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Ivan Izmestiev
Ivan Izmestiev
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Define the cross-ratio of four real or complex numbers as follows: $$[a,b,c,d] = \frac{(a-c)(b-d)}{(a-d)(b-c)}.$$ Then its logarithm has the same symmetries as the curvature tensor: $$\log[a,b,c,d] = -\log[b,a,c,d] = -\log[a,b,d,c] = \log[c,d,a,b].$$ Moreover, if $[a,b,c,d] = \lambda$, then $[b,c,a,d] = 1 - \lambda^{-1}$ and $[c,a,b,d] = (1-\lambda)^{-1}$, which implies an analog of the algebraic Bianchi identity: $$\log[a,b,c,d] + \log[b,c,a,d] + \log[c,a,b,d] = -1.$$$$\log[a,b,c,d] + \log[b,c,a,d] + \log[c,a,b,d] = \pi i.$$

Is there something behind these coincidences?

Define the cross-ratio of four real or complex numbers as follows: $$[a,b,c,d] = \frac{(a-c)(b-d)}{(a-d)(b-c)}.$$ Then its logarithm has the same symmetries as the curvature tensor: $$\log[a,b,c,d] = -\log[b,a,c,d] = -\log[a,b,d,c] = \log[c,d,a,b].$$ Moreover, if $[a,b,c,d] = \lambda$, then $[b,c,a,d] = 1 - \lambda^{-1}$ and $[c,a,b,d] = (1-\lambda)^{-1}$, which implies the algebraic Bianchi identity: $$\log[a,b,c,d] + \log[b,c,a,d] + \log[c,a,b,d] = -1.$$

Is there something behind these coincidences?

Define the cross-ratio of four real or complex numbers as follows: $$[a,b,c,d] = \frac{(a-c)(b-d)}{(a-d)(b-c)}.$$ Then its logarithm has the same symmetries as the curvature tensor: $$\log[a,b,c,d] = -\log[b,a,c,d] = -\log[a,b,d,c] = \log[c,d,a,b].$$ Moreover, if $[a,b,c,d] = \lambda$, then $[b,c,a,d] = 1 - \lambda^{-1}$ and $[c,a,b,d] = (1-\lambda)^{-1}$, which implies an analog of the algebraic Bianchi identity: $$\log[a,b,c,d] + \log[b,c,a,d] + \log[c,a,b,d] = \pi i.$$

Is there something behind these coincidences?

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Ivan Izmestiev
Ivan Izmestiev
  • 6.3k
  • 26
  • 50

Is there an explanation of analogies between the cross-ratio and the Riemann curvature tensor?

Define the cross-ratio of four real or complex numbers as follows: $$[a,b,c,d] = \frac{(a-c)(b-d)}{(a-d)(b-c)}.$$ Then its logarithm has the same symmetries as the curvature tensor: $$\log[a,b,c,d] = -\log[b,a,c,d] = -\log[a,b,d,c] = \log[c,d,a,b].$$ Moreover, if $[a,b,c,d] = \lambda$, then $[b,c,a,d] = 1 - \lambda^{-1}$ and $[c,a,b,d] = (1-\lambda)^{-1}$, which implies the algebraic Bianchi identity: $$\log[a,b,c,d] + \log[b,c,a,d] + \log[c,a,b,d] = -1.$$

Is there something behind these coincidences?