)}~~$$		Now the structure constants: $$f_{(ij)(kl)(mn)} = \langle[M_{ij}, M_{kl}], M_{mn} \rangle = \langle M_{ij}M_{kl}, M_{mn} \rangle - \langle M_{kl}M_{ij}, M_{mn} \rangle$$
			$$f_{(ij)(kl(mn)} = tr(M_{ij}M_{kl}M_{mn}) - tr(M_{kl}M_{ij}M_{mn}) = const\cdot\epsilon_{(ij)(kl)(mn)}$$

	(With an appropriate choice of signs and ordering of the basis. In $so(3)$, an appropriate ordering and choice of signs is $\mathcal{B} = \{M_{12}, M_{23}, -M_{13}\}$.		(With an appropriate choice of signs and ordering of the basis. In $so(3)$, an appropriate ordering and choice of signs is $\mathcal{B} = \{M_{12}, M_{23}, -M_{13}\}$.
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	[[File:so(N)_1.jpg]]		[[File:so(N)_1.jpg]]

	The diagram illustrated below goes to the following expression:		The diagram illustrated below goes to the following expression: $$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$
		⚫	$$I = \sum_{i,...,n,i',...,n'}\epsilon_{(ij)(kl)(mn)}\delta^{ii'}\delta^{jj'}\delta^{kk'}\delta^{ll'}\delta^{mm'}\delta^{nn'}$$

	$$I = \sum_{i,...,n~~,i',...,n'~~}f_{(ij)(kl)(mn~~)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n'~~)}$$ \\		$$I = \sum_{i,...,n}\epsilon_{(ij)(kl)(mn)}$$
⚫	$$I = \sum_{i,...,n,i',...,n'}\epsilon_{(ij)(kl)(mn)}\delta^{ii'}\delta^{jj'}\delta^{kk'}\delta^{ll'}\delta^{mm'}\delta^{nn'}$$\\
	$$I = \sum_{i,...,n}\epsilon_{(ij)(kl)(mn)}$$

	The last line exactly corresponds with the last illustration.		The last line exactly corresponds with the last illustration.

Cameron.martin at 22:56, 31 July 2018

2018-07-31T22:56:23Z

← Older revision		Revision as of 18:56, 31 July 2018
Line 9:		Line 9:
	Note that this also gives us the inverses $t^{(ij)(kl)} = const\cdot\delta^{ik}\delta^{jl}$.		Note that this also gives us the inverses $t^{(ij)(kl)} = const\cdot\delta^{ik}\delta^{jl}$.

		⚫	Now the structure constants: $$f_{(ij)(kl)(mn)} = \langle[M_{ij}, M_{kl}], M_{mn} \rangle = \langle M_{ij}M_{kl}, M_{mn} \rangle - \langle M_{kl}M_{ij}, M_{mn} \rangle$$ \\ $$f_{(ij)(kl(mn)} = tr(M_{ij}M_{kl}M_{mn}) - tr(M_{kl}M_{ij}M_{mn}) = const\cdot\epsilon_{(ij)(kl)(mn)}$$
	Now the structure constants:

⚫	$f_{(ij)(kl)(mn)} = \langle[M_{ij}, M_{kl}], M_{mn} \rangle = \langle M_{ij}M_{kl}, M_{mn} \rangle - \langle M_{kl}M_{ij}, M_{mn} \rangle$ \\
	$f_{(ij)(kl(mn)} = tr(M_{ij}M_{kl}M_{mn}) - tr(M_{kl}M_{ij}M_{mn}) = const\cdot\epsilon_{(ij)(kl)(mn)}$

	(With an appropriate choice of signs and ordering of the basis. In $so(3)$, an appropriate ordering and choice of signs is $\mathcal{B} = \{M_{12}, M_{23}, -M_{13}\}$.		(With an appropriate choice of signs and ordering of the basis. In $so(3)$, an appropriate ordering and choice of signs is $\mathcal{B} = \{M_{12}, M_{23}, -M_{13}\}$.
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	The diagram illustrated below goes to the following expression:		The diagram illustrated below goes to the following expression:

	$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$ \\		$$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$ \\
	$I = \sum_{i,...,n,i',...,n'}\epsilon_{(ij)(kl)(mn)}\delta^{ii'}\delta^{jj'}\delta^{kk'}\delta^{ll'}\delta^{mm'}\delta^{nn'}$\\		$$I = \sum_{i,...,n,i',...,n'}\epsilon_{(ij)(kl)(mn)}\delta^{ii'}\delta^{jj'}\delta^{kk'}\delta^{ll'}\delta^{mm'}\delta^{nn'}$$\\
	$I = \sum_{i,...,n}\epsilon_{(ij)(kl)(mn)}$		$$I = \sum_{i,...,n}\epsilon_{(ij)(kl)(mn)}$$

	The last line exactly corresponds with the last illustration.		The last line exactly corresponds with the last illustration.

Cameron.martin at 22:54, 31 July 2018

2018-07-31T22:54:08Z

← Older revision		Revision as of 18:54, 31 July 2018
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	Now the structure constants:		Now the structure constants:

	$$f_{(ij)(kl)(mn)} = \langle[M_{ij}, M_{kl}], M_{mn} \rangle = \langle M_{ij}M_{kl}, M_{mn} \rangle - \langle M_{kl}M_{ij}, M_{mn} \rangle$$ \\		$f_{(ij)(kl)(mn)} = \langle[M_{ij}, M_{kl}], M_{mn} \rangle = \langle M_{ij}M_{kl}, M_{mn} \rangle - \langle M_{kl}M_{ij}, M_{mn} \rangle$ \\
	$$f_{(ij)(kl(mn)} = tr(M_{ij}M_{kl}M_{mn}) - tr(M_{kl}M_{ij}M_{mn}) = const\cdot\epsilon_{(ij)(kl)(mn)}$$		$f_{(ij)(kl(mn)} = tr(M_{ij}M_{kl}M_{mn}) - tr(M_{kl}M_{ij}M_{mn}) = const\cdot\epsilon_{(ij)(kl)(mn)}$

	(With an appropriate choice of signs and ordering of the basis. In $so(3)$, an appropriate ordering and choice of signs is $\mathcal{B} = \{M_{12}, M_{23}, -M_{13}\}$.		(With an appropriate choice of signs and ordering of the basis. In $so(3)$, an appropriate ordering and choice of signs is $\mathcal{B} = \{M_{12}, M_{23}, -M_{13}\}$.
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	The diagram illustrated below goes to the following expression:		The diagram illustrated below goes to the following expression:

	$$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$ \\		$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$ \\
	$$I = \sum_{i,...,n,i',...,n'}\epsilon_{(ij)(kl)(mn)}\delta^{ii'}\delta^{jj'}\delta^{kk'}\delta^{ll'}\delta^{mm'}\delta^{nn'}$$\\		$I = \sum_{i,...,n,i',...,n'}\epsilon_{(ij)(kl)(mn)}\delta^{ii'}\delta^{jj'}\delta^{kk'}\delta^{ll'}\delta^{mm'}\delta^{nn'}$\\
	$$I = \sum_{i,...,n}\epsilon_{(ij)(kl)(mn)} $$		$I = \sum_{i,...,n}\epsilon_{(ij)(kl)(mn)}$

	The last line exactly corresponds with the last illustration.		The last line exactly corresponds with the last illustration.

Cameron.martin: Created page with "In this note, we compute and interpret the structure constants $f_{abc}$ of $so(N)$, as well as the two-index tensors $t_{ab}$ encoding the information from the metric. In oth..."

2018-07-31T22:52:50Z

Created page with "In this note, we compute and interpret the structure constants $f_{abc}$ of $so(N)$, as well as the two-index tensors $t_{ab}$ encoding the information from the metric. In oth..."

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In this note, we compute and interpret the structure constants $f_{abc}$ of $so(N)$, as well as the two-index tensors $t_{ab}$ encoding the information from the metric. In other words, we follow a similar process to the lecture, while disregarding the representation/skeleton edges.

Let $so(N) = \{Q \in gl(N) | Q^TQ = QQ^T = I,$and det$Q = 1\}$, with the commutator as its bracket (i.e. $[A,B] = AB - BA$), and the metric $\langle A, B \rangle = tr(AB)$.

Let $\{\pm M_{ij}\}_{i < j}$ be a basis for $so(N)$, where $(M_{ij})_{kl} = \delta_{ij}\delta_{jl} - \delta_{il}\delta_{jk}$.

With this, compute $$t_{(ij)(kl)} = \langle M_{ij}, M_{kl} \rangle = tr(M_{ij}M_{kl}) = const\cdot\delta_{ik}\delta_{jl}$$

Note that this also gives us the inverses $t^{(ij)(kl)} = const\cdot\delta^{ik}\delta^{jl}$.

Now the structure constants:

$$f_{(ij)(kl)(mn)} = \langle[M_{ij}, M_{kl}], M_{mn} \rangle = \langle M_{ij}M_{kl}, M_{mn} \rangle - \langle M_{kl}M_{ij}, M_{mn} \rangle$$ \\
$$f_{(ij)(kl(mn)} = tr(M_{ij}M_{kl}M_{mn}) - tr(M_{kl}M_{ij}M_{mn}) = const\cdot\epsilon_{(ij)(kl)(mn)}$$

(With an appropriate choice of signs and ordering of the basis. In $so(3)$, an appropriate ordering and choice of signs is $\mathcal{B} = \{M_{12}, M_{23}, -M_{13}\}$.

If we order the basis, we can associate an integer lying somewhere from 1 to $N(N-1)/2$ (the dimension of $so(N)$) to each pair of indices $(ij)$, so the expression $\epsilon_{(ij)(kl)(mn)}$ makes sense - namely, let $a$, $b$, and $c$ correspond to $(ij)$, $(kl)$, and $(mn)$, respectively, and let $\epsilon_{(ij)(kl)(mn)} = \epsilon_{abc}$, the usual totally antisymmetric tensor.

Thus, up to a constant, $t^{(ij)(kl)} = \delta^{ik}\delta^{jl}$ and $f_{(ij)(kl)(mn)} = \epsilon_{(ij)(kl)(mn)}$.

As in the $gl(N)$ case, we can represent the result $t^{(ij)(kl)} = \delta^{ik}\delta^{jl}$ as a splitting of two lines in the diagram, as in the image below. In addition, we can represent the result $f_{(ij)(kl)(mn)} = \epsilon_{(ij)(kl)(mn)}$ as a trivalent vertex becoming a sum of diagrams, over transpositions of certain lines.

[[File:so(N)_1.jpg]]

The diagram illustrated below goes to the following expression:

$$I = \sum_{i,...,n,i',...,n'}f_{(ij)(kl)(mn)}t^{(ij)(i'j')}t^{(kl)(k'l')}t^{(mn)(m'n')}$$ \\
$$I = \sum_{i,...,n,i',...,n'}\epsilon_{(ij)(kl)(mn)}\delta^{ii'}\delta^{jj'}\delta^{kk'}\delta^{ll'}\delta^{mm'}\delta^{nn'}$$\\
$$I = \sum_{i,...,n}\epsilon_{(ij)(kl)(mn)} $$

The last line exactly corresponds with the last illustration.

[[File:so(N)_2.jpg]]