Timeline for Understanding Penrose diagrammatical notation
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 18, 2015 at 18:42 | comment | added | user21349 | the juxtaposition of "coupons" seems to be reserved to the composition of morphisms A covariant (or mixed) tensor is a linear transformation, which is a morphism. For example, $M^a_b N^b_c$ can be interpreted as a notation for matrix multiplication, which is the composition of two morphisms. contraction contrarily implicates some sort of Einstein summation, am I right? Penrose's birdtracks notation is isomorphic to abstract index notation. Indices don't take integer values or imply a choice of basis. Summation only happens if you convert to concrete index notation. | |
| Aug 18, 2015 at 15:46 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
ImageShack to imgur
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| Apr 16, 2013 at 21:01 | answer | added | Aleks Kissinger | timeline score: 6 | |
| Sep 23, 2011 at 17:58 | comment | added | fosco | cxontraction is a bilinear operation, isn't it? But try to write down the sum $\theta_c^{af}\chi_{fde}^b+\gamma_c^{af}\chi_{fde}^b$; it is not at all immediate that this diagram is (isotopy?) equivalent to that I linked in my question... | |
| Sep 23, 2011 at 17:51 | comment | added | Kelly Davis | I have not looked at these diagrams in years, but the $f$ leg of $\chi^b_{fde}$ is split in two due to the $+$ between $\theta^{af}_c$ and $\gamma^{af}_c$. The lower $f$ index of $\chi^b_{fde}$, the triangle, is being contracted with the upper $f$ index of $\gamma^{af}_c$, the square, and the result is being summed with the result of contracting the lower $f$ index of $\chi^b_{fde}$, the triangle, with the upper $f$ index of $\theta^{af}_c$, the circle. This split corresponds to the sum $\theta^{af}_c + \gamma^{af}_c$. If there were another summand, the $f$ leg would split in to three pieces. | |
| Sep 23, 2011 at 17:45 | comment | added | j.c. | In your bottom picture, there is one f leg for the bracketed sum which arises from an f leg from $\theta$ and one from $\gamma$. That's where the apparent "splitting" comes from. | |
| Sep 23, 2011 at 17:32 | history | edited | fosco | CC BY-SA 3.0 |
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| Sep 23, 2011 at 17:10 | history | edited | fosco | CC BY-SA 3.0 |
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| Sep 23, 2011 at 16:23 | history | asked | fosco | CC BY-SA 3.0 |