Drorbn - User contributions [en]

Notes for AKT-140127/0:47:34

2018-08-24T20:22:20Z

Cameron.martin:

This note closely follows the introductory parts of "Vassiliev and Quantum Invariants of Braids" by Dror Bar-Natan (1996).

The notion of finite type invariant can be extended to different knot-like objects (braids, tangles, links, etc.). While it is an open question whether finite type invariants separate knots, it is known that finite type invariants separate braids. In this note, we describe the process of getting a chord diagram from an m-singular braid, as well as describing the framework in which one would perform a similar analysis to the one done in lecture on braids.

Given a braid $B$ with n strands and m double points, the chord diagram $D_B$ is given as illustrated below (with $n$ = 4 and $m$ = 3).

[[File:braid_chord_diagram.jpg|400px]]

It is clear how to do this for all $n$, $m$. The space of $m$-chord pure braid diagrams then forms a space $\mathcal{D}^{pb}_m$, and just as in the case of knots, given a type $m$ invariant $V$ of braids with $n$ strands with values in an abelian group $A$, there is a map $W_V: \mathcal{D}^{pb}_m \rightarrow A$ defined by $W_V(D) = V(B)$, where $D_B = D$. Then define $\mathcal{D}^{pb}$ as the completed direct sum of all $\mathcal{D}^{pb}_m$, and define $\mathcal{A}^{pb}$ as $\mathcal{D}^{pb}/\mathcal{I}$, where $\mathcal{I}$ is an ideal generated by relations among braid chord diagrams analogous to the FI and 4T relations of the familiar knot chord diagrams. If one continues with this analysis, one can find that finite type invariants in fact do separate braids. For a proof of this fact and much more detail regarding the above, see https://www.math.toronto.edu/drorbn/papers/glN/glN.pdf.

Notes for AKT-140127/0:47:34

2018-08-24T20:21:48Z

Cameron.martin: Created page with "This note closely follows the introductory parts of "Vassiliev and Quantum Invariants of Braids" by Dror Bar-Natan (1996). The notion of finite type invariant can be extended..."

This note closely follows the introductory parts of "Vassiliev and Quantum Invariants of Braids" by Dror Bar-Natan (1996).

The notion of finite type invariant can be extended to different knot-like objects (braids, tangles, links, etc.). While it is an open question whether finite type invariants separate knots, it is known that finite type invariants separate braid. In this note, we describe the process of getting a chord diagram from an m-singular braid, as well as describing the framework in which one would perform a similar analysis to the one done in lecture on braids.

Given a braid $B$ with n strands and m double points, the chord diagram $D_B$ is given as illustrated below (with $n$ = 4 and $m$ = 3).

[[File:braid_chord_diagram.jpg|400px]]

It is clear how to do this for all $n$, $m$. The space of $m$-chord pure braid diagrams then forms a space $\mathcal{D}^{pb}_m$, and just as in the case of knots, given a type $m$ invariant $V$ of braids with $n$ strands with values in an abelian group $A$, there is a map $W_V: \mathcal{D}^{pb}_m \rightarrow A$ defined by $W_V(D) = V(B)$, where $D_B = D$. Then define $\mathcal{D}^{pb}$ as the completed direct sum of all $\mathcal{D}^{pb}_m$, and define $\mathcal{A}^{pb}$ as $\mathcal{D}^{pb}/\mathcal{I}$, where $\mathcal{I}$ is an ideal generated by relations among braid chord diagrams analogous to the FI and 4T relations of the familiar knot chord diagrams. If one continues with this analysis, one can find that finite type invariants in fact do separate braids. For a proof of this fact and much more detail regarding the above, see https://www.math.toronto.edu/drorbn/papers/glN/glN.pdf.

File:Braid chord diagram.jpg

2018-08-24T20:20:53Z

Cameron.martin:

Notes for AKT-140120/0:22:11

2018-08-18T17:51:38Z

Cameron.martin:

Much like the concept of a type n knot invariant described in this lecture, there is such a thing as a type n invariant of "virtual knots". Virtual knots can be defined (among other ways) as equivalence classes of Gauss diagrams under a certain set of generalized Reidemeister moves. Refer to the note at time 31:27 of the video found at http://drorbn.net/dbnvp/AKT-140127.php, and Goussarev, Polyak, and Viro's paper (https://arxiv.org/abs/math/9810073) for more information on Gauss diagrams and the specific set of moves. Equivalently, one can think of virtual knots as quadrivalent planar graphs, where each vertex can either be the usual over/undercrossing pair as in regular knots, or a so-called "virtual crossing", at which the lines simply cross each other with no extra information. Analogous to the double point in the regular knot situation is the "semi-virtual crossing", given by an analogous relation (illustrated below).

[[File:semi_virtual_crossings.jpg|600px]]

An invariant $v: \mathcal{VK} \rightarrow A$ (where $\mathcal{VK}$ is the space of virtual knots and $A$ is some abelian group) of virtual knots is said to be of type $n$ if it vanishes on virtual knots with $n+1$ semi-virtual crossings. By virtue of equation (3) above, it can be shown that type $n$ virtual knot invariants, when restricted to honest knots, are knot invariants of type at least n. Namely, a type n virtual knot invariant evaluated on $n+1$ double points is a sum of evaluations of $v$ on $n+1$ semi-virtual crossings (by equation 3), and is therefore equal to 0.

Notes for AKT-140120/0:22:11

2018-08-18T17:50:06Z

Cameron.martin:

Much like the concept of a type n knot invariant described in this lecture, there is such a thing as a type n invariant of "virtual knots". Virtual knots can be defined (among other ways) as equivalence classes of Gauss diagrams under a certain set of generalized Reidemeister moves. Refer to the note at time 31:27 of the video found at http://drorbn.net/dbnvp/AKT-140127.php, and Goussarev, Polyak, and Viro's paper (https://arxiv.org/abs/math/9810073) for more information on Gauss diagrams and the specific set of moves. Equivalently, one can think of virtual knots as quadrivalent planar graphs, where each vertex can either be the usual over/undercrossing pair as in regular knots, or a so-called "virtual crossing", at which the lines simply cross each other with no extra information. Analogous to the double point in the regular knot situation is the "semi-virtual crossing", given by an analogous relation (illustrated below).

[[File:semi_virtual_crossings.jpg|600px]]

An invariant $v: \mathcal{VK} \rightarrow A$ (where $\mathcal{VK}$ is the space of virtual knots and $A$ is some abelian group) of virtual knots is said to be of type n if it vanishes on virtual knots with n+1 semi-virtual crossings. By virtue of equation (3) above, it can be shown that type n virtual knot invariants, when restricted to honest knots, are knot invariants of type at least n. Namely, a type n virtual knot invariant evaluated on n+1 double points is a sum of evaluations of $v$ on n+1 semi-virtual crossings (by equation 3), and is therefore equal to 0.

Notes for AKT-140120/0:22:11

2018-08-18T17:48:44Z

Cameron.martin: Created page with "Much like the concept of a type n knot invariant described in this lecture, there is such a thing as a type n invariant of "virtual knots". Virtual knots can be defined in man..."

Much like the concept of a type n knot invariant described in this lecture, there is such a thing as a type n invariant of "virtual knots". Virtual knots can be defined in many ways, one such way being to describe them as equivalence classes of Gauss diagrams under a certain set of generalized Reidemeister moves. Refer to the note at time 31:27 of the video found at http://drorbn.net/dbnvp/AKT-140127.php, and Goussarev, Polyak, and Viro's paper (https://arxiv.org/abs/math/9810073) for more information on Gauss diagrams and the specific set of moves. Equivalently, one can think of virtual knots as quadrivalent planar graphs, where each vertex can either be the usual over/undercrossing pair as in regular knots, or a so-called "virtual crossing", at which the lines simply cross each other with no extra information. Analogous to the double point in the regular knot situation is the "semi-virtual crossing", given by an analogous relation (illustrated below).

[[File:semi_virtual_crossings.jpg|600px]]

An invariant $v: \mathcal{VK} \rightarrow A$ (where $\mathcal{VK}$ is the space of virtual knots and $A$ is some abelian group) of virtual knots is said to be of type n if it vanishes on virtual knots with n+1 semi-virtual crossings. By virtue of equation (3) above, it can be shown that type n virtual knot invariants, when restricted to honest knots, are knot invariants of type at least n. Namely, a type n virtual knot invariant evaluated on n+1 double points is a sum of evaluations of $v$ on n+1 semi-virtual crossings (by equation 3), and is therefore equal to 0.

File:Semi virtual crossings.jpg

2018-08-18T17:48:09Z

Cameron.martin:

Notes for AKT-140127/0:31:27

2018-08-17T20:14:10Z

Cameron.martin:

Much of this note comes from the paper "Finite type invariants of classical and virtual knots" by Goussarov, Polyak, and Viro, which can be found here: https://arxiv.org/abs/math/9810073.

Similar to the process of obtaining an n-chord diagram from an n-singular knot, there is a so called "Gauss diagram" obtainable from a regular knot. I.e. there is a map $\mathcal{K} \rightarrow \mathcal{D}$ from the space of knots into the space of Gauss diagrams. A Gauss diagram is a chord diagram with orientations and signs on the chords. The Gauss diagram of a specific knot is obtained as shown in examples below (the right-handed trefoil and the figure eight knots).

[[File:Gauss_diagrams.jpg|600px]]

Note the additional information in the Gauss diagram about the overpasses and underpasses of the knot diagram (the orientations on the chords), as well as the sign of the crossings. Unfortunately, unlike the fact that any n-chord diagram is $D_K$ for some n-singular knot $K$, it is not true that any Gauss diagram is the Gauss diagram of a knot. This observation is intimately related to the theory of virtual knots, described in detail in the paper linked above.

Notes for AKT-140127/0:31:27

2018-08-17T20:13:40Z

Cameron.martin: Created page with "Much of this note comes from the paper "Finite type invariants of classical and virtual knots" by Goussarov, Polyak, and Viro, which can be found here: https://arxiv.org/abs/m..."

Much of this note comes from the paper "Finite type invariants of classical and virtual knots" by Goussarov, Polyak, and Viro, which can be found here: https://arxiv.org/abs/math/9810073. \par

Similar to the process of obtaining an n-chord diagram from an n-singular knot, there is a so called "Gauss diagram" obtainable from a regular knot. I.e. there is a map $\mathcal{K} \rightarrow \mathcal{D}$ from the space of knots into the space of Gauss diagrams. A Gauss diagram is a chord diagram with orientations and signs on the chords. The Gauss diagram of a specific knot is obtained as shown in examples below (the right-handed trefoil and the figure eight knots).

[[File:Gauss_diagrams.jpg|600px]]

Note the additional information in the Gauss diagram about the overpasses and underpasses of the knot diagram (the orientations on the chords), as well as the sign of the crossings. Unfortunately, unlike the fact that any n-chord diagram is $D_K$ for some n-singular knot $K$, it is not true that any Gauss diagram is the Gauss diagram of a knot. This observation is intimately related to the theory of virtual knots, described in detail in the paper linked above.

File:Gauss diagrams.jpg

2018-08-17T20:12:35Z

Cameron.martin:

Notes for AKT-140310/0:35:45

2018-08-08T19:20:52Z

Cameron.martin:

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 $so(N) = \{Q \in gl(N) | Q^TQ = QQ^T = I, detQ = 1\}$, 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 $(M_{ij})_{kl} = \delta_{ij}\delta_{jl} - \delta_{il}\delta_{jk}$.

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|600px]]

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

The last line exactly corresponds with the last illustration.

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

Notes for AKT-140310/0:35:45

2018-08-08T19:20:10Z

Cameron.martin:

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 $so(N) = \{Q \in gl(N) | Q^TQ = QQ^T = I, detQ = 1\}$, 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 $(M_{ij})_{kl} = \delta_{ij}\delta_{jl} - \delta_{il}\delta_{jk}$.

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|600px]]

[[File:notafile.jpg]]

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

The last line exactly corresponds with the last illustration.

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

Notes for AKT-140310/0:35:45

2018-08-08T19:19:50Z

Cameron.martin:

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 $so(N) = \{Q \in gl(N) | Q^TQ = QQ^T = I, detQ = 1\}$, 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 $(M_{ij})_{kl} = \delta_{ij}\delta_{jl} - \delta_{il}\delta_{jk}$.

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|600px]]

[[File:Example.jpg]]

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

The last line exactly corresponds with the last illustration.

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

Notes for AKT-140310/0:35:45

2018-07-31T23:00:46Z

Cameron.martin:

Notes for AKT-140310/0:35:45

2018-07-31T22:59:32Z

Cameron.martin:

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|600px]]

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

The last line exactly corresponds with the last illustration.

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

Notes for AKT-140310/0:35:45

2018-07-31T22:58:41Z

Cameron.martin:

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|100px]]

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

The last line exactly corresponds with the last illustration.

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

Notes for AKT-140310/0:35:45

2018-07-31T22:57:15Z

Cameron.martin:

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]]

Notes for AKT-140310/0:35:45

2018-07-31T22:56:23Z

Cameron.martin:

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]]

Notes for AKT-140310/0:35:45

2018-07-31T22:54:08Z

Cameron.martin:

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]]

Notes for AKT-140310/0:35:45

2018-07-31T22:52:50Z

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..."

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]]

File:So(N) 2.jpg

2018-07-31T22:50:52Z

Cameron.martin:

File:So(N) 1.jpg

2018-07-31T22:49:34Z

Cameron.martin:

Notes for AKT-140224/0:22:08

2018-06-13T02:54:47Z

Cameron.martin:

We will check that the prospective Lie algebra $g = F^2$ satisfies the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$, and thus is indeed a Lie algebra. In contrast with the notation given in lecture, let $\alpha$ and $\beta$ be the basis elements of $F^2$ over $F$, so that $F^2 = \{a\alpha + b\beta: a, b \in F\}$. Now, let $x = a_1\alpha + a_2\beta$, $y = b_1\alpha + b_2\beta$, and $z = c_1\alpha + c_2\beta$ be arbitrary elements in $F^2$.

Using bilinearity,

$[y,z] = b_1c_1[\alpha,\alpha] + b_1c_2[\alpha, \beta] + b_2c_1[\beta,\alpha] + b_2c_2[\beta,\beta]$

Since $[\alpha, \alpha] = [\beta,\beta] = 0$, $[\alpha, \beta] = \alpha$, and $[\beta,\alpha] = -\alpha$, this evaluates to $$[y,z] = (b_1c_2 - b_2c_1)\alpha$$ Similar calculations yield $$[z,x] = (c_1a_2 - c_2a_1)\alpha$$ and $$[x,y] = (a_1b_2 - a_2b_1)\alpha$$ Thus, $$[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = (a_2(b_1c_2 - b_2c_1) + b_2(c_1a_2-c_2a_1) + c_2(a_1b_2+a_2b_1))[\beta, \alpha]$$
$$= -\alpha(a_2b_1c_2 - a_2b_2c_1 + a_2b_2c_1 - a_1b_2c_2 + a_1b_2c_2 - a_2b_1c_2) = 0$$

As a result, the Jacobi identity holds.

Notes for AKT-140224/0:22:08

2018-06-13T02:54:03Z

Cameron.martin:

Notes for AKT-140224/0:22:08

2018-06-13T02:52:05Z

Cameron.martin:

We will check that the prospective Lie algebra $g = F^2$ satisfies the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$, and thus is indeed a Lie algebra. In contrast with the notation given in lecture, let $\alpha$ and $\beta$ be the basis elements of $F^2$ over $F$, so that $F^2 = \{a\alpha + b\beta: a, b \in F\}$. Now, let $x = a_1\alpha + a_2\beta$, $y = b_1\alpha + b_2\beta$, and $z = c_1\alpha + c_2\beta$ be arbitrary elements in $F^2$.

Using bilinearity,

$[y,z] = b_1c_1[\alpha,\alpha] + b_1c_2[\alpha, \beta] + b_2c_1[\beta,\alpha] + b_2c_2[\beta,\beta]$

Since $[\alpha, \alpha] = [\beta,\beta] = 0$, $[\alpha, \beta] = \alpha$, and $[\beta,\alpha] = -\alpha$, this evaluates to $$[y,z] = (b_1c_2 - b_2c_1)\alpha$$ Similar calculations yield $$[z,x] = (c_1a_2 - c_2a_1)\alpha$$ and $$[x,y] = (a_1b_2 - a_2b_1)\alpha$$ Thus, $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = (a_2(b_1c_2 - b_2c_1) + b_2(c_1a_2-c_2a_1) + c_2(a_1b_2+a_2b_1))[\beta, \alpha] = -\alpha(a_2b_1c_2 - a_2b_2c_1 + a_2b_2c_1 - a_1b_2c_2 + a_1b_2c_2 - a_2b_1c_2) = 0$, and the Jacobi identity holds.

Notes for AKT-140224/0:22:08

2018-06-13T02:51:27Z

Cameron.martin:

We will check that the prospective Lie algebra $g = F^2$ satisfies the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$, and thus is indeed a Lie algebra. In contrast with the notation given in lecture, let $\alpha$ and $\beta$ be the basis elements of $F^2$ over $F$, so that $F^2 = \{a\alpha + b\beta: a, b \in F\}$. Now, let $x = a_1\alpha + a_2\beta$, $y = b_1\alpha + b_2\beta$, and $z = c_1\alpha + c_2\beta$ be arbitrary elements in $F^2$.

Using bilinearity,

$[y,z] = b_1c_1[\alpha,\alpha] + b_1c_2[\alpha, \beta] + b_2c_1[\beta,\alpha] + b_2c_2[\beta,\beta]$

Since $[\alpha, \alpha] = [\beta,\beta] = 0$, $[\alpha, \beta] = \alpha$, and $[\beta,\alpha] = -\alpha$, this evaluates to $$[y,z] = (b_1c_2 - b_2c_1)\alpha$$. Similar calculations yield $$[z,x] = (c_1a_2 - c_2a_1)\alpha$$ and $$[x,y] = (a_1b_2 - a_2b_1)\alpha$$. Thus, $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = (a_2(b_1c_2 - b_2c_1) + b_2(c_1a_2-c_2a_1) + c_2(a_1b_2+a_2b_1))[\beta, \alpha] = -\alpha(a_2b_1c_2 - a_2b_2c_1 + a_2b_2c_1 - a_1b_2c_2 + a_1b_2c_2 - a_2b_1c_2) = 0$, and the Jacobi identity holds.

Notes for AKT-140224/0:22:08

2018-06-13T02:50:54Z

Cameron.martin:

We will check that the prospective Lie algebra $g = F^2$ satisfies the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$, and thus is indeed a Lie algebra. In contrast with the notation given in lecture, let $\alpha$ and $\beta$ be the basis elements of $F^2$ over $F$, so that $F^2 = \{a\alpha + b\beta: a, b \in F\}$. Now, let $x = a_1\alpha + a_2\beta$, $y = b_1\alpha + b_2\beta$, and $z = c_1\alpha + c_2\beta$, arbitrary elements in $F^2$.

Using bilinearity,

$[y,z] = b_1c_1[\alpha,\alpha] + b_1c_2[\alpha, \beta] + b_2c_1[\beta,\alpha] + b_2c_2[\beta,\beta]$

Since $[\alpha, \alpha] = [\beta,\beta] = 0$, $[\alpha, \beta] = \alpha$, and $[\beta,\alpha] = -\alpha$, this evaluates to $$[y,z] = (b_1c_2 - b_2c_1)\alpha$$. Similar calculations yield $$[z,x] = (c_1a_2 - c_2a_1)\alpha$$ and $$[x,y] = (a_1b_2 - a_2b_1)\alpha$$. Thus, $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = (a_2(b_1c_2 - b_2c_1) + b_2(c_1a_2-c_2a_1) + c_2(a_1b_2+a_2b_1))[\beta, \alpha] = -\alpha(a_2b_1c_2 - a_2b_2c_1 + a_2b_2c_1 - a_1b_2c_2 + a_1b_2c_2 - a_2b_1c_2) = 0$, and the Jacobi identity holds.

Notes for AKT-140224/0:22:08

2018-06-13T02:49:03Z

Cameron.martin: Created page with "We will check that the prospective Lie algebra $g = \mathds{F}^2$ satisfies the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$, and thus is indeed a Lie algebra. In c..."

We will check that the prospective Lie algebra $g = \mathds{F}^2$ satisfies the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$, and thus is indeed a Lie algebra. In contrast with the notation given in lecture, let $\alpha$ and $\beta$ be the basis elements of $\mathds{F}^2$ over $\mathds{F}$, so that $\mathds{F}^2 = \{a\alpha + b\beta: a, b \in \mathds{F}\}$. Now, let $x = a_1\alpha + a_2\beta$, $y = b_1\alpha + b_2\beta$, and $z = c_1\alpha + c_2\beta$, arbitrary elements in $\mathds{F}^2$.

Using bilinearity,

$[y,z] = b_1c_1[\alpha,\alpha] + b_1c_2[\alpha, \beta] + b_2c_1[\beta,\alpha] + b_2c_2[\beta,\beta]$

Since $[\alpha, \alpha] = [\beta,\beta] = 0$, $[\alpha, \beta] = \alpha$, and $[\beta,\alpha] = -\alpha$, this evaluates to $$[y,z] = (b_1c_2 - b_2c_1)\alpha$$. Similar calculations yield $$[z,x] = (c_1a_2 - c_2a_1)\alpha$$ and $$[x,y] = (a_1b_2 - a_2b_1)\alpha$$. Thus, $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = (a_2(b_1c_2 - b_2c_1) + b_2(c_1a_2-c_2a_1) + c_2(a_1b_2+a_2b_1))[\beta, \alpha] = -\alpha(a_2b_1c_2 - a_2b_2c_1 + a_2b_2c_1 - a_1b_2c_2 + a_1b_2c_2 - a_2b_1c_2) = 0$, and the Jacobi identity holds.

Notes for AKT-140117/0:16:43

2018-05-30T18:55:03Z

Cameron.martin: Created page with "The "power-line" problem is also known as the "catenary" problem (https://en.wikipedia.org/wiki/Catenary). As stated, we want to minimize the potential energy, so we must find..."

The "power-line" problem is also known as the "catenary" problem (https://en.wikipedia.org/wiki/Catenary). As stated, we want to minimize the potential energy, so we must find an expression for the total potential energy of the system, which will be done by integrating along the rope/wire/line. The infinitesimal potential energy is given by $dV = gydm$, where g is the acceleration due to gravity, y is the height, and $dm$ is the infinitesimal mass along a length of the rope. In addition, $dm = \frac{m}{s}dl$, where $dl = \sqrt{(dx)^2 + (dy)^2} = dx\sqrt{1+(\frac{dy}{dx})^2}$ and $s$ is the total arclength of the rope. If we let the rope span a length from $x = -L$ to $x = L$, the total potential gravitational energy is $V(y) = \int_{-L}^{L}\frac{mg}{s}y(x)\sqrt{1+(y'(x))^2}dx$. This is what we want to minimize. As usual, let $f(\epsilon) = V(y_c + \epsilon y_q)$, where $y_q(0) = y_q(L) = 0$, and set $\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon = 0}$. For convenience, use dot notation for derivatives, and compute to get

$\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon = 0} = \frac{mg}{s}\int_{-L}^{L}y_q\sqrt{1+\dot{y}_q^2}+y_c(1+\dot{y}_c^2)^{\frac{-1}{2}(\dot{y}_c\dot{y}_q)}dx = 0$

Thus, $0 = \frac{mg}{s}(I_1 + I_2)$, where $I_1 = \int_{-L}^L y_q\sqrt{1+\dot{y}_c^2}dx$ and $I_2 = \int_{-L}^L y_c\dot{y}_c\dot{y}_q(1+\dot{y}_c^2)^{\frac{-1}{2}}dx$.

Integrating $I_2$ by parts with $u = y_c\dot{y}_c(1+\dot{y}_c^2)^{\frac{-1}{2}}$ and $dv = \dot{y}_q dx$, and applying boundary conditions $y_q(-L) = y_q(L) = 0$, we obtain

$I_2 = \int_{-L}^L y_q\big(y_c\dot{y}_c^2\ddot{y}_c(1+\dot{y}_c^2)^{\frac{-3}{2}} - (\dot{y}_c^2 + y_c\ddot{y}_c)(1+\dot{y}_c^2)^{\frac{-1}{2}})\big)dx$

From $I_1 + I_2 = 0$, factoring out the $y_q$ and the fundamental lemma of the calculus of variations, we obtain an ODE (replacing $y_c$ with $y$):

$0 = (1+\dot{y}^2)^\frac{-3}{2}((1+\dot{y}^2)^2 - (\dot{y}^2 + y\ddot{y})(1+\dot{y}^2) + y\dot{y}^2\ddot{y})$

Dividing through by $(1+\dot{y}^2)^\frac{-3}{2}$, expanding, and simplifying, we obtain our final ODE:

$1 + \dot{y}^2 - y\ddot{y} = 0$

A solution to this is $y(x) = \frac1{\lambda}cosh(\lambda x + c)$, where $\lambda$ and c are determined by physical constants and the boundary values of $y$ (at $x = -L$ and $x = L$). It turns out that this ODE is the same one you get when solving the soap bubble problem (HW 2, problem 3), since the Lagrangians of the two systems are the same up to constants.

Notes for AKT-140120/0:34:12

2018-05-30T18:46:02Z

Cameron.martin:

The second crossing on this line should be an undercrossing (as stated), not an overcrossing (as drawn). Namely, in the blackboard shot below the third line from the top should be $q^{-1}J(+)-qJ(-)=\ldots$, and not as written.

Notes for AKT-140106/0:46:29

2018-05-30T18:42:54Z

Cameron.martin: Created page with "The unknot $0_1$ and the figure-eight knot $4_1$ both have 3 legal 3-colorings, i.e. $\lambda(0_1) = \lambda(4_1) = 3$. 3-coloring fails to distinguish the unknot from the fig..."

The unknot $0_1$ and the figure-eight knot $4_1$ both have 3 legal 3-colorings, i.e. $\lambda(0_1) = \lambda(4_1) = 3$. 3-coloring fails to distinguish the unknot from the figure-eight. See http://katlas.math.toronto.edu/wiki/The_Rolfsen_Knot_Table for more information on specific knots.

Notes for AKT-140120/0:34:12

2018-05-20T16:11:11Z

Cameron.martin:

The second crossing on this line should be an undercrossing (as stated), not an overcrossing (as drawn). This applies to the 50:12 blackboard shot as well.

Notes for AKT-140120/0:34:12

2018-05-20T16:10:10Z

Cameron.martin: Created page with "The second crossing on this line should be an undercrossing (as stated), not an overcrossing (as drawn)."

The second crossing on this line should be an undercrossing (as stated), not an overcrossing (as drawn).

Notes for AKT-140106/0:43:23

2018-05-15T15:53:53Z

Cameron.martin: Created page with "'''Claim:''' The number of legal 3-colorings of a knot diagram is always a power of 3. This is an expansion on the proof given by Przytycki (https://arxiv.org/abs/math/06081..."

'''Claim:''' The number of legal 3-colorings of a knot diagram is always a power of 3.

This is an expansion on the proof given by Przytycki (https://arxiv.org/abs/math/0608172).

We'll show that the set of legal 3-colorings <math>\mathcal{S}</math> forms a subgroup of <math>Z_3^r</math>, for some r, which suffices to prove the claim. First, label each of the segments of the given diagram 1 through n, and denote a 3-coloring of this diagram by <math>x = (x_1, x_2, ..., x_n)</math>, where each <math>x_n</math> is an element of the cyclic group of order 3 <math>Z_3 = <a|a^3=1></math> (each element representing a different colour). It is clear that <math>\mathcal{S}</math> is a subset of <math>Z_3^n</math>. To show it is a subgroup, we'll take <math>x = (x_1, x_2, ..., x_n), y = (y_1, y_2, ..., y_n) \in \mathcal{S}</math>, and show that <math>xy^{-1} = (x_1y_1^{-1}, x_2y_2^{-1}, ..., x_ny_n^{-1}) \in \mathcal{S}</math>. It suffices to restrict our attention to one crossing in the given diagram, so we can without loss of generality let n = 3.

First, we (sub)claim that a crossing (involving colours <math>x_1, x_2, x_3</math> is legal if and only if <math>x_1x_2x_3 = 1</math> in <math>Z_3</math>. Indeed, if the crossing is legal, either it is the trivial crossing in which case their product is clearly 1, or each <math>x_i</math> is distinct, in which case <math>x_1x_2x_3 = 1aa^2 = a^3 = 1</math>. Conversely, suppose <math>x_1x_2x_3 = 1</math>, and suppose <math>x_1 = x_2</math>. It suffices to show that <math>x_3 = x_1</math>. This follows by case checking: if <math>x_1 = 1</math>, then <math>1 = x_1x_2x_3 = x_3</math>; if <math>x_1 = a</math>, then <math>1=a^2x_3</math>, implying that <math>x_3 = a^{-2} = a</math>; and if <math>x_1 = a^2</math>, then <math>1 = a^4x_3 = ax_3</math>, implying that <math>x_3 = a^{-1} = a^2</math>. Thus, the subclaim is proven.

As a result, <math>xy^{-1} = (x_1y_1^{-1}, x_2y_2^{-1}, x_3y_3^{-1})</math> satisfies <math>x_1y_1^{-1}x_2y_2^{-1}x_3y_3^{-1} = (x_1x_2x_3)(y_3y_2y_1)^{-1} = 1</math> since both <math>x, y \in \mathcal{S}</math>. This implies that <math>xy^{-1} \in \mathcal{S}</math>, and hence shows that <math>\mathcal{S}</math> is a subgroup of <math>Z_3^n</math> for n = the number of line segments in the diagram. By Lagrange's theorem, the number of legal 3-colorings (the order of <math>\mathcal{S}</math>) is a power of 3.

Notes for AKT-140117/0:21:24

2018-05-14T22:26:17Z

Cameron.martin: A derivation of Lemma 3.4

This is a more detailed derivation of the result from Lemma 3.4.

Let <math>f(\epsilon) = \mathcal{L}(x_c + \epsilon x_q)</math>. This now becomes a single variable minimum/maximum problem. We set <math>\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon=0} = 0</math>, and solve for <math>x_c</math>. First, simplifying <math>f(\epsilon)</math>, we compute

<math>f(\epsilon) = \int_0^T dt(\frac{1}{2}
(\dot{x}_c + \epsilon\dot{x}_q)^2 - V(x_c+\epsilon x_q))</math>

<math>f(\epsilon) = \int_0^T dt(\frac{1}{2}\dot{x}_c^2 + \epsilon \dot{x}_c\dot{x}_q - V(x_c) - \epsilon x_q V'(x_c)</math> + higher order terms).

Thus,

<math>\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon=0} = \int_0^T dt(\dot{x}_c\dot{x}_q - x_q V'(x_c))</math>

Integrating by parts with <math>u = \dot{x}_c, v = x_q</math>, this is equal to

<math>\dot{x}_c\dot{x}_q \mid_{0}^{T} - \int_0^T x_q\ddot{x}_c dt - \int_0^T x_q V'(x_c)dt</math>

The first term is equal to 0 by boundary conditions of <math>x_q</math>, so we obtain the equality

<math>\int_0^T -x_q(\ddot{x}_c + V'(x_c))dt = 0</math>

<math>\Rightarrow \ddot{x}_c + V'(x_c) = 0</math>, exactly as stated in the conclusion of Lemma 3.4. Solving this ODE with initial conditions gives the desired result. Explicitly, the solution of this ODE (with <math>V(x) = \frac{1}{2}x^2</math>) is

<math>x_c(t) = Acos(t) + Bsin(t)</math>
Plugging in <math>t = 0</math> and <math>t = \pi/2</math>, we have

<math>x_0 = x_c(0) = A</math> and <math>x_n = x_c(\pi/2) = B</math>, implying that <math>x_c(t) = x_0cos t + x_nsin t</math>, as claimed.