Difference between revisions of "Decoding the Graph-Wall-Tome Connection"

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<div style="font-weight:bold;line-height:1.6;">What are $$R_{\mu v}$$ and $$R$$ geometrically?</div>
<div style="font-weight:bold;line-height:1.6;">What are $$R_{\mu v}$$ and $$R$$ geometrically?</div>
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[https://www.youtube.com/watch?v=UfThVvBWZxM&t=12m6s Explanation of $$R$$]
[https://www.youtube.com/watch?v=UfThVvBWZxM&t=12m6s Explanation of $$R$$]


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[[Further thoughts on the meaning of R]]
[[Further thoughts on the meaning of R]]


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<div style="font-weight:bold;line-height:1.6;">Further thoughts on the meaning of R</div>
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===== Computing length in non-orthogonal bases =====
First, just describing the length of a vector on a curved space is hard. It is given by:
$$Length^{squared} = g_{11}dX^{1}dX^{1} + g_{12}dX^{1}dX^{2} + g_{21}dX^{2}dX^{1} + g_{22}dX^{2}dX^{2}$$
Some notes:
* This is not Pythagorean theorem, because $$dX^{1}$$ and $$dX^{2}$$ are not perpendicular.
* Instead, looks like a formula to get the diagonal from two opposite vertices in a parallelogram.
* If $$dX^{1}$$ and $$dX^{2}$$ are perpendicular, then $$g_{12}$$ and $$g_{21}$$ would be 0, and we would get $$Length^{squared} = g_{11}(dX^{1})^{2} + g_{22}(dX^{2})^{2}$$
* See: [https://www.youtube.com/watch?v=UfThVvBWZxM&t=14m27s the video @ 14m27s]
===== Computing vector rotation due to parallel transport =====
Then, they show parallel transport when following a parallelogram, but over a curved 3D manifold. To compute the vector rotation by components, they show:
$$dV^{1} = dX^{1}dX^{2} (V^{1}R^{1}_{112} + V^{2}R^{1}_{212} + V^{3}R^{1}_{312})$$
$$dV^{2} = dX^{1}dX^{2} (V^{1}R^{2}_{112} + V^{2}R^{2}_{212} + V^{3}R^{2}_{312})$$
$$dV^{3} = dX^{1}dX^{2} (V^{1}R^{3}_{112} + V^{2}R^{3}_{212} + V^{3}R^{3}_{312})$$
or, using $$i$$ to summarize across all 3 components (difference vectors):
$$dV^{i} = dX^{1}dX^{2} (V^{1}R^{i}_{112} + V^{2}R^{i}_{212} + V^{3}R^{i}_{312})$$
or , using $$j$$ to index over all 3 components (original vector):
$$dV^{i} = dX^{1}dX^{2} \Sigma_{j} [(V^{j}R^{i}_{j12}]$$
See: [https://www.youtube.com/watch?v=UfThVvBWZxM&t=19m33s the video @ 19m33s]
Open questions:
* Why a parallelogram?
* How to properly overlay the parallelogram onto the 3d manifold, in order to do the parallel transport?
* How does this relate to the length computation above?
===== Putting it all together =====
Now, moving to 4D, we can compute $$R_{\mu v}$$ as:
$$R_{00} = R^{0}_{000} + R^{1}_{010} + R^{2}_{020} + R^{3}_{030}$$
$$R_{10} = R^{0}_{100} + R^{1}_{110} + R^{2}_{120} + R^{3}_{130}$$
$$R_{01} = R^{0}_{001} + R^{1}_{011} + R^{2}_{021} + R^{3}_{030}$$
etc.
Indexing i over all 4 component vectors / dimensions, we get:
$$R_{00} = \Sigma_{i} R^{i}_{0i0}$$
$$R_{10} = \Sigma_{i} R^{i}_{1i0}$$
$$R_{01} = \Sigma_{i} R^{i}_{0i1}$$
etc.
Summarizing on $$\mu$$, we get:
$$R_{\mu 0} = \Sigma_{i} R^{i}_{\mu i0}$$
$$R_{\mu 1} = \Sigma_{i} R^{i}_{\mu i1}$$
etc
Summarizing on $$v$$, we get:
$$R_{\mu v} = \Sigma_{i} R^{i}_{\mu iv}$$
Open questions:
* If we hadn't moved from 3D to 4D, what would this all have looked like?
* What does this have to do with the parallelogram?
* Why are there two indices?
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